March 6 Homework Solutions Math 151, Winter 2012 …

1 March 6 Homework Solutions math 151, Winter 2012 . Chapter 6 Problems (pages 287-291). Problem 31. According to the National Center for Health Statistics, percent of males and percent of females never eat breakfast. Suppose that random samples of 200 men and 200 women are chosen. Approximate the probability that (a) at least 110 of these 400 people never eat breakfast. Let M denote the number of men that never eat breakfast and W denote the number of women that never eat breakfast. Note that M is a binomial random variable with p = .252 and n = 200, and W is a binomial random variable with p = .236 and n = 200. We compute E[M ] = 200(.252) = , Var(M ) = 200(.252)(1 .252) , E[W ] = 200(.236) = , Var(M ) = 200(.236)(1 .236) Thus we may approximate M by a normal random variable with = and 2 and we may approximate W by a normal random variable with = and 2 We want to compute P {110 M + W }.

2 We may approximate M + W by a normal distribution with = + = and 2 + = Hence . 110 M + W 110 P {110 M + W } = P 1 . 1 ( ) = 1 .9251 = .0749. (b) the number of the women who never eat breakfast is at least as large as the number of the men who never eat breakfast. We want to compute P {M W } = P {M W 0}. We may approximate M W. by a normal distribution with = = and 2 = + = Hence . M W P {M W } = P {M W 0} = P .. 1 (.37) = 1 .6443 = .3557. Problem 44. If X1 , X2 , X3 are independent random variables that are uniformly distributed over (0, 1), compute the probability that the largest of the three is greater than the sum of the other two. 1. Note that if, for example, X1 X2 + X3 , then X1 is automatically the largest of the three. We want to compute P {X1 X2 + X3 } + P {X2 X1 + X3 } + P {X3 X1 + X2 }.

3 By symmetry, these three terms are all equal, so it suffices to compute the first term P {X1 X2 + X3 }. Recall from Example 3a on page 252 that . y if 0 y 1, fX2 +X3 (y) = 2 y if 1 < y 2, 0 otherwise.. Hence Z 1 Z x Z 1 Z x Z 1. 1 2 1. P {X1 X2 + X3 } = fX1 (x)fX2 +X3 (y)dydx = 1 y dydx = x dx = . 0 0 0 0 0 2 6. So the probability that the largest of the three is greater than the sum of the other two is 1 1 1 1. P {X1 X2 + X3 } + P {X2 X1 + X3 } + P {X3 X1 + X2 } = + + = . 6 6 6 2. Chapter 6 Theoretical Exercises (pages 291-293). Problem 19. Let X1 , X2 , X3 be independent and identically distributed continuous random variables. (a) Compute P {X1 > X2 |X1 > X3 }. We consider 7 mutually exclusive cases: the first six are X1 > X2 > X3 , X1 >. X3 > X2 , X2 > X1 > X3 , X2 > X3 > X1 , X3 > X1 > X2 , X3 > X2 > X1 , and the seventh case is where at least two of X1 , X2 , X3 are equal.

4 Since the Xi are continuous random variables, the probability that at least two of the Xi are equal is zero. Since X1 , X2 , X3 , are independent and identically distributed, the first six cases are all equally likely. This means that P (X1 > X2 > X3 ) = 1/6, and similarly for the other orderings. Hence P {X1 > X2 > X3 } + P {X1 > X3 > X2 } 2. P {X1 > X2 |X1 > X3 } = = . P {X1 > X2 > X3 } + P {X1 > X3 > X2 } + P {X2 > X1 > X3 } 3. (b) Compute P {X1 > X2 |X1 < X3 }. P {X3 > X1 > X2 } 1. P {X1 > X2 |X1 < X3 } = = . P {X2 > X3 > X1 } + P {X3 > X1 > X2 } + P {X3 > X2 > X1 } 3. 2. (c) Compute P {X1 > X2 |X2 > X3 }. P {X1 > X2 > X3 } 1. P {X1 > X2 |X1 > X3 } = = . P {X1 > X2 > X3 } + P {X2 > X1 > X3 } + P {X1 > X3 > X2 } 3. (d) Compute P {X1 > X2 |X2 < X3 }. P {X1 > X3 > X2 } + P {X3 > X1 > X2 } 2.

5 P {X1 > X2 |X1 > X3 } = = . P {X1 > X3 > X2 } + P {X3 > X1 > X2 } + P {X3 > X2 > X1 } 3. Problem 28. Show that the median of a sample of size 2n + 1 from a uniform distribution on (0, 1) has a beta distribution with parameters (n + 1, n + 1). Let X denote the median of the independent and identically distributed random variables X1 , .. , X2n+1 . Consider equation on page 272. By replacing n with 2n + 1. and by choosing j = n + 1, we get that the probability density function of X is (2n + 1)! fX (x) = (F (x))n (1 F (x))n f (x), n!n! where f is the common probability density function and F is the common cumulative distribution function of the Xi 's. Since X1 , .. , X2n+1 are uniformly distributed on (0, 1), this means that f (x) = 1 for 0 x 1 and F (x) = x for 0 x 1. Hence (. (2n+1)!)

