Chapter 6 Problems
2. (a) For all real numbers s 6= 1,
"∞
#
n
∞
X
1X k
1 X k
G(s) =
s =
s −1−
sk
n
n
k=1
k=0
k=n+1
1
sn+1
s − sn+1
1
−1−
=
.
=
n 1−s
1−s
n(1 − s)
If s = 1, then G(s) = 1, trivially.
(b) For all real numbers s 6= 1,
n
n
2n
X
1
s−n X k+n
s−n X j
sk =
s
=
s
2n + 1
2n + 1
2n + 1 j=0
k=−n
k=−n
1 − s2n+1
s−n − sn+1
s−n
=
.
=
2n + 1
1−s
(2n + 1)(1 − s)
G(s) =
If s = 1, then G(s) = 1, trivially.
(c) We write, for all k ≥ 1,
1
1
1
= −
.
k(k + 1)
k k+1
Therefore, as long as |s| < 1,
G(s) =
∞
X
sk
k=1
k
−
∞
X
sk
.
k+1
k=1
Now,
∞
X
sk
k=1
k
=
∞ Z
X
k=1
Z s
=
0
s
k−1
x
s
Z
dx =
0
0
1
dx = ln
1−x
∞
X
!
k−1
x
k=1
1
1−s
.
Also,
∞
∞
∞
X
sk
1 X sk+1
1 X sj
=
=
k+1
s
k+1
s j=2 j
k=1
k=1


∞
1 X sj
= 
− s
s j=1 j
1
1
=
ln
−s
s
1−s
1
1
= ln
− 1.
s
1−s
1
dx
Therefore, whenever |s| < 1,
1
1
G(s) = 1 + 1 −
ln
.
s
1−s
(d) If G converges [absolutely], then we can write it
G(s) =
∞
X
k=1
−1
X
sk
sk
+
.
2k(k + 1)
2k(k − 1)
k=−∞
The first sum converges if |s| ≤ 1, whereas the second if |s| ≥ 1.
Therefore, the only convergent values are for s = ±1. Moreover,
G(1) = 1 because after a change of variables,
∞
X
N
X
1
1
= lim
k(k + 1) N →∞
k(k + 1)
k=1
k=1
N
X
1
1
1
= lim
−
= lim 1 −
= 1.
N →∞
N →∞
k k+1
1+N
G(1) =
k=1
(e) Clearly,
G(s) =
∞
1 − c X k |k|
s c
1+c
k=−∞
=
=
∞
−1
1−c X
1 − c X c −k
(sc)k +
1+c
1+c
s
1−c
1+c
k=0
∞
X
(sc)k +
k=0
1−c
1+c
k=−∞
∞ X
j=1
c j
.
s
The first sum converges as long as |s| ≤ 1. But the second converges
only if |s| > c. Therefore, as long as c < |s| ≤ 1,


∞ j
X
1−c
1
1−c
c
G(s) =
×
+
− 1
1 + c 1 − sc 1 + c j=0 s
1−c
1
1−c
1
=
×
+
−1 .
1 + c 1 − sc 1 + c 1 − (c/s)
2
17. According to Example (6) on page 245,
n
1 − (λ/n)
G(s) =
1 − λs/n
n −n
λ
λs
= 1−
1−
n
n
→
e−λ
= e−λ(1−s) .
e−λs
This proves that the mass function of Xn converges to that of Poisson(λ)
as n → ∞. Also,
n
p
= pn (1 − qs)−n
G(s) =
1 − qs
G0 (s) = nqpn (1 − qs)−n−1
G00 (s) = n(n + 1)q 2 pn (1 − qs)−n−2 .
Therefore,
G0 (1) = nqpn (1 − q)−n−1 = nqpn p−n−1 =
G00 (1) = n(n + 1)q 2 pn p−n−2 =
nq
p
n(n + 1)q 2
.
p2
Consequently, E(X) = nq/p, and
Var(X) = G00 (1) + G0 (1) − [G0 (1)]2
(p. 237)
2
n(n + 1)q 2
nq
nq
=
+
−
p2
p
p
2
nq
nq
= 2 +
p
p
nq
nq q
+1 = 2.
=
p p
p
Chapter 7 Problems
1. Because (x − α)(β − x) = −x2 + (β − α)x − αβ, f (x) has to be zero unless x
lies between α and β. If α = β, then this cannot be done. If α < β, then
we choose c(α , β) so that
Z β
2
1
=
−x + (β − α)x − αβ dx
c(α, β)
α
β 3 − α3
(β − α)(β 2 − α2 )
=−
+
− αβ(β − α).
3
2
Else, if α > β, then c(α , β) is minus one times the reciprocal of the
preceding term.
3
9. Compute directly to find that
P{X > x + (a/x)}
a P X > x + X > x =
x
P{X > x}
R
2
e−y /2 dy
x+(a/x)
= R ∞ −y2 /2
e
dy
x
R
0
−y 2 /2
e
dy
x+(a/x)
∼
,
R∞
−y 2 /2 dy 0
e
x
where “prime” denotes d/dx. We apply the fundamental theorem of calculus to find that
0
2
Z ∞
x
−y 2 /2
e
dy = − exp −
.
2
x
Also, by the fundamental theorem of calculus and the change rule,
!0
!
Z ∞
2
(x + (a/x))
a 0
−y 2 /2
e
dy = − exp −
× x+
2
x
x+(a/x)
!
2
a
(x + (a/x))
= − exp −
× 1− 2
2
x
!
2
(x + (a/x))
∼ − exp −
2
2
2
x + 2a
x + 2a + (a2 /x2 )
∼ exp −
.
= exp −
2
2
The claim of the problem follows from these computations.
10. (a) F|X| (a) = P{|X| ≤ a} = P{−a ≤ X ≤ a} = Φ(a) − Φ(−a), provided
that a ≥ 0. Else, F|X| (a) = 0. By symmetry, Φ(a) = 1 − Φ(−a),
whence follows the claim.
(b) As before, F|X| (a) = F (a)−F (−a). Differentiate to find that f|X| (a) =
f (a) + f (−a), when the density f of X exists.
26. For any integer k ≥ 1, X ≥ k if and only log U/ log(1−p) ≥ k−1. Therefore,
log U
P{X ≥ k} = P
≥k−1
log q
= P {log U ≤ (k − 1) log q}
(because log q ≤ 0)
Z qk−1
= P U ≤ q k−1 =
dy = q k−1 .
0
Therefore, F (n) = 1 − P{X ≥ n + 1} = 1 − q n for all n ≥ 0. It follows that
f (n) = P{X = n} = F (n) − F (n − 1) = q n−1 − q n = q n−1 (1 − q) = pq n−1
for all n ≥ 0. This proves that X is geometric(p).
4