1.
Excercise 1.2.8
Solution
Ck , k = 1, 2, . . . is a sequence of increasing sets.
(a) Ck = {x : 1/k ≤ x ≤ 3 − 1/k}, k = 1, 2, 3, . . .
As k → ∞, 1/k → 0 and (3 − 1/k) → 3.
∞
[
lim Ck =
n→∞
Ck = (0, 3)
k=1
Note that the two points 0 and 3 are not in any of Ck for k = 1, 2, 3, . . .. Therefore,
these two points are not included in limn→∞ Ck above.
(b) Ck = {(x, y) : 1/k ≤ x2 + y 2 ≤ 4 − 1/k}
As k → ∞, 1/k → 0 and (4 − 1/k) → 4.
lim Ck =
n→∞
∞
[
Ck = {(x, y) : 0 < x2 + y 2 < 4}
k=1
2.
Excercise 1.2.9
Solution
Ck , k = 1, 2, . . . is a sequence of decreasing sets.
(a) Ck = {x : 2 − 1/k < x ≤ 2}
As k → ∞, (2 − 1/k) → 2. The l.h.s of the inequality is approaching 2 from left then.
By definition,
∞
\
lim =
n→∞
Ck = {2}
k=1
Since 2 is the only element that is in all Ck s.
(b) Ck = {x : 2 < x ≤ 2 + 1/k} As k → ∞, (2 + 1/k) → 2. The r.h.s of the inequality is
approaching 2 from right then.
By definition,
lim =
n→∞
∞
\
k=1
1
Ck = ∅
Since there is no single element could be found that is in all Ck s.
To see this, we could assmue there exists a sufficiently small δ > 0 such that x = 2+δ.
Certainly, x > 2 must hold then. However, we could always find another sufficiently
large k such that 1/k < δ. Then it follows that x = 2+δ > 2+1/k, which contradicts
our assumption.
(c) Ck = {(x, y) : 0 ≤ x2 + y 2 ≤ 1/k}
As k → ∞, 1/k → 0.
Note the right part of the inequalities above includes equality. Consequently, 0 is the
only element in the shrinking Ck sequence.
Hence,
lim =
n→∞
∞
\
Ck = {(0, 0)}
k=1
3.
Excercise 1.3.10
Solution
(a) Follow the hint, denote
Ci := {an exact match on the ith turn
Also, by the general inclusion-exclusion formula, we could get
pM = Pr {at least one exact match}
= P (C1 ∪ C2 ∪ . . . ∪ C52 )
52
52
52
52
=
P (Ci ) −
P (Ci ∩ Cj ) +
P (Ci ∩ Cj ∩ Ck ) − . . . −
P (∩52
i=1 Ci )
1
2
3
52
52
Further, define p2 = 52
2 P (Ci ∩ Cj ), p3 = 3 P (Ci ∩ Cj ∩ Ck ), . . .. Then,
pM = p1 − p2 + p3 − . . . − p52
1
1
1
= 1 − + − ... −
2! 3!
52!
as desired.
2
(?)
(b) Recall the Taylor expansion of ex ,
∞
X xi
1
ex = 1 + x + x2 + . . . =
2
i!
i=0
Evaluating the above equation at x = −1, we get
e−1 = 1 − 1 +
1
1
− + ...
2 3!
1 − e−1 = 1 −
1
1
+ − ...
2! 3!
Then we could get
Note that (?) is similar as the above equation, then we reach
1 − e−1 ≈ pM
4.
Excercise 1.3.15
Solution
The number of all possible outcomes:
The number of possible outcoms of two red chips:
The number of possible outcoms of two blue chips:
The number of possible outcoms of two 1s: 22 · 60
The number of possible outcoms of two 2s: 22 · 60
The number of possible outcoms of two 3s: 22 · 60
8
2
5
2 ·
3
2 ·
6
0
6
0
Then,
Pr {2 chips randomly drawn have either same color or same number}
6
6
6
5
3
2
2 · 0 + 2 · 0 +3· 2 · 0
=
8
2
5.
Excercise 1.4.8
Solution
Let D denote the event that a spring is defective; and I, II, III the events that
3
a spring is produced by Machcine I, II and III, respectively.
We know the following probabilities
P (I) = 1%,
P (II) = 4%,
P (III) = 2%,
P (D | I) = 30%
P (D | II) = 25%
P (D | III) = 45%
(a) Randomly draw a spring, find the probability that is defective.
P (D) = P (D | I) · P (I) + P (D | II) · P (II) + P (D | III) · P (III)
= 0.3 × 0.01 + 0.25 × 0.04 + 0.45 × 0.02
= 0.022
(b) Given a randomly drawn spring is defective, what the probability that is produced by
Machcine II ?
P (D | II) · P (II)
P (D)
0.04 × 0.25
=
0.022
= 5/11
P (II|D) =
6.
Excercise 1.4.19
Solution
Define A := {at least 4 draws to obtain a spade}.
Then we could have the following,
Ac = {at most 3 draws to obtain a spade}
= {obtain a spade at the 1st draw}+
{obtain a spade at the 2st draw}+
{obtain a spade at the 3st draw}
4
(a) Draw cards with replacement.
P (A) = 1 − P (Ac )
13 52 − 13 13
52 − 13 2 13
=1−
+
·
+(
) ·
52
52
52
52
52
27
=
64
(b) Draw cards without replacement.
P (A) = 1 − P (Ac )
13 52 − 13 13 52 − 13 38 13
+
·
+
·
·
=1−
52
52
51
52
51 50
13 3 13 3 38 13
=1−
+ ·
+ ·
·
52 4 51 4 51 50
7.
Excercise 1.4.32
Solution
Suppose that we could select p1 and p2 satisfies the following equation
P (zero hits) = P (one hit) = P (two hits)
Note that by assuming independence, and plugging p1 , p2 to the chain of equation above

