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Let $X,Y$ be independent integrable random variables and let $f :\mathbb{R}^2 \to \mathbb{R}$ be integrable. It makes intuitive sense that $$ \mathrm E [f(X,Y) \mid Y=y] = \mathrm E [f(X,y)], $$ but I'm having trouble showing this. Does this only hold if $Y$ is discrete? I'd like to show it rigorously using regular conditional probabilities, so starting out I would have $$ \mathrm E [f(X,Y) \mid Y=y] = \int_{\Omega} (f \circ (X,Y))(\omega)\, P^Y(\mathrm d\omega \mid y), $$ where $P^Y(\cdot \mid \cdot)$ is the regular conditional probability of $P$ given $Y$. I'm really not sure how to proceed form here, so any help is appreciated.

Update: Added that $X$ and $Y$ should be independent.

bcf
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  • Sorry, I'm not familiar with your notation. Do you mean you are conditioning on the $\sigma$-algebra created by the set $ { Y = y } $? – kummerer94 Jul 24 '15 at 21:55
  • @kummerer94 Not quite, as $\mathrm E [f(X,Y) \mid Y=y]$ is not a conditional expectation w.r.t. a sigma-algebra, but rather a function $y \mapsto \mathrm E [f(X,Y) \mid Y=y]$, which is defined to be this integral. I'm using notation from this answer: http://math.stackexchange.com/questions/496608/formal-definition-of-conditional-probability – bcf Jul 24 '15 at 21:59
  • Oh! Quite interesting, I mistook it completely. I will need to dig deeper into the definitions to help here. Thanks so far! – kummerer94 Jul 24 '15 at 22:01
  • @kummerer94 If you'd like to get your feet wet, perhaps you could help me with this seemingly simple problem? http://math.stackexchange.com/questions/1372960/mathrm-e-x-mid-x-x-x – bcf Jul 24 '15 at 22:09
  • The identity you are interested in holds only when $X$ and $Y$ are independent. // What is the source from which you copy this string of questions about conditional expectations? – Did Jul 24 '15 at 22:24
  • @Did The source did assume they were independent, thanks for noticing. I updated my question. – bcf Jul 25 '15 at 14:45
  • Bis: What is the source from which you copy this string of questions about conditional expectations? – Did Jul 25 '15 at 15:18
  • @Did It was from a solutions document for Shreve's Stochastic Calculus for Finance II. Why? – bcf Jul 25 '15 at 16:36

1 Answers1

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To flesh out Did's comment: let $P(X=1)=P(X=-1)=1/2$. Let $Y=-X$ and note that $X,Y$ are not independent. Then define $f(x,y)=x+y$.

$0=E[X+Y|Y=1]\neq E[X+1]=1.$

Alex R.
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