Chapter 4 Problems
3. Let Xequal the number of dots. Then, f(x) = 1
6if x=1, . . . , 6; else,
f(x) = 0. Consequently,
EX =1×1
6+···+6×1
6=7
2
E(X2) = 12×1
6+···+62×1
6=91
6.
Therefore
Var X=91
6−7
22
=35
12 =2.916 ··· .
4. If Xis uniform on {1 , . . . , n}, then Var X= (n2−1)/12; see “Lecture 12.”
11. Because E({X−a}2) = Px(x−a)2f(x)is assumed to converge nicely, we
can differentiate term by term to find that
d
da E{X−a}2=d
da X
x
(x2−2xa +a2)f(x)
=X
x
(−2x+2a)f(x)
=2aX
x
f(x) − 2X
x
xf(x)
=2a−2EX.
Set this equal to zero to find that a=EX. Also,
d2
da2E{X−a}2=X
x
2f(x) = 2,
which is positive. Positive second derivative means the minimum occurs
when the derivative is zero; that is, a=EX. But E({X−EX}2) = Var X.
18. By Theorem 4 (p. 121),
E1
X=∞
X
k=1
1
kpqk−1=p
q∞
X
k=1
1
kqk=p
q∞
X
k=1Zq
0
xk−1dx
=p
qZq
0
∞
X
k=1
xk−1dx =p
qZq
0
1
1−xdx
= − p
qln(1−q)=− p
1−pln(p) = 1
p−1ln(p) = ln p1
p−1.
1
