Question

(45) Two random variables X and Y have the joint probability density | 2, 0sxs1 and 0 s ys1 and x + y21 fxY (x, y) = 0, elsewhere Answer each of these independent questions about X, Y, carefully indicating the domain of all functions where needed. Parts a). - i). are 5 points each. a). Find E(Z), where Z is a new random variable defined by Z = XY b). A is the event {X >0.75}. Find P(A). c). B is the event {2X>Y}. Find P(B). d). Find f(x). e). Find frx(y 1x). f). Are X and Y independent? Why or why not? g). Find E(Y|X = x). h). Find and sketch the density fxia (x1A) where A is the event {X>0.75} as in part b). i). Find and sketch the density fria (y1A) where A is the event {X>0.75} as in part b).

Answer 1

The joint PDF is $\fn_jvn f_{X,Y}(x,y)=2;0\leqslant x\leqslant 1,0\leqslant y\leqslant 1,x+y\geqslant 1$

a) The expectation,

$\fn_jvn E(Z)=E(XY)\\ E(Z)=\int_{0}^{1}\int_{0}^{1-x}2xydydx\\ E(Z)=\int_{0}^{1}x(1-x)^2dx\\ E(Z)=\int_{0}^{1}x^2(1-x)dx\\ E(Z)=1/3-1/4\\ {\color{Blue} E(Z)=\frac{1}{12}}$

d) The marginal PDF of X is

$\fn_jvn f_X(x)=\int_{0}^{1-x}2dydx\\ {\color{Blue} f_X(x)=2(1-x);0\leqslant x\leqslant 1}$

b) The probability,

$\fn_jvn P(A)=P(X>0.5)\\ P(A)=\int_{0.75}^{1}2(1-x)dx\\ P(A)=0.5-(1-0.75^2)\\ {\color{Blue} P(A)=0.0625}$

c) The probability,

$\fn_jvn P(B)=P(2X>Y)\\ P(B)=\int_{0}^{1/3}\int_{0}^{2x}2dydx+\int_{1/3}^{1}\int_{0}^{1-x}2dydx\\ P(B)=\int_{0}^{1/3}4xdx+\int_{1/3}^{1}2(1-x)dx\\ P(B)=2/9+4/3-(1-1/9)\\ {\color{Blue} P(B)=2/3}$