QUESTION 5 The conditional pdf of X given Y = y is given by fxy (x | y) = e (0 (y))-0(y)xxn−1, F(n) x>0 where (y) is a function of y (a) Find E(X | Y = y) 1 (b) For given E(X | Y = y) = and fy (y) Be-By, y >0 = Y derive the unconditional ndf of Y [10] Tou (4) (A)

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