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MA2500/18 Section B (Answer THREE questions) 6. Let X and Y be jointly continuous random variables defined on the same prob- ability space, let fx.y denote their joint PDF, and let fx and fy respectively denote their marginal PDFs (a) Let z be a fixed value such that fx(x) >0. Write down expressions for 12] (i) the conditional PDF of Y given X = z, and (i) the conditional expectation of Y given X (b) State and prove the law of total expectation (c) Suppose that the joint PDF of X andY is 15] otherwise, where c is a constant (i) Show that c= 15 (ii) Show that E(X)5/12 andE(Y) (İİİ) Show that 3(Y、X) (iv) Verify that E [3(y|X] = E(Y) 5/8 4(2- 3X +X3)

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Answer #1

(i) The conditional PDF of Y given Xxis given by fx(x) (ii) The conditional expectation of Y given Xex is given by Sy S,Jx(Ca)where Sy is the support of Y

The law of total expectation states that if Y is a random variable whose expected value E(Y) is defined, and X is any random variable on the same probability space, then EY) EE(YX)e, the expected value of the conditional expected value of Y given X is the same as the expected value of Y. where Sy is the support of Y where Sx is the support of X fx(x) fx,y(r, y)dx-fy(y) S. = E(Y)

otherwise where c is a constant (i) It is known: J J0<rSys1 fx,y(x, y) dx dy

10 15 55 3 46 12 15 0 0 | 15 1 5 2 128 E(Y) = yfy(y)dy = PR

cr(1 - y2) 15r(1 y) 31-y 2 - 3x 34(2 3r)

1 3(12r2+3 1-212+45 E(Y|X)dz Thus it can be seen that ; EE(WX)]=E(Y) ENC

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