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4 Pages

### hw3_PS_Solution

Course: ORIE 3500, Fall 2008
School: Cornell
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Word Count: 493

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3500/5500 ORIE Fall Term 2008 Assignment 3-Solution 1. We have pX (r) = a + b r, r = 1, . . . , 5 and E(X) = 11/3. So 5 (a + b r) = 1 5a + 15b = 1 r=1 5 r(a + b r) = 11/3 15a + 55b = 11/3. r=1 (a) Solving the above two equations, a = 0 and b = 1/15. (b) 5 5 r=1 E(X ) = r=1 2 r (a + b r) = 2 r3 15 = 225/15 = 15. So var(X) = E(X 2 ) [E(X)]2 = 15 (11/3)2 = 14/9. 2. (a) Let X be the number of...

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3500/5500 ORIE Fall Term 2008 Assignment 3-Solution 1. We have pX (r) = a + b r, r = 1, . . . , 5 and E(X) = 11/3. So 5 (a + b r) = 1 5a + 15b = 1 r=1 5 r(a + b r) = 11/3 15a + 55b = 11/3. r=1 (a) Solving the above two equations, a = 0 and b = 1/15. (b) 5 5 r=1 E(X ) = r=1 2 r (a + b r) = 2 r3 15 = 225/15 = 15. So var(X) = E(X 2 ) [E(X)]2 = 15 (11/3)2 = 14/9. 2. (a) Let X be the number of students interviewed by the rst investigator who owns a car. Then X Binomial(5, 3/4). So P (X 4) = 1 P (x = 5) = 1 (3/4)5 = 781/1024. (b) Let p be the probability that an investigator nds at least one student who does not own a car. Then p = 781/1024. So N Binomial(20, p), pN (r) = 20 (781/1024)r (243/1024)20r , r = 0, 1, . . . , 20. r (c) E(N ) = 20p = 20 781/1024 = 3905/256 = 15.2539. 3. (a) The conditional PMF of X given N = n is pX|N =n (x) = P (X = x|N = n) = This is Binomial(n, 1/4). (b) X|N = 40 Binomial(40, 1/4). So E(X|N = n) = 40/4 = 10. n (1/4)x (3/4)nx , x = 0, 1, . . . , n. x 1 (c) For 0 x n < , the joint PMF will be pX,N (x, n) = P (X = x, N = n) = P (X = x|N = n)P (N = n) e20 20n n = (1/4)x (3/4)nx x n! e20 20n n! (1/4)x (3/4)nx = x!(n x)! n! 20 n x nx e 20 (1/4) (3/4) = x!(n x)! (d) pX (x) = P (X = x) = n=x P (X = x, N = n) e20 20n (1/4)x (3/4)nx x!(n x)! n=x = = y=0 e20 20x+y (1/4)x (3/4)y x!y! (using y = n x) e20 (1/4)x 20x = x! = e 20 x x y=0 (20 3/4)y y! (1/4) 203/4 20 e x! = e5 5x /x!. So X P oisson(5). 4. (a) Possible values of T1 and T2 are 1, 2, . . . and T1 < T2 . Their joint PMF will be pT1 ,T2 (x, y) = P (T1 = x, T2 = y) = 9 10 y2 1 10 2 , 1 x < y < . 2 (b) Using part (a), E(T2 ) = x=1 y=x+1 y(9/10)y2 (1/10)2 x1 = x=1 (9/10) (1/10) y=x+1 y(9/10)yx1 (1/10) (z + x)(9/10)z1 (1/10) z=1 = x=1 (9/10)x1 (1/10) (using z = y x) = x=1 (9/10)x1 (1/10)[(1/10)(1 9/10)2 + x(1/10)(1 9/10)1 ] (9/10)x1 (1/10)(10 + x) x=1 = = 10 + (1/10)(1 9/10)2 = 20. (c) We have for any x 1 P (T1 = x) = (9/10) So for any r, s 1 x1 (1/10) P (T1 > r) = x=r+1 (9/10)x1 (1/10) = (9/10)r . P (T1 > r + s) (9/10)r+s P (T1 > r + s|T1 > r) = = = (9/10)s . r P (T1 > r) (9/10) 5. The PMF of X is pX (x) = (a) We have E(X) = and hence var(X) = E(X ) = 2 1 , x = n, n + 1, . . . , n. 2n + 1 x = 0/(2n + 1) = 0, 2n + 1 x=n n n x2 2 n(n + 1)(2n + 1) n(n + 1) = = . 2n + 1 2n + 1 6 3 x=n 3 (b) Possible values of Y are 0, 1/n, 2/n, . . . , 1 and P (Y = j/n) = P (|X| = j) = P (X = j or X = j) = So 0 E(Y ) = + 2n + 1 n 1/(2n + 1) 2/(2n + 1) if j = 0 if j = 1, . . . , n. j=1 2 n(n + 1) n+1 j2 = = . n 2n + 1 n(2n + 1) 2 2n + 1 (c) Possible values of Z are 0, 12 , . . . , n2 and P (Z = j 2 ) = P (|X| = j) = So 2 n(n + 1)(2n + 1) n(n + 1) 2 0 + j2 = = . E(Z) = 2n + 1 j=1 2n + 1 2n + 1 6 3 n 1/(2n + 1) 2/(2n + 1) if j = 0 if j = 1, . . . , n. 4
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