Solutions Manual for
Probability and Random Processes for
Electrical and Computer Engineers
John A. Gubner
University of Wisconsin–Madison
File Generated July 13, 2007
CHAPTER 1
Problem Solutions
1. Ω = {1, 2, 3, 4, 5, 6}.
2. Ω = {0, 1, 2, . . . , 24, 25}.
3. Ω = [0, ∞). RTT > 10 ms is given by the event (10, ∞).
4.
(a) Ω = {(x, y) ∈ IR2 : x2 + y2 ≤ 100}.
(b) {(x, y) ∈ IR2 : 4 ≤ x2 + y2 ≤ 25}.
5.
(a) [2, 3] c = (−∞, 2) ∪ (3, ∞).
(b) (1, 3) ∪ (2, 4) = (1, 4).
(c) (1, 3) ∩ [2, 4) = [2, 3).
(d) (3, 6] \ (5, 7) = (3, 5].
6. Sketches:
y
y
y
1
1
x
x
−1
x
−1
B0
B1
B−1
y
y
y
3
x
C1
3
H3
1
x
x
J3
2
Chapter 1 Problem Solutions
y
y
3
3
3
U
H3
x
x
3
J3 = M 3
H3 U J3 = N 3
y
y
4
2
3
2
M2
7.
U
N 3 = M2
x
x
3 4
M4
U
N3
(a) [1, 4] ∩ [0, 2] ∪ [3, 5] = [1, 4] ∩ [0, 2] ∪ [1, 4] ∩ [3, 5] = [1, 2] ∪ [3, 4].
(b)
c
= [0, 1] c ∩ [2, 3] c
[0, 1] ∪ [2, 3]
h
i h
i
= (−∞, 0) ∪ (1, ∞) ∩ (−∞, 2) ∪ (3, ∞)
h
i
= (−∞, 0) ∩ (−∞, 2) ∪ (3, ∞)
h
i
∪ (1, ∞) ∩ (−∞, 2) ∪ (3, ∞)
= (−∞, 0) ∪ (1, 2) ∪ (3, ∞).
(c)
(d)
(e)
(f)
∞
\
(− 1n , n1 ) = {0}.
n=1
∞
\
1
) = [0, 3].
[0, 3 + 2n
n=1
∞
[
1
] = [5, 7).
[5, 7 − 3n
n=1
∞
[
[0, n] = [0, ∞).
n=1
Chapter 1 Problem Solutions
3
8. We first let C ⊂ A and show that for all B, (A ∩ B) ∪C = A ∩ (B ∪C). Write
A ∩ (B ∪C) = (A ∩ B) ∪ (A ∩C),
= (A ∩ B) ∪C,
by the distributive law,
since C ⊂ A ⇒ A ∩C = C.
For the second part of the problem, suppose (A ∩ B) ∪C = A ∩ (B ∪C). We must show
that C ⊂ A. Let ω ∈ C. Then ω ∈ (A ∩ B) ∪ C. But then ω ∈ A ∩ (B ∪ C), which
implies ω ∈ A.
9. Let I := {ω ∈ Ω : ω ∈ A ⇒ ω ∈ B}. We must show that A ∩ I = A ∩ B.
⊂: Let ω ∈ A ∩ I. Then ω ∈ A and ω ∈ I. Therefore, ω ∈ B, and then ω ∈ A ∩ B.
⊃: Let ω ∈ A ∩ B. Then ω ∈ A and ω ∈ B. We must show that ω ∈ I too. In other
words, we must show that ω ∈ A ⇒ ω ∈ B. But we already have ω ∈ B.
10. The function f : (−∞, ∞) → [0, ∞) with f (x) = x3 is not well defined because not all
values of f (x) lie in the claimed co-domain [0, ∞).
11.
(a) The function will be invertible if Y = [−1, 1].
(b) {x : f (x) ≤ 1/2} = [−π /2, π /6].
(c) {x : f (x) < 0} = [−π /2, 0).
12.
(a) Since f is not one-to-one, no choice of co-domain Y can make f : [0, π ] → Y
invertible.
(b) {x : f (x) ≤ 1/2} = [0, π /6] ∪ [5π /6, π ].
(c) {x : f (x) < 0} = ∅.
13. For B ⊂ IR,

X,



A,
−1
f (B) =
c,
A



∅,
0 ∈ B and 1 ∈ B,
1 ∈ B but 0 ∈
/ B,
0 ∈ B but 1 ∈
/ B,
0∈
/ B and 1 ∈
/ B.
14. Let f : X → Y be a function such that f takes only n distinct values, say y1 , . . . , yn .
Let B ⊂ Y be such that f −1 (B) is nonempty. By definition, each x ∈ f −1 (B) has the
property that f (x) ∈ B. But f (x) must be one of the values y1 , . . . , yn , say yi . Now
f (x) = yi if and only if x ∈ Ai := f −1 ({yi }). Hence,
f −1 (B) =
[
Ai .
i:yi ∈B
15.
(a) f (x) ∈ B c ⇔ f (x) ∈
/B⇔x∈
/ f −1 (B) ⇔ x ∈ f −1 (B) c .
(b) f (x) ∈
∞
[
n=1
Bn if and only if f (x) ∈ Bn for some n; i.e., if and only if x ∈ f −1 (Bn )
for some n. But this says that x ∈
∞
[
n=1
f −1 (Bn ).
4
Chapter 1 Problem Solutions
(c) f (x) ∈
∞
\
n=1
Bn if and only if f (x) ∈ Bn for all n; i.e., if and only if x ∈ f −1 (Bn )
for all n. But this says that x ∈
S
16. If B = i {bi } and C =
is countable.
S
i {ci },
∞
\
f −1 (Bn ).
n=1
put a2i := bi and a2i−1 := ci . Then A =
17. Since each Ci is countable, we can write Ci =
B :=
∞
[
Ci =
i=1
S
j ci j .
∞ [
∞
[
S
= B ∪C
i ai
It then follows that
{ci j }
i=1 j=1
is a doubly indexed sequence and is therefore countable as shown in the text.
S
18. Let A = m {am } be a countable set, and let B ⊂ A. We must show that B is countable.
If B = ∅, we’re done by definition. Otherwise, there is at least one element
of B in
S
A, say ak . Then put bn := an if an ∈ B, and put bn := ak if an ∈
/ B. Then n {bn } = B
and we see that B is countable.
19. Let A ⊂ B where A is uncountable. We must show that B is uncountable. We prove
this by contradiction. Suppose that B is countable. Then by the previous problem, A
is countable, contradicting the assumption that A is uncountable.
20. Suppose A is countable and B is uncountable. We must show that A∪B is uncountable.
We prove this by contradiction. Suppose that A ∪ B is countable. Then since B ⊂
A ∪ B, we would have B countable as well, contradicting the assumption that B is
uncountable.
21. MATLAB. OMITTED.
22. MATLAB. Intuitive explanation: Using only the numbers 1, 2, 3, 4, 5, 6, consider how
many ways there are to write the following numbers:
2
3
4
5
6
7
8
9
10
11
12
=
=
=
=
=
=
=
=
=
=
=
1+1
1+2 = 2+1
1+3 = 2+2 = 3+1
1+4 = 2+3 = 3+2 = 4+1
1+5 = 2+4 = 3+3 = 4+2 = 5+1
1+6 = 2+5 = 3+4 = 4+3 = 5+2 = 6+1
2+6 = 3+5 = 4+4 = 5+3 = 6+2
3+6 = 4+5 = 5+4 = 6+3
4+6 = 5+5 = 6+4
5+6 = 6+5
6+6
23. Take Ω := {1, . . . , 26} and put
P(A) :=
|A|
|A|
=
.
|Ω|
26
1 way,
2 ways,
3 ways,
4 ways,
5 ways,
6 ways,
5 ways,
4 ways,
3 ways,
2 ways,
1 way,
36 ways,
1/36
2/36
3/36
4/36
5/36
6/36
5/36
4/36
3/36
2/36
1/36
36/36
=
=
=
=
=
=
=
=
=
=
=
=
0.0278
0.0556
0.0833
0.1111
0.1389
0.1667
0.1389
0.1111
0.0833
0.0556
0.0278
1
Chapter 1 Problem Solutions
5
The event that a vowel is chosen is V = {1, 5, 9, 15, 21}, and P(V ) = |V |/26 = 5/26.
24. Let Ω := {(i, j) : 1 ≤ i, j ≤ 26 and i 6= j}. For A ⊂ Ω, put P(A) := |A|/|Ω|. The event
that a vowel is chosen followed by a consonant is
Bvc = (i, j) ∈ Ω : i = 1, 5, 9, 15, or 21 and j ∈ {1, . . . , 26} \ {1, 5, 9, 15, 21} .
Similarly, the event that a consonant is followed by a vowel is
Bcv = (i, j) ∈ Ω : i ∈ {1, . . . , 26} \ {1, 5, 9, 15, 21} and j = 1, 5, 9, 15, or 21 .
We need to compute
P(Bvc ∪ Bcv ) =
|Bvc | + |Bcv | 5 · (26 − 5) + (26 − 5) · 5 21
=
=
≈ 0.323.
|Ω|
650
65
The event that two vowels are chosen is
Bvv = (i, j) ∈ Ω : i, j ∈ {1, 5, 9, 15, 21} with i 6= j ,
and P(Bvv ) = |Bvv |/|Ω| = 20/650 = 2/65 ≈ .031.
25. MATLAB. The code for simulating the drawing of a face card is
% Simulation of Drawing a Face Card
%
n = 10000;
% Number of draws.
X = ceil(52*rand(1,n));
faces = (41 <= X & X <= 52);
nfaces = sum(faces);
fprintf(’There were %g face cards in %g draws.\n’,nfaces,n)
26. Since 9 pm to 7 am is 10 hours, take Ω := [0, 10]. The probability that the baby wakes
up during a time interval 0 ≤ t1 < t2 ≤ 10 is
P([t1 ,t2 ]) :=
Hence, P([2, 10] c ) = P([0, 2]) =
27. Starting with the equations
R2
0
Z t2
1
t1
10
dω .
1/10 d ω = 1/5.
SN = 1 + z + z2 + · · · + zN−2 + zN−1
zSN =
z + z2 + · · · + zN−2 + zN−1 + zN ,
subtract the second line from the first. Canceling common terms leaves
SN − zSN = 1 − zN ,
or
SN (1 − z) = 1 − zN .
If z 6= 1, we can divide both sides by 1 − z to get SN = (1 − zN )/(1 − z).
6
Chapter 1 Problem Solutions
28. Let x = p(1). Then p(2) = 2p(1) = 2x, p(3) = 2p(2) = 22 x, p(4) = 2p(3) = 23 x,
p(5) = 24 x, and p(6) = 25 x. In general, p(ω ) = 2ω −1 x and we can write
6
1 =
∑ p(ω )
ω =1
5
6
=
∑ 2ω −1 x
= x
ω =1
∑ 2ω
=
ω =0
1 − 26
x = 63x.
1−2
Hence, x = 1/63, and p(ω ) = 2ω −1 /63 for ω = 1, . . . , 6.
29.
(a) By inclusion–exclusion, P(A ∪ B) = P(A) + P(B) − P(A ∩ B), which can be rearranged as P(A ∩ B) = P(A) + P(B) − P(A ∪ B).
(b) Since P(A) = P(A ∩ B) + P(A ∩ B c ),
P(A ∩ B c ) = P(A) − P(A ∩ B) = P(A ∪ B) − P(B),
by part (a).
(c) Since B and A ∩ B c are disjoint,
P(B ∪ (A ∩ B c )) = P(B) + P(A ∩ B c ) = P(A ∪ B),
by part (b).
(d) By De Morgan’s law, P(A c ∩ B c ) = P([A ∪ B] c ) = 1 − P(A ∪ B).
30. We must check the four axioms of a probability measure. First,
P(∅) = λ P1 (∅) + (1 − λ )P2 (∅) = λ · 0 + (1 − λ ) · 0 = 0.
Second,
P(A) = λ P1 (A) + (1 − λ )P2 (A) ≥ λ · 0 + (1 − λ ) · 0 = 0.
Third,
∞
∞
∞
[
[
[
An = λ P1
P
An + (1 − λ )P2
An
n=1
n=1
∞
n=1
∞
n=1
= λ
=
n=1
∞
∑ P1 (An ) + (1 − λ ) ∑ P2 (An )
∑ [λ P1 (An ) + (1 − λ )P2 (An )]
n=1
∞
=
∑ P(An ).
n=1
Fourth, P(Ω) = λ P1 (Ω) + (1 − λ )P2 (Ω) = λ + (1 − λ ) = 1.
/ ∅S, µ (∅) = 0. Second, by definition, µ (A) ≥ 0. Third, for disjoint
31. First, since ω0 ∈
/ An for n 6= m. Then
An , suppose ω0 ∈ n An . Then ω0 ∈ Am for some
S m,and ω0 ∈
µ (Am ) = 1 and µ (An ) = 0 for n 6= m. Hence, µ n An = 1 and ∑n µ (An ) = µ (Am ) =
S
S
1. A similar analysis shows that if ω0 ∈
/ n An then µ n An and ∑n µ (An ) are both
zero. Finally, since ω0 ∈ Ω, µ (Ω) = 1.
Chapter 1 Problem Solutions
7
32. Starting with the assumption that for any two disjoint events A and B, P(A ∪ B) =
P(A) + P(B), we have that for N = 2,
N
N
[
An = ∑ P(An ).
(∗)
P
n=1
n=1
Now we must show that if (∗) holds for any N ≥ 2, then (∗) holds for N + 1. Write
N+1 N
[
[
An = P
An ∪ AN+1
P
n=1
n=1
N
[
= P
An + P(AN+1 ),
additivity for two events,
n=1
N
=
∑ P(An ) + P(AN+1 ),
by (∗),
n=1
N+1
=
∑ P(An ).
n=1
c ∩ · · · ∩ F c ⊂ F , it is easy to see that
33. Since An := Fn ∩ Fn−1
n
1
N
[
n=1
An ⊂
N
[
Fn .
n=1
S
The hard part is to show the reverse inclusion ⊃. Suppose ω ∈ Nn=1 Fn . Then ω ∈ Fn
for some n in the range 1, . . . , N. However, ω may belong to Fn for several values of
n since the Fn may not be disjoint. Let
k := min{n : ω ∈ Fn and 1 ≤ n ≤ N}.
/ Fn for n < k; in symbols,
In other words, 1 ≤ k ≤ N and ω ∈ Fk , but ω ∈
S
c
ω ∈ Fk ∩ Fk−1
∩ · · · ∩ F1c =: Ak .
Hence, ω ∈ Ak ⊂ Nn=1 An . The proof that
k := min{n : ω ∈ Fn and n ≥ 1}.
S∞
n=1 An
⊂
S∞
n=1 Fn
is similar except that
34. For arbitrary events Fn , let An be as in the preceding problem. We can then write
∞ ∞
∞
[
[
P
Fn = P
An = ∑ P(An ), since the An are disjoint,
n=1
n=1
n=1
N
∑ P(An ),
N→∞
= lim
n=1
N
[
= lim P
An
N→∞
n=1
N [
= lim P
Fn .
N→∞
n=1
by def. of infinite sum,
8
Chapter 1 Problem Solutions
35. For arbitrary events Gn , put Fn := Gnc . Then
∞
∞ \
[
P
Gn = 1 − P
Fn , by De Morgan’s law,
n=1
n=1
N [
Fn ,
= 1 − lim P
N→∞
by the preceding problem,
n=1
N
\
= 1 − lim 1 − P
Gn ,
N→∞
by De Morgan’s law,
n=1
N
\
Gn .
= lim P
N→∞
n=1
36. By the inclusion–exclusion formula,
P(A ∪ B) = P(A) + P(B) − P(A ∩ B) ≤ P(A) + P(B).
This establishes the union bound for N = 2. Now suppose the union bound holds for
some N ≥ 2. We must show it holds for N + 1. Write
N+1 N [
[
Fn = P
Fn ∪ FN+1
P
n=1
n=1
N [
≤ P
Fn + P(FN+1 ),
by the union bound for two events,
n=1
N
≤
=
∑ P(Fn ) + P(FN+1 ),
by the union bound for N events,
n=1
N+1
∑ P(Fn ).
n=1
37. To establish the union bound for a countable sequence of events, we proceed as folc ∩ · · · ∩ F c ⊂ F be disjoint with S∞ A = S∞ F . Then
lows. Let An := Fn ∩ Fn−1
n
n=1 n
n=1 n
1
∞ ∞
[
[
P
Fn = P
An
n=1
n=1
∞
=
∑ P(An ),
since the An are disjoint,
∑ P(Fn ),
since An ⊂ Fn .
n=1
∞
≤
n=1
38. Following the hint, we put Gn :=
S∞
k=n Bk
so that we can write
∞ ∞ ∞
N
\ [
\
\
P
Bk = P
Gn = lim P
Gn ,
n=1 k=n
n=1
N→∞
n=1
limit property of P,
Chapter 1 Problem Solutions
9
= lim P(GN ), since Gn ⊃ Gn+1 ,
N→∞
∞
[
= lim P
Bk , definition of GN ,
N→∞
k=N
∞
≤ lim
N→∞
∑ P(Bk ),
union bound.
k=N
This last limit must be zero since ∑∞
k=1 P(Bk ) < ∞.
39. In this problem, the probability of an interval is its length.
(a) P(A0 ) = 1, P(A1 ) = 2/3, P(A2 ) = 4/9 = (2/3)2 , and P(A3 ) = 8/27 = (2/3)3 .
(b) P(An ) = (2/3)n .
(c) Write
N
∞
\
\
An ,
An = lim P
P(A) = P
N→∞
n=1
limit property of P,
n=1
= lim P(AN ),
N→∞
since An ⊃ An+1 ,
= lim (2/3)N = 0.
N→∞
40. Consider the collection consisting of the empty set along with all unions of the form
S
i Aki for some finite subsequence of distinct elements from {1, . . . , n}. We first show
that this collection is a σ -field. First, it contains ∅ by definition. Second, since
A1 , . . . , An is a partition,
[
c
Aki
i
=
[
Ami ,
i
where mi is the subsequence {1, . . . , n} \ {ki }. Hence, the collection is closed under
complementation. Third,
∞ [
[
[
Akn,i =
Am j ,
i
n=1
j
where an integer l ∈ {1, . . . , n} is in {m j } if and only if kn,i = l for some n and some i.
This shows that the collection is a σ -field. Finally, since every element in our collection must be contained in every σ -field that contains A1 , . . . , An , our collection must
be σ (A1 , . . . , An ).
41. We claim that A is not a σ -field. Our proof is by contradiction: We assume A is a
σ -field and derive a contradiction. Consider the set
∞
\
[0, 1/2n ) = {0}.
n=1
Since [0, 1/2n ) ∈ Cn ⊂ An ⊂ A , the intersection must be in A since we are assuming
A is a σ -field. Hence, {0} ∈ A . Now, any set in A must belong to some An . By the
10
Chapter 1 Problem Solutions
preceding problem, every set in An must be a finite union of sets from Cn . However,
the singleton set {0} cannot be expressed as a finite union of sets from any Cn . Hence,
{0} ∈
/A.
42. Let Ai := X −1 ({xi }) for i = 1, . . . , n. By the problems mentioned in the hint, for any
subset B, if X −1 (B) 6= ∅, then
X −1 (B) =
[
i:xi ∈B
Ai ∈ σ (A1 , . . . , An ).
It follows that the smallest σ -field containing all the X −1 (B) is σ (A1 , . . . , An ).
43. (a) F = ∅, A, B, {3}, {1, 2}, {4, 5}, {1, 2, 4, 5}, Ω
(b) The corresponding probabilities are 0, 5/8, 7/8, 1/2, 1/8, 3/8, 1/2, 1.
(c) Since {1} ∈
/ F , P({1}) is not defined.
44. Suppose that a σ -field A contains an infinite sequence Fn of sets. If the sequence is
not disjoint, we can construct a new sequence An that is disjoint with each An ∈ A .
Let a = a1 , a2 , . . . be an infinite sequence of zeros and ones. Then A contains each
union of the form
[
Ai .
i:ai =1
Furthermore, since the Ai are disjoint, each sequence a gives a different union, and
we know from the text that the number of infinite sequences a is uncountably infinite.
45.
T
T
A ∈ α Aα , then A ∈ Aα
(a) First, since ∅ is in each Aα , ∅ ∈ α Aα . Second, if T
for each α , and so A c ∈ Aα for each α . Hence, A c ∈ α Aα . Third, if An ∈ A
S
for all n,
then for
each n and each α , An ∈ Aα . Then n An ∈ Aα for each α ,
S
T
and so n An ∈ α Aα .
(b) We first note that
A1 = {∅, {1}, {2}, {3, 4}, {2, 3, 4}, {1, 3, 4}, {1, 2}, Ω}
and
A2 = {∅, {1}, {3}, {2, 4}, {2, 3, 4}, {1, 2, 4}, {1, 3}, Ω}.
It is then easy to see that
A1 ∩ A2 = {∅, {1}, {2, 3, 4}, Ω}.
T
(c) First note that by part (a), A :C ⊂A A is a σ -field, and since C ⊂ A for each
T
A , the σ -field A :C ⊂A A contains C . Finally, if D is any σ -field that contains
C , then D is one of the A s in the intersection. Hence,
C ⊂
Thus
T
A :C ⊂A
\
A :C ⊂A
A ⊂ D.
A is the smallest σ -field that contains C .
Chapter 1 Problem Solutions
11
46. The union of two σ -fields is not always a σ -field. Here is an example. Let Ω :=
{1, 2, 3, 4}, and put
F := ∅, {1, 2}, {3, 4}, Ω
and G := ∅, {1, 3}, {2, 4}, Ω .
Then
F ∪G =
∅, {1, 2}, {3, 4}, {1, 3}, {2, 4}, Ω
is not a σ -field since it does not contain {1, 2} ∩ {1, 3} = {1}.
47. Let Ω denote the positive integers, and let A denote the collection of subsets A such
that either A or A c is finite.
(a) Let E denote the subset of even integers. Then E does not belong to A since
neither E nor E c (the odd integers) is a finite set.
(b) To show that A is closed under finite unions, we consider two cases. First
suppose that A1 , . . . , An are all finite. Then
n
[
n
i=1
and so
Sn
i=1 Ai
Ai ≤
∑ |Ai |
< ∞,
i=1
∈ A . In the second case, suppose that some A cj is finite. Then
Hence, the complement of
n
[
i=1
Ai
c
Sn
i=1 Ai
=
n
\
i=1
Aic ⊂ A cj .
is finite, and so the union belongs to A .
(c) A is not a σ -field. To see this, put Ai := {2i} for i = 1, 2, . . . . Then
E∈
/ A by part (a).
S∞
i=1 Ai
=
48. Let Ω be an uncountable set. Let A denote the collection of all subsets A such that
either A is countable or A c is countable. We show that A is a σ -field. First, the empty
set is countable. Second, if A ∈ A , we must show that A c ∈ A . There are two cases.
If A is countable, then the complement of A c is A, and so A c ∈ A . If A c is countable,
then A c ∈ A . Third, let A1S, A2 , . . . belong to A . There are two cases to consider. If
all An are countable, then n An is also countable by an earlier problem. Otherwise,
if some Amc is countable, then write
49.
∞
[
n=1
An
c
=
∞
\
n=1
Anc ⊂ Amc .
Since
the subset of a countable set is countable, we see that the complement of
S∞
A
n=1 n is countable, and thus the union belongs to A .
(b)
T∞
1
1
n=1 (a, b + n ), and since each (a, b + n ) ∈ B, (a, b] ∈ B.
T
1
1
1
1
Since {a} = ∞
n=1 (a− n , a+ n ), and since each (a− n , a+ n ) ∈ B, the singleton
(a) Since (a, b] =
{a} ∈ B.
12
Chapter 1 Problem Solutions
(c) Since by part (b), singleton sets are Borel sets, and since A is a countable union
of Borel sets, A ∈ B; i.e., A is a Borel set.
(d) Using part (a), write
λ (a, b] = λ
∞
\
(a, b +
n=1
= lim λ
N→∞
N
\
1
n)
(a, b + 1n ) ,
limit property of probability,
n=1
= lim λ (a, b + N1 ) ,
decreasing sets,
N→∞
= lim (b + N1 ) − a,
N→∞
characterization of λ ,
= b − a.
Similarly, using part (b), we can write
λ {a} = λ
∞
\
1
1
n,a+ n)
(a −
n=1
= lim λ
N→∞
N
\
(a −
n=1
= lim λ (a − N1 , a + N1 ) ,
N→∞
= lim 2/N,
N→∞
,
1
1
n,a+ n)
limit property of probability,
decreasing sets,
characterization of λ ,
= 0.
50. Let I denote the collection of open intervals, and let O denote the collection of open
sets. We need to show that σ (I ) = σ (O). Since I ⊂ O, every σ -field containing
O also contains I . Hence, the smallest σ -field containing O contains I ; i.e., I ⊂
σ (O). By the definition of the smallest σ -field containing I , it follows that σ (I ) ⊂
σ (O). Now, if we can show that O ⊂ σ (I ), then it will similarly follow that σ (O) ⊂
σ (I ). Recall that in the problem statement, it was shown that every open set U can
be written as a countable union of open intervals. This means U ∈ σ (I ). This proves
that O ⊂ σ (I ) as required.
51. MATLAB. Chips from S1 are 80% reliable; chips from S2 are 70% reliable.
52. Observe that
N(Od,S1 ) N(Ow,S1 ) = N(OS1 ) − N(Od,S1 ) = N(OS1 ) 1 −
N(OS1 )
and
N(Od,S2 ) .
N(Ow,S2 ) = N(OS2 ) − N(Od,S2 ) = N(OS2 ) 1 −
N(OS2 )
Chapter 1 Problem Solutions
13
53. First write
P(A|B ∩C) P(B|C) =
P(A ∩ [B ∩C]) P(B ∩C) P([A ∩ B] ∩C)
·
=
= P(A ∩ B|C).
P(B ∩C)
P(C)
P(C)
From this formula, we can isolate the equation
P(A|B ∩C) P(B|C) =
P([A ∩ B] ∩C)
.
P(C)
Multiplying through by P(C) yields P(A|B ∩C) P(B|C) P(C) = P(A ∩ B ∩C).
54.
(a) P(MM) = 140/(140 + 60) = 140/200 = 14/20 = 7/10 = 0.7. Then P(HT) =
1 − P(MM) = 0.3.
(b) Let D denote the event that a workstation is defective. Then
P(D) = P(D|MM)P(MM) + P(D|HT)P(HT)
= (.1)(.7) + (.2)(.3)
= .07 + .06 = 0.13.
(c) Write
P(MM|D) =
P(D|MM)P(MM)
.07
7
=
=
.
P(D)
.13
13
55. Let O denote the event that a cell is overloaded, and let B denote the event that a call
is blocked. The problem statement tells us that
P(O) = 1/3,
P(B|O) = 3/10,
and
P(B|O c ) = 1/10.
To find P(O|B), first write
P(O|B) =
P(B|O)P(O)
3/10 · 1/3
=
.
P(B)
P(B)
Next compute
P(B) = P(B|O)P(O) + P(B|O c )P(O c )
= (3/10)(1/3) + (1/10)(2/3) = 5/30 = 1/6.
We conclude that
P(O|B) =
1/10
6
3
=
=
= 0.6.
1/6
10
5
56. The problem statement tells us that P(R1 |T0 ) = ε and P(R0 |T1 ) = δ . We also know
that
1 = P(Ω) = P(T0 ∪ T1 ) = P(T0 ) + P(T1 ).
The problem statement tells us that these last two probabilities are the same; hence
they are both equal to 1/2. To find P(T1 |R1 ), we begin by writing
P(T1 |R1 ) =
P(R1 |T1 )P(T1 )
.
P(R1 )
14
Chapter 1 Problem Solutions
Next, we note that P(R1 |T1 ) = 1 − P(R0 |T1 ) = 1 − δ . By the law of total probability,
P(R1 ) = P(R1 |T1 )P(T1 ) + P(R1 |T0 )P(T0 )
= (1 − δ )(1/2) + ε (1/2) = (1 − δ + ε )/2.
So,
P(T1 |R1 ) =
(1 − δ )(1/2)
1−δ
=
.
(1 − δ + ε )/2
1−δ +ε
57. Let H denote the event that a student does the homework, and let E denote the event
that a student passes the exam. Then the problem statement tells us that
P(E|H c ) = .1,
P(E|H) = .8,
and
P(H) = .6.
We need to compute P(E) and P(H|E). To begin, write
P(E) = P(E|H)P(H) + P(E|H c )P(H c )
= (.8)(.6) + (.1)(1 − .6) = .48 + .04 = .52.
Next,
P(H|E) =
.48
12
P(E|H)P(H)
=
=
.
P(E)
.52
13
58. The problem statement tells us that
P(AF |CF ) = 1/3,
P(AF |CFc ) = 1/10,
and
P(CF ) = 1/4.
We must compute
P(CF |AF ) =
P(AF |CF )P(CF )
(1/3)(1/4)
1/12
=
=
.
P(AF )
P(AF )
P(AF )
To compute the denominator, write
P(AF ) = P(AF |CF )P(CF ) + P(AF |CFc )P(CFc )
= (1/3)(1/4) + (1/10)(1 − 1/4) = 1/12 + 3/40 = 19/120.
It then follows that
P(CF |AF ) =
10
1 120
·
=
.
12 19
19
59. Let F denote the event that a patient receives a flu shot. Let S, M, and R denote the
events that Sue, Minnie, or Robin sees the patient. The problem tells us that
P(S) = .2, P(M) = .4, P(R) = .4, P(F|S) = .6, P(F|M) = .3, and P(F|R) = .1.
We must compute
P(S|F) =
P(F|S)P(S)
(.6)(.2)
.12
=
=
.
P(F)
P(F)
P(F)
Chapter 1 Problem Solutions
15
Next,
P(F) = P(F|S)P(S) + P(F|M)P(M) + P(F|R)P(R)
= (.6)(.2) + (.3)(.4) + (.1)(.4) = .12 + .12 + .04 = 0.28.
Thus,
P(S|F) =
60.
12 100
3
·
= .
100 28
7
(a) Let Ω = {1, 2, 3, 4, 5} with P(A) := |A|/|Ω|. Without loss of generality, let 1
and 2 correspond to the two defective chips. Then D := {1, 2} is the event that
a defective chip is tested. Hence, P(D) = |D|/5 = 2/5.
(b) Your friend’s information tells you that of the three chips you may test, one is
defective and two are not. Hence, the conditional probability that the chip you
test is defective is 1/3.
(c) Yes, your intuition is correct. To prove this, we construct a sample space and
probability measure and compute the desired conditional probability. Let
Ω := {(i, j, k) : i < j and k 6= i, k 6= j},
where i, j, k ∈ {1, 2, 3, 4, 5}. Here i and j are the chips taken by the friend, and
k is the chip that you test. We again take 1 and 2 to be the defective chips. The
10 possibilities for i and j are
12 13 14 15
23 24 25
34 35
45
For each pair in the above table, there are three possible values of k:
345 245 235 235
145 135 134
125 124
123
Hence, there are 30 triples in Ω. For the probability measure we take P(A) :=
|A|/|Ω|. Now let Fi j denote the event that the friend takes chips i and j with
i < j. For example, if the friend takes chips 1 and 2, then from the second table,
k has to be 3 or 4 or 5; i.e.,
F12 = {(1, 2, 3), (1, 2, 4), (1, 2, 5)}.
The event that the friend takes two chips is then
T := F12 ∪ F13 ∪ F14 ∪ F15 ∪ F23 ∪ F24 ∪ F25 ∪ F34 ∪ F35 ∪ F45 .
Now the event that you test a defective chip is
D := {(i, j, k) : k = 1 or 2 and i < j with i, j 6= k}.
16
Chapter 1 Problem Solutions
We can now compute
P(D|T ) =
P(D ∩ T )
.
P(T )
Since the Fi j that make up T are disjoint, |T | = 10·3 = 30 and P(T ) = |T |/|Ω| =
1. We next observe that
D ∩ T = ∅ ∪ [D ∩ F13 ] ∪ [D ∩ F14 ] ∪ [D ∩ F15 ]
∪ [D ∩ F23 ] ∪ [D ∩ F24 ] ∪ [D ∩ F25 ]
∪ [D ∩ F34 ] ∪ [D ∩ F35 ] ∪ [D ∩ F45 ].
Of the above intersections, the first six intersections are singleton sets, and the
last three are pairs. Hence, |D∩T | = 6·1+3·2 and so P(D∩T ) = 12/30 = 2/5.
We conclude that P(D|T ) = P(D ∩ T )/P(T ) = (2/5)/1 = 2/5, which is the
answer in part (a).
Remark. The model in part (c) can be used to solve part (b) by observing
that the probability in part (b) is
P(D|F12 ∪ F13 ∪ F14 ∪ F15 ∪ F23 ∪ F24 ∪ F25 ),
which can be similarly evaluated.
61.
(a) If two sets A and B are disjoint, then by definition, A ∩ B = ∅.
(b) If two events A and B are independent, then by definition, P(A∩B) = P(A)P(B).
(c) If two events A and B are disjoint, then P(A ∩ B) = P(∅) = 0. In order for
them to be independent, we must have P(A)P(B) = 0; i.e., at least one of the
two events must have zero probability. If two disjoint events both have positive
probability, then they cannot be independent.
62. Let W denote the event that the decoder outputs the wrong message. Of course, W c
is the event that the decoder outputs the correct message. We must find P(W ) =
1 − P(W c ). Now, W c occurs if only the first bit is flipped, or only the second bit is
flipped, or only the third bit is flipped, or if no bits are flipped. Denote these disjoint
events by F100 , F010 , F001 , and F000 , respectively. Then
P(W c ) = P(F100 ∪ F010 ∪ F001 ∪ F000 )
= P(F100 ) + P(F010 ) + P(F001 ) + P(F000 )
= p(1 − p)2 + (1 − p)p(1 − p) + (1 − p)2 p + (1 − p)3
= 3p(1 − p)2 + (1 − p)3 .
Hence,
P(W ) = 1 − 3p(1 − p)2 − (1 − p)3 = 3p2 − 2p3 .
If p = 0.1, then P(W ) = 0.03 − 0.002 = 0.028.
63. Let Ai denote the event that your phone selects channel i, i = 1, . . . , 10. Let B j denote
the event that your neighbor’s phone selects channel j, j = 1, . . . , 10. Let P(Ai ) =
Chapter 1 Problem Solutions
17
P(B j ) = 1/10, and assume Ai and B j are independent. Then
10
10
[
P
[Ai ∪ B j ] = ∑ P(Ai ∩ B j ) =
i=1
i=1
10
10
∑ P(Ai )P(B j )
i=1
=
1
1
∑ 10 · 10
= 0.1.
i=1
64. Let L denote the event that the left airbag works properly, and let R denote the event
that the right airbag works properly. Assume L and R are independent with P(L c ) =
P(R c ) = p. The probability that at least one airbag works properly is
P(L ∪ R) = 1 − P(L c ∩ R c ) = 1 − P(L c )P(R c ) = 1 − p2 .
65. The probability that the dart never lands within 2 cm of the center is
∞
N
\
\
c
c
P
An = lim P
An ,
N→∞
n=1
limit property of P,
n=1
N
= lim
N→∞
= lim
N→∞
∏ P(Anc ),
independence,
n=1
N
∏ (1 − p)
n=1
= lim (1 − p)N = 0.
N→∞
66. Let Wi denote the event that you win on your ith play of the lottery. The probability
that you win at least once in n plays is
n
n n
\
[
c
= 1 − ∏ P(Wic ), by independence,
Wi
Wi = 1 − P
P
i=1
i=1
i=1
n
= 1 − ∏(1 − p) = 1 − (1 − p)n .
i=1
We need to choose n so that 1 − (1 − p)n > 1/2, which happens if and only if
1/2 > (1 − p)n
or
− ln 2 > n ln(1 − p) or
− ln 2
< n,
ln(1 − p)
where the last step uses the fact that ln(1 − p) is negative. For p = 10−6 , we need
n > 693147.
67. Let A denote the event that Anne catches no fish, and let B denote the event that Betty
catches no fish. Assume A and B are independent with P(A) = P(B) = p. We must
compute
P(A ∩ [A ∪ B])
P(A)
P(A|A ∪ B) =
=
,
P(A ∪ B)
P(A ∪ B)
where the last step uses the fact that A ⊂ A ∪ B. To compute the denominator, write
P(A ∪ B) = 1 − P(A c ∩ B c ) = 1 − P(A c )P(B c ) = 1 − (1 − p)2 = 2p − p2 = p(2 − p).
18
Chapter 1 Problem Solutions
Then
P(A|A ∪ B) =
p
1
=
.
p(2 − p)
2− p
68. We show that A and B \C are independent as follows. First, since C ⊂ B,
P(B) = P(C) + P(B \C).
Next, since A and B are independent and since A and C are independent,
P(A ∩ B) = P(A)P(B) = P(A)[P(C) + P(B \C)] = P(A ∩C) + P(A)P(B \C).
Again using the fact that C ⊂ B, we now write
P(A ∩ B) = P(A ∩ [C ∪ B \C]) = P(A ∩C) + P(A ∩ B \C).
It follows that P(A ∩ B \C) = P(A)P(B \C), which establishes the claimed independence.
69. We show that A, B, and C are mutually independent. To begin, note that P(A) =
P(B) = P(C) = 1/2. Next, we need to identify the events
A ∩ B = [0, 1/4)
A ∩C = [0, 1/8) ∪ [1/4, 3/8)
B ∩C = [0, 1/8) ∪ [1/2, 5/8)
A ∩ B ∩C = [0, 1/8)
so that we can compute
P(A ∩ B) = P(A ∩C) = P(B ∩C) = 1/4
and
P(A ∩ B ∩C) = 1/8.
We find that
P(A ∩ B) = P(A)P(B),
P(A ∩C) = P(A)P(C),
P(B ∩C) = P(B)P(C),
and
P(A ∩ B ∩C) = P(A)P(B)P(C).
70. From a previous problem we have that P(A ∩ C|B) = P(A|B ∩ C)P(C|B). Hence,
P(A ∩C|B) = P(A|B)P(C|B) if and only if P(A|B ∩C) = P(A|B).
71. We show that the probability of the complementary event is zero. By the union bound,
∞
∞ ∞
∞
\
[ \
c
c
Bk ≤ ∑ P
Bk .
P
n=1 k=n
n=1
k=n
We show that every term on the right is zero. Write
∞
N
\
\
c
c
P
Bk = lim P
Bk ,
k=n
N→∞
k=n
limit property of P,
Chapter 1 Problem Solutions
19
N
∏ P(Bkc ),
N→∞
= lim
independence,
k=n
N
= lim
N→∞
≤ lim
N→∞
∏ [1 − P(Bk )]
k=n
N
∏ exp[−P(Bk )],
the hint,
k=n
N
= lim exp − ∑ P(Bk )
N→∞
= exp − lim
N→∞
k=n
N
P(B
)
∑ k ,
since exp is continuous,
k=n
∞
= exp − ∑ P(Bk )
k=n
−∞
= e
= 0,
∞
where the second-to-last step uses the fact that ∑∞
k=1 P(Bk ) = ∞ ⇒ ∑k=n P(Bk ) = ∞.
72. There are 3 · 5 · 7 = 105 possible systems.
73. There are 2n n-bit numbers.
74. There are 100! different orderings of the 100 message packets. In order that the first
header packet to be received is the 10th packet to arrive, the first 9 packets to be
received must come from the 96 data packets, the 10th packet must come from the 4
header packets, and the remaining 90 packets can be in any order. More specifically,
there are 96 possibilities for the first packet, 95 for the second, . . . , 88 for the ninth, 4
for the tenth, and 90! for the remaining 90 packets. Hence, the desired probability is
75.
5
2
96 · · · 88 · 4 · 90!
96 · · · 88 · 4
90 · 89 · 88 · 4
=
=
= 0.02996.
100!
100 · · · 91
100 · 99 · 98 · 97
= 10 pictures are needed.
76. Suppose the player chooses distinct digits wxyz. The player wins if any of the 4! = 24
permutations of wxyz occurs. Since each permutation has probability 1/10 000 of
occurring, the probability of winning is 24/10 000 = 0.0024.
77. There are 83 = 56 8-bit words with 3 ones (and 5 zeros).
78. The probability that a random byte has 4 ones and 4 zeros is 84 /28 = 70/256 =
0.2734.
79. In the first case, since the prizes are different, order is important. Hence, there are
41 · 40 · 39 = 63 960 outcomes. In the second case, since the prizes are the same, order
is not important. Hence, there are 41
3 = 10 660 outcomes.
80. There are 52
the deck contains 13 spades, 13 hearts, 13
14 possible hands. Since
13 13 13
diamonds, and 13 clubs, there are 13
5 hands with 2 spades, 3 hearts, 4
4
3
2
20
Chapter 1 Problem Solutions
diamonds, and 5 clubs. The probability of such a hand is
13 13 13 13
2
3
4
52
14
5
= 0.0116.
81. All five cards are of the same suit if and only if they are all spades or all hearts or all
diamonds or all clubs. These are four disjoint events. Hence, the answer is four times
the probability of getting all spades:
13
1287
5
4 52
= 0.00198.
= 4
2 598 960
5
82. There are
n k1 ,...,km
such partitions.
83. The general result is
n
k0 , . . . , km−1
mn .
4 /10 000 =
When n = 4 and m = 10 and a player chooses xxyz, we compute 2,1,1
4
0.0012. For xxyy, we compute 2,2 /10 000 = 0.0006. For xxxy, we compute
4
3,1 /10 000 = 0.0004.
84. Two apples and three carrots corresponds to (0, 0, 1, 1, 0, 0, 0). Five apples corresponds to (0, 0, 0, 0, 0, 1, 1).
CHAPTER 2
Problem Solutions
1.
(a) {ω : X(ω ) ≤ 3} = {1, 2, 3}.
(b) {ω : X(ω ) > 4} = {5, 6}.
(c) P(X ≤ 3) = P(X = 1) + P(X = 2) + P(X = 3) = 3(2/15) = 2/5, and
P(X > 4) = P(X = 5) + P(X = 6) = 2/15 + 1/3 = 7/15.
2.
(a) {ω : X(ω ) = 2} = {1, 2, 3, 4}.
(b) {ω : X(ω ) = 1} = {41, 42, . . . , 52}.
(c) P(X = 1 or X = 2) = P({1, 2, 3, 4} ∪ {41, 42, . . . , 52}). Since these are disjoint
events, the probability of their union is 4/52 + 12/52 = 16/52 = 4/13.
3.
(a) {ω ∈ [0, ∞) : X(ω ) ≤ 1} = [0, 1].
(b) {ω ∈ [0, ∞) : X(ω ) ≤ 3} = [0, 3].
R
(c) P(X ≤ 1) = 01 e−ω d ω = 1 − e−1 . P(X ≤ 3) = 1 − e−3 , P(1 < X ≤ 3) = P(X ≤
3) − P(X ≤ 1) = e−1 − e−3 .
4. First, since X −1 (∅) = ∅, µ (∅) = P(X −1 (∅)) = P(∅) = 0. Second, µ (B) =
P(X −1 (B)) ≥ 0. Third, for disjoint Bn ,
∞
∞
∞
∞
∞
[
[
[
−1
−1
X (Bn ) = ∑ P(X −1 (Bn )) = ∑ µ (Bn ).
=P
Bn
Bn = P X
µ
n=1
n=1
n=1
n=1
n=1
Fourth, µ (IR) = P(X −1 (IR)) = P(Ω) = 1.
5. Since
P(Y > n − 1) =
∞
∞
k=n
k=n+1
∑ P(Y = k) = P(Y = n) + ∑
P(Y = k) = P(Y = n) + P(Y > n),
it follows that P(Y = n) = P(Y > n − 1) − P(Y > n).
6. P(Y = 0) = P({TTT,THH,HTH,HHT}) = 4/8 = 1/2, and
P(Y = 1) = P({TTH,THT,HTT,HHH}) = 4/8 = 1/2.
7. P(X = 1) = P(X = 2) = P(X = 3) = P(X = 4) = P(X = 5) = 2/15 and P(X = 6) =
1/3.
8. P(X = 2) = P({1, 2, 3, 4}) = 4/52 = 1/13. P(X = 1) = P({41, 42, . . . , 52}) = 12/52 =
3/13. P(X = 0) = 1 − P(X = 2) − P(X = 1) = 9/13.
9. The possible values of X are 0, 1, 4, 9, 16. We have P(X = 0) = P({0}) = 1/7, P(X =
1) = P({−1, 1}) = 2/7, P(X = 4) = P({−2, 2}) = 2/7, P(X = 9) = P({3}) = 1/7,
and P(X = 16) = P({4}) = 1/7.
21
22
Chapter 2 Problem Solutions
10. We have
P(X > 1) = 1 − P(X ≤ 1) = 1 − [P(X = 0) + P(X = 1)]
= 1 − [e−λ + λ e−λ ] = 1 − e−λ (1 + λ ).
When λ = 1, P(X > 1) = 1 − e−2 (2) = 1 − 2/e = 0.264.
11. The probability that the sensor fails to activate is
P(X < 4) = P(X ≤ 3) = P(X = 0) + · · · + P(X = 3) = e−λ (1 + λ + λ 2 /2! + λ 3 /3!).
If λ = 2, P(X < 4) = e−2 (1 + 2 + 2 + 4/3) = e−2 (19/3) = 0.857. The probability
that the sensor activates is 1 − P(X < 4) = 0.143.
12. Let {Xk = 1} correspond to the event that the kth student gets an A. This event has
probability P(Xk = 1) = p. Now, the event that only the kth student gets an A is
{Xk = 1 and Xl = 0 for l 6= k}.
Hence, the probability that exactly one student gets an A is
15
[
{Xk = 1 and Xl = 0 for l 6= k} =
P
k=1
=
15
∑ P({Xk = 1 and Xl = 0 for l 6= k})
k=1
15
∑ p(1 − p)14
k=1
= 15(.1)(.9)14 = 0.3432.
13. Let X1 , X2 , X3 be the random digits of the drawing. Then P(Xi = k) = 1/10 for k =
0, . . . , 9 since each digit has probability 1/10 of being chosen. Then if the player
chooses d1 d2 d3 , the probability of winning is
P {X1 = d1 , X2 = d2 , X3 = d3 } ∪ {X1 = d1 , X2 = d2 , X3 6= d3 }
∪{X1 = d1 , X2 6= d2 , X3 = d3 } ∪ {X1 6= d1 , X2 = d2 , X3 = d3 } ,
which is equal to .13 +3[.12 (.9)] = 0.028 since the union is disjoint and since X1 , X2 , X3
are independent.
14.
m
m
[
\
P
{Xk < 2} = 1 − P
{Xk ≥ 2}
k=1
k=1
m
m
= 1 − ∏ P(Xk ≥ 2) = 1 − ∏ [1 − P(Xk ≤ 1)]
k=1
k=1
−λ
= 1 − [1 − {e
−λ
+λe
m
}]
= 1 − [1 − e−λ (1 + λ )]m .
Chapter 2 Problem Solutions
15.
23
(a)
n
n
n
[
\
P
{Xi ≥ 2} = 1 − P
{Xi ≤ 1} = 1 − ∏ P(Xi ≤ 1)
i=1
i=1
i=1
n
= 1 − ∏[P(Xi = 0) + P(Xi = 1)]
i=1
−λ
= 1 − [e
n
\
(b) P
{Xi ≥ 1} =
i=1
n
\
{Xi = 1} =
(c) P
i=1
n
∏ P(Xi ≥ 1)
n
=
i=1
n
∏ P(Xi = 1)
+ λ e−λ ]n = 1 − e−nλ (1 + λ )n .
∏[1 − P(Xi = 0)]
i=1
= (1 − e−λ )n .
= (λ e−λ )n = λ n e−nλ .
i=1
16. For the geometric0 pmf, write
∞
∞
k=0
k=0
∑ (1 − p)pk = (1 − p) ∑ pk = (1 − p) ·
1
= 1.
1− p
For the geometric1 pmf, write
∞
∑ (1 − p)pk−1
k=1
∞
∞
k=1
n=0
= (1 − p) ∑ pk−1 = (1 − p) ∑ pn = (1 − p) ·
1
= 1.
1− p
17. Let Xi be the price of stock i, which is a geometric0 (p) random variable. Then
29
29
29
\
[
{Xi ≤ 10} = 1 − ∏[(1 − p)(1 + p + · · · + p10 )]
{Xi > 10} = 1 − P
P
i=1
i=1
i=1
1 − p11 29
= 1 − (1 − p)
= 1 − (1 − p11 )29 .
1− p
Substituting p = .7, we have 1 − (1 − .711 )29 = 1 − (.98)29 = 1 − .560 = 0.44.
18. For the first problem, we have
n
\
P(min(X1 , . . . , Xn ) > `) = P
{Xk > `} =
k=1
n
n
∏ P(Xk > `)
k=1
=
∏ p`
= pn` .
k=1
Similarly,
n
n
n
\
{Xk ≤ `} = ∏ P(Xk ≤ `) = ∏ (1− p` ) = (1− p` )n .
P(max(X1 , . . . , Xn ) ≤ `) = P
k=1
k=1
k=1
19. Let Xk denote the number of coins in the pocket of the kth student. Then the Xk are
independent and is uniformly distributed from 0 to 20; i.e., P(Xk = i) = 1/21.
24
Chapter 2 Problem Solutions
25
25
\
{Xk ≥ 5} = ∏ 16/21 = (16/21)25 = 1.12 × 10−3 .
(a) P
k=1
k=1
25
25
\
[
{Xk ≤ 18} = 1 − (1 − 2/21)25 = 0.918.
{Xk ≥ 19} = 1 − P
(b) P
k=1
k=1
(c) The probability that only student k has 19 coins in his or her pocket is
\
P {Xk = 10} ∩ {Xl 6= 19} = (1/21)(20/21)24 = 0.01477.
l6=k
Hence, the probability that exactly one student has 19 coins is
25
[
\
{Xk = 10} ∩ {Xl 6= 19} = 25(0.01477) = 0.369.
P
l6=k
k=1
20. Let Xi = 1 if block i is good. Then P(Xi = 1) = p and
P(Y = k) = P {X1 = 1}∩· · ·∩{Xk−1 = 1}∩{Xk = 0} = pk−1 (1− p),
k = 1, 2, . . . .
Hence, Y ∼ geometric1 (p).
21.
(a) Write
∞
P(X > n) =
∑
(1 − p)pk−1 =
k=n+1
∞
∞
∑ (1 − p)p`+n
`=0
= (1 − p)pn ∑ p` = (1 − p)pn
`=0
1
= pn .
1− p
(b) Write
P(X > n + k|X > n) =
pn+k
P(X > n + k, X > n) P(X > n + k)
=
= n = pk .
P(X > n)
P(X > n)
p
22. Since P(Y > k) = P(Y > n + k|Y > n), we can write
P(Y > k) = P(Y > n + k|Y > n) =
P(Y > n + k,Y > n)
P(Y > n + k)
=
.
P(Y > n)
P(Y > n)
Let p := P(Y > 1). Taking k = 1 above yields P(Y > n + 1) = P(Y > n)p. Then
with n = 1 we have P(Y > 2) = P(Y > 1)p = p2 . With n = 2 we have P(Y > 3) =
P(Y > 2)P(Y > 1) = p3 . In general then P(Y > n) = pn . Finally, P(Y = n) = P(Y >
n − 1) − P(Y > n) = pn−1 − pn = pn−1 (1 − p), which is the geometric1 (p) pmf.
23.
(a) To compute pX (i), we sum row i of the matrix. This yields pX (1) = pX (3) = 1/4
and pX (2) = 1/2. To compute pY ( j), we sum column j to get pY (1) = pY (3) =
1/4 and pY (2) = 1/2.
Chapter 2 Problem Solutions
25
(b) To compute P(X < Y ), we sum pXY (i, j) over i and j such that i < j. We have
P(X < Y ) = pXY (1, 2) + pXY (1, 3) + pXY (2, 3) = 0 + 1/8 + 0 = 1/8.
(c) We claim that X and Y are not independent. For example, pXY (1, 2) = 0 is not
equal to pX (1)pY (2) = 1/8.
24.
(a) To compute pX (i), we sum row i of the matrix. This yields pX (1) = pX (3) = 1/4
and pX (2) = 1/2. To compute pY ( j), we sum column j to get pY (1) = pY (3) =
1/6 and pY (2) = 2/3.
(b) To compute P(X < Y ), we sum pXY (i, j) over i and j such that i < j. We have
P(X < Y ) = pXY (1, 2) + pXY (1, 3) + pXY (2, 3) = 1/6 + 1/24 + 1/12 = 7/24.
(c) Using the results of part (a), it is easy to verify that pX (i)pY ( j) = pXY (i, j) for
i, j = 1, 2, 3. Hence, X and Y are independent.
25. To compute the marginal of X, write
pX (1) =
∞
e−3
3
3j
e−3 3
=
e = 1/3.
3
j=0 j!
4e−6
6
4e−6 6
6j
=
e = 2/3.
6
j=0 j!
∑
Similarly,
pX (2) =
∞
∑
Alternatively, pX (2) = 1 − pX (1) = 1 − 1/3 = 2/3. Of course pX (i) = 0 for i 6= 1, 2.
We clearly have pY ( j) = 0 for j < 0 and
pY ( j) =
6 j−1 e−6
3 j−1 e−3
+4
,
j!
j!
j ≥ 0.
Since pX (1)pY ( j) 6= pXY (1, j), X and Y are not independent.
26.
(a) For k ≥ 1,
∞
pX (k) =
(1 − p)pk−1 kn e−k
n!
n=0
∑
= (1 − p)pk−1 e−k
∞
kn
∑ n!
n=0
= (1 − p)pk−1 e−k ek = (1 − p)pk−1 ,
which we recognize as the geometric1 (p) pmf.
(b) Next,
∞
pY (0) =
∑ (1 − p)pk−1 e−k =
k=1
=
1− p ∞
∑ (p/e)k−1
e k=1
1− p ∞
1
1− p
1− p
∑ (p/e)m = e · 1 − p/e = e − p .
e m=0
26
Chapter 2 Problem Solutions
(c) Since pX (1)pY (0) = (1 − p)2 /(e − p) is not equal to pXY (1, 0) = (1 − p)/e, X
and Y are not independent.
27. MATLAB. Here is a script:
p = ones(1,51)/51;
k=[0:50];
i = find(g(k) >= -16);
fprintf(’The answer is %g\n’,sum(p(i)))
where
function y = g(x)
y = 5*x.*(x-10).*(x-20).*(x-30).*(x-40).*(x-50)/1e6;
28. MATLAB. If you modified your program for the preceding problem only by the way
you compute P(X = k), then you may get only 0.5001 = P(g(X) ≥ −16 and X ≤ 50).
Note that g(x) > 0 for x > 50. Hence, you also have to add P(X ≥ 51) = p51 = 0.0731
to 0.5001 to get 0.5732.
29. MATLAB. OMITTED.
30. MATLAB. OMITTED.
31. MATLAB. OMITTED.
32. E[X] = 2(1/3) + 5(2/3) = 12/3 = 4.
33. E[I(2,6) (X)] = ∑5k=3 P(X = k) = (1− p)[p3 + p4 + p5 ]. For p = 1/2, we get E[I(2,6) (X)] =
7/64 = 0.109375.
34. Write
∞
λ n e−λ
e−λ ∞ λ n+1
e−λ ∞ λ n
=
=
∑
∑ n!
n!
λ n=0 (n + 1)!
λ n=1
n=0
e−λ λ
1 − e−λ
e−λ ∞ λ n
−1 =
[e − 1] =
.
=
∑
λ n=0 n!
λ
λ
E[1/(X + 1)] =
1
∑ n+1
35. Since var(X) = E[X 2 ] − (E[X])2 , E[X 2 ] = var(X) + (E[X])2 . Hence, E[X 2 ] = 7 + 22 =
7 + 4 = 11.
36. Since Y = cX, E[Y ] = E[cX] = cm. Hence,
var(Y ) = E[(Y − cm)2 ] = E[(cX − cm)2 ] = E[c2 (X − m)2 ]
= c2 E[(X − m)2 ] = c2 var(X) = c2 σ 2 .
37. We begin with
E[(X +Y )3 ] = E[X 3 + 3X 2Y + 3XY 2 +Y 3 ]
= E[X 3 ] + 3E[X 2Y ] + 3E[XY 2 ] + E[Y 3 ]
= E[X 3 ] + 3E[X 2 ]E[Y ] + 3E[X]E[Y 2 ] + E[Y 3 ],
by independence.
Chapter 2 Problem Solutions
27
Now, as noted in the text, for a Bernoulli(p) random variable, X n = X, and so E[X n ] =
E[X] = p. Similarly E[Y n ] = q. Thus,
E[(X +Y )3 ] = p + 3pq + 3pq + q = p + 6pq + q.
38. The straightforward approach is to put
f (c) := E[(X − c)2 ] = E[X 2 ] − 2mc + c2
and differentiate with respect to c to get f 0 (c) = −2m + 2c. Solving f 0 (c) = 0 results
in c = m. An alternative approach is to write
E[(X − c)2 ] = E[{(X − m) + (m − c)}2 ] = σ 2 + (m − c)2 .
From this expression, it is obvious that c = m minimizes the expectation.
39. The two sketches are:
x /a
1/2
I[ a , )( x )
1
8
8
( x /a )
I[ a , )( x )
1
a
a
40. The right-hand side is easy: E[X]/2 = (3/4)/2 = 3/8 = 0.375. The left-hand side is
more work:
P(X ≥ 2) = 1 − P(X ≤ 1) = 1 − [P(X = 0) + P(X = 1)] = 1 − e−λ (1 + λ ).
For λ = 3/4, P(X ≥ 2) = 0.1734. So the bound is a little more than twice the value
of the probability.
41. The Chebyshev bound is (λ + λ 2 )/4. For λ = 3/4, the bound is 0.3281, which is
a little better than the Markov inequality bound in the preceding problem. The true
probability is 0.1734.
42. Comparing the definitions of ρXY and cov(X,Y ), we find ρXY = cov(X,Y )/(σX σY ).
Hence, cov(X,Y ) = σX σY ρXY . Since cov(X,Y ) := E[(X − mX )(Y − mY ), if Y = X,
we see that cov(X, X) = E[(X − mX )2 ] =: var(X).
43. Put
f (a) := E[(X − aY )2 ] = E[X 2 ] − 2aE[XY ] + a2 E[Y 2 ] = σX2 − 2aρσX σY + a2 σY2 .
Then
f 0 (a) = −2ρσX σY + 2aσY2 .
Setting this equal to zero and solving for a yields a = ρ (σX /σY ).
28
Chapter 2 Problem Solutions
44. Since P(X = ±1) = P(X = ±2) = 1/4, E[X] = 0. Similarly, since P(XY = ±1) = 1/4
and P(XY = ±4) = 1/4, E[XY ] = 0. Thus, E[XY ] = 0 = E[X]E[Y ] and we see that X
and Y are uncorrelated. Next, since X = 1 implies Y = 1, P(X = 1,Y = 1) = P(X =
1) = 1/4 while P(Y = 1) = P(X = 1 or X = −1) = 1/2. Thus,
P(X = 1)P(Y = 1) = (1/4)(1/2) = 1/8,
P(X = 1,Y = 1) = 1/4.
but
45. As discussed in the text, for uncorrelated random variables, the variance of the sum
is the sum of the variances. Since independent random variables are uncorrelated, the
same results holds for them too. Hence, for Y = X1 + · · · + XM ,
M
∑ var(Xk ).
var(Y ) =
k=1
We also have E[Y ] = ∑M
k=1 E[Xk ]. Next, since the Xk are i.i.d. geometric1 (p), E[Xk ] =
1/(1− p) and var(Xk ) = p/(1− p)2 . It follows that var(Y ) = M p/(1− p)2 and E[Y ] =
M/(1 − p). We conclude by writing
E[Y 2 ] = var(Y ) + (E[Y ])2 =
Mp
M2
M(p + M)
+
=
.
2
(1 − p)
(1 − p)2
(1 − p)2
46. From E[Y ] = E[dX − s(1 − X)] = d p − s(1 − p) = 0, we find that d/s = (1 − p)/p.
47.
(a) p = 1/1000.
(b) Since (1− p)/p = (999/1000)/(1/1000) = 999, the fair odds against are 999 :1.
(c) Since the fair odds of 999 :1 are not equal to the offered odds of 500 :1, the game
is not fair. To make the game fair, the lottery should pay $900 instead of $500.
48. First note that
Z ∞
1
1
tp
dt =





