1.
Chapter 6 Problems
P
(a) Because P{X < k} = k−1
j=0 f(j) for k > 1 and P{X < 0} = 0,




∞
∞
∞
∞
k−1
X
X
X
X
X


sk  f(j)
P{X < k}sk =
f(j) sk =
k=0
k=1
=
∞
X
j=0
(b) P{X > k} =
j=0
j=0
k=j+1
sj+1
s
f(j) =
G(s).
1−s
1−s
P∞
f(j) for all k > 1; therefore,


!
j
∞
∞
∞
∞
X
X
X
X
X
k
k
k

P{X > k}s =
f(j) s =
s f(j)
j=k
k=0
k=0
=
2.
j=0
∞
X
1 − sj+1
j=0
=
j=k
1−s
k=0

f(j) =
1 
1−s
∞
X
f(j) − s
j=0
∞
X

sj f(j)
j=0
1
(1 − sG(s)) .
1−s
(a) Here, n is a fixed integer > 1:
n
X
sk
1
G(s) =
=
n
n
k=1
n
X
!
k
s −1
k=0
1
=
n
1 − sn+1
−1
1−s
=
s − sn+1
.
n(1 − s)
(b) Again, n is a fixed integer > 1:
G(s) =
n
X
k=−n
n
n
X
X
sk
1
1
=
sk =
s−n
sk+n .
2n + 1
2n + 1
2n + 1
k=−n
k=−n
A change of variables [j := k + n] shows that
2n
s−n
s−n X j
1 − s2n+1
s−n + sn+1
G(s) =
s =
=
.
2n + 1
2n + 1
1−s
(2n + 1)(1 − s)
j=0
(c) Write f(k) = k−1 − (k + 1)−1 to find that
Z
∞
∞
∞ Zs
∞
X
X
X
sk X sk
1 s k
G(s) =
−
=
xk−1 dx −
x dx
k
k+1
s 0
k=1
k=1
k=1 0
k=1
Z ∞
Z Zs X
Zs
∞
1 s
1
1 sX k
dx
k−1
−
− 1 dx
=
x
dx −
x dx =
s 0
s 0 1−x
0 k=1
0 1−x
k=1
1
1
1
1
1
= ln
−
ln
−s =1−
− 1 ln
.
1−s
s
1−s
s
1−s
1
(d) This one is nonsense, sadly, because f(k) < 0 if k < 0; therefore, f is
not a probability mass function.
(e) Recall that Z denotes the collection of all integers [positive as well
as negative]. Therefore,
!
∞
−1
∞
X
X
1 − c X |k| k 1 − c
k
k
1+
(cs) +
(s/c)
c s =
G(s) =
1+c
1+c
k=1
k=−∞
k=−∞


