Given $(X,Y)$, 2-dimensional probability vector, and let $g: R^2 \rightarrow R, E[g(X,Y)^2 ] < \infty$ and $h:R \rightarrow R, E[h(X)^2] < \infty $, prove the following:


This is what I have done so far

$E[h(X){g(X,Y)]-E[h(x) E[g(X,Y)|X]}]=0$,

Then we have to show

$E[h(X){g(X,Y)]=E[h(x) E[g(X,Y)|X]}]$

On the right-hand side, $E[h(x)E[g(X,Y)|X]]=E[h(x)]E[E[g(X,Y)|X]]=E[h(X)]E[g(X,Y)]$ by the iteration of expectation

On the left-hand side,

$E[h(X)g(X,Y)]$ to be equal to $E[h(X)]E[g(X,Y)], \phantom 2 h(X), g(X,Y)$ have to be independent from each other but is it? How should I do this?


1 Answer 1


You assumed independence of $h(X)$ and $E[g(X,Y)|X]$ for the right hand side, which you don't know. The RHS due to conditional on $X$ is and then by iterative expectation is

\begin{align*} E\left[h(X)E[g(X,Y)|X] \right] & = E\left[E[h(X)g(X,Y)|X] \right]\\ & = E[h(X) g(X,Y)]\\ & = LHS. \end{align*}


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