∞
1
1
·
1 − p t p−1
1
, p 6= 1,
∞
lnt ,
p = 1.
1
For p > 1, the integral is equal to 1/(p − 1). For p ≤ 1, the integral is infinite.
For 0 < p ≤ 1, write
∞
1
∑ kp ≥
k=1
∞
∑
Z k+1
1
k=1 k
t
dt =
p
Z ∞
1
1
tp
dt = ∞.
p
For p > 1, it suffices to show that ∑∞
k=2 1/k < ∞. To this end, write
∞
1
∑ kp =
k=2
∞
∞
1
∑ (k + 1) p ≤ ∑
k=1
k=1
49. First write
E[X n ] =
∞
∑ kn
k=1
Z k+1
1
k
t
dt =
p
Z ∞
1
1
tp
∞
C−1
1
p
−1
=
C
.
∑
p
p
p−n
k
k=1 k
dt < ∞.
Chapter 2 Problem Solutions
29
By the preceding problem this last sum is finite for p − n > 1, or equivalently, n <
p − 1. Otherwise the sum is infinite; the case 1 ≥ p − n > 0 being handled by the
preceding problem, and the case 0 ≥ p − n being obvious.
50. If all outcomes are equally likely,
n
H(X) =
1
∑ pi log pi
=
i=1
1 n
∑ log n = log n.
n i=1
If X is a constant random variable with pi = 0 for i 6= j, then
n
H(X) =
1
∑ pi log pi
= p j log
i=1
1
= 1 log 1 = 0.
pj
51. Let P(X = xi ) = pi for i = 1, . . . , n. Then
n
E[g(X)] =
∑ g(xi )pi
and
i=1
n
g(E[X]) = g ∑ xi pi .
i=1
For n = 2, Jensen’s inequality says that
p1 g(x1 ) + p2 g(x2 ) ≥ g(p1 x1 + p2 x2 ).
If we put λ = p1 , then 1 − λ = p2 and the above inequality becomes
λ g(x1 ) + (1 − λ )g(x2 ) ≥ g(λ x1 + (1 − λ )x2 ),
which is just the definition of a convex function. Hence, if g is convex, Jensen’s
inequality holds for n = 2. Now suppose Jensen’s inequality holds for some n ≥ 2.
We must show it holds for n + 1. The case of n is
n
n
g(x
)p
≥
g
x
p
, if p1 + · · · + pn = 1.
i
i
i
i
∑
∑
i=1
i=1
Now suppose that p1 + · · · + pn+1 = 1, and write
n
∑ g(xi )pi = (1 − pn+1 ) ∑ g(xi )
n+1
i=1
i=1
pi
+ pn+1 g(xn+1 ).
1 − pn+1
Let us focus on the quantity in brackets. Since
n
pi
∑ 1 − pn+1
i=1
=
p1 + · · · + pn
1 − pn+1
=
= 1,
pn+1
1 − pn+1
Jensen’s inequality for n terms yields
n
pi
pi
∑ g(xi ) 1 − pn+1 ≥ g ∑ xi 1 − pn+1 .
i=1
i=1
n
30
Chapter 2 Problem Solutions
Hence,
n
g(x
)p
≥
(1
−
p
)g
i
i
n+1
∑
∑ xi
n+1
i=1
i=1
pi
1 − pn+1
+ pn+1 g(xn+1 ).
Now apply the two-term Jensen inequality to get
n
∑ g(xi )pi ≥ g (1 − pn+1 ) ∑ xi
n+1
i=1
n+1
= g ∑ pi xi .
i=1
pi
+ pn+1 xn+1
1 − pn+1
i=1
52. With X = |Z|α and g(x) = xβ /α , we have
E[g(X)] = E[X β /α ] = E[(|Z|α )β /α ] = E[|Z|β ]
and
E[X] = E[|Z|α ].
Then Jensen’s inequality tells us that
E[|Z|β ] ≥ (E[|Z|α ])β /α .
Raising both sides the 1/β power yields Lyapunov’s inequality.
53.
(a) For all discrete random variables, we have ∑i P(X = xi ) = 1. For a nonnegative
random variable, if xk < 0, we have
1 =
∑ P(X = xi )
i
≥
∑ I[0,∞) (xi )P(X = xi ) + P(X = xk )
= 1 + P(X = xk ).
i
From this it follows that 0 ≥ P(X = xk ) ≥ 0, and so P(X = xk ) = 0.
(b) Write
E[X] =
∑ xi P(X = xi )
i
=
∑
i:xi ≥0
xi P(X = xi ) +
∑
xi P(X = xk ).
k:xk <0
By part (a), the last sum is zero. The remaining sum is obviously nonnegative.
(c) By part (b), 0 ≤ E[X −Y ] = E[X] − E[Y ]. Hence, E[Y ] ≤ E[X].
CHAPTER 3
Problem Solutions
1. First, E[X] = G0X (1) =
1
6
+ 43 z
z=1
1
6
=
+
4
3
=
9
6
= 32 . Second, E[X(X − 1)] =
G00X (1) = 4/3. Third, from E[X(X −1)] = E[X 2 ]−E[X], we have E[X 2 ] = 4/3+3/2 =
17/6. Finally, var(X) = E[X 2 ] − (E[X])2 = 17/6 − 9/4 = (34 − 27)/12 = 7/12.
= 1/6, pX (1) = G0X (0) = 16 + 43 z
= 1/6,
2. pX (0) = GX (0) = 16 + 61 z + 23 z2
z=0
z=0
and pX (2) = G00X (0)/2 = (4/3)/2 = 2/3.
3. To begin, note that G0X (z) = 5((2 + z)/3)4 /3, and G00X (z) = 20((2 + z)/3)3 /9. Hence,
E[X] = G0X (1) = 5/3 and E[X(X − 1)] = 20/9. Since E[X(X − 1)] = E[X 2 ] − E[X],
E[X 2 ] = 20/9 + 5/3 = 35/9. Finally, var(X) = E[X 2 ] − (E[X])2 = 35/9 − 25/9 =
10/9.
4. For X ∼ geometric0 (p),
GX (z) =
∞
∞
∞
n=0
n=0
n=0
1
∑ zn P(X = n) = ∑ zn (1 − p)pn = (1 − p) ∑ (zp)n = (1 − p) 1 − pz .
Then
E[X] = G0X (1) =
Next,
1− p
p
(1 − pz)2
=
z=1
1− p
p
p=
.
(1 − p)2
1− p
E[X 2 ] − E[X] = E[X(X − 1)] = G00X (1) =
=
2p2
(1 − p)p · 2p
=
.
(1 − p)3
(1 − p)2
(1 − p)p
· 2p(1 − pz)
(1 − pz)4
z=1
This implies that
E[X 2 ] =
2p2
p
2p2 + p(1 − p)
p + p2
+
=
=
.
(1 − p)2 1 − p
(1 − p)2
(1 − p)2
Finally,
p + p2
p2
p
−
=
.
2
2
(1 − p)
(1 − p)
(1 − p)2
var(X) = E[X 2 ] − (E[X])2 =
For X ∼ geometric1 (p),
∞
GX (z) =
∑ zn P(X = n) =
n=1
∞
∑ zn (1 − p)pn−1 =
n=1
(1 − p)z
1 − p pz
=
.
=
p 1 − pz
1 − pz
31
1− p ∞
∑ (zp)n
p n=1
32
Chapter 3 Problem Solutions
Now
G0X (z) =
Hence,
1− p
(1 − pz)(1 − p) + (1 − p)pz
=
.
(1 − pz)2
(1 − pz)2
E[X] = G0X (1) =
Next,
G00X (z) =
We then have
1
.
1− p
(1 − p)
2p(1 − p)
· 2(1 − pz)p =
.
(1 − pz)4
(1 − pz)3
E[X 2 ] − E[X] = E[X(X − 1)] = G00X (1) =
and
E[X 2 ] =
Finally,
2p(1 − p)
2p
=
,
(1 − p)3
(1 − p)2
1
2p + 1 − p
1+ p
2p
+
=
=
.
2
2
(1 − p)
1− p
(1 − p)
(1 − p)2
var(X) = E[X 2 ] − (E[X])2 =
1
p
1+ p
−
=
.
(1 − p)2 (1 − p)2
(1 − p)2
5. Since the Xi are independent Poisson(λi ), we use probability generating functions to
find GY (z), which turns out to be Poisson(λ ) with λ := λ1 + · · · + λn . It then follows
that P(Y = 2) = λ 2 e−λ /2. It remains to write
GY (z) = E[zY ] = E[zX1 +···+Xn ] = E[zX1 · · · zXn ] =
= exp ∑ λi (z − 1) = eλ (z−1) ,
n
∏ E[zXi ]
i=1
n
=
∏ eλi (z−1)
i=1
n
i=1
which is the Poisson(λ ) pgf. Hence, Y ∼ Poisson(λ ) as claimed.
∞
k
6. If GX (z) = ∑∞
k=0 z P(X = k), then GX (1) = ∑k=0 P(X = k) = 1. For the particular
formula in the problem, we must have 1 = GX (1) = (a0 + a1 + a2 + · · · + an )m /D, or
D = (a0 + a1 + a2 + · · · + an )m .
7. From the table on the inside of the front cover of the text, E[Xi ] = 1/(1 − p). Thus,
n
E[Y ] =
∑ E[Xi ]
=
i=1
n
.
1− p
Second, since the variance of the sum of uncorrelated random variables is the sum of
the variances, and since var(Xi ) = p/(1 − p)2 , we have
n
var(Y ) =
∑ var(Xi )
i=1
=
np
.
(1 − p)2
Chapter 3 Problem Solutions
33
It then easily follows that
np
n2
n(p + n)
+
=
.
2
(1 − p)
(1 − p)2
(1 − p)2
E[Y 2 ] = var(Y ) + (E[Y ])2 =
Since Y is the sum of i.i.d. geometric1 (p) random variables, the pgf of Y is the product
of the individual pgfs. Thus,
n
(1 − p)z
(1 − p)z n
=
.
GY (z) = ∏
1 − pz
i=1 1 − pz
8. Starting with GY (z) = [(1 − p) + pz]n , we have
E[Y ] = GY0 (1) = n[(1 − p) + pz]n−1 p
z=1
= np.
Next,
E[Y (Y − 1)] = GY00 (1) = n(n − 1)[(1 − p) + pz]n−2 p2
z=1
= n(n − 1)p2 .
It then follows that
E[Y 2 ] = E[Y (Y − 1)] + E[Y ] = n(n − 1)p2 + np = n2 p2 − np2 + np.
Finally,
var(Y ) = E[Y 2 ] − (E[Y ])2 = n2 p2 − np2 + np − (np)2 = np(1 − p).
9. For the binomial(n, p) random variable,
n
GY (z) =
∑ P(Y = k)zk =
k=0
Hence,
n
∑ k pk (1 − p)n−k .
k=0
n
1 = GY (1) =
10. Starting from
n
∑ k pk (1 − p)n−k zk .
k=0
n
n
∑ k pk (1 − p)n−k = 1,
k=0
n
we follow the hint and replace p with a/(a + b). Note that 1 − p = b/(a + b). With
these substitutions, the above equation becomes
k n−k
n n n
a
b
n
ak bn−k
=
1
or
= 1.
∑
∑
k
n−k
a+b
a+b
k=0 k
k=0 k (a + b) (a + b)
Multiplying through by (a + b)n we have
n n
∑ k ak bn−k = (a + b)n .
k=0
34
Chapter 3 Problem Solutions
11. Let Xi = 1 if bit i is in error, Xi = 0 otherwise. Then Yn := X1 + · · · + Xn is the number
of errors in n bits. Assume the Xi are independent. Then
GYn (z) = E[zYn ] = E[zX1 +···+Xn ] = E[zX1 · · · zXn ]
= E[zX1 ] · · · E[zXn ], by independence,
= [(1 − p) + pz]n , since the Xi are i.i.d. Bernoulli(p).
We recognize this last expression
as the binomial(n, p) probability generating func
tion. Thus, P(Yn = k) = nk pk (1 − p)n−k .
12. Let Xi ∼ binomial(ni , p) denote the number of students in the ith room. Then Y =
X1 + · · · + XM is the total number of students in the school. Next,
GY (z) = E[zY ] = E[zX1 +···+XM ]
= E[zX1 ] · · · E[zXM ],
by independence,
M
=
∏[(1 − p) + pz]ni ,
since Xi ∼ binomial(ni , p),
i=1
= [(1 − p) + pz]n1 +···+nM .
Setting
n := n1 + · · · + nM , we see that Y ∼ binomial(n, p). Hence, P(Y = k) =
n k
p
(1
− p)n−k .
k
13. Let Y = X1 +· · ·+Xn , where the Xi are i.i.d. with P(Xi = 1) = 1− p and P(Xi = 2) = p.
Observe that Xi − 1 ∼ Bernoulli(p). Hence,
n
Y = n + ∑ (Xi − 1) .
i=1
|
{z
=: Z
}
Since Z is the sum of i.i.d. Bernoulli(p) random variables, Z ∼ binomial(n, p). Hence,
P(Y = k) = P(n + Z = k) = P(Z = k − n)
n
=
pk−n (1 − p)2n−k , k = n, . . . , 2n.
k−n
14. Let Xi be i.i.d. Bernoulli(p), where Xi = 1 means bit i is flipped. Then Y := X1 + · · · +
Xn (n = 10) is the number of bits flipped. A codeword cannot be decoded if Y > 2.
We need to find P(Y > 2). Observe that
GY (z) = E[zY ] = E[zX1 +···+Xn ] = E[zX1 · · · zXn ] =
n
∏ E[zXi ]
i=1
= [(1 − p) + pz]n .
This is the pgf of a binomial(n, p) random variable. Hence,
P(Y > 2) = 1 − P(Y ≤ 2) = 1 − [P(Y = 0) + P(Y = 1) + P(Y = 2)]
10
10
10 2
= 1−
(1 − p)10 +
p(1 − p)9 +
p (1 − p)8
0
1
2
= 1 − (1 − p)8 [(1 − p)2 + 10p(1 − p) + 45p2 ].
Chapter 3 Problem Solutions
35
15. For n = 150 and p = 1/100, we have
k P Binomial(n, p) = k
P Poisson(np) = k
0
1
2
3
4
5
0.2215
0.3355
0.2525
0.1258
0.0467
0.0138
0.2231
0.3347
0.2510
0.1255
0.0471
0.0141
16. If the Xi are i.i.d. with mean m, then
n 1 n
nm
1 n
1
X
E[X
]
=
m =
=
= m.
E[Mn ] = E
i
i
∑
∑
∑
n i=1
n i=1
n i=1
n
If X is any random variable with mean m, then E[cX] = cE[X] = cm, and
var(cX) = E[(cX − cm)2 ] = E[c2 (X − m)2 ] = c2 E[(X − m)2 ] = c2 var(X).
17. In general, we have
P(|Mn − m| < ε ) ≥ 0.9
⇔
P(|Mn − m| ≥ ε ) < 0.1.
By Chebyshev’s inequality,
P(|Mn − m| ≥ ε ) ≤
σ2
< 0.1
nε 2
if n > σ 2 /(.1)ε 2 . For σ 2 = 1 and ε = 0.25, we require n > 1/(.1)(.25)2 = 1/.00625 =
160 students. If instead ε = 1, we require n > 0.1 = 10 students.
18.
(a) E[Xi ] = E[IB (Zi )] = P(Zi ∈ B). Setting p := P(Zi ∈ B), we see that Xi = IB (Zi ) ∼
Bernoulli(p). Hence, var(Xi ) = p(1 − p).
(b) In fact, the Xi are independent. Hence, they are uncorrelated.
19. Mn = 0 if and only if all the Xi are zero. Hence,
n
\
{Xi = 0} =
P(Mn = 0) = P
i=1
n
∏ P(Xi = 0)
i=1
= (1 − p)n .
In particular, if p = 1/1000, then P(M100 = 0) = (1 − p)100 = 0.999100 = 0.905.
Hence, the chances are more than 90% that when we run a simulation, M100 = 0 and
we learn nothing!
20. If Xi = Z ∼ Bernoulli(1/2) for all i, then
Mn =
1 n
1 n
Xi = ∑ Z = Z,
∑
n i=1
n i=1
36
Chapter 3 Problem Solutions
and m = E[Xi ] = E[Z] = 1/2. So,
P(|Mn − m| ≥ 1/4) = P(|Z − 1/2| ≥ 1/4).
Now, Z − 1/2 = ±1/2, and |Z − 1/2| = 1/2 with probability one. Thus,
P(|Z − 1/2| ≥ 1/4) = P(1/2 ≥ 1/4) = 1 6→ 0.
21. From the discussion of the weak law in the text, we have
P(|Mn − m| ≥ εn ) ≤
σ2
.
nεn2
If nεn2 → ∞ as n → ∞, then probability on the left will go to zero as n → ∞.
22. We have from the example that with p := λ /(λ + µ ), pX|Z (i| j) = ij pi (1 − p) j−i
for i = 0, . . . , j. In other words, as a function of i, pX|Z (i| j) is a binomial( j, p) pmf.
Hence,
j
E[X|Z = j] =
∑ ipX|Z (i| j)
i=0
is just the mean of a binomial( j, p) pmf. The mean of such a pmf is j p. Hence,
E[X|Z = j] = j p = jλ /(λ + µ ).
23. The problem is telling us that P(Y = k|X = i) = nk pki (1 − pi )n−k . Hence
P(Y < 2|X = i) = P(Y = 0|X = i) + P(Y = 1|X = i) = (1 − pi )n + npi (1 − pi )n−1 .
24. The problem is telling us that P(X = k|Y = j) = λ jk e−λ j /k!. Hence,
P(X > 2|Y = j) = 1 − P(X ≤ 2|Y = j)
= 1 − [P(X = 0|Y = j) + P(X = 1|Y = j) + P(X = 2|Y = j)]
= 1 − [e−λ j + λ j e−λ j + λ j2 e−λ j /2]
= 1 − e−λ j [1 + λ j + λ j2 /2].
25. For the first formula, write
pX|Y (xi |y j ) :=
P(X = xi ,Y = y j ) P(X = xi )P(Y = y j )
=
= P(X = xi ) = pX (xi ).
P(Y = y j )
P(Y = y j )
Similarly, for the other formula,
pY |X (y j |xi ) :=
P(Y = y j , X = xi ) P(Y = y j )P(X = xi )
=
= P(Y = y j ) = pY (y j ).
P(X = xi )
P(X = xi )
Chapter 3 Problem Solutions
37
26. We use the law of total probability to write
∞
∑ P(X −Y = n|Y = k)P(Y = k)
P(T = n) = P(X −Y = n) =
k=0
∞
=
∑ P(X − k = n|Y = k)P(Y = k),
by the substitution law,
k=0
∞
=
∑ P(X = n + k|Y = k)P(Y = k)
k=0
∞
=
∑ P(X = n + k)P(Y = k),
by independence,
k=0
∞
=
∑ (1 − p)pn+k · (1 − q)qk
k=0
∞
= (1 − p)(1 − q)pn ∑ (pq)k =
k=0
(1 − p)(1 − q)pn
.
1 − pq
27. The problem is telling us that
µ n e−µ
n!
P(Y = n|X = 1) =
and
P(Y = n|X = 2) =
ν n e−ν
.
n!
The problem also tells us that P(X = 1) = P(X = 2) = 1/2. We can now write
P(X = 1|Y = 2) =
P(X = 1,Y = 2) P(Y = 2|X = 1)P(X = 1) (µ 2 e−µ /2)(1/2)
=
=
.
P(Y = 2)
P(Y = 2)
P(Y = 2)
It remains to use the law of total probability to compute
2
P(Y = 2) =
∑ P(Y = 2|X = i)P(X = i)
i=1
= [P(Y = 2|X = 1) + P(Y = 2|X = 2)]/2
= [µ 2 e−µ /2 + ν 2 e−ν /2]/2 = [µ 2 e−µ + ν 2 e−ν ]/4.
We conclude by writing
P(X = 1|Y = 2) =
1
µ 2 e−µ /4
=
.
[µ 2 e−µ + ν 2 e−ν ]/4
1 + (ν /µ )2 eµ −ν
28. Let X = 0 or X = 1 according to whether message zero or message one is sent. The
problem tells us that P(X = 0) = P(X = 1) = 1/2 and that
P(Y = k|X = 0) = (1 − p)pk
and
P(Y = k|X = 1) = (1 − q)qk ,
where q 6= p. We need to compute
P(X = 1|Y = k) =
(1 − q)qk (1/2)
P(Y = k|X = 1)P(X = 1)
=
.
P(Y = k)
P(Y = k)
38
Chapter 3 Problem Solutions
We next use the law of total probability to compute
P(Y = k) = [P(Y = k|X = 0) + P(Y = k|X = 1)]/2 = [(1 − p)pk + (1 − q)qk ]/2.
We can now compute
P(X = 1|Y = k) =
1
(1 − q)qk (1/2)
=
.
(1−p)pk
[(1 − p)pk + (1 − q)qk ]/2
1 + (1−q)qk
29. Let R denote the number of red apples in a crate, and let G denote the number of green
apples in a crate. The problem is telling us that R ∼ Poisson(ρ ) and G ∼ Poisson(γ )
are independent. If T = R + G is the total number of apples in the crate, we must
compute
P(T = k|G = 0)P(G = 0)
P(G = 0|T = k) =
.
P(T = k)
We first use the law of total probability, substitution, and independence to write
P(T = k|G = 0) = P(R+G = k|G = 0) = P(R = k|G = 0) = P(R = k) = ρ k e−ρ /k!.
We also note from the text that the sum of two independent Poisson random variables
is a Poisson random variable whose parameter is the sum of the individual parameters.
Hence, P(T = k) = (ρ + γ )k e−(ρ +γ ) /k!. We can now write
ρ k e−ρ /k! · e−γ
=
P(G = 0|T = k) =
(ρ + γ )k e−(ρ +γ ) /k!
ρ
ρ +γ
k
.
30. We begin with
n −λ
e
1
· λ n!
P(Y = 1|X = n)P(X = n)
= n+1
.
P(X = n|Y = 1) =
P(Y = 1)
P(Y = 1)
Next, we compute
∞
P(Y = 1) =
=
=
∑ P(Y = 1|X = n)P(X = n)
n=0
e−λ
λ
e−λ
λ
∞
λ n+1
∑ (n + 1)!
=
n=0
[eλ − 1] =
e−λ
λ
1 − e−λ
λ
∞
∑
k=1
∞
=
1
∑ n+1 ·
n=0
λk
k!
=
e−λ
λ
λ n e−λ
n!
λk
−
1
∑
k=0 k!
∞
.
We conclude with
n −λ
n −λ
e
1
· λ n!
·λ e
λ n+1
P(X = n|Y = 1) =
.
= n+1 −λn!
= λ
P(Y = 1)
(1 − e )/λ
(e − 1)(n + 1)!
1
n+1
Chapter 3 Problem Solutions
39
31. We begin with
P(Y = k|X = n)P(X = n)
=
P(X = n|Y = k) =
P(Y = k)
n k
n−k · λ n e−λ /n!
k p (1 − p)
P(Y = k)
.
Next,
n k
p (1 − p)n−k λ n e−λ /n!
P(Y = k) = ∑ P(Y = k|X = n)P(X = n) = ∑
n=0
n=k k
∞
∞
∞
=
pk λ k e−λ
k!
pk λ k e−λ
[(1 − p)λ ]n−k
=
(n − k)!
k!
n=k
=
(pλ )k e−pλ
pk λ k e−λ (1−p)λ
=
e
.
k!
k!
∑
Note that Y ∼ Poisson(pλ ). We continue with
P(X = n|Y = k) =
=
n k
n−k · λ n e−λ /n!
k p (1 − p)
P(Y = k)
=
[(1 − p)λ ]n−k e−(1−p)λ
.
(n − k)!
∞
[(1 − p)λ ]m
m!
m=0
∑
n k
n−k · λ n e−λ /n!
k p (1 − p)
(pλ )k e−pλ /k!
32. First write
P {X > k} ∩ {max(X,Y ) > k}
P(X > k)
,
P X > k max(X,Y ) > k =
=
P max(X,Y ) > k
P max(X,Y ) > k
since {X > k} ⊂ {max(X,Y ) > k}. We next compute
P max(X,Y ) > k = 1 − P max(X,Y ) ≤ k = 1 − P(X ≤ k)P(Y ≤ k).
If we put θk := P(X ≤ k) and use the fact that X and Y have the same pmf, then
1 − θk
1
1 − θk
=
=
.
P X > k max(X,Y ) > k =
(1 − θk )(1 + θk )
1 + θk
1 − θk2
With n = 100 and p = .01, we compute
θ1 = P(X ≤ 1) = P(X = 0) + P(X = 1) = .99100 + .9999 = .366 + .370 = .736.
It follows that the desired probability is 1/(1 + θ1 ) = 1/1.736 = 0.576.
33.
(a) Observe that
P(XY = 4) = P(X = 1,Y = 4) + P(X = 2,Y = 2) + P(X = 4,Y = 1)
= (1 − p)(1 − q)[pq4 + p2 q2 + p4 q].
40
Chapter 3 Problem Solutions
(b) Write
pZ ( j) =
∑ pY ( j − i)pX (i)
i
∞
=
∑ pY ( j − i)pX (i),
since pX (i) = 0 for i < 0,
∑ pY ( j − i)pX (i),
since pY (k) = 0 for k < 0,
i=0
j
=
i=0
j
j
= (1 − p)(1 − q) ∑ pi q j−i = (1 − p)(1 − q)q j ∑ (p/q)i .
i=0
i=0
Now, if p = q,
pZ ( j) = (1 − p)2 p j ( j + 1).
If p 6= q,
pZ ( j) = (1 − p)(1 − q)q j
1 − (p/q) j+1
q j+1 − p j+1
= (1 − p)(1 − q)
.
1 − p/q
q− p
34. For j = 0, 1, 2, 3,
j
∑ (1/16)
pZ ( j) =
= ( j + 1)/16.
i=0
For j = 4, 5, 6,
3
∑
pZ ( j) =
(1/16) = (7 − j)/16.
i= j−3
For other values of j, pZ ( j) = 0.
35. We first write
P(Y = j|X = 1)
P(X = 0)
≥
P(Y = j|X = 0)
P(X = 1)
as
λ1j e−λ1 / j!
λ0j e−λ0 / j!
≥
1− p
.
p
We can further simplify this to
λ j
1
λ0
≥
1 − p λ1 −λ0
e
.
p
Taking logarithms and rearranging, we obtain
h
1 − p i.
ln(λ1 /λ0 ).
j ≥ λ1 − λ0 + ln
p
Observe that the right-hand side is just a number (threshold) that is computable from
the problem data. If we observe Y = j, we compare j to the threshold. If j is greater
than or equal to this number, we decide X = 1; otherwise, we decide X = 0.
Chapter 3 Problem Solutions
36. We first write
as
41
P(Y = j|X = 1)
P(X = 0)
≥
P(Y = j|X = 0)
P(X = 1)
(1 − q1 )q1j
(1 − q0 )q0j
We can further simplify this to
q j
1
q0
≥
≥
1− p
.
p
(1 − p)(1 − q0 )
.
p(1 − q1 )
Taking logarithms and rearranging, we obtain
h (1 − p)(1 − q ) i.
0
j ≤ ln
ln(q1 /q0 ),
p(1 − q1 )
since q1 < q0 implies ln(q1 /q0 ) < 0.
37. Starting with P(X = xi |Y = y j ) = h(xi ), we have
P(X = xi ,Y = y j ) = P(X = xi |Y = y j )P(Y = y j ) = h(xi )pY (y j ).
If we can show that h(xi ) = pX (xi ), then it will follow that X and Y are independent.
Now observe that the sum over j of the left-hand side reduces to P(X = xi ) = pX (xi ).
The sum over j of the right-hand side reduces to h(xi ). Hence, pX (xi ) = h(xi ) as
desired.
38. First write
pXY (1, j) = (1/3)3 j e−3 / j!
and
pXY (2, j) = (4/6)6 j e−6 / j!
Notice that 3 j e−3 / j! is a Poisson(3) pmf and 6 j e−6 / j! is a Poisson(6) pmf. Hence,
pX (1) = ∑∞j=0 pXY (1, j) = 1/3 and pX (2) = ∑∞j=0 pXY (2, j) = 2/3. It then follows
that pY |X ( j|1) is Poisson(3) and pY |X ( j|2) is Poisson(6). With these observations, it
is clear that
E[Y |X = 1] = 3 and E[Y |X = 2] = 6,
and
E[Y ] = E[Y |X = 1](1/3) + E[Y |X = 2](2/3) = 3(1/3) + 6(2/3) = 1 + 4 = 5.
To obtain E[X|Y = j], we first compute
pY ( j) = pXY (1, j) + pXY (2, j) = (1/3)3 j e−3 / j! + (2/3)6 j e−6 / j!
and
pX|Y (1| j) =
and
pX|Y (2| j) =
1
(1/3)3 j e−3 / j!
=
(1/3)3 j e−3 / j! + (2/3)6 j e−6 / j!
1 + 2 j+1 e−3
(2/3)6 j e−6 / j!
(1/3)3 j e−3 / j! + (2/3)6 j e−6 / j!
=
1
.
1 + 2−( j+1) e3
42
Chapter 3 Problem Solutions
We now have
E[X|Y = j] = 1 ·
=
1
1 + 2 j+1 e−3
1
+2·
2 j+1 e−3
1 + 2 j+2 e−3
=
.
1 + 2 j+1 e−3
1 + 2 j+1 e−3
+2·
1 + 2 j+1 e−3
1
1 + 2−( j+1) e3
39. Since Y is conditionally Poisson(k) given X = k, E[Y |X = k] = k. Hence,
∞
∞
E[Y ] =
∑ E[Y |X = k]P(X = k)
∑ kP(X = k)
=
= E[X] =
k=1
k=1
1
,
1− p
since X ∼ geometric1 (p). Next
∞
E[XY ] =
∞
∑ E[XY |X = k]P(X = k)
=
∑ kE[Y |X = k]P(X = k)
=
k=1
∞
=
k=1
∞
k=1
= var(X) + (E[X])2 =
∑ E[kY |X = k]P(X = k), by substitution,
∑ k2 P(X = k)
= E[X 2 ]
k=1
p
1
1+ p
+
=
.
2
2
(1 − p)
(1 − p)
(1 − p)2
Since E[Y 2 |X = k] = k + k2 ,
E[Y 2 ] =
∞
∑ E[Y 2 |X = k]P(X = k)
k=1
∞
=
∑ (k + k2 )P(X = k)
= E[X] + E[X 2 ]
k=1
1+ p
2
1
+
=
.
=
1 − p (1 − p)2
(1 − p)2
Finally, we can compute
var(Y ) = E[Y 2 ] − (E[Y ])2 =
2
1
1
−
=
.
(1 − p)2 (1 − p)2
(1 − p)2
40. From the solution of the example, it is immediate that E[Y |X = 1] = λ and E[Y |X =
0] = λ /2. Next,
E[Y ] = E[Y |X = 0](1 − p) + E[Y |X = 1]p = (1 − p)λ /2 + pλ .
Similarly,
E[Y 2 ] = E[Y 2 |X = 0](1 − p) + E[Y 2 |X = 1]p
= (λ /2 + λ 2 /4)(1 − p) + (λ + λ 2 )p.
To conclude, we have
var(Y ) = E[Y 2 ] − (E[Y ])2 = (λ /2 + λ 2 /4)(1 − p) + (λ + λ 2 )p − [(1 − p)λ /2 + pλ ]2 .
Chapter 3 Problem Solutions
43
41. Write
E[(X + 1)Y 2 ] =
1
1
∑ E[(X + 1)Y 2 |X = i]P(X = i)
=
i=0
i=0
1
=
∑ E[(i + 1)Y 2 |X = i]P(X = i)
∑ (i + 1)E[Y 2 |X = i]P(X = i).
i=0
Now, since given X = i, Y is conditionally Poisson(3(i + 1)),
E[Y 2 |X = i] = (λ + λ 2 )
λ =3(i+1)
= 3(i + 1) + 9(i + 1)2 .
It now follows that
E[(X + 1)Y 2 ] =
1
∑ (i + 1)[3(i + 1) + 9(i + 1)2 ]P(X = i)
i=0
1
=
∑ (i + 1)2 [3 + 9(i + 1)]P(X = i)
i=0
= 12(1/3) + 84(2/3) = 4 + 56 = 60.
42. Write
∞
∞
E[XY ] =
∑ E[XY |X = n]P(X = n)
=
∑ E[nY |X = n]P(X = n)
n=0
∞
n=0
∞
λ n e−λ
n!
n=0
n=0
X +1−1
1
X
= E
= 1−E
.
= E
X +1
X +1
X +1
=
∑ nE[Y |X = n]P(X = n)
=
1
∑ nn+1
By a problem in the previous chapter, this last expectation is equal to (1 − e−λ )/λ .
Hence,
1 − e−λ
E[XY ] = 1 −
.
λ
43. Write
∞
E[XY ] =
∞
∑ E[XY |X = n]P(X = n)
=
∑ nE[Y |X = n]P(X = n)
=
n=1
∞
=
n=1
∑ E[nY |X = n]P(X = n)
n=1
∞
n
∑ n 1 − q P(X = n)
n=1
i
1
1 h
=
var(X) + (E[X])2
E[X 2 ] =
1−q
1−q
p
1
1+ p
1
+
=
.
=
2
2
1 − q (1 − p)
(1 − p)
(1 − q)(1 − p)2
44
Chapter 3 Problem Solutions
44. Write
E[X 2 ] =
∞
∑ E[X 2 |Y = k]P(Y = k)
k=1
∞
=
∑ (k + k2 )P(Y = k)
= E[Y +Y 2 ]
n=1
= E[Y ] + E[Y 2 ] = m + (r + m2 ) = m + m2 + r.
45. Using probability generating functions, we see that
GV (z) = E[zX+Y ] = E[zX zY ] = E[zX ]E[zY ]
= [(1 − p) + pz]n [(1 − p) + pz]m = [(1 − p) + pz]n+m .
Thus, V ∼ binomial(n + m, p). We next compute
P(V = 10|X = 4) = P(X +Y = 10|X = 4) = P(4 +Y = 10|X = 4)
m 6
= P(Y = 6|X = 4) = P(Y = 6) =
p (1 − p)m−6 .
6
46. Write
GY (z) = E[zY ] =
∞
∑ E[zY |X = k]P(X = k)
∞
=
k=1
∞
=
∑ (ez−1 )k P(X = k)
k=1
= GX (ez−1 ) =
∑ ek(z−1) P(X = k)
k=1
(1 − p)ez−1
.
1 − pez−1
CHAPTER 4
Problem Solutions
1. Let Vi denote the input voltage at the ith sampling time. The problem tells us that the
Vi are independent and uniformly distributed on [0, 7]. The alarm sounds if Vi > 5 for
i = 1, 2, 3. The probability of this is
3
3
\
{Vi > 5} = ∏ P(Vi > 5).
P
i=1
i=1
Now, P(Vi > 5) =
8/343 = 0.0233.
2. We must solve
R∞
t
R7
5
(1/7) dt = 2/7. Hence, the desired probability is (2/7)3 =
f (x) dx = 1/2 for t. Now,
Z ∞
t
2x−3 dx = −
√
Solving 1/t 2 = 1/2, we find that t = 2.
1
x2
∞
=
t
1
.
t2
R
3. To find c, we solve 01 cx−1/2 dx = 1. The left-hand side of this equation is 2cx1/2 |10 =
R
2c. Solving 2c = 1 yields c = 1/2. For the median, we must solve t1 (1/2)x−1/2 dx =
1/2 or x1/2 |t1 = 1/2. We find that t = 1/4.
4.
(a) For t ≥ 0, P(X > t) =
R∞
t
λ e−λ x dx = −e−λ x |t∞ = e−λ t .
(b) First, P(X > t + ∆t|X > t) = P(X > t + ∆t, X > t)/P(X > t). Next, observe that
{X > t + ∆t} ∩ {X > t} = {X > t + ∆t},
and so P(X > t + ∆t|X > t) = P(X > t + ∆t)/P(X > t) = e−λ (t+∆t) /e−λ t =
e−λ ∆t .
5. Let Xi denote the voltage output by regulator i. Then the Xi are i.i.d. exp(λ ) random
variables. Now put
10
Y :=
∑ I(v,∞) (Xi )
i=1
so that Y counts the number of regulators that output more than v volts. We must
compute P(Y = 3). Now, the I(v,∞) (Xi ) are i.i.d. Bernoulli(p) random variables, where
p = P(Xi > v) =
Z ∞
v
λ e−λ x dx = −e−λ x
∞
= e−λ v .
v
Next, we now from the previous chapter that a sum of n i.i.d. Bernoulli(p) random
variables is a binomial(n, p). Thus,
n 3
10 −3λ v
(1 − e−λ v )7 = 120e−3λ v (1 − e−λ v )7 .
P(Y = 3) =
p (1 − p)n−3 =
e
3
3
45
46
Chapter 4 Problem Solutions
R∞
−2λ .
λ e−λ x dx = −e−λ x |∞
2 =e
T
(a) P min(X1 , . . . , Xn ) > 2 = P ni=1 {Xi > 2} = ∏ni=1 P(Xi > 2) = e−2nλ .
6. First note that P(Xi > 2) =
2
(b) Write
P max(X1 , . . . , Xn ) > 2 = 1 − P max(X1 , . . . , Xn ) ≤ 2
n
\
= 1−P
{Xi ≤ 2}
i=1
n
= 1 − ∏ P(Xi ≤ 2) = 1 − [1 − e−2λ ]n .
i=1
7.
(a) P(Y ≤ 2) =
R2
0
µ e−µ y dy = 1 − e−2µ .
(b) P(X ≤ 12,Y ≤ 12) = P(X ≤ 12)P(Y ≤ 12) = (1 − e−12λ )(1 − e−12µ ).
(c) Write
P({X ≤ 12} ∪ {Y ≤ 12}) = 1 − P(X > 12,Y > 12)
= 1 − P(X > 12)P(Y > 12)
= 1 − e−12λ e−12µ = 1 − e−12(λ +µ ) .
8.
(a) Make the change of variable y = λ x p , dy = λ px p−1 dx to get
Z ∞
0
p
λ px p−1 e−λ x dx =
Z ∞
0
∞
e−y dy = −e−y
= 1.
0
(b) The same change of variables also yields
P(X > t) =
Z ∞
t
p
λ px p−1 e−λ x dx =
Z ∞
λt p
p
e−y dy = e−λ t .
(c) The probability that none of the Xi exceeds 3 is
n
n
\
p
{Xi ≤ 3} = ∏ P(Xi ≤ 3) = [1 − P(X1 > 3)]n = [1 − e−λ 3 ]n .
P
i=1
i=1
The probability that at least one of them exceeds 3 is
n
n
[
\
p
P
{Xi > 3} = 1 − P
{Xi ≤ 3} = 1 − [1 − e−λ 3 ]n .
i=1
9.
i=1
√
/ 2π , f 0 (x) < 0 for x > 0 and f 0 (x) > 0 for x < 0.
√
2
(b) Since f 00 (x) = (x2 −1)e−x /2 / 2π , we see that f 00 (x) > 0 for |x| > 1 and f 00 (x) <
0 for |x| < 1.
(a) Since f 0 (x) = −xe−x
(c) Rearrange ex
2 /2
2 /2
≥ x2 /2 to get e−x
2 /2
≤ 2/x2 → 0 as |x| → ∞.
Chapter 4 Problem Solutions
47
10. Following the hint, write f (x) = ϕ ((x − m)/σ )/σ , where ϕ is the standard normal
density. Observe that f 0 (x) = ϕ 0 ((x − m)/σ )/σ 2 and f 00 (x) = ϕ 00 ((x − m)/σ )/σ 3 .
(a) Since the argument of ϕ 0 is positive for x > m and negative for x < m, f (x) is
decreasing for x > m and increasing for x < m. Hence, f has a global maximum
at x = m.
(b) Since the absolute value of the argument of ϕ 00 is greater than one if and only if
|x − m| > σ , f (x) is concave for |x − m| < σ and convex for |x − m| > σ .
11. Since ϕ is bounded, limσ →∞ ϕ ((x − m)/σ )/σ = 0. Hence, limσ →∞ f (x) = 0. For
x 6= m, we have
2 . i
h 2
exp − x−m
σ
2σ
2
√
≤ √
→ 0
f (x) =
2 = √
2π σ
2π (x − m)2
2π σ x−m
σ
√
as σ → 0. Otherwise, since f (m) = [ 2π σ ]−1 , limσ →0 f (m) = ∞.
12. (a) f (x) = ∑n pn fn (x) is obviously nonnegative. Also,
Z ∞
−∞
f (x) dx =
Z ∞
∑ pn fn (x) dx =
−∞ n
∑ pn
n
Z ∞
−∞
fn (x) dx =
∑ pn
= 1.
n
3/4
1/2
(c)
1/4
0
=
n
3/4
(b)
∑ pn · 1
1/2
1/4
x
0
13. Clearly, (g ∗ h)(x) =
1
2
R∞
0
3
x
0
1
2
3
−∞ g(y)h(x − y) dy ≥
0 since g and h are nonnegative. Next,
g(y)h(x − y) dy dx
(g ∗ h)(x) dx =
−∞
−∞
−∞
Z ∞
Z ∞
g(y)
h(x − y) dx dy
=
−∞
−∞
Z ∞
Z ∞
Z ∞
g(y) dy = 1.
=
g(y)
h(θ ) d θ dy =
−∞
−∞
−∞
|
{z
}
Z ∞
Z ∞ Z ∞
=1
14.
R
(a) Let p > 1. On Γ(p) = 0∞ x p−1 e−x dx, use integration by parts with u = x p−1 and
dv = e−x dx. Then du = (p − 1)x p−2 dx, v = −e−x , and
∞
Γ(p) = −x p−1 e−x +(p − 1)
{z 0}
|
=0
Z ∞
0
x(p−1)−1 e−x dx = (p − 1)Γ(p − 1).
48
Chapter 4 Problem Solutions
√
R
(b) On Γ(1/2) = 0∞ x−1/2 e−x dx,√make the change of variable x = y2 /2 or y = 2x.
Then dx = y dy and x−1/2 = 2/y. Hence,
Z ∞√
√ Z ∞ −y2 /2
√ √ Z ∞ e−y2 /2
2 −y2 /2
√
y dy = 2
dy = 2 2π
e
Γ(1/2) =
dy
e
y
0
0
0
2π
√ √
√
1
= 2 2π · = π .
2
(c) By repeatedly using the recursion formula in part (a), we have
2n − 1
2n − 3
2n − 1
2n − 1 2n − 3
2n + 1
=
Γ
=
·
Γ
Γ
2
2
2
2
2
2
..
.
2n − 1 2n − 3
5 3 1
=
·
· · · · · Γ(1/2)
2
2
2 2 2
2n − 1 2n − 3
5 3 1 √
=
·
··· · · · π
2
2
2 2 2
(2n − 1)!! √
=
π.
2n
(d) First note that g p (y) = 0 for y ≤ 0, and similarly for gq (y). Hence, in order to
have g p (y)gq (x − y) > 0, we need y > 0 and x − y > 0, or equivalently, x > y > 0.
Of course, if x ≤ 0 this does not happen. Thus, (g p ∗ gq )(x) = 0 for x ≤ 0. For
x > 0, we follow the hint and write
(g p ∗ gq )(x) =
=
Z ∞
−∞
Z x
0
=
g p (y)gq (x − y) dy
g p (y)gq (x − y) dy
1
Γ(p)Γ(q)
Z x
0
y p−1 e−y · (x − y)q−1 e−(x−y) dy
Z
xq−1 e−x x p−1
y (1 − y/x)q−1 dy
=
Γ(p)Γ(q) 0
Z 1
xq e−x
=
(xθ ) p−1 (1 − θ )q−1 d θ ,
Γ(p)Γ(q) 0
=
x p+q−1 e−x
Γ(p)Γ(q)
Z 1
0
ch. of var. θ = y/x,
θ p−1 (1 − θ )q−1 d θ .
(∗)
Now, the left-hand side is a convolution of densities, and is therefore a density
by Problem 13. In particular, this means
that the left-hand side integrates to
R
one. On the right-hand side, note that 0∞ x p+q−1 e−x dx = Γ(p + q). Hence,
integrating the above equation with respect to x from zero to infinity yields
1 =
Γ(p + q)
Γ(p)Γ(q)
Z 1
0
θ p−1 (1 − θ )q−1 d θ .
Solving for the above integral and substituting the result into (∗), we find that
(g p ∗ gq )(x) = g p+q (x).
Chapter 4 Problem Solutions
15.
R
∞
(a) In −∞
fλ (x) dx =
λ dx to get
R∞
−∞ λ
49
f (λ x) dx, make the change of variable y = λ x, dy =
Z ∞
−∞
fλ (x) dx =
Z ∞
−∞
f (y) dy = 1.
(b) Observe that
g1,λ (x) = λ
(λ x)0 e−λ x
= λ e−λ x ,
0!
which we recognize as the exp(λ ) density.
(c) The desired probability is
Z ∞
Pm (t) :=
t
R
λ
(λ x)m−1 e−λ x
dx.
(m − 1)!
Note that P1 (t) = t∞ λ e−λ x dx = e−λ t . For m > 1, apply integration by parts
with u = (λ x)m−1 /(m − 1)! and dv = λ e−λ x dx. Then
Pm (t) =
(λ t)m−1 e−λ t
+ Pm−1 (t).
(m − 1)!
Applying this result recursively, we find that
(λ t)m−1 e−λ t (λ t)m−2 e−λ t
+
+ · · · + e−λ t .
(m − 1)!
(m − 2)!
Pm (t) =
(d) We have
g 2m+1 , 1 (x) =
2
2
1
2
( 21 x)m−1/2 e−x/2
(1/2)m (1/2)1/2 xm−1/2 e−x/2
=
(2m−1)!! √
Γ((2m + 1)/2)
π
m
2
xm−1/2 e−x/2
√ .
=
(2m − 1) · · · 5 · 3 · 1 · 2π
16.
(a) We see that b1,1√
(x) = 1 is the uniform(0, 1) density, b2,2 (x) = 6x(1 − x), and
b1/2,1 (x) = 1/(2 x ).
2
b
(x)
1/2,1
b (x)
2,2
1.5
b (x)
1
1,1
0.5
0
0
0.25
0.5
x
0.75
1
50
Chapter 4 Problem Solutions
(b) From Problem 14(d) and its hint, we have
g p+q (x) = (g p ∗ gq )(x) =
x p+q−1 e−x
Γ(p)Γ(q)
Z 1
0
θ p−1 (1 − θ )q−1 d θ .
Integrating the left and right-hand sides with respect to x from zero to infinity
yields
Z
Γ(p + q) 1 p−1
θ
1 =
(1 − θ )q−1 d θ ,
Γ(p)Γ(q) 0
which says that the beta density integrates to one.
17. Starting with
Γ(p) Γ(q) = Γ(p + q)
Z 1
0
u p−1 (1 − u)q−1 du,
make the change of variable u = sin2 θ , du = 2 sin θ cos θ d θ . We obtain
Γ(p) Γ(q) = Γ(p + q)
Z 1
0
= Γ(p + q)
u p−1 (1 − u)q−1 du
Z π /2
0
= 2Γ(p + q)
(sin2 θ ) p−1 (1 − sin2 θ )q−1 · 2 sin θ cos θ d θ
Z π /2
(sin θ )2p−1 (cos θ )2q−1 d θ .
0
Setting p = q = 1/2 on both sides yields
Γ(1/2)2 = 2
√
and it follows that Γ(1/2) =
Z π /2
1 dθ = π ,
0
π.
18. Starting with
Γ(p) Γ(q) = Γ(p + q)
Z 1
0
u p−1 (1 − u)q−1 du,
make the change of variable u = sin2 θ , du = 2 sin θ cos θ d θ . We obtain
Γ(p) Γ(q)
=
Γ(p + q)
=
Z 1
0
u p−1 (1 − u)q−1 du
Z π /2
0
= 2
(sin2 θ ) p−1 (1 − sin2 θ )q−1 · 2 sin θ cos θ d θ
Z π /2
(sin θ )2p−1 (cos θ )2q−1 d θ .
0
Setting p = (n + 1)/2 and q = 1/2 on both sides yields
n+1 √
π
Γ
Z π /2
2
= 2
sinn θ d θ ,
n+2
0
Γ
2
and the desired result follows.
Chapter 4 Problem Solutions
51
19. Starting with the integral definition of B(p, q), make the change of variable u = 1 −
e−θ , which implies both du = e−θ d θ and 1 − u = e−θ . Hence,
B(p, q) =
Z 1
0
u p−1 (1 − u)q−1 du =
=
Z ∞
0
(1 − e−θ ) p−1 (e−θ )q−1 e−θ d θ
0
(1 − e−θ ) p−1 e−qθ d θ .
Z ∞
20. We first use the fact that the density is even and then makepthe change of variable
eθ = 1 + x2 /ν , which implies both eθ d θ = 2x/ν dx and x = ν (eθ − 1). Thus,
Z ∞
Z ∞
x2 −(ν +1)/2
x2 −(ν +1)/2
dx = 2
dx
1+
1+
ν
ν
−∞
0
Z ∞
1
= 2
(eθ )−(ν +1)/2 · ν2 eθ p
dθ
0
ν (eθ − 1)
√ Z ∞ θ −ν /2 θ 1/2 p
(e )
(e ) / eθ − 1 d θ
= ν
0
√ Z ∞ θ −ν /2
(e )
(1 − e−θ )−1/2 d θ
= ν
0
√ Z∞
= ν
(1 − e−θ )1/2−1 e−θ ν /2 d θ .
0
√
By the preceding problem, this is equal to ν B(1/2, ν /2), and we see that Student’s t
density integrates to one.
21.
(a) Using Stirling’s formula,
1+ν 1 + ν ν /2
√ 1 + ν (1+ν )/2−1/2 −(1+ν )/2
2π
Γ
e
e−1/2
2 ≈
2
2
=
√ √ ν ν /2−1/2 −ν /2
√ ν ν /2−1/2
√
ν
e
ν 2π
ν
νΓ
2
2
2
1 + ν ν /2 (ν /2)1/2
1
√
=
e−1/2 = [(1 + 1/ν )ν ]1/2 √
ν
ν
2 e1/2
1
1
→ (e1 )1/2 √
= √ .
1/2
2e
2
(b) First write
1+
2
2
x2 1/2
x2 ν 1/2 x2 (ν +1)/2
= 1+
1+
→ [ex ]1/2 11/2 = ex /2 .
ν
ν
ν
It then follows that
1+ν 2 −(ν +1)/2
Γ
1 + xν
1
2 ·
=
fν (x) =
√
√
√ ν
2 (ν +1)/2
ν B( 12 , ν2 )
π 1 + xν
νΓ
2
2
→ √
e−x /2
1
√
.
2 /2 =
x
2π e
2π
52
Chapter 4 Problem Solutions
22. Making the change of variable t = 1/(1 + z) as suggested in the hint, note that it is
equivalent to 1 + z = t −1 , which implies dz = −t −2 dt. Thus,
Z 1
Z 1
Z ∞
p−1
1
1 − t p−1 p+q−2
z p−1
p+q dt
t
dt
dz
=
−
1
t
=
p+q
2
t
t
t
0
0
0 (1 + z)
Z 1
=
0
(1 − t) p−1t q−1 dt = B(q, p) = B(p, q).
Hence, fZ (z) integrates to one.
Z ∞
23. E[X] =
1
x·
2
dx =
x3
Z ∞
1
2x−2 dx =
−2
x
∞
= 2.
1
24. If the input-output relation has n levels, then the distance from −Vmax to +Vmax should
be n∆; i.e., n∆ = 2Vmax , or ∆ = 2Vmax /n. Next, we have from the example in the text
that the performance is ∆2 /12, and we need ∆2 /12 < ε , or
1 2Vmax 2
< ε.
12
n
√
Solving this for n yields Vmax / 3ε < n = 2b . Taking natural logarithms, we have
√ .
ln 2.
b > ln Vmax / 3ε
25. We use the change of variable x = z − m as follows:
E[Z] =
=
26. E[X 2 ] =
Z ∞
1
Z ∞
−∞
Z ∞
x2 ·
−∞
Z ∞
z fZ (z) dz =
−∞
x f (x) dx + m
2
dx =
x3
Z ∞
2
x
1
Z ∞
−∞
z f (z − m) dz =
Z ∞
−∞
f (x) dx = E[X] + m = 0 + m = m.
∞
dx = 2 ln x
1
= 2(∞ − 0) = ∞.
27. First note that since Student’s t density is even, E[|X|k ] =
tional to
Z ∞
0
xk
dx =
(1 + x2 /ν )(ν +1)/2
Z 1
0
(x + m) f (x) dx
xk
dx +
(1 + x2 /ν )(ν +1)/2
R∞
k
−∞ |x| f ν (x) dx
Z ∞
1
is propor-
xk
dx
(1 + x2 /ν )(ν +1)/2
With regard to this last integral, observe that
Z ∞
1
xk
dx ≤
2
(1 + x /ν )(ν +1)/2
Z ∞
1
xk
dx = ν (ν +1)/2
2
(x /ν )(ν +1)/2
Z ∞
1
dx
,
xν +1−k
the range of
which is finite if ν + 1 − k > 1, or k < ν . Next, instead of breaking
√
integration at one, we break it at the solution of x2 /ν = 1, or x = ν . Then
Z ∞
√
ν
xk
dx ≥
2
(1 + x /ν )(ν +1)/2
Z ∞
√
ν
xk
dx =
2
(x /ν + x2 /ν )(ν +1)/2
which is infinite if ν + 1 − k ≤ 1, or k ≥ ν .
Z ∞
ν (ν +1)/2
√
2
ν
dx
,
xν +1−k
Chapter 4 Problem Solutions
53
28. Begin with E[Y 4 ] = E[(Z + n)4 ] = E[Z 4 + 4Z 3 n + 6Z 2 n2 + 4Zn3 + n4 ]. The moments
of the standard normal were computed in an example in this chapter. Hence E[Y 4 ] =
3 + 4 · 0 · n + 6 · 1 · n2 + 4 · 0 · n3 + n4 = 3 + 6n2 + n4 .
29. E[X n ] =
Z ∞
xn
0
30.
x p−1 e−x
1
dx =
Γ(p)
Γ(p)
Z ∞
0
Γ(n + p)
.
Γ(p)
x(n+p)−1 e−x dx =
(a) First write
E[X] =
Z ∞
0
−x2 /2
x · xe
1
dx =
2
Z ∞
2 −x2 /2
x e
−∞
√ Z
2
2π ∞ 2 e−x /2
dx =
x √
dx,
2 −∞
2π
where the last integral is√the second moment of a standard normal density, which
2π p
is one. Hence, E[X] =
= π /2.
2
(b) For higher-order moments, first write
Z ∞
E[X n ] =
0
xn · xe−x
2 /2
dx =
Z ∞
0
xn+1 e−x
2 /2
dx.
2
Now
√ make the change of variable t = x /2, which implies x =
dt/ 2t. Hence,
E[X n ] =
Z ∞
0
[(2t)1/2 ]n+1 e−t
= 2n/2
Z ∞
0
√
2t, or dx =
dt
21/2t 1/2
t [(n/2)+1]−1 e−t dt = 2n/2 Γ(1 + n/2).
31. Let Xi denote the flow on link i, and put Yi := I(β ,∞) (Xi ) so that Yi = 1 if the flow
on link i is greater than β . Put Z := ∑ni=1 Yi so that Z counts the number of links
with flows greater than β . The buffer overflows if Z > 2. Since the Xi are i.i.d.,
so are the Yi . Furthermore, the Yi are Bernoulli(p), where p = P(Xi > β ). Hence,
Z ∼ binomial(n, p). Thus,
P(Z > 2) = 1 − P(Z ≤ 2)
n 2
n
n
p (1 − p)n−2
p(1 − p)n−1 +
(1 − p)n +
= 1−
2
1
0
= 1 − (1 − p)n−2 [(1 − p)2 + np(1 − p) + 21 n(n − 1)p2 ].
In remains to compute
p = P(Xi > β ) =
Z ∞
β
xe−x
2 /2
dx = −ex
2 /2
∞
= e−β
2 /2
.
β
32. The key is to use the change of variable θ = λ x p , which implies both d θ = λ px p−1 dx
and x = (θ /λ )1/p . Hence,
E[X n ] =
Z ∞
0
p
xn · λ px p−1 e−λ x dx =
n/p
= (1/λ )
Z ∞
0
Z ∞
0
[(θ /λ )1/p ]n e−θ d θ
θ [(n/p)+1]−1 e−θ d θ = Γ(1 + n/p) λ n/p .
54
Chapter 4 Problem Solutions
33. Write
Z ∞
x1/2 e−x dx =
0
√
= Γ(3/2) = (1/2)Γ(1/2) = π /2.
E[Y ] = E[(X 1/4 )2 ] = E[X 1/2 ] =
Z ∞
0
x3/2−1 e−x dx
34. We have
n
n
\
[
{Xi ≥ µ /2}
{Xi < µ /2} = 1 − P
P
i=1
i=1
n
= 1 − ∏ P(Xi ≥ µ /2)
i=1
= 1−
Z
∞
µ /2
−λ x
λe
dx
n
,
with λ := 1/µ ,
= 1 − (e−λ µ /2 )n = 1 − e−n/2 .
35. Let Xi ∼ exp(λ ) be i.i.d., where λ = 1/20. We must compute
5
5
[
\
P
{Xi > 25} = 1 − P
{Xi ≤ 25}
i=1
i=1
5
= 1 − ∏ P(Xi ≤ 25)
i=1
= 1−
Z
25
−λ x
λe
0
dx
5
= 1 − (1 − e−25λ )5 = 1 − (1 − e−5/4 )5 = 0.815.
36. The first two calculations are
h(X) =
Z 2
(1/2) log 2 dx = log 2 and
0
h(X) =
Z 1/2
2 log(1/2) dx = log(1/2).
0
For the third calculation, note that − ln f (x) = 21 [(x − m)/σ ]2 + 12 ln 2πσ 2 . Then
Z ∞
f (x) [(x − m)/σ ]2 + ln 2πσ 2 dx
−∞
1
X −m 2
+ ln 2πσ 2 = 21 {1 + ln 2πσ 2 } =
=
E
2
σ
h(X) =
1
2
37. The main difficulty is to compute
Z ∞
−∞
x2n (1 + x2 /ν )−(ν +1)/2 dx.
1
2
ln 2πσ 2 e.
Chapter 4 Problem Solutions
55
First use the fact that the integrand is even and then makep
the change of variable
eθ = 1 + x2 /ν , which implies both eθ d θ = 2x/ν dx and x = ν (eθ − 1). Thus,
Z ∞
−∞
x2n (1 + x2 /ν )−(ν +1)/2 dx = ν
= ν
Z ∞
x2n−1 (1 + x2 /ν )−(ν +1)/2
0
Z ∞ p
0
2x
dx
ν
( ν (eθ − 1) )2n−1 (eθ )−(ν +1)/2 eθ d θ
= ν n+1/2
= ν n+1/2
Z ∞
0
(eθ − 1)n−1/2 e−θ (ν +1)/2 eθ d θ
0
(1 − e−θ )n−1/2 e−θ (ν −2n)/2 d θ
Z ∞
Z ∞
(1 − e−θ )(n+1/2)−1 e−θ (ν −2n)/2 d θ
= ν n+1/2 B n + 1/2, (ν − 2n)/2 , by Problem 19.
= ν n+1/2
0
Hence,
E[X 2n ] = ν n+1/2 B n + 1/2, (ν − 2n)/2 · √
= ν n+1/2
38. From MX (s) = eσ
2 s2 /2
ν −2n
Γ( 2n+1
2 )Γ( 2 )
Γ( ν +1
2 )
1
ν B( 21 , ν2 )
ν −2n
2n+1
Γ( ν +1
n Γ( 2 )Γ( 2 )
2 )
·√
=
.
ν
ν Γ( 12 )Γ( ν2 )
Γ( 12 )Γ( ν2 )
, we have MX0 (s) = MX (s)σ 2 s and then
MX00 (s) = MX (s)σ 4 s2 + MX (s)σ 2 .
Since MX (0) = 1, we have MX00 (1) = σ 2 .
2
39. Let M(s) := es /2 denote the moment generating function of the standard normal random variable. For the N(m, σ 2 ) moment generating function, we use the change of
variable y = (x − m)/σ , dy = dx/σ to write
Z ∞
−∞
esx
2
2
Z ∞
Z ∞
)2 ]
exp[− 21 ( x−m
e−y /2
e−y /2
√ σ
dx =
es(σ y+m) √
dy = esm
dy
esσ y √
−∞
−∞
2π σ
2π
2π
= esm M(sσ ) = esm+σ
−s
40. E[esY ] = E[es ln(1/X) ] = E[eln X ] = E[X −s ] =
41. First note that
|x| =
Z ∞
−∞
0
x, x ≥ 0,
−x, x < 0.
Then the Laplace(λ ) mgf is
E[esX ] =
Z 1
esx · λ2 e−λ |x| dx
2 s2 /2
.
x−s dx =
x1−s
1−s
1
=
0
1
.
1−s
56
Chapter 4 Problem Solutions
Z ∞
Z 0
λ
sx λ x
sx −λ x
dx +
=
e e dx
e e
2 0
−∞
Z ∞
Z 0
λ
x(λ +s)
−x(λ −s)
=
dx +
dx .
e
e
2 0
−∞
Of these last two integrals, the one on the left is finite if λ > Re s, while the second is
finite if Re s > −λ . For both of them to be finite, we need −λ < Re s < λ . For such
s both integrals are easy to evaluate. We get
1
1
2λ
λ
λ
λ2
sX
MX (s) := E[e ] =
+
=
· 2 2 = 2 2.
2 λ −s λ +s
2 λ −s
λ −s
Now, MX0 (s) = 2sλ 2 /(λ 2 − s2 )2 , and so the mean is MX0 (0) = 0. We continue with
MX00 (s) = 2λ 2
(λ 2 − s2 )2 + 4s2 (λ 2 − s2 )
(λ 2 − s2 )4 .
Hence, the second moment is MX00 (0) = 2/λ 2 . Since the mean is zero, the second
moment is also the variance.
42. Since X is a nonnegative random variable, for s ≤ 0, sX ≤ 0 and esX ≤ 1. Hence, for
s ≤ 0, MX (s) = E[esX ] ≤ E[1] = 1 < ∞. For s > 0, we show that MX (s) = ∞. We use
the fact that for z > 0,
∞ n
z3
z
≥
.
ez = ∑
3!
n=0 n!
Then for s > 0, sX > 0, and we can write
Z
(sX)3
2s3 ∞ x3
s3
MX (s) = E[esX ] ≥ E
E[X 3 ] =
dx = ∞.
=
3!
3!
3! 1 x3
43. We apply integration by parts with u = x p−1 /Γ(p) and dv = e−x(1−s) dx. Then du =
x p−2 /Γ(p − 1) dx and v = −e−x(1−s) /(1 − s). Hence,
Mp (s) =
Z ∞
esx
x p−1 e−x
dx =
Γ(p)
x p−1
e−x(1−s) ∞
0
= −
Γ(p)
·
1−s
0
+
Z ∞ p−1
x
e−x(1−s) dx
0
Γ(p)
1
1−s
Z ∞
0
x p−2 −x(1−s)
e
dx.
Γ(p − 1)
The last term is Mp−1 (s)/(1 − s). The other term is zero if p > 1 and Re s < 1.
44.
(a) In this case, we use the change of variable t = x(1 − s), which implies x =
t/(1 − s) and dx = dt/(1 − s). Hence,
Z ∞
Z
∞
x p−1 e−x
1
dx =
x p−1 e−x(1−s) dx
Γ(p)
Γ(p) 0
0
Z ∞
1
t p−1 −t dt
=
e
Γ(p) 0 1 − s
1−s
Z
1 p
1 p 1
∞
p−1 −t
·
t e dt =
.
=
1−s
Γ(p) 0
1−s
|
{z
}
Mp (s) =
esx
=1
Chapter 4 Problem Solutions
57
(b) From MX (s) = (1 − s)−p , we find MX0 (s) = p(1 − s)−p−1 , MX00 (s) = p(p + 1)(1 −
s)−p−2 , and so on. The general result is
Γ(n + p)
(1 − s)−p−n .
Γ(p)
(n)
MX (s) = p(p + 1) · · · (p + [n − 1])(1 − s)−p−n =
Hence, the Taylor series is
∞
sn
(n)
∑ n! MX
MX (s) =
∞
(0) =
n=0
45.
sn Γ(n + p)
.
Γ(p)
n=0
∑ n! ·
(a) Make the change of variable t = λ x or x = t/λ , dx = dt/λ . Thus,
E[esX ] =
Z ∞
esx
0
λ (λ x) p−1 e−λ x
dx =
Γ(p)
Z ∞
e(s/λ )t
0
t p−1 e−t
dt,
Γ(p)
which is the moment generating function of g p evaluated at s/λ . Hence,
E[esX ] =
λ p
1 p
,
=
1 − s/λ
λ −s
and the characteristic function is
λ p
p
1
E[e jν X ] =
=
.
1 − jν /λ
λ − jν
λ m
λ m
, and the chf is
.
λ −s
λ − jν
1 k/2
(c) The chi-squared with k degrees of freedom mgf is
, and the chf is
1 − 2s
1 k/2
.
1 − 2 jν
(b) The Erlang(m, λ ) mgf is
46. First write
Z ∞
2
−x /2
2e
esx √
dx =
−∞
2π
√
If we let (1 − 2s) = 1/σ 2 ; i.e., σ = 1/ 1 − 2s, then
2
MY (s) = E[esY ] = E[esX ] =
MY (s) = σ
Z ∞ −x2 (1−2s)/2
e
−∞
√
2π
dx.
Z ∞ −(x/σ )2 /2
e
−∞
1
√
.
dx = σ = √
1 − 2s
2π σ
47. First observe that
2
esx e−(x−m)
2 /2
2
2
2
2
2
= e−(x −2xm+m −2sx )/2 = e−[x (1−2s)−2xm]/2 e−m /2
2
2
2
2
= e−(1−2s){x −2xm/(1−2s)+[m/(1−2s)] −[m/(1−2s)] }/2 e−m /2
2
2
2
= e−(1−2s){x−[m/(1−2s)]} /2 em /[2(1−2s)] e−m /2
= e−(1−2s){x−[m/(1−2s)]}
2 /2
2 /(1−2s)
esm
.
58
Chapter 4 Problem Solutions
√
If we now let 1 − 2s = 1/σ 2 , or σ = 1/ 1 − 2s, and µ = m/(1 − 2s), then
sX 2
sY
E[e ] = E[e
] =
2 /(1−2s)
= esm
Z ∞
−(x−m)2 /2
sx2 e
√
e
−∞
2π
sm2 /(1−2s)
dx = e
σ
2
σ =
esm /(1−2s)
√
.
1 − 2s
Z ∞ −[(x−µ )/σ ]2 /2
e
−∞
√
2π σ
dx
48. ϕY (ν ) = E[e jνY ] = E[e jν (aX+b) ] = E[e j(ν a)X ]e jν b = ϕX (aν )e jν b .
49. The key observation is that
|ν | =
ν , ν ≥ 0,
−ν , ν < 0.
It then follows that
fX (x) =
=
=
=
=
50.
Z
1 ∞ −λ |ν | − jν x
e
e
dν
2π −∞
Z ∞
Z
1
1 0 λ ν − jν x
e e
e−λ ν e− jν x d ν +
dν
2π 0
2π −∞
Z ∞
Z 0
1
e−ν (λ + jx) d ν +
eν (λ − jx) d ν
2π 0
−∞
∞
0 1
1
−1 −ν (λ + jx)
ν (λ − jx)
+
e
e
2π λ + jx
λ − jx
0
−∞
1
1
1
2λ
1
λ /π
=
= 2
+
.
2
2
2π λ + jx λ − jx
2π λ + x
λ + x2
2 2
e−x /2
d e−x /2
√
= −x √
(a)
= −x f (x).
dx
2π
2π
Z
Z ∞
Z ∞
d ∞ jν x
e f (x) dx = j
e jν x x f (x) dx = − j
e jν x f 0 (x) dx.
(b) ϕX0 (ν ) =
d ν −∞
−∞
−∞
(c) In this last integral, let u = e jν x and dv = f 0 (x) dx. Then du = jν e jν x dx, v =
f (x), and the last integral is equal to
e jν x f (x)
|
{z
=0
∞
−∞
}
− jν
Z ∞
−∞
e jν x f (x) dx = − jνϕX (ν ).
(d) Combining (b) and (c), we have ϕX0 (ν ) = − j[− jνϕX (ν )] = −νϕX (ν ).
(e) If K(ν ) := ϕX (ν )eν
K 0 (ν ) = ϕX0 (ν )eν
2 /2
2 /2
, then
+ ϕX (ν ) · ν eν
2 /2
= −νϕX (ν )eν
2 /2
+ ϕX (ν ) · ν eν
2 /2
= 0.
Chapter 4 Problem Solutions
59
(f) By the mean-value theorem of calculus, for every ν , there is a ν0 between 0 and
ν such that K(ν ) − K(0) = K 0 (ν0 )(ν − 0). Since the derivative is zero, we have
2
K(ν ) = K(0) = ϕX (0) = 1. It then follows that ϕX (ν ) = e−ν /2 .
51. Following the hints, we first write
d x p e−x
px p−1 e−x − x p e−x
d
xg p (x) =
=
= pg p (x) − xg p (x) = (p − x)g p (x).
dx
dx Γ(p)
Γ(p)
In
ϕX0 (ν ) =
d
dν
Z ∞
0
e jν x g p (x) dx = j
Z ∞
0
e jν x xg p (x) dx,
apply integration by parts with u = xg p (x) and dv = e jν x dx. Then du is given above,
v = e jν x /( jν ), and
Z
xg p (x)e jν x ∞ 1 ∞ jν x
0
e (p − x)g p (x) dx
−
ϕX (ν ) = j
jν
jν 0
0
{z
}
|
=0
Z∞
Z
1
1 ∞ jν x
jν x
= − p
e g p (x) dx −
e ( jx)g p (x) dx
ν
j 0
0
1
1
= − pϕX (ν ) − ϕX0 (ν ) = −(p/ν )ϕX (ν ) + (1/ jν )ϕX0 (ν ).
ν
j
Rearrange this to get
ϕX0 (ν )(1 − 1/ jν ) = −(p/ν )ϕX (ν ),
and multiply through by − jν to get
ϕX0 (ν )(− jν + 1) = j pϕX (ν ).
Armed with this, the derivative of K(ν ) := ϕX (ν )(1 − jν ) p is
K 0 (ν ) = ϕX0 (ν )(1 − jν ) p + ϕX (ν )p(1 − jν ) p−1 (− j)
= (1 − jν ) p−1 [ϕX0 (ν )(1 − jν ) − j pϕX (ν )] = 0.
By the mean-value theorem of calculus, for every ν , there is a ν0 between 0 and
ν such that K(ν ) − K(0) = K 0 (ν0 )(ν − 0). Since the derivative is zero, we have
K(ν ) = K(0) = ϕX (0) = 1. It then follows that ϕX (ν ) = 1/(1 − jν ) p .
52. We use the formula cov(X, Z) = E[XZ] − E[X]E[Z]. The mean of an exp(λ ) random
variable is 1/λ . Hence, E[X] = 1. Since Z := X + Y , E[Z] = E[X] + E[Y ]. Since
the Laplace random variable has zero mean, E[Y ] = 0. Hence, E[Z] = E[X] = 1.
Next, E[XZ] = E[X(X + Y )] = E[X 2 ] + E[XY ] = E[X 2 ] + E[X]E[Y ] by independence.
Since E[Y ] = 0, E[XZ] = E[X 2 ] = var(X) + (E[X])2 = 1 + 12 = 2, where we have used
the fact that the variance of an exp(λ ) random variable is 1/λ 2 . We can now write
cov(X, Z) = 2 − 1 = 1. Since Z is the sum of independent, and therefore uncorrelated,
random variables, var(Z) = var(X + Y ) = var(X) + var(Y ) = 1 + 2 = 3, where we
have used the fact that the variance of a Laplace(λ ) random variable is 2/λ 2 .
60
Chapter 4 Problem Solutions
53. Since Z = X +Y , where X ∼ N(0, 1) and Y ∼ Laplace(1) are independent, we have
var(Z) = var(X +Y ) = var(X) + var(Y ) = 1 + 2 = 3.
54. Write
MZ (s) = E[esZ ] = E[es(X−Y ) ] = E[esX e−sY ] = E[esX ]E[e−sY ] = MX (s)MY (−s).
If MX (s) = MY (s) = λ /(λ − s), then
MZ (s) =
λ
λ
λ
λ
λ2
·
=
·
= 2 2,
λ − s λ − (−s)
λ −s λ +s
λ −s
which is the Laplace(λ ) mgf.
55. Because the Xi are independent, we can write
n
n
n
sXi
sYn
s(X1 +···+Xn )
MYn (s) := E[e ] = E[e
] = E ∏e
= ∏ E[esXi ] = ∏ MXi (s).
i=1
i=1
(∗)
i=1
2 2
(a) For Xi ∼ N(mi , σi2 ), MXi (s) = esmi +σi s /2 . Hence,
n
n
n
smi +σi2 s2 /2
2
σ 2 s2 /2
2
MYn (s) = ∏ e
= exp s ∑ mi + ∑ σi s /2 = e|sm+{z
},
i=1
i=1
i=1
N(m,σ 2 ) mgf
provided we put m := m1 + · · · + mn and σ 2 := σ12 + · · · + σn2 .
(b) For Cauchy random variables, we must observe that the moment generating
function exists only for s = jν . Equivalently, we must use characteristic functions. In this case, (∗) becomes
ϕYn (ν ) := E[e jνYn ] =
n
∏ ϕXi (ν ).
i=1
Now, the Cauchy(λi ) chf is ϕXi (ν ) = e−λi |ν | . Hence,
n n
−λi |ν |
ϕYn (ν ) = ∏ e
= exp − ∑ λi |ν | =
i=1
i=1
provided we put λ := λ1 + · · · + λn .
−λ |ν |
|e {z } ,
Cauchy(λ ) chf
(c) For Xi ∼ gamma(pi , λ ), the mgf is MXi (s) = [λ /(λ − s)] pi . Hence,
n
MYn (s) =
∏
i=1
λ
λ −s
pi
=
λ
λ −s
p1 +···+pn
=
p
λ
,
λ −s
| {z }
gamma(p, λ ) mgf
provided we put p := p1 + · · · + pn .
Chapter 4 Problem Solutions
61
56. From part (c) of the preceding problem, Y ∼ gamma(rp, λ ). The table inside the back
cover of the text gives the nth moment of a gamma random variable. Hence,
E[Y n ] =
Γ(n + rp)
.
λ n Γ(rp)
57. Let Ti denote the time to transmit packet i. Then the time to transmit n packets is T :=
T1 + · · · + Tn . We need to find the density of T . Since the Ti are exponential, we can
apply the remark in the statement of Problem 55(c) to conclude that T ∼ Erlang(n, λ ).
Hence,
λ (λ t)n−1 e−λ t
fT (t) =
, t ≥ 0.
(n − 1)!
n
n
1
1
= ∑ ln . By Problem 40, each term is an exp(1)
58. Observe that Y = ln ∏
X
X
i
i
i=1
i=1
random variable. Hence, by the remark in the statement of Problem 55(c), Y ∼
Erlang(n, 1); i.e.,
yn−1 e−y
fY (y) =
, y ≥ 0.
(n − 1)!
59. Consider the characteristic function,
ϕY (ν ) = E[e
n
=
jν Y
−λ |νβi |
∏e
i=1
] = E[e
jν (β1 X1 +···+βn Xn )
n
=
−λ βi |ν |
∏e
i=1
] = E
n
∏e
j(νβi )Xi
i=1
n
=
n = exp −λ ∑ βi |ν | .
∏ E[e j(νβi )Xi ]
i=1
i=1
This is the chf of a Cauchy random variable with parameter λ ∑ni=1 βi . Hence,
fY (y) = λ
π
∑ni=1 βi
.
2
λ ∑ni=1 βi + y2
60. We need to compute P(|X −Y | ≤ 2). If we put Z := X −Y , then we need to compute
P(|Z| ≤ 2). We first find the density of Z using characteristic functions. Write
ϕZ (ν ) = E[e jν (X−Y ) ] = E[e jν X e− jνY ] = E[e jν X ]E[e j(−ν )Y ] = e−|ν | e−|−ν | = e−2|ν | ,
which is the chf of a Cauchy(2) random variable. Since the Cauchy density is even,
Z 2
1 2 2
1
2 π
1
−1 z
+
tan
P(|Z| ≤ 2) = 2
fZ (z) dz = 2
= tan−1 (1) = · = .
π
2
2
π
π
4
2
0
0
61. Let X := U +V +W be the sum of the three voltages. The alarm sounds if X > x. To
find P(X > x), we need the density of X. Since U, V , and W are i.i.d. exp(λ ) random
variables, by the remark in the statement of Problem 55(c), X ∼ Erlang(3, λ ). By
Problem 15(c),
2
(λ x)k e−λ x
P(X > x) = ∑
.
k!
k=0
62
Chapter 4 Problem Solutions
62. Let Xi ∼ Cauchy(λ ) be the i.i.d. line loads. Let Y := X1 + · · · + Xn be the total load.
The substation shuts down if Y > `. To find P(Y > `), we need to find the density of
Y . By Problem 55(b), Y ∼ Cauchy(nλ ), and so
Z ∞
y 1 ∞
1
tan−1
+
P(Y > `) =
fY (y) dy =
π
nλ
2 `
`
` 1 1
1
1
`
= − tan−1
+
.
= 1−
tan−1
π
nλ
2
2 π
nλ
63. Let the Ui ∼ uniform[0, 1] be the i.i.d. efficiencies of the extractors. Let Xi = 1 if
extractor i operates with efficiency less than 0.25; in symbols, Xi = I[0,0.25) (Ui ), which
is Bernoulli(p) with p = 0.25. Then Y := X1 + · · · + X13 is the number of extractors
operating at less than 0.25 efficiency. The outpost operates normally if Y < 3. We
must compute P(Y < 3). Since Y is the sum of i.i.d. Bernoulli(p) random variables,
Y ∼ binomial(13, p). Thus,
P(Y < 3) = P(Y = 0) + P(Y = 1) + P(Y = 2)
13 2
13
13 0
12
13
p (1 − p)11
p(1 − p) +
p (1 − p) +
=
2
1
0
= (1 − p)11 [(1 − p)2 + 13p(1 − p) + 78p2 ] = 0.3326.
64. By the remark in the statement of Problem 55(c), R = T + A is chi-squared with
k = 2 degrees of freedom. Since the number of degrees of freedom is even, R is
Erlang(k/2, 1/2) = Erlang(1, 1/2) = exp(1/2). Hence,
P(R > r) =
Z ∞
r
65.
(1/2)e−x/2 dx = e−r/2 .
(a) Since c2n+k is a density, it integrates to one. So,
Z ∞
0
ck,λ 2 (x) dx =
2
Z ∞ ∞
(λ 2 /2)n e−λ /2
∑
n!
0 n=0
∞
=
(λ 2 /2)n e−λ
n!
n=0
∑
∞
=
∑
2 /2
2
(λ 2 /2)n e−λ /2
n=0 |
n!
{z
Z ∞
|0
c2n+k (x) dx
c2n+k (x) dx
{z
}
=1
= 1.
}
Poisson(λ 2 /2) pmf
(b) The mgf is
Z ∞
esx ck,λ 2 (x) dx
2
Z ∞ ∞
(λ 2 /2)n e−λ /2
sx
c2n+k (x) dx
=
e ∑
n!
0
n=0
Mk,λ 2 (s) =
0
Chapter 4 Problem Solutions
∞
(λ 2 /2)n e−λ
= ∑
n!
n=0
2 /2
Z ∞
0
63
esx c2n+k (x) dx
(2n+k)/2
1
1 − 2s
2
∞
e−λ /2
1 λ 2 /2 n
=
∑
(1 − 2s)k/2 n=0 n! 1 − 2s
h
i
1
2
exp (λ 2 /2) 1−2s
−1
e−λ /2 (λ 2 /2)/(1−2s)
=
=
e
(1 − 2s)k/2
(1 − 2s)k/2
h
i
2s
exp (λ 2 /2) 1−2s
exp[sλ 2 /(1 − 2s)]
=
.
=
(1 − 2s)k/2
(1 − 2s)k/2
∞
(λ 2 /2)n e−λ
= ∑
n!
n=0
(c) If we first note that
d
sλ 2
ds 1 − 2s
=
s=0
2 /2
(1 − 2s)λ 2 − sλ 2 (−2)
1 − 2s
= λ 2,
s=0
α (s) + β (s)
, where
(1 − 2s)k
2
2
0
α (0) = λ and β (0) = k. Hence, E[X] = Mk,λ 2 (0) = λ + k.
then it is easy to show that Mk,0 λ 2 (s) has the general form
(d) The usual mgf argument gives
MY (s) = E[esY ] = E[es(X1 +···+Xn ) ] =
n
∏ Mki ,λi2 (s)
i=1
n
exp[sλi2 /(1 − 2s)]
= ∏
(1 − 2s)ki /2
i=1
=
exp[s(λ12 + · · · + λn2 )/(1 − 2s)]
.
(1 − 2s)(k1 +···+kn )/2
If we put k := k1 + · · · + kn and λ 2 := λ12 + · · · + λn2 , we see that Y is noncentral
chi-squared with k degrees of freedom and noncentrality parameter λ 2 .
(e) We first consider
eλ
√
√
x + e−λ x
2
√
1 ∞ (λ x )n
= ∑
[1 + (−1)n ] =
2 n=0 n!
∞
=
λ 2n xn
2n (λ 2 /2)n (x/2)n
∑ 1 · 3 · 5 · · · (2n − 1) · n!
n=0
∞
=
λ 2n xn
n=0 (2n)!
∑ 1 · 3 · 5 · · · (2n − 1) · 2n · n!
n=0
∞
=
∞
∑
∑
n=0
√
π (λ 2 /2)n (x/2)n
,
Γ( 2n+1
2 ) · n!
by Problem 14(c).
64
Chapter 4 Problem Solutions
We can now write
2
e−(x+λ )/2 eλ
√
·
2π x
√
√
x + e−λ x
2
2
e−(x+λ )/2 ∞
√
=
∑
2π x n=0
∞
=
(λ 2 /2)n e−λ
n!
n=0
∑
∞
=
(λ 2 /2)n e−λ
∑
n!
n=0
√
π (λ 2 /2)n (x/2)n
Γ( 2n+1
2 ) · n!
2 /2
·
(1/2)(x/2)n−1/2 e−x/2
Γ( 2n+1
2 )
2 /2
c2n+1 (x) = c1,λ 2 (x).
R
66. First, P(X ≥ a) = a∞ 2x−3 dx = 1/a2 , while, using the result of Problem 23, the
Markov bound is E[X]/a = 2/a. Thus, the true probability is 1/a2 , but the bound
is 2/a, which decays much more slowly for large a.
67. We begin by noting that P(X ≥ a) = e−a , E[X] = 1, and E[X 2 ] = 2. Hence, the Markov
bound is 1/a, and the Chebyshev bound is 2/a2 . To find the Chernoff bound, we must
minimize h(s) := e−sa MX (s) = e−sa /(1 − s) over 0 < s < 1. Now,
h0 (s) =
(1 − s)(−a)e−sa + e−sa
.
(1 − s)2
Solving h0 (s) = 0, we find s = (a − 1)/a, which is positive only for a > 1. Hence, the
Chernoff bound is valid only for a > 1. For a > 1, the Chernoff bound is
h((a − 1)/a) =
e−a·(a−1)/a
= ae1−a .
1 − (a − 1)/a
(a) It is easy to see that the Markov bound is smaller than the Chebyshev bound
for 0 < a < 2. However, note that the Markov bound is greater than one for
0 < a < 1, and the Chebyshev bound is greater than one for 0 < a < 2.
(b) MATLAB.
2
0
10
Chebyshev 2/a2
Markov 1/a
−2
1.5
10
2
Chebyshev 2/a
−4
10
1
1−a
Chernoff ae
Chernoff ae1−a
−6
10
0.5
Markov 1/a
P(X > a)
0
1
2
3
a
4
5
P(X > a)
−8
10
6
6
8
10
12
a
14
16
18
20
The Markov bound is the smallest on [1, 2]. The Chebyshev bound is the smallest from a = 2 to a bit more than a = 5. Beyond that, the Chernoff bound is the
smallest.
CHAPTER 5
Problem Solutions
1. For x ≥ 0,
Z x
F(x) =
0
x
λ e−λ t dt = −e−λ t
0
= 1 − e−λ x .
For x < 0, F(x) = 0.
2. For x ≥ 0,
F(x) =
Z x
t
0
λ
e−(t/λ )
2
2 /2
dt = −e−(t/λ )
2 /2
x
0
= 1 − e−(x/λ )
2 /2
.
For x < 0, F(x) = 0.
3. For x ≥ 0,
F(x) =
Z x
0
p
λ pt p−1 e−λ t dt = −e−λ t
p
x
0
p
= 1 − e−λ x .
For x < 0, F(x) = 0.
4. For x ≥ 0, first write
F(x) =
Z xr
0
2 t 2 −(t/λ )2 /2
dt =
e
π λ3
Z x/λ r
2
0
π
θ 2 e−θ
2 /2
dθ ,
where we have used the change of variable θ = t/λ . Next use integration by parts
2
with u = θ and dv = θ e−θ /2 d θ . Then
r Z x/λ
x/λ
2
−θ 2 /2
−θ 2 /2
F(x) =
e
+
dθ
−θ e
π
0
0
r
Z x/λ −θ 2 /2
2 x −(x/λ )2 /2
e
√
+2
e
dθ
= −
πλ
0
2π
r
2 x −(x/λ )2 /2
+ 2[Φ(x/λ ) − 1/2]
= −
e
πλ
r
2 x −(x/λ )2 /2
= 2Φ(x/λ ) − 1 −
.
e
πλ
For x < 0, F(x) = 0.
5. For y > 0, F(y) = P(Y ≤ y) = P(eZ ≤ y) = P(Z ≤ ln y) = FZ (ln y). Then fY (y) =
fZ (ln y)/y for y > 0. Since Y := eZ > 0, fY (y) = 0 for y ≤ 0.
65
66
Chapter 5 Problem Solutions
6. To begin, write FY (y) = P(Y ≤ y) = P(1 − X ≤ y) = P(1 − y ≤ X) = 1 − FX (1 − y).
Thus, fY (y) = − fX (1 − y) · (−1) = fX (1 − y). In the case of X ∼ uniform(0, 1),
fY (y) = fX (1 − y) = I(0,1) (1 − y) = I(0,1) (y),
since 0 < 1 − y < 1 if and only if 0 < y < 1.
7. For y > 0,
FY (y) = P(Y ≤ y) = P(ln(1/X) ≤ y) = P(1/X ≤ ey ) = P(X ≥ e−y ) = 1 − FX (e−y ).
Thus, fY (y) = − fX (e−y ) · (−e−y ) = e−y , since fX (e−y ) = I(0,1) (e−y ) = 1 for y > 0.
Since Y := ln(1/X) > 0, fY (y) = 0 for y ≤ 0.
8. For y ≥ 0,
FY (y) = P(λ X P ≤ y) = P(X p ≤ y/λ ) = P(X ≤ (y/λ )1/p ) = FX ((y/λ )1/p ).
Thus, fY (y) = fX ((y/λ )1/p ) · 1p (y/λ )(1/p)−1 /λ . Using the formula for the Weibull
density, we find that fY (y) = e−y for y ≥ 0. Since Y := λ X p ≥ 0, fY (y) = 0 for y < 0.
Thus, Y ∼ exp(1).
√
2
9. For y ≥ 0, write FY (y) = P( X ≤ y) = P(X ≤ y2 ) = FX (y2 ) = 1 − e−y . Thus,
2
fY (y) = −e−y · (−2y) =
√ 2
y
√ e−(y/(1/ 2)) /2 ,
(1/ 2)2
√
which is the Rayleigh(1/ 2 ) density.
10. Recall that the moment generating function of X ∼ N(m, σ 2 ) is MX (s) = E[esX ] =
2 2
esm+s σ /2 . Thus,
E[Y n ] = E[(eX )n ] = E[enX ] = MX (n) = enm+n
2 σ 2 /2
.
√
√
√
√
11. For y > 0, FY (y) = P(X 2 ≤ y) = P(− y ≤ X ≤ y) = FX ( y) − FX (− y). Thus,
√
√
fY (y) = fX ( y)( 12 y−1/2 ) − fX (− y)(− 12 y−1/2 ).
Since fX is even,
e−y/2
√
,
fY (y) = y−1/2 fX ( y) = √
2π y
y > 0.
12. For y > 0,
FY (y) = P((X + m)2 ≤ y)
√
√
= P(− y ≤ X + m ≤ y)
√
√
= P(− y − m ≤ X ≤ y − m)
√
√
= FX ( y − m) − FX (− y − m).
Chapter 5 Problem Solutions
67
Thus,
√
√
fY (y) = fX ( y − m)( 21 y−1/2 ) − fX (− y − m)(− 12 y−1/2 )
√
√
2
2
1
e−( y−m) /2 + e−(− 2−m) /2
= √
2
2π y
√
√
2
2
1
2
= √
e−(y−2 ym+m )/2 + e−(y+2 ym+m )/2
2π y
√ √
2
e−(y+m )/2 em y + e−m y
= √
, y > 0.
2
2π y
13. Using the example mentioned in the hint, we have
n
FXmax (z) =
∏ FXk (z) = F(z)n
n
and
k=1
FXmin (z) = 1− ∏ [1−FXk (z)] = 1−[1−F(z)]n .
k=1
14. Let Z := max(X,Y ). Since X and Y are i.i.d., we have from the preceding problem
that FZ (z) = FX (z)2 . Hence,
fZ (z) = 2FX (z) fX (z) = 2(1 − e−λ z ) · λ e−λ z ,
z ≥ 0.
Next,
E[Z] =
Z ∞
0
= 2
z fZ (z) dz = 2
Z ∞
0
= 2
Z ∞
z · λ e−λ z dz −
1
3
1
−
=
.
λ 2λ
2λ
0
λ ze−λ z (1 − e−λ z ) dz
Z ∞
0
z · (2λ )e−(2λ )z dz
15. Use the laws of total probability and substitution and the fact that conditioned on
X = m, Y ∼ Erlang(m, λ ). In particular, E[Y |X = m] = m/λ . We can now write
∞
E[XY ] =
∞
∑ E[XY |X = m]P(X = m)
=
∑ mE[Y |X = m]P(X = m)
=
∑ E[mY |X = m]P(X = m)
m=0
∞
=
m=0
∞
∑ m(m/λ )P(X = m)
m=0
m=0
=
µ + µ2
1
E[X 2 ] =
,
λ
λ
since X ∼ Poisson(µ ).
16. The problem statement tells us that P(Y > y|X = n) = e−ny . Using the law of total
probability and the pgf of X ∼ Poisson(λ ), we have
∞
P(Y > y) =
∑ P(Y > y|X = n)P(X = n)
n=0
−yX
= E[e
−y
∞
=
∑ e−ny P(X = n)
n=0
λ (e−y −1)
] = GX (e ) = e
.
68
17.
Chapter 5 Problem Solutions
(a) Using the law of substitution, independence, and the fact that Y ∼ N(0, 1), write
FZ|X (z|i) = P(Z ≤ z|X = i) = P(X +Y ≤ z|X = i) = P(Y ≤ z − i|X = i)
= P(Y ≤ z − i) = Φ(z − i).
Next,
1
FZ (z) =
∑ FZ|X (z|i)P(X = i)
= (1 − p)Φ(z) + pΦ(z − 1),
i=0
and so
fZ (z) =
(1 − p)e−z
2 /2
+ pe−(z−1)
√
2π
2 /2
.
√
(b) From part (a), it is easy to see that fZ|X (z|i) = exp[−(z − i)2 /2]/ 2π . Hence,
fZ|X (z|1)
P(X = 0)
≥
fZ|X (z|0)
P(X = 1)
becomes
exp[−(z − 1)2 /2]
1− p
≥
,
exp[−z2 /2]
p
or ez−1/2 ≥ (1 − p)/p. Taking logarithms, we can further simplify this to
z ≥
1
1− p
+ ln
.
2
p
18. Use substitution and independence to write
FZ|A,X (z|a, i) = P(Z ≤ z|A = a, X = i) = P(X/A +Y ≤ z|A = a, X = i)
= P(Y ≤ z − i/a|A = a, X = i) = P(Y ≤ z − i/a) = Φ(z − i/a).
19.
(a) Write
√
FZn (z) = P(Zn ≤ z) = P( Yn ≤ z) = P(Yn ≤ z2 ) = FYn (z2 ).
For future reference, note that
fZn (z) = fYn (z2 ) · (2z).
Since Yn is chi-squared with n degrees of freedom, i.e., gamma(n/2, 1/2),
fYn (y) =
1
2
(y/2)n/2−1 e−y/2
,
Γ(n/2)
and so
fZn (z) = z
(z2 /2)n/2−1 e−z
Γ(n/2)
y > 0,
2 /2
,
z > 0.
(b) When n = 1, we obtain the folded normal density,
fZ1 (z) = z
(z2 /2)−1/2 e−z
Γ(1/2)
2 /2
2
e−z /2
= 2√ ,
2π
z > 0.
Chapter 5 Problem Solutions
69
(c) When n = 2, we obtain the Rayleigh(1) density,
fZ2 (z) = z
(z2 /2)0 e−z
Γ(1)
2 /2
= ze−z
2 /2
,
z > 0.
(d) When n = 3, we obtain the Maxwell density,
(z2 /2)1/2 e−z
fZ3 (z) = z
Γ(3/2)
2 /2
2
z2 e−z /2
= √ ·1 1 =
2 2 Γ( 2 )
r
2 2 −z2 /2
z e
,
π
z > 0.
(e) When n = 2m, we obtain the Nakagami-m density,
fZ2m (z) = z
(z2 /2)m−1 e−z
Γ(m)
2 /2
=
2
2
z2m−1 e−z /2 ,
2m Γ(m)
z > 0.
20. Let Z := X1 +
√· · · + Xn . By Problem 55(a) in Chapter 4, Z ∼ N(0, n), and it follows
that V := Z/ n ∼ N(0, 1). We can now write Y = Z 2 = nV 2 . By Problem 11, V 2 is
chi-squared with one degree of freedom. Hence,
FY (y) = P(nV 2 ≤ y) = P(V 2 ≤ y/n),
and
21.
e−(y/n)/2
e−(y/n)/2 1
,
fY (y) = fV 2 (y/n)/n = p
· = √
2π ny
2π y/n n
y > 0.
(a) Since FY (y) = P(Y ≤ y) = P(X 1/q ≤ y) = P(X ≤ yq ) = FX (yq ),
q
q
q−1
fY (y) = fX (y ) · (qy
q−1
) = qy
q
(yq ) p−1 e−y
qyqp−1 e−y
·
=
,
Γ(p)
Γ(p)
y > 0.
q
(b) Since q > 0, as y → 0, yq → 0, and e−y → 1. Hence, the behavior of fY (y)
as y → 0 is determined by the behavior of y p−1 . For p > 1, p − 1 > 0, and
so y p−1 → 0. For p = 1, y p−1 = y0 = 1. For 0 < p < 1, p − 1 < 0 and so
y p−1 = 1/y1−p → ∞. Thus,