(1 − p1 )(1 − p2 ) = (1 − p1 )p2 + (1 − p2 )p1
(1 − p )p + (1 − p )p = p p
1
2
2
1
1 2
Solving for the above system, we could get
p1 = 1/3p2 ,
3p22 − 3p2 + 1 = 0
It then turns out that we couldn’t get a real root for p2 . Thus, it is not possible to select
p1 , p2 satisfying the condition.
8.
Find pmf of Y = X 3 , where
pX (x) =


1 2
2
0
x = 1, 2, . . .
otherwise
5
Excercise 1.6.8
Solution
First, identify the support of Y , SY = {1, 8, 27, . . .}.
pY (y) = P (Y = y)
X
=
P (X = x)
x∈g −1 (y)
X
=
pX (x)
x∈g −1 (y)
√
= pX ( 3 y)
√
3y
1
=
2
Hence, we could have the pmf of Y as
pY (y) =


1
2
√
3y
For y ∈ SY
0
otherwise
9.
Excercise 1.7.9
Solution
Median of distribution.
(a)
p(x) =


4!
x!(4−x)!
1 x
4
3 4−x
4
0
x = 0, 1, 2, 3, 4
otherwise
Computing P (X ≤ x) and P (X < x) for x = 1, 2, 3, 4, it easy to see that
3
3 4
3
4! 1
3
3
189
1
+
=
=
>
4
3! 4
4
4
256
2
81
1
P (X < 1) = p(0) =
<
256
2
P (X ≤ 1) = p(0) + p(1) =
Then the median for this distribution is 0.
6
(b)

3x2
f (x) =
0
0<x<1
otherwise
Since point probability of continuous r.v. is 0, then
P (X ≤ x) = P (X < x) =⇒ P (X ≤ x) = 1/2
Calculate the cdf,
F (x) = P (X ≤ x)
Z x
3t2 dt
=
0
x
= t3 0
= x3
Then, by solving x3 = 1/2 yields the median for this distribution is
p
3
1/2.
(c)
f (x) =
1
, −∞ < x < ∞
π(1 + x2 )
Calculate the cdf as
F (X) = P (X ≤ x)
Z x
1
=
dt
2
−∞ π(1 + t )
Z
1 x
1
=
dt
π −∞ 1 + t2
1
= [arctan t]x−∞
π
1
1
= arctan x +
π
2
Then, by solving
1
π
arctan x +
1
2
=
1
2
yields the median for this distribution is x = 0.
7
10.
Excercise 1.7.21
Solution

2xe−x2
f (x) =
0
0<x<∞
otherwise
Let X , Y denote the support of the r.v. X and Y, respectively.
Note the transformation Y = g(X) = X 2 is monotone on X ; the support transformation
is then
X = {x : 0 < x < ∞},
Y = {y : 0 < y < ∞}
1
Identify the inverse as g −1 (y) = y 2 , for y ∈ Y.
Compute the Jacobian of transformation
d −1 1 − 1
J = g (y) = y 2
dy
2
Now we could calculate the pdf of Y within the support Y
fY (y) = fX (g −1 (y)) · J = e−y
Then, we could get the pdf of Y as

e−y
fY (y) =
0
0<y<∞
otherwise
11.
Excercise 1.8.10
Solution

2x
f (x) =
0
0<x<1
otherwise
8
(a) E[1/x]
Compute the expectation directly as
1
E[ ] =
X
Z
1
0
Z
1
=
0
1
· f (x) dx
x
1
· 2x dx
x
=2
(b) Compute the cdf and pdf of Y = 1/X
Identify the inverse as g −1 (y) = 1/y and Y = (1, ∞), also note that g(·) is monotonically decreasing.
Also get the cdf of X as FX (x) =
Rx
0
2t dt = x2 , then
FY (y) = 1 − FX (g −1 (y)) = 1 − y −2
d 1
−2
Compute the Jacobian of transformation as J = dy
y = y , then the cdf is
fY (y) =


2
y3
0
1<y<∞
otherwise
(c) Calculate the expectation by definition
∞
Z
y · fY (y) dx
EY =
Z−∞
∞
=
y · 2y −3 dy
1
∞
= −2 · y −1 1
=2
9