Let X and Y be independent continuous random variables

Chapter , Problem 35

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Let X and Y be independent continuous random variables with PDFs j x and jy, respectively, and let Z = X + Y. (a) Show that fZlx (z I x) = fy (z - x). Hint: Write an expression for the conditional CDF of Z given X, and differentiate. (b) Assume that X and Y are exponentially distributed with parameter A. Find the conditional PDF of X, given that Z = z. (c) Assume that X and Y are normal random variables with mean zero and variances (7 ; and (7 , respectively. Find the conditional PDF of X, given that Z = z. Solution. (a) We have P(Z :::; z I X = x) = P(X + Y :::; z I X = x) = P(x + Y :::; z I X = x) = P(x + Y :::; z) = P(Y :::; z - x) , where the third equality follows from the independence of X and Y. By differentiating both sides with respect to z, the result follows. (b) We have, for 0 :::; x :::; z, f ( I) - fZlx (z I x)fx (x) XIZ x z - fz(z) fy (z - x) f x (x) fz(z) Ae->'(z-x) Ae->'x A 2 e->'z fz(z) - fz(z) . Since this is the same for all x, it follows that the conditional distribution of X is uniform on the interval [0, z] , with PDF fXlz(x I z) = liz. (c) We have f ( I ) - fy (z - x)fx (x) XIZ x z - fz(z) We focus on the terms in the exponent. By completing the square, we find that the negative of the exponent is of the form (z - X) 2 x 2 (7 ; + (7 ( Z(7; ) 2 Z 2 ( (7 ; "":""-----,,...:- ) + - - x - + - I - ---;