0,
p > 1,

q/Γ(1/q), p = 1,
lim fY (y) =

y→0
∞,
0 < p < 1.
(c) We begin with the given formula
q
fY (y) =
λ q(λ y) p−1 e−(λ y)
,
Γ(p/q)
y > 0.
(i) Taking q = p and replacing λ with λ 1/p yields
fY (y) = λ 1/p p(λ 1/p y) p−1 e−(λ
p
= λ py p−1 e−λ y ,
which is the Weibull(p, λ ) density.
1/p y) p
= λ 1/p pλ 1−1/p y p−1 e−λ y
p
70
Chapter 5 Problem Solutions
√
(ii) Taking p = q = 2 and replacing λ with 1/( 2 λ ) yields
√
√
√
2
2
fY (y) = 2/( 2 λ )[y/( 2 λ )]e−[y/( 2 λ )] = (y/λ 2 )e−(y/λ ) /2 ,
which is the required Rayleigh density.
√
(iii) Taking p = 3, q = 2, and replacing λ with 1/( 2 λ ) yields
√
√
√
2
2
2/( 2 λ )[y/( 2 λ )]2 e−[y/( 2 λ )]
(y2 /λ 3 )e−(y/λ ) /2
√ 1 1
fY (y) =
=
Γ(3/2)
2 · 2 Γ( 2 )
r
2
2 y −(y/λ )2
=
e
,
π λ3
which is the required Maxwell density.
(d) In
Z ∞
E[Y n ] =
q
yn
0
λ q(λ y) p−1 e−(λ y)
dy,
Γ(p/q)
make the change of variable t = (λ y)q , dt = q(λ y)q−1 λ dy. Then
Z
∞
q
1
(λ y)n (λ y) p−q e−(λ y) λ q(λ y)q−1 dy
n
Γ(p/q)λ 0
Z ∞
1
=
(t 1/q )n+p−q e−t dt
Γ(p/q)λ n 0
Z ∞
Γ((n + p)/q)
1
t (n+p)/q−1 e−t dt =
.
=
Γ(p/q)λ n 0
Γ(p/q)λ n
E[Y n ] =
(e) We use the same change of variable as in part (d) to write
FY (y) =
=
q
Z y
λ q(λ θ ) p−1 e−(λ θ )
Γ(p/q)
0
1
Γ(p/q)
Z (λ y)q
0
x
e−t
2 /2
Z (λ y)q
2 /2
dt = −
e−t
t
2
2
<
2 /2
−
e−x /2
.
x
Z ∞ −t 2 /2
e
t2
x
Z ∞ −t 2 /2
e
x
t2
dt
The next step is to write the integral in (∗) as
Z ∞ −t 2 /2
e
x
(t 1/q ) p−q e−t dt
dt so that du = −1/t 2 dt and v =
∞
x
e−x /2
=
−
x
t2
0
t (p/q)−1 e−t dt = G p/q ((λ y)q ).
22. Following the hint, let u = t −1 and dv = te−t
2
−e−t /2 . Then
Z ∞
1
Γ(p/q)
dθ =
dt =
Z ∞
1
x
t3
· te−t
2 /2
dt
dt
(∗)
Chapter 5 Problem Solutions
71
and apply integration by parts with u = t −3 and dv = te−t
2
and v = −e−t /2 . Then
Z ∞ −t 2 /2
e
t2
x
2
e−t /2
dt = − 3
t
∞
x
2
=
e−x /2
−3
x3
−3
2 /2
dt so that du = −3/t 4 dt
Z ∞ −t 2 /2
e
t4
x
Z ∞ −t 2 /2
e
x
t4
dt
dt.
Substituting this into (∗), we find that
Z ∞
x
2
−t 2 /2
e
2
≥
23.
2
e−x /2 e−x /2
dt =
−
+3
x
x3
2
e−x /2 e−x /2
−
.
x
x3
Z ∞ −t 2 /2
e
x
t4
dt
(a) Write
E[FXc (Z)] =
=
=
=
=
=
=
Z ∞
FXc (z) fZ (z) dz
−∞
Z ∞ Z ∞
fX (x) dx fZ (z) dz
I(z,∞) (x) fX (x) dx fZ (z) dz
−∞
Z ∞
fX (x)
I(z,∞) (x) fZ (z) dz dx
−∞
Z ∞
fX (x)
I(−∞,x) (z) fZ (z) dz dx
−∞
Z x
fZ (z) dz dx
fX (x)
−∞
z
Z ∞ Z ∞
−∞
Z ∞
−∞
Z ∞
−∞
Z ∞
−∞
Z ∞
−∞
−∞
fX (x)FZ (x) dx = E[FZ (X)].
(b) From Problem 15(c) in Chapter 4, we have
m−1
FZ (z) = 1 −
∑
k=0
(λ z)k e−λ z
,
k!
for z ≥ 0,
and FZ (z) = 0 for z < 0. Hence,
m−1
(λ X)k e−λ X
E[FZ (X)] = E I[0,∞) (X) 1 − ∑
k!
k=0
m−1
= P(X ≥ 0) −
∑
k=0
λk
E[X k e−λ X I[0,∞) (X)].
k!
72
Chapter 5 Problem Solutions
(c) Let X ∼ N(0, 1) so that E[Q(Z)] = E[FXc (Z)] = E[FZ (X)]. Next, since Z ∼
exp(λ ) = Erlang(1, λ ), we can use the result of part (b) to write
E[FZ (X)] = P(X ≥ 0) − E[e−λ X I[0,∞) (X)]
1
= −
2
Z ∞
0
−x2 /2
−λ x e
2
=
1 eλ /2
−√
2
2π
2
=
1 eλ /2
−√
2
2π
2
=
√
dx
2π
e
1 eλ /2
−√
2
2π
Z ∞
0
Z ∞
0
Z ∞
λ
e−(x
2 +2xλ +λ 2 )/2
e−(x+λ )
e−t
2 /2
2 /2
dx
dx
dt =
2
1
− eλ /2 Q(λ ).
2
√
(d) Put Z := σ Y and make the following observations. First, for z ≥ 0,
√
FZ (z) = P(σ Y ≤ z) = P(σ 2Y ≤ z2 ) = P(Y ≤ (z/σ )2 ) = FY ((z/σ )2 ),
and FZ (z) = 0 for z < 0. Second, since Y is chi-squared with 2m-degrees of
freedom, Y ∼ Erlang(m, 1/2). Hence,
m−1
FZ (z) = 1 −
∑
k=0
((z/σ )2 /2)k e−(z/σ )
k!
2 /2
,
for z ≥ 0.
Third, with X ∼ N(0, 1),
√
E[Q(σ Y )] = E[Q(Z)] = E[FXc (Z)] = E[FZ (X)]
is equal to
2
m−1
((X/σ )2 /2)k e−(X/σ ) /2
,
E I[0,∞) (X) 1 − ∑
k!
k=0
which simplifies to
m−1
P(X ≥ 0) −
∑
k=0
2
1
E[X 2k e−(X/σ ) /2 I[0,∞) (X)].
σ 2k 2k k!
e 2 := (1 + 1/σ 2 )−1 , we have
Now, P(X ≥ 0) = 1/2, and with σ
2k −(X/σ )2 /2
E[X e
I[0,∞) (X)] =
Z ∞
0
−x2 /2
2k −(x/σ )2 /2 e
x e
Z
√
dx
2π
∞
2
σe
= √
x2k e−(x/σe ) /2 dx
2π σe 0
Z ∞
2
σe
1
x2k e−(x/σe ) /2 dx
=
·√
2
e −∞
2π σ
σe
e 2k .
=
· 1 · 3 · 5 · · · (2k − 1) · σ
2
Putting this all together, the desired result follows.
Chapter 5 Problem Solutions
73
(e) If m = 1 in part (d), then Z defined in solution of part (d) is Rayleigh(σ ) by the
same argument as in the solution of Problem 19(b). Hence, the desired result
follows by taking m = 1 in the result of part (d).
(f) Since the Vi are independent exp(λi ), the mgf of Y := V1 + · · · +Vm is
m
λk
∏ λk − s
MY (s) =
m
=
k=1
λk
∑ ck λk − s ,
k=1
where the ck are the result of expansion by partial fractions. Inverse transforming term by term, we find that
m
∑ ck · λk e−λk y ,
fY (y) =
k=1
y ≥ 0.
Using this, we can write
m
∑ ck (1 − e−λk y ),
FY (y) =
k=1
Next, put Z :=
y ≥ 0.
√
Y , and note that for z ≥ 0,
√
FZ (z) = P( Y ≤ z) = P(Y ≤ z2 ) = FY (z2 ).
Hence,
m
FZ (z) =
2
∑ ck (1 − e−λk z ),
k=1
z ≥ 0.
We can now write that E[Q(Z)] = E[FXc (Z)] = E[FZ (X)] is equal to
m
−λk X 2
E I[0,∞) (X) ∑ ck (1 − e
) .
k=1
ek2 := (1 + 2λk )−1 ,
Now observe that with σ
−λk X 2
E[I[0,∞) (X)e
] =
Z ∞
0
−x2 /2
−λk x2 e
e
1
√
dx =
2
2π
Z
2
Z ∞
−∞
σek
1 ∞ e−(1+2λk )x /2
√
dx =
2 −∞
2
2π
1
σek
.
= p
=
2
2 1 + 2λk
=
−x2 /2
−λk x2 e
e
√
2π
dx
Z ∞ −(x/σek )2 /2
e
−∞
√
ek
2π σ
dx
Putting this all together, the desired result follows.
24. Recall that the noncentral chi-squared density with k degrees of freedom and noncentrality parameter λ 2 is given by
∞
ck,λ 2 (t) :=
(λ 2 /2)n e−λ
n!
n=0
∑
2 /2
c2n+k (t),
t > 0,
74
Chapter 5 Problem Solutions
where c2n+k denotes the central chi-squared density with 2n + k degrees of freedom.
Hence,
Ck,λ 2 (x) =
Z x
0
ck,λ 2 (t) dt =
∞
=
25.
(λ 2 /2)n e−λ
∑
n!
n=0
2
Z x ∞
(λ 2 /2)n e−λ /2
∑
n!
0 n=0
2 /2
Z x
0
c2n+k (t) dt
∞
c2n+k (t) dt =
(λ 2 /2)n e−λ
∑
n!
n=0
2 /2
C2n+k (x).
(a) Begin by writing
√
FZn (z) = P( Yn ≤ z) = P(Yn ≤ z2 ) = FYn (z2 ).
Then fZn (z) = fYn (z2 ) · 2z. Next observe that
∞
(mz/2)2`+n/2−1
∑ `!Γ(` + (n/2) − 1 + 1)
In/2−1 (mz) =
∞
=
`=0
(mz/2)2`+n/2−1
.
`=0 `!Γ(` + n/2)
∑
From Problem 65 in Chapter 4,
2 /2
∞
fYn (y) :=
(m2 /2)` e−m
`!
`=0
∑
2 /2
∞
=
(m2 /2)` e−m
∑
`!
`=0
c2`+n (y)
· 21
(y/2)(2`+n)/2−1 e−y/2
.
Γ((2` + n)/2)
So,
2 +z2 )/2
fZn (z) = fYn (z2 ) · 2z = 2ze−(m
2 +z2 )/2
= ze−(m
=
· 21
∞
(m2 /2)` (z2 /2)`+n/2−1
`!Γ(` + n/2)
`=0
∑
∞
(mz)2`+n/2−1 m−n/2+1 (1/2)2`+n/2−1 zn/2−1
`!Γ(` + n/2)
`=0
∑
zn/2 −(m2 +z2 )
e
In/2−1 (mz).
mn/2−1
(b) Obvious.
(c) Begin by writing
FYn (y) = P(Zn2 ≤ y) = P(Zn ≤
√
y ) = FZn (y1/2 ).
Then
fYn (y) = fZn (y1/2 ) · y−1/2 /2
(y1/2 )n/2 −(m2 +y)/2
√
e
In/2−1 (m y )y−1/2
m√n/2−1
y n/2−1
2
√
= 12
e−(m +y)/2 In/2−1 (m y ).
m
=
1
2
Chapter 5 Problem Solutions
75
(d) First write
Z ∞
t n/2 −(m2 +t 2 )/2
e
In/2−1 (mt) dt
z mn/2−1
Z ∞
(mt)n/2 −(m2 +t 2 )/2
e
=
In/2−1 (mt) dt.
mn−1
z
FZcn (z) =
Now apply integration by parts with
dv = mte−t
u = (mt)n/2−1 In/2−1 (mt) and
2 /2
2 /2
e−m
/mn−1 dt.
Then
2 /2
v = −me−m
e−t
2 /2
/mn−1 , and by the hint, du = (mt)n/2−1 In/2−2 (mt) · m dt.
Thus,
2
FZcn (z) = −(mt)n/2−1 In/2−1 (mt)
2
e−(m +t )/2
mn−2
Z ∞
(mt)n/2−1 −(m2 +t 2 )/2
I
e
+
=
z n/2−1
m
z
n/2−2 (mt) dt
mn−3
z
∞
2 +z2 )/2
e−(m
In/2−1 (mz) + FZn−2 (z).
(e) Using induction, this is immediate from part (d).
e z) = e−z
(f) It is easy to see that Q(0, z) = Q(0,
2 /2
. We then turn to
∂ e
∂ −(m2 +z2 )/2 ∞
Q(m, z) =
e
∑ (m/z)k Ik (mz)
∂m
∂m
k=0
∞ −(m2 +z2 )/2
k
−2k
k
= e
−m(m/z)
I
(mz)
+
z
(mz)
I
(mz)z
k
k−1
∑
−(m2 +z2 )/2
= e
k=0
∞ ∑
k
−m(m/z) Ik (mz) + (m/z) Ik−1 (mz)z
k=0
∞
2
2
−(m +z )/2
k
∑ (m/z)
= ze
2 +z2 )/2
= ze−(m
k
k=0
∞
Ik−1 (mz) − (m/z)Ik (mz)
∑ (m/z)k Ik−1 (mz) − (m/z)k+1 Ik (mz)
k=0
−(m2 +z2 )/2
= ze
2 +z2 )/2
I−1 (mz) = ze−(m
I1 (mz).
To conclude, we compute
∂Q
∂
=
∂m
∂m
=
Z ∞
z
Z ∞
z
2 +t 2 )/2
te−(m
2 +t 2 )/2
−mte−(m
I0 (mt) dt
I0 (mt) dt +
Z ∞
z
2 +t 2 )/2
te−(m
I−1 (mt) · t dt.
76
Chapter 5 Problem Solutions
Write this last integral as
Z ∞
z
2 +t 2 )/2
(mt)I1 (mt) · (t/m)e−(m
dt.
2
2
Now apply integration by parts with u = (mt)I1 (mt) and dv = te−(t +m )/2 /m dt.
2 2
Then du = (mt)I0 (mt)m dt and v = −e−(m +t )/2 /m, and the above integral is
equal to
Z
2 +z2 )/2
ze−(m
∞
I1 (mz) +
z
2 +t 2 )/2
mte−(m
I0 (mt) dt.
2 +z2 )/2
Putting this all together, we find that ∂ Q/∂ m = ze−(m
∞
26. Recall that Iν (x) :=
I1 (mz).
(x/2)2`+ν
∑ `!Γ(` + ν + 1) .
`=0
(a) Write
∞
(x/2)2`
1
1
Iν (x)
+∑
→
=
ν
(x/2)
Γ(ν + 1) `=1 `!Γ(` + ν + 1)
Γ(ν + 1)
as x → 0.
Now write
zn/2 −(m2 +z2 )/2
e
In/2−1 (mz)
mn/2−1
In/2−1 (mz)
2
2
zn/2
(mz/2)n/2−1
= n/2−1 e−(m +z )/2
m
(mz/2)n/2−1
In/2−1 (mz)
2
2
zn−1
= n/2−1 e−(m +z )/2
.
2
(mz/2)n/2−1
fZn (z) =
Thus,

2
0,
n > 1,

e−m /2
2 p
lim fZn (z) = n/2−1
lim zn−1 =
e−m /2 2/π , n = 1,

z→0
2
Γ(n/2) z→0
∞,
0 < n < 1.
(b) First note that
∞
Iν −1 (x) =
(x/2)2`+ν −1
=
`=0 `!Γ(` + ν )
∑
∞
(` + ν )(x/2)2`+ν −1
.
`!Γ(` + ν + 1)
`=0
∑
Second, with the change of index ` = k + 1,
∞
Iν +1 (x) =
(x/2)2k+ν +1
∑ k!Γ(k + ν + 2) =
k=0
∞
`(x/2)2`+ν −1
=
= ∑
`=1 `!Γ(` + ν + 1)
∞
(x/2)2`+ν −1
∑ (` − 1)!Γ(` + ν + 1)
`=1
∞
`(x/2)2`+ν −1
∑ `!Γ(` + ν + 1) .
`=0
Chapter 5 Problem Solutions
77
It is now easy to see that
∞
(2` + ν )(x/2)2`+ν −1
∑ `!Γ(` + ν + 1) = 2Iν0 (x).
`=0
Iν −1 (x) + Iν +1 (x) =
It is similarly easy to see that
∞
ν (x/2)2`+ν −1
∑ `!Γ(` + ν + 1)
Iν −1 (x) − Iν +1 (x) =
= 2(ν /x)Iν (x).
`=0
R
(c) To the integral I˜n (x) := (2π )−1 −ππ ex cos θ cos(nθ ) d θ , apply integration by parts
with u = ex cos θ and dv = cos nθ d θ . Then du = ex cos θ (−x sin θ ) d θ and v =
sin(nθ )/n. We find that
Z
1 x cos θ sin nθ π
x π x cos θ
e
sin nθ sin θ d θ .
I˜n (x) =
e
+
2π
n −π n −π
|
{z
}
=0
We next use the identity sin A sin B = 12 [cos(A − B) − cos(A + B)] to get
x ˜
I˜n (x) =
[In−1 (x) − I˜n+1 (x)].
2n
(d) Since I˜0 (x) := (1/π )
Then
1
I˜0 (x) =
π
=
1
π
R π x cos θ
d θ , make the change of variable t = θ − π /2.
0 e
Z π /2
−π /2
ex cos(t+π /2) dt =
Z π /2 ∞
(−x sint)k
∑
k!
−π /2 k=0
1 ∞ x2`
=
∑ (2`)!
π `=0
1
π
Z π /2
dt =
−π /2
∞
e−x sint dt
(−x)k
1
∑
π k=0 k!
Z π /2
Z π /2
|
Z
sink t dt
{z
}
−π /2
= 0 for k odd
π /2
2 ∞ x2`
sin2` t dt
sin2` t dt =
∑
π `=0 (2`)! 0
−π /2
2` + 1 √
π
Γ
2 ∞ x2`
2
=
, by Problem 18 in Ch. 4.
·
∑ (2`)!
2` + 2
π `=0
2Γ
2
Now,
(2`)! = 1 · 3 · 5 · · · (2` − 1) · 2 · 4 · 6 · · · 2` = 1 · 3 · 5 · · · (2` − 1) · 2` `!,
and from Problem 14(c) in Chapter 4,
2` + 1
1 · 3 · 5 · · · (2` − 1) √
Γ
=
π.
2
2`
Hence,
∞
I˜0 (x) =
(x/2)2`
∑ `!Γ(` + 1)
`=0
=: I0 (x).
78
Chapter 5 Problem Solutions
(e) Begin by writing
Z
1 π x cos θ
e
cos([n ± 1]θ ) d θ
2π −π
Z π
1
=
ex cos θ [cos nθ cos θ ∓ sin nθ sin θ ] d θ .
2π −π
In±1 (x) =
Then
1
2 [In−1 (x) + In+1 (x)]
Since
In0 (x)
=
1
2π
Z π
ex cos θ cos nθ cos θ d θ .
−π
Z
Z
1 π x cos θ
∂ 1 π x cos θ
e
cos(nθ ) d θ =
e
cos(nθ ) cos θ d θ ,
=
∂ x 2π −π
2π −π
we see that 12 [In−1 (x) + In+1 (x)] = In0 (x).
27. MATLAB.
0.5
0.4
0.3
0.2
0.1
0
0
1
2
3
4
28. See previous problem solution for graph. The probabilities are:
k
0
1
2
3
4
29.
P(X = k)
0.0039
0.0469
0.2109
0.4219
0.3164
(a) Sketch of f (t):
1
1/2
1/3
1/6
0
1
Chapter 5 Problem Solutions
(b) P(X = 0) =
(c) We have
R
{0}
f (t) dt = 1/2, and P(X = 1) =
Z 1−
P(0 < X < 1) =
0+
Z 1−
=
0+
Z 1−
f (t) dt =
1 −t
3 e dt
=
0+
Z 1
0
79
R
{1}
f (t) dt = 1/6.
1 −t
1
1
3 e u(t) + 2 δ (t) + 6 δ (t − 1) dt
1 −t
3 e dt
= − 31 e−t
1
0
=
1 − e−1
3
and
P(X > 1) =
Z ∞
f (t) dt =
1+
Z ∞
1
1 −t
3 e dt
= − 31 e−t
∞
1
= e−1 /3.
(d) Write
P(0 ≤ X ≤ 1) = P(X = 0) + P(0 < X < 1) + P(X = 1)
=
1 1 − e−1 1
+
+ = 1 − e−1 /3
2
3
6
and
P(X > 1) = P(X = 1) + P(X > 1) = 1/6 + e−1 /3 =
1 + 2e−1
.
6
(e) Write
E[X] =
=
Z ∞
−∞
t f (t) dt =
Z ∞
0
1
3 E[exp
t −t
3 e dt + 0 · P(X
RV w/λ = 1] + 1/6 = 1/3 + 1/6 = 1/2.
30. For the first part of the problem, we have E[eX ] =
1 R1 x
1 0
2 e + 2 0 e dx
= 0) + 1 · P(X = 1)
= 1/2 + ex /2
write
P(X = 0|X ≤ 1/2) =
1
0
R∞
x
−∞ e
· 21 [δ (x) + I(0,1] (x)] dx =
= 1/2 + (e − 1)/2 = e/2. For the second part, first
P({X = 0} ∩ {X ≤ 1/2})
P(X = 0)
=
.
P(X ≤ 1/2)
P(X ≤ 1/2)
Since P(X = 0) = 1/2 and
P(X ≤ 1/2) =
Z 1/2
−∞
1
2 [δ (x) + I(0,1] (x)] dx
we have P(X = 0|X ≤ 1/2) =
1/2
3/4
= 12 + 21
Z 1/2
0
dx = 21 + 41 = 34 ,
= 2/3.
R
∞
31. The approach is to find the density and then compute E[X] = −∞
x fX (x) dx. The catch
is that the cdf has a jump at x = 1/2, and so the density has an impulse there. Put

 2x, 0 < x < 1/2,
1, 1/2 < x < 1,
f˜X (x) :=

0, otherwise.
80
Chapter 5 Problem Solutions
Then the density is fX (x) = f˜X (x) + 41 δ (x − 1/2). Hence,
E[X] =
Z ∞
−∞
x fX (x) dx =
=
=
Z 1/2
0
x · 2x dx +
2 1 3
+ 12 [1 −
3 2
1
3
1
12 + 8 + 8 =
Z 1
x · 1 dx + 41 · 12
1/2
1
1 2
2 ]+ 8
11
1
24 + 8 =
7/12.
√
R∞ √
x fX (x) dx. The
32. The approach is to find the density and then compute E[ X ] = −∞
catch is that the cdf has a jump at x = 4, and so the density has an impulse there. Put

 x−1/2 /8, 0 < x < 4,
f˜X (x) :=
1/20,
4 < x < 9,

0,
otherwise.
Then the density is fX (x) = f˜X (x) + 14 δ (x − 4). To begin, we compute
Z ∞
−∞
x1/2 f˜X (x) dx =
Z 4
1/8 dx +
0
Z 9
x1/2 /20 dx = 1/2 + x3/2 /30
4
9
4
= 1/2 + (27 − 8)/30 = 1/2 + 19/30 = 17/15.
The complete answer is
√
√
E[ X ] = 17/15 + 4/4 = 17/15 + 1/2 = 34/30 + 15/30 = 49/30.
33. First note that for y < 0,
Z y
−∞
and for y ≥ 0,
Z y
−∞
e−|t| dt =
Hence,
Z 0
−∞
et dt +
e−|t| dt =
Z y
0
Z y
−∞
et dt = ey ,
e−t dt = 1 + (−e−t )
y
0
= 1 + 1 − e−y = 2 − e−y .
 y
y < 0,
 e /4,
(2 − e−y )/4 + 1/3,
0 ≤ y < 7,
FY (y) =

(2 − e−y )/4 + 1/3 + 1/6, y ≥ 7,
 y
y < 0,
 e /4,
5/6 − e−y /4, 0 ≤ y < 7,
=

1 − e−y /4, y ≥ 7.
1
0.75
0.5
0.25
0
−2 0 2 4 6 8
Chapter 5 Problem Solutions
81
34. Let {Y = 0} = {loose connection}, and {Y = 1} = {Y = 0} c . Then P(Y = 0) =
P(Y = 1) = 1/2. Using the law of total probability,
1
FX (x) = P(X ≤ x) =
∑ P(X ≤ x|Y = i)P(Y = i)
i=0
Z
1
= 2 P(X ≤ x|Y = 0) +
x
−∞
I(0,1] (t) dt .
Since P(X = 0|Y = 0) = 1, we see that

1,
x ≥ 1,

1
FX (x) =
(1
+
x),
0 ≤ x < 1,
2
0,
x < 0.
Since there is a jump at x = 0, we must be careful in computing the density. It is
fX (x) =
1
1
1/2
1/2
0
1
cdf
35.
1
2 [I(0,1) (x) + δ (x)].
0
1
density
(a) Recall that as x varies from +1 to −1, cos−1 x varies
from 0 to π . Hence,
FX (x) = P(cos Θ ≤ x) = P Θ ∈ [−π , −θx ] ∪ [θx , π ] , where θx := cos−1 x. Since
Θ ∼ uniform[−π , π ],
FX (x) =
cos−1 x
π − θx −θx − (−π )
π − θx
+
= 2
= 1−
.
2π
2π
2π
π
(b) Recall that as y varies from +1 to −1, sin−1 y varies from π /2 to −π /2. For
y ≥ 0,
FY (y) = P(sin Θ ≤ y) = P Θ ∈ [−π , θy ] ∪ [π − θy , π ]
=
θy θy − (−π )
π + 2θy
1 sin−1 y
+
=
= +
,
2π
2π
2π
2
π
and for y < 0,
FY (y) = P(sin Θ ≤ y) = P(Θ ∈ [−π − θy , θy ])
=
2θy + π
1 sin−1 y
= +
.
2π
2
π
82
Chapter 5 Problem Solutions
(c) Write
fX (x) = −
and
fY (y) =
(d) First write
−1
1
1/π
·√
= √
,
π
1 − x2
1 − x2
1/π
1
1
= p
.
·p
2
π
1−y
1 − y2
Y + 1
FZ (z) = P(Z ≤ z) = P
≤ z = P(Y ≤ 2z − 1) = FY (2z − 1).
2
Then differentiate to get
fZ (z) = fY (2z − 1) · 2 =
=
Γ( 12 + 21 )
Γ( 12 )Γ( 21 )
1
2
p
= √ √ p
2
π π z(1 − z)
π 1 − (2z − 1)
z1/2−1 (1 − z)1/2−1 ,
which is the beta density with p = q = 1/2.
36. The cdf is
FY (y) =
and the density is
37. The cdf is
and the density is



Z
1,
√
−1+ 1+y
y ≥ 3,
1/4 dx, −1 ≤ y < 3,
√


 −1− 1+y
0,
y < −1,

y ≥ 3,
 √1,
1
=
1
+
y,
−1 ≤ y < 3,
2
0,
y < −1,

 √1
, −1 < y < 3,
fY (y) =
4 1+y

0,
otherwise.

1,


p


 2(3 − 2/y )
,
FY (y) =
6


1/3,



0,
fY (y) =
√
y ≥ 2,
1/2 ≤ y < 2,
0 ≤ y < 1/2,
y < 0,
1
1
2 −3/2
y
I(1/2,2) (y) + δ (y) + δ (y − 2).
6
3
3
Chapter 5 Problem Solutions
83
38. For 0 ≤ y ≤ 1, we first compute the cdf
p
FY (y) = P(Y ≤ y) = P( 1 − R2 ≤ y) = P(1 − R2 ≤ y2 ) = P(1 − y2 ≤ R2 )
p
√
p
1
2 − 1 − y2
2
√
= P( 1 − y ≤ R) =
= 1 − √ (1 − y2 )1/2 .
2
2
We then differentiate to get the density

p y
, 0 < y < 1,
2(1 − y2 )
fY (y) =

0,
otherwise.
√
39. The first thing to note is that for 0 ≤ R ≤ 2, 0 ≤ R2 ≤ 2. It is then easy√to see that
the minimum value of Z = [R2 (1 − R2 /4)]−1 occurs when R2 = 2 or R = 2. Hence,
the random variable Z takes values in the range [1, ∞). So, for z ≥ 1, we write
1
2
2
FZ (z) = P 2
≤
z
=
P
R
(1
−
R
/4)
≥
1/z
R (1 − R2 /4)
= P (R2 /4)(1 − R2 /4) ≥ 1/(4z) .
Put y := 1/(4z) and observe that x(1 − x) ≥ y if and only if 0 ≥ x2 − x + y, with
equality if and only if
√
1 ± 1 − 4y
.
x =
2
Since we will have x = R2 /4 ≤ 1/2, we need the negative root. Thus,
√
p
1 − 1 − 4y
2
2
= P R ≥ 2[1 − 1 − 1/z ]
FZ (z) = P (R /4) ≥
2
q
p
√
q
2 − 2[1 − 1 − 1/z ]
p
√
= P R ≥ 2[1 − 1 − 1/z ] =
2
q
p
= 1 − 1 − 1 − 1/z.
Differentiating, we obtain
p
p
−1/2 d
1
[1 − 1 − 1/z ]
1 − 1 − 1/z
2
dz
p
−1/2
1
1
1
= − 1 − 1 − 1/z
· − (1 − 1/z)−1/2 · 2
2
2
z
p
1
= 2 [(1 − 1 − 1/z )(1 − 1/z)]−1/2 .
4z
√
√
√
40. First note that as R varies from 0 to 2, T varies from π to 2 π . For π ≤ t ≤ 2 π ,
write
2
π
π
2
FT (t) = P(T ≤ t) = P p
≤ t = P 2 ≤ 1 − R /4
t
1 − R2 /4
fZ (z) = −
84
Chapter 5 Problem Solutions
= P(R2 /4 ≤ 1 − π 2 /t 2 ) = P(R2 ≤ 4[1 − π 2 /t 2 ])
q
√
= P R ≤ 2 1 − π 2 /t 2 = 2(1 − π 2 /t 2 )1/2 .
Differentiating, we find that
√
√
π2 2
π2 2
2 2 −1/2
√
fT (t) =
.
=
(1 − π /t )
t3
t2 t2 − π2
For the second part of the problem, observe that as R varies between 0 and
varies between 1 and e−π . For m in this range, note that ln m < 0, and write
√
2
FM (m) = P(M ≤ m) = P(e−π (R/2)/ 1−R /4 ≤ m)
.q
= P π (R/2)
1 − R2 /4 ≥ − ln m
.
= P π 2 (R2 /4) (1 − R2 /4) ≥ (− ln m)2
= P (R2 /4) ≥ 1/[1 + {π /(− ln m)}2 ]
= P R ≥ 2[1 + {π /(− ln m)}2 ]−1/2
√
= 1 − 2[1 + {π /(− ln m)}2 ]−1/2 ,
√
2, M
Differentiating, we find that
√ 2
√ 2
2π
2π
2 −3/2
.
[1
+
{
π
/(−
ln
m)}
]
=
fM (m) =
m(− ln m)3
m[(− ln m)2 + π 2 ]3/2
41.
(a) For X ∼ uniform[−1, 1],
FY (y) =
and
fY (y) =
(b) For X ∼ uniform[−1, 2],
and


1,
y ≥ 1,
√
(y + y )/2, 0 ≤ y < 1,

0,
y < 0,
√
1
2 [1 + 1/(2 y )],
0,




1,
(y + 1)/3,
√
FY (y) =
(y + y )/3,



0,
fY (y) =



1/3,
y ≥ 2,
1 ≤ y < 2,
0 ≤ y < 1,
y < 0,
√
1
3 [1 + 1/(2 y )],
0,
0 < y < 1,
otherwise.
1<y<2
0 < y < 1,
otherwise.
Chapter 5 Problem Solutions
(c) For X ∼ uniform[−2, 3],
1,
(y + 2)/5,
√
FY (y) =
(y + y )/5,



0,
and
fY (y) =
y ≥ 2,
1 ≤ y < 2,
0 ≤ y < 1,
y < 0,
1
√
[I
(y) + {1 + 1/(2 y )}I(0,1) (y) + δ (y − 1) + δ (y − 2)].
5 (1,2)
(d) For X ∼ exp(λ ),


1,
y ≥ 2,
P(X ≤ y) + P(X ≥ 3), 0 ≤ y < 2,

0,
y < 0,

1,
y ≥ 2,

=
(1 − e−λ y ) + e−3λ , 0 ≤ y < 2,

0,
y < 0,
FY (y) =
and
42.




85
fY (y) = λ e−λ y I(0,2) (y) + e−3λ δ (y) + (e−2λ − e−3λ )δ (y − 2).
(a) If X ∼ uniform[−1, 1], then Y = g(X) = 0, and so
FY (y) = u(y) (the unit step function),
(b) If X ∼ uniform[−2, 2],
FY (y) =
and



fY (y) =
(c) We have
FY (y) =
and
fY (y) =
(d) If X ∼ Laplace(λ ),
1,
1
2 (y + 1),
0,
and
fY (y) = δ (y).
y ≥ 1,
0 ≤ y < 1,
y < 0,
1
[I
(y) + δ (y)].
2 (0,1)


1,
y ≥ 1,
(y + 1)/3, 0 ≤ y < 1,

0,
y < 0,
1
3 [I(0,1) (y) + δ (y) + δ (y − 1)].

y ≥ 1,

 R y+1 1,
λ −λ x
dx, 0 ≤ y < 1,
2 0 2e
FY (y) =


0,
y < 0,

1,
y ≥ 1,

=
1 − e−λ (y+1) , 0 ≤ y < 1,

0,
y < 0,
86
Chapter 5 Problem Solutions
and
43.
fY (y) = λ e−λ (y+1) I(0,1) (y) + (1 − e−λ )δ (y) + e−2λ δ (y − 1).
(a) If X ∼ uniform[−3, 2],
FY (y) =
and
fY (y) =







FY (y) =
and
fY (y) =







FY (y) =
and
fY (y) =
fY (y) =
1,
1 1/3
+ y + 2),
4 (y
1
[y
+
2
−
(−y)1/2 ],
4
0,
y ≥ 1,
0 ≤ y < 1,
−1 ≤ y < 0,
y < −1,
√
2/3
1
)]I(0,1) (y) + 41 [1 + 1/(2 −y )]I(−1,0) (y).
4 [1 + 1/(3y
(c) If X ∼ uniform[−1, 1],
and
0,
y ≥ 1,
0 ≤ y < 1,
−1 ≤ y < 0,
y < −1,
√
2/3
1
)]I(0,1) (y) + 15 [1 + 1/(2 −y )]I(−1,0) (y) + 15 δ (y − 1).
5 [1 + 1/(3y
(b) If X ∼ uniform[−3, 1],
44. We have
1,
1 1/3
+ y + 2),
5 (y
1
[y
+
2
−
(−y)1/2 ],
5







1,
1 1/3
+ 1),
2 (y
1
1/2 ],
[1
−
(−y)
2
0,
1
1
I(0,1) (y) + √ I(−1,0) (y).
4 −y
6y2/3

0,


 1 [y + 1 + √y + 1],
6
FY (y) =
√
1


 6 [3 + y + 1 ],
1,
1
1
6 δ (y − 1) + [ 6
45. We have
y ≥ 1,
0 ≤ y < 1,
−1 ≤ y < 0,
y < −1,
y < −1,
−1 ≤ y < 1,
1 ≤ y < 8,
y ≥ 8,
1
1
+ 12
(y + 1)−1/2 ]I(−1,1) (y) + 12
(y + 1)−1/2 I(1,8) (y).
FY (y) =


fY (y) =
√
1
1
4 δ (y) + 4 [2y + 1/(2 y) + 1]I(0,1) (y).
and
0,
y < 0,
√
(y2 + y + y + 1)/4, 0 ≤ y < 1,

1,
y ≥ 1,
Chapter 5 Problem Solutions
46. We have
FY (y) =
and







1,
1
2
6 [4 + 3y − y ],
1
2
6 [3 + 2y − y ],
0,
87
y ≥ 1,
0 ≤ y < 1,
−1 ≤ y < 0,
y < −1,
fY (y) = [ 12 − 13 y]I(0,1) (y) + 31 [1 − y]I(−1,0) (y) + 61 δ (y).
47. We have
FY (y) =


fY (y) =
1
3
and

1
3 [1 + y +
1,
y ≥ 1,
p
y/(2 − y) ], 0 ≤ y < 1,
0,
y < 0,
y −1/2
1
δ (y) + 1 +
.
2−y
(2 − y)2
48. Observe that g is a periodic sawtooth function with period one. Also note that since
0 ≤ g(x) < 1, Y = g(X) takes values in [0, 1).
(a) We begin by writing
∞
∞
k=0
k=0
FY (y) = P(Y ≤ y) = P(g(X) ≤ y) = ∑ P(k ≤ X ≤ k+y) = ∑ FX (k+y)−FX (k).
When X ∼ exp(1), we obtain, for 0 ≤ y < 1,
∞
FY (y) =
∑ (1 − e−k e−y ) − (1 − e−k )
k=0
∞
=
∑ e−k (1 − e−y )
k=0
=
1 − e−y
.
1 − e−1
Differentiating, we get
fY (y) =
e−y
,
1 − e−1
0 ≤ y < 1.
We say that Y has a “truncated” exponential density.
(b) When X ∼ uniform[0, 1), we obtain Y = X ∼ uniform[0, 1).
(c) Suppose X ∼ uniform[ν , ν + δ ), where ν = m + δ for some integer m ≥ 0 and
some 0 < δ < 1. For 0 ≤ y < δ
FY (y) = (m + 1 + y) − (m + 1) = y,
and for δ ≤ y < 1,
FY (y) = [(m + 1 + δ ) − (m + 1)] + [(m + y) − (m + δ )] = y.
Since FY (y) = y for 0 ≤ y < 1, Y ∼ uniform[0, 1).
88
Chapter 5 Problem Solutions
49. In the derivation of Property (vii) of cdfs, we started with the formula
(−∞, x0 ) =
∞
[
(−∞, x0 − n1 ].
∞
[
(−∞, x0 − n1 ).
n=1
However, we can also write
(−∞, x0 ) =
n=1
Hence,
∞
[
1
G(x0 ) = P(X < x0 ) = P
{X < x0 − n } = lim P(X < x0 − N1 ) = lim G(x0 − N1 ).
N→∞
n=1
N→∞
Thus, G is left continuous. In a similar way, we can adapt the derivation of Property (vi) of cdfs to write
∞
\
{X < x0 + 1n } = lim P(X < x0 + 1n ) = lim G(x0 + 1n ) = G(x0 +).
P(X ≤ x0 ) = P
N→∞
n=1
N→∞
To conclude, write
P(X = x0 ) = P(X ≤ x0 ) − P(X < x0 ) = G(x0 +) − G(x0 ).
50. First note that since FY (t) is right continuous, so is 1−FY (t) = P(Y > t). Next, we use
the assumption P(Y > t + ∆t|T > t) = P(Y > ∆t) to show that with h(t) := P(Y > t),
h(t + ∆t) = h(t) + h(∆t). To this end, write
h(∆t) = ln P(Y > ∆t) = ln P(Y > t + ∆t|Y > t) = ln
= ln
P({Y > t + ∆t} ∩ {Y > t})
P(Y > t)
P(Y > t + ∆t)
= ln P(Y > t + ∆t) − ln P(Y > t) = h(t + ∆t) − h(t).
P(Y > t)
Rewrite this result as h(t + ∆t) = h(t) + h(∆t). Then with ∆t = t, we have h(2t) =
2h(t). With ∆t = 2t, we have h(3t) = h(t) + h(2t) = h(t) + 2h(t) = 3h(t). In general,
h(nt) = nh(t). In a similar manner we can show that
t
t
h(t) = h
= mh(t/m),
+···+
m
m
and so h(t/m) = h(t)/m. We now have that for rational a = n/m, h(at) = h(n(t/m)) =
nh(t/m) = (n/m)h(t) = ah(t). For general a ≥ 0, let ak ↓ a with ak rational. Then by
the right continuity of h,
h(at) = lim h(ak t) = lim ak h(t) = ah(t).
k→∞
k→∞
We can now write
h(t) = h(t · 1) = th(1).
Chapter 5 Problem Solutions
89
Thus,
t · h(1) = h(t) = ln P(Y > t) = ln(1 − FY (t)),
and we have 1 − FY (t) = eh(1)t , which implies Y ∼ exp(−h(1)). Of couse, −h(1) =
− ln P(Y > 1) = − ln[1 − FY (1)].
51. We begin with
1
1 n Xi − m
= √
E[Yn ] = E √ ∑
n i=1
σ
nσ
n
∑
i=1
E[Xi ] − m = 0.
For the variance, we use the fact that since independent random variables are uncorrelated, the variance of the sum is the sum of the variances. Thus,
n
n
n
Xi − m
var(Xi )
σ2
√
var(Yn ) = ∑ var
= ∑
=
= 1.
∑
2
2
σ n
i=1
i=1 nσ
i=1 nσ
52. Let Xi denote the time to transmit the ith packet, where Xi has mean m and variance
σ 2 . The total time to transmit n packets is Tn := X1 + · · · + Xn . The expected total
time is E[Tn ] = nm. Since we do not know the distribution of the Xi , we cannot know
the distribution of Tn . However, we use the central limit theorem to approximate
P(Tn > 2nm). Note that the sample mean Mn = Tn /n. Write
P(Tn > 2nm) = P( 1n Tn > 2m) = P(Mn > 2m) = P(Mn − m > m)
Mn − m
m
m
√ >
√
√
= P
= P Yn >
σ/ n
σ/ n
σ/ n
√
√
= 1 − FYn (m n/σ ) ≈ 1 − Φ(m n/σ ),
by the central limit theorem.
53. Let Xi = 1 if bit i is in error, and Xi = 0 otherwise. Then P(Xi = 1) = p. Although
the problem does not say so, let us assume that the Xi are independent. Then Mn =
1 n
n ∑i=1 Xi is the fraction of bits in error. We cannot reliably decode of Mn > t. To
approximate the probability that we cannot reliably decode, write
t −m
t −m
Mn − m
t −m
√ >
√
√
√
= P Yn >
= 1 − FYn
P(Mn > t) = P
σ/ n
σ/ n
σ/ n
σ/ n
t −m
t−p
√
≈ 1−Φ
,
= 1−Φ p
σ/ n
p(1 − p)/n
since m = E[Xi ] = p and σ 2 = var(Xi ) = p(1 − p).
54. If the Xi are i.i.d. Poisson(1), then Tn := X1 + · · · + Xn is Poisson(n). Thus,
nk e−n
1
1
1 k−n·1 2
√
√ .
≈ √ exp −
k!
2 1· n
1· n
2π
√
√
Taking k = n, we obtain nn e−n ≈ n!/ 2π n or n! ≈ 2π nn+1/2 e−n .
P(Tn = k) =
90
Chapter 5 Problem Solutions
55. Recall that gn is the density of X1 + · · · + Xn . If the Xi are i.i.d. uniform[−1, 1], then
gn is the convolution of (1/2)I[−1,1] (x) with itself n times. From graphical considerations, it is clear that gn (x) = 0 for |x| > n; i.e., xmax = n.
56. To begin, write
√ ϕYn (ν ) = E e jν (X1 +···+Xn )/ n = E
n
∏ e j(ν /
√
n )Xi
i=1
n
=
∏
i=1
√
2
ν 2 /2 n
= cosn (ν / n ) ≈ 1 −
→ e−ν /2 .
n
57.
h
√
√ i
1 jν / n
1 − jν / n
e
e
+
2
2
(a) MTTF = E[T ] = n (from Erlang in table).
(b) The desired probability is
R(t) := P(T > t) =
Z ∞ n−1 −τ
τ
e
t
(n − 1)!
dτ .
R
Let Pn (t) denote the above integral. Then P1 (t) = t∞ e−τ d τ = e−τ . For n > 1,
apply integration by parts with u = τ n−1 /(n − 1)! and dv = e−τ d τ . Then
Pn (t) =
t n−1 e−t
+ Pn−1 (t).
(n − 1)!
Applying this result recursively, we find that
Pn (t) =
t n−1 e−t t n−2 e−t
+
+ · · · + e−t ,
(n − 1)! (n − 2)!
which is the desired result.
(c) The failure rate is
r(t) =
t n−1
t n−1 e−t /(n − 1)!
fT (t)
.
=
=
n−1 k
n−1
R(t)
t
t k −t
(n − 1)! ∑
∑ e
k=0 k!
k=0 k!
For n = 2, r(t) = t/(1 + t):
1
0
0
58.
2
4
6
8
10
√
(a) Let λ =√1. For p = 1/2, r(t) = 1/(2 2). For p = 1, r(t) = 1. For p = 3/2,
r(t) = 3 t/2. For p = 2, r(t) = 2t. For p = 3, r(t) = 3t 2 .
Chapter 5 Problem Solutions
p=3
2
p=2
p = 3/2
p=1
p = 1/2
1
0
0
91
1
(b) We have from the text that
Zt
Zt
p
p−1
d τ = e−λ t .
R(t) = exp − r(τ ) d τ = exp − λ pτ
0
0
(c) The MTTF is
E[T ] =
Z ∞
R(t) dt =
0
Z ∞
0
p
e−λ t dt.
Now make the change of variable θ = λ t p , or t = (θ /λ )1/p . Then
Z ∞
Z ∞
θ 1/p−1 d θ
1
θ 1/p−1 e−θ d θ
E[T ] =
e−θ
=
λ
λp
pλ 1/p 0
0
Γ(1/p + 1)
1
Γ(1/p) =
.
=
1/p
pλ
λ 1/p
(d) Using the result of part (b),
p
fT (t) = −R0 (t) = λ t p−1 e−λ t ,
59.
t > 0.
(a) Write
Z
R(t) = exp −
∞
r(τ ) d τ
0
Zt
−1
= exp − pτ d τ
t0
= exp[−p(lnt − lnt0 )] = exp[ln(t0 /t) p ] = (t0 /t) p ,
t ≥ t0 .
(b) If t0 = 1 and p = 2, R(t) = 1/t 2 has the form
1
0
0
1
R(t) dt =
Z ∞
2
3
4
5
(c) For p > 1, the MTTF is
E[T ] =
Z ∞
0
t0
(t0 /t) p dt = t0p
t 1−p 1− p
∞
t0
=
t0
.
p−1
92
Chapter 5 Problem Solutions
(d) For the density, write
pt0
p t0 p
·
= p+1
,
t
t
t
p
fT (t) = r(t)R(t) =
60.
t ≥ t0 .
(a) A sketch of r(t) = t 2 − 2t + 2 for t ≥ 0 is:
6
4
2
0
0
1
2
3
(b) We first compute
Z t
r(τ ) d τ =
Z t
0
0
τ 2 − 2τ + 2 d τ =
2
1 3
3 t − t + 2t.
Then
fT (t) = r(t)e−
61.
Rt
0 r(τ ) d τ
1 3 −t 2 +2t)
= [t 2 − 2t + 2]e−( 3 t
,
t ≥ 0.
(a) If T ∼ uniform[1, 2], then for 0 ≤ t < 1, R(t) = P(T > t) = 1, and for t ≥ 2,
R(t) = P(T > t) = 0. For 1 ≤ t < 2,
R(t) = P(T > t) =
Z 2
t
1 d τ = 2 − t.
The complete formula and sketch are

 1, 0 ≤ t < 1,
2 − t, 1 ≤ t < 2,
R(t) =

0, t ≥ 2.
1
0
0
1
2
3
(b) The failure rate is
r(t) = −
d
1
d
ln R(t) = − ln(2 − t) =
,
dt
dt
2−t
1 < t < 2.
Chapter 5 Problem Solutions
93
(c) Since T ∼ uniform[1, 2], the MTTF is E[T ] = 1.5.
62. Write
R(t) := P(T > t) = P(T1 > t, T2 > t) = P(T1 > t)P(T2 > t) = R1 (t)R2 (t).
63. Write
R(t) := P(T > t) = P({T1 > t} ∪ {T2 > t}) = 1 − P(T1 ≤ t, T2 ≤ t)
= 1 − P(T1 ≤ t)P(T2 ≤ t) = 1 − [1 − R1 (t)][1 − R2 (t)]
= 1 − [1 − R1 (t) − R2 (t) + R1 (t)R2 (t)]
= R1 (t) + R2 (t) − R1 (t)R2 (t).
64. We follow the hint and write
E[Y n ] = E[T ] =
Z ∞
P(T > t) dt =
0
Z ∞
P(Y n > t) dt =
0
Z ∞
P(Y > t 1/n ) dt.
0
We then make the change of variable y = t 1/n , or t = yn , dt = nyn−1 dy, to get
E[Y n ] =
Z ∞
0
P(Y > y) · nyn−1 dy.
CHAPTER 6
Problem Solutions
1. Since the Xi are uncorrelated with common mean m and common variance σ 2 ,
n
1
E[Sn2 ] =
E ∑ Xi2 − nE[Mn ]
n−1
i=1
n
oi
1 h
n(σ 2 + m2 ) − n var(Mn ) + (E[Mn ])2
=
n−1
1 =
n(σ 2 + m2 ) − n{σ 2 /n + m2 }
n−1
1 (n − 1)σ 2 + nm2 − nm2 = σ 2 .
=
n−1
2.
(a) The mean of a Rayleigh(λ ) random variable is λ
Then
p
π /2. Consider
p
λn := Mn / π /2.
p
p
p
p
π /2 = λ .
E[λn ] = E[Mn / π /2 ] = E[Mn ]/ π /2 = λ π /2
Thus, λn is unbiased. Next,
p
p
p
p
λn = Mn / π /2 → E[Mn ]
π /2 = λ π /2
π /2 = λ ,
and we see that λn is strongly consistent.
(b) MATLAB. Add the line of code lambdan=mean(X)/sqrt(pi/2).
p
(c) MATLAB. Since Mn ≈ λ π /2, we solve for π and put
πn := 2(Mn /λ )2 .
Since λ = 3, add the line of code pin=2*(mean(X)/3)ˆ 2.
3.
(a) The mean of a gamma(p, λ ) random variable is p/λ . We put
pn := λ Mn .
Then E[pn ] = λ E[Mn ] = λ · p/λ = p. Also pn = λ Mn → λ (p/λ ) = p. Thus, pn
is unbiased and strongly consistent.
(b) MATLAB. In this problem λ = 1/2 and p = k/2, or k = 2p. We use kn := 2pn =
2(λ Mn ) = 2((1/2)Mn ) = Mn . We therefore add the line of code kn=mean(X).
94
Chapter 6 Problem Solutions
4.
95
(a) Since the mean of a noncentral chi-squared random variable with k degrees of
freedom and noncentrality parameter λ 2 is k + λ 2 , we put
λn2 := Mn − k.
Then E[λn2 ] = E[Mn − k] = E[Mn ] − k = (k + λ 2 ) − k = λ 2 , and we see that
λn2 is an unbiased estimator of λ 2 . Next, since λn2 = Mn − k → E[Mn ] − k =
(k + λ 2 ) − k = λ 2 , the estimator is strongly consistent.
(b) MATLAB. Since k = 5, add the line of code lambda2n=mean(X)-5.
5.
(a) Since the mean of a gamma(p, λ ) random variable is p/λ , we put λn := p/Mn .
Then λn = p/Mn → p/E[Mn ] = p/(p/λ ) = λ , and we see that λn is a strongly
consistent estimator of λ .
(b) MATLAB. Since p = 3, add the line of code lambdan=3/mean(X).
6.
(a) Since the variance of a Laplace(λ ) random variable is 2/λ 2 , we put
λn :=
q
2/Sn2 .
Since Sn2 converges to the variance, we have λn →
that λn is a strongly consistent estimator of λ .
p
2/(2/λ 2 ) = λ , and we see
(b) MATLAB. Add the line of code lambdan=sqrt(2/var(X)).
7.
(a) The mean of a gamma(p, λ ) random variable is p/λ . The second moment is
p(p + 1)/λ 2 . Hence, the variance is
p(p + 1) p2
p
− 2 = 2.
2
λ
λ
λ
Thus, Mn ≈ p/λ and Sn2 ≈ p/λ 2 . Solving for p and λ suggests that we put
λn :=
Mn
Sn2
and
pn := λn Mn .
Now, Mn → p/λ and Sn2 → p/λ 2 . It follows that λn → (p/λ ) (p/λ 2 ) = λ and
then pn → λ · (p/λ ) = p. Hence, λn is a strongly consistent estimator of λ , and
pn is a strongly consistent estimator of p.
(b) MATLAB. Add the code
Mn = mean(X)
lambdan = Mn/var(X)
pn = lambdan*Mn
8. Using results from the problem referenced in the hint, we have
Γ (q + p)/q
Γ(1 + p/q)
p/q
=
=
.
E[X q ] =
Γ(p/q)λ q
Γ(p/q)λ q
λq
96
Chapter 6 Problem Solutions
This suggests that we put
λn :=
Then
λn →
p/q
1
n
q
∑ni=1 Xi
p/q
(p/q)/λ q
1/q
1/q
.
= λ,
and we see that λn is a strongly consistent estimator of λ .
9. In the preceding problem E[X q ] = (p/q)/λ q . Now consider
Γ (2q + p)/q
Γ(2 + p/q)
(1 + p/q)(p/q)
2q
=
=
.
E[X ] =
2q
2q
Γ(p/q)λ
Γ(p/q)λ
λ 2q
In this equation, replace λ q by (p/q)/E[X q ] and solve for (p/q). Thus,
p/q =
(E[X q ])2
.
var(X q )
This suggests that we first put
Xnq :=
and then
pn := q
and
λn :=
pn /q
Xnq
1/q
1
n−1
h

