Let $X$ and $Y$ be independent identically distributed random variables. How can one show that for all $A \in \sigma\{X+Y\}$, we have $\mathbb{E}[X:A] = \mathbb{E}[Y:A]$?
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Is the column the sign of conditioning (|)? – zoli May 05 '18 at 12:19
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Any $A \in \sigma (X+Y)$ is of the form $A=(X+Y)^{-1} (E)$ for some Borel set $E$ in $\mathbb R$. Now write the integrals as integrals over the product space and apply the transformation $(x,y) \to (y,x)$. Since $X,Y$ is i.i.d. this transformation does not change the product measure.
Kavi Rama Murthy
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Thanks, this seems to be the standard approach. For some reason I found no similar questions (despite searching) until I had posted this. Here are further elaborations for reference: https://math.stackexchange.com/questions/139405/conditional-expectation-of-book-shiryaev-page-233/139407#139407 https://math.stackexchange.com/questions/78546/conditional-expectation-for-a-sum-of-iid-random-variables-e-xi-mid-xi-eta-e?rq=1 Thanks! – Daven May 05 '18 at 12:36