How do I calculate $E[X^2Y]$ for discrete (and DEPENDENT) random variable given the joint pmf table?

Question

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!

Answer 1

Use $E(X^2Y)=\sum_i\sum_jP(X=i, Y=j) i^2j$.

Answer 2

You can do that, but then you must be precise:$$\mathsf{Cov}(X^2,Y)=\mathbb E(X^2-\mathbb EX^2)(Y-\mathbb EY)=\sum_x\sum_y(x^2-\mu_{X^2})(y-\mu_Y)f(x,y)\tag1$$

Note that in $(1)$ we have $\mu_{X^2}=\mathbb EX^2$ and not $\mu_X^2=(\mathbb EX)^2$.