= 
1 n q
∑ Xi
n i=1
(Xnq )2
i
q
q
∑ni=1 (Xi )2 − n(Xn )2
1/q
Xnq
h
i
q 2
q 2
1
n
(X
)
−
n(X
)
∑
n
i=1 i
n−1
.
10. MATLAB. OMITTED.
11. MATLAB. The required script can be created using the code from Problem 2 followed
by the lines
global lambdan
lambdan = mean(X)/sqrt(pi/2)
followed by the script from Problem 10 modifed as follows: The chi-squared statistic
Z can be computed by inserting the lines
p = CDF(b) - CDF(a);
Z = sum((H-n*p).ˆ2./(n*p))
Chapter 6 Problem Solutions
97
after the creation of the right edge sequence in the script given in Problem 10, where
CDF is the function
function y = CDF(t) % Rayleigh CDF
global lambdan
y = zeros(size(t));
i = find(t>0);
y(i) = 1-exp(-(t(i)/lambdan).ˆ2/2);
In addition, the line defining y in the script from Problem 10 should be changed to
y=PDF(t), where PDF is the function
function y = PDF(t) % Rayleigh density
global lambdan
y = zeros(size(t));
i = find(t>0);
y(i) = (t(i)/lambdanˆ2).*exp(-(t(i)/lambdan).ˆ2/2);
Finally, the chi-squared statistic Z should be compared with zα = 22.362, since α =
0.05 and since there are m = 15 bins and r = 1 estimated parameter, the degrees of
freedom parameter is k = m − 1 − r = 15 − 1 − 1 = 13 in the chi-squared table in the
text.
12. MATLAB. Similar to the solution of Problem 11 except that it is easier to use the
M ATLAB function chi2cdf or gamcdf to compute the required cdfs for evaluating
the chi-squared statistic Z. For the same reasons as in Problem 11, zα = 22.362.
13. MATLAB. Similar to the solution of Problem 11 except that it is easier to use the
M ATLAB function ncx2cdf to compute the required cdfs for evaluating the chisquared statistic Z. For the same reasons as in Problem 11, zα = 22.362.
14. MATLAB. Similar to the solution of Problem 11 except that it is easier to use the M ATLAB function gamcdf to compute the required cdfs for evaluating the chi-squared
statistic Z. For the same reasons as in Problem 11, zα = 22.362.
15. MATLAB. Similar to the solution of Problem 11. For the same reasons as in Problem 11, zα = 22.362.
16. Since
E[H j ] = E
I
(X
)
∑ [e j ,e j+1 ) i =
n
i=1
n
n
∑ P(e j ≤ Xi < e j+1 )
i=1
H j − np j
E √
np j
=
∑ pj
= np j ,
i=1
= 0.
Since the Xi are i.i.d., the I[e j ,e j+1 ) (Xi ) are i.i.d. Bernoulli(p j ). Hence,
n
E[(H j − np j )2 ] = var(H j ) = var ∑ I[e j ,e j+1 ) (Xi )
i=1
n
=
∑ var
i=1
I[e j ,e j+1 ) (Xi ) = n · p j (1 − p j ),
98
Chapter 6 Problem Solutions
and so
E
H j − np j
√
np j
2 = 1 − p j.
17. If f is an even density, then
F(−x) =
Z −x
−∞
f (t) dt = −
Z x
∞
f (−θ ) d θ =
Z ∞
x
f (θ ) d θ = 1 − F(x).
√
18. The width of any confidence interval is w = 2σ y/ n. If σ = 2 and n = 100,
w99% =
2 · 2 · 2.576
= 1.03.
10
To make w99% < 1/4 requires
2σ y
√ < 1/4
n
or
n > (8σ y)2 = (16 · 2.576)2 = 1699.
2
19. First observe that with Xi = m+Wi , E[X
√i ] = m, and var(Xi ) = var(Wi ) = 4. So, σ = 4.
For 95% confidence interval, σ yα /2 / n = 2 · 1.960/10 = 0.392, and so
m = 14.846 ± 0.392
with 95% probability.
The corresponding confidence interval is [14.454, 15.238].
20. Write
1 n
P(|Mn − m| ≤ δ ) = P(−δ ≤ Mn − m ≤ δ ) = P −δ ≤ ∑ (m +Wi ) − m ≤ δ
n i=1
n
n
1
= P −δ ≤ ∑ Wi ≤ δ = P −nδ ≤ ∑ Wi ≤ nδ
n i=1
i=1
| {z }
Cauchy(n)
2
= tan−1 (nδ /n),
π
which is equal to 2/3 if and only if tan−1 (δ ) = π /3, or δ =
√
3.
21. MATLAB. OMITTED.
√
22. We use the formula m = Mn ± yα /2 Sn / n = 10.083 ± (1.960)(0.568)/10 to get
m = 10.083 ± 0.111
with 95% probability,
and the confidence intervale is [9.972, 10.194].
√
23. We use the formula m = Mn ± yα /2 Sn / n = 4.422 ± (1.812)(0.957)/10 to get
m = 4.422 ± 0.173
with 93% probability,
and the confidence interval is [4.249, 4.595].
Chapter 6 Problem Solutions
99
24. We have√Mn = number defective/n = 10/100 = 0.1. We use the formula m = Mn ±
yα /2 Sn / n = 0.1 ± (1.645)(.302)/10 to get
m = 0.1 ± 0.0497
with 90% probability.
The number of defectives is 10 000m, or
number of defectives = 1000 ± 497
with 90% probability.
Thus, we are 90% sure that the number of defectives is between 503 and 1497 out of
a total of 10 000 units.
√
25. We have Mn = 1559/3000. We use the formula m = Mn ± yα /2 Sn / n = 0.520 ±
√
(1.645)(.5)/ 3000 to get
m = 0.520 ± 0.015
with 90% probability,
and the confidence interval is [0.505, 0.535]. Hence, the probability is at least 90%
that more than 50.5% of the voters will vote for candiate A. So we are 90% sure that
candidate A will win. The 99% confidence interval is given by
m = 0.520 ± 0.024
with 99% probability,
and the confidence interval is [0.496, 0.544]. Hence, we are not 99% sure that candidate A will win.
26. We have Mn =
√ = 0.0960. We use the formula m =
√ number defective/n = 48/500
Mn ± yα /2 Sn / n = 0.0960 ± (1.881)(.295)/ 500 to get
m = 0.0960 ± 0.02482
with 94% probability.
The number of defectives is 100 000m, or
number of defectives = 9600 ± 2482
with 94% probability.
Thus, we are 94% sure that the number of defectives is between 7118 and 12 082 out
of a total of 100 000 units.
√
27. (a) We have Mn = 6/100. We use the formula m = Mn ± yα /2 Sn / n = 0.06 ±
(2.170)(.239)/10 to get
m = 0.06 ± 0.0519
with 97% probability.
We are thus 97% sure that p = m lies in the interval [0.0081, 0.1119]. Thus, we
are not 97% sure that p < 0.1.
√
(b) We have Mn = 71/1000. We use the formula m = Mn ± yα /2 Sn / n = 0.071 ±
√
(2.170)(.257)/ 1000 to get
m = 0.071 ± 0.018
with 97% probability.
We are thus 97% sure that p = m lies in the interval [0.053, 0.089]. Thus, we are
97% sure that p < 0.089 < 0.1.
100
28.
Chapter 6 Problem Solutions
(a) Let Ti denote the time to transmit the ith packet. Then we need to compute
n
n
\
[
{Ti ≤ t} = 1 − FT1 (t)n = 1 − (1 − e−t/µ )n .
{Ti > t} = 1 − P
P
i=1
i=1
(b) Using the notation from part (a), T = T1 + · · · + Tn . Since the Ti are i.i.d.
exp(1/µ ), T is Erlang(n, 1/µ ) by Problem 55(c) in Chapter 4 and the remark
following it. Hence,
fT (t) = (1/µ )
(t/µ )n−1 e−t/µ
,
(n − 1)!
t ≥ 0.
(c) We have
yα /2 Sn
1.960(1.798)
µ = Mn ± √
= 1.994 ±
= 1.994 ± 0.352
n
10
and confidence interval [1.642, 2.346] with 95% probability.
29. MATLAB. OMITTED.
30. By the hint,
∑ni=1 Xi is Gaussian with mean nm and variance nσ 2 . Since Mn =
n
∑i=1 Xi /n, it is easy to see that Mn is still Gaussian, and its mean is (nm)/n = m.
Gaussian but with mean zero
Its variance is (nσ 2 )/n2 = σ 2 /n. Next, Mn − m remains
p
and the same variance σ 2 /n. Finally, (M
σ
/n
remains Gaussian and with
−
m)/
n
p
mean zero, but its variance is (σ 2 /n)/( σ 2 /n )2 = 1.
√
31. We use the fomula m = Mn ± yα /2 Sn / n, where in this Gaussian case, yα /2 is taken
from the tables using Student’s t distribution with n = 10. Thus,
yα /2 Sn
2.262 · 1.904
√
m = Mn ± √
= 14.832 ±
= 14.832 ± 1.362,
n
10
and the confidence interval is [13.470, 16.194] with 95% probability.
32. We use [nVn2 /u, nVn2 /`], where u and ` are chosen from the appropriate table. For a
95% confidence interval, ` = 74.222 and u = 129.561. Thus,
2
nVn nVn2
100(4.413) 100(4.413)
=
= [3.406, 5.946].
,
,
u
`
129.561
74.222
33. We use [(n − 1)Sn2 /u, (n − 1)Sn2 /`], where u and ` are chosen from the appropriate
table. For a 95% confidence interval, ` = 73.361 and u = 128.422. Thus,
(n − 1)Sn2 (n − 1)Sn2
99(4.736) 99(4.736)
,
=
,
= [3.651, 6.391].
u
`
128.422
73.361
34. For the two-sided test at the 0.05 significance level, we compare |Zn | with yα /2 =
1.960. Since |Zn | = 1.8 ≤ 1.960 = yα /2 , we accept the null hypothesis. For the
one-sided test of m > m0 at the 0.05 significance level, we compare Zn with −yα =
−1.645. Since it is not the case that Zn = −1.80 > −1.645 = −yα , we do not accept
the null hypothesis.
Chapter 6 Problem Solutions
101
35. Suppose Φ(−y) = α . Then by Problem 17, Φ(−y) = 1 − Φ(y), and so 1 − Φ(y) = α ,
or Φ(y) = 1 − α .
36.
(a) Since Zn = 1.50 ≤ yα = 1.555, the Internet service provider accepts the null
hypothesis.
(b) Since Zn = 1.50 > −1.555 = −yα , we accept the null hypothesis; i.e., we reject
the claim of the Internet service provider.
37. The computer vendor would take the null hypothesis to be m ≤ m0 . To give the vendor
the benefit of the doubt, the consumer group uses m ≤ m0 as the null hypothesis. To
accept the null hypothesis would require Zn ≤ yα . Only by using the sigificance level
of 0.10, which has yα = 1.282, can the consumer group give the benefit of the doubt
to the vendor and still reject the vendor’s claim.
38. Giving itself the benfit of the doubt, the company uses the null hypothesis m > m0 and
uses a 0.05 significance level. The null hypothesis will be accepted if Zn > −yα =
−1.645. Since Zn = −1.6 > −1.645, the company believes it has justified its claim.
39. Write
n
e(b
g) =
∑ |Yk − (baxk + bb)|2
k=1
n
=
=
∑ |Yk − (baxk + [Y − abx])|2
k=1
∑ |(Yk −Y ) + ab(xk − x)|2
k=1
n h
∑
k=1
40. Write
n
=
a(xk − x)(Yk −Y ) + ab2 (xk − x)2
(Yk −Y )2 − 2b
i
= SYY − 2b
aSxY + ab2 Sxx
S xY
2
= SYY − 2b
aSxY + ab
Sxx = SYY − abSxY = SYY − SxY
/Sxx .
Sxx
E[Y |X = x] = E[g(X) +W |X = x] = E[g(x) +W |X = x]
= g(x) + E[W |X = x] = g(x) + E[W ] = g(x).
41. MATLAB. OMITTED.
42. MATLAB. OMITTED.
43. MATLAB. If z = c/t q , then ln z = ln c − q lnt. If y = ln z and x = lnt, then y = (−q)x +
ln c. If y ≈ a(1)x + a(2), then q = −a(1) and c = exp(a(2)). Hence, the two lines of
code that we need are
qhat = -a(1)
chat = exp(a(2))
44. Obvious.
102
Chapter 6 Problem Solutions
45. Write
e−(z−s) /2
e−z /2 . s2 /2
√
=
fZe (z) = esz fZ (z)/MZ (s) = esz √
.
e
2π
2π
2
2
e = t, put s = t. Then Ze ∼ (t, 1).
To make E[Z]
46. Write
Then
fZe (z) = esz fZ (z)/MZ (s) = esz
e =
E[Z]
λ (λ z) p−1 e−λ z . λ p
,
Γ(p)
λ −s
z > 0.
Z ∞
λ −p Z ∞ λ (λ z) p−1 e−z(λ −s)
(λ − s) p z p−1 e−z(λ −s)
z
dz =
z
dz,
λ −s
Γ(p)
Γ(p)
0
0
e = p/(λ − s). To make
which is the mean of a gamma(p, λ − s) density. Hence, E[Z]
e
E[Z] = t, we need p/(λ − s) = t or s = λ − p/t.
47. MATLAB. OMITTED.
48. First,
Then
.
pZe (zi ) = eszi pZ (zi ) [(1 − p) + pes ].
pZe (1) = es p/[(1 − p) + pes ] and
pZe (0) = (1 − p)/[(1 − p) + pes ].
CHAPTER 7
Problem Solutions
1. We have
FZ (z) = P(Z ≤ z) = P(Y − X ≤ z) = P((X,Y ) ∈ Az ),
where
Az := {(x, y) : y − x ≤ z} = {(x, y) : y ≤ x + z}.
y
z
−z
x
2. We have
h
i
FZ (z) = P(Z ≤ z) = P(Y /X ≤ z) = P {Y /X ≤ z} ∩ {X < 0} ∪ {X > 0}
+
= P((X,Y ) ∈ D−
z ∪ Dz ) = P((X,Y ) ∈ Az ),
where
+
Az := D−
z ∪ Dz ,
and
D−
z := {(x, y) : y/x ≤ z and x < 0} = {(x, y) : y ≥ zx and x < 0},
and
z
D+
z := {(x, y) : y/x ≤ z and x > 0} = {(x, y) : y ≤ zx and x > 0}.
lin
eo
fs
lo
pe
y
+
−
Dz
Dz
3.
(a) R := (a, b] × (c, d] is
103
x
104
Chapter 7 Problem Solutions
d
c
a
b
(b) A := (−∞, a] × (−∞, d] is
d
c
a
b
(c) B := (−∞, b] × (−∞, c] is
d
c
a
b
(d) C := (a, b] × (−∞, c] is
d
c
a
b
(e) D := (−∞, a] × (c, d] is
d
c
a
b
Chapter 7 Problem Solutions
105
(f) A ∩ B is
d
c
a
b
4. Following the hint and then observing that R and A ∪ B are disjoint, we have
P (X,Y ) ∈ (−∞, b] × (−∞, d] = P (X,Y ) ∈ R + P (X,Y ) ∈ A ∪ B .
(∗)
Next, by the inclusion–exclusion formula,
P (X,Y ) ∈ A ∪ B = P (X,Y ) ∈ A + P (X,Y ) ∈ B − P (X,Y ) ∈ A ∩ B
= FXY (a, d) + FXY (b, c) − FXY (a, c).
Hence, (∗) becomes
FXY (b, d) = P (X,Y ) ∈ R + FXY (a, d) + FXY (b, c) − FXY (a, c),
5.
which is easily rearranged to get the rectangle formula,
P (X,Y ) ∈ R = FXY (b, d) − FXY (a, d) − FXY (b, c) + FXY (a, c).
(a) {(x, y) : |x| ≤ y ≤ 1} is NOT a product set.
(b) {(x, y) : 2 < x ≤ 4, 1 ≤ y < 2} = (2, 4] × [1, 2).
(c) {(x, y) : 2 < x ≤ 4, y = 1} = (2, 4] × {1}.
(d) {(x, y) : 2 < x ≤ 4} = (2, 4] × IR.
(e) {(x, y) : y = 1} = IR × {1}.
(f) {(1, 1), (2, 1), (3, 1)} = {1, 2, 3} × {1}.
(g) The union of {(1, 3), (2, 3), (3, 3)} and the set in (f) is equal to {1, 2, 3} × {1, 3}.
(h) {(1, 0), (2, 0), (3, 0), (0, 1), (1, 1), (2, 1), (3, 1)} is NOT a product set.
6. We have

 1, x ≥ 2,
x − 1, 1 ≤ x < 2,
FX (x) =

0, x < 1,
and
FY (y) =
(
1− e
−y −e−2y
y
0,
, y ≥ 0,
y < 0,
where the quotient involving division by y is understood as taking its limiting value
of one when y = 0. Since FX (x)FY (y) 6= FXY (x, y) when 1 ≤ x ≤ 2 and y > 0, X and Y
are NOT independent.
106
Chapter 7 Problem Solutions
7. We have

 1, x ≥ 3,
2/7, 2 ≤ x < 3,
FX (x) =

0, x < 2,
FY (y) =
and
1
−2y − 5e−3y ],
7 [7 − 2e
0,
y ≥ 0,
y < 0.
8. Let y > 0. First compute
(y + 1){1 + e−x(y+1) (−x)} − {y + e−x(y+1) }(1)
∂ y + e−x(y+1)
=
∂y
y+1
(y + 1)2
=
(y + 1) − x(y + 1)e−x(y+1) − y − e−x(y+1)
(y + 1)2
=
1 − e−x(y+1) {1 + x(y + 1)}
.
(y + 1)2
Then compute
{1 + x(y + 1)}e−x(y+1) (y + 1) − e−x(y+1) (y + 1)
∂ ∂ y + e−x(y+1)
=
∂x ∂y
y+1
(y + 1)2
=
xe−x(y+1) (y + 1)2
= xe−x(y+1) ,
(y + 1)2
x, y > 0.
9. The first step is to recognize that fXY (x, y) factors into
2
exp[−|y − x| − x2 /2]
e−x /2 e−|y−x|
√
fXY (x, y) =
= √
·
.
2
2 2π
2π
When integrating this last factor with respect to y, make the change of variable θ =
y − x to get
fX (x) =
Z ∞
2
−∞
e−x /2
fXY (x, y) dy = √
2π
2
e−x /2
= √
2π
Z ∞
0
2
Z ∞
−∞
2
1 −|y−x|
dy
2e
e−x /2
= √
2π
e−x /2
e−θ d θ = √
.
2π
Z ∞
−∞
1 −|θ |
dθ
2e
Thus, X ∼ N(0, 1).
10. The first step is to factor fXY (x, y) as
2
fXY (x, y) =
2
4 e−(x−y) /2
4e−(x−y) /2
√
= 5· √
.
5
y
y 2π
2π
Regarding this last factor a function of x, it is an N(y, 1) density. In other words, when
integrated with respect to x, the result is one. In symbols,
fY (y) =
Z ∞
−∞
fXY (x, y) dx =
4
y5
Z ∞ −(x−y)2 /2
e
−∞
√
2π
dx =
4
,
y5
y ≥ 1.
Chapter 7 Problem Solutions
107
11. We first analyze U := max(X,Y ). Then
FU (u) = P(max(X,Y ) ≤ u) = P(X ≤ u and Y ≤ u) = FXY (u, u),
and the density is
fU (u) =
∂ FXY (x, y)
∂x
+
x=u, y=u
∂ FXY (x, y)
∂y
.
x=u, y=u
If X and Y are independent, then
FU (u) = FX (u)FY (u),
and
fU (u) = fX (u)FY (u) + FX (u) fY (u).
If in addition X and Y have the same density, say f , (and therefore the same cdf, say
F), then
FU (u) = F(u)2 , and fU (u) = 2F(u) f (u).
We next analyze V := min(X,Y ). Using the inclusion–exclusion formula,
FV (v) = P(min(X,Y ) ≤ v) = P(X ≤ v or Y ≤ v)
= P(X ≤ v) + P(Y ≤ v) − P(X ≤ v and Y ≤ v)
= FX (v) + FY (v) − FXY (v, v).
The density is
fV (v) = fX (v) + fY (v) −
∂ FXY (x, y)
∂x
x=v, y=v
−
∂ FXY (x, y)
∂y
.
x=v, y=v
If X and Y are independent, then
FV (v) = FX (v) + FY (v) − FX (v)FY (v),
and
fV (v) = fX (v) + fY (v) − fX (v)FY (v) − FX (v) fY (v).
If in addition X and Y have the same density f and cdf F, then
FV (v) = 2F(v) − F(v)2 ,
and
fV (v) = 2[ f (v) − F(v) f (v)].
12. Since X ∼ gamma(p, 1) and Y ∼ gamma(q, 1) are independent, we have from Problem 55(c) in Chapter 4 that Z ∼ gamma(p + q, 1). Since p = q = 1/2, we further have
Z ∼ gamma(1, 1) = exp(1). Hence, P(Z > 1) = e−1 .
13. We have from Problem 55(b) in Chapter 4 that Z ∼ Cauchy(λ + µ ). Since λ = µ =
1/2, Z ∼ Cauchy(1). Thus,
P(Z ≤ 1) =
1π 1
3
1
1
+ = .
tan−1 (1) + =
π
2
π4 2
4
108
Chapter 7 Problem Solutions
14. First write
FZ (z) =
Z ∞ Z z−y
−∞
−∞
fXY (x, y) dx dy.
It then follows that
fZ (z) =
∂
FZ (z) =
∂z
Z ∞
Z
∂ z−y
−∞
∂z
fXY (x, y) dx dy =
−∞
Z ∞
−∞
fXY (z − y, y) dy.
15. First write
−
FZ (z) = P((X,Y ) ∈ Az ) = P((X,Y ) ∈ B+
z ∪ Bz )
ZZ
=
fXY (x, y) dx dy +
B+
z
Z ∞ Z z/y
=
−∞
0
ZZ
fXY (x, y) dx dy
B−
z
Z
fXY (x, y) dx dy +
0
−∞
Z
∞
z/y
fXY (x, y) dx dy.
Then
∂
FZ (z) =
∂z
fZ (z) =
=
Z ∞
0
=
Z ∞
0
fXY (z/y, y)/y dy −
Z 0
fXY (z/y, y)/|y| dy +
Z ∞
−∞
−∞
Z 0
−∞
fXY (z/y, y)/y dy
fXY (z/y, y)/|y| dy
fXY (z/y, y)/|y| dy.
16. For the cdf, write
FZ (z) =
Z ∞ Z z+x
−∞
−∞
fXY (x, y) dy dx.
Then
fZ (z) =
∂
FZ (z) =
∂z
Z
Z ∞
∂ z+x
−∞
∂z
−∞
fXY (x, y) dy dx =
Z ∞
−∞
fXY (x, z + x) dx.
17. For the cdf, write
FZ (z) = P(Z ≤ z) = P(Y /X ≤ z)
= P(Y ≤ Xz, X > 0) + P(Y ≥ Xz, X < 0)
Z 0 Z ∞
Z ∞ Z xz
fXY (x, y) dy dx.
fXY (x, y) dy dx +
=
−∞
−∞
0
xz
Then
fZ (z) =
18.
∂
FZ (z) =
∂z
(a) The region D:
Z ∞
0
fXY (x, xz)x dx −
Z 0
−∞
fXY (x, xz)x dx =
Z ∞
−∞
fXY (x, xz)|x| dx.
Chapter 7 Problem Solutions
1
109
D
0
−1
−1
0
1
(b) Since fXY (x, y) = Kxn ym for (x, y) ∈ D and since D contains negative values of
x, we must have n even in order that the density be nonnegative. In this case,
the integral of the density over the region D must be one. Hence, K must be
such that
m+1 1
ZZ
Z 1 Z 1
Z 1
n m
n y
dx
1 =
fXY (x, y) dy dx =
Kx y dy dx =
Kx
m + 1 |x|
−1
|x|
−1
D
Z 1
Z 1
Kxn
K
=
xn − |x|n+m+1 dx
[1 − |x|m+1 ] dx =
m + 1 −1
−1 m + 1
Z 1
2K
1
1
2K
=
−
xn − xn+m+1 dx =
m+1 0
m+1 n+1 n+m+2
2K
=
.
(n + 1)(n + m + 2)
Hence, K = (n + 1)(n + m + 2)/2.
(c) A sketch of Az with z = 0.3:
A
1
z
0
−1 Az
−1
0
1
0
1
(d) A sketch of Az ∩ D with z = 0.3:
1
0
−1
−1
(e)
P((X,Y ) ∈ Az ) =
Z √z Z 1
z
z/x
Z 1 Z 1
Kxn ym dy dx + √
Kxn ym dy dx
z
x
110
Chapter 7 Problem Solutions
Z √z
=
Kxn
z
K
m+1
=
K
=
m+1
K
=
m+1
Z
Z 1
Z 1
ym dy dx + √ Kxn
ym dy dx
z/x
√
z
n
Z
z
Z
√
z
1
z
x [1 − (z/x)m+1 ] dx +
n
m+1 n−m−1
n
x dx −
Z
K
1 − zn
−
m+1 n+1
z
=
x
z
x −z
z
Z
1
dx +
x
Z
√
z
z
√
z
z
Z 1
xn [1 − xm+1 ] dx
√
z
Z 1
√
m+1 n−m−1
x
n
z
x −x
dx −
zm+1 xn−m−1 dx −
n+m+1
Z 1
√ x
z
dx
n+m+1
dx
1 − z(n+m+2)/2
.
n+m+2
If n 6= m, the remaining integral is equal to
√
( z )n+m+2 − zn+1
.
n−m
Otherwise, the integral is equal to
−
zm+1
ln z.
2
19. Let X ∼ uniform[0, w] and Y ∼ uniform[0, h]. We need to compute P(XY ≥ λ wh).
Before proceeding, we make a few observations. First, since X ≥ 0, we can write for
z > 0,
Z ∞Z ∞
P(XY ≥ z) =
0
z/x
fXY (x, y) dy dx =
Z ∞
0
fX (x)
Z ∞
z/x
fY (y) dy dx.
Since Y ∼ uniform[0, h], the inner integral will be zero if z/x > h. Since z/x ≤ h if
and only if x ≥ z/h,
P(XY ≥ z) =
Z ∞
z/h
fX (x)
Z h
1
z/x
h
dy dx =
Z ∞
z/h
z
fX (x) 1 −
dx.
xh
We can now write
Z ∞
Z ∞
λ wh λw
dx =
dx
fX (x) 1 −
fX (x) 1 −
xh
x
λ wh/h
λw
Z w w
1
λw
1−
dx = (1 − λ ) − λ ln
= (1 − λ ) + λ ln λ .
=
x
λw
λw w
P(XY ≥ λ wh) =
20. We first compute
E[XY ] = E[cos Θ sin Θ] =
=
1
2 E[sin 2Θ]
cos(−2π ) − cos(2π )
= 0.
8π
=
1
4π
Z π
−π
sin 2θ d θ
Chapter 7 Problem Solutions
111
Similarly,
E[X] = E[cos Θ] =
1
2π
1
2π
Z π
Z π
−π
cos θ d θ =
sin(π ) − sin(−π )
0−0
=
= 0,
2π
2π
and
E[Y ] = E[sin Θ] =
−π
sin θ d θ =
cos(−π ) − cos(π )
(−1) − (−1)
=
= 0.
2π
2π
To prove that X and Y are not independent, we argue by contradiction. However,
before we begin, observe that since (X,Y ) satisfies
X 2 +Y 2 = cos2 Θ + sin2 Θ = 1,
(X,Y ) always lies on the unit circle. Now consider the square of side one centered at
the origin,
S := {(x, y) : |x| ≤ 1/2, and |y| ≤ 1/2}.
Since this region lies strictly inside the unit circle, P((X,Y ) ∈ S) = 0. Now, to obtain
a contradiction suppose that X and Y are independent. Then fXY (x, y) = fX (x) fY (y),
where fX and fY are both arcsine densities by Problem 35 in Chapter 5. Hence, for
|x| < 1 and |y| < 1, fXY (x, y) ≥ fXY (0, 0) = 1/π 2 . We can now write
P((X,Y ) ∈ S) =
ZZ
S
fXY (x, y) dx dy ≥
ZZ
1/π 2 dx dy = 1/π 2 > 0,
S
which is a contradiction.
21. If E[h(X)k(Y )] = E[h(X)]E[k(Y )] for all bounded continuous functions h and k, then
we may specialize this equation to the functions h(x) = e jν1 x and k(y) = e jν2 y to show
that the joint characteristic function satisfies
ϕXY (ν1 , ν2 ) := E[e j(ν1 X+ν2Y ) ] = E[e jν1 X e jν2Y ] = E[e jν1 X ]E[e jν2Y ] = ϕX (ν1 )ϕY (ν2 ).
Since the joint characteristic function is the product of the marginal characteristic
functions, X and Y are independent.
22.
(a) Following the hint, let D denote the half-plane D := {(x, y) : x > x0 },
distance of point from origin is x0 / cos θ
y
D
θ
x0
x
112
Chapter 7 Problem Solutions
and write
P(X > x0 ) = P(X > x0 ,Y ∈ IR) = P((X,Y ) ∈ D),
where X and Y are both N(0, 1). Then
ZZ
P((X,Y ) ∈ D) =
D
2
e−(x +y
2π
2 )/2
dx dy.
Now convert to polar coordinates and write
P((X,Y ) ∈ D) =
Z π /2 Z ∞
=
2
−π /2 x0 / cos θ
1
2π
=
1
2π
=
1
π
Z π /2
−π /2
Z π /2
−π /2
Z π /2
e−r /2
r dr d θ
2π
−e−r
2 /2
∞
dθ
x0 / cos θ
exp[−(x0 / cos θ )2 /2] d θ
exp
0
−x02
dθ .
2 cos2 θ
(b) In the preceding integral, make the change of variable θ = π /2 − t, d θ = −dt.
Then cos θ becomes cos(π /2 − t) = sint, and the preceding integral becomes
Z
−x02
1 π /2
dt.
exp
π 0
2 sin2 t
23. Write
Z ∞
−∞
fY |X (y|x) dy =
Z ∞
fXY (x, y)
−∞
fX (x)
dy =
1
fX (x)
Z ∞
−∞
fXY (x, y) dy =
fX (x)
= 1.
fX (x)
24. First write
P((X,Y ) ∈ A|x < X ≤ x + ∆x) =
=
=
=
P (X,Y ) ∈ A ∩ x < X ≤ x + ∆x
P(x < X ≤ x + ∆x)
P (X,Y ) ∈ A ∩ (X,Y ) ∈ (x, x + ∆x] × IR
P(x < X ≤ x + ∆x)
n
h
io
P (X,Y ) ∈ A ∩ (x, x + ∆x] × IR
ZZ
P(x < X ≤ x + ∆x)
fXY (t, y) dy dt
A∩ (x,x+∆x]×IR
Z x+∆x
x
fX (τ ) d τ
Chapter 7 Problem Solutions
=
ZZ
=
Z x+∆x Z ∞
113
IA (t, y)I(x,x+∆x]×IR (t, y) fXY (t, y) dy dt
Z x+∆x
x
fX (τ ) d τ
IA (t, y) fXY (t, y) dy dt
−∞
Z x+∆x
x
fX (τ ) d τ
x
Z
Z
1 x+∆x ∞
∆x
=
x
1
∆x
IA (t, y) fXY (t, y) dy dt
−∞
Z x+∆x
fX (τ ) d τ .
x
It now follows that
lim P((X,Y ) ∈ A|x < X ≤ x + ∆x) =
∆t→∞
=
Z ∞
IA (x, y) fXY (x, y) dy
Z ∞
IA (x, y) fY |X (y|x) dy.
−∞
−∞
fX (x)
25. We first compute, for x > 0,
xe−x(y+1)
= xe−xy , y > 0.
e−x
As a function of y, this is an exponential density with parameter x. This is very
different from fY (y) = 1/(y + 1)2 . We next compute, for y > 0,
fY |X (y|x) =
fX|Y (x|y) =
xe−x(y+1)
= (y + 1)2 xe−(y+1)x ,
1/(y + 1)2
x > 0.
As a function of x this is an Erlang(2, y + 1) density, which is not the same as fX ∼
exp(1).
26. We first compute, for x > 0,
xe−x(y+1)
= xe−xy , y > 0.
e−x
As a function of y, this is an exponential density with parameter x. Hence,
fY |X (y|x) =
E[Y |X = x] =
Z ∞
0
y fY |X (y|x) dy = 1/x.
We next compute, for y > 0,
fX|Y (x|y) =
xe−x(y+1)
= (y + 1)2 xe−(y+1)x ,
1/(y + 1)2
x > 0.
As a function of x this is an Erlang(2, y + 1) density. Hence,
E[X|Y = y] =
Z ∞
0
x fX|Y (x|y) dx = 2/(y + 1).
114
Chapter 7 Problem Solutions
27. Write
Z ∞
−∞
fX|Y (x|y) dx fY (y) dy
−∞ B
Z ∞ Z ∞
=
IB (x) fX|Y (x|y) dx fY (y) dy
−∞ −∞
Z ∞
Z ∞
=
IB (x)
fXY (x, y) dy dx
P(X ∈ B|Y = y) fY (y) dy =
Z ∞ Z
−∞
=
Z
B
−∞
fX (x)dx = P(X ∈ B).
28. For z ≥ 0,
fZ (z) =
2 )/(2σ 2 )
Z √z e−(z−y
√
√
− z
e−z/(2σ
=
πσ 2
=
p
2)
z − y2
Z √z
0
−z/(2σ 2 )
e
πσ 2
2
e−(y/σ ) /2
e−z/(2σ
· √
dy =
2πσ 2
2πσ
2πσ
p
1
z − y2
dy =
e−z/(2σ
πσ 2
2)
Z 1
0
2
2)
Z √z
√
− z
p
1
z − y2
dy
1
√
dt
1 − t2
e−z/(2σ )
,
2σ 2
[sin−1 (1) − sin−1 (0)] =
which is an exponential density with parameter 1/(2σ 2 ).
29. For z ≥ 0,
Z ∞
fZ (z) =
0
= 2
x · λ e−λ xz · λ e−λ x dx +
Z ∞
0
0
y · λ e−λ yz · λ e−λ y dy
x · λ e−λ xz · λ e−λ x dx
2λ 2
λz+λ
=
Z ∞
Z ∞
0
x · [λ z + λ ]e−x[λ z+λ ] dx.
Now, this last integral is the expectation of an exponential density with parameter
λ z + λ . Hence,
fZ (z) =
2λ 2
1
2
,
·
=
λz+λ λz+λ
(z + 1)2
z ≥ 0.
30. Using the law of total probability, substitution, and independence, we have
P(X ≤ Y ) =
=
Z ∞
Z0 ∞
0
= λ
P(X ≤ Y |X = x) fX (x) dx =
P(Y ≥ x) fX (x) dx =
Z ∞
0
e−x(λ +µ ) dx =
Z ∞
0
Z ∞
0
P(x ≤ Y |X = x) fX (x) dx
e−µ x · λ e−λ x dx
λ −(λ +µ ) −e
λ +µ
∞
0
=
λ
.
λ +µ
Chapter 7 Problem Solutions
115
31. Using the law of total probability, substitution, and independence, we have
P(Y / ln(1 + X 2 ) > 1) =
=
Z ∞
−∞
Z 2
1
=
Z 2
P(Y / ln(1 + X 2 ) > 1|X = x) fX (x) dx
P(Y > ln(1 + x2 )|X = x) · 1 dx
2
P(Y > ln(1 + x )) dx =
1
=
Z 2
eln(1+x
2 )−1
dx =
1
Z 2
1
= tan−1 (2) − tan−1 (1).
Z 2
1
2
e− ln(1+x ) dx
1
dx
1 + x2
32. First find the cdf using the law of total probability and substitution. Then differentiate
to obtain the density.
(a) For Z = eX Y ,
FZ (z) = P(Z ≤ z) = P(eX Y ≤ z) =
=
Z ∞
−∞
Z ∞
−∞
P(eX Y ≤ z|X = x) fX (x) dx
P(Y ≤ ze−x |X = x) fX (x) dx =
Then
fZ (z) =
and so
Z ∞
−∞
fZ (z) =
Z ∞
−∞
FY |X (ze−x |x) fX (x) dx.
fY |X (ze−x |x)e−x fX (x) dx,
Z ∞
−∞
fXY (x, ze−x )e−x dx.
(b) Since Z = |X + Y | ≥ 0, we know that FZ (z) and fZ (z) are zero for z < 0. For
z ≥ 0, write
FZ (z) = P(Z ≤ z) = P(|X +Y | ≤ z) =
=
=
=
=
Then
Z ∞
−∞
Z ∞
−∞
Z ∞
−∞
Z ∞
−∞
P(|X +Y | ≤ z|X = x) fX (x) dx
P(|x +Y | ≤ z|X = x) fX (x) dx
P(−z ≤ x +Y ≤ z|X = x) fX (x) dx
P(−z − x ≤ Y ≤ z − x|X = x) fX (x) dx
Z ∞
−∞
FY |X (z − x|x) − FY |X (−z − x|x) fX (x) dx.
fZ (z) =
=
=
Z ∞
−∞
Z ∞
−∞
Z ∞
−∞
fY |X (z − x|x) − fY |X (−z − x|x)(−1) fX (x) dx
fY |X (z − x|x) + fY |X (−z − x|x) fX (x) dx
fXY (x, z − x) + fXY (x, −z − x) dx.
116
33.
Chapter 7 Problem Solutions
(a) First find the cdf of Z using the law of total probability, substitution, and independence. Then differentiate to obtain the density. Write
FZ (z) = P(Z ≤ z) = P(Y /X ≤ z) =
=
=
=
=
Z ∞
−∞
Z 0
−∞
Z 0
−∞
Z 0
−∞
Z ∞
−∞
P(Y /X ≤ z|X = x) fX (x) dx
P(Y /x ≤ z|X = x) fX (x) dx =
P(Y /x ≤ z) fX (x) dx +
P(Y ≥ zx) fX (x) dx +
Z ∞
Z
[1 − FY (zx)] fX (x) dx +
0
∞
Z ∞
−∞
P(Y /x ≤ z) fX (x) dx
P(Y /x ≤ z) fX (x) dx
P(Y ≤ zx) fX (x) dx
0
Z ∞
0
FY (zx) fX (x) dx.
Then
fZ (z) =
=
=
Z 0
−∞
Z 0
−∞
Z ∞
−∞
− fY (zx)x fX (x) dx +
fY (zx)|x| fX (x) dx +
Z ∞
0
Z ∞
0
fY (zx)x fX (x) dx
fY (zx)|x| fX (x) dx
fY (zx) fX (x)|x| dx.
(b) Using the result of part (a) and the fact that the integrand is even, write
Z
Z ∞ −|zx|2 /(2σ 2 ) −(x/σ )2 /2
e
e
2 ∞ x −(x/σ )2 [1+z2 ]/2
√
dx.
|x| dx =
e
2π 0 σ 2
−∞
2πσ
2πσ
√
Now make the change of variable θ = (x/σ ) 1 + z2 to get
fZ (z) =
fZ (z) =
√
1/π
1 + z2
Z ∞
0
θ e−θ
2 /2
dθ =
1/π −θ 2 /2 −e
1 + z2
∞
0
=
1/π
,
1 + z2
which is the Cauchy(1) density.
(c) Using the result of part (a) and the fact that the integrand is even, write
fZ (z) =
=
Z ∞
−∞
λ −λ |zx| λ −λ |x|
|x| dx
2e
2e
λ 2 /2
λ (|z| + 1)
Z ∞
0
=
λ2
2
Z ∞
0
xe−xλ (|z|+1) dx
x · λ (|z| + 1)e−λ (|z|+1)x dx,
where this last integral is the mean of an exponential density with parameter
λ (|z| + 1). Hence,
fZ (z) =
1
1
λ 2 /2
·
=
.
λ (|z| + 1) λ (|z| + 1)
2(|z| + 1)2
Chapter 7 Problem Solutions
117
(d) For z > 0, use the result of part (a) and the fact that the integrand is even, write
fZ (z) =
Z ∞
−∞
2
e−x /2
1
√
|x| dx
2 I[−1,1] (zx)
2π
2
1
= √ (−e−x /2 )
2π
Z 1/z −x2 /2
e
=
x√
0
dx
2π
2
1/z
0
1 − e−1/(2z )
√
.
2π
=
√
The same formula holds for z < 0, and it is easy to check that fZ (0) = 1/ 2π .
(e) Since Z ≥ 0, for z ≥ 0, we use the result from part (a) to write
Z ∞
fZ (z) =
0
= z
zx −(zx/λ )2 /2 x −(x/λ )2 /2
e
e
x dx
λ2
λ2
Z ∞
0
θ 3 e−θ
z
2
(z + 1)2
=
2 (z2 +1)/2
Z ∞
0
dθ = z
Z ∞
0
t 3 e−t
2 /2
dt =
= z
Z ∞
0
√
t
z2 + 1
2z
2
(z + 1)2
Z ∞
0
x 3 −(x/λ )2 (z2 +1)/2
λ e
3
e−t
2 /2
√
dx
λ
dt
z2 + 1
se−s ds.
This last integral is the mean of an exp(1) density, which is one. Hence fZ (z) =
2z/(z2 + 1)2 .
34. For the cdf, use the law of total probability, substitution, and independence to write
Z ∞
FZ (z) = P(Z ≤ z) = P(Y / ln X ≤ z) =
=
=
Z ∞
0
0
P(Y / ln x ≤ z|X = x) fX (x) dx =
0
P(Y ≥ z ln x) fX (x) dx +
Z 1
Z ∞
1
P(Y / ln X ≤ z|X = x) fX (x) dx
Z ∞
0
P(Y / ln x ≤ z) fX (x) dx
P(Y ≤ z ln x) fX (x) dx.
Then
fZ (z) =
=
Z 1
Z0 ∞
− fY (z ln x)(ln x) fX (x) dx +
0
Z ∞
1
fY (z ln x)(ln x) fX (x) dx
fX (x) fY (z ln x)| ln x| dx.
35. Use the law of total probability, substitution, and independence to write
E[e(X+Z)U ] =
=
=
Z 1/2
−1/2
Z 1/2
−1/2
1
1−u
E[e(X+Z)U |U = u] du =
E[e(X+Z)u ] du =
1/2
=
−1/2
Z 1/2
−1/2
Z 1/2
−1/2
E[e(X+Z)u |U = u] du
E[eXu ]E[eZu ] du =
Z 1/2
1
du
2
−1/2 (1 − u)
1
2
4
1
−
= 2− = .
1 − 1/2 1 + 1/2
3
3
118
Chapter 7 Problem Solutions
36. Use the law of total probability and substitution to write
E[X 2Y ] =
=
Z 2
1
E[X 2Y |Y = y] dy =
Z 2
1/y dy = 2 ln 2.
1
y · (2/y2 ) dy = 2
Z 2
E[X 2 y|Y = y] dy =
1
Z 2
Z 2
1
yE[X 2 |Y = y] dy
1
37. Use the law of total probability and substitution to write
E[X nY r ] =
Z ∞
0
Z ∞
E[X nY r |Y = y] fY (y) dy =
Z ∞
0
Z ∞
E[X n yr |Y = y] fY (y) dy
Γ(n + p)
fY (y) dy
yn Γ(p)
Z
Γ(n + p)
Γ(n + p) (r − n)!
Γ(n + p) ∞ r−n
y fY (y) dy =
E[Y r−n ] =
· r−n .
=
Γ(p)
Γ(p)
Γ(p)
λ
0
=
0
38.
yr E[X n |Y = y] fY (y) dy =
yr
0
(a) Use the law of total probability, substitution, and independence to find the cdf.
Write
FY (y) = P(Y ≤ y) = P(eVU ≤ y) =
=
Z ∞
0
=
Z ∞
0
vU
P(e
Z ∞
P(eVU ≤ y|V = v) fV (v) dv
0
≤ y|V = v) fV (v) dv =
Z ∞
0
P(evU ≤ y) fV (v) dv
P(U ≤ 1v ln y) fV (v) dv.
Then
fY (y) =
Z ∞
0
1
1
vy fU ( v
ln y) fV (v) dv.
To determine when fU ( 1v ln y) is nonzero, we consider the cases y > 1 and y < 1
separately. For y > 1, 1v ln y ≥ 0 for all v ≥ 0, and 1v ln y ≤ 1/2 for v ≥ 2 ln y.
Thus, fU ( 1v ln y) = 1 for v ≥ 2 ln y, and we can write
fY (y) =
Z ∞
2 ln y
1
yv
· ve−v dv =
e−2 ln y
1
= 3.
y
y
For y < 1, fU ( 1v ln y) = 1 for −1/2 ≤ 1v ln y, or v ≥ −2 ln y. Thus,
fY (y) =
Z ∞
−2 ln y
1
yv
· ve−v dv =
1 2 ln y
e
= y.
y
Putting this all together, we have

 1/y3 , y ≥ 1,
y, 0 ≤ y < 1,
fY (y) =

0, y < 0.
Chapter 7 Problem Solutions
119
(b) Using the density of part (a),
Z 1
E[Y ] =
y2 dy +
Z ∞
1
1
0
y2
1
4
+1 = .
3
3
dy =
(c) Using the law of total probability, substitution, and independence, we have
VU
E[e
] =
=
39.
Z 1/2
VU
E[e
−1/2
Z 1/2
−1/2
|U = u] du =
Z 1/2
Z 1/2
−1/2
E[eVu |U = u] du
1
1
du =
2
1−u
−1/2 (1 − u)
E[eVu ] du =
1/2
=
−1/2
4
.
3
(a) This problem is interesting because the answer does not depend on the random
variable X. Assuming X has a density fX (x), first write
E[cos(X +Y )] =
=
Z ∞
−∞
Z ∞
−∞
E[cos(X +Y )|X = x] fX (x) dx
E[cos(x +Y )|X = x] fX (x) dx.
Now use the conditional density of Y given X = x to write
E[cos(x +Y )|X = x] =
Z x+π
x−π
cos(x + y)
dy
=
2π
Z 2x+π
2x−π
cos θ
dθ
= 0,
2π
since we are integrating cos θ over an interval of length 2π . Thus, E[cos(X +
Y )] = 0 as well.
(b) Write
P(Y > y) =
=
Z ∞
−∞
Z 2
1
Z 2
P(Y > y|X = x) fX (x) dx =
e−xy dx =
P(Y > y|X = x) dx
1
e−y − e−2y
.
y
(c) Begin in the usual way by writing
E[XeY ] =
=
Z ∞
−∞
Z ∞
−∞
E[XeY |X = x] fX (x) dx =
Z ∞
−∞
E[xeY |X = x] fX (x) dx
xE[eY |X = x] fX (x) dx.
Now observe that
E[eY |X = x] = E[esY |X = x]
s=1
= es
2 x2 /2
s=1
= ex
2 /2
.
Then continue with
E[XeY ] =
Z ∞
−∞
xex
2 /2
fX (x) dx =
e49/2 − e9/2
=
.
4
1
4
Z 7
3
xex
2 /2
dx =
1 x2 /2 e
4
7
3
120
Chapter 7 Problem Solutions
(d) Write
E[cos(XY )] =
=
Z ∞
−∞
Z 2
E[cos(XY )|X = x] fX (x) dx =
E[Re(e jxY )|X = x] dx = Re
1
= Re
Z 2
1
e−x
2 (1/x)/2
Z 2
dx =
1
Z 2
1
Z 2
E[cos(xY )|X = x] dx
1
E[e jxY |X = x] dx
e−x/2 dx = 2(e−1/2 − e−1 ).
40. Using the law of total probability, substitution, and independence,
MY (s) = E[esY ] = E[esZX ] =
=
Z ∞
0
=
Z ∞
=
0
0
E[esZX |Z = z] fZ (z) dz
E[eszX |Z = z] fZ (z) dz =
e(sz)
2 σ 2 /2
0
Z ∞
Z ∞
ze−(1−s
Z ∞
fZ (z) dz =
2 σ 2 )z2 /2
Z ∞
E[eszX ] fZ (z) dz
0
e(sz)
2 σ 2 /2
0
ze−z
2 /2
dz
dz.
√
Now make the change of variable t = z 1 − s2 σ 2 to get
Z ∞
MY (s) =
0
2
dt
t
1/σ 2
1
√
=
.
e−t /2 √
=
2
2
1−s σ
1/σ 2 − s2
1 − s2 σ 2
1 − s2 σ 2
Hence, Y ∼ Laplace(1/σ ).
41. Using the law of total probability and substitution,
Z ∞
n m
E[X Y ] =
0
E[X Y |Y = y] fY (y) dy =
0
ym E[X n |Y = y] fY (y) dy =
Z ∞
=
Z ∞
n m
Z 0∞
0
E[X n ym |Y = y] fY (y) dy
ym 2n/2 yn Γ(1 + n/2) fY (y) dy
= 2n/2 Γ(1 + n/2)E[Y n+m ] = 2n/2 Γ(1 + n/2)
42.
(n + m)!
.
β n+m
(a) We use the law of total probability, substitution, and independence to write
Z ∞
FZ (z) = P(Z ≤ z) = P(X/Y ≤ z) =
=
=
Z ∞
Z0 ∞
0
0
P(X/y ≤ z|Y = y) fY (y) dy =
P(X ≤ zy) fY (y) dy =
Z ∞
0
P(X/Y ≤ z|Y = y) fY (y) dy
Z ∞
0
P(X ≤ zy|Y = y) fY (y) dy
FX (zy) fY (y) dy.
Differentiating, we have
fZ (z) =
Z ∞
0
fX (zy)y · fY (y) dy =
Z ∞
0
λ
(λ zy) p−1 e−λ zy
(λ y)q−1 e−λ y
·λ
· y dy.
Γ(p)
Γ(q)
Chapter 7 Problem Solutions
121
Making the change of variable w = λ y, we obtain
fZ (z) =
=
Z ∞
(zw) p−1 e−zw wq−1 e−w
·
Γ(p)
0
z p−1
Γ(p)Γ(q)
Z ∞
0
Γ(q)
· w dw
w p+q−1 e−w(1+z) dw.
Now make the change of variable θ = w(1 + z) so that d θ = (1 + z) dw and
w = θ /(1 + z). Then
p+q−1
Z ∞
z p−1
θ
dθ
fZ (z) =
e−θ
Γ(p)Γ(q) 0 1 + z
(1 + z)
=
=
z p−1
Γ(p)Γ(q)(1 + z) p+q
z p−1
B(p, q)(1 + z) p+q
.
Z ∞
|
0
θ p+q−1 e−θ d θ
{z
}
= Γ(p+q)
(b) Starting with V := Z/(1 + Z), we first write
Z
FV (v) = P
≤ v = P(Z ≤ v + vZ) = P(Z(1 − v) ≤ v)
1+Z
= P(Z ≤ v/(1 − v)).
Differentiating, we have
v 1
v (1 − v) + v
= fZ
.
fV (v) = fZ
2
1 − v (1 − v)
1 − v (1 − v)2
Now apply the formula derived in part (a) and use the fact that
1+
to get
fV (v) =
1
v
=
1−v
1−v
[v/(1 − v)] p−1
v p−1 (1 − v)q−1
1
=
,
·
1 p+q (1 − v)2
B(p, q)
B(p, q)( 1−v )
which is the beta density with parameters p and q.
43. Put q := (n − 1)p and Zi := ∑ j6=i X j , which is gamma(q, λ ) by Problem 55(c) in Chapter 4. Now observe that Yi = Xi /(Xi + Zi ), which has a beta density with parameters p
and q := (n − 1)p by Problem 42(b).
44. Using the law of total probability, substitution, and independence,
p
p
FZ (z) = P(Z ≤ z) = P(X/ Y /k ≤ z) = P(X ≤ z Y /k )
Z ∞
Z ∞
p
p
P(X ≤ z y/k |Y = y) fY (y) dy
=
P(X ≤ z Y /k |Y = y) fY (y) dy =
=
Z0 ∞
0
P(X ≤ z
p
0
y/k ) fY (y) dy.
122
Chapter 7 Problem Solutions
Then
fZ (z) =
Z ∞
0
p
p
fX (z y/k ) y/k fY (y) dy
Z ∞ −(z2 y/k)/2 p
e
1
(y/2)k/2−1 e−y/2
√
dy
y/k 2
Γ(k/2)
0
2π
Z ∞
2
1
= √ √
(y/2)k/2−1/2 e−y(1+z /k)/2 dy.
2 π k Γ(k/2) 0
=
Now make the change of variable θ = y(1 + z2 /k)/2, d θ = (1 + z2 /k)/2 dy to get
1
fZ (z) = √ √
π k Γ(k/2)
=
=
Z ∞
0
θ
1 + z2 /k
k/2−1/2
1
√
Γ(1/2) k Γ(k/2)(1 + z2 /k)k/2+1/2
e−θ
Z ∞
0
dθ
1 + z2 /k
θ k/2+1/2−1 e−θ d θ
(1 + z2 /k)−(k+1)/2
1
√
Γ(k/2 + 1/2) = √
,
Γ(1/2) k Γ(k/2)(1 + z2 /k)k/2+1/2
k B(1/2, k/2)
which is the required Student’s t density with k degrees of freedom.
45. We use the law of total probability, substitution, and independence to write
FZ (z) = P(Z ≤ z) = P(X/Y ≤ z) =
=
=
Z ∞
Z0 ∞
0
Z ∞
0
P(X/Y ≤ z|Y = y) fY (y) dy
P(X/y ≤ z|Y = y) fY (y) dy =
P(X ≤ zy) fY (y) dy =
Z ∞
0
Z ∞
0
P(X ≤ zy|Y = y) fY (y) dy
FX (zy) fY (y) dy.
Differentiating, we have
fZ (z) =
Z ∞
0
fX (zy)y · fY (y) dy =
Z ∞
r
λr
0
r
(λ zy) p−1 e−(λ zy)
(λ y)q−1 e−(λ y)
·λr
· y dy.
Γ(p/r)
Γ(q/r)
Making the change of variable w = λ y, we obtain
fZ (z) =
=
r
Z ∞
(zw) p−1 e−(zw)
r
0
Γ(p/r)
r2 z p−1
Γ(p/r)Γ(q/r)
Z ∞
0
r
·r
wq−1 e−w
· w dw
Γ(q/r)
r
r
w p+q−1 e−w (1+z ) dw.
Now make the change of variable θ = wr (1 + zr ) so that d θ = rwr−1 (1 + zr ) dw and
w = (θ /[1 + zr ])1/r . Then
fZ (z) =
r2 z p−1
Γ(p/r)Γ(q/r)
Z ∞
0
θ
1 + zr
(p+q−1)/r
e−θ
dθ
r(1 + zr )
(r−1)/r
θ
1+zr
Chapter 7 Problem Solutions
=
123
Z
∞
rz p−1
θ (p+q)/r−1 e−θ d θ
(p+q)/r
r
Γ(p/r)Γ(q/r)(1 + z )
0
|
{z
}
= Γ((p+q)/r)
=
rz p−1
B(p/r, q/r)(1 + zr )(p+q)/r
46. For 0 < z ≤ 1,
fZ (z) =
1
4
Z z
0
.
y−1/2 (z − y)−1/2 dy =
1
4z
Z z
0
p
1
(y/z)(1 − (y/z))
dy.
Now make the change of variable t 2 = y/z, 2t dt = dy/z to get
fZ (z) =
Next, for 1 < z ≤ 2,
Z 1
1
2
Z 1
0
√
1
1 − t2
dt =
1 −1
sin t
2
Z
1
0
= π /4.
√
1
1
1 1/ z
√
dt
dy =
√
2
1−1/z
z−1
(y/z)(1 − (y/z))
1 − t2
√
p
1 −1 √
1 −1 1/ z
=
= sin t √
sin (1/ z ) − sin−1 ( 1 − 1/z ) .
2
2
1−1/z
fZ (z) =
1
4z
p
Putting this all together yields

π /4, p
 0 < z ≤ 1,
√
−1
−1
1
fZ (z) =
sin
(1/
z
)
−
sin
(
1
−
1/z
)
, 1 < z ≤ 2,
2
0,
otherwise.
47. Let ψ denote the N(0, 1) density. Using
v − ρu
1
fUV (u, v) = ψρ (u, v) = ψ (u) · p
ψ p
,
1 − ρ2
1 − ρ2
{z
}
|
N(ρ u, 1−ρ 2 ) density in v
we see that
Z ∞
Z ∞
v − ρu
fU (u) =
fUV (u, v) dv =
ψ (u) · p
ψ p
2
−∞
−∞
1−ρ
1 − ρ2
Z ∞
1
v − ρu
p
ψ p
dv = ψ (u).
= ψ (u)
−∞
1 − ρ2
1 − ρ2
|
{z
}
1
density in v integrates to one
Similarly writing
u − ρv
fUV (u, v) = ψρ (u, v) = p
ψ p
2
1−ρ
1 − ρ2
1
· ψ (v),
dv
124
Chapter 7 Problem Solutions
we have
Z ∞
48. Using
Z ∞
u − ρv
p
fV (v) =
fUV (u, v) du =
ψ p
2
−∞
−∞
1−ρ
1 − ρ2
Z ∞
1
v − ρu
p
ψ p
= ψ (v)
du = ψ (v).
−∞
1 − ρ2
1 − ρ2
ψρ (u, v) = ψ (u) · p
we can write
1
· ψ (v) du
v − ρu
ψ p
,
1 − ρ2
1 − ρ2
1
1
x − mX y − mY
ψρ
,
σX σY
σX
σY
!
! y−m
x−mX
Y
1
1
x − mX
σY − ρ σX
p
ψ
ψ
=
·p
.
σX σY
σX
1 − ρ2
1 − ρ2
fXY (x, y) =
Then in
to get
R∞
−∞ f XY (x, y) dy,
fX (x) =
Z ∞
−∞
(∗)
make the change of variable v = (y − mY )/σY , dv = dy/σY
!
Z ∞
X
v − ρ x−m
1
1
x − mX
σX
p
fXY (x, y) dy =
ψ
ψ p
dv
σX
σX
−∞
1 − ρ2
1 − ρ2
|
{z
}
1
x − mX
=
ψ
.
σX
σX
density in v integrates to one
Thus, fX ∼ N(mX , σX2 ). Using this along with (∗), we obtain
1
fXY (x, y)
p
ψ
=
fY |X (y|x) =
fX (x)
σY 1 − ρ 2
y−mY
σY
p
X
− ρ x−m
σX
!
1 − ρ2
!
y − [mY + σσYX ρ (x − mX )]
1
p
p
=
ψ
.
σY 1 − ρ 2
σY 1 − ρ 2
Thus, fY |X (· |x) ∼ N mY + σσYX ρ (x − mX ), σY2 (1 − ρ 2 ) . Proceeding in an analogous
way, using
u − ρv
1
ψρ (u, v) = p
ψ p
· ψ (v),
1 − ρ2
1 − ρ2
we can write
1
x − mX y − mY
fXY (x, y) =
ψρ
,
σX σY
σX
σY
=
σX σY
1
p
1 − ρ2
ψ
x−mX
σX
p
Y
− ρ y−m
σY
1 − ρ2
!
y − mY
·ψ
.
σY
(∗∗)
Chapter 7 Problem Solutions
Then in
to get
R∞
−∞ f XY (x, y) dx,
fY (y) =
Z ∞
−∞
125
make the change of variable u = (x − mX )/σX , du = dx/σX
!
Z ∞
Y
u − ρ y−m
1
1
y − mY
σY
p
ψ
ψ p
du
fXY (x, y) dx =
σY
σY
−∞
1 − ρ2
1 − ρ2
|
{z
}
1
y − mY
=
.
ψ
σY
σY
density in u integrates to one
Thus, fY ∼ N(mY , σY2 ). Using this along with (∗∗), we obtain
1
fXY (x, y)
p
=
ψ
fX|Y (x|y) =
fY (y)
σX 1 − ρ 2
x−mX
σX
p
Y
− ρ y−m
σY
1 − ρ2
!
x − [mX + σσYX ρ (y − mY )]
1
p
p
ψ
.
=
σX 1 − ρ 2
σX 1 − ρ 2
Thus, fX|Y (· |y) ∼ N mX + σσYX ρ (y − mY ), σX2 (1 − ρ 2 ) .
!
49. From the solution of Problem 48, we have that
σY
fY |X (· |x) ∼ N mY +
ρ (x − mX ), σY2 (1 − ρ 2 )
σX
and
σX
fX|Y (· |y) ∼ N mX +
ρ (y − mY ), σX2 (1 − ρ 2 ) .
σY
Hence,
E[Y |X = x] =
Z ∞
y fY |X (y|x) dx = mY +
σY
ρ (x − mX ),
σX
E[X|Y = y] =
Z ∞
x fX|Y (x|y) dx = mX +
σX
ρ (y − mY ).
σY
and
−∞
−∞
50. From Problem 48, we know that fX ∼ N(mX , σX2 ). Hence, E[X] = mX and E[X 2 ] =
var(X) + m2X = σX2 + m2X . To compute cov(X,Y ), we use the law of total probability
and substitution to write
cov(X,Y ) = E[(X − mX )(Y − mY )] =
=
=
Z ∞
−∞
Z ∞
−∞
Z ∞
Z ∞
−∞
E[(X − mX )(Y − mY )|Y = y] fY (y) dy
E[(X − mX )(y − mY )|Y = y] fY (y) dy
(y − mY )E[(X − mX )|Y = y] fY (y) dy
(y − mY ) E[X|Y = y] − mX fY (y) dy
−∞
Z ∞
nσ
o
X
=
(y − mY )
ρ (y − mY ) fY (y) dy
σY
−∞
=
126
Chapter 7 Problem Solutions
Z
∞
σX
σX
ρ
ρ · E[(Y − mY )2 ]
(y − mY )2 fY (y) dy =
σY
σY
−∞
σX
ρ · σY2 = σX σY ρ .
=
σY
=
It then follows that
51.
cov(X,Y )
= ρ.
σX σY
(a) Using the results of Problem 47, we have
fU (u) =
=
Z ∞
fUV (u, v) dv =
Z ∞
fUV (u, v) du =
−∞
1
2
Z ∞
ψρ1 (u, v) + ψρ2 (u, v) dv
Z ∞
ψρ1 (u, v) + ψρ2 (u, v) du
−∞
1
[ψ (u) + ψ (u)] = ψ (u),
2
and
fV (v) =
=
−∞
1
2
−∞
1
[ψ (v) + ψ (v)] = ψ (v).
2
Thus, fU and fV are N(0, 1) densities.
(b) Write
ρ := E[UV ] =
=
Z ∞Z ∞
−∞
Z ∞Z ∞
−∞ −∞
uv fUV (u, v) du dv
1
uv · [ψρ1 (u, v) + ψρ2 (u, v)] du dv
2
−∞
Z ∞ Z ∞
Z ∞Z ∞
1
=
uvψρ1 (u, v) du dv +
uvψρ2 (u, v) du dv
2 −∞ −∞
−∞ −∞
ρ1 + ρ 2
.
=
2
(c) If indeed
fUV (u, v) =
1
[ψρ (u, v) + ψρ2 (u, v)]
2 1
is a bivariate normal density, then
exp
fUV (u, v) =
In particular then,
−1
[u2 − 2ρ uv + v2 ]
2(1−ρ 2 )
2π
p
1−ρ
2
.
1
[ψρ (u, u) + ψρ2 (u, u)],
2 1
#
"
2
2
1 e−u /(1+ρ1 )
e−u /(1+ρ2 )
q
=
+ q
.
2 2π 1 − ρ 2 2π 1 − ρ 2
fUV (u, u) =
or
2
e−u /(1+ρ )
p
2π 1 − ρ 2
1
2
Chapter 7 Problem Solutions
127
Since t := u2 ≥ 0 is arbitrary, part (iii) of the hint tells us that
2π
which is false.
p
1
1−ρ
=
2
−1
−1
q
q
=
= 0,
2
4π 1 − ρ 1
4π 1 − ρ 22
(d) First observe that
fV |U (v|u) =
fUV (u, v)
1
=
fU (u)
2
ψρ1 (u, v)
ψρ2 (u, v)
+
.
ψ (u)
ψ (u)
| {z } | {z }
N(ρ1 u,1−ρ12 )
Hence,
Z ∞
−∞
N(ρ2 u,1−ρ22 )
1
[(1 − ρ12 ) + (ρ1 u)2 + (1 − ρ22 ) + (ρ2 u)2 ]
2
1
= [2 − ρ12 − ρ22 + (ρ12 + ρ22 )u2 ],
2
v2 fV |U (v|u) dv =
which depends on u unless ρ1 = ρ2 = 0.
52. For u0 , v0 ≥ 0, let D := {(u, v) : u ≥ u0 , v ≥ v0 }. Then
P(U > u0 ,V > v0 ) =
ZZ
ψρ (u, v) du dv
D
=
Z tan−1 (v0 /u0 ) Z ∞
v0 / sin θ
0
+
Z π /2
Z ∞
ψρ (r cos θ , r sin θ )r dr d θ
tan−1 (v0 /u0 ) u0 / cos θ
Now, since
exp
ψρ (r cos θ , r sin θ ) =
ψρ (r cos θ , r sin θ )r dr d θ . (##)
−r2
[1 − ρ sin 2θ ]
2(1−ρ 2 )
2π
(#)
p
1 − ρ2
,
we can express the anti-derivative of rψρ (r cos θ , r sin θ ) with respect to r in closed
form as
p
− 1 − ρ2
−r2
[1 − ρ sin 2θ ] .
exp
2π (1 − ρ sin 2θ )
2(1 − ρ 2 )
Hence, the double integral in (#) reduces to
Z tan−1 (v0 /u0 )
0
h(v20 , θ ) d θ .
The double integral in (##) reduces to
p
Z π /2
−u20
1 − ρ2
exp
[1
−
ρ
sin
2
θ
]
dθ .
2(1 − ρ 2 ) cos2 θ
tan−1 (v0 /u0 ) 2π (1 − ρ sin 2θ )
128
Chapter 7 Problem Solutions
Applying the change of variable t = π /2 − θ , dt = −d θ , we obtain
Z π /2−tan−1 (v0 /u0 )
0
hρ (u20 ,t) dt.
53. If ρ = 0 in Problem 52, then U and V are independent. Taking u0 = v0 = x0 ,
Q(x0 )2 =
Z π /4
0
= 2
h0 (x02 , θ ) d θ +
Z π /4
0
2
h0 (x02 , θ ) d θ
Z π /4
exp[−x02 /(2 sin θ )]
2π
0
54. Factor
fXY Z (x, y, z) =
1
dθ =
π
Z π /4
0
−x02
exp
dθ .
2 sin2 θ
2 exp[−|x − y| − (y − z)2 /2]
√
,
z5 2π
as
z ≥ 1,
2
4 e−(y−z) /2 1 −|x−y|
.
· √
· 2e
z5
2π
Now, the second factor on the right is an N(z, 1) density in the variable y, and the
third factor is a Laplace(1) density that has been shifted to have mean y. Hence, the
integral of the third factor with respect to x yields one, and we have by inspection that
fXY Z (x, y, z) =
2
fY Z (y, z) =
4 e−(y−z) /2
· √
,
z5
2π
z ≥ 1.
We then easily see that
fX|Y Z (x|y, z) :=
fXY Z (x, y, z)
=
fY Z (y, z)
1 −|x−y|
,
2e
z ≥ 1.
Next, since the right-hand factor in the formula for fY Z (y, z) is an N(z, 1) density in y,
if we integrate this factor with respect to y, we get one. Thus,
4
fZ (z) = 5 , z ≥ 1.
z
We can now see that
2
fY Z (y, z)
e−(y−z) /2
√
=
,
fZ (z)
2π
fY |Z (y|z) :=
z ≥ 1.
55. To find fXY (x, y), first write
fXY Z (x, y, z) :=
=
=
=
e−(x−y)
2 /2
e−(x−y)
2 /2
√ 2
2 ) /2
e−(x−y)
2 /2
√ 2
2 ) /2
e−(x−y)
2 /2
√ 2
2 ) /2
e−(y−z)
(2π )3/2
e−(y/
2π
e−(y/
√
2π 2
e−(y/
√
2π 2
2 /2
e−z
2 /2
=
e−(x−y)
e−(z−y/2)
· √
2π
2 /2
2
e−(z−y/2) e−y
(2π )3/2
2
2
e−(z−y/2)
√
·√
2π / 2
√
e−[(z−y/2)/(1/ 2 )]
√
√
·
2π / 2
2 /2
.
2 /4
Chapter 7 Problem Solutions
129
Now the right-hand factor is an N(y/2, 1/2) density in the variable z. Hence, its
integral with respect to z is one. We thus have
e−(x−y)
fXY (x, y) =
2 /2
e−(y/
√
2π 2
√ 2
2 ) /2
√
2
2
e−(y/ 2 ) /2 e−(x−y) /2
= √ √ · √
,
2π
2π 2
which shows that Y ∼ N(0, 2), and given Y = y, X is conditionally N(y, 1). Thus,
E[Y ] = 0 and
var(Y ) = 2.
Next,
E[X] =
Z ∞
−∞
Z ∞
E[X|Y = y] fY (y) dy =
−∞
y fY (y) dy = E[Y ] = 0,
and
var(X) = E[X 2 ] =
Z ∞
−∞
E[X 2 |Y = y] fY (y) dy =
Z ∞
= 1 + E[Y 2 ] = 1 + var(Y ) = 1 + 2 = 3.
−∞
(1 + y2 ) · fY (y) dy
Finally,
E[XY ] =
=
Z ∞
−∞
Z ∞
−∞
E[XY |Y = y] fY (y) dy =
yE[X|Y = y] fY (y) dy =
Z ∞
−∞
Z ∞
−∞
E[Xy|Y = y] fY (y) dy
y2 fY (y) dy = E[Y 2 ] = var(Y ) = 2.
56. First write
E[XY ] =
=
=
Z ∞Z ∞
−∞ −∞
Z ∞Z ∞
−∞ −∞
Z ∞Z ∞
−∞ −∞
E[XY |Y = y, Z = z] fY Z (y, z) dy dz
E[Xy|Y = y, Z = z] fY Z (y, z) dy dz
yE[X|Y = y, Z = z] fY Z (y, z) dy dz.
Since fX|Y Z (· |y, z) ∼ N(y, z2 ), the preceding conditional expectation is just y. Hence,
E[XY ] =
Z ∞Z ∞
−∞ −∞
y2 fY Z (y, z) dy dz = E[Y 2 ] =
Z ∞
−∞
E[Y 2 |Z = z] fZ (z) dz.
Since fY |Z (· |z) ∼ exp(z), the preceding conditional expectation is just 2/z2 . Thus,
E[XY ] =
Z ∞
2
−∞
z2
fZ (z) dz =
Z 2
2
1
3
6
· z2 dz = .
z2 7
7
A similar analysis yields
E[Y Z] =
=
Z ∞
−∞
Z ∞
−∞
E[Y Z|Z = z] fZ (z) dz =
zE[Y |Z = z] fZ (z) dz =
Z ∞
−∞
Z ∞
−∞
E[Y z|Z = z] fZ (z) dz
z(1/z) fZ (z) dz =
Z ∞
−∞
fZ (z) dz = 1.
130
Chapter 7 Problem Solutions
57. Write
E[XY Z] =
=
=
Z ∞Z ∞
−∞ −∞
Z ∞Z ∞
−∞ −∞
Z ∞Z ∞
−∞ −∞
E[XY Z|Y = y, Z = z] fY Z (y, z) dy dz
E[Xyz|Y = y, Z = z] fY Z (y, z) dy dz
yzE[X|Y = y, Z = z] fY Z (y, z) dy dz.
Since fX|Y Z (· |y, z) is a shifted Laplace density with mean y, the preceding conditional
expectation is just y. Hence,
Z ∞Z ∞
E[XY Z] =
2
2
y z fY Z (y, z) dy dz = E[Y Z] =
−∞ −∞
Z ∞
2
=
−∞
E[Y z|Z = z] fZ (z) dy =
Z ∞
−∞
Z ∞
−∞
E[Y 2 Z|Z = z] fZ (z) dy
zE[Y 2 |Z = z] fZ (z) dy.
Since fY |Z (· |z) ∼ N(z, 1), the preceding conditional expectation is just 1 + z2 . Thus,
E[XY Z] =
Z ∞
−∞
z(1 + z2 ) fZ (z) dz =
= 4(1/3 + 1) = 16/3.
Z ∞
1
[z + z3 ] · 4/z5 dz = 4
Z ∞
1
z−4 + z−2 dz
58. Write
E[XY Z] =
=
=
=
=
=
Z ∞Z ∞
−∞ −∞
Z ∞Z ∞
−∞ −∞
Z ∞Z ∞
−∞ −∞
Z ∞Z ∞
−∞ −∞
Z ∞
2
−∞
Z ∞
−∞
E[XY Z|X = x,Y = y] fXY (x, y) dx dy
E[xyZ|X = x,Y = y] fXY (x, y) dx dy
xyE[Z|X = x,Y = y] fXY (x, y) dx dy
x2 y fXY (x, y) dx dy = E[X 2Y ] =
E[x Y |X = x] fX (x) dx =
x3 fX (x) dx =
Z 2
Z ∞
−∞
Z ∞
−∞
E[X 2Y |X = x] fX (x) dx
x2 E[Y |X = x] fX (x) dx
x3 dx = 15/4.
1
59. We use the law of total probability, substitution, and independence to write
N
ϕY (ν ) = E[e jνY ] = E[e jν ∑i=1 Xi ] =
∞
n=1
∞
=
∑ E[e jν ∑i=1 Xi |N = n]P(N = n)
n=1
∞
=
n
n
∑ E[e jν ∑i=1 Xi ]P(N = n)
n=1
N
∑ E[e jν ∑i=1 Xi |N = n]P(N = n)
Chapter 7 Problem Solutions
∞
=
=
n
∑ E ∏e
i=1
n=1
∞ n
∑ ∏ E[e
n=1 i=1
∞
=
jν Xi
131
P(N = n)
jν Xi
] P(N = n)
∑ ϕX (ν )n P(N = n)
= GN (ϕX (ν )).
n=1
Now, if N ∼ geometric1 (p), GN (z) = [(1 − p)z]/[1 − pz], and if X ∼ exp(λ ), ϕX (ν ) =
λ /(λ − jν ). Then
ϕY (ν ) =
(1 − p)λ
(1 − p)ϕX (ν ) (1 − p)λ /(λ − jν )
(1 − p)λ
=
=
=
,
1 − pϕX (ν )
1 − pλ /(λ − jν )
(λ − jν ) − pλ
(1 − p)λ − jν
which is the exp((1 − p)λ ) characteristic function. Thus, Y ∼ exp((1 − p)λ ).
CHAPTER 8
Problem Solutions
1. We have



230 280 330
10 40  20 50  7 8 9 =  340 410 480 
4 5 6
450 540 630
30 60

and

230 280 330
tr  340 410 480  = 1270.
450 540 630

2. MATLAB. See the answer to the previous problem.
3. MATLAB. We have


7 4
A0 =  8 5  .
9 6
4. Write
r
r
tr(AB) = ∑ (AB)ii = ∑
i=1
i=1
5.
n
∑ Aik Bki
k=1
n
=
r
∑ ∑ Bki Aik
k=1
i=1
n
=
∑ (BA)kk = tr(BA).
k=1
(a) Write
0
tr(AB ) =
r
0
∑ (AB )ii
i=1
r
=
n
0
∑ ∑ Aik (B )ki
i=1
k=1
r
=
n
∑ ∑ Aik Bik .
i=1 k=1
(b) If tr(AB0 ) = 0 for all B, then in particular, it is true for B = A; i.e.,
0 = tr(AA0 ) =
r
n
∑ ∑ A2ik ,
i=1 k=1
which implies Aik = 0 for all i and k. In other words, A is the zero matrix of size
r × n.
6. Following the hint, we first write
0 ≤ kx − λ yk2 = hx − λ y, x − λ yi = kxk2 − 2λ hx, yi + λ 2 kyk2 .
Taking λ = hx, yi/kyk2 yields
0 ≤ kxk2 − 2
|hx, yi|2
|hx, yi|2 |hx, yi|2
+
kyk2 = kxk2 −
,
2
4
kyk
kyk
kyk2
132
Chapter 8 Problem Solutions
133
which can be rearranged to get |hx, yi|2 ≤ kxk2 kyk2 . Conversely, suppose |hx, yi|2 =
kxk2 kyk2 . There are two cases to consider. If y 6= 0, then reversing the above sequence
of observations implies 0 = kx − λ yk2 , which implies x = λ y. On the other hand, if
y = 0 and if
|hx, yi| = kxk kyk,
then we must have kxk = 0; i.e., x = 0 and y = 0, and in this case x = λ y for all λ .
7. Consider the i j component of E[XB]. Since
(E[XB])i j = E[(XB)i j ] = E ∑ Xik Bk j =
∑ E[Xik ]Bk j
=
k
k
∑(E[X])ik Bk j
k
= (E[X]B)i j
holds for all i j, E[XB] = E[X]B.
8. tr(E[X]) =
∑(E[X])ii =
i
∑ E[Xii ] = E
i
∑ Xii
i
= E[tr(X)].
9. Write
E[kX − E[X]k2 ] = E[(X − E[X])0 (X − E[X])], which is a scalar,
n
o
= tr E[(X − E[X])0 (X − E[X])]
h n
oi
= E tr (X − E[X])0 (X − E[X]) , by Problem 8,
oi
h n
= E tr (X − E[X])(X − E[X])0 , by Problem 4,
n
o
= tr E[(X − E[X])(X − E[X])0 ] , by Problem 8,
n
= tr(C) =
∑ Cii
n
∑ var(Xi ).
=
i=1
i=1
0
10. Since E [X,Y, Z]0 = E[X], E[Y ], E[Z] , and

