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Let $X$ and $Y$ be independent and identically distributed random variables. Show that $$E[X\mid X+Y] = \frac{X+Y} 2$$

Michael Hardy
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Cnine
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  • Do you know that $E[X|X+Y] = E[Y|X+Y]$ in that case? – Lukas Betz Dec 03 '16 at 17:22
  • oh okay, because its IID? I didn't really know that – Cnine Dec 03 '16 at 17:25
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    People usually refer to it as being obvious. I asked for a rigorous proof on this site a while ago. You might want to have a look but I guess you would be allowed to just use this result. http://math.stackexchange.com/questions/1945550/explain-why-ex-1x-1x-2-ex-2x-1x-2-if-x-1-x-2-are-i-i-d – Lukas Betz Dec 03 '16 at 17:27
  • I looked at it, but you see that Michael's post below makes it even easier – Cnine Dec 03 '16 at 18:24
  • No it doesn't. It uses the same identity I proposed just without proof. This is fine of course, you can use that if you want. But thats not easier but just skipping part of the proof. – Lukas Betz Dec 03 '16 at 18:25
  • I see, I guess I didn't get it at first. I am hoping I can use it without the proof. – Cnine Dec 03 '16 at 18:29
  • Probably the reason for the votes to close the question as "off-topic" is that it's phrased in language suitable for assigning homework. That tends to be frowned on here. – Michael Hardy Dec 03 '16 at 18:34

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Symmetry shows that $$\operatorname{E}(X\mid X+Y) = \operatorname{E}(Y\mid X+Y).$$ And clearly $$\operatorname{E}(X\mid X+Y) + \operatorname{E}(Y\mid X+Y) = \operatorname{E}(X+Y\mid X+Y) = X+Y.$$ So these two things are equal to each other and their sum is $X+Y$.

Michael Hardy
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