MATH 5010–001 SUMMER 2003
SOLUTIONS TO ASSIGNMENT 4
Problems from pp. 171–180
1.
The possible values of X are −2, −1, 0, 1, 2, 4, and the corresponding probabilities
are:
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P {X = −2} =
P {X = −1} =
P {X = 0} =
P {X = 1} =
P {X = 2} =
P {X = 4} =
17.
•
•
•
2
14 ,
2
8 2
1 1
14 ,
2
2
2
14 ,
2
4 8
1 1
14 ,
2 4 2
1 1
14 ,
2
4
2
14 .
2
(a) You may recall from your reading that P {X = i} = F (i) − F (i−). This gives:
1
1
P {X = 1} = F (1) − F (1−) = 12 + 1−1
4 − 4 = 4 ;
1
2−1
= 16 .
P {X = 2} = F (2) − F (2−) = 11
12 − 2 + 4
1
P {X = 3} = F (3) − F (3−) = 1 − 11
12 = 12 .
17. (b) P { 12 < X < 32 } = F ( 32 −) − F ( 12 ) =
1
2
+
(3/2)−1
4
−
(1/2)
4
= 12 .
18. Note that X is a binomial with parameters n = 4 and p = 12 . Therefore, P {X − 2 =
4
for k = −2, −1, 0, 1, 2.
k} = P {X = 2 + k} = k4 (1/2)2+k (1/2)n−(2+k) = 2−4 2+k
24. (a) If A chooses 1, and if the winnings of B are written as X, then P {X = 1} = p
and P {X = −3/4} = 1 − p. Therefore, in this case, E(X) = p − (3/4)(1 − p) = 74 p − 34 .
24. (b) If A chooses 2, then P {X = 2} = 1 − p and P {X = −3/4} = p. Therefore,
p.
E(X) = 2(1 − p) − (3/4)p = 2 − 11
4
Since B does not know what A is going to choose, B chooses the p that makes the
two expected-value lines of (a) and (b) intersect. (See the diagram given during the
lectures.) That is, the p for which 74 p − 34 = 2 − 11
p; i.e., p = 11
.
4
18
As for player A, she expects to win q − 34 q if B chooses 1, and − 34 q + 2(1 − q) if B
chooses 2. Therefore, the optimal choice for A is also where these two lines meet and
that is q = 11
18 .
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Theoretical Problems from pp. 180–184
13. Recall that
n k
k = 0, . . . , n.
P {X = k} =
p (1 − p)n−k ,
k
As far as the variable p is concerned, nk is a constant. So we are asked to maximize
f (p) = pk (1 − p)n−k .
Of course,
f 0 (p) = kpk−1 (1 − p)n−k − (n − k)pk (1 − p)n−k−1 .
(1)
So we need to analyse f and f 0 . We do this with some care:
Case 1: k = 0. In this case, f 0 (p) = −n(1 − p)n−1 is always negative. This means that f is
decreasing, so the maximzer is p = 0.
Case 2: k = n. In this case, f 0 (p) = npn−1 is always positive. This means that f is increasing,
so the maximizer is p = 1.
Case 3: 1 ≤ k ≤ n − 1. In this case, f 0 (p) = 0 yields the equation, kpk−1 (1 − p)n−k =
(n − k)pk (1 − p)n−k−1 . Because the maximizer is some p strictly inside (0, 1), we
can divide by pk−1 (1 − p)n−k−1 and get p = (k/n). To check that this is indeed a
maximizer, need to check that f 00 (k/n) < 0. But
f 00 (p) = k(k − 1)pk−2 (1 − p)n−k
− 2k(n − k)pk−1 (1 − p)n−k−1
+ (n − k)(n − k − 1)pk (1 − p)n−k−2 .
To check that f 00 (k/n) < 0, we can divide f 00 (p) by pk−2 (1 − p)n−k−2 , and then let
p := (k/n) to check for the following inequality:
2
2
k
k
k
k
− 2k(n − k)
< 0?
k(k − 1) 1 −
1−
+ (n − k)(n − k − 1)
n
n
n
n
A few lines of calculations show that the left-hand side is equal to −k + (k 2 /n). So
f 00 (kn) < 0 if and only if (k 2 /n) < k; i.e., if k 2 < nk. Because 1 ≤ k ≤ n − 1, this is
the case and we are done.
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