 

E[X]2 E[X]E[Y ] E[X]E[Z]
E[X 2 ] E[XY ] E[XZ]
cov([X,Y, Z]0 ) =  E[Y X] E[Y 2 ] E[Y Z]  −  E[Y ]E[X] E[Y ]2 E[Y ]E[Z]  ,
E[Z]E[X] E[Z]E[Y ] E[Z]2
E[ZX] E[ZY ] E[Z 2 ]
we begin by computing all the entries of these matrices. For the mean vector,
E[Z] =
Z 2
1
z·
3 2
7 z dz
=
3
7
Z 2
z3 dz =
1
3
7
· 41 z4
2
1
= 3 · 15/28 = 45/28.
Next,
E[Y ] =
Z 2
1
E[Y |Z = z] fZ (z) dz =
Z 2
1
1
z
·
3 2
7 z dz
=
3
7
Z 2
1
z dz =
3
7
· 12 z2
Since E[U] = 0 and since U and Z are independent,
E[X] = E[ZU +Y ] = E[Z]E[U] + E[Y ] = E[Y ] = 9/14.
2
1
= 9/14.
134
Chapter 8 Problem Solutions
Thus, the desired mean vector is
E [X,Y, Z]0 = [9/14, 9/14, 45/28]0 .
We next compute the correlations. First,
E[Y Z] =
Z 2
1
=
Z 2
1
E[Y Z|Z = z] fZ (z) dz =
zE[Y |Z = z] fZ (z) dz =
Next,
Z 2
E[Z 2 ] =
1
z2 · 37 z2 dz =
3
7
Z 2
1
E[Y z|Z = z] fZ (z) dz
Z 2
z(1/z) fZ (z) dz =
Z 2
3
7
1
z4 dz =
1
· 51 z5
2
1
Z 2
1
fZ (z) dz = 1.
= 93/35.
Again using the fact that E[U] = 0 and independence,
E[XZ] = E[(ZU +Y )Z] = E[Z 2 ]E[U] + E[Y Z] = E[Y Z] = 1.
Now,
E[Y 2 ] =
Z 2
1
E[Y 2 |Z = z] fZ (z) dz =
We can now compute
Z 2
1
(2/z2 ) · 73 z2 dz = 6/7.
E[XY ] = E[(ZU +Y )Y ] = E[ZY ]E[U] + E[Y 2 ] = 6/7,
and
E[X 2 ] = E[(ZU +Y )2 ] = E[Z 2 ]E[U 2 ] + 2E[U]E[ZY ] + E[Y 2 ]
= E[Z 2 ] + E[Y 2 ] = 93/35 + 6/7 = 123/35.
We now have that

 
81/196 81/196 405/392
123/35 6/7 1
cov([X,Y, Z]0 ) =  6/7 6/7 1  −  81/196 81/196 405/392 
405/392 405/392 2025/784
1
1 93/35


3.1010 0.4439 −0.0332
=  0.4439 0.4439 −0.0332  .
−0.0332 −0.0332 0.0742

0
11. Since E [X,Y, Z]0 = E[X], E[Y ], E[Z] , and

 

E[X 2 ] E[XY ] E[XZ]
E[X]2 E[X]E[Y ] E[X]E[Z]
cov([X,Y, Z]0 ) =  E[Y X] E[Y 2 ] E[Y Z]  −  E[Y ]E[X] E[Y ]2 E[Y ]E[Z]  ,
E[ZX] E[ZY ] E[Z 2 ]
E[Z]E[X] E[Z]E[Y ] E[Z]2
we compute all the entries of these matrices. To make this job easier, we first factor
fXY Z (x, y, z) =
2 exp[−|x − y| − (y − z)2 /2]
√
,
z5 2π
z ≥ 1,
Chapter 8 Problem Solutions
135
as fX|Y Z (x|y, z) fY |Z (y|z) fZ (z) by writing
2
e−(y−z) /2 4
√
fXY Z (x, y, z) =
· 5 , z ≥ 1.
z
2π
We then see that as a function of x, fX|Y Z (x|y, z) is a shifted Laplace(1) density. Similarly, as a function of y, fY |Z (y|z) is an N(z, 1) density. Thus,
1 −|x−y|
·
2e
E[X] =
Z ∞Z ∞
−∞ −∞
= E[Y ] =
Z ∞
E[X|Y = y, Z = z] fY Z (y, z) dy dz =
Z ∞
Z ∞
E[Y |Z = z] fZ (z) dz =
−∞
−∞
Z
Z ∞Z ∞
−∞ −∞
y fY Z (y, z) dy dz
z fZ (z) dz = E[Z]
∞ 4
4
z · 5 dz =
dz = 4/3.
4
z
1
1 z
Thus, E [X,Y, Z]0 = [4/3, 4/3, 4/3]0 . We next compute
=
E[X 2 ] =
=
where
E[Y 2 ] =
Now,
Z ∞
−∞
Z ∞Z ∞
−∞ −∞
Z ∞Z ∞
−∞ −∞
E[X 2 |Y = y, Z = z] fY Z (y, z) dy dz
(2 + y2 ) fY Z (y, z) dy dz = 2 + E[Y 2 ],
E[Y 2 |Z = z] fZ (z) dz =
Z ∞
Z ∞
−∞
Z ∞
4
4
dz =
z5
1
Thus, E[Y 2 ] = 3 and E[X 2 ] = 5. We next turn to
E[Z 2 ] =
E[XY ] =
=
=
=
We also have
E[XZ] =
=
=
=
=
Z ∞Z ∞
−∞ −∞
Z ∞Z ∞
−∞ −∞
Z ∞Z ∞
−∞ −∞
Z ∞Z ∞
−∞ −∞
Z ∞Z ∞
−∞ −∞
Z ∞Z ∞
−∞ −∞
Z ∞Z ∞
−∞ −∞
Z ∞
−∞
Z ∞
−∞
z2
(1 + z2 ) fZ (z) dz = 1 + E[Z 2 ].
z3
1
dz = 2.
E[XY |Y = y, Z = z] fY Z (y, z) dy dz
E[Xy|Y = y, Z = z] fY Z (y, z) dy dz
yE[X|Y = y, Z = z] fY Z (y, z) dy dz
y2 fY Z (y, z) dy dz = E[Y 2 ] = 3.
E[XZ|Y = y, Z = z] fY Z (y, z) dy dz
E[Xz|Y = y, Z = z] fY Z (y, z) dy dz
yz fY Z (y, z) dy dz = E[Y Z]
E[Y Z|Z = z] fZ (z) dz =
Z ∞
−∞
z2 fZ (z) dz = E[Z 2 ] = 2.
zE[Y |Z = z] fZ (z) dz
136
Chapter 8 Problem Solutions
We now have that



5 3 2
1 1 1
16
cov([X,Y, Z]0 ) =  3 3 2  −  1 1 1 
9
2 2 2
1 1 1




3.2222 1.2222 0.2222
29 11 2
1
11 11 2  =  1.2222 1.2222 0.2222  .
=
9
0.2222 0.2222 0.2222
2 2 2

0
12. Since E [X,Y, Z]0 = E[X], E[Y ], E[Z] , and


 
E[X]2 E[X]E[Y ] E[X]E[Z]
E[X 2 ] E[XY ] E[XZ]
cov([X,Y, Z]0 ) =  E[Y X] E[Y 2 ] E[Y Z]  −  E[Y ]E[X] E[Y ]2 E[Y ]E[Z]  ,
E[Z]E[X] E[Z]E[Y ] E[Z]2
E[ZX] E[ZY ] E[Z 2 ]
we compute all the entries of these matrices. We begin with
E[Z] =
=
Z ∞Z ∞
−∞ −∞
Z ∞Z ∞
−∞ −∞
E[Z|Y = y, X = x] fXY (x, y) dy dx
x fXY (x, y) dy dx = E[X] = 3/2.
Next,
E[Y ] =
Z ∞
−∞
E[Y |X = x] fX (x) dx =
Z ∞
−∞
x fX (x) dx = E[X] = 3/2.
We now compute
E[XY ] =
=
Z ∞
−∞
Z ∞
−∞
E[XY |X = x] fX (x) dx =
xE[Y |X = x] fX (x) dx =
Z ∞
−∞
Z ∞
E[xY |X = x] fX (x) dx
x2 fX (x) dx = E[X 2 ]
−∞
2
= var(X) + E[X]2 = 1/12 + (3/2) = 7/3.
Then
E[XZ] =
=
Z ∞Z ∞
−∞ −∞
Z ∞Z ∞
−∞ −∞
E[XZ|Y = y, X = x] fXY (x, y) dy dx
xE[Z|Y = y, X = x] fXY (x, y) dy dx
=
Z ∞Z ∞
x2 fXY (x, y) dy dx = E[X 2 ] = 7/3,
E[Y Z] =
Z ∞Z ∞
E[Y Z|Y = y, X = x] fXY (x, y) dy dx
and
=
=
−∞ −∞
−∞ −∞
Z ∞Z ∞
−∞ −∞
Z ∞Z ∞
−∞ −∞
yE[Z|Y = y, X = x] fXY (x, y) dy dx
xy fXY (x, y) dy dx = E[XY ] = 7/3.
Chapter 8 Problem Solutions
137
Next,
E[Y 2 ] =
Z ∞
E[Z 2 ] =
Z ∞Z ∞
and
=
−∞
E[Y 2 |X = x] fX (x) dx =
−∞ −∞
Z ∞Z ∞
−∞ −∞
Z ∞
−∞
2x2 fX (x) dx = 2E[X 2 ] = 14/3,
E[Z 2 |Y = y, X = x] fXY (x, y) dy dx
(1 + x2 ) fXY (x, y) dy dx = 1 + E[X 2 ] = 1 + 7/3 = 10/3.
We now have that



7 7 7
3 2 1 1
1
1 1
cov([X,Y, Z]0 ) =  7 14 7  −
3
2
1 1
7 7 10



0.0833
1 1 1
1 
1 29 1  =  0.0833
=
12
0.0833
1 1 13

1
1
1

0.0833 0.0833
2.4167 0.0833  .
0.0833 1.0833
0
13. Since E [X,Y, Z]0 = E[X], E[Y ], E[Z] , and


 
E[X]2 E[X]E[Y ] E[X]E[Z]
E[X 2 ] E[XY ] E[XZ]
cov([X,Y, Z]0 ) =  E[Y X] E[Y 2 ] E[Y Z]  −  E[Y ]E[X] E[Y ]2 E[Y ]E[Z]  ,
E[Z]E[X] E[Z]E[Y ] E[Z]2
E[ZX] E[ZY ] E[Z 2 ]
we compute all the entries of these matrices. In order to do this, we first note that
Z ∼ N(0, 1). Next, as a function of y, fY |Z (y|z) is an N(z, 1) density. Similarly, as a
function of x, fX|Y Z (x|y, z) is an N(y, 1) density. Hence, E[Z] = 0,
E[Y ] =
Z ∞
−∞
E[Y |Z = z] fZ (z) dz =
and
E[X] =
=
Z ∞Z ∞
−∞ −∞
Z ∞Z ∞
−∞ −∞
Z ∞
−∞
z fZ (z) dz = E[Z] = 0,
E[X|Y = y, Z = z] fY Z (y, z) dy dz
y fY Z (y, z) dy dz = E[Y ] = 0.
We next compute
E[XY ] =
=
=
Z ∞Z ∞
−∞ −∞
Z ∞Z ∞
−∞ −∞
Z ∞Z ∞
−∞ −∞
= E[Y 2 ] =
E[XY |Y = y, Z = z] fY Z (y, z) dy dz
E[Xy|Y = y, Z = z] fY Z (y, z) dy dz
yE[X|Y = y, Z = z] fY Z (y, z) dy dz =
Z ∞
−∞
E[Y 2 |Z = z] fZ (z) dz =
= 1 + E[Z 2 ] = 2.
Z ∞
−∞
Z ∞Z ∞
−∞ −∞
y2 fY Z (y, z) dy dz
(1 + z2 ) fZ (z) dz
138
Chapter 8 Problem Solutions
Then
E[Y Z] =
=
Z ∞
−∞
Z ∞
−∞
E[Y Z|Z = z] fZ (z) dz =
Z ∞
−∞
zE[Y |Z = z] fZ (z) dz
z2 fZ (z) dz = E[Z 2 ] = 1,
and
E[XZ] =
=
=
Z ∞Z ∞
−∞ −∞
Z ∞Z ∞
−∞ −∞
Z ∞Z ∞
−∞ −∞
E[XZ|Y = y, Z = z] fY Z (y, z) dy dz
zE[X|Y = y, Z = z] fY Z (y, z) dy dz
yz fY Z (y, z) dy dz = E[Y Z] = 1.
Next,
E[Y 2 ] =
Z ∞
−∞
E[Y 2 |Z = z] fZ (z) dz =
Z ∞
−∞
(1 + z2 ) fZ (z) dz = 1 + E[Z 2 ] = 2,
and
E[X 2 ] =
=
Z ∞Z ∞
−∞ −∞
Z ∞Z ∞
−∞ −∞
E[X 2 |Y = y, Z = z] fY Z (y, z) dy dz
(1 + y2 ) fY Z (y, z) dy dz = 1 + E[Y 2 ] = 3.
Since E[Z 2 ] = 1, we now have that


3 2 1
cov([X,Y, Z]0 ) =  2 2 1  .
1 1 1
14. We first note that Z ∼ N(0, 1). Next, as a function of y, fY |Z (y|z) is an N(z, 1) density.
Similarly, as a function of x, fX|Y Z (x|y, z) is an N(y, 1) density. Hence,
E[e j(ν1 X+ν2Y +ν3 Z) ] =
=
=
=
Z ∞Z ∞
−∞ −∞
Z ∞Z ∞
−∞ −∞
Z ∞Z ∞
−∞ −∞
Z ∞Z ∞
−∞ −∞
2
= e−ν1 /2
2
E[e j(ν1 X+ν2Y +ν3 Z) |Y = y, Z = z] fY Z (y, z) dy dz
E[e j(ν1 X+ν2 y+ν3 z) |Y = y, Z = z] fY Z (y, z) dy dz
E[e jν1 X |Y = y, Z = z]e jν2 y e jν3 z fY Z (y, z) dy dz
2
e jν1 y−ν1 /2 e jν2 y e jν3 z fY Z (y, z) dy dz
Z ∞Z ∞
−∞ −∞
e j(ν1 +ν2 )y e jν3 z fY Z (y, z) dy dz
= e−ν1 /2 E[e j(ν1 +ν2 )Y e jν3 Z ]
Chapter 8 Problem Solutions
2
= e−ν1 /2
2
= e−ν1 /2
2
= e−ν1 /2
2
Z ∞
−∞
Z ∞
−∞
Z ∞
139
E[e j(ν1 +ν2 )Y e jν3 Z |Z = z] fZ (z) dz
e jν3 z E[e j(ν1 +ν2 )Y |Z = z] fZ (z) dz
e jν3 z e j(ν1 +ν2 )z−(ν1 +ν2 )
−∞
= e−ν1 /2 e−(ν1 +ν2 )
2 /2
2
2 /2
2
2
= e−ν1 /2 e−(ν1 +ν2 )
Z ∞
−∞
2 /2
fZ (z) dz
e j(ν1 +ν2 +ν3 )z fZ (z) dz
E[e j(ν1 +ν2 +ν3 )z ]
= e−ν1 /2 e−(ν1 +ν2 ) /2 e−(ν1 +ν2 +ν3 )
2
2
2
= e−[ν1 +(ν1 +ν2 ) +(ν1 +ν2 +ν3 ) ]/2 .
2 /2
15. RY := E[YY 0 ] = E[(AX)(AX)0 ] = E[AXX 0 A0 ] = AE[XX 0 ]A0 = ARX A0 .
16. First, since R = E[XX 0 ], we see that R0 = E[XX 0 ]0 = E[(XX 0 )0 ] = E[XX 0 ] = R. Thus,
R is symmetric. Next, define the scalar Y := c0 X. Then
0 ≤ E[Y 2 ] = E[YY 0 ] = E[(c0 X)(c0 X)0 ] = E[c0 XX 0 c] = c0 E[XX 0 ]c = c0 Rc,
and we see that R is positive semidefinite.
17. Use the Cauchy–Schwarz inequality to write
(CXY )i j = E[(Xi − mX,i )(Y j − mY, j )]
q
q
≤ E[(Xi − mX,i )2 ]E[(Y j − mY, j )2 ] = (CX )ii (CY ) j j .
18. If
P =
then
U
V
:= P0
X
Y
=
cos θ − sin θ
,
sin θ cos θ
cos θ sin θ
− sin θ cos θ
X
Y
=
X cos θ +Y sin θ
.
−X sin θ +Y cos θ
We now have to find θ such that E[UV ] = 0. Write
E[UV ] = E[(X cos θ +Y sin θ )(−X sin θ +Y cos θ )]
= E[(Y 2 − X 2 ) sin θ cos θ + XY (cos2 θ − sin2 θ )]
=
σY2 − σX2
sin 2θ + E[XY ] cos 2θ ,
2
which is zero if and only if
tan 2θ =
Hence,
2E[XY ]
.
σX2 − σY2
1 −1 2E[XY ]
.
θ = tan
2
σX2 − σY2
140
19.
Chapter 8 Problem Solutions
(a) Write (ei e0i )mn = (ei )m (e0i )n , which equals one if and only if m = i and n = i.
Hence, ei e0i must be all zeros except at position i i where it is one.
(b) Write
 0 
e1
e1 e4 e5
 e04 
E 0E =
e5
= e1 e01 + e4 e04 + e5 e05
= diag(1, 0, 0, 0, 0) + diag(0, 0, 0, 1, 0) + diag(0, 0, 0, 0, 1)
= diag(1, 0, 0, 1, 1).
20.
(a) We must solve P0CX P = diagonal. With X := U +V , we have
CX = E[XX 0 ] = E[(U +V )(U +V )0 ] = CU + E[UV 0 ] + E[VU 0 ] +CV
= QMQ0 + I = Q(M + I)Q0 .
Hence, Q0CX Q = M + I, which is diagonal. The point here is that we may take
P = Q.
(b) We now put Y := P0 X = QX = Q(U +V ). Then
CY = E[YY 0 ] = E[Q(U +V )(U 0 +V 0 )Q0 ]
= Q{CU + E[UV 0 ] + E[VU 0 ] +CV }Q0 = QCU Q0 + QIQ0 = M + I.
21. Starting with u = x + y and v = x − y, we have
x =
u+v
2
#
"
u−v
.
2
y =
and
We can now write
dH =
"
∂x
∂u
∂y
∂u
∂x
∂v
∂y
∂v
=
1/2
1/2
1/2 −1/2
#
,
det dH = −1/2,
and
| det dH | = 1/2,
and so
fUV (u, v) = fXY
u+v u−v 1
,
· .
2
2
2
22. Starting with u = xy and v = y/x, write y = xv. Then u = x2 v, and x = (u/v)1/2 . We
also have y = (u/v)1/2 v = (uv)1/2 . Then
∂x
∂u
∂y
∂u
√
= (1/2)/ uv,
p
= (1/2) v/u,
∂x
∂v
∂y
∂v
√
= (−1/2) u/v3/2 ,
p
= (1/2) u/v.
Chapter 8 Problem Solutions
In other words,
dH =
and so
"
#
√
√
(1/2)/ uv (−1/2) u/v3/2
,
p
p
(1/2) v/u (1/2) u/v
1
1
1
=
+
.
4v 4v
2|v|
| det dH | =
Thus,
fUV (u, v) = fXY
For fU , write
fU (u) =
Z ∞
0
141
p
√ 1
,
u/v, uv
2v
fUV (u, v) dv =
Z ∞
0
fXY
u, v > 0.
p
√ 1
u/v, uv
dv
2v
√
Now make the change of variable y = uv, or y2 = uv. Then 2y dy = u dv, and
fU (u) =
Z ∞
0
fXY
u 1
, y dy.
y
y
For fV , write
fV (v) =
Z ∞
0
fUV (u, v) du =
Z ∞
0
fXY
p
p
√ 1
u/v, uv
du.
2v
This time make the change of variable x = u/v or x2 = u/v. Then 2x dx = du/v,
and
Z ∞
fV (v) =
fXY (x, vx)x dx.
0
23. Starting with u = x and v = y/x, we find that y = xv = uv. Hence,
" ∂x ∂x #
#
"
1 0
∂u ∂v
dH = ∂ y ∂ y =
, det dH = u, and | det dH | = |u|.
v
u
∂u ∂v
Then
fUV (u, v) = fXY (u, uv)|u| =
λ −λ |u| λ −λ |uv|
· 2e
· |u|,
2e
and
fV (v) =
=
Z ∞
−∞
−λ (1+|v|)|u|
λ2
du
4 |u|e
λ
2(1 + |v|)
Z ∞
0
=
λ2
2
Z ∞
0
ue−λ (1+|v|)u du
u · λ (1 + |v|)e−λ (1+|v|)u du.
Now, this last integral is the mean of an exponential density with parameter λ (1+|v|).
Hence,
1
1
λ
·
=
.
fV (v) =
2(1 + |v|) λ (1 + |v|)
2(1 + |v|)2
142
Chapter 8 Problem Solutions
24. Starting with u =
√
√
−2 ln x cos(2π y) and v = −2 ln x sin(2π y), we have
u2 + v2 = (−2 ln x)[cos2 (2π y) + sin2 (2π y)] = −2 ln x.
Hence, x = e−(u
2 +v2 )/2
. We also have
v
= tan(2π y) or
u
1
tan−1 (v/u).
2π
y =
We can now write
∂x
∂u
∂y
∂u
= −ue−(u
=
1
1
2π 1+(u/v)2
In other words,
and so
2 +v2 )/2

dH = 
·
∂x
∂v
∂y
∂v
,
−v
,
u2
−ue−(u
2 +v2 )/2
1
1
2π 1+(u/v)2
2
· −v
u2
= −ve−(u
=
2 +v2 )/2
1
1
2π 1+(u/v)2
−ve−(u
2 +v2 )/2
1
1
2π 1+(u/v)2
2
2
· 1u
·
,
1
u.

,
2
1 e−(u +v )/2 v2
−1 e−(u +v )/2
| det dH| =
−
·
2π 1 + (v/u)2 2π 1 + (v/u)2 u2
2
=
e−(u +v
2π
2 )/2
2
1
(v/u)2
+
2
1 + (v/u)
1 + (v/u)2
2
e−u /2 e−v /2
√
.
= √
2π
2π
2
2
We next use the formula fUV (u, v) = fXY (x, y) · | det dH|, where x = e−(u +v )/2 and
y = tan−1 (v/u)/(2π ). Fortunately, since these formulas for x and y lie in (0, 1],
fXY (x, y) = I(0,1] (x)I(0,1] (y) = 1, and we see that
2
2
e−u /2 e−v /2
√
fUV (u, v) = 1 · | det dH| = √
.
2π
2π
√
2
Integrating out v shows that fU (u) = e−u /2 / 2π , and integrating out v shows that
√
2
fV (v) = e−v /2 / 2π . It now follows that fUV (u, v) = fU (u) fV (v), and we see that U
and V are independent.
25. Starting with u = x + y and v = x/(x + y) = x/u, we see that v = x/u, or x = uv. Next,
from y = u − x = u − uv, we get y = u(1 − v). We can now write
∂x
∂u
∂y
∂u
= v,
= 1 − v,
In other words,
dH =
∂x
∂v
∂y
∂v
= u,
= −u.
v
u
,
1 − v −u
Chapter 8 Problem Solutions
143
and so
| det dH| = | − uv − u(1 − v)| = |u|.
We next write
fUV (u, v) = fXY (uv, u(1 − v))|u|.
If X and Y are independent gamma RVs, then for u > 0 and 0 < v < 1,
λ (λ uv) p−1 e−λ uv λ (λ u(1 − v))q−1 e−λ u(1−v)
·
·u
Γ(p)
Γ(q)
fUV (u, v) =
λ (λ u) p+q−1 e−λ u Γ(p + q) p−1
·
v (1 − v)q−1 ,
Γ(p + q)
Γ(p)Γ(q)
=
which we recognize as the product of a gamma(p + q, λ ) and a beta(p, q) density.
Hence, it is easy to integrate out either u or v and show that fUV (u, v) = fU (u) fV (v),
where fU ∼ gamma(p + q, λ ) and fV ∼ beta(p, q). Thus, U and V are independent.
26. Starting with u = x + y and v = x/y, write x = yv and then u = yv + y = y(v + 1). Solve
for y = u/(v + 1) and x = u − y = u − u/(v + 1) = uv/(v + 1). Next,
∂x
∂u
∂y
∂u
∂x
∂v
∂y
∂v
= v/(v + 1),
= 1/(v + 1),
In other words,
dH =
= u/(v + 1)2 ,
= −u/(v + 1)2 .
v/(v + 1) u/(v + 1)2
,
1/(v + 1) −u/(v + 1)2
and so
| det dH | =
−u(v + 1)
−uv
|u|
u
=
=
−
.
(v + 1)3 (v + 1)3
(v + 1)3
(v + 1)2
We next write
fUV (u, v) = fXY
uv
u
|u|
,
.
v + 1 v + 1 (v + 1)2
When X ∼ gamma(p, λ ) and Y ∼ gamma(q, λ ) are independent, then U and V are
nonnegative, and
fUV (u, v) = λ
u
[λ uv/(v + 1)] p−1 e−λ uv/(v+1)
[λ u/(v + 1)]q−1 e−λ u/(v+1)
·λ
·
Γ(p)
Γ(q)
(v + 1)2
= λ
v p−1
(λ u) p+q−1 e−λ u Γ(p + q)
·
·
Γ(p + q)
Γ(p)Γ(q) (v + 1) p+q
= λ
v p−1
(λ u) p+q−1 e−λ u
·
,
Γ(p + q)
B(p, q)(v + 1) p+q
which shows that U and V are independent with the required marginal densities.
144
Chapter 8 Problem Solutions
27. From solution of the Example at the end of the section,
fR,Θ (r, θ ) = fXY (r cos θ , r sin θ )r.
What is different in this problem is that X and Y are correlated Gaussian random
variables; i.e.,
2
2
2
e−(x −2ρ xy+y )/[2(1−ρ )]
p
fXY (x, y) =
.
2π 1 − ρ 2
Hence,
fR,Θ (r, θ ) =
=
re−(r
2 cos2 θ −2ρ (r cos θ )(r sin θ )+r2 sin2 θ )/[2(1−ρ 2 )]
2π
re−r
p
1 − ρ2
2 (1−ρ sin 2θ )/[2(1−ρ 2 )]
2π
p
1 − ρ2
.
To find the density of Θ, we must integrate this with respect to r. Notice that the
2
2
integrand is proportional to re−λ r /2 , whose anti-derivative is −e−λ r /2 /λ . Here we
2
have λ = (1 − ρ sin 2θ )/(1 − ρ ). We can now write
fΘ (θ ) =
=
Z ∞
0
2π
fR,Θ (r, θ ) dr =
p
1 − ρ2
1
p
2π 1 − ρ 2
1 − ρ 2 (1 − ρ sin 2θ )
=
Z ∞
0
re−λ r
2 /2
p
dr =
1 − ρ2
.
2π (1 − ρ sin 2θ )
1
1
p
·
2
λ
2π 1 − ρ
28. Since X ∼ N(0, 1), we have E[X] = E[X 3 ] = 0, E[X 4 ] = 3, and E[X 6 ] = 15. Since
W ∼ N(0, 1), E[W ] = 0 too. Hence, E[Y ] = E[X 3 + W ] = 0. It then follows that mY ,
mX and b = mX − Amy are zero. We next compute
CXY = E[XY ] = E[X(X 3 +W )] = E[X 4 ] + E[X]E[W ] = 3 + 0 = 3,
and
CY = E[Y 2 ] = E[X 6 + 2X 3W +W 2 ] = 15 + 0 + 1 = 16.
Hence, ACY = CXY implies A = 3/16, and then Xb = A(Y − mY ) + mX = (3/16)Y .
29. First note that since X and W are zero mean, so is Y . Next,
CXY = E[XY ] = E[X(X +W )] = E[X 2 ] + E[XW ] = E[X 2 ] + E[X]E[W ] = E[X 2 ] = 1,
and
CY = E[Y 2 ] = E[(X +W )2 ] = E[X 2 + 2XW +W 2 ]
= E[X 2 ] + 2E[X]E[W ] + E[W 2 ] = 1 + 0 + 2/λ 2 = 1 + 2/λ 2 .
Then A = CXY /CY = 1/[1 + 2/λ 2 ], and
Xb =
λ2
Y.
2+λ2
Chapter 8 Problem Solutions
145
30. We first have E[Y ] = E[GX +W ] = GmX + 0 = GmX . Next,
CXY = E[(X − mX )(Y − mY )0 ] = E[(X − mX )(GX +W − GmX )0 ]
= E[(X − mX )(G{X − mX } +W )0 ] = CX G0 +CXW = CX G0 ,
and
CY = E[(Y − mY )(Y − mY )0 ] = E[(G{X − mX } +W )(G{X − mX } +W )0 ]
= GCX G0 + GCXW +CW X G0 +CW = GCX G0 +CW ,
since X and W are uncorrelated. Solving ACY = CXY implies
A = CX G0 (GCX G0 +CW )−1
and
Xb = CX G0 (GCX G0 +CW )−1 (Y − GmX ) + mX .
31. We begin with the result of Problem 30 that
A = CX G0 (GCX G0 +CW )−1 .
Following the hint, we make the identifications α = CW , γ = CX , β = G, and δ = G0 .
Then
A = CX G0 (α + β γδ )−1
−1
−1
−1
−1
= CX G0 [CW
−CW
G(CX−1 + G0CW
G)−1 G0CW
]
−1
−1
−1
−1
= CX G0CW
−CX G0CW
G(CX−1 + G0CW
G)−1 G0CW
−1
−1
−1
= [CX −CX G0CW
G(CX−1 + G0CW
G)−1 ]G0CW
−1
−1
−1
−1
= [CX (CX−1 + G0CW
G) −CX G0CW
G](CX−1 + G0CW
G)−1 G0CW
−1
−1
−1
−1
= [I +CX G0CW
G −CX G0CW
G](CX−1 + G0CW
G)−1 G0CW
−1
−1
= (CX−1 + G0CW
G)−1 G0CW
.
32. We begin with
Xb = A(Y − mY ) + mX ,
where
ACY = CXY .
Next, with Z := BX, we have mZ = BmX and
CZY = E[(Z − mZ )(Y − mY )0 ] = E[B(X − mX )(Y − mY )0 ] = BCXY .
e Y = CZY . Starting with ACY = CXY , multiply this equation by B to
We must solve AC
e := BA solves the required equation. Hence,
get (BA)CY = BCXY = CZY . We see that A
the linear MMSE estimate of X based on Z is
b
(BA)(Y − mY ) + mZ = (BA)(Y − mY ) + BmX = B{A(Y − mY ) + mX } = BX.
33. We first show that the orthogonality condition
E[(CY )0 (X − AY )] = 0,
for all C,
146
Chapter 8 Problem Solutions
implies A is optimal. Write
E[kX − BY k2 ] = E[k(X − AY ) + (AY − BY )k2 ]
= E[k(X − AY ) + (A − B)Y )k2 ]
= E[kX − AY k2 ] + 2E[{(A − B)Y }0 (X − AY )] + E[k(A − B)Y k2 ]
= E[kX − AY k2 ] + E[k(A − B)Y k2 ]
≥ E[kX − AY k2 ],
where the cross terms vanish by taking C = A − B in the orthogonality condition.
Next, rewrite the orthogonality condition as
E[(CY )0 (X − AY )] = tr{E[(CY )0 (X − AY )]} = E[tr{(CY )0 (X − AY )}]
= E[tr{(X − AY )(CY )0 }] = tr{E[(X − AY )(CY )0 ]}
= tr{(RXY − ARY )C0 }.
Now, this expression must be zero for all C, including C = RXY − ARY . However,
since tr(DD0 ) = 0 implies D = 0, we conclude that the optimal A solves ARY = RXY .
Next, the best constant estimator is easily found by writing
E[kX − bk2 ] = E[k(X − mX ) + (mX − b)k2 ] = E[kX − mX k2 ] + kmX − bk2 .
Hence, the optimal value of b is b = mX .
34. To begin, write
b
b 0 ] = E[{(X − mX ) − A(Y − mY )}{(X − mX ) − A(Y − mY )}0 ]
E[(X − X)(X
− X)
= CX − ACY X −CXY A0 + ACY A0 .
(∗)
We now use the fact that ACY = CXY . If we multiply ACY = CXY on the right by A0 ,
we obtain
ACY A0 = CXY A0 .
Furthermore, since ACY A0 is symmetric, we can take the transpose of the above expression and obtain
ACY A0 = ACY X .
By making appropriate substitutions in (∗), we find that the error covariance is also
given by
CX − ACY X , CX −CXY A0 , and CX − ACY A0 .
35. Write
n
o
b 2 ] = E[(X − X)
b 0 (X − X)]
b = tr E[(X − X)
b 0 (X − X)]
b
E[kX − Xk
b 0 (X − X)}]
b
b
b 0 }]
= E[tr{(X − X)
= E[tr{(X − X)(X
− X)
b
b 0 ]} = tr{CX − ACY X }.
= tr{E[(X − X)(X
− X)
Chapter 8 Problem Solutions
147
36. MATLAB. We found


−0.0622 0.0467 −0.0136 −0.1007
 0.0489 −0.0908 −0.0359 −0.1812 

A = 
 −0.0269 0.0070 −0.0166 0.0921 
0.0619 0.0205 −0.0067 0.0403
and MSE = 0.0806.
37. Write X in the form X = [Y 0 , Z 0 ]0 , where Y := [X1 , . . . , Xm ]0 . Then
0 0 Y
CY CY Z
C1 C2
Y Z
CX = E
=
=
,
Z
CZY CZ
C20 C3
and
CXY = E
Solving ACY = CXY becomes
AC1 =
Y
CY
C1
0
Y =
=
.
Z
CZY
C20
C1
,
C20
or
A =
I
.
C20 C1−1
The linear MMSE estimate of X = [Y 0 , Z 0 ]0 is
Y
I
.
Y
=
AY =
C20 C1−1Y
C20 C1−1
In other words, Yb = Y and Zb = C20 C1−1Y . Note that the matrix required for the linear
MMSE estimate of Z based on Y is the solution of BCY = CZY or BC1 = C20 ; i.e.,
B = C20 C1−1 . Next, the error covariance for estimating X based on Y is
I
C1 C2
b
b 0 ] = CX − ACY X = C10 C2 −
E[(X − X)(X
− X)
C2 C3
C20 C1−1
C1
C2
C1 C2
−
=
C20 C3
C20 C20 C1−1C2
0
0
,
=
0 C3 −C20 C1−1C2
and the MSE is
b 2 ] = tr(CX − ACY X ) = tr(C3 −C20 C−1C2 ).
E[kX − Xk
1
38. Since P0 decorrelates Y , the covariance matrix CZ of Z := P0Y is diagonal. Writing
ÃCZ = CXZ in component form, and using the fact that CZ is diagonal,
∑ Ãik (CZ )k j
= (CXZ )i j
k
becomes
Ãi j (CZ ) j j = (CXZ )i j
148
Chapter 8 Problem Solutions
If (CZ ) j j 6= 0, then Ãi j = (CXZ )i j /(CZ ) j j . If (CZ ) j j = 0, then Ãi j (CZ ) j j = (CXZ )i j
can be solved only if (CXZ )i j = 0, which we now show to be the case by using the
Cauchy–Schwarz inequality. Write
|(CXZ )i j | = E[(Xi − (mX )i )(Z j − (mZ ) j )]
q
≤ E[(Xi − (mX )i )2 ]E[(Z j − (mZ ) j )2 ]
q
= (CX )ii (CZ ) j j .
Hence, if (CZ ) j j = 0 then (CXZ )i j = 0, and any value of Ãi j solves Ãi j (CZ ) j j =
(CXZ )i j . Now that we have shown that we can always solve ÃCZ = CXZ , observe
that this equation is equivalent to
Ã(P0CY P) = CXY P or
(ÃP0 )CY = CXY .
Thus, A = ÃP0 solves the original problem.
39. Since X has the form X = [Y 0 , Z 0 ]0 , if we take G = [I, 0] and W ≡ 0, then
I 0 Y
GX +W =
= Y.
Z
40. Write
1 n 2
E
∑ Xk
n k=1
41. Write
E
1 n
∑ Xk Xk0
n k=1
=
=
1 n
1 n
E[Xk2 ] = ∑ σ 2 = σ 2 .
∑
n k=1
n k=1
1 n
1 n
E[Xk Xk0 ] = ∑ C = C.
∑
n k=1
n k=1
42. MATLAB. Additional code:
Mn = mean(X,2)
MnMAT = kron(ones(1,n),Mn);
Chat = (X-MnMAT)*(X-MnMAT)’/(n-1)
43. We must first find fY |X (y|x). To this end, use substitution and independence to write
P(Y ≤ y|X = x) = P(X +W ≤ y|X = x) = P(x +W ≤ y|X = x) = P(W ≤ y − x).
Then fY |X (y|x) = fW (y − x) = (λ /2)e−λ |y−x| . For fixed y, the maximizing value of x
is x = y. Hence, gML (y) = y.
44. By the same argument as in the solution of Problem 43, fY |X (y|x) = (λ /2)e−λ |y−x| .
When X ∼ exp(µ ) and we maximize over x, we must impose the constraint x ≥ 0.
Hence,
y, y ≥ 0,
gML (y) = argmax λ2 e−λ |y−x| =
0, y < 0.
x≥0
Chapter 8 Problem Solutions
When X ∼ uniform[0, 1],
gML (y) = argmax λ2 e−λ |y−x|
0≤x≤1
149

 y, 0 ≤ y ≤ 1,
1, y > 1,
=

0, y < 0.
45. By the same argument as in the solution of Problem 43, fY |X (y|x) = (λ /2)e−λ |y−x| .
For the MAP estimator, we must maximize
fY |X (y|x) fX (x) =
λ −λ |y−x|
· µ e−µ x ,
2e
x ≥ 0.
By considering separately the cases x ≤ y and x > y,
( µλ
−λ y e(λ −µ )x , 0 ≤ x ≤ y,
2 e
fY |X (y|x) fX (x) =
µλ λ y −(λ +µ )x
, x > y, x ≥ 0.
2 e e
When y ≥ 0, observe that the two formulas agree at x = y and have the common value
(µλ /2)e−µ y ; in fact, if λ > µ , the first formula is maximized at x = y, while the
second formula is always maximized at x = y. If y < 0, then only the second formula
is valid, and its region of validity is x ≥ 0. This formula is maximized at x = 0. Hence,
for λ > µ ,
y, y ≥ 0,
gMAP (y) =
0, y < 0.
We now consider the case λ ≤ µ . As before, if y < 0, the maximizing value of x is
zero. If y ≥ 0, then the maximum value of fY |X (y|x) fX (x) for 0 ≤ x ≤ y occurs at
x = 0 with a maximum value of (µλ /2)e−λ y . The maximum value of fY |X (y|x) fX (x)
for x ≥ y occurs at x = y with a maximum value of (µλ /2)e−µ y . For λ < µ ,
max{(µλ /2)e−λ y , (µλ /2)e−µ y } = (µλ /2)e−λ y ,
which corresponds to x = 0. Hence, for λ < µ ,
gMAP (y) = 0,
−∞ < y < ∞.
46. From the formula
fXY (x, y) = (x/y2 )e−(x/y)
2 /2
· λ e−λ y ,
x, y > 0,
we see that Y ∼ exp(λ ), and
p that given Y = y, X ∼ Rayleigh(y). Hence, the MMSE
estimator is E[X|Y = y] = π /2 y. To compute the MAP estimator, we must solve
argmax(x/y2 )e−(x/y)
2 /2
.
x≥0
We do this by differentiating with respect to x and setting the derivative equal to zero.
Write
2
∂
e−(x/y) /2
x2
2 −(x/y)2 /2
=
(x/y )e
1− 2 .
∂x
y2
y
Hence,
gMAP (y) = y.
150
Chapter 8 Problem Solutions
47. Suppose that
E[(X − g1 (Y ))h(Y )] = 0
and
E[(X − g2 (Y ))h(Y )] = 0.
Subtracting the second equation from the first yields
E[{g2 (Y ) − g1 (Y )}h(Y )] = 0.
Since h is an arbitrary bounded function, put h(y) := sgn[g2 (y) − g1 (y)], where

 1, x > 0,
0, x = 0,
sgn(x) :=

−1, x < 0.
Note also that x · sgn(x) = |x|. Then (∗) becomes E[|g2 (Y ) − g1 (Y )|] = 0.
(∗)
CHAPTER 9
Problem Solutions
1. We first compute
σ12 σ1 σ2 ρ
detC = det
σ1 σ2 ρ σ22
p
√
and detC = σ1 σ2 1 − ρ 2 . Next,
C−1 =
1
detC
= σ12 σ22 − (σ1 σ2 ρ )2 = σ12 σ22 (1 − ρ 2 ),
−σ1 σ2 ρ
σ22
−σ1 σ2 ρ
σ12
and

1
−ρ
 σ 2 (1 − ρ 2 ) σ1 σ2 (1 − ρ 2 ) 
,
1
= 


−ρ
1
2
2
2
σ1 σ2 (1 − ρ ) σ2 (1 − ρ )



x
ρy
−
2
2
 σ (1 − ρ 2 ) σ1 σ2 (1 − ρ ) 
x
x y C−1
= x y  1 −ρ x

y
y
+ 2
2
2
σ1 σ2 (1 − ρ ) σ1 (1 − ρ )
x2
y2
ρ xy
ρ xy
−
−
+
σ12 (1 − ρ 2 ) σ1 σ2 (1 − ρ 2 ) σ1 σ2 (1 − ρ 2 ) σ12 (1 − ρ 2 )
y 2 i
1 h x 2
xy
=
ρ
+
−
2
,
1 − ρ 2 σ1
σ1 σ2
σ2
=
and the result follows.
2. Here is the plot:
n=4
1
n=1
0
−4 −2
3.
0
2
4
(a) First write c1 X + c2Y = c1 X + c2 (3X) = (c1 + 3c2 )X, which is easily seen to be
N(0, (c1 + 3c2 )2 ). Thus, X and Y are jointly Gaussian.
(b) Observe that E[XY ] = E[X(3X)] = 3E[X 2 ] = 3 and E[Y 2 ] = E[(3X)2 ] = 9E[X 2 ] =
9. Since X and Y have zero means,
E[X 2 ] E[XY ]
1 3
cov([X,Y ]0 ) =
=
.
3 9
E[Y X] E[Y 2 ]
151
152
Chapter 9 Problem Solutions
(c) The conditional cdf of Y given X = x is
FY |X (y|x) = P(Y ≤ y|X = x) = P(3X ≤ y|X = x).
By substitution, this last conditional probability is P(3x ≤ y|X = x). The event
{3x ≤ y} is deterministic and therefore independent of X. Hence, we can drop
the conditioning and get
FY |X (y|x) = P(3x ≤ y).
If 3x ≤ y, then {3x ≤ y} = Ω, and {3x ≤ y} = ∅ otherwise. Hence, the above
probability is just u(y − 3x).
4. If Y = ∑ni=1 ci Xi , its characteristic function is
n
jν ci Xi
jν (∑ni=1 ci Xi )
] = E ∏e
=
ϕY (ν ) = E[e
i=1
n
=
2 2
∏ e j(ν ci )mi −(ν ci ) σi /2
n
n
∏ E[e jν ci Xi ]
n
=
i=1
= e jν (∑i=1 ci mi )−ν
2 ( n c2 σ 2 )/2
∑i=1 i i
∏ ϕXi (ν ci )
i=1
,
i=1
which is the characteristic function of an N(∑ni=1 ci mi , ∑ni=1 c2i σi2 ) random variable.
5. First, E[Y ] = E[AX + b] = AE[X] + b = Am + b. Second,
E[{Y − (Am + b)}{Y − (Am + b)}0 ] = E[A(X − m)(X − m)0 A0 ] = ACA0 .
6. Write out
Y1 = X1
Y2 = X1 + X2
Y3 = X1 + X2 + X3
..
.
In general, Yn = Yn−1 + Xn , or Xn = Yn − Yn−1 , which we can write in matrix-vector
notation as





X1
Y1
1 0 ···
0
 X2 


 −1 1 0
0 



  Y2 
 X3 

 0 −1 1 0

0   Y3 




 ..
 = 
.
  ..
..
 .



 .
.





 Xn−1 

 0

Yn−1 
0 −1 1 0
Xn
Yn
0
· · · 0 −1 1
{z
}
|
=: A
Since X = AY and Y is Gaussian, so is X.
7. Since X ∼ N(0,C), the scalar Y := ν 0 X is also Gaussian and has zero mean. Hence,
E[(ν 0 XX 0 ν )k ] = E[(Y 2 )k ] = E[Y 2k ] = (2k − 1) · · · 5 · 3 · 1 · (E[Y 2 ])k . Now observe that
E[Y 2 ] = E[ν 0 XX 0 ν ] = ν 0Cν and the result follows.
Chapter 9 Problem Solutions
153
8. We first have
E[Y j ] = E[X j − X] = m − E
1 n
∑ Xi
n i=1
= m−
1 n
∑ m = 0.
n i=1
Next, since E[XY j ] = E[X(X j − X)], we first compute
n
1
1 n n
2
2
2
E[ X X ] = 2 ∑ ∑ E[Xi X j ] = 2 ∑ (σ + m ) + ∑ m
n i=1 j=1
n i=1
i6= j
o
o
n
n
1
1
σ2
+ m2 ,
= 2 n(σ 2 + m2 ) + n(n − 1)m2 = 2 nσ 2 + n2 m2 =
n
n
n
and
E[ X X j ] =
o
1n 2
1 n
σ2
2
2
+
m
)
+
(n
−
1)m
=
E[X
X
]
=
(
+ m2 .
σ
i j
∑
n i=1
n
n
It now follows that E[ X Y j ] = 0.
9. Following the hint, in the expansion of E[(ν 0 X)2k ], the sum of all the coefficients of
νi1 · · · νi2k is (2k)!E[Xi1 · · · Xi2k ]. The corresponding sum of coeficients in the expansion of
(2k − 1)(2k − 3) · · · 5 · 3 · 1 · (ν 0Cν )k
is
(2k − 1)(2k − 3) · · · 5 · 3 · 1 · 2k k!
∑
j1 ,..., j2k
C j1 j2 · · ·C j2k−1 j2k ,
where the sum is over all j1 , . . . , j2k that are permutations of i1 , . . . , i2k and such that
the product C j1 j2 · · ·C j2k−1 j2k is distinct. Since
(2k − 1)(2k − 3) · · · 5 · 3 · 1 · 2k k! = (2k − 1)(2k − 3) · · · 5 · 3 · 1 · (2k)(2k − 2) · · · 4 · 2
= (2k)!,
Wick’s Theorem follows.
10. Write
E[X1 X2 X3 X4 ] =
∑
C j1 j2 C j3 j4 = C12C34 +C13C24 +C14C23 .
j 1 , j2 , j3 , j4
E[X1 X32 X4 ] = C13C34 +C14C33 .
2
E[X12 X22 ] = C11C22 +C12
.
11. Put a := [a1 , . . . , an ]0 . Then Y = a0 X, and
0
0
ϕY (η ) = E[e jηY ] = E[e jη (a X) ] = E[e j(η a) X ] = ϕX (η a)
0
0
0
0
= e j(η a) m−(η a) C(η a)/2 = e jη (a m)−(a Ca)η
2 /2
,
which is the characteristic function of a scalar N(a0 m, a0Ca) random variable.
154
Chapter 9 Problem Solutions
0
0
12. With X = [U 0 ,W 0 ]0 and ν = [α 0 , β 0 ]0 , we have ϕX (ν ) = e jν m−ν Cν /2 , where
mU
ν 0m = α 0 β 0
= α 0 mU + β 0 mW ,
mW
and
0
ν Cν =
Thus,
α0
β0
0 0 Sα
S 0
α
= α β
= α 0 Sα + β 0 T β .
Tβ
β
0 T
0
ϕX (ν ) = e j(α mU +β
0 m )−(α 0 Sα +β 0 T β )/2
W
0
0
= e jα mU −α Sα /2 e jβ
0 m −β 0 T β /2
W
,
which has the required form ϕU (α )ϕW (β ) of a product of Gaussian characteristic
functions.
13.
(a) Since X is N(0,C) and Y := C−1/2 X, Y is also normal. It remains to find the
mean and covariance of Y . We have E[Y ] = E[C−1/2 X] = C−1/2 E[X] = 0 and
E[YY 0 ] = E[C−1/2 XX 0C−1/2 ] = C−1/2 E[XX 0 ]C−1/2 = C−1/2CC−1/2 = I. Hence,
Y ∼ N(0, I).
(b) Since the covariance matrix of Y is diagonal, the components of Y are uncorrelated. Since Y is also Gaussian, the components of Y are independent. Since the
covariance matrix of Y is the indentity, each Yk ∼ N(0, 1). Hence, each Yk2 is chisquared with one degree of freedom by Problem 46 in Chapter 4 or Problem 11
in Chapter 5.
(c) By the Remark in Problem 55(c) in Chapter 4, V is chi-squared with n degrees
of freedom.
14. Since
Z := det
X −Y
Y X
= X 2 +Y 2 ,
where X and Y are independent N(0, 1), observe that X 2 and Y 2 are chi-squared with
one degree of freedom. Hence, Z is chi-squared with two degrees of freedom, which
is the same as exp(1/2).
15. Begin with
fX (x) =
1
(2π )n
Z
0
IRn
0
0
e− jν x e jν m−ν Cν /2 d ν =
1
(2π )n
Z
IRn
0
0
e− j(x−m) ν e−ν Cν /2 d ν .
Now make the multivariate change of variable ζ = C1/2 ν , d ζ = detC1/2 d ν . Then
Z
0 −1/2
0
1
dζ
e− j(x−m) C ζ e−ζ ζ /2
(2π )n IRn
detC1/2
Z
−1/2
0
0
1
dζ
e− j{C (x−m)} ζ e−ζ ζ /2 √
=
.
(2π )n IRn
detC
fX (x) =
Chapter 9 Problem Solutions
Put t = C−1/2 (x − m) so that
fX (x) =
1
(2π )n
Z
IRn
0
e− jt ζ e−ζ
0 ζ /2
155
Z
n 1 ∞ − jti ζi −ζ 2 /2
1
dζ
i
√
e
=√
d
e
ζ
i .
∏
detC
detC i=1 2π −∞
2
Observe that e−ζi /2 is the characteristic function of a scalar N(0, 1) random variable.
Hence,
n −ti2 /2
1
e
1
√
√
fX (x) = √
exp[−t 0t/2].
=
∏
n/2
detC i=1 2π
(2π )
detC
Recalling that t = C−1/2 (x − m) yields
fX (x) =
16. First observe that
exp[− 12 (x − m)0C−1 (x − m)]
√
.
(2π )n/2 detC
X Y
Z := det
U V
= XV −YU.
Then consider the conditional cumulative distribution function,
FZ|UV (z|u, v) = P(Z ≤ z|U = u,V = v) = P(XV −YU ≤ z|U = u,V = v)
= P(Xv −Yu ≤ z|U = u,V = v).
Since [X,Y ]0 and [U,V ]0 are jointly Gaussian and uncorrelated, they are independent.
Hence, we can drop the conditioning and get
FZ|UV (z|u, v) = P(Xv −Yu ≤ z).
Next, since X and Y are independent and N(0, 1), Xv −Yu ∼ N(0, u2 + v2 ). Hence,
fZ|UV (· |u, v) ∼ N(0, u2 + v2 ).
17. We first use the fact that A solves ACY = CXY to show that (X − mX ) − A(Y − mY ) and
Y are uncorrelated. Write
E[{(X − mX ) − A(Y − mY )}Y 0 ] = CXY − ACY = 0.
We next show that (X − mX ) − A(Y − mY ) and Y are jointly Gaussian by writing them
as an affine transformation of the Gaussian vector [X 0 ,Y 0 ]0 ; i.e.,
(X − mX ) − A(Y − mY )
I −A
X
AmY − mX
=
+
.
Y
0 I
Y
0
It now follows that (X − mX ) − A(Y − mY ) and Y are independent. Using the hints on
substitution and independence, we compute the conditional characteristic function,
0
0
0
E[eν X |Y = y] = E e jν [(X−mX )−A(Y −mY )] · e jν [mX +A(Y −mY )] Y = y
0
0
= e jν [mX +A(y−mY )] E e jν [(X−mX )−A(Y −mY )] Y = y
0
0
= e jν [mX +A(y−mY )] E e jν [(X−mX )−A(Y −mY )] .
156
Chapter 9 Problem Solutions
This last expectation is the characteristic function of the zero-mean Gaussian random
vector (X − mX ) − A(Y − mY ). To compute its covariance matrix first observe that
since ACY = CXY , we have ACY A0 = CXY A0 . Then
E[{(X − mX ) − A(Y − mY )}{(X − mX ) − A(Y − mY )}0 ]
= CX −CXY A0 − ACY X + ACY A0
= CX − ACY X .
We now have
0
0
0
E[eν X |Y = y] = e jν [mX +A(y−mY )] e−ν [CX −ACY X ]ν /2 .
$
Thus, given Y = y, X is conditionally N mX + A(y − mY ),CX − ACY X .
18. First observe that
X Y
Z := det
U V
= XV −YU.
Then consider the conditional cumulative distribution function,
FZ|UV (z|u, v) = P(Z ≤ z|U = u,V = v) = P(XV −YU ≤ z|U = u,V = v)
= P(Xv −Yu ≤ z|U = u,V = v).
Since [X,Y,U,V ]0 is Gaussian, given U = u and V = v, [X,Y ]0 is conditionally
u
,C[X,Y ]0 − AC[U,V ]0 ,[X,Y ]0 ,
N A
v
where A solves AC[U,V ]0 = C[X,Y ]0 ,[U,V ]0 . We now turn to the conditional distribution of
Xv −Yu. Since the conditional distribution of [X,Y ]0 is Gaussian, so is the conditional
distribution of the linear combination Xv − Yu. Hence, all we need to find are the
conditional mean and the conditional variance of Xv −Yu; i.e.,
X
U = u,V = v
E[Xv −Yu|U = u,V = v] = E v u
Y
X
U = u,V = v
= v u E
Y
u
= v u A
,
v
and
E
2
X
u
v u
U = u,V = v
−A
Y
v
2
U
X
v u
U = u,V = v
−A
= E
V
Y
0
v
X
U
X
U
U = u,V = v
= v u E
−A
−A
u
Y
V
Y
V
v
= v u C[X,Y ]0 − AC[U,V ]0 ,[X,Y ]0
,
u
Chapter 9 Problem Solutions
157
where the last step uses the fact that [X,Y ]0 − A[U,V ]0 is independent of [U,V ]0 . If
[X,Y ]0 and [U,V ]0 are uncorrelated, i.e., C[X,Y ]0 ,[U,V ]0 = 0, then A = 0 solves AC[U,V ]0 =
0; in this case, the conditional mean is zero, and the conditional variance simplifies to
v
v
v u C[X,Y ]0
= v2 + u2 .
= v u I
u
u
19. First write
Z − E[Z] = (X + jY ) − (mX + jmY ) = (X − mX ) + j(Y − mY ).
Then
cov(Z) := E[(Z − E[Z])(Z − E[Z])∗ ]
= E[{(X − mX ) + j(Y − mY )}{(X − mX ) + j(Y − mY )}∗ ]
= E[{(X − mX ) + j(Y − mY )}{(X − mX ) − j(Y − mY )}]
= var(X) − j cov(X,Y ) + j cov(Y, X) + var(Y ) = var(X) + var(Y ).
20.
(a) Write
K := E[(Z − E[Z])(Z − E[Z])H ]
= E[{(X − mX ) + j(Y − mY )}{(X − mX ) + j(Y − mY )}H ]
= E[{(X − mX ) + j(Y − mY )}{(X − mX )H − j(Y − mY )H }]
= CX − jCXY + jCY X +CY = (CX +CY ) + j(CY X −CXY ).
(b) If CXY = −CY X , the (CXY )ii = −(CY X )ii implies
E[(Xi − (mX )i )(Yi − (mY )i )] = −E[(Yi − (mY )i )(Xi − (mX )i )]
= −E[(Xi − (mX )i )(Yi − (mY )i )].
Hence, E[(Xi − (mX )i )(Yi − (mY )i )] = 0, and we see that Xi and Yi are uncorrelated.
(c) By part (a), if K is real, then CY X = CXY . Circular symmetry implies CY X =
−CXY . It follows that CXY = −CXY , and then CXY = 0; i.e., X and Y are uncorrelated.
21. First,
fX (x) =
0
e−x (2I)x/2
(2π )n/2 (1/2)n/2
Then
fXY (x, y) = fX (x) fY (y) =
22.
fY (y) =
and
0
e−(x x+y
πn
0 y)
0
e−y (2I)y/2
.
(2π )n/2 (1/2)n/2
H
=
e−(x+ jy) (x+ jy)
.
πn
(a) Immediate from Problem 20(a).
(b) Since ν 0 Qν is a scalar, (ν 0 Qν )0 = ν 0 Qν . Since Q0 = −Q, (ν 0 Qν )0 = ν 0 Q0 ν =
−ν 0 Qν . Thus, ν 0 Qν = −ν 0 Qν , and it follows that ν 0 Qν = 0.
158
Chapter 9 Problem Solutions
(c) Begin by observing that if K = R + jQ and w = ν + jθ , then
Kw = (R + jQ)(ν + jθ ) = Rν + jRθ + jQν − Qθ .
Next,
wH (Kw) = (ν 0 − jθ 0 )(Rν + jRθ + jQν − Qθ )
= ν 0 Rν + jν 0 Rθ + jν 0 Qν − ν 0 Qθ − jθ 0 Rν + θ 0 Rθ + θ 0 Qν − jθ 0 Qθ
= ν 0 Rν + jν 0 Rθ + jν 0 Qν + θ 0 Qν − jν 0 Rθ + θ 0 Rθ + θ 0 Qν − jθ 0 Qθ
= ν 0 Rν + θ 0 Rθ + 2θ 0 Qν ,
where we have used the result of parts (a) and (b).
23. First,
AZ = (α + jβ )(X + jY ) = (α X − β Y ) + j(β X + α Y ).
Second,
α −β
β α
X
Y
=
αX − βY
.
β X + αY
Now assume that circular symmetry holds; i.e., CX = CY and CXY = −CY X . Put
U := α X − β Y and V := β X + α Y . Assuming zero means to simplify the notation,
CU = E[(α X − β Y )(α X − β Y )0 ] = α CX α 0 − β CY X α 0 − α CXY β 0 + β 0CY β
= α CX α 0 − β CY X α 0 + α CY X β 0 + β 0CX β .
Similarly,
CV = E[(β X + α Y )(β X + α Y )0 ] = β CX β 0 + α CY X β 0 + β CXY α 0 + α CY α 0
= β CX β 0 + α CY X β 0 − β CY X α 0 + α CX α 0 .
Hence, CU = CV . It remains to compute
CUV = E[(α X − β Y )(β X + α Y )0 ] = α CX β 0 + α CXY α 0 − β CY X β 0 − β CY α 0
= α CX β 0 − α CY X α 0 − β CY X β 0 − β CX α 0
and
CVU = E[(β X + α Y )(α X − β Y )0 ] = β CX α 0 − β CXY β 0 + α CY X α 0 − α CY β 0
= β CX α 0 + β CY X β 0 + α CY X α 0 − α CX β 0 ,
which shows that CU = −CV . Thus, if Z is circularly symmetric, so is AZ.
24. To begin, note that with R = [X 0 ,U 0 ]0 and I = [Y 0 ,V 0 ]0 ,
CX CXU
CY CYV
CR =
, CI =
,
CUX CU
CVY CV
and
CRI =
CXY CXV
CUY CUV
,
CIR =
CY X CYU
.
CV X CVU
Also, Θ is circularly symmetric means CR = CI and CRI = −CIR .
Chapter 9 Problem Solutions
159
(a) We assume zero means to simplify the notation. First,
KZW = E[ZW H ] = E[(X + jY )(U + jV )H ] = E[(X + jY )(U H − jV H )]
= CXU − jCXV + jCYU +CYV
= 2(CXU − jCXV ),
since Θ is circularly symmetric.
Second,
CZeWe =
CXU CXV
CYU CYV
=
CXU CXV
,
−CXV CXU
since Θ is circularly symmetric.
It is now clear that KZW = 0 if and only if CZeWe = 0.
(b) Assuming zero means again, we compute
KW = E[WW H ] = E[(U + jV )(U + jV )H ] = E[(U + jV )(U H − jV H )]
= CU − jCUV + jCVU +CV = 2(CU − jCUV ).
We now see that AKW = KZW becomes
2(α + jβ )(CU − jCUV ) = 2(CXU − jCXV )
or
(α CU + β CUV ) + j(β CU − α CUV ) = CXU − jCXV .
We also have
CWe =
CU CUV
CVU CV
(∗)
e e = C e e becomes
so that AC
W
ZW
CU CUV
α −β
= CZeWe
CVU CV
β α
or
or
α CU − β CVU α CUV − β CV
β CU + α CVU β CUV + α CV
α CU + β CUV α CUV − β CU
β CU − α CUV β CUV + α CU
=
= CZeWe
CXU CXV
,
−CXV CXU
which is equivalent to (∗).
e solves AC
e e = C e e . Hence, by
(c) If A solves AKW = KZW , then by part (b), A
W
ZW
e = w,
e
Problem 17, given W
ew
e ee .
e − mWe ),CZe − AC
Ze ∼ N mZe + A(
WZ
e e e is equivalent to
Next, CZe − AC
WZ
CX CXY
α −β
CUX CUY
−
,
CY X CY
β α
CV X CVY
160
Chapter 9 Problem Solutions
which, by the circular symmetry of Θ, becomes
CX CXY
CUX CUY
α −β
−
,
−CXY CX
−CUY CUX
β α
or
CX CXY
−CXY CX
which is equivalent to
−
α CUX + β CUY α CUY − β CUX
,
β CUX − α CUY β CUY + α CUX
2(CX − jCXY ) − (α + jβ ) · 2(CUX − jCUY ),
which is exactly KZ − AKW Z . Thus, given W = w,
$
Z ∼ N mZ + A(w − mW ), KZ − AKW Z .
25. Let Z = X + jY with X and Y independent N(0, 1/2) as in the text.
(a) Since X and Y are zero mean,
cov(Z) = E[ZZ ∗ ] = E[X 2 +Y 2 ] = 12 + 12 = 1.
√
√
√
√
(b) First write 2|Z|2 = 2(X 2 +Y 2 ) = ( 2 X)2 + ( 2Y )2 . Now, 2 X and 2Y are
both N(0, 1). Hence, their squares are chi-squared with one degree of freedom
by Problem 46 in Chapter 4 or Problem 11 in Chapter 5. Hence, by Problem 55(c) in Chapter 4 and the remark following it, 2|Z|2 is chi-squared with
two degrees of freedom.
26. With X ∼ N(mr , 1) and Y ∼ N(mi , 1), it follows either from Problem 47 in Chapter 4
or from Problem 12 in Chapter 5 that X 2 and Y 2 are noncentral chi-squared with one
degree of freedom and respective noncentrality parameters m2r and m2i . Since X and Y
are independent, it follows from Problem 65 in Chapter 4 that X 2 +Y 2 is noncentral
chi-squared with two degrees of freedom and noncentrality√parameter m2r + m2i . It is
now immediate from Problem 26 in Chapter 5 that |Z| = X 2 +Y 2 has the orginal
Rice density.
27.
(a) The covariance matrix of W is
E[WW H ] = E[K −1/2 ZZ H K −1/2 ] = K −1/2 E[ZZ H ]K −1/2 = K −1/2 KK −1/2 = I.
Hence,
H
fW (w) =
e−w w
=
πn
2
e−|wk |
∏ π .
k=1
n
(b) By part (a), the Wk = Uk + jVk are i.i.d. N(0, 1) with
2
2
√
2
2
e−|wk |
e−(uk +vk )
e−[(uk /(1/ 2 )) +(vk /(1/
fWk (w) =
= $p
$p
2 =
2
π
2π /2
2π /2
√
e−[uk /(1/ 2 )]
p
=
2π /2
2 /2
√
2 /2
e−[vk /(1/ 2 )]
p
·
2π /2
Hence, Uk and Vk are independent N(0, 1/2).
√ 2
2 )) ]/2
= fUkVk (uk , vk ).
Chapter 9 Problem Solutions
(c) Write
n
2kW k2 =
161
√
√
2Uk )2 + ( 2Vk )2 .
∑(
k=1
√
√
Since 2Uk and 2Vk are independent N(0, 1), their squares are chi-squared
with one degree
√ of freedom
√ by Problem 46 in Chapter 4 or Problem 11 in Chapter 5. Next, ( 2Uk )2 + ( 2Vk )2 is chi-squared with two degrees of freedom by
Problem 55(c) in Chapter 4 and the remark following it. Similarly, since the Wk
are indepdendent, 2kW k2 is chi-squared with 2n degrees of freedom.
28.
(a) Write
0 = (u + v)0 M(u + v) = u0 Mu + v0 Mu + u0 Mv + v0 Mv = 2v0 Mu,
since M 0 = M. Hence, v0 Mu = 0.
(b) By part (a) with v = Mu we have
0 = v0 Mu = (Mu)0 Mu = kMuk2 .
Hence, Mu = 0 for all u, and it follows that M must be the zero matrix.
29. We have from the text that
is equal to
ν0 θ 0
CX CXY
CY X CY
ν
θ
ν 0CX ν + ν 0CXY θ + θ 0CY X ν + θ 0CY θ ,
which, upon noting that ν 0CXY θ is a scalar and therefore equal to its transpose, simplifies to
ν 0CX ν + 2θ 0CY X ν + θ 0CY θ .
(∗)
We also have from the text (via Problem 22) that
wH Kw = ν 0 (CX +CY )ν + θ 0 (CX +CY )θ + 2θ 0 (CY X −CXY )ν .
If (∗) is equal to wH Kw/2 for all ν and all θ , then in particular, this must hold for all
ν when θ = 0. This implies
ν 0CX ν = ν 0
CX +CY
ν
2
or
ν0
CX −CY
ν = 0.
2
Since ν is arbitrary, (CX −CY )/2 = 0, or CX = CY . This means that we can now write
wH Kw/2 = ν 0CX ν + θ 0CY θ + θ 0 (CY X −CXY )ν .
Comparing this with (∗) shows that
2θ 0CY X ν = θ 0 (CY X −CXY )ν
or
θ 0 (CY X +CXY )ν = 0.
Taking θ = ν arbitrary and noting that CY X +CXY is symmetric, it follows that CY X +
CXY = 0, and so CXY = −CY X .
162
30.
Chapter 9 Problem Solutions
(a) Since Γ is 2n × 2n, det(2Γ) = 22n det Γ. From the hint it follows that det Γ =
(det K)2 /22n .
(b) Write
VV −1 = (A + BCD)[A−1 − A−1 B(C−1 + DA−1 B)−1 DA−1 ]
= (A + BCD)A−1 [I − B(C−1 + DA−1 B)−1 DA−1 ]
= (I + BCDA−1 )[I − B(C−1 + DA−1 B)−1 DA−1 ]
= I + BCDA−1 − B(C−1 + DA−1 B)−1 DA−1
− BCDA−1 B(C−1 + DA−1 B)−1 DA−1
= I + BCDA−1
− B[I +CDA−1 B](C−1 + DA−1 B)−1 DA−1
= I + BCDA−1
− BC[C−1 + DA−1 B](C−1 + DA−1 B)−1 DA−1
= I + BCDA−1 − BCDA−1 = I.
(c) To begin, write
ΓΓ−1 =
CX −CY X
CY X CX
∆−1
CX−1CY X ∆−1
−1
−1
∆−1
−∆ CY X CX
CX ∆−1 +CY X ∆−1CY X CX−1
CY X ∆−1 −CY X ∆−1
−1
−1
−1
CY X ∆ −CX ∆ CY X CX CY X CX−1CY X ∆−1 +CX ∆−1
CX ∆−1 +CY X ∆−1CY X CX−1
0
=
CY X ∆−1 −CX ∆−1CY X CX−1 (CY X CX−1CY X +CX )∆−1
CX ∆−1 +CY X ∆−1CY X CX−1 0
=
.
CY X ∆−1 −CX ∆−1CY X CX−1 I
=
Using the hint that
∆−1 = CX−1 −CX−1CY X ∆−1CY X CX−1 ,
we easily obtain
CX ∆−1 = I −CY X ∆−1CY X CX−1 ,
from which it follows that
−1
ΓΓ
=
I
CY X ∆−1 −CX ∆−1CY X CX−1
0
.
I
To show that the lower-left block is also zero, use the hint to write
CY X ∆−1 −CX ∆−1CY X CX−1
= CY X [CX−1 −CX−1CY X ∆−1CY X CX−1 ] −CX ∆−1CY X CX−1
= CY X CX−1 −CY X CX−1CY X ∆−1CY X CX−1 −CX ∆−1CY X CX−1
= CY X CX−1 − [CY X CX−1CY X +CX ]∆−1CY X CX−1
= CY X CX−1 − ∆∆−1CY X CX−1 = 0.
Chapter 9 Problem Solutions
163
(d) Write
KK −1 = 2(CX + jCY X )(∆−1 − jCX−1CY X ∆−1 )/2
= CX ∆−1 +CY X CX−1CY X ∆−1 + j(CY X ∆−1 −CY X ∆−1 )
= ∆∆−1 = I.
(e) We begin with
x0
y0
−1
Γ
0 0
x
∆−1
CX−1CY X ∆−1
x
= x y
y
y
−∆−1CY X CX−1
∆−1
−1
−1
∆ x +CX CY X ∆−1 y
= x0 y0
−∆−1CY X CX−1 x + ∆−1 y
= x 0 ∆−1 x + x 0CX−1CY X ∆−1 y − y 0 ∆−1CY X CX−1 x + y 0 ∆−1 y.
Now, since each of the above terms on the third line is a scalar, each term is
equal to its transpose. In particular,
y 0 ∆−1CY X CX−1 x = x 0CX−1CXY ∆−1 y = −x 0CX−1CY X ∆−1 y.
Hence,
1
2
x
x 0 y 0 Γ−1
=
y
0 −1
−1
0 −1
1 0 −1
2 (x ∆ x + 2x CX CY X ∆ y + y ∆ y).
(∗)
We next compute
zH K −1 z =
=
=
1 0
2 (x −
1 0
2 (x −
1 0
2 (x −
jy 0 )(∆−1 − jCX−1CY X ∆−1 )(x + jy)
jy 0 )[(∆−1 x +CX−1CY X ∆−1 y) + j(∆−1 y −CX−1CY X ∆−1 x)]
jy 0 )[(∆−1 x +CX−1CY X ∆−1 y) + j(∆−1 y − ∆−1CY X CX−1 x)]
by the hint that CX−1CY X ∆−1 = ∆−1CY X CX−1 . We continue with
zH K −1 z =
=
0 −1
−1
−1
0 −1
−1
−1
1
2 [{x (∆ x +CX CY X ∆ y) + y (∆ y − ∆ CY X CX x)}
+ j{x 0 (∆−1 y − ∆−1CY X CX−1 x) − y 0 (∆−1 x +CX−1CY X ∆−1 y)}]
0 −1
0 −1
−1
0 −1
0 −1
−1
1
2 [{x ∆ x + x CX CY X ∆ y + y ∆ y − y ∆ CY X CX x}
+ j{x 0 ∆−1 y − x 0 ∆−1CY X CX−1 x − y 0 ∆−1 x − y 0CX−1CY X ∆−1 y}].
We now use the fact that since each of the terms in the last line is a scalar, it is
equal to its transpose. Also CXY = −CY X . Hence,
zH K −1 z =
0 −1
0 −1
−1
0 −1
1
2 [{x ∆ x + 2x CX CY X ∆ y + y ∆ y}
− j{x 0 ∆−1CY X CX−1 x + y 0CX−1CY X ∆−1 y}].
Since
x 0 ∆−1CY X CX−1 x = (x 0 ∆−1CY X CX−1 x)0 = −x 0CX−1CY X ∆−1 x = −x 0 ∆−1CY X CX−1 x,
and similarly for y 0CX−1CY X ∆−1 y, the two imaginary terms above are zero.
CHAPTER 10
Problem Solutions
1. Write
mX (t) := E[Xt ] = E[g(t, Z)]
= g(t, 1)P(Z = 1) + g(t, 2)P(Z = 2) + g(t, 3)P(Z = 3)
= p1 a(t) + p2 b(t) + p3 c(t),
and
RX (t, s) := E[Xt Xs ] = E[g(t, Z)g(s, Z)]
= g(t, 1)g(s, 1)p1 + g(t, 2)g(s, 2)p2 + g(t, 3)g(s, 3)p3
= a(t)a(s)p1 + b(t)b(s)p2 + c(t)c(s)p3 .
2. Imitating the derivation of the Cauchy–Schwarz inequality for random variables in
Chapter 2 of the text, write
0 ≤
=
Z ∞
−∞
Z ∞
−∞
|g(θ ) − λ h(θ )|2 d θ
2
|g(θ )| d θ − λ
−λ∗
Z ∞
−∞
λ =
h(θ )g(θ )∗ d θ
Z ∞
−∞
|h(θ )|2 d θ .
R∞
∗
−∞ g(θ )h(θ ) d θ
R∞
2
−∞ |h(θ )| d θ
to get
=
−∞
g(θ )h(θ )∗ d θ + |λ |2
Then put
0 ≤
Z ∞
R∞
2
g(θ )h(θ )∗ d θ
−∞
|g(θ )|2 d θ − R ∞
2
−∞
−∞ |h(θ )| d θ
2
2
R∞
R∞
∗
∗
−∞ g(θ )h(θ ) d θ
−∞ g(θ )h(θ ) d θ
− R∞
+ R
2
2
∞
2
−∞ |h(θ )| d θ
−∞ |h(θ )| d θ
Z ∞
Z ∞
−∞
2
|g(θ )| d θ −
R∞
∗
−∞ g(θ )h(θ ) d θ
R∞
2
−∞ |h(θ )| d θ
Rearranging yields the desired result.
164
2
.
Z ∞
−∞
|h(θ )|2 d θ
Chapter 10 Problem Solutions
165
3. Write
CX (t1 ,t2 ) = E[(Xt1 − mX (t1 ))(Xt2 − mX (t2 ))]
= E[Xt1 Xt2 ] − mX (t1 )E[Xt2 ] − E[Xt1 ]mX (t2 ) + mX (t1 )mX (t2 )
= E[Xt1 Xt2 ] − mX (t1 )mX (t2 ) − mX (t1 )mX (t2 ) + mX (t1 )mX (t2 )
= RX (t1 ,t2 ) − mX (t1 )mX (t2 ).
Similarly,
CXY (t1 ,t2 ) = E[(Xt1 − mX (t1 ))(Yt2 − mY (t2 ))]
= E[Xt1 Yt2 ] − mX (t1 )E[Yt2 ] − E[Xt1 ]mY (t2 ) + mX (t1 )mY (t2 )
= E[Xt1 Xt2 ] − mX (t1 )mY (t2 ) − mX (t1 )mY (t2 ) + mX (t1 )mY (t2 )
= RXY (t1 ,t2 ) − mX (t1 )mY (t2 ).
4. Write
0 ≤ E
n
=
n
∑ ci Xti
i=1
n
2
= E
n
∑ ci Xti
i=1
n
∑ ck Xtk
k=1
∗ n
=
n
∑ ∑ ci E[Xti Xtk ]c∗k
i=1 k=1
∑ ∑ ci RX (ti ,tk )c∗k .
i=1 k=1
5. Since Xt has zero mean, var(Xt ) = RX (t,t) = t. Thus, Xt ∼ N(0,t), and
2
e−x /(2t)
√
.
2π t
fXt (x) =
6. First note that by making the change of variable k = n − i, we have
∞
Yn =
∑
h(k)Xn−k .
k=−∞
(a) Write
mY (n) = E[Yn ] = E
∞
∑
h(k)Xn−k
k=−∞
∞
=
∑
h(k)E[Xn−k ]
∑
h(k)mX (n − k).
k=−∞
∞
=
k=−∞
(b) Write
E[XnYm ] = E Xn
∞
∑
k=−∞
h(k)Xm−k
∞
=
∑
h(k)E[Xn Xm−k ]
∑
h(k)RX (n, m − k).
k=−∞
∞
=
k=−∞
166
Chapter 10 Problem Solutions
(c) Write
E[YnYm ] = E
∞
∑
l=−∞
∞
∑
=
l=−∞
h(l)Xn−l Ym =
h(l)RXY (n − l, m) =
∞
∑
h(l)E[Xn−l Ym ]
l=−∞
∞
∑
l=−∞
h(l)
∞
∑
k=−∞
h(k)RX (n − l, m − k) .
7. Let Xt = cos(2π f t + Θ), where Θ ∼ uniform[−π , π ].
(a) Consider choices t1 = 0 and t2 = −(π /2)/(2π f ). Then Xt1 = cos(Θ) and Xt2 =
sin(Θ), which are not jointly continuous.
(b) Write
E[g(Xt )] = E[g(cos(2π f t + Θ))] =
=
Z π +2π f t
−π +2π f t
g(cos(τ ))
Z π
dτ
=
2π
−π
g(cos(2π f t + θ ))
Z π
−π
g(cos(τ ))
dθ
2π
dτ
,
2π
since the integrand has period 2π and the range of integration has length 2π .
Thus, E[g(Xt )] does not depend on t.
8.
(a) Using independence and a trigonometric identity, write
E[Yt1 Yt2 ] = E[Xt1 Xt2 cos(2π f t1 + Θ) cos(2π f t2 + Θ)]
= E[Xt1 Xt2 ]E[cos(2π f t1 + Θ) cos(2π f t2 + Θ)]
= 21 RX (t1 − t2 )E cos(2π f [t1 − t2 ]) + cos(2π f [t1 + t2 ] + 2Θ)
= 21 RX (t1 − t2 ) cos(2π f [t1 − t2 ]) + E cos(2π f [t1 + t2 ] + 2Θ)
{z
}
|
=0
=
1
2 RX (t1 − t2 ) cos(2π
f [t1 − t2 ]).
(b) A similar argument yields
E[Xt1 Yt2 ] = E[Xt1 Xt2 cos(2π f t2 + Θ)] = E[Xt1 Xt2 ] E[cos(2π f t2 + Θ)] = 0.
|
{z
}
=0
(c) It is clear that Yt is zero mean. Together with part (a) it follows that Yt is WSS.
9. By Problem 7(b), FXt (x) = P(Xt ≤ x) = E[I(−∞,x] (Xt )] does not depend on t, and so
we can restrict attention to the case t = 0. Since
√ X0 = cos(Θ) has the arcsine density
of Problem 35 in Chapter 5, f (x) = (1/π )/ 1 − x2 for |x| < 1.
10. Second-order strict stationarity means that for every two-dimensional set B, for every
t1 ,t2 , and ∆t, P((Xt1 +∆t , Xt2 +∆t ) ∈ B) does not depend on ∆t. In particular, this is true
whenever B has the form B = A × IR for any one-dimensional set A; i.e.,
P((Xt1 +∆t , Xt2 +∆t ) ∈ B) = P((Xt1 +∆t , Xt2 +∆t ) ∈ A × IR) = P(Xt1 ∈ A, Xt2 ∈ IR)
= P(Xt1 ∈ A).
does not depend on ∆t. Hence Xt is first-order strictly stationary.
Chapter 10 Problem Solutions
11.
167
(a) If p1 = p2 = 0 and p3 = 1, then Xt = c(t) = −1 with probability one. Then
E[Xt ] = −1 does not depend on t, and E[Xt1 Xt2 ] = (−1)2 = 1 depends on t1
and t2 only through their difference t1 − t2 ; in fact the correlation function is a
constant function of t1 − t2 . Thus, Xt is WSS.
(b) If p1 = 1 and p2 = p3 = 0, then Xt = e−|t| with probability one. Then E[Xt ] =
e−|t| depends on t. Hence, Xt is not WSS.
(c) First, the only way to have X0 = 1 is to have Xt = a(t) = e−|t| , which requires
Z = 1. Hence,
P(X0 = 1) = P(Z = 1) = p1 .
Second, the only way to have Xt ≤ 0 for 0 ≤ t ≤ 0.5 is to have Xt = c(t) = −1,
which requires Z = 3. Hence,
P(Xt ≤ 0, 0 ≤ t ≤ 0.5) = P(Z = 3) = p3 .
Third, the only way to have Xt ≤ 0 for 0.5 ≤ t ≤ 1 is to have Xt = b(t) = sin(2π t)
or Xt = c(t) = −1. Hence,
P(Xt ≤ 0, 0.5 ≤ t ≤ 1) = P(Z = 2 or Z = 3) = p2 + p3 .
12. With Yk := q(Xk , Xk+1 , . . . , Xk+L−1 ), write
E[e j(ν1Y1+m +···+νnYn+m ) ] = E[e j{ν1 q(X1+m ,...,Xm+L )+···+νn q(Xn+m ,...,Xn+m+L−1 )} ].
The exponential on the right is just a function of X1+m , . . . , Xn+L−1+m . Since the Xk
process is strictly stationary, the expectation on the right is unchanged if we replace
X1+m , . . . , Xn+L−1+m by X1 , . . . , Xn+L−1 ; i.e., the above right-hand side is equal to
E[e j{ν1 q(X1 ,...,XL )+···+νn q(Xn ,...,Xn+L−1 )} ] = E[e j(ν1Y1 +···+νnYn ) ].
Hence, the Yk process is strictly stationary.
13. We begin with
E[g(X0 )] = E[X0 I[0,∞) (X0 )] =
Z ∞
−∞
xI[0,∞) (x) ·
λ −λ |x|
dx
2e
=
1
2
Z ∞
0
x · λ e−λ x dx,
which is just 1/2 times the expectation of an exp(λ ) random variable. We thus see
that E[g(X0 )] = 1/(2λ ). Next, for n 6= 0, we compute
Z ∞
2
Z
∞
2
1
e−x /2
dx = √
xe−x /2 dx
E[g(Xn )] =
xI[0,∞) (x) · √
−∞
2π
2π 0
2 ∞
1
1
= √
−e−x /2
= √ .
0
2π
2π
Hence, E[g(X0 )] 6= E[g(Xn )] for n 6= 0, and it follows that Xk is not strictly stationary.
168
Chapter 10 Problem Solutions
14. First consider the mean function,
E[q(t + T )] =
1
T0
Z T0
q(t + θ ) d θ =
0
1
T0
Z t+T0
q(τ ) d τ =
t
1
T0
Z T0
q(τ ) d τ ,
0
where we have used the fact that since q has period T0 , the integral of q over any interval of length T0 yields the same result. The second thing to consider is the correlation
function. Write
Z
1 T0
q(t1 + θ )q(t2 + θ ) d θ
T0 0
Z
1 t2 +T0
=
q(t1 + τ − t2 )q(τ ) d τ
T0 t2
Z
1 T0
=
q([t1 − t2 ] + τ )q(τ ) d τ ,
T0 0
E[q(t1 + T )q(t2 + T )] =
where we have used the fact that as a function of τ , the product q([t1 − t2 ] + τ )q(τ )
has period T0 . Since the mean function does not depend on t, and since the correlation
function depends on t1 and t2 only through their difference, Xt is WSS.
15. For arbitrary functions h, write
E[h(Xt1 +∆t , . . . , Xtn +∆t )] = E h q(t1 + ∆t + T ), . . . , q(tn + ∆t + T )
Z
1 T0
h q(t1 + ∆t + θ ), . . . , q(tn + ∆t + θ ) d θ
T0 0
Z
1 ∆t+T0
=
h q(t1 + τ ), . . . , q(tn + τ ) d τ
T0 ∆t
Z
1 T0
=
h q(t1 + τ ), . . . , q(tn + τ ) d τ ,
T0 0
=
where the last step follows because we are integrating a function of period T0 over an
interval of length T0 . Hence,
E[h(Xt1 +∆t , . . . , Xtn +∆t )] = E[h(Xt1 , . . . , Xtn )],
and we see that Xt is strictly stationary.
16. First write E[Yn ] = E[Xn − Xn−1 ] = E[Xn ] − E[Xn−1 ] = 0 since E[Xn ] does not depend
on n. Next, write
E[YnYm ] = E[(Xn − Xn−1 )(Xm − Xm−1 )]
= RX (n − m) − RX ([n − 1] − m) − RX (n − [m − 1]) + RX (n − m)
= 2RX (n − m) − RX (n − m − 1) − RX (n − m + 1),
which depends on n and m only through their difference. Hence, Yn is WSS.
√
2
17. From the Fourier transform table, SX ( f ) = 2π e−(2π f ) /2 .
18. From the Fourier transform table, SX ( f ) = π e−2π | f | .
Chapter 10 Problem Solutions
19.
169
(a) Since correlation functions are real and even, we can write
SX ( f ) =
Z ∞
RX (τ )e− j2π f τ d τ =
−∞
Z ∞
= 2
0
Z ∞
−∞
RX (τ ) cos(2π f τ ) d τ
RX (τ ) cos(2π f τ ) d τ = 2 Re
Z ∞
0
RX (τ )e− j2π f τ d τ .
(b) OMITTED.
(c) The requested plot is at the left; at the right the plot is focused closer to f = 0.
2
2
1
1
0
−6 −3
0
3
0
−1 −0.5
6
0
0.5
1
(d) The requested plot is at the left; at the right the plot is focused closer to f = 0.
4
4
2
2
0
−6 −3
20.
0
3
0
−1 −0.5
6
0
0.5
1
(a) RX (n) = E[Xk+n Xk ] = E[Xk Xk+n ] = RX (−n).
(b) Since RX (n) is real and even, we can write
∞
SX ( f ) =
∑
∞
RX (n)e− j2π f n =
∑
n=−∞
n=−∞
∞
∞
=
∑
n=−∞
RX (n) cos(2π f n) − j
∞
=
∑
RX (n)[cos(2π f n) − j sin(2π f n)]
RX (n) cos(2π f n),
∑
n=−∞
|
odd function of n
}|
{
z
RX (n) sin(2π f n)
{z
=0
n=−∞
which is a real and even function of f .
21.
(a) Since correlation functions are real and even, we can write
∞
SX ( f ) =
∑
n=−∞
RX (n)e− j2π f n
}
170
Chapter 10 Problem Solutions
∞
= RX (0) + ∑ RX (n)e− j2π f n +
−1
∑
n=1
∞
n=−∞
∞
n=1
∞
k=1
RX (n)e− j2π f n
= RX (0) + ∑ RX (n)e− j2π f n + ∑ RX (−k)e j2π f k
= RX (0) + 2 ∑ RX (n) cos(2π f n)
n=1
∞
= RX (0) + 2 Re ∑ RX (n)e− j2π f n .
n=1
(b) OMITTED.
(c) Here is the plot:
2
1
0
−0.5
0
0.5
0
0.5
(d) Here is the plot:
4
2
0
−0.5
22. Write
Z ∞
−∞
h(−t)e− j2π f t dt =
=
Z ∞
−∞
Z ∞
−∞
h(θ )e− j2π f (−θ ) d θ
h(θ )e j2π f θ d θ
∗
h(θ )∗ e− j2π f θ d θ
−∞
Z ∞
∗
− j2π f θ
dθ ,
h(θ )e
=
=
Z
∞
since h is real,
−∞
∗
= H( f ) .
23. We begin with SXY ( f ) = H( f )∗ SX ( f ) = ( j2π f )∗ SX ( f ) = − j2π f SX ( f ). It then follows that
2
2
d
d
RXY (τ ) = − RX (τ ) = − e−τ /2 = τ e−τ /2 .
dτ
dτ
Chapter 10 Problem Solutions
171
Similarly, since SY ( f ) = |H( f )|2 SX ( f ) = −( j2π f )( j2π f )SX ( f ),
RY (τ ) = −
2
d2
d
d −τ 2 /2
= e−τ /2 (1 − τ 2 ).
τe
RX (τ ) =
RXY (τ ) =
2
dτ
dτ
dτ
24. Since RX (τ ) = 1/(1 + τ 2 ), we have from the transform table that SX ( f ) = π e−2π | f | .
Similarly, since h(t) = 3 sin(π t)/(π t), we have from the transform table that H( f ) =
3I[−1/2,1/2] ( f ). We can now write
SY ( f ) = |H( f )|2 SX ( f ) = 9I[−1/2,1/2] ( f ) · π e−2π | f | = 9π e−2π | f | I[−1/2,1/2] ( f ).
√
2
2π e−(2π f ) /2 .
√
√
2
2
2
(a) SXY ( f ) = H( f )∗ SX ( f ) = e−(2π f ) /2 ]∗ 2π e−(2π f ) /2 = 2π e−(2π f ) .
25. First note that since RX (τ ) = e−τ
2 /2
, SX ( f ) =
(b) Writing
√ 2
2
1 √ √
SXY ( f ) = √ 2π 2e−( 2) (2π f ) /2 ,
2
we have from the transform table that
√ 2
2
1
1
RXY (τ ) = √ e−(τ / 2) /2 = √ e−τ /4 .
2
2
(c) Write
2
1
E[Xt1 Yt2 ] = RXY (t1 − t2 ) = √ e−(t1 −t2 ) /4 .
2
√
√
2
2
2
(d) SY ( f ) = |H( f )|2 SX ( f ) = e−(2π f ) · 2π e−(2π f ) /2 = 2π e−3(2π f ) /2 .
(e) Writing
√ 2
2
1 √ √
SY ( f ) = √ 2π 3e−( 3) (2π f ) /2 ,
3
we have from the transform table that
√ 2
2
1
1
RY (τ ) = √ e−(τ / 3) /2 = √ e−τ /6 .
3
3
26. We have from the transform table that SX ( f ) = [sin(π f )/(π f )]2 . The goal is to
choose a filter H( f ) so that RY (τ ) = sin(πτ )/(πτ ); i.e., so that SY ( f ) = I[−1/2,1/2] ( f ).
Thus, the formula SY ( f ) = |H( f )|2 SX ( f ) becomes
I[−1/2,1/2] ( f ) = |H( f )|2
We therefore take


