Chapter 4 Problems
5.
(a)
i. Recall first that c = 1/(e2 − 1) here. Therefore,
EX =
∞
∞
1 X
2x
1 X 2x
x
·
=
e2 − 1 x=1
x!
e2 − 1 x=1 (x − 1)!
∞
=
2 X 2j
2
e −1
j!
[j = x − 1]
j=0
=
2e2
,
e2 − 1
the last identity following from the following Taylor–McLaurin
1 2
formula: et = 1 + t + 2!
t + ···.
ii. Recall that c = (1 − p)/p. Therefore,
∞
∞
X
1−p X x
xpx−1
xp = (1 − p)
p x=1
x=1
0
∞
X
d
1
1
= (1 − p)
px = (1 − p)
=
.
dp x=0
1−p
1−p
EX =
iii. Recall that c = ln(1/(1 − p)). Therefore,
X
∞
px
1
x
px
EX = ln
= ln
x
1
−
p
x=1
x=1
X
∞
1
1
p
x
0
= ln
p −p
= ln
.
1−p
1
−
p
1
−
p
x=0
1
1−p
X
∞
iv. Recall that c = π2 /6. Therefore,
EX =
∞
∞
π2 X
1
π2 X 1
x· 2 =
= +∞.
6 x=1
x
6 x=1 x
Note that this is a well-defined sum, but it also diverges.
v. Recall that c = 1. Therefore,
EX =
∞
X
∞
x·
x=1
X 1
1
=
= +∞.
x(x + 1) x=1 x + 1
10. Note that for all integers k > 0,
f(k + 1)
e−λ λk+1 /(k + 1)!
λ
=
=
.
f(k)
e−λ λk /k!
k+1
1
Whenever k + 1 > λ, f(k + 1) < f(k). This shows that
max f(x) = max f(k).
x
16k6[λ]
But if k < λ, i.e., k + 1 6 λ, then f(k + 1)/f(k) > 1. This proves that
maxx f(x) = f([λ]).
12. Let X1 and X2 be independent random variables with respective mass
functions f1 and f2 . We toss an independent p-coin; if it comes up heads,
then we set Y := X1 ; else, Y := X2 . Then,
P{Y = x} = P{Y = x | tails} · P{tails} + P{Y = x | heads} · P{heads}
= P{X2 = x}(1 − p) + P{X1 = x}p = (1 − p)f2 (x) + pf1 (x),
whence it follows that the mass function of Y is f3 .
13. By the definition of conditional probabilities,
P{X > n + k | X > n} =
P{X > n + k , X > n}
.
P{X > n}
The numerator is equal to P{X > n + k} [why?]. Therefore,
P∞
j−1
j=n+k+1 p(1 − p)
P{X > n + k | X > n} = P∞
i−1
i=n+1 p(1 − p)
P∞
(1 − p)l
= Pl=n+k
.
∞
m
m=n (1 − p)
P
m
Because ∞
= p−1 (1 − p)N ,
m=N (1 − p)
P{X > n + k | X > n} =
(1 − p)n+k
= (1 − p)k = P{X > k}.
(1 − p)n
14. I am assuming that a 6 b and a and b are integers between 1 and m.
Now,
P{X = k , a 6 X 6 b}
P{X = k | a 6 X 6 b} =
.
P{a 6 X 6 b}
If k is not an integer between a and b, then P{X = k , a 6 X 6 b} = 0;
else, it is equal to P{X = k} = 1/m. The denominator is
b
X
P{X = j} =
j=a
n
X
1
b−a+1
=
.
m
m
j=a
Therefore,
P{X = k | a 6 X 6 b} =
1/m
1
=
,
(b − a + 1)/m
b−a+1
if k is an integer between a and b; and P{X = k | a 6 X 6 b} = 0, otherwise. This all means that given that X is between a and b, the conditional
distribution of X is uniform on {a , . . . , b}.
2