Answer to Question #320989 in Statistics and Probability for kevin

Question

Let X and Y have joint probability distribution function f(x, y) = ( 2x+y /12) , (x, y) = (0, 1); (0, 2); (1, 2); (1, 3) 0, elsewhere

Find

(i) the covariance between X and Y.

(ii) the joint probability generating function of X and Y.

Answer 1

$f\left( x,y \right) =\left\{ \begin{array}{c} \frac{1}{12},\left( x,y \right) =\left( 0,1 \right)\\ \frac{1}{6},\left( x,y \right) =\left( 0,2 \right)\\ \frac{1}{3},\left( x,y \right) =\left( 1,2 \right)\\ \frac{5}{12},\left( x,y \right) =\left( 1,3 \right)\\\end{array} \right. \\i:EX=0\cdot \frac{1}{12}+0\cdot \frac{1}{6}+1\cdot \frac{1}{3}+1\cdot \frac{5}{12}=\frac{3}{4}\\EY=1\cdot \frac{1}{12}+2\cdot \frac{1}{6}+2\cdot \frac{1}{3}+3\cdot \frac{5}{12}=\frac{7}{3}\\EXY=0\cdot \frac{1}{12}+0\cdot \frac{1}{6}+2\cdot \frac{1}{3}+3\cdot \frac{5}{12}=\frac{23}{12}\\cov\left( X,Y \right) =\frac{23}{12}-\frac{3}{4}\cdot \frac{7}{3}=\frac{1}{6}\\ii:\\M_{XY}\left( t,s \right) =Ee^{tX+sY}=e^s\cdot \frac{1}{12}+e^{2s}\cdot \frac{1}{6}+e^{t+2s}\cdot \frac{1}{3}+e^{t+3s}\cdot \frac{5}{12}$