sin(π f )
πf
2
πf
, | f | ≤ 1/2,
H( f ) =
sin(π f )

0,
| f | > 1/2.
.
172
Chapter 10 Problem Solutions
27. Since Yt and Zt are responses of LTI systems to a WSS input, Yt and Zt are individually
WSS. If we can show that E[Yt1 Zt2 ] depends on t1 and t2 only throught their difference,
then Yt and Zt will be J-WSS. We show this to be the case. Write
Z ∞
Z ∞
E[Yt1 Zt2 ] = E
h(θ )Xt1 −θ d θ
g(τ )Xt2 −τ d τ
−∞
=
=
−∞
Z ∞Z ∞
−∞ −∞
Z ∞Z ∞
−∞ −∞
h(θ )g(τ )E[Xt1 −θ Xt2 −τ ] d τ d θ
h(θ )g(τ )RX ([t1 − θ ] − [t2 − τ ]) d τ d θ ,
which depends on t1 − t2 as claimed.
28. Observe that if h(t) := δ (t) − δ (t − 1), then
R∞
−∞ h(τ )Xt−τ d τ
= Xt − Xt−1 .
(a) Since Yt := Xt − Xt−1 is the response of an LTI system to a WSS input, Xt and Yt
is J-WSS.
(b) Since H( f ) = 1 − e− j2π f ,
|H( f )|2 = (1 − e− j2π f )(1 − e j2π f ) = 2 − e j2π f − e− j2π f
= 2−2
e j2π f + e− j2π f
= 2[1 − cos(2π f )],
2
we have
SY ( f ) = |H( f )|2 SX ( f ) = 2[1 − cos(2π f )] ·
29. In Yt =
Rt
t−3 Xτ
4[1 − cos(2π f )]
2
=
.
1 + (2π f )2
1 + (2π f )2
d τ , make the change of variable θ = t − τ , d θ = −d τ to get
Yt =
Z 3
0
Xt−θ d θ =
Z ∞
−∞
I[0,3] (θ )Xt−θ d θ .
This shows that Yt is the response to Xt of the LTI system with impulse h(t) = I[0,3] (t).
Hence, Yt is WSS.
30. Apply the formula
H(0) =
(1/π )[(sint)/t]2 .
with h(t) =
and we find that
1
1 =
π
31.
Z ∞
sint 2
−∞
t
Z
∞
−∞
h(t)e− j2π f t dt
f =0
Then from the table, H( f ) = (1 − π | f |)I[−1/π ,1/π ] ( f ),
dt,
which is equivalent to π =
Z ∞
sint 2
−∞
t
dt.
(a) Following the hint, we have
∞
E[XnYm ] =
∑
k=−∞
∞
h(k)RX (n, m − k) =
∑
k=−∞
h(k)RX (n − m + k).
Chapter 10 Problem Solutions
173
(b) Similarly,
∞
∑
E[YnYm ] =
l=−∞
∞
=
∑
h(l)
∑
h(l)
l=−∞
∞
=
h(l)
h(k)RX (n − l, m − k)
∞
∑
k=−∞
∞
h(k)RX (n − l − [m − k])
∑
k=−∞
∞
h(k)RX ([n − m] − [l − k]) .
∑
k=−∞
l=−∞
(c) By part (a),
∞
∑
RX (n) =
h(k)RX (n + k),
k=−∞
and so
∞
SXY ( f ) =
∑
RX (n)e
∑
h(k)
∑
h(k)
n=−∞
∞
=
k=−∞
∞
=
=
=
∞
∑
− j2π f n
RX (n + k)e
∑
RX (m)e− j2π f (m−k)
m=−∞
j2π f k
h(k)e
∑
h(k)e− j2π f k
k=−∞
∗
∞
h(k)RX (n + k) e− j2π f n
∑
∑
k=−∞
∞
∑
=
∞
n=−∞ k=−∞
n=−∞
∞
k=−∞
∞
∞
− j2π f n
− j2π f m
∑
RX (m)e
m=−∞
∗
SX ( f ),
since h(k) is real,
= H( f ) SX ( f ).
(d) By part (b),
∞
∑
RY (n) =
l=−∞
and so
∑
∑
h(l)
∑
h(l)
∑
h(l)e− j2π f l
∞
SY ( f ) =
n=−∞
∞
=
∞
∑
l=−∞
∞
=
l=−∞
h(l)
l=−∞
∞
l=−∞
∞
=
h(l)
∞
∑
k=−∞
∞
∑
k=−∞
∞
∑
h(k)
∑
h(k)
h(k)RX (n − [l − k]) ,
h(k)RX (n − [l − k]) e− j2π f n
∑
RX (n − [l − k])e− j2π f n
∑
RX (m)e− j2π f (m+[l−k])
n=−∞
∞
k=−∞
∞
k=−∞
m=−∞
∞
∑
k=−∞
h(k)e j2π f k
∞
∑
m=−∞
= H( f )H( f )∗ SX ( f ) = |H( f )|2 SX ( f ).
RX (m)e− j2π f m
174
Chapter 10 Problem Solutions
32. By the hint,
Z
1 T
|x(t)|2 dt =
2T 0
We first observe that since
τ
T0 ≤ T0 + =
n
nE0
1
+
2T
2T
Z τ
0
|x(t)|2 dt.
T
T0
≤ T0 +
→ T0 ,
n
n
T /n → T0 . It follows that
nE0
E0
E0
.
=
→
2T
2(T /n)
2T0
We next show that the integral on the right goes to zero. Write
1
2T
Z τ
0
|x(t)|2 dt ≤
1
2T
Z T0
0
|x(t)|2 dt =
E0
→ 0.
2T
A similar argument shows that
1
2T
Z 0
−T
|x(t)|2 dt → E0 /(2T0 ).
Putting this all together shows that (1/2T )
33. Write
E
Z
∞
−∞
Xt2 dt
R
=
Z ∞
−∞
RT
−T
|x(t)|2 dt → E0 /T0 .
E[Xt2 ] dt =
Z ∞
−∞
RX (0) d τ = ∞.
34. If the function q(W ) := 0W S1 ( f ) − S2 ( f ) d f is identically zero, then so is its derivative, q0 (W ) = S1 ( f ) − S2 ( f ). But then S1 ( f ) = S2 ( f ) for all f ≥ 0.
35. If h(t) = I[−T,T ] (t), and white noise is applied to the corresponding system, the cross
power spectral density of the input and output is
SXY ( f ) = H( f )∗ N0 /2 = 2T
sin(2π T f ) N0
· ,
2π T f
2
which is real and even, but not nonnegative. Similarly, if h(t) = e−t I[0,∞) (t),
SXY ( f ) =
N0 /2
,
1 + j2π f
which is complex valued.
36. First write
2
SY ( f ) = |H( f )| N0 /2 =
(1 − f 2 )2 N0 /2, | f | ≤ 1,
0,
| f | > 1.
Then
PY =
Z ∞
−∞
SY ( f ) d f =
N0
2
2
i1
h
= N0 f − 32 f 3 + 15 f 5
0
Z 1
0
(1 − 2 f 2 + f 4 ) d f
= N0 [1 − 23 + 51 ] = 8N0 /15.
Chapter 10 Problem Solutions
175
2
/2 has power spectral density S ( f ) = e− f
37. First note that RX (τ ) = e−(2πτ ) p
X
Using the definition of H( f ) = | f | I[−1,1] ( f ),
E[Yt2 ] = RY (0) =
Z ∞
Z ∞
2 /2
Z 1
√
/ 2π .
SY ( f ) d f =
|H( f )|2 SX ( f ) d f =
| f |SX ( f ) d f
−∞
−∞
−1
r
r
Z 1
Z
2 1 − f 2 /2
2 − f 2 /2 1
= 2
f SX ( f ) d f =
−e
fe
df =
π 0
π
0
0
r
2
.
= (1 − e−1/2 )
π
38. First observe that
2
SY ( f ) = |H( f )| SX ( f ) =
sin π f
πf
2
N0 /2.
This is the Fourier transform of
RY (τ ) = (1 − |τ |)I[−1,1] (τ ) N20 .
Then PY = RY (0) = N0 /2.
39.
(a) First note that
H( f ) =
1
1/(RC)
=
.
1/(RC) + j2π f
1 + j(2π f )RC
Then
SXY ( f ) = H( f )∗ SX ( f ) =
N0 /2
.
1 − j(2π f )RC
(b) The inverse Fourier transform of SXY ( f ) = H( f )∗ N0 /2 is
RXY (τ ) = h(−τ )∗ N0 /2 = h(−τ )N0 /2,
where the last step follows because h is real. Hence,
RXY (τ ) =
N0 τ /(RC)
u(−τ ).
e
2RC
(c) E[Xt1 Yt2 ] = RXY (t1 − t2 ) = (N0 /(2RC))e(t1 −t2 )/(RC) u(t2 − t1 ).
(d) SY ( f ) = |H( f )|2 SX ( f ) =
N0 /2
.
1 + (2π f RC)2
(e) Since
N0 /2
N0
SY ( f ) =
=
·
1 + (2π f RC)2
2(RC)2
we have that
RY (τ ) =
2(1/(RC))
1 2
+ (2π f )2
RC
N0 −|τ |/(RC)
e
.
4RC
RC
,
2
176
Chapter 10 Problem Solutions
(f) PY = RY (0) = N0 /(4RC).
40. To begin, write E[Yt+1/2Yt ] = RY (1/2). Next, since the input has power spectral density N0 /2 and since h(t) = 1/(1 + t 2 ) has transform H( f ) = π e−2π | f | , we can write
SY ( f ) = |H( f )|2 SX ( f ) = |π e−2π | f | |2 N20 = π 2 e−4π | f | N20 =
π N0
2
· π e−2π (2)| f | .
From the transform table, we conclude that
RY (τ ) =
π N0
π N0
2
·
=
,
2 4 + τ2
4 + τ2
and so E[Yt+1/2Yt ] = RY (1/2) = 4π N0 /17.
41. Since H( f ) = sin(π T f )/(π T f ), we can write
SY ( f ) = |H( f )|2
Then
RY (τ ) =
sin π T f 2 N0
N0
= T
.
2
πT f
2T
N0
(1 − |τ |/T )I[−T,T ] (τ ).
2T
42. To begin, write
−t
Yt = e
Z t
θ
−∞
e Xθ d θ =
Z t
−∞
−(t−θ )
e
Xθ d θ =
Z ∞
−∞
e−(t−θ ) u(t − θ )Xθ d θ ,
where u is the unit-step function. We then see that Yt is the response to Xt of the
LTI system with impulse response h(t) := e−t u(t). Hence, we know from the text
that Xt and Yt are jointly wide-sense stationary. Next, since SX ( f ) = N0 /2, RX (τ ) =
(N0 /2)δ (τ ). We then compute in the time domain,
RXY (τ ) =
Z ∞
−∞
h(−α )RX (τ − α ) d α =
N0
2
N0 τ
e u(−τ ).
=
2
Z ∞
−∞
h(−α )δ (τ − α ) d α =
Next,
SXY ( f ) = H( f )∗ SX ( f ) =
and
SY ( f ) = |H( f )|2 SX ( f ) =
N0 /2
,
1 − j2π f
N0 /2
.
1 + (2π f )2
It then follows that RY (τ ) = (N0 /4)e−|τ | .
43. Consider the impulse response
∞
h(τ ) :=
∑
n=−∞
hn δ (τ − n).
N0
h(−τ )
2
Chapter 10 Problem Solutions
177
Then
Z ∞
−∞
h(τ )Xt−τ d τ =
=
Z ∞ ∞
∑
hn δ (τ − n) Xt−τ d τ
−∞ n=−∞
Z ∞
∞
∑
hn
n=−∞
−∞
Xt−τ δ (τ − n) d τ =
∞
∑
hn Xt−n =: Yt .
n=−∞
(a) Since Yt is the response of the LTI system with impulse response h(t) to the
WSS input Xt , Xt and Yt are J-WSS.
(b) Since Yt is the response of the LTI system with impulse response h(t) to the
WSS input Xt , SY ( f ) = |H( f )|2 SX ( f ), where
Z ∞ ∞
Z ∞
δ
(
τ
−
n)
e− j2π f τ d τ
h
H( f ) =
h(τ )e− j2π f τ d τ =
∑ n
−∞
∞
=
∑
n=−∞
hn
Z ∞
−∞
−∞ n=−∞
δ (τ − n)e− j2π f τ d τ =
∞
∑
hn e− j2π f n
n=−∞
has period one. Hence, P( f ) = |H( f )|2 is real, nonnegative, and has period one.
44. When the input power spectral density is SW ( f ) = 3, the output power spectral density
2
is |H( f )|2 · 3. We are also told that this output power spectral density is equal to e− f .
2
2
2
Hence, |H( f )|2 · 3 = e− f , or |H( f )|2 = e− f /3. Next, if SX ( f ) = e f I[−1,1] ( f ), then
2
2
SY ( f ) = |H( f )|2 SX ( f ) = (e− f /3) · e f I[−1,1] ( f ) = (1/3)I[−1,1] ( f ). It then follows
that
1 sin(2πτ )
2 sin(2πτ )
RY (τ ) = · 2
= ·
.
3
2πτ
3
2πτ
45. Since H( f ) = GI[−B,B] ( f ) and Yt is the response to white noise, the output power
spectral density is SY ( f ) = G2 I[−B,B] ( f ) · N0 /2, and so
RY (τ ) =
sin(2π Bτ )
G2 N0
sin(2π Bτ )
· 2B
= G2 BN0 ·
.
2
2π Bτ
2π Bτ
Note that
RY (k∆t) = RY (k/(2B)) = G2 BN0 ·
sin(2π Bk/(2B))
sin(π k)
= G2 BN0 ·
,
2π Bk/(2B)
πk
which is G2 BN0 for k = 0 and zero otherwise. It is obvious that the Xi are zero mean.
Since E[Xi X j ] = RY (i − j), and the Xi are uncorrelated with variance E[Xi2 ] = RY (0) =
G2 BN0 .
46.
∗ ])∗ =
(a) First write RX (τ ) = E[Xt+τ Xt∗ ]. Then RX (−τ ) = E[Xt−τ Xt∗ ] = (E[Xt Xt−
τ
∗
RX (τ ) .
(b) Since
SX ( f ) =
Z ∞
−∞
RX (τ )e− j2π f τ d τ ,
178
Chapter 10 Problem Solutions
we can write
SX ( f )∗ =
Z ∞
−∞
RX (τ )∗ e j2π f τ d τ =
=
Z ∞
−∞
Z ∞
−∞
RX (−t)∗ e− j2π f t dt
RX (t)e− j2π f t dt,
by part (a),
= SX ( f ).
Since SX ( f ) is equal to its complex conjugate, SX ( f ) is real.
(c) Write
Z
E[Xt1 Yt∗2 ] = E Xt1
=
Z ∞
∞
∗ h(θ )Xt2 −θ d θ
−∞
=
h(θ )∗ RX ([t1 − t2 ] + θ ) d θ =
−∞
(d) By part (c),
RXY (τ ) =
Z ∞
Z ∞
h(θ )∗ E[Xt1 Xt∗2 −θ ] d θ
−∞
Z ∞
h(−β )∗ RX ([t1 − t2 ] − β ) d β .
−∞
h(−β )∗ RX (τ − β ) d β ,
−∞
h(−·)∗ and
which is the convolution of
RX . Hence, the transform of this equation is the product of the transform of h(−·)∗ and SX . We just have to observe
that
Z ∞
∗
Z ∞
Z ∞
h(−β )∗ e− j2π f β d β =
h(t)e− j2π f t dt = H( f )∗ .
h(t)∗ e j2π f t dt =
−∞
−∞
−∞
Hence, SXY ( f ) = H( f )∗ SX ( f ). Next, since
Z ∞
∗
RY (τ ) = E[Yt+τ Yt ] = E
h(θ )Xt+τ −θ d θ Yt∗
=
Z ∞
−∞
−∞
h(θ )E[Xt+τ −θ Yt∗ ] d θ
=
Z ∞
−∞
h(θ )RXY (τ − θ ) d θ ,
is a convolution, its transform is
SY ( f ) = H( f )SXY ( f ) = H( f )H( f )∗ SX ( f ) = |H( f )|2 SX ( f ).
47.
(a) RX (τ ) =
(b) Write
Z ∞
−∞
RX (τ ) =
SX ( f )e j2π f τ d f =
Z ∞
−∞
Z ∞
−∞
δ ( f )e j2π f τ d f = e j2π 0τ = 1.
[δ ( f − f0 ) + δ ( f + f0 )]e j2π f τ d f = e j2π f0 τ + e− j2π f0 τ
= 2 cos(2π f0 τ ).
(c) First write
SX ( f ) = e− f
2 /2
h
1 2
2
1 √ i 2π
= e−( 2π ) (2π f ) /2 ·
· 2π √
2π
2π
i√
h
√
1
2
2
1
= e−( 2π ) (2π f ) /2 ·
· 2π 2π .
2π
Chapter 10 Problem Solutions
179
From the Fourier transform table with σ = 1/(2π ),
RX (τ ) =
√
2π e−(2πτ )
2 /2
.
(d) From the transform table with λ = 1/(2π ),
RX (τ ) =
2
1
1/(2π )
=
.
·
π (1/(2π ))2 + τ 2
1 + (2πτ )2
48. Write
E[Xt2 ] = RX (0) =
49.
Z
∞
SX ( f )e j2π f τ d f
−∞
=
τ =0
Z ∞
SX ( f ) d f =
−∞
Z W
−W
1 d f = 2W.
(a) e− f u( f ) is not even.
2
(b) e− f cos( f ) is not nonnegative.
(c) (1 − f 2 )/(1 + f 4 ) is not nonnegative.
(d) 1/(1 + j f 2 ) is not real valued.
50.
(a) Since sin τ is odd, it is NOT a valid correlation function.
(b) Since the Fourier transform of cos τ is [δ ( f − 1) + δ ( f + 1)]/2, which is real,
even, and nonnegative, cos τ IS a valid correlation function.
√
2
2
(c) Since the Fourier transform of e−τ /2 is 2π e−(2π f ) /2 , which is real, even, and
2
nonnegative, e−τ /2 IS a valid correlation function.
(d) Since the Fourier transform of e−|τ | is 2/[1 + (2π f )2 ], which is real, even, and
nonnegative, e−|τ | IS a valid correlation function.
(e) Since the value of τ 2 e−|τ | at τ = 0 is less than the value for other values of τ ,
τ 2 e−|τ | is NOT a valid correlation function.
(f) Since the Fourier transform of I[−T,T ] (τ ) is (2T ) sin(2π T f )/(2π T f ) is not nonnegative, I[−T,T ] (τ ) is NOT a valid correlation function.
51. Since R0 (τ ) is a correlation function, S0 ( f ) is real, even, and nonnegative. Since
R(τ ) = R0 (τ ) cos(2π f0 τ ),
S( f ) =
1
2 [S0 ( f
− f0 ) + S0 ( f + f0 )],
which is obviously real and nonnegative. It is also even since
S(− f ) =
1
2 [S0 (− f − f 0 ) + S0 (− f + f 0 )]
1
2 [S0 ( f + f 0 ) + S0 ( f − f 0 )], since
=
= S( f ).
S0 is even,
Since S( f ) is real, even, and nonnegative, R(τ ) is a valid correlation function.
180
Chapter 10 Problem Solutions
52. First observe that the Fourier transform of R(τ ) = R(τ − τ0 ) + R(τ + τ0 ) is S( f ) =
2S( f ) cos(2π f τ0 ). Hence, the answer cannot be (a) because it is possible to have
S( f ) > 0 and cos(2π f τ0 ) < 0 for some values of f . Let S( f ) = I[−1/(4τ0 ),1/(4τ0 )] ( f ),
which is real, even, and nonnegative. Hence, its inverse transform, which we denote
by R(τ ), is a correlation function. In this case, S( f ) = 2S( f ) cos(2π f τ0 ) ≥ 0 for all f ,
and is real and even too. Hence, for this choice of R(τ ), R(τ ) is a correlation function.
Therefore, the answer is (b).
53. To begin, write
R(τ ) =
Z ∞
−∞
S( f )e
j2π f τ
df =
Z ∞
−∞
S( f )[cos(2π f τ ) − j sin(2π f τ )] d f .
Since S is real and even, the integral of S( f ) sin(2π f τ ) is zero, and we have
R(τ ) =
Z ∞
−∞
S( f ) cos(2π f τ ) d f ,
which is a real and even function of τ . Finally,
|R(τ )| =
=
Z ∞
Z
−∞
∞
−∞
S( f )e j2π f τ d f ≤
|S( f )| d f =
Z ∞
−∞
Z ∞
−∞
S( f )e j2π f τ d f
S( f )d f = R(0).
54. Let S0 ( f ) denote the Fourier transform of R0 (τ ), and let S( f ) denote the Fourier
transform of R(τ ).
(a) The derivation in the text showing that the transform of a correlation function
is real and even uses only the fact that correlation functions are real and even.
Hence, S0 ( f ) is real and even. Furthermore, since R is the convolution of R0
with itself, S( f ) = S0 ( f )2 , which is real, even, and nonnegative. Hence, R(τ ) is
a correlation function.
(b) If R0 (τ ) = I[−T,T ] (τ ), then
sin(2π T f )
S0 ( f ) = 2T
2π T f
and
sin(2π T f )
S( f ) = 2T · 2T
2π T f
Hence, R(τ ) = 2T · 1 − |τ |/(2T ) I[−2T,2T ] (τ ).
2
55. Taking α = N0 /2 as in the text,
h(t) = v(t0 − t) = sin(t0 − t)I[0,π ] (t0 − t).
Then h is causal for t0 ≥ π .
56. Since v(t) = e(t/
√ 2
√ √
2 ) /2 , V ( f ) = 2π 2e−2(2π f )2 /2
√
2
= 2 π e−(2π f ) . Then
√
2
V ( f )∗ e− j2π f t0
2 π e−(2π f ) e− j2π f t0
H( f ) = α
= α
2
SX ( f )
e−(2π f ) /2
√
√
√
2
2
= 2α π e−(2π f ) /2 e− j2π f t0 = 2 α · 2π e−(2π f ) /2 e− j2π f t0 ,
√
2
and it follows that h(t) = 2 α e−(t−t0 ) /2 .
.
Chapter 10 Problem Solutions
181
57. Let v0 (n) := ∑k h(n − k)v(k) and Yn := ∑k h(n − k)Xk . The SNR is v0 (n0 )2 /E[Yn20 ]. We
have
Z
Z
1/2
E[Yn20 ] =
−1/2
1/2
SY ( f ) d f =
−1/2
|H( f )|2 SX ( f ) d f .
Let V ( f ) := ∑k v(k)e− j2π f k . Then
|v0 (n0 )|2 =
=
≤
Z 1/2
2
−1/2
H( f )V ( f )e j2π f n0 d f
Z 1/2
p
V ( f )e j2π f n0
H( f ) SX ( f ) · p
df
−1/2
SX ( f )
Z 1/2
−1/2
|H( f )|2 SX ( f ) d f
2
Z 1/2
|V ( f )∗ e− j2π f n0 |2
SX ( f )
−1/2
d f,
with equality if and only if
p
V ( f )∗ e− j2π f n0
H( f ) SX ( f ) = α p
SX ( f )
(#)
for some constant α . It is now clear that the SNR is upper bounded by
Z 1/2
|V ( f )∗ e− j2π f n0 |2
−1/2
SX ( f )
df
and that the SNR equals the bound if and only if (#) holds with equality for some
constant α . Hence, the matched filter transfer function is
H( f ) = α
V ( f )∗ e− j2π f n0
.
SX ( f )
58. We begin by writing
E[Ut ] = E[Vt + Xt ] = E[Vt ] + E[Xt ],
which does not depend on t since Vt and Xt are each individually WSS. Next write
E[Ut1 Vt2 ] = E[(Vt1 + Xt1 )Vt2 ] = RV (t1 − t2 ) + RXV (t1 − t2 ).
(∗)
Now write
E[Ut1 Ut2 ] = E[Ut1 (Vt2 + Xt2 ] = E[Ut1 Vt2 ] + E[Ut1 Xt2 ].
By (∗), the term E[Ut1 Vt2 ] depends on t1 and t2 only through their difference. Since
E[Ut1 Xt2 ] = E[(Vt1 + Xt1 )Xt2 ] = RV X (t1 − t2 ) + RX (t1 − t2 ),
it follows that E[Ut1 Ut2 ] depends on t1 and t2 only through their difference. Hence, Ut
and Vt are J-WSS.
182
Chapter 10 Problem Solutions
59. The assumptions in the problem imply that Vt and Xt are J-WSS, and by the preceding
problem, it follows that Ut and Vt are J-WSS. We can therefore apply the formulas
for the Wiener filter derived in the text. It just remains to compute the quantities used
in the formulas. First,
RVU (τ ) = E[Vt+τ Ut ] = E[Vt+τ (Vt + Xt )] = RV (τ ) + RXV (τ ) = RV (τ ),
which implies SVU ( f ) = SV ( f ). Similarly,
RU (τ ) = E[Ut+τ Ut ] = E[(Vt+τ + Xt+τ )Ut ]
= RVU (τ ) + E[Xt+τ (Vt + Xt )]
= RV (τ ) + RXV (τ ) + RX (τ ) = RV (τ ) + RX (τ ),
and so SU ( f ) = SV ( f ) + SX ( f ). We then have
H( f ) =
SV ( f )
SVU ( f )
=
.
SU ( f )
SV ( f ) + SX ( f )
60. The formula for RV (τ ) implies SV ( f ) = (1 − | f |)I[−1,1] ( f ). We then have
(1 − | f |)I[−1,1] ( f )
SV ( f )
=
SV ( f ) + SX ( f )
(1 − | f |)I[−1,1] ( f ) + 1 − I[−1,1] ( f )
H( f ) =
(1 − | f |)I[−1,1] ( f )
= I[−1,1] ( f ),
1 − | f |I[−1,1] ( f )
=
and so
h(t) = 2
sin(2π t)
.
2π t
61. To begin, write
E[|Vt − Vbt |2 ] = E[(Vt − Vbt )(Vt − Vbt )] = E[(Vt − Vbt )Vt ] − E[(Vt − Vbt )Vbt ]
= E[(Vt − Vbt )Vt ], by the orthogonality principle,
= E[Vt2 ] − E[Vbt Vt ] = RV (0) − E[Vbt Vt ] =
Next observe that
Z
E[Vbt Vt ] = E
=
=
=
∞
−∞
Z ∞
−∞
Z ∞
−∞
Z ∞
Z
h(θ )Ut−θ d θ Vt =
∞
−∞
Z ∞
−∞
h(θ )E[Vt Ut−θ ] d θ
h(θ )RVU (θ ) d θ =
Z ∞
H( f )SVU ( f )∗ d f ,
by Parseval’s formula,
−∞
SVU ( f )
SVU ( f )∗ d f =
−∞ SU ( f )
h(θ )RVU (θ )∗ d θ ,
Z ∞
|SVU ( f )|2
−∞
SV ( f ) d f − E[Vbt Vt ].
SU ( f )
since RVU (θ ) is real,
d f.
Putting these two observations together yields
E[|Vt − Vbt |2 ] =
Z ∞
−∞
SV ( f ) d f −
Z ∞
|SVU ( f )|2
−∞
SU ( f )
df =
Z ∞
−∞
SV ( f ) −
|SVU ( f )|2
d f.
SU ( f )
Chapter 10 Problem Solutions
183
62. Denote the optimal estimator by Vbn = ∑∞
k=−∞ h(k)Un−k , and denote any other estimah̃(k)U
.
The
discrete-time
orthogonality principle says that if
tor by Ven = ∑∞
n−k
k=−∞
E (Vn − Vbn )
∞
∑
h̃(k)Un−k
k=−∞
= 0
(∗)
for every h̃, then h is optimal in that E[|Vn − Vbn |2 ] ≤ E[|Vn − Ven |2 ] for every h̃. To establish the orthogonality principle, assume the above equation holds for every choice
of h̃. Then we can write
E[|Vn − Ven |2 ] = E[|(Vn − Vbn ) + (Vbn − Ven )|2 ]
= E[|Vn − Vbn |2 + 2(Vn − Vbn )(Vbn − Ven ) + |Vbn − Ven |2 ]
= E[|Vn − Vbn |2 ] + 2E[(Vn − Vbn )(Vbn − Ven )] + E[|Vbn − Ven |2 ].
(∗∗)
Now observe that
E[(Vn − Vbn )(Vbn − Ven )] = E (Vn − Vbn )
= E (Vn − Vbn )
∞
∑
∞
k=−∞
∞ ∑
k=−∞
h(k)Un−k −
∑
h̃(k)Un−k
k=−∞
h(k) − h̃(k) Un−k
= 0,
by (∗).
We can now continue (∗∗) writing
E[|Vn − Ven |2 ] = E[|Vn − Vbn |2 ] + E[|Vbn − Ven |2 ] ≥ E[|Vn − Vbn |2 ],
and thus h is the filter that minimizes the mean-squared error.
The next task is to characterize the filter h that satisfies the orthogonality condition
for every choice of h̃. Write the orthogonality condition as
0 = E (Vn − Vbn )
∞
=
∑
k=−∞
∞
=
∑
k=−∞
∞
∑
k=−∞
h̃(k)Ut−k
= E
E[h̃(k)(Vn − Vbn )Ut−k ] =
∞
∑
k=−∞
∞
∑
k=−∞
h̃(k)[RVU (k) − RVU
b (k)].
h̃(k)(Vn − Vbn )Ut−k
h̃(k)E[(Vn − Vbn )Ut−k ]
Since this must hold for all h̃, take h̃(k) = RVU (k) − RVU
b (k) to get
∞
∑
k=−∞
RVU (k) − RVU
b (k)
2
= 0.
Thus, the orthogonality condition holds for all h̃ if and only if RVU (k) = RVU
b (k) for
all k.
184
Chapter 10 Problem Solutions
b
The next task is to analyze RVU
b . Recall that Vn is the response of an LTI system to
input Un . Applying the result of Problem 31(a) with X replaced by U and Y replaced
by Vb , we have, also using the fact that RU is even,
∞
RVU
b (m) = RU Vb (−m) =
∑
k=−∞
h(k)RU (m − k).
Taking discrete-time Fourier transforms of
∞
yields
RVU (m) = RVU
b (m) =
SVU ( f ) = H( f )SU ( f ),
∑
k=−∞
h(k)RU (m − k)
and so H( f ) =
SVU ( f )
.
SU ( f )
63. We have
2λ /[λ 2 + (2π f )2 ]
2λ
SV ( f )
=
=
SV ( f ) + SX ( f )
2λ /[λ 2 + (2π f )2 ] + 1
2λ + λ 2 + (2π f )2
λ
2A
=
·
,
A A2 + (2π f )2
√
where A := 2λ + λ 2 . Hence, h(t) = (λ /A)e−A|t| .
H( f ) =
64. To begin, write
K( f ) =
λ + j2π f
λ
1
=
+ j2π f
.
A + j2π f
A + j2π f
1 + j2π f
Then
d −At
e u(t)
dt
= λ e−At u(t) − Ae−At u(t) + e−At δ (t)
= (λ − A)e−At u(t) + δ (t),
k(t) = λ e−At u(t) +
since e−At δ (t) = δ (t) for both t = 0 and for t 6= 0. This is a causal impulse response.
65. Let Zt := Vt+∆t . Then the causal Wiener filter for Zt yields the prediction or smoothing
filter for Vt+∆t . The Wiener–Hopf equation for Zt is
RZU (τ ) =
Z ∞
0
h∆t (θ )RU (τ − θ ) d θ ,
τ ≥ 0.
Now, RZU (τ ) = E[Zt+τ Ut ] = E[Vt+τ +∆t Ut ] = RVU (τ + ∆t), and so we must solve
RVU (τ + ∆t) =
Z ∞
0
h∆t (θ )RU (τ − θ ) d θ ,
τ ≥ 0.
Chapter 10 Problem Solutions
185
For white noise with RU (τ ) = δ (τ ), this reduces to
RVU (τ + ∆t) =
Z ∞
0
h∆t (θ )δ (τ − θ ) d θ = h∆t (τ ),
τ ≥ 0.
If h(t) = RVU (t)u(t) denotes the causal Wiener filter, then for ∆t ≥ 0 (prediction), we
can write
h∆t (τ ) = RVU (τ + ∆t) = h(τ + ∆t), τ ≥ 0.
If ∆t < 0 (smoothing), we can write h∆t (τ ) = h(τ + ∆t) only for τ ≥ −∆t. For 0 ≤
τ < −∆t, h(τ + ∆t) = 0 while h∆t (τ ) = RVU (τ + ∆t).
66. By the hint, the limit of the double sums is the desired double integral. If we can show
that each of these double sums is nonnegative, then the limit will also be nonnegative.
To this end put Zi := Xti e− j2π f ti ∆t i . Then
n n
∗ n
2
n n
= E ∑ Zi
= ∑ ∑ E[Zi Zk∗ ]
0 ≤ E ∑ Zi
∑ Zk
n
=
i=1
i=1
n
i=1 k=1
k=1
∑ ∑ E[Xti Xtk ]e− j2π f ti e j2π f tk ∆t i ∆t k
i=1 k=1
n n
=
∑ ∑ RX (ti − tk )e− j2π f ti e j2π f tk ∆t i ∆t k .
i=1 k=1
67.
(a) The Fourier transform of CY (τ ) = e−|τ | is 2/[1 + (2π f )2 ], which
is continuous
1 RT
at f = 0. Hence, we have convergence in mean square of 2T
Y
−T t dt to E[Yt ].
(b) The Fourier transform of CY (τ ) = sin(πτ )/(πτ ) is I[−1/2,1/2] ( f ), which is con1 RT
tinuous at f = 0. Hence, we have convergence in mean square of 2T
−T Yt dt to
E[Yt ].
68. We first point out that this is not a question about mean-square convergence. Write
1
2T
Z T
−T
cos(2π t + Θ) dt =
sin(2π T + Θ) − sin(2π (−T ) + Θ)
.
2T · 2π
Since | sin x | ≤ 1, we can write
1
2T
Z T
−T
cos(2π t + Θ) dt ≤
2
→ 0,
4π T
and so the limit in question exists and is equal to zero.
69. As suggested by the hint, put Yt := Xt+τ Xt . It will be sufficient if Yt is WSS and if the
Fourier transform of the covariance function of Yt is continuous at the origin. First,
since Xt is WSS, the mean of Yt is
E[Yt ] = E[Xt+τ Xt ] = RX (τ ),
which does not depend on t. Before examining the correlation function of Yt , we
assume that Xt is fourth-order strictly stationary so that
E[Yt1 Yt2 ] = E[Xt1 +τ Xt1 Xt2 +τ Xt2 ]
186
Chapter 10 Problem Solutions
must be unchanged if on the right-hand side we subtract t2 from every subscript expression to get
E[Xt1 +τ −t2 Xt1 −t2 Xτ X0 ].
Since this depends on t1 and t2 only through their difference, we see that Yt is WSS if
Xt is fourth-order strictly stationary. Now, the covariance function of Yt is
C(θ ) = E[Xθ +τ Xθ Xτ X0 ] − RX (τ )2 .
If the Fourier transform of this function of θ is continuous at the origin, then
1
2T
Z T
−T
Xt+τ Xt dt → RX (τ ).
70. As suggested by the hint, put Yt := IB (Xt ). It will be sufficient if Yt is WSS and if
the Fourier transform of the covariance function of Yt is continuous at the origin. We
assume at the outset that Xt is second-order strictly stationary. Then the mean of Yt is
E[Yt ] = E[IB (Xt )] = P(Xt ∈ B),
which does not depend on t. Similarly,
E[Yt1 Yt2 ] = E[IB (Xt1 )IB (Xt2 )] = P(Xt1 ∈ B, Xt2 ∈ B)
must be unchanged if on the right-hand side we subtract t2 from every subscript to get
P(Xt1 −t2 ∈ B, X0 ∈ B).
Since this depends on t1 and t2 only through their difference, we see that Yt is WSS if
Xt is second-order strictly stationary. Now, the covariance function of Yt is
C(θ ) = P(Xθ ∈ B, X0 ∈ B) − P(Xt ∈ B)2 .
If the Fourier transform of this function of θ is continuous at the origin, then
1
2T
Z T
−T
IB (Xt ) dt → P(Xt ∈ B).
71. We make the following definition and apply the hints:
Z
1 T
RXY (τ + θ , θ ) d θ
T →∞ 2T −T
Z
Z ∞
1 T
= lim
h(α )RX (τ + θ , θ − α ) d α d θ
T →∞ 2T −T
−∞
Z
Z ∞
1 T
RX (τ + θ , θ − α ) d θ d α
=
h(α ) lim
T →∞ 2T −T
−∞
Z ∞
Z
1 T −α
=
RX (τ + α + β , β ) d β d α
h(α ) lim
T →∞ 2T −T −α
−∞
Z
Z ∞
1 T
RX (τ + α + β , β ) d β d α .
=
h(α ) lim
T →∞ 2T −T
−∞
|
{z
}
RXY (τ ) := lim
= RX (τ +α )
Chapter 10 Problem Solutions
187
72. Write
Z
1 T
RY (τ + θ , θ ) d θ
T →∞ 2T −T
Z ∞
Z
1 T
h(β )RXY (τ + θ − β , θ ) d β d θ
= lim
T →∞ 2T −T
−∞
Z ∞
Z
1 T
=
h(β ) lim
RXY ([τ − β ] + θ , θ ) d θ d β .
T →∞ 2T −T
−∞
|
{z
}
RY (τ ) := lim
= RXY (τ −β )
73. Let SXY ( f ) denote the Fourier transform of RXY (τ ), and let SY ( f ) denote the Fourier
transform of RY (τ ). Then by the preceding two problems,
SY ( f ) = H( f )SXY ( f ) = H( f )H(− f )SX ( f ) = |H( f )|2 SX ( f ),
where, since h is real, H(− f ) = H( f )∗ .
74. This is an instance of Problem 32.
CHAPTER 11
Problem Solutions
1. With λ = 3 and t = 10,
P(Nt = 0) = e−λ t = e−3·10 = e−30 = 9.358 × 10−14 .
2. With λ = 12 per minute and t = 20 seconds, λ t = 4. Thus,
(λ t)2 (λ t)3
−λ t
P(Nt > 3) = 1 − P(Nt ≤ 3) = 1 − e
1 + λt +
+
2!
3!
3
2
4
4
= 1 − e−4 (5 + 8 + 32/3)
= 1 − e−4 1 + 4 + +
2
6
= 1 − e−4 (39/3 + 32/3) = 1 − e−4 (71/3) = 0.5665.
3.
(a) P(N5 = 10) =
(2 · 5)10 e−2·5
= 0.125.
10!
(b) We have
5
\
{Ni − Ni−1 = 2} =
P
i=1
5
5
∏ P(Ni − Ni−1 = 2)
i=1
−2 5
−10
=
∏
i=1
22 e−2
2!
= 1.453 × 10−3 .
= (2e ) = 32e
4. Let Nt denote the number of crates sold through time t (in days). Then Ni − Ni−1 is
the number of crates sold on day i, and so
5
\
P
{Ni − Ni−1 ≥ 3} =
i=1
5
∏ P(Ni − Ni−1 ≥ 3)
5
=
i=1
∏
i=1
h
i
1 − P(Ni − Ni−1 ≤ 2)
5 h
i
= ∏ 1 − e−λ 1 + λ + λ 2 /2!
i=1
i5
= 1 − e−3 1 + 3 + 9/2
= 0.06385.
h
5. Let Nt denote the number of fishing rods sold through time t (in days). Then Ni − Ni−1
is the number of crates sold on day i, and so
5
5
\
[
{Ni − Ni−1 ≤ 2}
{Ni − Ni−1 ≥ 3} = 1 − P
P
i=1
i=1
5
= 1 − ∏ P(Ni − Ni−1 ≤ 2)
i=1
188
Chapter 11 Problem Solutions
189
5 = 1 − ∏ e−λ (1 + λ + λ 2 /2!)
i=1
5
= 1 − e−2 (1 + 2 + 4/2!) = 1 − e−10 · 55 = 0.858.
6. Since the average time between hit songs is 7 months, the rate is λ = 1/7 per month.
(a) Since a year is 12 months, we write
P(N12 > 2) = 1 − P(N12 ≤ 2) = 1 − e−12λ [1 + 12λ + (12λ )2 /2!]
= 1 − e−12/7 [1 + 12/7 + (12/7)2 /2] = 0.247.
(b) Let Tn denote the time of the nth hit song. Since Tn = X1 + · · · + Xn , E[Tn ] =
nE[X1 ] = 7n. For n = 10, E[T10 ] = 70 months.
7.
(a) Since N0 ≡ 0, Nt = Nt − N0 . Since (0,t] ∩ (t,t + ∆t] = ∅, Nt − N0 and Nt+∆t − Nt
are independent.
(b) Write
P(Nt+∆t = k + `|Nt = k) = P(Nt+∆t − Nt = `|Nt = k),
by substitution,
= P(Nt+∆t − Nt = `|Nt − N0 = k), since N0 ≡ 0,
= P(Nt+∆t − Nt = `), by independent increments.
(c) Write
P(Nt+∆t = k + `|Nt = k)P(Nt = k)
P(Nt+∆t = k + `)
P(Nt+∆t − Nt = `)P(Nt = k)
, by part (b),
=
P(Nt+∆t = k + `)
P(Nt = k|Nt+∆t = k + `) =
=
=
(λ ∆t)` e−λ ∆t (λ t)k e−λ t
·
`!
k!
[λ (t+∆t)]k+` e−λ (t+∆t)
(k+`)!
k k+`
k
t
t + ∆t
∆t
t + ∆t
(d) In part (c), put ` = n − k and put p = t/(t + ∆t). Then
n k
P(Nt = k|Nt+∆t = n) =
p (1 − p)n−k ,
k
`
.
k = 0, . . . , n.
8. The nth customer arrives at time Tn ∼ Erlang(n, λ ). Hence,
E[Tn ] =
Γ(1 + n)
nΓ(n)
n
=
= .
λ Γ(n)
λ Γ(n)
λ
Alternatively, since Tn = X1 +· · ·+Xn , where the Xi are i.i.d. exp(λ ), E[Tn ] = nE[Xi ] =
n/λ .
190
9.
Chapter 11 Problem Solutions
(a) E[Xi ] = 1/λ = 0.5 weeks.
(b) P(N2 = 0) = e−λ ·2 = e−4 = 0.0183.
(c) E[N12 ] = λ · 12 = 24 snowstorms.
(d) Write
12
12
[
\
P
{Ni − Ni−1 ≥ 5} = 1 − P
{Ni − Ni−1 ≤ 4}
i=1
i=1
= 1 − [e−λ (1 + λ + λ 2 /2 + λ 3 /6 + λ 4 /24)]12
= 1 − [e−2 (1 + 2 + 2 + 4/3 + 2/3)]12
= 1 − [7e−2 ]12 = 1 − 0.523 = 0.477.
10. First observe that since 1/λ = 2 months, λ = 1/2 per month.
(a) P(N4 = 0) = e−λ ·4 = e−4/2 = e−2 = 0.135.
(b) Write
4
4
[
\
P
{Ni − Ni−1 ≥ 2} = 1 − P
{Ni − Ni−1 ≤ 1}
i=1
i=1
4
= 1 − ∏ P(Ni − Ni−1 ≤ 1)
i=1
4
= 1 − ∏[e−λ ·1 + e−λ ·1 (λ · 1)]
i=1
= 1 − [e−λ (1 + λ )]4 = 1 − [ 32 e−1/2 ]4
81 −2
= 1 − 16
e = 0.315.
11. We need to find var(Tn ) = var(X1 + · · · + Xn ). Since the Xi are independent, they
are uncorrelated, and so the variance of the sum is the sum of the variances. Since
Xi ∼ exp(λ ), var(Tn ) = n var(Xi ) = n/λ 2 . An alternative approach is to use the fact
that Tn ∼ Erlang(n, λ ). Since the moments of Tn are available,
var(Tn ) = E[Tn2 ] − (E[Tn ])2 =
(1 + n)n n 2
n
= 2.
−
2
λ
λ
λ
12. To begin, use the law of total probability, substitution, and independence to write
Yt
E[z ] = E[z
=
Nln(1+tU)
Z 1
] =
Z 1
E[zNln(1+tu) ] du =
0
=
Z 1
0
E[z
0
Nln(1+tU)
Z 1
0
(1 + tu)z−1 du.
|U = u] du =
Z 1
0
E[zNln(1+tu) |U = u] du
exp[(z − 1) ln(1 + tu)] du =
Z 1
0
eln(1+tu)
z−1
du
Chapter 11 Problem Solutions
191
To compute this integral, we need to treat the cases z = 0 and z 6= 0 separately. We
find that

ln(1 + t)

