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how can i show that:

$E[XY \vert X ] = XE[Y \vert X]$ for two random variables $X$ and $Y$

sorry this must be wrong what i meant was

$E[ E[XY \vert X ] ]= E [XE[Y \vert X]]$

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2 Answers 2

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Ok I'll give you a hint. Does $E[3Y|X]=3E[Y|X]$? Can you prove this?

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I am using the nomenclature from this article.

Let $f(x,y)$ be the joint probability density of $X$ and $Y$. Then $$ g(x)=\int f(x,y)\,\mathrm{d}y $$ is the probability density of $X$ and $$ h(y|x)=\frac{f(x,y)}{g(x)} $$ is the conditional density of $Y$ given $X$. Then $$ \mathrm{E}[X\mathrm{E}[Y|X]]=\int x\left(\int y\,h(y|x)\,\mathrm{d}y\right)\,g(x)\,\mathrm{d}x $$ and $$ \mathrm{E}[\mathrm{E}[XY|X]]=\int\left(\int xy\,h(y|x)\,\mathrm{d}y\right)\,g(x)\,\mathrm{d}x $$

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  • $\begingroup$ "E[Y|X] is the expected value of Y for a given X." This is not the definition of E[Y|X] (and absurd in many cases). $\endgroup$
    – Did
    Mar 29, 2014 at 15:43
  • $\begingroup$ Sorry, that was my understanding, which is obviously flawed. $\endgroup$
    – robjohn
    Mar 29, 2014 at 15:46
  • $\begingroup$ @Did: is what I have above incorrect? $\endgroup$
    – robjohn
    Mar 29, 2014 at 16:30
  • $\begingroup$ The new version is "Since X is fixed for the evaluation of the expectation and since expectation is linear". Unfortunately I don't know what "X is fixed for the evaluation of the expectation" means and I don't see how "expectation is linear" is related. Why not being rigorous, why not using the definition? $\endgroup$
    – Did
    Mar 29, 2014 at 16:51
  • $\begingroup$ @Did: I am learning while doing. I think I have properly converted my ideas to the case of continuous random variables. $\endgroup$
    – robjohn
    Mar 30, 2014 at 0:40

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