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Statistics 116 - Fall 2004 Theory of Probability Assignment # 8 Solution show (and briefly explain) all of your work Q. 1) (Ross # 6.39) Choose a number at random from the set of numbers f1; 2; 3; 4; 5g. Now choose a number at random from the subset f1; : : : ; X g. Call this second number Y . (a) Find the joint p.m.f. of X and Y . (b) Find the conditional mass function of X given that Y = i. Do it for i = 1; 2; 3; 4; 5. (c) Are X and Y independent? Why? Solution: (a) P fX = j; Y = ig = 15 1j . (b) P fX = j jY = ig = P51=51j=5k , for i j k=i 5. (c) No, because the distribution of random variable Y depends on the value of random variable X . Q. 2) (Ross # 6.43) The joint density of X and Y is y )e x ; f (x; y) = c(x 2 2 0 x < 1; x y x Find the conditional distribution of Y given X = x, i.e. nd fY jX (y j!x). Solution: fY jX (yjx) = x3 (x y ), x < y < x FY jX (yjx) = x3 (x y y =3 + 2x =3), x < y < x. 3 2 2 4 3 2 3 3 4 1 Q. 3) (Ross # 7.3) If X and Y are independent uniform (0,1) random variables, show that E (jX Solution: E (jX Y j ) = Z 1 0 RR 1 0 jx 2 ; ( + 1)( + 2) Y j ) = 1 0 jx yj dydx. yj dy = = Z Zx ! x 0 jx u du + 0 = (x+1 + (1 Hence E (jX Q. 4) Y j ) = 1 a+1 Z 1 yj dy + Z 1 x > 0: Z 1 x jx yj dy = u du = 0 x) )=(a + 1): +1 (x+1 + (1 x) )dx = +1 0 2 : (a + 1)(a + 2) Ross # 7.25) Let X ; X ; : : : be a sequence of independent and identi1 2 cally distributed continuous random variables. Dene the random variable N 2 as the rst point in which the sequence (X1 ; X2 ; : : : ) stops decreasing, i.e. if N = n, then X 1 X X Xn < Xn : 2 3 1 (a) Compute the p.m.f. of N by rst computing P (N (b) Show that E (N ) = e. n). (a) P fN ng = P fX1 X2 ::: Xn g = n1! 1 1 (b) E [N ] = 1 n=1 P fN ng = n=1 n! = e P Q. 5) P (Ross # 7.40) The joint density function of X and Y is given by 1 f (x; y) = e y x=y ; y ( + ) x > 0; y > 0: (a) Find E (X ): (b) Find E (Y ): (c) Show that Cov(X; Y ) = 1: R (a) fY (y ) = e y y1 e x=y dx = e y . (b) In addition, the conditional distribution of X given Y = y is exponential with mean y . Hence, E [Y ] = 1, E [X ] = E [E [X jY ]] = E [Y ] = 1. (c) Since, E [XY ] = E [E [XY jY ]] = E [Y E [X jY ]] = E [Y ] = 2 (the last equality follows because Y is exponential with mean 1). Hence, Cov(X; Y ) = 2 1 = 1. 2 2

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