,
z = 0,

t
Yt
E[z ] =
z

 (1 + t) − 1 , z 6= 0.
tz
13. Denote the arrival times of Nt by T1 , T2 , . . . , and let Xk := Tk − Tk−1 denote the interarrival times. Similarly, denote the arrival times of Mt by S1 , S2 , . . . . (As it turns out,
we do not need the interarrival times of Mt .) Then for arbitrary k > 1, we use the law
of total probability, substitution, and independence to compute
P(MTk − MTk−1 = m)
=
=
=
Z ∞Z ∞
Z0 ∞ Zθ∞
Z0 ∞ Zθ∞
0
θ
P(MTk − MTk−1 = m|Tk = t, Tk−1 = θ ) fTk Tk−1 (t, θ ) dt d θ
P(Mt − Mθ = m|Tk = t, Tk−1 = θ ) fTk Tk−1 (t, θ ) dt d θ
P(Mt − Mθ = m) fTk Tk−1 (t, θ ) dt d θ
Z ∞Z ∞
[µ (t − θ )]m e−µ (t−θ )
fTk Tk−1 (t, θ ) dt d θ
m!
[µ (Tk − Tk−1 )]m e−µ (Tk −Tk−1 )
= E
m!
Z ∞
(µ Xk )m e−µ Xk
(µ x)m e−µ x
= E
=
· λ e−λ x dx
m!
m!
0
Z ∞
µ mλ
xm · (λ + µ )e−(λ +µ )x dx
=
(λ + µ )m! 0
|
{z
}
=
0
θ
mth moment of exp(λ + µ ) density
m!
µ mλ
λ
=
·
=
m
(λ + µ )m! (λ + µ )
λ +µ
µ
λ +µ
m
,
which is a geometric0 (µ /(λ + µ )) pmf in m.
14. It suffices to show that the probability generating function GMt (z) has the form eµ (z−1)
for some µ > 0. We use the law of total probability, substitution, and independence
to write
∞
Nt
Nt
∑ E[z∑i=1 Yi |Nt = n]P(Nt = n)
GMt (z) = E[zMt ] = E[z∑i=1 Yi ] =
n=0
∞
=
=
n
∑ E[z∑i=1 Yi |Nt = n]P(Nt = n)
n=0
∞ n
∑
n=0
Yi
∏ E[z ] P(Nt = n) =
i=1
n
∑ E[z∑i=1 Yi ]P(Nt = n)
n=0
∞
∑ [pz + (1 − p)]n P(Nt = n)
n=0
λ t([pz+(1−p)]−1)
= GNt pz + (1 − p) = e
Thus, Mt ∼ Poisson(pλ t).
∞
=
= e pλ t(z−1) .
192
Chapter 11 Problem Solutions
t
Vi denote the total energy throught time t, with Mt = 0 for Nt = 0. We
15. Let Mt = ∑Ni=1
use the law of total probability, substitution, and independence to write
∞
∑ E[Mt |Nt = n]P(Nt = n) =
E[Mt ] =
n=0
∞
∞
∑E
n=1
= ∑ E ∑ Vi Nt = n P(Nt = n) =
n=1
∞
n
i=1
∑ nE[V1 ]P(Nt = n)
=
n=1
∞
N
=
n
P(Nt = n)
V
t
i
∑
Nt
i=1
∑
n=1
∞
∑ E[Vi ] P(Nt = n)
n
i=1
= E[V1 ] ∑ nP(Nt = n) = E[V1 ]E[Nt ] = E[V1 ](λ t).
n=1
The average time between lightning strikes is 1/λ minutes.
16. We have
2.132(1.96)
= 5.170 ± 0.418 with 95% probability.
10
λ = 5.170 ±
In other words, λ ∈ [4.752, 5.588] with 95% probability.
17. To begin, observe that
∞
∞
Y = ∑ g(Ti ) = ∑
i=1 k=1
i=1
n
∞
n
k=1
i=1
k=1
n
∑ gk I(tk−1 ,tk ] (Ti ) = ∑ gk ∑ I(tk−1 ,tk ] (Ti ) = ∑ gk (Ntk − Ntk−1 )
is a sum of independent random variables. Then
n
n
E[Y ] =
∑ gk E[Ntk − Ntk−1 ]
k=1
=
∑ gk λ (tk − tk−1 )
k=1
=
Z ∞
g(τ )λ d τ ,
0
since g is piecewise constant. Next,
n
ϕY (ν ) = E[e jνY ] = E[e jν ∑k=1 gk (Ntk −Ntk−1 ) ] =
n
∏ E[e jν gk (Ntk −Ntk−1 ) ]
k=1
n
i
= ∏ exp λ (tk − tk−1 )(e jν gk − 1) = exp ∑ λ (tk − tk−1 )(e jν gk − 1)
n
k=1
= exp
Z
h
∞
0
k=1
jν g(τ )
e
− 1 λ dτ ,
since e jν g(τ ) − 1 is piecewise constant with values e jν gk − 1 (or the value zero if τ
does not lie in any (tk−1 ,tk ]). We now compute the correlation,
n
n
E[Y Z] = E
∑ gk (Ntk − Ntk−1 ) ∑ hl (Ntl − Ntl−1 )
k=1
=
l=1
∑ gk hl E[(Ntk − Ntk−1 )(Ntl − Ntl−1 )] + ∑ gk hk E[(Ntk − Ntk−1 )2 ]
k6=l
=
∑ gk hl λ
k6=l
k
2
(tk − tk−1 )(tl − tl−1 ) + ∑ gk hk [λ (tk − tk−1 ) + λ 2 (tk − tk−1 )2 ]
k
Chapter 11 Problem Solutions
=
∑ gk hl λ 2 (tk − tk−1 )(tl − tl−1 ) + ∑ gk hk λ (tk − tk−1 )
=
k,l
=
193
k
∑ gk λ (tk − tk−1 ) ∑ hl λ (tl − tl−1 ) + ∑ gk hk λ (tk − tk−1 )
k
Z ∞
g(τ )λ d τ
Z ∞
l
h(τ )λ d τ +
0
0
Z ∞
k
g(τ )h(τ )λ d τ .
0
Hence,
cov(Y, Z) = E[Y Z] − E[Y ]E[Z] =
Z ∞
g(τ )h(τ )λ d τ .
0
18. The key is to use g(τ ) = h(t − τ ) in the preceding problem. It then immediately
follows that
Z ∞
Z ∞
jν h(t−τ )
E[Yt ] =
h(t − τ )λ d τ , ϕYt (ν ) = exp
e
− 1 λ dτ ,
0
0
and
cov(Yt ,Ys ) =
Z ∞
0
h(t − τ )h(s − τ )λ d τ .
19. MATLAB. Replace the line
X = -log(rand(1))/lambda; % Generate exp(lambda) RV
with
X = randn(1)ˆ2; % Generate chi-squared RV
20. If Nt is a Poisson process of rate λ , then FX is the exp(λ ) cdf. Hence,
P(X1 < Y1 ) =
Z ∞
0
FX (y) fY (y) dy =
Z ∞
0
[1 − e−λ y ] fY (y) dy = 1 − MY (−λ ).
If Mt is a Poisson process of rate µ , then MY is the exp(µ ) mgf, and
P(X1 < Y1 ) = 1 −
µ
µ
λ
=
.
= 1−
µ − (−λ )
µ +λ
λ +µ
21. Since Tn := X1 + · · · + Xn is the sum of i.i.d. random variables, var(Tn ) = n var(X1 ).
Since X1 ∼ uniform[0, 1], var(X1 ) = 1/12, and var(Tn ) = n/12.
22. In the case of a Poisson process, Tk is Erlang(k, λ ). Hence,
∞ ∞
k−1
(λ t)l e−λ t
(λ t)l e−λ t
= ∑ ∑
∑ Fk (t) = ∑ 1 − ∑ l!
l!
k=1
k=1 l=k
k=1
l=0
∞ ∞
∞
l
−
l
λ
t
(λ t) e
(λ t) e−λ t ∞
= ∑ ∑
I[k,∞) (l) = ∑
∑ I[k,∞) (l)
l!
l!
k=1 l=0
l=0
k=1
∞
∞
∞
=
∑l
l=0
(λ t)l e−λ t
= E[Nt ] = λ t.
l!
194
Chapter 11 Problem Solutions
23. To begin, write
E[Nt |X1 = x] = E
∞
∑ I[0,t] (Tn ) X1 = x
n=1
∞
∞
∑ E I[0,t] (X1 + · · · + Xn ) X1 = x
=
n=1
= ∑ E I[0,t] (x + X2 + · · · + Xn ) X1 = x .
n=1
(a) If x > t, then x + X2 + · · · + Xn > t, and so I[0,t] (x + X2 + · · · + Xn ) = 0. Thus,
E[Nt |X1 = x] = 0 for x > t.
(b) First, for n = 1 and x ≤ t, I[0,t] (x) = 1. Next, for n ≥ 2 and x ≤ t, I[0,t] (x + X2 +
· · · + Xn ) = I[0,t−x] (X2 + · · · + Xn ). So,
∞ E[Nt |X1 = x] = 1 + ∑ E I[0,t−x] (X2 + · · · + Xn )
n=2
∞
= 1 + ∑ E I[0,t−x] (X1 + · · · + Xn )
n=1
= 1 + E[Nt−x ],
where we have used fact that the Xi are i.i.d.
(c) By the law of total probability,
E[Nt ] =
=
=
Z ∞
0
E[Nt |X1 = x] f (x) dx
0
E[Nt |X1 = x] f (x) dx +
Z t
Z t
0
Z ∞
t
E[Nt |X1 = x] f (x) dx
{z
}
|
= 0 for x>t
Z t
1 + E[Nt−x ] f (x) dx = F(t) + E[Nt−x ] f (x) dx.
0
24. With the understanding that m(t) = 0 for t < 0, we can write the renewal equation as
m(t) = F(t) +
Z ∞
0
m(t − x) f (x) dx,
where the last term is a convolution. Hence, taking the Laplace transform of the
renewal equation yields
M(s) :=
Z ∞
m(t)est dt =
0
Z ∞
0
F(t)est dt + M(s)MX (s).
Using integration by parts, we have
Z ∞
0
Thus,
F(t)est dt =
F(t)est
s
∞
0
−
1
s
Z ∞
f (t)est dt.
0
1
M(s) = − MX (s) + M(s)MX (s),
s
Chapter 11 Problem Solutions
195
which we can rearrange as
1
M(s)[1 − MX (s)] = − MX (s),
s
or
1
MX (s)
1
1 λ
λ /(λ − s)
λ
M(s) = − ·
= − ·
= − ·
= 2.
s 1 − MX (s)
s 1 − λ /(λ − s)
s −s
s
It follows that m(t) = λ tu(t).
25. For 0 ≤ s < t < ∞, write
Z t
Z s Z t
Z s
E[Xτ Xθ ] d τ d θ
Xθ d θ =
E[Vt Vs ] = E
Xτ d τ
0
0
0
0
Z s Z t
Z s Z t
2
=
δ (τ − θ ) d τ d θ
RX (τ − θ ) d τ d θ = σ
0
= σ
2
Z s
0
0
0
2
d θ = σ s.
0
26. For 0 ≤ s < t < ∞, write
E[Wt Ws ] = E[(Wt −Ws )Ws ] + E[Ws2 ]
= E[(Wt −Ws )(Ws −W0 )] + σ 2 s
= E[Wt −Ws ]E[Ws −W0 ] + σ 2 s = 0 · 0 + σ 2 s = σ 2 s.
27. For t2 > t1 , write
E[Yt1 Yt2 ] = E[eWt1 eWt2 ] = E[eWt2 −Wt1 e2Wt1 ] = E[eWt2 −Wt1 e2(Wt1 −W0 ) ]
= E[eWt2 −Wt1 ]E[e2(Wt1 −W0 ) ] = eσ
2 (t
2 −t1 )/2
e4σ
2t
1 /2
= eσ
2 (t
2 +3t1 )/2
.
28. Since cov(Wti ,Wt j ) = σ 2 min(ti ,t j ),

t1
 t1


cov(X) = σ 2  t1
 ..
 .
t1
t2
t2
..
.
t1
t2
t3
..
.
···
···
···
..
.
t1 t2 t3

t1
t2 

t3 
.
.. 
. 
tn
29. Let 0 ≤ t1 < · · · < tn+1 < ∞ and 0 ≤ s1 < · · · < sm+1 < ∞, and suppose that
n
g(τ ) =
∑ gi I(ti ,ti+1 ] (τ )
m
and
h(τ ) =
i=1
∑ h j I(s j ,s j+1 ] (τ ).
j=1
Denote the distinct points of {ti } ∪ {s j } in increasing order by θ1 < · · · < θ p . Then
p
g(τ ) =
∑ g̃k I(θk ,θk+1 ] (τ )
k=1
p
and
h(τ ) =
∑ h̃k I(θk ,θk+1 ] (τ ),
k=1
196
Chapter 11 Problem Solutions
where the g̃k are taken from the gi , and the h̃k are taken from the h j . We can now
write
Z ∞
0
g(τ ) dWτ +
Z ∞
0
p
h(τ ) dWτ =
p
∑ g̃k (Wθk+1 −Wθk ) + ∑ h̃k (Wθk+1 −Wθk )
k=1
p
=
=
k=1
∑ [g̃k + h̃k ](Wθk+1 −Wθk )
k=1
Z ∞
0
[g(τ ) + h(τ )] dWτ .
30. Following the hint, we first write
2 Z ∞
2 Z ∞
Z ∞
= E
g(τ ) dWτ
E
g(τ ) dWτ −
h(τ ) dWτ
0
0
0
Z ∞
Z ∞
− 2E
g(τ ) dWτ
h(τ ) dWτ
0
+E
= σ2
− 2E
∞
0
0
h(τ ) dWτ
2 = E
Z
∞
0
∞
0
Z ∞
0
= σ2
Z ∞
0
− 2σ
2
31.
g(τ ) dWτ
Z
Z ∞
0
∞
0
h(τ )2 d τ
h(τ ) dWτ .
[g(τ ) − h(τ )] dWτ
g(τ )2 d τ + σ 2
Z ∞
Z ∞
h(τ )2 d τ
0
g(τ )h(τ ) d τ .
0
∞
g(τ )h(τ ) d τ .
0
(a) Following the hint,
Yt :=
Z t
0
g(τ ) dWτ =
Z ∞
g(τ )I[0,t] (τ ) dWτ ,
2
2
0
it follows that
E[Yt2 ]
= σ
2
Z ∞
0
2 [g(τ ) − h(τ )]2 d τ
Comparing these two formulas, we see that
Z ∞
Z ∞
Z
h(τ ) dWτ = σ 2
E
g(τ ) dWτ
0
h(τ ) dWτ
2 g(τ )2 d τ + σ 2
Z
= σ2
0
∞
0
Z ∞
0
Second, we write
Z ∞
Z
g(τ ) dWτ −
E
0
Z
[g(τ )I[0,t] (τ )] d τ = σ
Z t
0
g(τ )2 d τ .
Chapter 11 Problem Solutions
(b) For t1 ,t2 ≥ 0, write
Z
E[Yt1 Yt2 ] = E
∞
0
= σ2
g(τ )I[0,t1 ] (τ ) dWτ
Z ∞
g(τ )I[0,t1 ] (τ )
0
= σ2
Z ∞
0
Z ∞
0
197
g(τ )I[0,t2 ] (τ ) dWτ
g(τ )I[0,t2 ] (τ ) d τ
g(τ )2 I[0,t1 ] (τ )I[0,t2 ] (τ ) d τ = σ 2
Z min(t1 ,t2 )
g(τ )2 d τ .
0
32. By independence of V and the Wiener process along with the result of the previous
problem,
RY (t1 ,t2 ) = e−λ (t1 +t2 ) E[V 2 ] + σ 2 e−λ (t1 +t2 )
Z min(t1 ,t2 )
e2λ τ d τ .
0
If t1 ≤ t2 , this last integral is equal to
Z t1
e2λ τ d τ =
0
1 2 λ t1
(e
− 1),
2λ
and so
σ 2 σ 2 −λ (t2 −t1 )
.
E[Yt1 Yt2 ] = e−λ (t1 +t2 ) q2 −
e
+
2λ
2λ
Similarly, if t2 < t1 ,
σ 2 σ 2 −λ (t1 −t2 )
E[Yt1 Yt2 ] = e−λ (t1 +t2 ) q2 −
.
e
+
2λ
2λ
In either case, we can write
σ 2 σ 2 −λ |t1 −t2 |
E[Yt1 Yt2 ] = e−λ (t1 +t2 ) q2 −
.
e
+
2λ
2λ
33. Write
E[Yt1 Yt2 ] =
e−λ (t1 +t2 )
e−λ (t1 +t2 ) 2
E[We2λ t1 We2λ t2 ] =
· σ min(e2λ t1 , e2λ t2 ).
2λ
2λ
For t1 ≤ t2 , this reduces to
σ 2 −λ (t2 −t1 )
e−λ (t1 +t2 ) 2 2λ t1
.
·σ e
=
e
2λ
2λ
If t2 < t1 , we have
σ 2 −λ (t1 −t2 )
e−λ (t1 +t2 ) 2 2λ t2
·σ e
=
e
.
2λ
2λ
We conclude that
E[Yt1 Yt2 ] =
σ 2 −λ |t1 −t2 |
e
.
2λ
198
34.
Chapter 11 Problem Solutions
(a) P(t) := E[Yt2 ] = E
Z
t
0
g(τ ) dWτ
2 Z t
=
g(τ )2 d τ .
0
(b) If g(t) is never zero, then for 0 ≤ t1 < t2 < ∞,
P(t2 ) − P(t1 ) =
Z t2
g(τ )2 d τ > 0.
t1
Thus, P(t1 ) < P(t2 ).
(c) First,
E[Xt ] = E[YP−1 (t) ] = E
Z
P−1 (t)
0
g(τ ) dWτ
= 0
since Wiener integrals have zero mean. Second,
Z P−1 (t)
2 Z P−1 (t)
2
E[Xt ] = E
g(τ ) dWτ
g(τ )2 d τ = P(P−1 (t)) = t.
=
0
35.
0
(a) For t > 0, E[Wt2 ] = E[(Wt −W0 )2 ] = t.
(b) For s < 0, E[Ws2 ] = E[(W0 −Ws )2 ] = −s.
(c) From parts (a) and (b) we see that no matter what the sign of t, E[Wt2 ] = |t|.
Whether t > s or s < t, we can write
|t − s| = E[(Wt −Ws )2 ] = E[Wt2 ] − 2E[Wt Ws ] + E[Ws2 ],
and so
E[Wt2 ] + E[Ws2 ] − |t − s|
|t| + |s| − |t − s|
=
.
2
2
E[Wt Ws ] =
36.
(a) Write
P(X = xk ) = P((X,Y ) ∈ {xk } × IR) =
=
Z ∞
−∞
∑
i
Z ∞
−∞
=1
z }| {
I{xk } (xi ) IIR (y) fXY (xi , y) dy
Z
∑ I{xk } (xi ) fXY (xi , y) dy =
∞
−∞
i
fXY (xk , y) dy.
(b) Write
P(Y ∈ C) = P (X,Y ) ∈ IR ×C =
=
Z ∞
−∞
∑
i
Z ∞
−∞
IIR (xi )IC (y) fXY (xi , y) dy
Z IC (y) ∑ fXY (xi , y) dy =
∑ fXY (xi , y) dy.
i
C
i
(c) First write
P(Y ∈ C|X = xk ) =
P((X,Y ) ∈ {xk } ×C)
P(X = xk ,Y ∈ C)
=
.
P(X = xk )
P(X = xk )
Chapter 11 Problem Solutions
Then since
P((X,Y ) ∈ {xk } ×C) =
∑
i
Z ∞
−∞
we have
Z
I{xk } (xi )IC (y) fXY (xi , y) dy =
C
P(Y ∈ C|X = xk ) =
199
fXY (xk , y) dy
=
pX (xk )
Z C
Z
C
fXY (xk , y) dy,
fXY (xk , y)
dy.
pX (xk )
(d) Write
Z ∞
−∞
I
(x
)p
(x
|y)
∑ B i X|Y i fY (y) dy
Z ∞
P(X ∈ B|Y = y) fY (y) dy =
−∞
i
∑ IB (xi )
=
i
∑ IB (xi )
=
i
∑ IB (xi )
=
i
∑
=
i
Z ∞
−∞
Z ∞
fXY (xi , y)
fY (y)
−∞
Z ∞
−∞
Z ∞
−∞
fY (y) dy
fXY (xi , y) dy
IIR (y) fXY (xi , y) dy
IB×IR (xi , y) fXY (xi , y) dy
= P((X,Y ) ∈ B × IR)
= P(X ∈ B,Y ∈ IR) = P(X ∈ B).
37. The key observations are that since F is nondecreasing,
F −1 (U) ≤ x ⇒ U ≤ F(x),
and since F −1 is nondecreasing,
U ≤ F(x) ⇒ F −1 (U) ≤ x.
Hence, {F −1 (U) ≤ x} = {U ≤ F(x)}, and with X := F −1 (U) we can write
P(X ≤ x) = P(F −1 (U) ≤ x) = P(U ≤ F(x)) =
38.
Z F(x)
0
(a) The cdf is
F(x)
1
3/4
x/2
1/2
1/4
0
1
2
1
2
x
1 du = F(x).
200
Chapter 11 Problem Solutions
(b) For 1/2 ≤ u < 1, Bu = [2u, ∞), and G(u) = 2u. For 1/4 < u < 1/2, Bu = [1, ∞),
and G(u)
√ = 1. For u = 1/4,
√Bu = [1/2, ∞), and G(u) = 1/2. For 0 ≤ u < 1/4,
Bu = [ u, ∞), and G(u) = u. Hence,
G(u)
2
2u
1
1/2
0
1
4
1
2
3
4
u
1
39. Suppose G(u) ≤ x. Since F is nondecreasing, F(G(u)) ≤ F(x). By the definition of
G(u), F(G(u)) ≥ u. Thus, F(x) ≥ F(G(u)) ≥ u. Now suppose u ≤ F(x). Then by the
definition of G(u), G(u) ≤ x.
40. With X := G(U), write
P(X ≤ x) = P(G(U) ≤ x) = P(U ≤ F(x)) =
Z F(x)
1 du = F(x).
0
41. To compute G(u), we used the M ATLAB function
function x = G(u)
i1 = find(u <= .25);
i2 = find(.25 < u & u <= .5);
i3 = find(.5 < u & u <= 1);
x(i1) = sqrt(u(i1));
x(i2) = 1;
x(i3) = 2*u(i3);
We obtained the histogram
3
2
1
0
0
0.5
1
2
This is explained by noting that the density corresponding to F(x) is impulsive:
Chapter 11 Problem Solutions
201
f (x)
1
3/4
1/2
1/4
2
x
0
1
2
∑
IB ( jm , . . . , jn )pm,n ( jm , . . . , jn ).
1
42. For an integer-valued process,
µm,n (B) =
jm ,..., jn
If B = {im } × · · · × {in }, then
µm,n+1 (B × IR) =
=
∑
IB×IR ( jm , . . . , jn , j)pm,n+1 ( jm , . . . , jn , j)
jm ,..., jn , j
∑ pm,n+1 (im , . . . , in , j),
j
µm−1,n (IR × B) =
=
∑
IIR×B ( j, jm , . . . , jn )pm−1,n ( j, jm , . . . , jn )
j, jm ,..., jn
∑ pm−1,n ( j, im , . . . , in ),
j
and µm,n (B) = pm,n (im , . . . , in ). Hence, if the conditions
µm,n+1 (B × IR) = µm,n (B) and
µm−1,n (IR × B) = µm,n (B)
hold for all B and we take B = {im } × · · · × {in }, then we obtain
∑ pm,n+1 (im , . . . , in , j)
= pm,n (im , . . . , in )
∑ pm−1,n ( j, im , . . . , in )
= pm,n (im , . . . , in ).
j
and
j
Conversely, suppose these two formulas hold. Fix an arbitrary B ⊂ IRn−m+1 . If we
multiply both formulass by IB (im , . . . , in ) and sum over all im , . . . , in , we obtain the
original consistency conditions.
43. For the first condition, write
∑ pm,n+1 (im , . . . , in , j)
j
=
∑ q(im )r(im+1 |im ) · · · r(in |in−1 )r( j|in )
j
= q(im )r(im+1 |im ) · · · r(in |in−1 )
|
{z
}
= pm,n (im ,...,in )
∑ r( j|in ) .
j
| {z }
=1
202
Chapter 11 Problem Solutions
For the second condition, write
∑ pm−1,n ( j, im , . . . , in )
=
∑ q( j)r(im | j)r(im+1 |im ) · · · r(in |in−1 )
=
j
j
|
|
44. If
∑ q( j)r(im | j) r(im+1 |im ) · · · r(in |in−1 ) .
j
{z
= q(im )
}
{z
= pm,n (im ,...,in )
}
µn (Bn ) = µn+1 (Bn × IR)
=
=
Z ∞
−∞
Z
···
···
Bn
Z
Z ∞
IBn (x1 , . . . , xn )IIR (y) fn+1 (x1 , . . . , xn , y) dy dxn · · · dx1
−∞
Z ∞
−∞
fn+1 (x1 , . . . , xn , y) dy dxn · · · dx1 ,
then necessarily the quantity in square brackets is the joint density fn (x1 , . . . , xn ).
Conversely, if the quantity in square brackets is equal to fn (x1 , . . . , xn ), we can write
µn+1 (Bn × IR) =
=
=
Z ∞
−∞
Z
Z
···
···
Bn
···
Bn
Z ∞
−∞
IBn (x1 , . . . , xn )IIR (y) fn+1 (x1 , . . . , xn , y) dy dxn · · · dx1
fn+1 (x1 , . . . , xn , y) dy dxn · · · dx1
Z Z ∞
Z
−∞
fn (x1 , . . . , xn )dxn · · · dx1 = µn (Bn ).
45. The key is to write
µt1 ,...,tn+1 (Bn,k )
=
=
Z ∞
···
−∞
Z
Z ∞
···
Bn
IBn (x1 , . . . , xk−1 , xk+1 , . . . , xn )IIR (xk ) ft1 ,...,tn+1 (x1 , . . . , xn+1 ) dxn+1 · · · dx1
ft1 ,...,tn+1 (x1 , . . . , xn+1 ) dxk dxn+1 · · · dxk+1 dxk−1 · · · dx1 .
−∞
Z Z ∞
−∞
Then if µt1 ,...,tn+1 (Bn,k ) = µn (Bn ), we must have
ft1 ,...,tk−1 ,tk+1 ,...,tn (x1 , . . . , xk−1 , xk+1 , . . . , xn ) =
Z ∞
−∞
ft1 ,...,tn+1 (x1 , . . . , xn+1 ) dxk
and conversely.
46. By the hint, [Wt1 , . . . ,Wtn ]0 is a linear transformation of a vector of independent Gaussian increments, which is a Gaussian vector. Hence, [Wt1 , . . . ,Wtn ]0 is a Gaussian
vector. Since n and the times t1 < · · · < tn are arbitrary, Wt is a Gaussian process.
Chapter 11 Problem Solutions
203
47. Let X = [Wt1 −W0 , . . . ,Wtn −Wtn−1 ]0 , and let Y = [Wt1 , . . . ,Wtn ]0 . Then Y = AX, where
A denotes the matrix given in the statement of Problem 46. Since the components of
X are independent,
2
n
e−xi /[2(ti −ti−1 )]
,
fX (x) = ∏ p
2π (ti − ti−1 )
i=1
where it is understood that t0 := 0. Using the example suggested in the hint, we have
fX (x)
| det A|
fY (y) =
.
x=A−1 y
Fortunately, det A = 1, and by definition, Xi = Wti −Wti−1 . Hence,
2
e−(wi −wi−1 ) /[2(ti −ti−1 )]
∏ p2π (t − t ) ,
i
i−1
i=1
n
ft1 ,...,tn (w1 , . . . , wn ) =
where it is understood that w0 := 0.
48. By the hint, it suffices to show that C is positive semidefinite. Write
a0Ca =
n
n
∑ ∑ ai akCik
n
=
i=1 k=1
n
∑ ∑ ai ak R(ti − tk )
i=1 k=1
n n
=
∑ ai ak
∑
=
Z ∞
i=1 k=1
−∞
Z ∞
−∞
n
S( f )
S( f )e j2π f (ti −tk ) d f
∑ ai e j2π f ti
i=1
2
d f ≥ 0.
CHAPTER 12
Problem Solutions
1. If P(A|X = i,Y = j, Z = k) = h(i), then
P(A, X = i,Y = j, Z = k) = P(A|X = i,Y = j, Z = k)P(X = i,Y = j, Z = k)
= h(i)P(X = i,Y = j, Z = k).
Summing over j yields
P(A, X = i, Z = k) = h(i)P(X = i, Z = k),
(∗)
and thus h(i) = P(A|X = i, Z = k). Next, summing (∗) over k yields
P(A, X = i) = h(i)P(X = i),
and then h(i) = P(A|X = i) as well.
2. Write
X1 = g(X0 , Z1 )
X2 = g(X1 , Z2 ) = g(g(X0 , Z1 ), Z2 )
X3 = g(X2 , Z3 ) = g(g(g(X0 , Z1 ), Z2 ), Z3 )
..
.
In general, Xn is a function of X0 , Z1 , . . . , Zn , and thus (X0 , . . . , Xn ) is a function of
(X0 , Z1 , . . . , Zn ), which is independent of Zn+1 . Now observe that
P(Xn+1 = in+1 |Xn = in , . . . , X0 = i0 ) = P(g(Xn , Zn+1 ) = in+1 |Xn = in , . . . , X0 = i0 )
= P(g(in , Zn+1 ) = in+1 |Xn = in , . . . , X0 = i0 )
= P(g(in , Zn+1 ) = in+1 ),
where we have used substitution and independence. Since
P(Xn+1 = in+1 |Xn = in , . . . , X0 = i0 )
does not depend on in−1 , . . . , i0 , Xn is a Markov chain.
3. Write
P(A ∩ B|C) =
P(A ∩ B ∩C)
P(A ∩ B ∩C) P(B ∩C)
=
·
= P(A|B ∩C)P(B|C).
P(C)
P(B ∩C)
P(C)
204
Chapter 12 Problem Solutions
205
4. Write
P(X0 = i, X1 = j, X2 = k, X3 = l)
= P(X3 = l|X2 = k, X1 = j, X0 = i)
· P(X2 = k|X1 = j, X0 = i)P(X1 = j|X0 = i)P(X0 = i)
= pkl p jk pi j νi .
5. The first equation we write is
π0 =
∑ πk pk0
π0 = π0 p00 + π1 p10 = π0 (1 − a) + π1 b.
as
k
This tells us that π1 = (a/b)π0 . Then we use the fact that π0 + π1 = 1 to write π0 +
(a/b)π0 = 1, or π0 = 1/(1 + a/b) = b/(a + b). We also have π1 = (a/b)π0 = a/(a +
b).
6. The state transition diagram is
1/2
0
1/2
1/2
1/4
1/2
1
2
3/4
The stationary distribution is π0 = 5/12, π1 = 1/3, π2 = 1/4.
7. The state transition diagram is
0
1/2
1/4
1/4
1/2
1
3/4
2
3/4
The stationary distribution is π0 = 1/5, π1 = 7/10, π2 = 1/10.
206
Chapter 12 Problem Solutions
8. The state transition diagram is
1/2
1/2
0
1
9/10
1/10
1/10
1/2
3
2
9/10
1/2
The stationary distribution is π0 = 9/28, π1 = 5/28, π2 = 5/28, π3 = 9/28.
9. The first equation is
π0 = π0 p00 + π1 p10 = π0 (1 − a) + π1 b,
which tells us that π1 = (a/b)π0 . The next equation is
π1 = π0 p01 + π1 p11 + π2 p21 = π0 a + π1 (1 − [a + b]) + π2 b,
which tells us that
(a + b)π1 = π0 a + π2 b or
(a + b)(a/b)π0 = π0 a + π2 b,
from which it follows that π2 = (a/b)2 π0 . Now suppose that πi = (a/b)i π0 holds for
i = 0, . . . , j < N. Then from
π j = π j−1 p j−1, j + π j p j j + π j+1 p j+1, j = π j−1 a + π j (1 − [a + b]) + π j+1 b,
we obtain
(a + b)π j = π j−1 a + π j+1 b
or
(a + b)(a/b) j π0 = (a/b) j−1 a + π j+1 b,
from which it follows that π j+1 = (a/b) j+1 π0 . To find π0 , we use the fact that π0 +
· · · + πN = 1. Using the finite geometric series formula, we have
N
1 =
∑ (a/b) j π0
= π0
j=0
Hence,
πj =
1 − a/b a j
,
1 − (a/b)N+1 b
1 − (a/b)N+1
.
1 − a/b
j = 0, . . . , N,
a 6= b.
If a = b, then π j = π0 for j = 0, . . . , N implies π j = 1/(N + 1).
Chapter 12 Problem Solutions
207
10. Let π and πb denote the two solutions in the example. For 0 ≤ λ ≤ 1, put
πej := λ π j + (1 − λ )πbj ≥ 0.
Then
∑ πej = ∑ λ π j + (1 − λ )πbj
j
j
= λ ∑ π j + (1 − λ ) ∑ πbj = λ · 1 + (1 − λ ) · 1 = 1.
Hence, πe is a valid pmf. Finally, note that
∑ πei pi j
i
=
j
j
∑[λ πi + (1 − λ )πbi ]pi j
i
= λ π j + (1 − λ )πbj = πej .
= λ ∑ πi pi j + (1 − λ ) ∑ πbi pi j
i
i
11. MATLAB. OMITTED.
12. Write
∞
E[T1 ( j)|X0 = i] =
∑ kP(T1 ( j) = k|X0 = i) + ∞ · P(T1 ( j) = ∞|X0 = i)
k=1
∞
=
k=1
13.
(k)
∑ k fi j
+ ∞(1 − fi j ).
(a) First note that {T1 = 2} = {X2 = j, X1 6= j} and that
h
{T2 = 5} = {X5 = j} ∩
{X4 = j, X3 6= j, X2 6= j, X1 6= j}
∪{X4 6= j, X3 = j, X2 6= j, X1 6= j}
∪{X4 6= j, X3 6= j, X2 = j, X1 =
6 j} i
∪{X4 6= j, X3 6= j, X2 6= j, X1 = j} .
Then
{T2 = 5} ∩ {T1 = 2} = {X5 = j, X4 6= j, X3 6= j, X2 = j, X1 6= j}.
It now follows that
P(T2 = 5|T1 = 2, X0 = i) = P(X5 = j, X4 6= j, X3 6= j|X2 = j, X1 6= j, X0 = i).
(b) Write
P(X5 = j, X4 6= j, X3 6= j, X2 = j, X1 6= j, X0 = i)
as
[
{X5 = j, X4 6= j, X3 6= j, X2 = j, X1 = l, X0 = i} ,
P
l6= j
which is equal to
∑ P(X5 = j, X4 6= j, X3 6= j|X2 = j, X1 = l, X0 = i)P(X2 = j, X1 = l, X0 = i).
l6= j
208
Chapter 12 Problem Solutions
By the Markov property, this simplifies to
∑ P(X5 = j, X4 6= j, X3 6= j|X2 = j)P(X2 = j, X1 = l, X0 = i),
l6= j
which is just
P(X5 = j, X4 6= j, X3 6= j|X2 = j)P(X2 = j, X1 6= j, X0 = i).
It now follows that P(X5 = j, X4 6= j, X3 6= j|X2 = j, X1 6= j, X0 = i) is equal to
P(X5 = j, X4 6= j, X3 6= j|X2 = j).
(c) Write
P(X5 = j, X4 6= j, X3 6= j|X2 = j) =
∑
P(X5 = j, X4 = k, X3 = l|X2 = j)
∑
P(X3 = j, X2 = k, X1 = l|X0 = j)
k6= j,l6= j
=
k6= j,l6= j
= P(X3 = j, X2 6= j, X1 6= j|X0 = j).
14. Write
∞
\
P(V ( j) = ∞|X0 = i) = P
{V ( j) ≥ L} X0 = i
L=1
M
\
{V ( j) ≥ L} X0 = i ,
= lim P
M→∞
= lim P(V ( j) ≥ M|X0 = i),
decreasing events.
M→∞
15. First write
E[V ( j)|X0 = i] = E
∞
∑ I{ j} (Xn ) X0 = i
n=1
∞
=
limit property of P,
L=1
∑ P(Xn = j|X0 = i)
n=1
=
∞
=
∑ E[I{ j} (Xn )|X0 = i]
n=1
∞
(n)
∑ pi j
.
(∗)
n=1
Next,
P(V ( j) = ∞|X0 = i) = lim P(V ( j) ≥ L|X0 = i),
L→∞
= lim fi j ( f j j )L−1 ,
L→∞
from preceding problem,
from the text.
Now, if j is recurrent ( f j j = 1), the above limit is fi j . If fi j > 0, then V ( j) = ∞ with
positive probability, and so the expectation on the left in (∗) must be infinite, and
hence, so is the sum on the far right in (∗). If fi j = 0, then for any n = 1, 2, . . . , we
can write
∞
[
0 = fi j := P(T1 ( j) < ∞|Xi = 0) ≥ P
{Xm = j} Xi = 0
m=1
(n)
≥ P(Xn = j|X0 = i) = pi j ,
Chapter 12 Problem Solutions
209
(n)
and it follows that ∑∞
n=1 pi j = 0.
In the transient case ( f j j < 1), we have from the text that
P(V ( j) ≥ L|X0 = i) = fi j ( f j j )L−1 .
Using the preceding problem along with f j j < 1, we have
P(V ( j) = ∞|X0 = i) = lim fi j ( f j j )L−1 = 0.
L→∞
Now, we also have from the text that
P(V ( j) = L|X0 = i) =
fi j ( f j j )L−1 (1 − f j j ), L ≥ 1,
1 − fi j ,
L = 0,
where we use the fact that the number of visits to state j is zero if and only if we never
visit state j, and this happens if and only if T1 ( j) = ∞. We can therefore write
∞
∑ L · P(V ( j) = L|X0 = i)
E[V ( j)|X0 = i] =
=
L=0
fi j
< ∞,
1− fjj
and it follows that the last sum in (∗) is finite too.
16. From mλ − 1 ≤ bmλ c ≤ mλ ,
1
1
1
≤
.
≤
mλ
bmλ c
mλ − 1
Then
17.
1
m
1
1
≤
≤
→ .
λ
bmλ c
λ − 1/m
λ
(a) Write h( j) = IS ( j) = ∑nl=1 I{sl } ( j). Then
1 m n
1 m
lim
lim
∑ h(Xk ) = m→∞
∑ ∑ I{sl } (Xk ) =
m→∞ m
m k=1
l=1
k=1
n
=
lim Vm (sl )/m
∑ m→∞
∑ IS ( j)π j
j
=
1
m
l=1
k=1
n
=
l=1
=
n
lim
I (Xk )
∑ m→∞
m ∑ {sl }
∑ πs l ,
by Theorems 3 and 4,
l=1
∑ h( j)π j .
j
(b) With h( j) = ∑nl=1 cl ISl ( j), where each Sl is a finite subset of states, we can write
n
1 m n
1 m
1 m
cl ISl (Xk ) = ∑ cl lim
h(Xk ) = lim
lim
∑
∑
∑
∑ ISl (Xk )
m→∞ m
m→∞ m
m→∞ m
k=1
k=1 l=1
k=1
l=1
n
= ∑ cl ∑ ISl ( j)π j , by part (a)
l=1
=
∑
j
j
c
I
(
j)
∑ l Sl π j =
n
l=1
∑ h( j)π j .
j
210
Chapter 12 Problem Solutions
(c) If h( j) = 0 for all but finitely many states, then there are at most finitely many
distinct nonzero values that h( j) can take, say c1 , . . . , cn . Put Sl := { j : h( j) =
cl }. By the assumption about h, each Sl is a finite set. Furthermore, we can
write
n
∑ cl ISl ( j).
h( j) =
l=1
Hence, the desired result is immediate by part (b).
18. MATLAB. OMITTED.
19. MATLAB. OMITTED.
20. Suppose k ∈ Ai ∩ A j . Since k ∈ Ai , i ↔ k. Since k ∈ A j , j ↔ k. Hence, i ↔ j. Now,
if l ∈ Ai , l ↔ i ↔ j and we see that l ∈ A j . Similarly, if l ∈ A j , l ↔ j ↔ i and we see
that l ∈ Ai . Thus, Ai = A j .
21. Yes. As pointed out in the discussion at the end of the section, by combining Theorem 8 and Theorem 6, the chain has a unique stationary distribution. Now, by Theorem 4, all states are positive recurrent, which by definition means that the expected
time to return to a state is finite.
22. Write
(λ ∆t)n e−λ ∆t
λ
=
lim (λ ∆t)n−1 e−λ ∆t = 0,
∆t↓0
∆t · n!
n! ∆t↓0
gi,i+n = lim
for n ≥ 2.
23. The state transition diagram is
0
1
5
2
4
1
2
3
The stationary distribution is π0 = 11/15, π1 = 1/5, π2 = 1/15.
24. MATLAB. Change the line
A=P-In;
to
A=P;.
25. The forward equation is
j+1
p0i j (t) =
∑ pik (t)gk j
k
=
∑
pik gk j
k= j−1
= pi, j−1 (t)λ j−1 − pi j (t)[λ j + µ j ] + pi, j+1 (t)µ j+1 .
Chapter 12 Problem Solutions
211
The backward equation is
p0i j (t) =
∑ gik pk j (t)
i+1
=
k
∑
gik pi j (t)
k=i−1
= µi pi−1, j (t) − (λi + µi )pii (t) + λi pi+1, j (t).
The chain is conservative because gi,i−1 +gi,i+1 = µi + λi = −[−(λi + µi )] = −gii < ∞.
26. We use the formula 0 = ∑k πk gk j . For j = 0, we have
0 =
∑ πk gk0
= π0 (−λ0 ) + π1 µ1 ,
k
which implies π1 = (λ0 /µ1 )π0 . For j = 1, we have
0 =
∑ πk gk1
= π0 λ0 + π1 [−(λ1 + µ1 )] + π2 µ2
k
= π0 λ0 + (λ0 /µ1 )π0 [−(λ1 + µ1 )] + π2 µ2 = π0 (λ0 λ1 /µ1 ) + π2 µ2 ,
which implies π2 = π0 λ0 λ1 /(µ1 µ2 ). Now suppose that πi = π0 λ0 · · · λi−1 /(µ1 · · · µi )
for i = 1, . . . , j. Then from
0 =
∑ πk gk j
k
=
= π j−1 λ j−1 − π j (λ j + µ j ) + π j+1 µ j+1
λ0 · · · λ j−2
λ0 · · · λ j−1
π0 λ j−1 −
π0 (λ j + µ j ) + π j+1 µ j+1 ,
µ1 · · · µ j−1
µ1 · · · µ j
and it follows that
π j+1 =
Now, let
λ0 · · · λ j
π0 .
µ1 · · · µ j+1
∞
B :=
λ0 · · · λ j−1
< ∞.
j=1 µ1 · · · µ j
∑
From the condition ∑∞j=0 π j = 1, we have
∞
λ0 · · · λ j−1
= π0 (1 + B).
j=1 µ1 · · · µ j
1 = π 0 + π 1 + · · · = π0 + π 0 ∑
It then follows that π0 = 1/(1 + B), and
λ0 · · · λ j−1
1
πj =
,
1 + B µ1 · · · µ j
j ≥ 1.
If λi = λ and µi = µ , then B = ∑∞j=1 (λ /µ ) j , which is finite if and only if λ < µ . In
in this case,
∞ k
1
λ
1+B = ∑
=
,
µ
1
+
λ /µ
k=0
and
π j = (1 − λ /µ )(λ /µ ) j ∼ geometric0 (λ /µ ).
212
Chapter 12 Problem Solutions
27. The solution is very similar the that of the preceding problem. In this case, put
N
λ0 · · · λ j−1
< ∞,
j=1 µ1 · · · µ j
∑
BN :=
and then π0 = 1/(1 + BN ), and
1
πj =
1 + BN
λ0 · · · λ j−1
,
µ1 · · · µ j
j = 1, . . . , N.
If λi = λ and µi = µ , then
N
∑ (λ /µ )k
1 + BN =
=
k=0
and
1 − (λ /µ )N+1
,
1 − λ /µ
1 − λ /µ
(λ /µ ) j ,
1 − (λ /µ )N+1
πj =
j = 0, . . . , N.
If λ = µ , then π j = 1/(N + 1).
28. To begin, write
mi (t) := E[Xt |X0 = i] =
∑ jP(Xt = j|X0 = i)
=
j
∑ j pi j (t).
j
Then
m0i (t) =
∑ j p0i j (t) =
j
∑j
j
∑ pik (t)gk j
k
=
p
(t)
jg
ik
k
j
∑
∑
k
h
i
= ∑ pik (t) (k − 1)gk,k−1 + kgkk + (k + 1)gk,k+1
k
=
h
i
(k − 1)(k µ ) − k(kλ + α + k µ ) + (k + 1)(kλ + α )
∑ pik (t)
k
=
j
∑ pik (t)[−kµ + kλ + α ]
k
= −µ mi (t) + λ mi (t) + α .
We must solve
m0i (t) + (µ − λ )mi (t) = α ,
If µ 6= λ , it is readily verified that
mi (t) = i −
mi (0) = i.
α
α
e−(µ −λ )t +
µ −λ
µ −λ
solves the equation. If µ = λ , then mi (t) = α t + i solves m0i (t) = α with mi (0) = i.
29. To begin, write
p0i j (t) =
∑ pik (t)gk j
k
= pi, j−1 (t)g j−1, j + pi j (t)g j j = λ pi, j−1 (t) − λ pi j (t). (∗)
Chapter 12 Problem Solutions
213
Observe that for j = i, pi,i−1 (t) = 0 since the chain can not go from state i to a lower
state. Thus,
p0ii (t) = −λ pii (t).
Also, pii (0) = P(Xt = i|X0 = i) = 1. Thus, pii (t) = e−λ t , and it is easily verified that
(λ t)n e−λ t
,
n!
pi,i+n (t) =
n = 0, 1, 2, . . . ,
solves (∗). Thus, Xt is a Poisson process of rate λ with X0 = i instead of X0 = 0.
30.
(a) The assumption that D̂ > −∞ implies that the πk are not all zero, in which case,
the πk would not sum to one and would not be a pmf. Aside from this, it is clear
that πk ≥ 0 for all k and that ∑k πk = D̂/D̂ = 1. Next, since p j j = 0, write
∑ πk pk j
k
−1
∑ π̂k gk j
D̂ k6= j
k6= j
−1
=
π̂
π̂
−
g
g
j j j = π̂ j g j j /D̂ = π j .
∑ k kj
D̂
k
| {z }
=
gk j
∑ [π̂k gkk /D̂] −gkk
=
=0
(b) The assumption that Ď > −∞ implies that the πk are not all zero. Aside from
this, it is clear that πk ≥ 0 for all k and that ∑k πk = Ď/Ď = 1. Next, write
∑ πk gk j
k
=
gk j
−1
π̌k
∑
−g
Ď k
kk
− π̌ j
∑[(π̌k /gkk )/Ď]gk j
k
−1
=
Ď
−1
=
Ď
= 0.
gk j
∑ π̌k −gkk
k6= j
∑ π̌k pk j
|
k
{z
= π̌ j
}
=
−π̌ j ,
since p j j = 0,
(c) Since π̂k and π̌k are pmfs, they sum to one. Hence, if gii = g for all i, D̂ = g
and Ď = 1/g. In (a), πk = π̂k gkk /D̂ = π̂k g/g = π̂k . In (b), πk = (π̌k /gkk )/Ď =
(π̌k /g)/(1/g) = π̌k .
31. Following the hints, write
P(T > t + ∆t|T > t, X0 = i) = P(Xs = i, 0 ≤ s ≤ t + ∆t|Xs = i, 0 ≤ s ≤ t)
= P(Xs = i,t ≤ s ≤ t + ∆t|Xs = i, 0 ≤ s ≤ t)
= P(Xs = i,t ≤ s ≤ t + ∆t|Xt = i)
= P(Xs = i, 0 ≤ s ≤ ∆t|X0 = i)
= P(T > ∆t|X0 = i).
214
Chapter 12 Problem Solutions
The exponential parameter is
lim
∆t↓0
1 − P(T > ∆t|X0 = i)
1 − P(Xs = i, 0 ≤ s ≤ ∆t|X0 = i)
= lim
∆t↓0
∆t
∆t
1 − P(X∆t = i|X0 = i)
= lim
=: −gii .
∆t↓0
∆t
32. Using substitution in reverse, write
P(Wt ≤ y|Ws = x,Wsn−1 = xn−1 , . . . ,Ws0 = x0 )
= P(Wt − x ≤ y − x|Ws = x,Wsn−1 = xn−1 , . . . ,Ws0 = x0 )
= P(Wt −Ws ≤ y − x|Ws = x,Wsn−1 = xn−1 , . . . ,Ws0 = x0 ).
Now, using the fact that W0 ≡ 0, this last conditional probability is equal to
P(Wt −Ws ≤ y − x|Ws −Wsn−1 = x − xn−1 , . . . ,Ws1 −Ws0 = x1 − x0 ,Ws0 −W0 = x0 ).
Since the Wiener process has independent increments that are Gaussian, this last expression reduces to
√
Z y−x
exp[−{θ /[σ t − s ]}2 /2]
p
dθ .
P(Wt −Ws ≤ y − x) =
−∞
2πσ 2 (t − s)
Since this depends on x but not on xn−1 , . . . , x0 ,
P(Wt ≤ y|Ws = x,Wsn−1 = xn−1 , . . . ,Ws0 = x0 ) = P(Wt ≤ y|Ws = x).
Hence, Wt is a Markov process.
33. Write
∑ P(X = x|Y = y, Z = z)P(Y = y|Z = z)
y
=
∑
y
P(X = x,Y = y, Z = z) P(Y = y, Z = z)
·
P(Y = y, Z = z)
P(Z = z)
=
1
P(X = x,Y = y, Z = z)
P(Z = z) ∑
y
=
P(X = x, Z = z)
= P(X = x|Z = z).
P(Z = z)
34. Using the law of total conditional probability, write
Pt+s (B) = P(Xt+s ∈ B|X0 = x)
=
=
=
=
Z ∞
−∞
Z ∞
−∞
Z ∞
−∞
Z ∞
−∞
P(Xt+s ∈ B|Xs = z, X0 = x) fXs |X0 (z|x) dz
P(Xt+s ∈ B|Xs = z) fs (x, z) dz
P(Xt ∈ B|X0 = z) fs (x, z) dz
Pt (z, B) fs (x, z) dz.
CHAPTER 13
Problem Solutions
1. First observe that E[|Xn | p ] = n p/2 P(U ≤ 1/n) = n p/2 · (1/n) = n p/2−1 , which goes to
zero if and only if p < 2. Thus, 1 ≤ p < 2.
2
var(Nt )
λt
λ
Nt
1
−λ
= 2 =
→ 0.
2. E
= 2 E[|Nt − λ t|2 ] =
t
t
t2
t
t
3. Given ε > 0, let n ≥ N imply |C(k)| ≤ ε /2. Then for such n,
1 n−1
1 N−1
1 n
C(k)
=
C(k)
+
∑
∑
∑ C(k)
n k=0
n k=0
n k=N
and
1 N−1
1 n−1
1 N−1
1 n
C(k) ≤
|C(k)| + ∑ ε /2 ≤
∑
∑
∑ |C(k)| + ε /2.
n k=0
n k=0
n k=N
n k=0
For large enough n, the right-hand side will be less that ε .
4. Applying the hint followed by the Cauchy–Schwarz inequality shows that
1 n−1
∑ C(k) ≤
n k=0
q
E[(X1 − m)2 ]E[(Mn − m)2 ].
Hence, if Mn converges in mean square to m, the left-hand side goes to zero as n → ∞.
5. Starting with the hint, we write
E[ZIA ] = E[(Z − Zn )IA ] + E[Zn IA ]
≤ E[Z − Zn ] + nE[IA ], since Zn ≤ Z and Zn ≤ n,
= E[|Z − Zn |] + nP(A).
Since Zn converges in mean to Z, given ε > 0, we can let n ≥ N imply E[|Z − Zn |] ≤
ε /2. Then in particular we have
E[ZIA ] < ε /2 + NP(A).
Hence, if 0 < δ < ε /(2N),
P(A) < δ
implies
E[ZIA ] < ε /2 + N ε /(2N) = ε .
6. Following the hint, we find that given ε > 0, there exists a δ > 0 such that
P(U ≤ ∆x) = ∆x < δ
implies
E[ f (x +U)I{U≤∆x} ] =
215
Z ∆x
0
f (x + t) dt < ε .
216
Chapter 13 Problem Solutions
Now make the change of variable θ = x + t to get
Z ∆x
f (x + t) dt =
Z x+∆x
x
0
f (θ ) d θ = F(x + ∆x) − F(x).
For 0 < −∆x < δ , take A = {U > 1 + ∆x} and Z = f (x − 1 + U). Then P(U >
1 + ∆x) = −∆x implies
E[ f (x + 1 +U)I{U>1+Dx} ] =
Z 1
1+∆x
f (x − 1 + t) dt < ε .
In this last integral make the change of variable θ = x − 1 + t to get
Z x
x+∆x
f (θ ) d θ = F(x) − F(x + ∆x).
Thus, F(x + ∆x) − F(x) > −ε . We can now write that given ε > 0, there exists a δ
such that |∆x| < δ implies |F(x + ∆x) − F(x)| < ε .
7. Following the hint, we can write
h1
i
p
q
1
1
1
exp ln(|X|/α ) p + ln(|Y |/β )q ≤ eln(|X|/α ) + eln(|Y |/β )
p
q
p
q
or
i
h
1 |X| p 1 |Y | q
exp ln(|X|/α ) + ln(|Y |/β ) ≤
+
p α
q β
or
|XY |
1 |X| p 1 |Y |q
≤
+
.
αβ
p αp
q βq
Hence,
1 αp 1 βq
1 E[|X| p ] 1 E[|Y |q ]
E[|XY |]
+
=
+
= 1.
≤
αβ
p αp
q βq
p αp q βq
The hint assumes neither α nor β is zero or infinity. However, if either α or β is zero,
then both sides of the inequality are zero. If neither is zero and one of them is infinity,
then the right-hand side is infinity and the inequality is trivial.
8. Following the hint, with X = |Z|α , Y = 1, and p = β /α , we have
E[|Z|α ] = E[|XY |] ≤ E[|X| p ]1/p E[|Y |q ]1/q = E[(|Z|α )β /α ]α /β · 1 = E[|Z|β ]α /β .
Raising both sides to the 1/α yields E[|Z|α ]1/α ≤ E[|Z|β ]1/β .
9. By Lyapunov’s inequality, E[|Xn − X|α ]1/α ≤ E[|Xn − X|β ]1/β . Raising both sides to
the α power yields E[|Xn − X|α ] ≤ E[|Xn − X|β ]α /β . Hence, if E[|Xn − X|β ] → 0, then
E[|Xn − X|α ] → 0 too.
10. Following the hint, write
E[|X +Y | p ] = E[|X +Y | |X +Y | p−1 ]
≤ E[|X| |X +Y | p−1 ] + E[|Y | |X +Y | p−1 ]
≤ E[|X| p ]1/p E[(|X +Y | p−1 )q ]1/q + E[|Y | p ]1/p E[(|X +Y | p−1 )q ]1/q ,
Chapter 13 Problem Solutions
217
where 1/q := 1 − 1/p. Hence, 1/q = (p − 1)/p and q = p/(p − 1). Now divide the
above inequality by E[|X +Y | p ](p−1)/p to get
E[|X +Y | p ]1/p ≤ E[|X| p ]1/p + E[|Y | p ]1/p .
11. For a Wiener process, Wt −Wt0 ∼ N(0, σ 2 |t −t0 |). To simplify the notation, put σ̃ 2 :=
σ 2 |t − t0 |. Then
Z ∞
2
e−(x/σ̃ ) /2
|x| √
dx = 2
−∞
2π σ̃
Z ∞ −(x/σ̃ )2 /2
e
x √
dx
2π σ̃
r
Z ∞
2σ̃ −t 2 /2 ∞
2σ̃
|t − t0 |
−t 2 /2
= 2σ
dt = √
= √
−e
,
te
2π
0
0
2π
2π
E[|Wt −Wt0 |] =
0
which goes to zero as |t − t0 | → 0.
12. Let t0 > 0 be arbitrary. Since E[|Nt − Nt0 |2 ] = λ |t − t0 | + λ 2 |t − t0 |2 , it is clear that as
t → t0 , E[|Nt − Nt0 |2 ] → 0. Hence, Nt is continuous in mean square.
13. First write
R(t, s) − R(τ , θ ) = R(t, s) − R(t, θ ) + R(t, θ ) − R(τ , θ )
= E[Xt (Xs − Xθ )] + E[(Xt − Xτ )Xθ ].
Then by the Cauchy–Schwarz inequality,
q
q
|R(t, s) − R(τ , θ )| ≤ E[Xt2 ]E[(Xs − Xθ )2 ] + E[(Xt − Xτ )2 ]E[Xθ2 ],
which goes to zero as (t, s) → (τ , θ ). Note that we need the boundedness of E[Xt2 ] for
t near τ .
14.
(a) E[|Xt+T − Xt |2 ] = R(0) − 2R(T ) + R(0) = 0.
(b) Write
|R(t + T ) − R(t)| = E[(Xt+T − Xt )X0 ] ≤
q
E[|Xt+T − Xt |2 ]E[X02 ],
which is zero by part (a). Hence, R(t + T ) = R(t) for all t.
15. Write
k(Xn +Yn ) − (X +Y )k p = k(Xn − X) + (Yn −Y )k p ≤ kXn − Xk p + kYn −Y k p → 0.
16. Let t0 be arbitrary Since
kZt − Zt0 k p = k(Xt +Yt ) − (Xt0 +Yt0 )k p
= k(Xt − Xt0 ) + (Yt −Yt0 )k p
≤ kXt − Xt0 k p + kYt −Yt0 k p ,
it is clear that if Xt and Yt are continuous in mean of order p, then so is Zt := Xt +Yt .
218
Chapter 13 Problem Solutions
17. Let ε > 0 be given. Since kXn − Xk p → 0, there is an N such that for n ≥ N, kXn −
Xk p < ε /2. Thus, for n, m ≥ N,
kXn − Xm k p = k(Xn − X) + (X − Xm )k p ≤ kXn − Xk p + kX − Xm k p < ε /2 + ε /2 = ε .
18. Suppose Xn is Cauchy in L p . With ε = 1, there is an N such that for all n, m ≥ N,
kXn − Xm k p < 1. In particular, with m = N we have from
kXn k p − kXN k p ≤ kXn − XN k p
that
kXn k p ≤ kXn − XN k p + kXN k p < 1 + kXN k p ,
To get a bound that also holds for n = 1, . . . , N − 1, write
for n ≥ N.
kXn k p ≤ max(kX1 k p , . . . , kXN−1 k p , 1 + kXN k p ).
19. Since Xn converges, it is Cauchy and therefore bounded by the preceding two problems. Hence, we can write kXn k p ≤ B < ∞ for some constant B. Given ε > 0, let
n ≥ N imply kXn − Xk p < ε /(2kY kq ) and kYn −Y kq < ε /(2B). Then
kXnYn − XY k1 = E[|XnYn − XnY + XnY − XY |]
≤ E[|Xn (Yn −Y )|] + E[|(Xn − X)Y |]
≤ kXn k p kYn −Y kq + kXn − Xk p kY kq ,
ε
ε
+
kY kq = ε .
< B·
2B 2kY kq
by Hölder’s inequality,
20. If Xn converges in mean of order p to both X and Y , write
kX −Y k p = k(X − Xn ) + (Xn −Y )k p ≤ kX − Xn k p + kXn −Y k p → 0.
Since kX −Y k p = 0, E[|X −Y | p ] = 0.
21. For the rightmost inequality, observe that
kX −Y k p = kX + (−Y )k p ≤ kXk p + k −Y k p = kXk p + kY k p .
For the remaining inequality, first write
kXk p = k(X −Y ) +Y k p ≤ kX −Y k p + kY k p ,
from which it follows that
kXk p − kY k p ≤ kX −Y k p .
(∗)
Similarly, from
kY k p = k(Y − X) + Xk p ≤ kY − Xk p + kXk p ,
it follows that
kY k p − kXk p ≤ kY − Xk p = kX −Y k p .
From (∗) and (∗∗) it follows that
kXk p − kY k p ≤ kX −Y k p .
(∗∗)
Chapter 13 Problem Solutions
219
22. By taking pth roots, we see that
lim E[|Xn | p ] = E[|X| p ]
(#)
n→∞
is equivalent to kXn k p → kXk p . Since
kXk p − kY k p ≤ kX −Y k p ,
we see that convergence in mean of order p implies (#).
23. Write
kX +Y k2p + kX −Y k2p = hX +Y, X +Y i + hX −Y, X −Y i
= hX, Xi + 2hX,Y i + hY,Y i + hX, Xi − 2hX,Y i + hY,Y i
= 2(kXk2p + kY k2p ).
24. Write
|hXn ,Yn i − hX,Y i| = |hXn ,Yn i − hX,Yn i + hX,Yn i − hX,Y i|
≤ khXn − X,Y i| + |hX,Yn −Y i|
≤ kXn − Xk2 kYn k2 + kXk2 kYn −Y k2 → 0.
Here we have used the Cauchy–Schwarz inequality and the fact that since Yn converges, it is bounded.
25. As in the example, for n > m, we can write
kYn −Ym k22 ≤
n
n
∑ ∑
k=m+1 l=m+1
|hk | |hl | |hXk , Xl i|.
In this problem, hXk , Xl i = 0 for k 6= l. Hence,
kYn −Ym k22 ≤
n
∑
k=m+1
B|hk |2 ≤ B
n
∑
k=m+1
|hk |2 → 0
2
as n > m → ∞ on account of the assumption that ∑∞
k=1 |hk | < ∞.
26. Put Yn := ∑nk=1 hk Xk . It suffices to show that Yn is Cauchy in L p . Write, for n > m,
n
n
kYn −Ym k p =
∑
hk Xk
p
k=m+1
≤
∑
k=m+1
|hk | kXk k p = B1/p
n
∑
k=m+1
as n > m → ∞ on account of the assumption that ∑∞
k=1 |hk | < ∞.
27. Write
E[Y Z] = E
n
=
ν
X
(t
−
t
)
(s
−
s
)
X
∑ τi i i−1 ∑ θ j j j−1
n
i=1
ν
j=1
∑ ∑ R(τi , θ j )(ti − ti−1 )(s j − s j−1 ).
i=1 j=1
|hk | → 0
220
Chapter 13 Problem Solutions
28. First note that if
ν
n
Y :=
∑ g(τi )Xτi (ti − ti−1 )
and
Z :=
i=1
then
∑ g(θ j )Xθ j (s j − s j−1 ),
j=1
ν
n
∑ ∑ g(τi )R(τi , θ j )g(θ j )(ti − ti−1 )(s j − s j−1 ).
E[Y Z] =
i=1 j=1
Hence, given finer and finer partitions, with Ym defined analogously to Y above, we
see that
E[|Ym −Yk |2 ] = E[Ym2 ] − 2E[YmYk ] + E[Yk2 ]
→
Z bZ b
a
+
Z
a
g(t)R(t, s)g(s) dt ds − 2
a
bZ b
Z bZ b
a
g(t)R(t, s)g(s) dt ds
a
g(t)R(t, s)g(s) dt ds = 0.
a
Thus, Ym is Cauchy in L2 , and there exists a Y ∈ L2 with kYm −Y k2 → 0. Furthermore,
since Ym converges in mean square to Y , E[Ym2 ] → E[Y 2 ], and it is clear that
E[Ym2 ] →
Z bZ b
a
g(t)R(t, s)g(s) dt ds.
a
29. Consider the formula
Z T
0
R(t − s)ϕ (s) ds = λ ϕ (t),
0 ≤ t ≤ T.
(∗)
Since R is defined for all t, we can extend the definition of ϕ on the right-hand side in
the obvious way. Furthermore, since R has period T , so will the extended definition
of ϕ . Hence, both R and ϕ have Fourier series representations, say
R(t) =
∑ rn e j2π nt/T
and
ϕ (s) =
n
∑ ϕn e j2π ns/T .
n
Substituting these into (∗) yields
∑ rn e j2π nt/T
n
Z T
|
0
e− j2π ns/T ϕ (s) ds =
{z
}
∑ λ ϕn e j2π nt/T .
n
= T ϕn
It follows that Trn ϕn = λ ϕn . Now, if ϕ is an eigenfunction, ϕ cannot be the zero
function. Hence, there is at least one value of n with ϕn 6= 0. For all n with ϕn 6= 0,
λ = Trn . Thus,
ϕ (t) =
∑ ϕn e j2π nt/T .
n:rn =λ /T
Chapter 13 Problem Solutions
30. If
Rb
a
221
R(t, s)ϕ (s) ds = λ ϕ (t) then
Z bZ b
0 ≤
a
a
Z b
=
R(t, s)ϕ (t)ϕ (s) dt ds =
Z b
ϕ (t)[λ ϕ (t)] dt = λ
a
Z b
a
Z b
ϕ (t)
R(t, s)ϕ (s) ds dt
a
ϕ (t)2 dt.
a
Hence, λ ≥ 0.
31. To begin, write
λk
Z b
a
Z b Z b
λk ϕk (t)ϕm (t) dt =
R(t, s)ϕk (s) ds ϕm (t) dt
a
a
a
Z b
Z b
R(s,t)ϕm (t) dt ds, since R(t, s) = R(s,t),
ϕk (s)
=
ϕk (t)ϕm (t) dt =
Z b
=
Z b
a
a
a
ϕk (s) · λm ϕm (s) ds = λm
We can now write
(λk − λm )
If λk 6= λm , we must have
32.
(a) Write
Z T
Rb
a
Z b
a
Z b
a
ϕk (s)ϕm (s) ds.
ϕk (t)ϕm (t) dt = 0.
ϕk (t)ϕm (t) dt = 0.
R(t, s)g(s) ds =
0
Z T ∞
∑ λk ϕk (t)ϕk (s) g(s) ds
0
k=1
∞
=
∑ λk ϕk (t)
k=1
Z T
0
g(s)ϕk (s) ds =
∞
∑ λk gk ϕk (t).
k=1
(b) Write
Z T
R(t, s)ϕ (s) ds =
Z T
0
0
=
Z T
0
=
∞
R(t, s) g(s) − ∑ gk ϕk (s) ds
k=1
∞
R(t, s)g(s) ds − ∑ gk
k=1
∞
∞
k=1
k=1
Z T
|0
∑ λk gk ϕk (t) − ∑ λk gk ϕk (t)
R(t, s)ϕk (s) ds
{z
}
= λk ϕk (t)
= 0.
(c) Write
E[Z 2 ] =
Z TZ T
0
0
ϕ (t)R(t, s)ϕ (s) dt ds =
Z T
0
Z
ϕ (t)
0
T
R(t, s)ϕ (s) ds dt = 0.
222
Chapter 13 Problem Solutions
33. Consider the equation
λ ϕ (t) =
Z T
0
=
Z t
0
−t
= e
e−|t−s| ϕ (s) ds
(#)
Z T
e−(t−s) ϕ (s) ds +
Z t
t
t
s
e ϕ (s) ds + e
e−(s−t) ϕ (s) ds
Z T
t
0
e−s ϕ (s) ds.
Differentiating yields
λ ϕ 0 (t) = −e−t
= −e−t
Z t
0
Z t
es ϕ (s) ds + e−t et ϕ (t) + et
es ϕ (s) ds + et
Z T
t
0
Z T
t
e−s ϕ (s) ds + et (−e−t ϕ (t))
e−s ϕ (s) ds.
(##)
Differentiating again yields
λ ϕ 00 (t) = e−t
=
Z T
0
Z t
0
es ϕ (s) ds − e−t et ϕ (t) + et
Z T
t
e−s ϕ (s) ds + et (−e−t ϕ (t))
e−|t−s| ϕ (s) ds − 2ϕ (t)
= λ ϕ (t) − 2ϕ (t),
by (#),
= (λ − 2)ϕ (t).
We can now write
ϕ 00 (t) = (1 − 2/λ )ϕ (t) = −(2/λ − 1)ϕ (t).
p
For 0 < λ < 2, put µ := 2/λ − 1. Then ϕ 00 (t) = −µ 2 ϕ (t). Hence, we must have
ϕ (t) = A cos µ t + B sin µ t
for some constants A and B.
34. The required orthogonality principle is that
L
E (Xt − Xbt ) ∑ ci Ai = 0
i=1
for all constants ci , where
Xbt =
L
∑ cbj A j .
j=1
In particular, we must have E[(Xt − Xbt )Ai ] = 0. Now, we know from the text that
E[Xt Ai ] = λi ϕi (t). We also have
E[Xbt Ai ] =
L
∑ cbj E[A j Ai ]
j=1
= cbi λi .
Chapter 13 Problem Solutions
Hence, cbi = ϕi (t), and we have
Xbt =
223
L
∑ Ai ϕi (t).
i=1
35. Since kYn − Y k2 → 0, by the hint, E[Yn ] → E[Y ] and E[Yn2 ] → E[Y 2 ]. Since gn is
piecewise constant, we know that E[Yn ] = 0, and so E[Y ] = 0 too. Next, an argument
analogous to the one in Problem 21 tells us that if kgn − gk → 0, then kgn k → kgk.
Hence,
2
E[Y ] =
lim E[Yn2 ]
n→∞
= lim σ
n→∞
2
Z ∞
0
2
gn (t) dt = σ
2
Z ∞
g(t)2 dt.
0
36. First write
kY − Ye k2 = kY −Yn k2 + kYn − Yen k2 + kYen − Ye k2 ,
where the first and last terms on the right go to zero. As for the middle term, write
2 Z ∞
Z ∞
gen (t) dWt
gn (t) dWt −
kYn − Yen k22 = E
= E
Z
0
We can now write
0
0
∞
[gn (t) − gen (t)] dWt
2 = σ2
Z ∞
0
[gn (t) − gen (t)]2 dt.
kYn − Yen k2 = σ kgn − gen k ≤ σ kgn − gk + σ kg − gen k → 0.
37. We know from our earlier work that the Wiener integral
is linear on piecewise-conR
R
stant functions. To analyze theRgeneral case, let Y = 0∞ g(t) dWt and Z = 0∞ h(t) dWt .
We must show that aY + bZ = 0∞ ag(t) + bh(t) dWt . Let gn (t) and hn (t) be piecewiseconstant
functions suchR that kgn − gk → 0 and khn − hk → 0 and such that Yn :=
R∞
∞
g
(t)
dW
n
t and Zn := 0 hn (t) dWt converge in mean square to Y and Z, respectively.
0
Now observe that
aYn + bZn = a
Z ∞
0
gn (t) dWt + b
Z ∞
0
hn (t) dWt =
Z ∞
0
agn (t) + bhn (t) dWt ,
(∗)
since gn and hn are piecewise constant. Next, since
k(agn + bhn ) − (ag + bh)k ≤ |a| kgn − gk + |b| khn − hk → 0,
R
it follows that the right-hand side of (∗) converges in mean square to 0∞ ag(t) +
bh(t) dWt . Since the left-hand side of (∗) converges in mean square to aY + bZ, the
desired result follows.
38. We must find all values of β for which E[|Xt /t|2 ] → 0. First compute
Z t
2 Z t
t 2β +1
2
β
E[Xt ] = E
=
.
τ dWτ
τ 2β d τ =
2β + 1
0
0
Then t 2β +1 /t 2 = t 2β −1 → 0 if and only if 2β − 1 < 0; i.e., 0 < β < 1/2.
224
Chapter 13 Problem Solutions
39. Using the law of total probability, substitution, and independence, write
2 2
Z T
Z ∞ Z T
2
n
n
=
E
T = t fT (t) dt
τ dWτ
τ dWτ
E[YT ] = E
0
=
Z ∞ Z t
E
0
0
τ n dWτ
2
0
0
T = t fT (t) dt =
Z ∞ Z t
E
0
0
τ n dWτ
2 fT (t) dt.
Now use properties of the Wiener process to write
Z ∞ Z t
Z ∞ 2n+1
t
E[T 2n+1 ]
2
2n
E[YT ] =
τ d τ fT (t) dt =
fT (t) dt =
2n + 1
0
0
0 2n + 1
=
40.
(2n)!
(2n + 1)!/λ 2n+1
= 2n+1 .
2n + 1
λ
(a) Write
g(t + ∆t) − g(t) = E[ f (Wt+∆t )] − E[ f (Wt )]
= E[ f (Wt+∆t ) − f (Wt )]
≈ E[ f 0 (Wt )(Wt+∆t −Wt )] + 21 E[ f 00 (Wt )(Wt+∆t −Wt )2 ]
= E[ f 0 (Wt −W0 )(Wt+∆t −Wt )]
+ 12 E[ f 00 (Wt −W0 )(Wt+∆t −Wt )2 ]
= E[ f 0 (Wt −W0 )] · 0 + 21 E[ f 00 (Wt −W0 )] · ∆t.
It then follows that g0 (t) = 12 E[ f 00 (Wt )].
(b) If f (x) = ex , then g0 (t) = 21 E[eWt ] = 12 g(t). In this case, g(t) = et/2 , since g(0) =
E[eW0 ] = 1.
(c) We have by direct calculation that g(t) = E[eWt ] = es
2 t/2
s=1
= et/2 .
41. Let C be the ball of radius r, C := {Y ∈ L p : kY k p ≤ r}. For X ∈
/ C, i.e., kXk p > r, we
show that
r
Xb =
X.
kXk p
b p = r so that Xb ∈ C as
To begin, note that the proposed formula for Xb satisfies kXk
required. Now observe that
b p =
kX − Xk
X−
r
X
kXk p
p
= 1−
r
kXk p = kXk p − r.
kXk p
Next, for any Y ∈ C,
kX −Y k p ≥ kXk p − kY k p
b
Thus, no Y ∈ C is closer to X than X.
= kXk p − kY k p
≥ kXk p − r
b p.
= kX − Xk
Chapter 13 Problem Solutions
225
b i = 0 for
42. Suppose that Xb and Xe are both elements of a subspace M and that hX − X,Y
e
all Y ∈ M and hX − X,Y i = 0 for all Y ∈ M. Then write
e 22 = hXb − X,
e Xb − Xi
e = h(Xb − X) + (X − X),
e Xb − Xei = 0 + 0 = 0.
kXb − Xk
| {z }
∈M
43. For XbN is to be the projection of XbM onto N, it is sufficient that the orthogonality
principle be satisfied. In other words, it suffices to show that
hXbM − XbN ,Y i = 0,
Observe that
for all Y ∈ N.
hXbM − XbN ,Y i = h(XbM − X) + (X − XbN ),Y i = −hX − XbM ,Y i + hX − XbN ,Y i.
Now, the last term on the right is zero by the orthogonality principle for the projection
of X onto N, since Y ∈ N. To show that hX − XbM ,Y i = 0, observe that since N ⊂ M,
Y ⊂ N implies Y ⊂ M. By the orthogonality principle for the projection of X onto M,
hX − XbM ,Y i = 0 for Y ∈ M.
44. In the diagram, M is the disk and N is the horizontal line segment. The ◦ is XbM , the
projection of X onto the disk M. The is XbN , the projection of X onto line segment
N. The × is ([
XbM )N , the projection of the circle XbM onto the line segment N. We see
[
that (Xb ) 6= Xb .
M N
N
X
XbM
XbN
d
(XbM )N
45. Suppose that gn (Y ) ∈ M and gn (Y ) converges in mean square to some X. We must
show that X ∈ M. Since gn (Y ) converges, it is Cauchy. Writing
kgn (Y ) − gm (Y )k22 = E[|gn (Y ) − gm (Y )|2 ]
=
Z ∞
−∞
|gn (y) − gm (y)|2 fY (y) dy = kgn − gm kY ,
we see that gn is Cauchy in G, which is complete. Hence, there exists a g ∈ G with
kgn − gkY → 0. We claim X = g(Y ). Write
kg(Y ) − Xk2 = kg(Y ) − gn (Y ) + gn (Y ) − Xk2
≤ kg(Y ) − gn (Y )k2 + kgn (Y ) − Xk2 ,
where the last term goes to zero by assumption. Now observe that
kg(Y ) − gn (Y )k22 = E[|g(Y ) − gn (Y )|2 ] =
Z ∞
−∞
|g(y) − gn (y)|2 fY (y) dy
= kg − gn kY2 → 0.
Thus, X = g(Y ) ∈ M as required.
226
Chapter 13 Problem Solutions
R
46. We claim that the required projection is 01 f (t) dWt . Note that this is an element of M
since
Z 1
Z ∞
f (t)2 dt ≤
f (t)2 dt < ∞.
0
0
Consider the orthogonality condition
Z ∞
Z 1
Z
Z 1
E
f (t) dWt −
f (t) dWt
g(t) dWt = E
0
0
0
∞
f (t) dWt
1
Z
1
0
g(t) dWt .
Now put
f˜(t) :=
f (t), t > 1,
0, 0 ≤ t ≤ 1,
g̃(t) :=
and
so that the orthgonality condition becomes
Z ∞
Z ∞
Z
E
f˜(t) dWt
g̃(t) dWt
=
0
0
∞
0, t > 1,
g(t), 0 ≤ t ≤ 1,
f˜(t)g̃(t) dt,
0
which is zero since f˜(t)g̃(t) = 0 for all t ≥ 0.
47. The function g(t) will be optimal if the orthogonality condition
Z ∞
Z ∞
g̃(τ ) dWτ = 0
g(τ ) dWτ
E X−
0
0
R∞
holds for all g̃ with 0 g̃(τ )2 d τ . In particular, this must be true for g̃(τ ) = I[0,t] (τ ). In
this case, the above expectation reduces to
Z∞
Z ∞
Z ∞
E X
I[0,t] (τ ) dWτ − E
I[0,t] (τ ) dWτ .
g(τ ) dWτ
0
Now, since
0
0
Z ∞
0
I[0,t] (τ ) dWτ =
Z t
0
dWτ = Wt ,
we have the further simplification
E[XWt ] − σ 2
Z ∞
0
g(τ )I[0,t] (τ ) d τ = E[XWt ] − σ 2
Z t
g(τ ) d τ .
0
Since this must be equal to zero for all t ≥ 0, we can differentiate and obtain
g(t) =
48.
1 d
· E[XWt ].
σ 2 dt
(a) Using the methods of Chapter 5, it is not too hard to show that

y ≥ 1/4,


√ 1,

[2 − 1 −
4y
]/2,
0 ≤ y < 1/4,
√
FY (y) =
1
−
4y
]/2,
−2 ≤ y < 0,
[3/2
−
(1/2)



0,
y < −2.
Chapter 13 Problem Solutions
227
It then follows that

√
 1/ √1 − 4y, 0 < y < 1/4,
1/(2 1 − 4y ), −2 < y < 0,
fY (y) =

0,
otherwise.
(b) We must find a gb(y) such that
E[v(X)g(Y )] = E[b
g(Y )g(Y )],
for all bounded g.
For future reference, note that
E[v(X)g(Y )] = E[v(X)g(X(1 − X))] =
1
2
Z 1
−1
v(x)g(x(1 − x)) dx.
Now, by considering the problem of solving g(x) = y for x in the two cases
0 ≤ y ≤ 1/4 and −2 ≤ y < 0 suggests that we try
√
 √
 1 v 1+ 1−4y + 1 v 1− 1−4y , 0 ≤ y ≤ 1/4,
2
2
2
2
√
gb(y) =
1− 1−4y

v
,
−2 ≤ y < 0.
2
To check, we compute
E[b
g(Y )g(Y )] = E[b
g(X(1 − X))g(X(1 − X))]
Z 0 Z 1/2 Z 1 1
=
+
+
gb(x(1 − x))g(x(1 − x)) dx
2 −1
0
1/2
Z 0
1
v(x)g(x(1 − x)) dx
=
2 −1
+
Z 1/2
v(1 − x) + v(x)
2
0
+
Z 1
v(x) + v(1 − x)
1/2
1
=
2
=
49.
Z
Z 1
−1
2
g(x(1 − x)) dx
g(x(1 − x)) dx
0
−1
v(x)g(x(1 − x)) dx +
Z 1
0
v(x)g(x(1 − x)) dx
v(x)g(x(1 − x)) dx.
(a) Using the methods of Chapter 5, it is not too hard to show that
(
FΘ (sin−1 (y)) + 1 − FΘ (π − sin−1 (y)), 0 ≤ y ≤ 1,
FY (y) =
FΘ (sin−1 (y)) − FΘ (−π − sin−1 (y)), −1 ≤ y < 0.
Hence,

fΘ (sin−1 (y)) + fΘ (π − sin−1 (y))



p
, 0 ≤ y < 1,

1 − y2
fY (y) =
f (sin−1 (y)) + fΘ (−π − sin−1 (y))


 Θ
p
, −1 < y < 0.

