Math 5010-1, Spring 2005
Assignment 5
Problems
#31, p. 175. Let R denote the event that it will rain tomorrow. The metereologist truly feels that
P (R) = p∗ . Now, suppose we are told that P (R) = p (when, in fact, it is p∗ ). Then,
we give weight 1 − (1 − p)2 if it rains, and 1 − p2 otherwise. So, if W denotes the
weight we assigned to the metereologist, then
E[W ] = 1 − (1 − p)2 P (R) + 1 − p2 P (Rc )
= 1 − (1 − p)2 p∗ + 1 − p2 (1 − p∗ ).
Call this h(p). The metereologist’s goal is to report a p that maximizes h(p). We can
use calculus to do this: Remembering that p is the variable and p∗ ∈ [0, 1] is a fixed
constant, we have
h0 (p) = 2(1 − p)p∗ − 2p(1 − p∗ ) = 2p∗ − 2p,
h00 (p) = −2.
So the answer is p = p∗ . Please note the correction to the previous solution.
#33, p. 175. Let M denote the number of papers that the paperboy buys each day. Also let X
denote the paperboy’s demand, so that X is Bin(n = 10, p = 1/3). The paperboy’s
profit is the following random variable:
Z = 15 min(X, M ) − 10M.
[The min has to be there, because the paperboy can sell at most M copies now that
he has bought M copies.] Therefore,
f (M ) = E[Z] = 15E [min (X, M )] − 10M.
(1)
This is too hard to compute. So it pays to be tricky: Note that min(X, M ) ≤ X.
Therefore, E[min(X, M )] ≤ E[X] = np = 10/3. Therefore, f (M ) ≤ 15(10/3) − 10M =
50 − 10M . Also, min(X, M ) ≤ M , so that f (M ) ≤ 15M − 10M = 5M . Therefore,
f (M ) ≤ min (50 − 10M, 5M ) = g(M ).
Therefore, max f ≤ max g, which happens where 50 − 10M = 5M (why?). Equivalently,
where M = 50/15. Because M is an integer the closest is M = 3. That is, we have proved
that maxM f (M ) ≤ g(3) = 15. To finish we must show that maxM f (M ) is not too far
below 15. If so, then M = 3 is (nearly) optimal. But
i 10−i
10 i 10−i
X
2
1
2
1
E [min (X, M )] =
min(i, M )
≥M
3
3
3
3
i=0
10
X
i=M
= M P {X ≥ M } .
1
Therefore,
f (M ) ≥ 15M P {X ≥ M } − 10M.
In particular, maxM f (M ) ≥ 45P {X ≥ 3} − 30. Note that
P {X ≥ 3} = 1 − P {X ≤ 2}
0 10 1 9 2 8
2
10
1
2
10
1
2
10
1
−
−
=1−
3
3
1
3
3
2
3
3
0
10
9
2 8
2
1
2
1
2
=1−
− 10
− 45
3
3
3
3
3
≥ 0.7.
Therefore, maxM f (M ) ≥ 45 × 0.7 − 30 = 1.5 > 1.
#1, p. 228.
R1
(a) Set c −1 (1 − x2 ) dx = 1 to find that c = 3/4.
(b) If a ≤ −1 then F (a) = 0. If a > 1, then F (a) = 1. If −1 < a < 1 then
Z
3
1
3 a
(1 − x2 ) dx = (1 + a) − (a3 + 1).
F (a) =
4 −1
4
4
#4, p. 228.
R∞
(a) P {X > 20} = 20 (10/x2 ) dx = 1/2.
(b) If a ≤ 10 then F (a) = 0. If a > 10 then
Z a
1
1
10
10
dx = 10
−
=1− .
F (a) =
2
10 a
a
10 x
(c) Let N denote the number of types of devices that will funciton
for at least 15
R∞
hours. Then N is Bin(n = 6, p), where p = P {X > 15} = 15 (10/x2 ) dx = (2/3).
Therefore, we are asked to find
P {N ≥ 3} =
6 i 6−i
X
1
6
2
i=3
#6, p. 228.
(a) Recall that E[X] =
3
i
3
.
R∞
xf (x) dx. Therefore,
Z
Z ∞
1 ∞ 2 −x/2
E[X] =
x e
dx = 2
y 2 e−y dy.
4 0
0
R 0
R 0
Recall integration by parts: uv = (uv) − vu . Therefore, apply this with
u(y) = y 2 and v(y) = −e−y ; u0 (y) = 2y and v 0 (y) = e−y . Whence we obtain
Z ∞
Z ∞
Z ∞
∞
2 −y
2 −y −y
y e dy = −y e + 2
ye dy = 2
ye−y dy.
0
−∞
0
2
0
0
Apply integration by parts (again!) to find that the last integral is one. I.e.,
E[X] = 4.
(b) We saw in #1 that c = 3/4. Therefore,
3
E[X] =
4
Z
1
x(1 − x2 ) dx = 0,
−1
by symmetry (for instance).
(c) This one is a bit tricky:
Z
∞
E[X] = 5
5
1
x 2 dx = 5
x
∞
Z
5
dx
= ∞.
x
#16, p. 229. Let X denote the number of years we have to wait until the first year with rainfall
over 40 inches. Then X has a geometric distribution with parameter p = P {Z > 15},
where Z is normally distributed with parameters µ = 40 and σ 2 = 16. We are asked
to find
∞
∞
X
X
i−1
P {X ≥ 10} =
p(1 − p)
=p
(1 − p)i−1 = (1 − p)10 .
i=10
i=10
It remains to find p. But recalling that (Z − 40)/4 is standard normal, we have
15 − 40
p = P {Z > 15} = P N (0, 1) >
= P {N (0, 1) > −6.25} ≈ 1.
4
Theoretical Exercises
#19, p. 186.
(a) Let X denote the number of games played this way. The possible values for X
are 5, 6, 7, · · ·. Now P {X = 5} = P {game #5 is a loss} = 1 − p. More generally,
if k ≥ 5, then
P {X = k} = P {games #5 through k are wins; game #(k + 1) is a loss}
= pk−5 (1 − p).
Therefore,
E[X] =
∞
X
kp
k−5
(1 − p) = (1 − p)
k=5
= (1 − p)
= (1 − p)
∞
X
kpk−5
k=5
∞
X
k=5
∞
X
(k − 5)p
k−5
+ (1 − p)5
∞
X
k=5
ipi + 5(1 − p)
i=1
∞
X
i=0
3
pi .
pk−5
The first sum can be written as
p
∞
X
i−1
ip
i=1
The second sum is
d
=p
dp
P∞
i=0
∞
X
!
p
i
i=0
R∞
−∞
1
1−p
=
p
.
(1 − p)2
pi = 1/(1 − p). Therefore,
E[X] =
#1, p. 232. We know that
d
=p
dp
p
+ 5.
1−p
f (x) dx = 1. Thus,
Z
√
a ∞ 3 −y2
1=
ax e
y e
dx = 2
dy
(y = x b)
b 0
0
Z ∞
a
1 −1/2
3/2 −z
= 2
z e
z
dz
(z = y 2 )
b 0
2
Z ∞
a
a
= 2
ze−z dz = 2 .
2b 0
2b
Z
∞
3 −bx2
Therefore, a = 2b2 .
4