Computing $E[HXY] $ when H is distributed binomialy on XY

Question

Consider X, Y are independent variables both distributed geometrically with parameter $\frac{1}{2}$, H is distributed binomialy with parameters XY, $\frac{1}{2}$. I need to find E[HXY].

I understand that $E[H|XY = m] = \frac{1}{2}m$. So, using the law of total expectation I can say that $E[HXY] = E[E[HXY|XY]] = E[\frac{1}{2}(XY)^2]$.

From here it I am sure how to calculate the result, so I am asking if this is a correct solution, because even though I know the final expression leads to the correct result, I am unsure if I used the law of total expectation correctly here.

Thanks.

Answer 1

Indeed.

Because $H\mid XY\sim\mathcal {Bin}(XY, 1/2)$, therefore $\mathsf E(H\mid XY) = \tfrac 12XY$.

Hence by the Law of Iterated Expextation: $\mathsf E(HXY)~{=\mathsf E(\mathsf E(H\mid XY)XY)\\ = \tfrac 12\mathsf E(X^2Y^2)\\ \ddots}$