1 − y2
228
Chapter 13 Problem Solutions
(b) Consider
E[v(X)g(Y )] = E[v(cos Θ)g(sin Θ)] =
Z π
−π
v(cos θ )g(sin θ ) fΘ (θ ) d θ .
Next, write
Z π /2
0
v(cos θ )g(sin θ ) fΘ (θ ) d θ =
Z 1
v(cos θ )g(sin θ ) fΘ (θ ) d θ =
Z 0
0
and
Z 0
−π /2
p
fΘ (sin−1 (y))
dy
v( 1 − y2 )g(y) p
1 − y2
p
fΘ (sin−1 (y))
v( 1 − y2 )g(y) p
dy.
−1
1 − y2
Similarly, we can write with a bit more work
Z π
π /2
v(cos θ )g(sin θ ) fΘ (θ ) d θ =
=
Z π /2
0
Z π /2
0
=
Z 1
0
and
Z −π /2
−π
v(cos(π − t))g(sin(π − t)) fΘ (π − t) dt
v(− cost)g(sint) fΘ (π − t) dt
p
fΘ (π − sin−1 (y))
p
v(− 1 − y2 )g(y)
dy,
1 − y2
v(cos θ )g(sin θ ) fΘ (θ ) d θ
=
=
=
Z 0
−π /2
Z 0
−π /2
v(cos(−π − t))g(sin(−π − t)) fΘ (−π − t) dt
v(− cost)g(sint) fΘ (−π − t) dt
Z 0
p
fΘ (−π − sin−1 (y))
p
v(− 1 − y2 )g(y)
dy.
−1
1 − y2
Putting all this together, we see that
E[v(X)g(Y )]
=
p
Z 1 p
v( 1 − y2 ) fΘ (sin−1 (y)) + v(− 1 − y2 ) fΘ (π − sin−1 (y))
0
fΘ (sin−1 (y)) + fΘ (π − sin−1 (y))
· g(y) fY (y) dy
p
Z 0 p
v( 1 − y2 ) fΘ (sin−1 (y)) + v(− 1 − y2 ) fΘ (−π − sin−1 (y))
+
fΘ (sin−1 (y)) + fΘ (−π − sin−1 (y))
−1
· g(y) fY (y) dy.
Chapter 13 Problem Solutions
229
We conclude that
E[v(X)|Y = y]
 √
√
−1
−1
2
2

 v( 1−y ) fΘ (sin −1(y))+v(− 1−y−1) fΘ (π −sin (y)) , 0 < y < 1,
fΘ (sin (y))+ fΘ (π −sin (y))
√
√
=
−1
−1
2
2

 v( 1−y ) fΘ (sin −1(y))+v(− 1−y )−1fΘ (−π −sin (y)) , −1 < y < 0.
f (sin (y))+ f (−π −sin (y))
Θ
Θ
50. Since X ≥ 0 and g(y) = IB (E[X|Y ]) ≥ 0, E[Xg(Y )] ≥ 0. On the other hand,
E[X|Y ]I(−∞,−1/n) (E[X|Y ]) ≤
−1
I
(E[X|Y ]),
n (−∞,−1/n)
and so
0 ≤ E[Xg(Y )] = E[E[X|Y ]g(Y )] ≤
−1
P(E[X|Y ] < −1/n) < 0,
n
which is a contradiction. Hence, P(E[X|Y ] < 0) = 0.
51. Write
E[Xg(Y )] = E[(X + − X − )g(Y )]
= E[X + g(Y )] − E[X − g(Y )]
= E E[X + |Y ]g(Y ) − E E[X − |Y ]g(Y )
= E E[X + |Y ] − E[X − |Y ] g(Y ) .
By uniqueness, we conclude that E[X|Y ] = E[X + |Y ] − E[X − |Y ].
52. Following the hint, we begin with
E[X|Y ] = E[X + |Y ] − E[X − |Y ] ≤ E[X + |Y ] + E[X − |Y ] = E[X + |Y ] + E[X − |Y ].
Then
E E[X|Y ] ≤ E E[X + |Y ] + E E[X − |Y ] = E[X + ] + E[X − ] < ∞.
53. To show that E[h(Y )X|Y ] = h(Y )E[X|Y ], we have to show that the right-hand side
satisfies the characterizing equation of the left-hand side. Since the characterizing
equation for the left-hand side is
E[{h(Y )X}g(Y )] = E E[h(Y )X|Y ]g(Y ) , for all bounded g,
we must show that
E[{h(Y )X}g(Y )] = E {h(Y )E[X|Y ]}g(Y ) ,
for all bounded g.
The only other thing we know is the characterizing equation for E[X|Y ], which is
E[Xg(Y )] = E E[X|Y ]g(Y ) , for all bounded g.
(∗)
230
Chapter 13 Problem Solutions
Since g in the above formula is an arbitrary and bounded function, and since h is
also bounded, we can rewrite the above formula by replacing g(Y ) with g(Y )h(Y ) for
arbitrary bounded g. We thus have
E[X{g(Y )h(Y )}] = E E[X|Y ]{g(Y )h(Y )} ,
for all bounded g,
which is equivalent to (∗).
54. We must show that E[X|q(Y )] satisfies the characterizing
equation
of E E[X|Y ] q(Y ) .
To write down the characterizing equation for E E[X|Y ] q(Y ) , it is convenient to use
the notation Z := E[X|Y ]. Then the characterizing equation for E[Z|q(Y )] is
E[Zg(q(Y ))] = E E[Z|q(Y )]g(q(Y )) ,
for all bounded g.
We must show that this equation holds when E[Z|q(Y )] is replaced by E[X|q(Y )]; i.e.,
we must show that
E[Zg(q(Y ))] = E E[X|q(Y )]g(q(Y )) ,
for all bounded g.
Replacing Z by its definition, we must show that
E[E[X|Y ]g(q(Y ))] = E E[X|q(Y )]g(q(Y )) ,
for all bounded g.
(∗∗)
We begin with the characterizing equation for E[X|Y ], which is
E[Xh(Y )] = E E[X|Y ]h(Y ) ,
for all bounded h.
Since h is bounded and arbitrary, we can replace h(Y ) by g(q(Y )) for arbitrary bounded g. Thus,
E[Xg(q(Y ))] = E E[X|Y ]g(q(Y )) ,
for all bounded g.
We next consider the characterizing equation of E[X|q(Y )], which is
E[Xg(q(Y ))] = E E[X|q(Y )]g(q(Y )) ,
for all bounded g.
Combining these last two equations yields (∗∗).
55. The desired result,
E[{h(Y )X}g(Y )] = E[{h(Y )E[X|Y ]}g(Y )],
can be rewritten as
E[(X − E[X|Y ])g(Y )h(Y )] = 0,
where g is a bounded function and h(Y ) ∈ L2 . But then g(Y )h(Y ) ∈ L2 , and therefore this last equation must hold by the orthogonality principle since E[X|Y ] is the
projection of X onto M = {v(Y ) : E[v(Y )2 ] < ∞}.
Chapter 13 Problem Solutions
231
56. Let hn (Y ) be bounded and converge to h(Y ). Then for bounded g,
h
i
E[{h(Y )X}g(Y )] = E lim hn (Y )Xg(Y )
n→∞
= lim E[X{hn (Y )g(Y )}]
n→∞
= lim E E[X|Y ]{hn (Y )g(Y )}
n→∞
h
i
= E lim hn (Y )E[X|Y ]g(Y )
n→∞
= E {h(Y )E[X|Y ]}g(Y ) .
By uniqueness, E[h(Y )X|Y ] = h(Y )E[X|Y ].
57. First write
n
Y = E[Y |Y ] = E[X1 + · · · + Xn |Y ] =
∑ E[Xi |Y ].
i=1
By symmetry, we must have E[Xi |Y ] = E[X1 |Y ] for all i. Then Y = nE[X1 |Y ], or
E[X1 |Y ] = Y /n.
58. Write
E[Xn+1 |Y1 , . . . ,Yn ] = E[Yn+1 +Yn + · · · +Y1 |Y1 , . . . ,Yn ]
= E[Yn+1 |Y1 , . . . ,Yn ] +Yn + · · · +Y1
= E[Yn+1 ] + Xn , by indep. & def. of Xn ,
= Xn ,
59. For n ≥ 1,
since E[Yn+1 ] = 0.
E[Xn+1 ] = E E[Xn+1 |Yn , . . . ,Y1 ] = E[Xn ],
where the second equality uses the definition of a martingale. Hence, E[Xn ] = E[X1 ]
for n ≥ 1.
60. For n ≥ 1,
E[Xn+1 ] = E E[Xn+1 |Yn , . . . ,Y1 ] ≤ E[Xn ],
where the inequality uses the definition of a supermartingale. Since Xn ≥ 0, E[Xn ] ≥ 0.
Hence, 0 ≤ E[Xn ] ≤ E[X1 ] for n ≥ 1.
61. Since Xn+1 := E[Z|Yn+1 , . . . ,Y1 ],
E[Xn+1 |Yn , . . . ,Y1 ] = E E[Z|Yn+1 , . . . ,Y1 ] Yn , . . . ,Y1
= E[Z|Yn , . . . ,Y1 ],
=: Xn .
by the smoothing property,
62. Since Xn := w(Yn ) · · · w(Y1 ), observe that Xn is a function of Y1 , . . . ,Yn and that Xn+1 =
w(Yn+1 )Xn . Then
E[Xn+1 |Yn , . . . ,Y1 ] = E[w(Yn+1 )Xn |Yn , . . . ,Y1 ] = Xn E[w(Yn+1 )|Yn , . . . ,Y1 ]
= Xn E[w(Yn+1 )],
by independence.
232
Chapter 13 Problem Solutions
It remains to compute
E[w(Yn+1 )] =
Z ∞
−∞
w(y) f (y) dy =
Z ∞ ˜
f (y)
−∞
f (y)
f (y) dy =
Z ∞
−∞
f˜(y) dy = 1.
Hence, E[Xn+1 |Yn , . . . ,Y1 ] = Xn ; i.e., Xn is a martingale with respect to Yn .
63. To begin, write
f˜Yn+1 ···Y1 (yn+1 , . . . , y1 )
fYn+1 ···Y1 (yn+1 , . . . , y1 )
f˜Y |Y ···Y (yn+1 |yn , . . . , y1 ) f˜Yn ···Y1 (yn , . . . , y1 )
·
.
= n+1 n 1
fYn+1 |Yn ···Y1 (yn+1 |yn , . . . , y1 ) fYn ···Y1 (yn , . . . , y1 )
wn+1 (y1 , . . . , yn+1 ) =
If we put
ŵn+1 (yn+1 , . . . , y1 ) :=
f˜Yn+1 |Yn ···Y1 (yn+1 |yn , . . . , y1 )
,
fYn+1 |Yn ···Y1 (yn+1 |yn , . . . , y1 )
then Xn+1 := wn+1 (Y1 , . . . ,Yn+1 ) = ŵn+1 (Yn+1 , . . . ,Y1 )Xn , where Xn is a function of
Y1 , . . . ,Yn . We can now write
E[Xn+1 |Yn , . . . ,Y1 ] = E[ŵn+1 (Yn+1 , . . . ,Y1 )Xn |Yn , . . . ,Y1 ]
= Xn E[ŵn+1 (Yn+1 , . . . ,Y1 )|Yn , . . . ,Y1 ].
We now show that this last factor is equal to one. Write
E[ŵn+1 (Yn+1 ,Yn , . . . ,Y1 )|Yn = yn , . . . ,Y1 = y1 ]
= E[ŵn+1 (Yn+1 , yn . . . , y1 )|Yn = yn , . . . ,Y1 = y1 ]
=
=
Z ∞
−∞
Z ∞
−∞
ŵn+1 (y, yn , . . . , y1 ) fYn+1 |Yn ···Y1 (y|yn , . . . , y1 ) dy
f˜Yn+1 |Yn ···Y1 (y|yn , . . . , y1 ) dy = 1.
64. Since Wk depends on Yk and Yk−1 , and since Xn = W1 + · · · + Wn , Xn depends on
Y0 , . . . ,Yn . Note also that Xn+1 = Xn +Wn+1 . Now write
E[Xn+1 |Yn , . . . ,Y0 ] = Xn + E[Wn+1 |Yn , . . . ,Y0 ].
Next,
Z ∞
E[Wn+1 |Yn = yn , . . . ,Y0 = y0 ] = E Yn+1 −
zp(z|Yn ) dz Yn = yn , . . . ,Y0 = y0
−∞
Z ∞
= E Yn+1 −
zp(z|yn ) dz Yn = yn , . . . ,Y0 = y0
−∞
= E[Yn+1 |Yn = yn , . . . ,Y0 = y0 ] −
=
Z ∞
−∞
Z ∞
−∞
z fYn+1 |Yn ···Y0 (z|yn , . . . , y0 ) dz −
zp(z|yn ) dz
Z ∞
−∞
zp(z|yn ) dz
Chapter 13 Problem Solutions
=
=
Z ∞
−∞
Z ∞
−∞
233
z fYn+1 |Yn (z|yn ) dz −
zp(z|yn ) dz −
Z ∞
−∞
Z ∞
−∞
zp(z|yn ) dz
zp(z|yn ) dz = 0.
Hence, E[Xn+1 |Yn , . . . ,Y0 ] = Xn ; i.e., Xn is a martingale with respect to Yn .
65. First write
E[Yn+1 |Xn = in , . . . , X0 = i0 ] = E ρ Xn+1 Xn = in , . . . , X0 = i0
=
∑ ρ j P(Xn+1 = j|Xn = in , . . . , X0 = i0 )
j
=
∑ ρ j P(Xn+1 = j|Xn = in ).
j
Next, for 0 < i < N,
∑ ρ j P(Xn+1 = j|Xn = i)
j
= ρ i−1 (1 − a) + ρ i+1 a
1 − a i−1
1 − a i+1
(1 − a) +
a
a
a
1 − a 2 i
1 − a i−1 h
1−a+
a
=
a
a
1 − a i−1 h
(1 − a)2 i
=
1−a+
a
a
1−ai
(1 − a)i h
1+
= ρ i.
=
ai−1
a
=
If Xn = 0 or Xn = 1, then Xn+1 = Xn , and so
E[ρ Xn+1 |Xn , . . . , X0 ] = E[ρ Xn |Xn , . . . , X0 ] = ρ Xn = Yn .
Hence, in all cases, E[Yn+1 |Xn , . . . , X0 ] = Yn , and so Yn is a martingale with respect to
Xn .
66. First, since Xn is a submartingale, it is clear that
An+1 := An + (E[Xn+1 |Yn , . . . ,Y1 ] − Xn ) ≥ 0
and is a function of Y1 , . . . ,Yn . To show that Mn is a martingale, write
E[Mn+1 |Yn , . . . ,Y1 ] = E[Xn+1 − An+1 |Yn , . . . ,Y1 ]
= E[Xn+1 |Yn , . . . ,Y1 ] − An+1
h
i
= E[Xn+1 |Yn , . . . ,Y1 ] − An + (E[Xn+1 |Yn , . . . ,Y1 ] − Xn )
= Xn − An = Mn .
67. Without loss of generality, assume f1 ≤ f2 . Then
E[Z f1 Z ∗f2 ] = E[Z f1 (Z f2 − Z f1 )∗ ] + E[|Z f1 |2 ]
234
Chapter 13 Problem Solutions
= E[(Z f1 − Z−1/2 )(Z f2 − Z f1 )∗ ] + E[|Z f1 |2 ]
= E[Z f1 − Z−1/2 ]E[(Z f2 − Z f1 )∗ ] + E[|Z f1 |2 ]
= 0 · 0 + E[|Z f1 |2 ] =
Z f1
−1/2
S(ν ) d ν .
68. Using the geometric series formula,
n+1
1 n j2π f n
1 e j2π f − e j2π f
e
= ·
∑
n k=1
n
1 − e j2π f
=
=
e j2π f 1 − e j2π f n
·
n
1 − e j2π f
e j2π f e jπ f n e− jπ f n − e jπ f n
e j2π f e jπ f n sin(π f n)
· jπ f · − jπ f
=
· jπ f ·
.
n
e
e
− e jπ f
n
e
sin(π f )
69. Following the hint, write
E[|Y |2 ] = E[YY ∗ ] = E
N
N
N
∑
dn Xn
n=−N
n=−N k=−N
N
N
n=−N k=−N
=
Z 1/2
Z 1/2
k=−N
N
∑ ∑
S( f )
−1/2
∗ dn dk∗ R(n − k)
S( f )e j2π f (n−k) d f
2
N
−1/2
∑
dk Xk
n=−N k=−N
dn dk∗
∑ ∑
=
N
N
dn dk∗ E[Xn Xk ] =
∑ ∑
=
∑
dn e j2π f n d f .
n=−N
Hence, if ∑Nn=−N dn e j2π f n = 0, then E[|Y |2 ] = 0.
70. Write
E[T (G0 )T (H0 )] = E
N
gn Xn
n=−N
N
=
∑
N
∑ ∑
n=−N k=−N
=
=
71.
Z 1/2
−1/2
Z 1/2
−1/2
gn h∗n
S( f )
N
∑
k=−N
Z 1/2
−1/2
N
∑
n=−N
hk Xk
gn e
∗ N
=
N
∑ ∑
n=−N k=−N
gn h∗n R(n − k)
S( f )e j2π f (n−k) d f
j2π f n
N
∑
k=−N
hk e
j2π f k
∗
df
S( f )G0 ( f )H0 ( f )∗ d f .
(a) Write
en ) + T (G
en ) −Y k2
kT (G) −Y k2 = kT (G) − T (Gn ) + T (Gn ) − T (G
en )k2 + kT (G
en ) −Y k2 .
≤ kT (G) − T (Gn )k2 + kT (Gn ) − T (G
Chapter 13 Problem Solutions
235
On the right-hand side, the first and third terms go to zero. To analyze the middle
term, write
en )k2 = kT (Gn − G
en )k2 = kGn − G
en k
kT (Gn ) − T (G
en k → 0.
≤ kGn − Gk + kG − G
(b) To show T is norm preserving on L2 (S), let Gn → G with T (Gn ) → T (G). Then
by Problem 21, kT (Gn )k2 → kT (G)k2 , and similarly, kGn k → kGk. Now write
kT (G)k2 = lim kT (Gn )k2 = lim kGn k2 ,
n→∞
n→∞
since Gn is a trig. polynomial,
= kGk.
(c) To show T is linear on L2 (S), fix G, H ∈ L2 (S), and let Gn and Hn be trigonometric polynomials converging to G and H, respectively. Then
α Gn + β Hn → α G + β H,
(∗)
and we can write
kT (α G + β H) − {α T (G) + β T (H)}k2
= kT (α G + β H) − T (α Gn + β Hn )k2
+ kT (α Gn + β Hn ) − {α T (Gn ) + β T (Hn )}k2
+ k{α T (Gn ) + β T (Hn )} − {α T (G) + β T (H)}k2 .
Now, the first term on the right goes to zero on account of (∗) and the defintion
of T . The second term on the right is equal to zero because T is linear on
trigonometric polynomials. The third term goes to zero upon observing that
k{α T (Gn ) + β T (Hn )} − {α T (G) + β T (H)}k2 ≤ |α | kT (Gn ) − T (G)k2
+ |β | kT (Hn ) − T (H)k2 .
(d) Using parts (b) and (c), kT (G) − T (H)k2 = kT (G − H)k2 = kG − Hk implies
that T is actually uniformly continuous.
72. It suffices to show that I[−1/2, f ] ∈ L2 (S). Write
Z 1/2
−1/2
73.
2
I[−1/2, f ] (ν ) S(ν ) d ν =
Z f
S(ν ) d ν ≤
−1/2
Z 1/2
S(ν ) d ν = R(0) = E[Xn2 ] < ∞.
−1/2
(a) We know that for trigonometric polynomials, E[T (G)] = 0. Hence, if Gn is a
sequence of trigonometric polynomials converging to G in L2 (S), then T (Gn ) →
T (G) in L2 , and then E[T (Gn )] → E[T (G)].
(b) For trigonometric polynomials G and H, we have
hT (G), T (H)i := E[T (G)T (H)∗ ] =
Z 1/2
−1/2
G( f )H( f )∗ S( f ) d f .
236
Chapter 13 Problem Solutions
If Gn and Hn are sequences of trigonometric polynomials converging to G and
H in L2 (S), then we can use the result of Problem 24 to write
Z 1/2
hT (G), T (H)i = lim hT (Gn ), T (Hn )i = lim
n→∞
n→∞ −1/2
Z 1/2
=
−1/2
Gn ( f )Hn ( f )∗ S( f ) d f
G( f )H( f )∗ S( f ) d f .
(c) For −1/2 ≤ f1 < f2 ≤ f3 < f4 ≤ 1/2, write
E[(Z f2 − Z f1 )(Z f4 − Z f3 )∗ ] = E[T (I( f1 , f2 ] )T (I( f3 , f4 ] )∗ ]
Z 1/2
=
74.
(a) We take
L2 (SY ) :=
−1/2
Z 1/2
G:
−1/2
I( f1 , f2 ] ( f )I( f3 , f4 ] ( f )S( f ) d f = 0.
|G( f )|2 SY ( f ) d f < ∞ .
(b) Write
∑
∞
∑
gn
n=−N
N
gn
n=−N
=
∑
gnYn =
n=−N
=
∞
N
N
TY (G0 ) =
Z 1/2
−1/2
∑
hk
k=−∞
Z 1/2
−1/2
∑
hk Xn−k
k=−∞
e j2π f (n−k) dZ f
G0 ( f )H( f ) dZ f .
(c) For G ∈ L2 (SY ), let Gn be a sequence of trigonometric polynomials converging
to G in L2 (SY ). Since TY is continuous,
TY (G) = lim TY (Gn ) = lim
n→∞
Z 1/2
n→∞ −1/2
Gn ( f )H( f ) dZ f
= lim T (Gn H) = T (GH),
=
n→∞
Z 1/2
−1/2
since T is continuous,
G( f )H( f ) dZ f .
(d) Using part (c),
V f := TY (I[−1/2, f ] ) =
=
Z 1/2
−1/2
Z f
−1/2
I[−1/2, f ] (ν )H(ν ) dZν
H(ν ) dZν .
(∗)
Chapter 13 Problem Solutions
237
(e) A slight generalization of (∗) establishes the result for piecewise-constant functions. For general G ∈ L2 (SY ), approximate G by a sequence of piecewiseconstant functions Gn and write
Z 1/2
−1/2
G( f ) dV f = lim
Z 1/2
n→∞ −1/2
Gn ( f ) dV f = lim
n→∞ −1/2
= lim T (Gn H) = T (GH) =
n→∞
Z 1/2
Z 1/2
−1/2
Gn ( f )H( f ) dZ f
G( f )H( f ) dZ f .
CHAPTER 14
Problem Solutions
1. We must show that P(|Xn | ≥ ε ) → 0. Since Xn ∼ Cauchy(1/n) has an even density,
we can write
P(|Xn | ≥ ε ) = 2
Z ∞
ε
1/(nπ )
dx = 2
(1/n)2 + x2
Z ∞
1/π
nε
1 + y2
dy → 0.
2. Since |cn − c| → 0, given ε > 0, there is an N such that for n ≥ N, |cn − c| < ε . For
such n,
{ω ∈ Ω : |cn − c| ≥ ε } = ∅,
and so P(|cn − c| ≥ ε ) = 0.
3. We show that Xn converges in probability to zero. Observe that Xn takes only the
values n and zero. Hence, for
√ 0 < ε < 1, |Xn | ≥ ε if and only if Xn = n, which
happens if and only if U ≤ 1/ n. We can now write
√
√
P(|Xn | ≥ ε ) = P(U ≤ 1/ n ) = 1/ n → 0.
4. Write
P(|Xn | ≥ ε ) = P(|V | ≥ ε cn ) = 2
Z ∞
ε cn
fV (v) dv → 0,
since cn → ∞.
5. Using the hint,
∞
[
P(X 6= Y ) = P
{|X −Y | ≥ 1/k} = lim P(|X −Y | ≥ 1/K).
K→∞
k=1
We now show that the above limit is zero. To begin, observe that
|Xn − X| < 1/(2K)
and
|Xn −Y | < 1/(2K)
imply
|X −Y | = |X − Xn + Xn −Y | ≤ |X − Xn | + |Xn −Y | < 1/(2K) + 1/(2K) = 1/K.
Hence, |X −Y | ≥ 1/K implies |Xn − X| ≥ 1/(2K) or |Xn −Y | ≥ 1/(2K), and we can
write
P(|X −Y | ≥ 1/K) ≤ P {|Xn − X| ≥ 1/(2K)} ∪ {|Xn −Y | ≥ 1/(2K)}
≤ P {|Xn − X| ≥ 1/(2K)} + P {|Xn −Y | ≥ 1/(2K)} ,
which goes to zero as n → ∞
238
Chapter 14 Problem Solutions
239
6. Let ε > 0 be given. We must show that for every η > 0, for all sufficiently large n,
P(|Xn − X| ≥ ε ) < η . Without loss of generality, assume η < ε . Let n be so large that
P(|Xn − X| ≥ η ) < η . Then since η < ε ,
P(|Xn − X| ≥ ε ) ≤ P(|Xn − X| ≥ η ) < η .
7.
(a) Observe that
P(|X| > α ) = 1 − P(|X| ≤ α ) = 1 − P(−α ≤ X ≤ α )
= 1 − [FX (α ) − FX ((−α )−)] = 1 − FX (α ) + FX ((−α )−)
≤ 1 − FX (α ) + FX (−α ).
Now, since FX (x) → 1 as x → ∞ and FX (x) → 0 as x → −∞, for large α , 1 −
FX (α ) < ε /8 and FX (−α ) < ε /8, and then P(|X| > α ) < ε /4. Similarly, for
large β , P(|Y | > β ) < ε /4.
(b) Observe that if the four conditions hold, then
|Xn | = |Xn − X + X| ≤ |Xn − X| + |X| < δ + α < 2α ,
and similarly, |Yn | ≤ 2β . Now that both (Xn ,Yn ) and (X,Y ) lie in the rectangle,
|g(Xn ,Yn ) − g(X,Y )| < ε .
(c) By part (b), observe that
{|g(Xn ,Yn ) − g(X,Y )| ≥ ε } ⊂ {|Xn − X| ≥ δ } ∪ {|Yn −Y | ≥ δ }
∪ {|X| > α } ∪ {|Y | > β }.
Hence,
P(|g(Xn ,Yn ) − g(X,Y )| ≥ ε ) ≤ P(|Xn − X| ≥ δ ) + P(|Yn −Y | ≥ δ )
+ P(|X| > α ) + P(|Y | > β )
< ε /4 + ε /4 + ε /4 + ε /4 = ε .
8.
(a) Since the Xi are i.i.d., the Xi2 are i.i.d. and therefore uncorrelated and have common mean E[Xi2 ] = σ 2 + m2 and common variance
2 2
E Xi2 − E[Xi2 ]
= E[Xi4 ] − E[Xi2 ] = γ 4 − (σ 2 + m2 )2 .
By the weak law of large numbers, Vn converges in probability to E[Xi2 ] = σ 2 +
m2 .
(b) Observe that Sn2 = g(Mn ,Vn )n/(n − 1), where g(m, v) := v − m2 is continuous.
Hence, by the preceding problem, g(Mn ,Vn ) converges in probability to g(m, v)
= (σ 2 + m2 ) − m2 = σ 2 . Next, by Problem 2, n/(n − 1) converges in probability
to 1, and then the product [n/(n − 1)]g(Mn ,Vn ) converges in probability to 1 ·
g(m, v) = σ 2 .
240
9.
Chapter 14 Problem Solutions
(a) Since Xn converges in probability to X, with ε = 1 we have P(|Xn − X| ≥ 1) → 0
as n → ∞. Now, if |X − Xn | < 1, then |X| − |Xn | < 1, and it follows that
|X| < 1 + |Xn | ≤ 1 + |Y |.
Equivalently,
Hence,
|X| ≥ |Y | + 1 implies
|Xn − X| ≥ 1.
P(|X| ≥ |Y | + 1) ≤ P(|Xn − X| ≥ 1) → 0.
(b) Following the hint, write
E[|Xn − X|] = E[|Xn − X|IAn ] + E[|Xn − X|IAnc ]
≤ E |Xn | + |X| IAn + ε P(Anc )
≤ E Y + |X| IAn + ε
≤ ε + ε,
since for large n, P(An ) < δ implies E Y + |X| IAn < ε . Hence, E[|Xn − X|] <
2ε .
10.
(a) Suppose g(x) is bounded, nonnegative, and g(x) → 0 as x → 0. Then given
ε > 0, there exists a δ > 0 such that g(x) < ε /2 for all |x| < δ . For |x| ≥ δ ,
we use the fact that g is bounded to write g(x) ≤ G for some positive, finite G.
Since Xn converges to zero in probability, for large n, P(|Xn | ≥ δ ) < ε /(2G).
Now write
E[g(Xn )] = E[g(Xn )I[0,δ ) (|Xn |)] + E[g(Xn )I[δ ,∞) (|Xn |)]
≤ E[(ε /2)I[0,δ ) (|Xn |)] + E[GI[δ ,∞) (|Xn |)]
ε
= P(|Xn | < δ ) + G P(|Xn | ≥ δ )
2
ε
ε
< +G
= ε.
2
2G
(b) By applying part (a) to the function g(x) = x/(1 + x), it follows that if Xn converges in probability to zero, then
|Xn |
= 0.
lim E
n→∞
1 + |Xn |
Now we show that if the above limit holds, then Xn must converge in probability
to zero. Following the hint, we use the fact that g(x) = 1/(1 + x) is an increasing
function for x ≥ 0. Write
E[g(Xn )] = E[g(Xn )I[ε ,∞) (|Xn |)] + E[g(Xn )I[0,ε ) (|Xn |)]
≥ E[g(Xn )I[ε ,∞) (|Xn |)]
≥ E[g(ε )I[ε ,∞) (|Xn |)]
= g(ε )P(|Xn | ≥ ε ).
Thus, if g(x) is nonnegative and nondecreasing, if E[g(Xn )] → 0, then Xn converges in distribution to zero.
Chapter 14 Problem Solutions
241
11. First note that for the constant random variable Y ≡ c, FY (y) = I[c,∞) (y). Similarly,
for Yn ≡ cn , FYn (y) = I[cn ,∞) (y). Since the only point at which FY is not continuous is
y = c, we must show that I[cn ,∞) (y) → I[c,∞) (y) for all y 6= c. Consider a y with c < y.
For all sufficiently large n, cn will be very close to c — so close that cn < y, which
implies FYn (y) = 1 = FY (y). Now consider y < c. For all sufficiently large n, cn will
be very close to c — so close that y < cn , which implies FYn (y) = 0 = FY (y).
12. For 0 < c < ∞, FY (y) = P(cX ≤ y) = FX (y/c), and FYn (y) = FX (y/cn ). Now, y is a
continuity point of FY if and only if y/c is a continuity point of FX . For such y, since
y/cn → y/c, FX (y/cn ) → FX (y/c). For c = 0, Y ≡ 0, and FY (y) = I[0,∞) (y). For y 6= 0,
+∞, y > 0,
y/cn →
−∞, y < 0,
and so
FYn (y) = FX (y/cn ) →
which is exactly FY (y) for y 6= 0.
13. Since Xt ≤ Yt ≤ Zt ,
1, y > 0,
0, y < 0,
{Zt ≤ y} ⊂ {Yt ≤ y} ⊂ {Xt ≤ y},
we can write
P(Zt ≤ y) ≤ P(Yt ≤ y) ≤ P(Xt ≤ y),
or FZt (y) ≤ FYt (y) ≤ FXt (y), and it follows that
lim FZt (y) ≤ lim FYt (y) ≤ lim FXt (y) .
t→∞
t→∞
| {z }
| {z }
t→∞
= F(y)
Thus, FYt (y) → F(y).
= F(y)
14. Since Xn converges in mean to X, the inequality
E[Xn ] − E[X] = E[Xn − X] ≤ E[|Xn − X|]
shows that E[Xn ] → E[X]. We now need the following implications:
conv. in mean ⇒ conv. in probability ⇒ conv. in distribution ⇒ ϕXn (ν ) → ϕX (ν ).
Since Xn is exponential, we also have
ϕXn (ν ) =
1/E[Xn ]
1/E[X]
→
.
1/E[Xn ] − jν
1/E[X] − jν
Since limits are unique, the above right-hand side must be ϕX (ν ), which implies X is
an exponential random variable.
15. Since Xn and Yn each converge in distribution to constants x and y, respectively, they
also converge in probability. Hence, as noted in the text, Xn +Yn converges in probability to x + y. Since convergence in probability implies convergence in distribution,
Xn +Yn converges in distribution to x + y.
242
Chapter 14 Problem Solutions
16. Following the hint, we note that each Yn is a finite linear combination of independent
Gaussian increments. Hence, each Yn is Gaussian. Since Y is the mean-square limit
of the Gaussian Yn , the distribution of RY is also Gaussian by the example cited in
theRhint. Furthermore, since each Yn = 0∞ gn (τ ) dWτ , Yn has zero mean and variance
σ 2 0∞ gn (τ )2 d τ = σ 2 kgn k2 . By the cited example, Y has zero mean. Also, since
kgn k − kgk ≤ kgn − gk → 0,
we have
var(Y ) = E[Y 2 ] = lim E[Yn2 ] = lim var(Yn )
n→∞
n→∞
= lim σ 2 kgn k2 = σ 2 kgk2 = σ 2
n→∞
Z ∞
g(τ )2 d τ .
0
17. For any constants c1 , . . . , cn , write
n
∑ ci Xti
i=1
n
=
∑ ci
i=1
Z ∞
0
g(ti , τ ) dWτ =
∑ ci g(ti , τ ) dWτ ,
Z ∞ n
0
which is normal by the previous problem.
|
i=1
{z
=: g(τ )
}
18. The plan is to show that the increments are Gaussian and uncorrelated. It will then
follow that the increments are independent. For 0 ≤ u < v ≤ s < t < ∞, write
 
Xu

Xv − Xu
−1 1 0 0 
 Xv  .
=

Xt − Xs
0 0 −1 1
Xs 
Xt
By writing
Xt :=
Z t
0
g(τ ) dWτ =
Z ∞
0
g(τ )I[0,t] (τ ) dWτ =
Z ∞
0
h(t, τ ) dWτ ,
where h(t, τ ) := g(τ )I[0,t] (τ ), we see that by the preceding problem, the vector on the
right above is Gaussian, and hence, so are the increments in the vector on the left.
Next,
Z t
Z v
g(τ ) dWτ
E[(Xt − Xs )(Xv − Xu )] = E
g(τ ) dWτ
s
u
Z ∞
Z ∞
= E
g(τ )I(s,t] (τ ) dWτ
g(τ )I(u,v] (τ ) dWτ
0
= σ
2
Z ∞
0
0
2
g(τ ) I(s,t] (τ )I(u,v] (τ ) d τ = 0.
19. Given a linear combination ∑nk=1 ck Ak , put g(τ ) := ∑nk=1 ck ϕk (τ ). Then
Zb n
Z b
Z b
n
n
∑ ck Ak = ∑ ck Xτ ϕk (τ ) d τ = Xτ ∑ ck ϕk (τ ) d τ = g(τ )Xτ d τ ,
k=1
k=1
a
a
k=1
a
Chapter 14 Problem Solutions
243
which is a mean-square limit of sums of the form
∑ g(τi )Xτi (ti − ti−1 ).
i
Since Xτ is a Gaussian process, these sums are Gaussian, and hence, so is their meansquare limit.
20. If MXn (s) → MX (s), then this holds when s = jν ; i.e., ϕXn (ν ) → ϕX (ν ). But this
implies that Xn converges in distribution to X. Similarly, if GXn (z) → GX (z), then
this holds when z = e jν ; i.e., ϕXn (ν ) → ϕX (ν ). But this implies Xn converges in
distribution to X.
21.
(a) Write
pn (k) = FXn (k + 1/2) − FXn (k − 1/2)
→ FX (k + 1/2) − FX (k − 1/2) = p(k),
where we have used the fact that since X is integer valued, k ±1/2 is a continuity
point of FX .
(b) The continuity points of FX are the noninteger values of x. For such x > 0,
suppose k < x < k + 1. Then
k
k
i=0
i=0
FXn (x) = P(Xn ≤ x) = P(Xn ≤ k) = ∑ pn (i) → ∑ p(i) = P(X ≤ k) = FX (x).
22. Let
n k
n!
pk (1 − pn )n−k
p (1 − pn )n−k =
pn (k) := P(Xn = k) =
k!(n − k)! n
k n
and p(k) := P(X = k) = λ k e−λ /k!. Next, by Stirling’s formula, as n → ∞,
√
2π nn+1/2 e−n
(n − k)!
qn :=
→ 1,
→ 1 and rn (k) := √
n!
2π (n − k)n−k+1/2 e−(n−k)
and so qn rn (k) → 1 as well. If we can show that pn (k)qn rn (k) → p(k), then
lim pn (k) = lim
n→∞
n→∞
lim pn (k)qn rn (k)
p(k)
pn (k)qn rn (k)
= n→∞
=
= p(k).
qn rn (k)
lim qn rn (k)
1
n→∞
Now write
lim pn (k)qn rn (k) = lim pn (k)
n→∞
n→∞
=
√
(n − k)!
2π nn+1/2 e−n
·√
n!
2π (n − k)n−k+1/2 e−(n−k)
e−k
nn+1/2
lim pkn (1 − pn )n−k ·
k! n→∞
(n − k)n−k+1/2
244
Chapter 14 Problem Solutions
=
=
=
=
nn+1/2
e−k
lim pkn (1 − pn )n−k · n−k+1/2
k! n→∞
n
(1 − k/n)n−k+1/2
(npn )k
e−k
(1 − pn )n−k
lim
·
k! n→∞ (1 − k/n)−k+1/2 (1 − k/n)n
[1 − (npn )/n]n (1 − pn )−k
e−k
(npn )k
·
lim
k! n→∞ (1 − k/n)−k+1/2
(1 − k/n)n
λ k e−λ
e−k λ k e−λ · 1
·
· −k =
= p(k).
k! 1
k!
e
Note that since npn → λ , pn → 0, and so (1 − pn )−k → 1.
23. First write
npn (z − 1) n
.
GXn (z) = [(1 − pn ) + pn z]n = [1 + pn (z − 1)]n = 1 +
n
Since npn (z − 1) → λ (z − 1), GXn (z) → eλ (z−1) = GX (z).
24.
(a) Here are the sketches:
I(−∞,a] (t)
1
a
b
I(−∞,b] (t)
1
ga,b (t)
1
t
a
b
t
a
b
t
(b) From part (a) we can write
I(−∞,a] (Y ) ≤ ga,b (Y ) ≤ I(−∞,b] (Y ),
from which it follows that
E[I(−∞,a] (Y )] ≤ E[ga,b (Y )] ≤ E[I(−∞,b] (Y )],
|
{z
}
{z
}
|
= P(Y ≤a)
or
= P(Y ≤b)
FY (a) ≤ E[ga,b (Y )] ≤ FY (b).
(c) Since FXn (x) ≤ E[gx,x+∆x (Xn )],
lim FXn (x) ≤ lim E[gx,x+∆x (Xn )] = E[gx,x+∆x (X)] ≤ FX (x + ∆x).
n→∞
n→∞
(d) Similarly,
FX (x − ∆x) ≤ E[gx−∆x,x (X)] = lim E[gx−∆x,x (Xn )] ≤ lim FXn (x).
n→∞
n→∞
(e) If x is a continuity point of FX , then given any ε > 0, for sufficiently small ∆x,
FX (x) − ε < FX (x − ∆x) and
FX (x + ∆x) < FX (x) + ε .
Chapter 14 Problem Solutions
245
Combining this with parts (c) and (d), we obtain
FX (x) − ε ≤ lim FXn (x) ≤ lim FXn (x) < FX (x) + ε .
n→∞
n→∞
Since ε > 0 is arbitrary, the liminf and the limsup are equal, limn→∞ FXn (x) exists
and is equal to FX (x).
25. If Xn converges in distribution to zero, then Xn converges in probability to zero, and
by Problem 10,
|Xn |
E
= 0.
1 + |Xn |
Conversely, if the above limit holds, then by Problem 10, Xn converges in probability
to zero, which implies convergence in distribution to zero.
26. Observe that fn (x) = n f (nx) implies

 1, x > 0,
F(0), x = 0,
Fn (x) = F(nx) →

0, x < 0.
So, for x 6= 0, Fn (x) → I[0,∞) (x), which is the cdf of X ≡ 0. In other words, Xn converges in distribution to zero, which implies convergence in probability to zero.
27. Since Xn converges in mean square to X, Xn converges in distribution to X. Since
2
g(x) := x2 e−x is a bounded continuous function, E[g(Xn )] → E[g(X)].
28. Let 0 < δ < 1 be given. For large t, we have |u(t) − 1| < δ , or
−δ < u(t) − 1 < δ
or
1 − δ < u(t) < 1 + δ .
Hence, for z > 0,
P(Zt ≤ z(1 − δ )) ≤ P(Zt ≤ zu(t)) ≤ P(Zt ≤ z(1 + δ )).
Rewrite this as
FZt (z(1 − δ )) ≤ P(Zt ≤ zu(t)) ≤ FZt (z(1 + δ )).
Then
F(z(1 − δ )) = lim FZt (z(1 − δ )) ≤ lim P(Zt ≤ zu(t))
t→∞
t→∞
and
lim P(Zt ≤ zu(t)) ≤ lim FZt (z(1 + δ )) = FZ (z(1 + δ )).
t→∞
t→∞
Since 0 < δ < 1 is arbitrary, and since F is continuous, we must have
lim P(Zt ≤ zu(t)) = F(z).
t→∞
The case for z < 0 is similar.
246
Chapter 14 Problem Solutions
29. Rewrite P(c(t)Zt ≤ z) as P(Zt ≤ z/c(t)) = P(Zt ≤ (z/c)(c/c(t)). Then if we put
u(t) := c/c(t), we have by the preceding problem that P(c(t)Zt ≤ z) → F(z/c).
30. First write FXt (x) = P(Zt + s(t) ≤ x) = P(Zt ≤ x − s(t)). Let ε > 0 be given. Then
s(t) → 0 implies that for large t, |s(t)| < ε , or
−ε < s(t) < ε
− ε < −s(t) < ε .
or
Then
FZt (x − ε ) = P(Zt ≤ x − ε ) ≤ FXt (x) ≤ P(Zt ≤ x + ε ) = FZt (x + ε ).
It follows that
F(x − ε ) ≤ lim FXt (x) ≤ lim FXt (x) ≤ F(x + ε ).
t→∞
t→∞
Since F is continuous and ε > 0 is arbitrary,
lim FXt (x) = F(x).
t→∞
31. Starting with Nbtc ≤ Nt ≤ Ndte , it is easy to see that
Nbtc
t
Xt := p
−λ
λ /t
Ndte
t
≤ Yt ≤ p
−λ
λ /t
=: Zt .
According to Problem 13, it suffices to show that Xt and Zt converge in distribution
to N(0, 1) random variables. By the preceding two problems, the distribution limit of
Zt is the same as that of c(t)Zt + s(t) if c(t) → 1 and s(t) → 0. We take
q
t dte
c(t) := q = t dte → 1
t dte
and
λ t dte
t dte − 1 q
λ
√
dte → 0.
= λ
s(t) := q − q λ
λ dte
λ dte
Finally, observe that
Ndte
−λ
dte
c(t)Zt + s(t) = q ,
λ dte
goes through the values of Yn and therefore converges in distribution to an N(0, 1)
random variable. It is similar to show that the distribution limit of Xt is also N(0, 1).
32.
(a) Let G := {Xn → X}. For ω ∈ G,
1
1
→
.
1 + Xn (ω )2
1 + X(ω )2
Since P(G c ) = 0, 1/(1 + Xn2 ) converges almost surely to 1/(1 + X 2 ).
Chapter 14 Problem Solutions
247
(b) Since almost sure convergence implies convergence in probability, which implies convergence in distribution, we have 1/(1 + Xn2 ) converging in distribution
to 1/(1 + X 2 ). Since g(x) = 1/(1 + x2 ) is bounded and continuous, E[g(Xn )] →
E[g(X)]; i.e.,
1
1
lim E
=
E
.
n→∞
1 + Xn2
1 + X2
33. Let GX := {Xn → X} and GY := {Yn → Y }, where P(GXc ) = P(GYc ) = 0. Let G :=
{g(Xn ,Yn ) → g(X,Y )}. We must show that P(G c ) = 0. Our plan is to show that
GX ∩ GY ⊂ G. It follows that G c ⊂ GXc ∪ GYc , and we can then write
P(G c ) ≤ P(GXc ) + P(GYc ) = 0.
For ω ∈ GX ∩ GY , (Xn (ω ),Yn (ω )) → (X(ω ),Y (ω )). Since g(x, y) is continuous, for
such ω , g(Xn (ω ),Yn (ω )) → g(X(ω ),Y (ω )). Thus, GX ∩ GY ⊂ G.
c ) = 0. Put G :=
34. Let GX := {Xn → X} and GXY := {X = Y }, where P(GXc ) = P(GXY
Y
c
{Xn → Y }. We must show that P(GY ) = 0. Our plan is to show that GX ∩ GXY ⊂ GY .
c , and we can then write
It follows that GYc ⊂ GXc ∪ GXY
c
P(GYc ) ≤ P(GXc ) + P(GXY
) = 0.
For ω ∈ GX ∩ GXY , Xn (ω ) → X(ω ) and X(ω ) = Y (ω ). Hence, for such ω , Xn (ω ) →
Y (ω ); i.e., ω ∈ GY . Thus, GX ∩ GXY ⊂ GY .
35. Let GX := {Xn → X} and GY := {Xn → Y }, where P(GXc ) = P(GYc ) = 0. Put GXY :=
c ) = 0. Our plan is to show that G ∩ G ⊂ G .
{X = Y }. We must show that P(GXY
X
Y
XY
c
c
c
It follows that GXY ⊂ GX ∪ GY , and we can then write
c
P(GXY
) ≤ P(GXc ) + P(GYc ) = 0.
For ω ∈ GX ∩ GY , Xn (ω ) → X(ω ) and Xn (ω ) → Y (ω ). Since limits of sequences of
numbers are unique, for such ω , X(ω ) = Y (ω ); i.e., ω ∈ GXY . Thus, GX ∩ GY ⊂ GXY .
T
c
36. Let GX := {Xn → X}, GY := {Yn → Y }, and GI := ∞
n=1 {Xn ≤ Yn }, where P(GX ) =
c
c
P(GY ) = P(GI ) = 0. This last equality follows because
P(GIc ) ≤
∞
∑ P(Xn > Yn )
= 0.
n=1
Let G := {X ≤ Y }. We must show that P(G c ) = 0. Our plan is to show that GX ∩
GY ∩ GI ⊂ G. It follows that G c ⊂ GXc ∪ GYc ∪ GIc , and we can then write
P(G c ) ≤ P(GXc ) + P(GYc ) + P(GIc ) = 0.
For ω ∈ GX ∩ GY ∩ GI , Xn (ω ) → X(ω ), Yn (ω ) → Y (ω ), and for all n, Xn (ω ) ≤ Yn (ω ).
By properties of sequences of real numbers, for such ω , we must have X(ω ) ≤ Y (ω ).
Thus, GX ∩ GY ∩ GI ⊂ G.
37. Suppose Xn converges almost surely to X, and Xn converges in mean to Y . Then Xn
converges in probability to X and to Y . By Problem 5, X = Y a.s.
248
Chapter 14 Problem Solutions
38. If X(ω ) > 0, then cn X(ω ) → ∞. If X(ω ) < 0, then cn X(ω ) → −∞. If X(ω ) = 0, then
cn X(ω ) = 0 → 0. Hence,

 +∞, if X(ω ) > 0,
0, if X(ω ) = 0,
Y (ω ) =

−∞, if X(ω ) < 0.
39. By a limit property of probability, we can write
∞ ∞
∞
\ [
[
P
{Xn = j} X0 = i = lim P
{Xn = j} X0 = i
M→∞
N=1 n=N
n=M
∞
∑ P(Xn = j|X0 = i)
M→∞
≤ lim
∞
∑
M→∞
= lim
n=M
(n)
pi j ,
n=M
(n)
which must be zero since the pi j are summable.
40. Following the hint, write
∞
\
E[S]
P(S = ∞) = P
{S > n} = lim P(S > N) ≤ lim
= 0.
N→∞
N→∞ N
n=1
41. Since Sn converges in mean to S, by the problems referred to in the hint,
n
n
∞
k=1
k=1
k=1
lim ∑ |hk |E[|Xk |] ≤ B1/p ∑ |hk | < ∞.
∑ E[|hk ||Xk |] = n→∞
n→∞
E[S] = lim E[Sn ] = lim
n→∞
42. From the inequality
Z n+1
1
t
n
dt ≥
p
Z n+1
n
1
1
dt =
,
p
(n + 1)
(n + 1) p
we obtain
∞ >
Z ∞
1
1
which implies
t
∞
dt =
p
p
∑∞
n=1 1/n
∑
Z n+1
1
n=1 n
t
dt ≥
p
∞
1
∑ (n + 1) p =
n=1
∞
1
∑ np ,
n=2
< ∞.
43. Write
n n
n n
1
E
X
X
X
X
i
j
m
l
∑
∑
∑
∑
n4
i=1
j=1
m=1
l=1
n
n
n
n
n
1
= 4 ∑ E[Xi4 ] + ∑ ∑ E[Xi Xi Xl Xl ] + ∑ ∑ E[Xi X j Xi X j ]
n i=1
i=1 l=1,l6=i
i=1 j=1, j6=i
n
n
n
n
3
+ ∑ ∑ E[Xi X j X j Xi ] + 4 ∑ ∑ E[Xi X j ]
i=1 j=1, j6=i
i=1 j=1, j6=i | {z }
E[Mn4 ] =
=0
1
= 4 [nγ + 3n(n − 1)σ 4 ].
n
Chapter 14 Problem Solutions
249
44. It suffices to show that ∑∞
n=1 P(|Xn | ≥ ε ) < ∞. To this end, write
P(|Xn | ≥ ε ) = 2
Then
Z ∞
ε
∞
∑ P(|Xn | ≥ ε ) =
n=1
n −nx
dx
2e
= −e−nx
∞
∞
ε
= e−nε .
e−ε
< ∞.
1 − e−ε
∑ (e−ε )n =
n=1
45. First use the Rayleigh cdf to write
P(|Xn | ≥ ε ) = P(Xn ≥ ε ) = e−(nε )
2 /2
.
Then
∞
∑ P(|Xn | ≥ ε )
∞
=
n=1
2 2
∑ e−ε n /2
n=1
∞
≤
2
∑ (e−ε /2 )n
=
n=1
1
< ∞.
2
1 − e−ε /2
Thus, Xn converges almost surely to zero.
46. For n > 1/ε , write
Z ∞
p − 1 −p
P(|Xn | ≥ ε ) =
x dx =
p−1
ε
Then
∞
∑ P(|Xn | ≥ ε )
n>1
=
∑
n<1/ε
n
P(|Xn | ≥ ε ) +
1
.
n p−1 ε p−1
1
ε p−1
∞
1
< ∞
p−1
n
n>1/ε
∑
since p − 1 > 1. Thus, Xn converges almost surely to zero.
47. To begin, first observe that
E[|Xn − X|] = E[|n2 (−1)nYn |] = n2 E[Yn ] = n2 pn .
Also observe that
P(|Xn − X| ≥ ε ) = P(n2Yn ≥ ε ) = P(Yn ≥ ε /n2 ) = P(Yn = 1) = pn .
In order to have Xn converge almost surely to X, it is sufficient to consider pn such
that ∑∞
n=1 pn < ∞.
2
1/2 6→ 0. For this choice of p ,
(a) If pn = 1/n3/2 , then ∑∞
n
n=1 pn < ∞, but n pn = n
Xn converges almost surely but not in mean to X.
2
(b) If pn = 1/n3 , then ∑∞
n=1 pn < ∞ and n pn = 1/n → 0. For this choice of pn , Xn
converges almost surely and in mean to X.
48. To apply the weak law of large numbers of this section to (1/n) ∑ni=1 Xi2 requires only
that the Xi2 be i.i.d. and have finite mean; there is no second-moment requirement on
Xi2 (which would be a requirement on Xi4 ).
250
Chapter 14 Problem Solutions
49. By writing
1 n
Nn
= ∑ Nk − Nk−1 ,
n
n k=1
which is a sum of i.i.d. Poisson(λ ) random variables, we see that Nn /n converges
almost surely to λ by the strong law of large numbers. We next observe that Nbtc ≤
Nt ≤ Ndte . Then
Nbtc
Ndte
Nt
≤
≤
,
t
t
t
and it follows that
Nbtc
Ndte
Nt
≤
≤
,
dte
t
btc
and then
Ndte dte
Nt
btc Nbtc
·
≤
.
≤
·
dte btc
t
dte btc
|{z} |{z}
|{z} |{z}
→1
→λ
→λ
→1
Hence Nt /t converges almost surely to λ .
50.
(a) By the strong law of large numbers, for ω not in a set of probability zero,
1 n
∑ Xk (ω ) → µ .
n k=1
Hence, for ε > 0, for all sufficiently large n,
1 n
∑ Xk (ω ) − µ < ε ,
n k=1
which implies
1 n
∑ Xk (ω ) − µ < ε ,
n k=1
from which it follows that
n
∑ Xk (ω )
< n(µ + ε ).
k=1
(b) Given M > 0, let ε > 0 and choose n in part (a) so that both n(µ + ε ) > M and
n ≥ M hold. Then
n
Tn (ω ) =
∑ Xk (ω )
< n(µ + ε ).
k=1
Now, for t > n(µ + ε ) ≥ Tn (ω ),
∞
Nt (ω ) =
∑ I[0,t] (Tk (ω ))
k=1
≥ n ≥ M.
Chapter 14 Problem Solutions
251
(c) As noted in the solution of part (a), the strong law of large numbers implies
Tn
1 n
= ∑ Xk → µ
n
n k=1
a.s.
Hence, n/Tn → 1/µ a.s.
(d) First observe that
Nt
Nt
≥
,
TNt
t
YNt =
and so
Nt
1
= lim YNt ≥ lim .
t→∞
t→∞ t
µ
Next,
YNt +1 =
and so
Nt 1
Nt + 1
Nt + 1
=
+ ,
≤
TNt +1
t
t
t
1
Nt
+ 0.
= lim YNt +1 ≤ lim
µ
t→∞
t→∞ t
Hence,
lim
t→∞
Nt
1
= .
t
µ
51. Let Xn := nI(0,1/n] (U), where U ∼ uniform(0, 1]. Then for every ω , U(ω ) ∈ (0, 1]. For
n > 1/U(ω ), or U(ω ) > 1/n, Xn (ω ) = 0. Thus, Xn (ω ) → 0 for every ω . However,
E[Xn ] = nP(U ≤ 1/n) = 1.
Hence, Xn does not converge in mean to zero.
52. Suppose Dε contains n points, say a ≤ x1 < · · · < xn ≤ b. Then
n
nε <
∑ G(xk +) − G(xk −)
k=1
n−1
n
= G(xn +) + ∑ G(xk +) − ∑ G(xk −) − G(x1 −)
k=1
n−1
k=2
n
≤ G(b) + ∑ G(xk+1 ) − ∑ G(xk−1 ) − G(a)
k=1
n−1
k=2
= G(b) + ∑ [G(xk+1 ) − G(xk )] − G(a)
k=1
≤ G(b) + G(xn ) − G(x1 ) − G(a) ≤ 2[G(b) − G(a)].
53. Write
∞
∞
[
[
P U∈
Kn = P
{U ∈ Kn } ≤
n=1
n=1
∞
∞
∑ P(U ∈ Kn )
n=1
≤
Since ε > 0 is arbitrary, the probability in question must be zero.
2ε
∑ 2n
n=1
= 2ε .
CHAPTER 15
Problem Solutions
1. Using the formula FWt (x) = FW1 (t −H x), we see that
FW1 (∞) = 1, x > 0,
lim FWt (x) =
FW1 (−∞) = 0, x < 0,
t↓0
which is the cdf of the zero random variable for x 6= 0. Hence, Wt converges in
distribution to the zero random variable.
Next,
X(ω ) := lim t H W1 (ω ) =
t→∞
Thus,
P(X = ∞) = P(W1 > 0) = 1 − FW1 (0),


∞, if W1 (ω ) > 0,
0, if W1 (ω ) = 0,

−∞, if W1 (ω ) < 0.
P(X = −∞) = P(W1 < 0) = FW1 (0−),
and
P(X = 0) = P(W1 = 0) = FW1 (0) − FW1 (0−).
2. Since the joint characteristic function of a zero-mean Gaussian process is completely
determined by the covariance matrix, we simply observe that
E[Wλ t1 Wλ t2 ] = σ 2 min(λ t1 , λ t2 ) = λ σ 2 min(t1 ,t2 ),
and
E[(λ 1/2Wt1 )(λ 1/2Wt2 )] = λ σ 2 min(t1 ,t2 ).
3. Fix τ > 0, and consider the process Zt := Wt − Wt−τ . Since the Wiener process is
Gaussian with zero mean, so is the process Zt . Hence, it suffices to consider the
covariance
E[Zt1 Zt2 ] = E[(Wt1 −Wt1 −τ )(Wt2 −Wt2 −τ )].
The time intervals involved do not overlap if t1 < t2 − τ or if t2 < t1 − τ . Hence,
0,
|t2 − t1 | > τ ,
E[Zt1 Zt2 ] =
σ 2 (|t2 − t1 | + τ ), |t2 − t1 | ≤ τ ,
which depends on t1 and t2 only through their difference.
4. For H = 1/2, qH (θ ) = I(0,∞) (θ ), and CH = 1. So,
BH (t) − BH (s) =
Z ∞
−∞
[I(−∞,t) (τ ) − I(−∞,s) (τ )] dWτ = Wt −Ws .
252
Chapter 15 Problem Solutions
253
5. It suffices to show that
Z ∞
1
[(1 + θ )H−1/2 − θ H−1/2 ]2 d θ < ∞.
Consider the function f (t) := t H−1/2 . By the mean-value theorem of calculus,
(1 + θ )H−1/2 − θ H−1/2 = f 0 (tˆ),
for some tˆ ∈ (θ , θ + 1).
Since f 0 (t) = (H − 1/2)t H−3/2 ,
(1 + θ )H−1/2 − θ H−1/2 ≤ |H − 1/2|/θ 3/2−H .
Then
Z ∞
1
[(1 + θ )H−1/2 − θ H−1/2 ]2 d θ ≤ (H − 1/2)2
Z ∞
1
1
θ 3−2H
d θ < ∞,
since 3 − 2H > 1.
6. The expression
M −µ
n
≤ y = 1−α
P
1−H
σ /n
says that
yσ
with probability 1 − α .
n1−H
Hence, the width of the confidence interval is
µ = Mn ±
2yσ
2yσ
= √ · nH−1/2 .
1−H
n
n
7. Suppose E[Yk2 ] = σ 2 k2H for k = 1, . . . , n. Substituting this into the required formula
yields
2
E[Yn+1
] − σ 2 n2H − σ 2 n2H − σ 2 (n − 1)2H = 2C(n),
which we are assuming is equal to
σ 2 (n + 1)2H − 2n2H + (n − 1)2H .
2 ] = σ 2 (n + 1)2H .
It follows that E[Yn+1
8.
(a) Clearly,
is zero mean. Also,
(m)
Xeν :=
(m) (m) E Xeν Xen
=
νm
∑
νm
∑
(Xk − µ )
k=(ν −1)m+1
nm
∑
k=(ν −1)m+1 l=(n−1)m+1
C(k − l)
254
Chapter 15 Problem Solutions
νm
∑
=
m
∑ C(k − [i + (n − 1)m])
k=(ν −1)m+1 i=1
m m
∑ ∑ C([ j + (ν − 1)m] − [i + (n − 1)m])
=
i=1 j=1
m m
∑ ∑ C( j − i + m(ν − n)).
=
i=1 j=1
(m)
Thus, Xeν
is WSS.
(b) From the solution of part (a), we see that
Ce(m) (0) =
m
m
∑ ∑ C( j − i)
i=1 j=1
= mC(0) + 2
m−1
∑ C(k)(m − k)
k=1
2 2H
= E[Ym2 ] = σ m
,
by Problem 7.
Thus, C(m) (0) = σ 2 m2H /m2 = σ 2 m2H−2 .
9. Starting with
σ∞2 C(m) (n)
|n + 1|2H − 2|n|2H + |n − 1|2H ,
=
m→∞ m2H−2
2
lim
put n = 0 to get
C(m) (0)
= σ∞2 .
m→∞ m2H−2
lim
Then observe that
(m)
E[(X1 − µ )2 ]
C(m) (0)
=
=
m2H−2
m2H−2
1 m
E
∑ (Xk − µ )
m k=1
m2H−2
2
.
10. Following the hint, observe that
2
2
2
2
Ce(m) (n)
1 E[Y(n+1)m ] − E[Ynm ] − E[Ynm ] − E[Y(n−1)m ]
=
m2H
2
m2H
2
2
2
E[Y(n+1)m ]
E[Y(n−1)m
] 1
2H E[Ynm ]
2H
(n + 1)2H
−
2n
+
(n
−
1)
=
2
[(n + 1)m]2H
(nm)2H
[(n − 1)m]2H
→
σ∞2
[(n + 1)2H − 2n2H + (n − 1)2H ].
2
11. If the equation cited in the text holds, then in particular,
Ce(m) (0)
= σ∞2 ,
m→∞ m2H−2
lim
Chapter 15 Problem Solutions
255
and
Ce(m) (n)/m2H−2
Ce(m) (n)
= lim
m→∞
m→∞ C
m→∞ C
e(m) (0)
e(m) (0)/m2H−2
1
= [|n + 1|2H − 2|n|2H + |n − 1|2H ].
2
Conversely, if both conditions hold,
lim ρ (m) (n) = lim
Ce(m) (0)ρ (m) (n)
Ce(m) (n)
= lim
2H−2
m→∞
m→∞ m
m2H−2
(m)
Ce (0)
= lim 2H−2 lim ρ (m) (n)
m→∞ m
m→∞
2
σ
= ∞ [|n + 1|2H − 2|n|2H + |n − 1|2H ].
2
lim
12. To begin, write
S( f ) = 1 − e− j2π f
= 2 sin(π f )
−2d
−2d
Then
= e− jπ f [e jπ f − e− jπ f ]
−2d
= 2j
−d
= 4 sin2 (π f ) .
C(n) =
=
e jπ f − e− jπ f
2j
Z 1/2 −1/2
−d
4 sin2 (π f ) e j2π f n d f
Z 1/2 −1/2
= 2
−d
4 sin2 (π f )
cos(2π f n) d f
Z 1/2 0
Z π
−d
cos(2π f n) d f
4 sin2 (π f )
−d
dν
cos(nν )
4 sin2 (ν /2)
2π
0
Z
−d
1 π
2
=
cos(nν ) d ν .
4 sin (ν /2)
π 0
= 2
Next, as suggested in the hint, apply the change of variable θ = 2π − ν to
1
π
to get
Z 2π π
−d
cos(nν ) d ν
4 sin2 (ν /2)
Z
−d
1 π
4 sin2 ([2π − θ ]/2)
cos(n[2π − θ ]) d θ ,
π 0
which, using a trigonometric identity, is equal to (∗). Thus,
Z
−d
1 2π cos(nν ) d ν
4 sin2 (ν /2)
2π 0
Z
−d
1 π
4 sin2 (t)
cos(2nt) dt.
=
π 0
C(n) =
−2d
(∗)
256
Chapter 15 Problem Solutions
By the formula provided in the hint,
cos(nπ )Γ(2 − 2d)2 p−1 21−p
(1 − 2d)Γ((2 − 2d + 2n)/2)Γ((2 − 2d − 2n)/2)
(−1)n Γ(1 − 2d)
.
=
Γ(1 − d + n)Γ(1 − d − n)
C(n) =
13. Following hint (i), let u = sin2 θ and dv = θ α −3 d θ . Then du = 2 sin θ cos θ d θ and
v = θ α −2 /(α − 2). Hence,
Z r
θ α −3 sin2 θ d θ =
ε
θ α −2 sin2 θ
α −2
r
ε
−
1
α −2
Z r
ε
θ α −2 sin 2θ d θ .
Next,
1
2−α
Z r
ε
θ
α −2
Z
2r
1
dt
(t/2)α −2 sint
sin 2θ d θ =
2 − α 2ε
2
Z 2r
α
−1
(1/2)
=
t α −2 sint dt
2−α
2ε
Z 2r
2r
21−α t α −1
1
sint +
=
t α −1 cost dt .
2−α α −1
1 − α 2ε
2ε
Now write
Z 2r
2ε
t α −1 cost dt = Re
Z 2r
2ε
t α −1 e− jt dt → Re e− jαπ /2 Γ(α ) = cos(απ /2)Γ(α )
as ε → 0 and r → ∞ by hint (iii). To obtain the complete result, observe that
θ α −2 sin2 θ = θ α
sin θ 2
θ
and t α −1 sint = t α
sint
t
both tend to zero as their arguments tend to zero or to infinity.
14.
(a) First observe that
∞
Q(− f ) =
Hence,
and then
1
∑ |i − f |2H+1 =
i=1
−1
1
∑ | − l − f |2H+1 =
l=−∞
h
S( f ) = Q(− f ) +
1
| f |2H+1
−1
1
.
2H+1
i=−∞ |i + f |
∑
i
+ Q( f ) sin2 (π f ),
S( f )
sin2 (π f )
2H−1
2
=
|
f
|
[Q(−
f
)
+
Q(
f
)]
sin
(
→ π 2.
π
f
)
+
| f |1−2H
f2
Chapter 15 Problem Solutions
257
(b) We have
Z 1/2
−1/2
S( f ) d f = σ 2 = π 2 ·
=
4 cos([1 − H]π )Γ(2 − 2H)
(2π )2−2H (2H − 1)2H
(2π )2H cos(π H)Γ(2 − 2H)
.
2H(1 − 2H)
15. Write
(−1)n Γ(1 − 2d)
Γ(n + 1 − d)Γ(1 − d − n)
(−1)n Γ(1 − 2d)
=
Γ(n + 1 − d)(−1)n Γ(d)Γ(1 − d)/Γ(n + d)
Γ(n + d)
Γ(1 − 2d)
·
.
=
Γ(1 − d)Γ(d) Γ(n + 1 − d)
C(n) =
Now, with ε = 1 − d, observe that
Γ(n + 1 − ε )
(n + 1 − ε )n+1−ε −1/2 e−(n+1−ε )
∼
Γ(n + ε )
(n + ε )n+ε −1/2 e−(n+ε )
= e1−2d
nn+1/2−ε [1 + (1 − ε )/n)]n+1/2−ε
nn+ε −1/2 (1 + ε /n)n+ε −1/2
= e1−2d n2d−1
[1 + (1 − ε )/n)]n+1/2−ε
.
(1 + ε /n)n+ε −1/2
Thus,
1−2d Γ(n + 1 − ε )
n
Γ(n + ε )
1−2d
→ e
1/2−ε
e1−ε
1−2d 1/2−ε
= e1/2−d .
e
ε −1/2 = e
eε
Thus, α = 1 − 2d and c = e1/2−d Γ(1 − 2d)/[Γ(1 − d)Γ(d)].
16. Evaluating the integral, we have
In
n1−α
=
n1−α − k1−α
1 − (k/n)1−α
1
=
→
.
(1 − α )n1−α
1−α
1−α
Now, given a small ε > 0, for large n, In /n1−α > 1/(1 − α ) − ε , which implies
In /n−α > n(1/(1 − α ) − ε ) → ∞. With Bn as in the hint, we have from Bn + n−α −
k−α ≤ In ≤ Bn that
1
In
Bn
Bn
k−α
+
≤ 1−α ≤ 1−α
+
α
α
1−
1−
n
n n
n
n
or
In
1−
n α
Thus, Bn /n1−α → 1/(1 − α ).
≤
Bn
In
1 k−α
≤ 1−α − − 1−α .
α
1−
n
n
n n
258
Chapter 15 Problem Solutions
17. We begin with the inequality
n−1
∑ ν 1−α ≤
ν =k
| {z }
Z n
n−1
t 1−α dt ≤ ∑ (ν + 1)1−α .
k
ν =k
| {z }
|
{z
}
=: In
=: Bn
= Bn −k1−α +n1−α
Next, since In = (n2−α − k2−α )/(2 − α ),
In
n2−α
1 − (k/n)2−α
1
→
.
2−α
2−α
=
Also,
In
n2−α
≤
and
Bn
k1−α 1
− 2−α +
2−
α
n
n
n
In
Bn
≤ 2−α
n2−α
n
imply Bn /n2−α → 1/(2 − α ) as required.
18. Observe that
∞
∑ |hn |
∞
=
n=q
n
∑ ∑
αk bn−k
n=q k=n−q
∞ ∞
≤
=
∑ ∑ |αk ||bn−k |I[n−q,n] (k)
n=q k=0
∞
∞
∑ |αk | ∑ |bn−k |I[0,q] (n − k)
k=0
≤ M
n=q
∞
|b
|
∑ i ∑ (1 + δ /2)−k
q
i=0
q
= M ∑ |bi |
i=0
k=0
1
< ∞.
1 − 1/(1 + δ /2)
19. Following the hint, we first compute
∑ αm−nYn = ∑ αm−n ∑ ak Xn−k =
n
n
=
k
∑ Xl ∑ αm−n an−l
l
n
=
∑ αm−n ∑ an−l Xl
n
l
∑ Xl ∑ αm−(ν +l) aν ,
l
ν
|
{z
= δ (m−l)
}
where the reduction to the impulse follows because the convolution of αn and an
corresponds in the z-transform domain to the product [1/A(z)] · A(z) = 1, and 1 is the
transform of the unit impulse. Thus, ∑n αm−nYn = Xm .
Chapter 15 Problem Solutions
259
Next,
∑ αm−n
∑ αm−nYn =
n
n
∑ bk Zn−k
k
∑ Zl ∑ αm−n bn−l
=
n
l
=
∑ αm−n ∑ bn−l Zl
=
n
l
∑ Zl ∑ αm−(l+k) bk
l
k
|
{z
= hm−l
}
since this last convolution corresponds in the z-transform domain to the product
[1/A(z)] · B(z) =: H(z).
20. Since Xn is WSS, E[|Xn |2 ] is a finite constant. Since ∑∞
k=0 |hk | < ∞, we have by an
example in Chapter 13 or by Problem 26 in Chapter 13 that ∑m
k=0 hk Xn−k converges in
mean square as m → ∞. By another example in Chapter 13,
∞
∞
E[Yn ] = E ∑ hk Xn−k = ∑ hk E[Xn−k ].
k=0
k=0
If E[Xn−k ] = µ , then
∞
E[Yn ] =
∑ hk µ
k=0
is finite and does not depend on n. Next, by the continuity of the inner product
(Problem 24 in Chapter 13),
∞
m
m
E Xl ∑ hk Xn−k
= lim E Xl ∑ hk Xn−k
= lim ∑ hk E[Xl Xn−k ]
m→∞
k=0
m→∞
k=0
= lim
m→∞
∑ hk RX (l − n + k)
k=0
∞
m
=
k=0
∑ hk RX ([l − n] + k).
k=0
Similarly,
E[YnYl ] = E
= lim
m→∞
∞
=
m
∑ hk Xn−k Yl = lim E ∑ hk Xn−k Yl
∞
m→∞
k=0
m
∑ hk E[Xn−kYl ]
= lim
k=0
∑ hk RXY ([n − l] − k).
k=0
Thus, Xn and Yn are J-WSS.
k=0
m
m→∞
∑ hk RXY (n − k − l)
k=0