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Sorry if the title is confusing. I am studying this property: $$ \mathbb{E}[\mathbb{E}(Y|W)|X] = \mathbb E[Y|X] $$ where $X=f(W)$, and I don't really know if I am proving it right.

Here's my proof:
$$ \begin{align} \mathbb{E}[\mathbb{E}(Y|W)|X] &= \mathbb{E}\left[ \int_\mathbb R y \frac{f_{WXY}(w,x,y)} {f_{WX}(w,x)} dy \big|X \right]\\ &=\int_\mathbb R\int_\mathbb R y \frac{f_{WXY}(w,x,y)} {f_{WX}(w,x)} dy \frac {f_{WX}(w,x)} {f_X(x)} dw \\ \end{align} $$ and so the $f_{WX}$'s cancel out and we have $$ \begin{align} &=\int_\mathbb R\int_\mathbb R y \frac{f_{WXY}}{f_X}dydw = \int_\mathbb R y\frac{f_{XY}}{f_X}dy=\mathbb E(Y|X) \end{align} $$ Am I getting anything wrong here?
What makes me unsure about my proof is the first step, i.e. writing $$ \mathbb{E}(Y|W) = \int_\mathbb R y \frac{f_{WXY}(w,x,y)} {f_{WX}(w,x)} dy $$ But I am not able to prove it otherwise.

Thank you.

WinnieXi
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  • Instead of working with probability densites, which don't have to exist in the first place (and I don't see what kind of definition of conditional expectations you are using, even if they exist), you could go to the defining property of conditional expectations. The basic idea is: Since the $\sigma$-algebra generated by $f(W)$ is coarser than the one generated by $W$, you obtain the same object whether you first condition on the finer $\sigma$-algebra and then on the coarser one, or on the coarser $\sigma$-algebra directly. – IljaKlebanov Mar 21 '23 at 22:38

1 Answers1

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It is enough to apply the definition of conditional expectation: \begin{gather*} E\biggl[E\Bigl[E\big[Y\mid W\big]\Big| X\Bigr]I_A(X)\biggr]= E\Bigl[E\big[Y\mid W\big]I_A(X)\Bigr]=\\ =E\Bigl[E\big[Y\mid W\big]I_A(f(W))\Bigr]= E\Bigl[E\big[Y\mid W\big]I_{f^{-1}(A)}(W)\Bigr]=\\ =E\big[YI_{f^{-1}(A)}(W)\big]=E\big[YI_A(f(W)\big]=E\big[YI_A(X)\big] \end{gather*}

Speltzu
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