∞
X
1−c
cs
cs
1−c
c/s
j
=
1+
.
1+
+
(c/s)
=
+
1+c
1 − cs
1+c
1 − cs 1 − (c/s)
j=1
Is it really the case that
wise all is lost.
3.
P
k∈Z
f(k) = 1? This is important, for other-
(a) Let N be a Poisson(λ) r.v., and recall that GN (s) = exp{−λ(1 − s)}
(Example 12, p. 238). Therefore,
exp {−λ(1 − GX (s))} = GN (GX (s)).
This and Theorem 13 (p. 242) together show that exp{−λ(1 − GX (s))}
P
is the generating function of N
i=0 Xi , where the Xi ’s are independent random variable, all independent of Y; it is convergent if and
only if GX (s) < ∞. [Question: Does it matter that Theorem 13 is
P
about N
i=1 Xi , that is the sum starts from one there? The answer is
“no,” but you should sort this out.]
(b) Since 0 < s < 1, G(s) := sin(πs/2) is positive increasing [as a function of s], and at most one. The Taylor series for sin is
∞
sin x = x −
X (−x)j+1
x3 x5
+
··· =
.
3!
5!
(2j + 1)!
j=1
Therefore,
G(s) =
∞
X
(−πs/2)j+1
j=1
(2j + 1)!
.
If this were a probability generating function, then the coefficient of
sj+1 is f(j+1); but that produces nonsense; for example plug in j = 1
to obtain f(2) = −π/(2 × 5!) < 0, which cannot be.
(c) Throughout, I assume that r is a positive integer.
If p = 0 then G(s) = 1 for all s. Therefore, G is the generating
function for X ≡ 0 [that is, f(0) = 1 and f(s) = 0 for s 6= 0]. If p = 1,
then q = 0 and so G(s) = 0 for all s. In that case, G(1) = 0 6= 1, and
so G is not a p.g.f.
2
Next consider the interesting case that 0 < p < 1. Let X1 , . . . , Xr be
independent random variables each with p.g.f.,
q
GXj (s) =
.
1 − ps
According to Example 4 (p. 233), if |s| < p−1 , then GXj are well
defined. Let Y = X1 + · · · + Xr to find that as long as |s| < p−1 ,
GY (s) = E sX1 +···+Xr = E sX1 × · · · × E sXr
r
q
= GX1 (s) × · · · × GXr (s) =
.
1 − ps
See also Example 6 (p. 245).
(d) This is a Binomial(r , p) example; see Example 11, p. 238.
(e) Yes as long as |s| < 1; see Example 2 (p. 259).
(f) According to Example 16 (p. 242), if
P{X = k} =
pk
k log(1/q)
(k > 1),
and 0 < p < 1 with q := 1 − p, then
GX (s) =
log(1 − sp)
.
log q
And of course this is defined only when 1 − sp > 0; i.e., when s <
p−1 . Therefore, if α log(1 + βs) is a p.g.f., then β = −p and α =
1/ log q, with s < p−1 .
Chapter 7 Problems
1. I will assume that α < β; this can be done without incurring a loss in
generality.
If α < x < β, then we want
Zβ
Zβ
1 = c(α , β) (x − α)(β − x) dx = c(α , β)
−αβ + (α + β)x − x2 dx
α
α
β2 − α2 β3 − α3
−
= c(α , β) −αβ(β − α) + (α + β)
2
3
2
1
β
+
αβ
+
α2
2
= c(α , β)(β − α) −αβ + (α + β) −
2
3
2
α + β2 β2 + αβ + α2
−
= c(α , β)(β − α)
2
3
(α − β)2
1
= c(α , β)(β − α) ×
= c(α , β)(β − α)3 .
6
6
Therefore, c(α , β) = 6(β − α)−3 . If x is not between α and β then f ≡ 0.
3
6. We want
Za
exp −x − e−x dx
F(a) =
for all real numbers a.
−∞
Change variables [z := e−x ]: For all a ∈ R,
Z exp(−a)
e−w dw = exp(−e−a ).
F(a) =
−∞
9. First of all,
a
P X>x+
x
P{X > x + ax−1 }
X>x =
.
P{X > x}
The numerator and the denominator both tend to zero as x → ∞. Therefore, by the l’Hôpital’s rule,
a
lim P X > x +
x→∞
x
d
−1
}
dx P{X > x + ax
.
d
x→∞
dx P{X > x}
X > x = lim
Now,
d
d
1
2
(1 − FX (x)) = −fX (x) = − √ e−x /2 .
P{X > x} =
dx
dx
2π
Also,
d
d a a a
1 − FX x +
= −1 + 2 fX x +
.
P{X > x + ax−1 } =
dx
dx
x
x
x
Since
2
a
1
a2
x + 2a + (a/x)2
fX x +
= √ exp −
= fX (x) exp −a − 2 ,
x
2
2x
2π
it follows that
a
lim P X > x +
x→∞
x
−fX (a + ax−1 )
a2
X > x = lim
= lim exp −a − 2 = e−a .
x→∞
x→∞
−fX (x)
2x
4