6 N n!n! x (1 x)n for 0 x 1. fX (x) =. 0 otherwise. Now, consider the Beta distribution on page 218. Note that Z 1. B(n + 1, n + 1) = xn (1 x)n dx 0. Z 1. n = xn+1 (1 x)n 1 dx after integrating by parts n+1 0. = .. Z 1. n! = x2n dx after integrating by parts (2n) .. (n + 1) 0. n! =. (2n + 1)(2n) .. (n + 1). n!n! = . (2n + 1)! Hence (. (2n+1)! n n!n! x (1 x)n for 0 x 1. fX (x) =. 0 otherwise (. 1. B(n+1,n+1). xn (1 x)n for 0 x 1. =. 0 otherwise. 3. So X has a Beta distribution with parameters (n + 1, n + 1). Chapter 7 Problems (pages 373-379). Problem 1. A player throws a fair die and simultaneously flips a fair coin. If the coin lands heads, then she wins twice, and if tails, then one-half of the value that appears on the die. De- termine her expected winnings. Let X denote the value on the die, let Y be 1 if the coin lands heads and 0 if the coin lands tails, and let g(X, Y ) denote the winnings.))

7 Then her expected winnings are 6 X 1 6 . X X 1. E[winnings] = g(x, y)p(x, y) = 2x p(x, 0) + x p(x, 1). x=1 y=0 x=1. 2. 6 6. X 1 1 1 5 X 5. = 2x + x = x= 21 = x=1. 12 2 12 24 x=1 24. Problem 12. A group of n men and n women is lined up at random. (a) Find the expected number of men who have a woman next to them. Label the people in order 1 though 2n and let Xi = 1 if the i-th person is a man standing next to a woman and Xi = 0 otherwise. We want to compute " 2n # 2n 2n X X X. E Xi = E[Xi ] = P {Xi = 1}. i=1 i=1 i=1. Note X1 = 1 only if the first person is male and the second is female. There are 2n(2n 1) ways to choose the first two people. There are n ways to choose the first person to be male and n ways to choose the second person to be female, and hence n2 ways we can have Xi = 1.

8 Hence n2 n P {X1 = 1} = = . 2n(2n 1) 4n 2. n Similarly P {X2n = 1} = 4n 2.. Now let's find P {Xi = 1} for 1 < i < 2n. There are 2n(2n 1)(2n 2) ways to choose the (i 1)-th, i-th, and (i + 1)-th person. We have Xi = 1 if the three people chosen are female male female, female male male, or male male female. Hence there are (n)(n)(n 1) + (n)(n)(n 1) + (n)(n 1)(n) = 3n2 (n 1). ways we can have Xi = 1. Hence 3n2 (n 1) 3n P {Xi = 1} = = . 2n(2n 1)(2n 2) 8n 4. 4. Hence " 2n # 2n X X n 3n 3n2 n E Xi = P {Xi = 1} = 2 + (2n 2) = . i=1 i=1. 4n 2 8n 4 4n 2. (b) Repeat part (a), but now assuming that the group is randomly seated at a round table. Label the people in order 1 though 2n and let Xi = 1 if the i-th person is a man standing next to a woman and Xi = 0 otherwise. We want to compute " 2n # 2n 2n X X X.

9 E Xi = E[Xi ] = P {Xi = 1}. i=1 i=1 i=1. Since the group is now seated in a circle, for 1 i 2n we have 3n P {Xi = 1} = . 8n 4. This computation is the same as the one that we used in part (a) for 1 < i < 2n. Hence " 2n # 2n X X 3n 3n2. E Xi = P {Xi = 1} = 2n = . i=1 i=1. 8n 4 4n 2. Problem 19. A certain region is inhabited by r distinct types of a certain species of insect. Each insect caught will, independently of the types of the previous catches, be of type i with probability r X. Pi , i = 1, .. , r, Pi = 1. i=1. (a) Compute the mean number of insects that are caught before the first type 1 catch. Let X denote the number of insects caught before the first type 1 catch. Then P {X = x} = (1 P1 )x P1 . Hence . X. E[X] = x(1 P1 )x P1. x=0.. X. = P1 x(1 P1 )x x=0. 1 P1. = P1 . P12.. X z using the formula nz n = for |z| < 1.

10 N=1. (1 z)2. 1 P1. = . P1. 5. (b) Compute the mean number of types of insects that are caught before the first type 1. catch. Let Xi denote the number of insects of type i caught before the first type 1 catch. Let g(0) = 0 and let g(x) = 1 for positive integers x > 0. We want to compute " r # r r r X X X X. E g(Xi ) = E[g(Xi )] = P {g(Xi ) = 1} = P {Xi 1}. i=2 i=2 i=2 i=2. Let X denote the number of insects caught before the first type 1 catch as in part (a). Then . X. P {Xi 1} = P {Xi 1, X = x}. x=0. since the events{Xi 1, X = x} are mutually exclusive and have union{Xi 1}.. X. = (P {X = x} P {Xi = 0, X = x}). x=0. X . = ((1 P1 )x P1 (1 P1 Pi )x P1 ). x=0. P1 P1. = . P1 P1 + Pi using the formula for a geometric series, twice Pi = . P1 + Pi Hence " #. r r r X X X Pi E g(Xi ) = P {Xi 1} =.