Question

8. Let the joint density function of X and Y be 36 if 1 < x = y < 6 f(x, y) = { if 1<x<y < 6. What is the correlation coefficient of X and Y ?

Answer 1

$E(X)=\int_{0}^{6}\int_{X}^{6}Xf(X,Y)dYdX=\int_{0}^{6}\int_{X}^{6}(2/36)XdYdX=\int_{0}^{6}(2/36)(6X-X^2)dX=(2/36)(108-72)=2$

$E(Y)=\int_{0}^{6}\int_{0}^{Y}Yf(X,Y)dXdY=\int_{0}^{6}\int_{0}^{Y}(2/36)YdXdY=\int_{0}^{6}(2/36)Y^2dY=(2/36)(72)=4$

$E(XY)=\int_{0}^{6}\int_{X}^{6}XYf(X,Y)dYdX=\int_{0}^{6}\int_{X}^{6}(2/36)XYdYdX=\int_{0}^{6}(1/36)(36X-X^3)dX=18-2=9$

$E(X^2)=\int_{0}^{6}\int_{X}^{6}X^2f(X,Y)dYdX=\int_{0}^{6}\int_{X}^{6}(2/36)X^2dYdX=\int_{0}^{6}(2/36)(6X^2-X^3)dX=(24-18)=6$

$E(Y^2)=\int_{0}^{6}\int_{0}^{Y}Y^2f(X,Y)dXdY=\int_{0}^{6}\int_{0}^{Y}(2/36)Y^2dXdY=\int_{0}^{6}(2/36)Y^3dY=18$

Cov(X,Y) = E(XY) - E(X).E(Y)

= 9 - 8 = 1

Var(X) = E(X²) - (E(X))²

= 6 - 4 = 2

Var(Y) = E(Y²) - (E(Y))²

^{= 18 - 16 = 2}

Correlation of X and Y is $\rho$XY = Cov(X,Y) / (V(X).V(Y) )^1/2

= 1/2

Note : while doing integration we excluded f(X,Y) for x =y, as integration is 0 for measure 0 set