So far I have found the marginal pmf for X and Y and proved that the random variable are not independent by showing that;
$P(X=i, y=j)\neq P(X=i)P(Y=j)$
The next question asks to calculate;
$E[X^2Y]$
To calculate this I'm using the following property;
$Cov(X,Y) =E(XY)-E(X)E(Y)$
So in relation to my question;
$E(X^2Y) = Cov(X^2,Y)+E(X^2)E(Y)$
I have calculated $E(X^2)$ & $E(Y)$
To calculate the $Cov(X^2,Y)$
can I use the following formula for discrete RV;
$Cov(X,Y)=\sum \sum_{(x,y)\in S} (x-\mu_x)(y-\mu_y)f(x,y) $
where f(x,y) is the joint pmf
and simply replace the expectation of x with the second moment of x?
ie.
$Cov(X^2,Y)=\sum \sum_{(x,y)\in S} (x^2-\mu_x^2)(y-\mu_y)f(x,y) $
Any help much appreciated!
Best Answer
Use $E(X^2Y)=\sum_i\sum_jP(X=i, Y=j) i^2j$.