# Chapter 4 Problems . Consequently,

```Chapter 4 Problems
3. Let X equal the number of dots. Then, f(x) = 16 if x = 1, . . . , 6; else,
f(x) = 0. Consequently,
1
7
1
+ &middot;&middot;&middot; + 6 &times;
=
EX = 1 &times;
6
6
2
1
1
91
E(X2 ) = 12 &times;
+ &middot; &middot; &middot; + 62 &times;
=
.
6
6
6
Therefore
91
−
Var X =
6
2
35
7
=
= 2.916 &middot; &middot; &middot; .
2
12
4. If X is uniform on {1 , . . . , n}, then Var X = (n2 − 1)/12; see “Lecture 12.”
P
11. Because E({X − a}2 ) = x (x − a)2 f(x) is assumed to converge nicely, we
can differentiate term by term to find that
d X 2
d
E {X − a}2 =
(x − 2xa + a2 )f(x)
da
da x
X
=
(−2x + 2a)f(x)
x
= 2a
X
f(x) − 2
x
X
xf(x)
x
= 2a − 2EX.
Set this equal to zero to find that a = EX. Also,
X
d2
E {X − a}2 =
2f(x) = 2,
2
da
x
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),
E
X
∞
∞
∞ Z
1 k−1 p X 1 k p X q k−1
1
=
pq
=
q =
x
dx
X
k
q
k
q
k=1
k=1
k=1 0
Z ∞
Z
p q X k−1
p q 1
=
x
dx =
dx
q 0
q 0 1−x
k=1
1 p
p
1
= − ln(1 − q) = −
ln(p) =
ln(p) = ln p p−1 .
q
1−p
p−1
1
29.
(a) “No tail in the first n tosses” means “all heads in the first n tosses.”
Therefore, probab. = pn .
(b) pn = (1 − p)n−1 p.
P
P∞
n−1
(c) E(number) = ∞
, which is equal to
n=1 npn = p
n=1 n(1 − p)
0
∞
1
1
d X
n
= .
(1 − p) = −p
−p
dp n=0
p
p
2
```