Independence and conditional expectation

probability-theory random-variables conditional-expectation independence

Solution 1:

Assume that ${\rm E}[h(X)\mid Y]={\rm E}[h(X)]$ holds for any bounded Borel-measurable function $h$ and let $f,g$ be bounded Borel-measurable functions. Then $$ \begin{align} {\rm E}[f(X)g(Y)]&={\rm E}[{\rm E}[f(X)g(Y)\mid Y]]={\rm E}[g(Y){\rm E}[f(X)\mid Y]]\\ &={\rm E}[g(Y){\rm E}[f(X)]]={\rm E}[g(Y)]{\rm E}[f(X)] \end{align} $$ which shows that $X$ and $Y$ are independent.

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