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I want to prove that if $X_1,X_2$ are i.i.d. random variables then $E[X_1| X_1+X_2] = E[X_2|X_1+X_2]$.

I see that this is intuitive but I think it is by no means trivial yet everybody just states this property as though it was completely obvious. Do I really miss something obvious here?

In my attempt to prove this I didn't get so far. It suffices to show $E[1_AX_1] = E[1_AX_2]$ for $A=\{X_1+X_2\in B\}$ where $B$ is any Borel set. I don't see how this is trivial and would appreciate help to let me see how it works.

Did
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Lukas Betz
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    The key argument is that conditional expectations depend only on distributions, in the following sense: if the distributions of $(U,V)$ and $(U',V')$ coincide, with $U$ integrable (or equivalently, $U'$ integrable), and if $E(U\mid V)=g(V)$, then $E(U'\mid V')=g(V')$. If you never saw this result, you could try to prove it (there is no real difficulty here), then to apply it to $(U,V)=(X_1,X_1+X_2)$ and $(U',V')=(X_2,X_1+X_2)$, to get the result in your post. More generally, for every measurable $h$ such that $h(X_1)$ is integrable, $$E(h(X_1)\mid X_1+X_2)=E(h(X_2)\mid X_1+X_2)$$ – Did Sep 28 '16 at 21:30
  • I see how this is a solution to my question even though I haven't figured out the proof of the result you stated yet but I know what I have to be working on now. Thank you very much. – Lukas Betz Sep 28 '16 at 21:51
  • @LeBtz Did is great and clarifies a lot :-) To split hairs, though, I thought the problem was how to show that $(X_1, X_1 + X_2)$ and $(X_2, X_1 + X_2)$ are identically distributed; from then on the equality of the conditional expectations has been established (though it's always nice to have a more general theorem clearly stated) – Ant Sep 28 '16 at 21:57
  • @Ant I wasn't aware of the theorem Did mentioned. I was wondering why [2] of your answer was helpful and looking in some books to find an answer before Dids comment. Now I have two distinct steps that I can worry about (and which look more managable than my question without any clues). One from your answer and one from Dids comment. So thanks to both of you :) – Lukas Betz Sep 28 '16 at 22:03
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    @Did I managed to show the mentioned theorem on conditional expectations in general now as well as [2] from Ant. Thanks again to you two, this was just the help I needed. – Lukas Betz Sep 28 '16 at 22:38
  • See also this quation. – Gordon Sep 30 '16 at 14:59
  • https://math.stackexchange.com/q/78546/321264 – StubbornAtom May 11 '21 at 14:30

1 Answers1

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The conditional expectation $E[X \mid Y]$ is defined by the property that for every $Z \in \sigma(Y)$, then $$E[XZ] = E[ZE[X \mid Y]]$$

You can think of it as a "projection"[1]; it tells you that in the "space" generated by $Y$ (in this case the sigma algebra $\sigma(Y)$), $X$ and $E[X \mid Y]$ act in the same way on elements of $Y$; like the vectors $(1, 1, 0)$ and $(1,1,4)$ behave in the same way in the $\mathbb R^2$ plane.

Now you know that $$E[X_1Z] = E[Z E[X_1 \mid X_1 + X_2]$$

for every $Z \in \sigma(X_1 + X_2)$. But the left hand side is also equal to $E[X_2Z]$ (since $X_1, X_2$ are iid[2]). Hence

$$E[Z E[X_1 \mid X_1 + X_2]] = E[ZX_2]$$

Looking again at the definition, this tells you that

$$E[X_1 \mid X_1 + X_2] = E[X_2 \mid X_1 + X_2]$$

Another way to do this is looking at the symmetry of the problem; how you going to distinguish $X_1$ and $X_2$? Calling $X_1$ $X_2$ and viceversa leaves everything the same, so clearly the two conditional expectations have to be equal

[1]: Indeed, if we restrict ourselves to functions in $L^2$, then the inner product is precisely $(X,Y) = E[XY]$, so the definition is exactly analogous to the usual definition of projections on euclidean space. Or for elements of Hilbert spaces, for what matter. And in fact $L^2$ is an hilbert space. For functions not in $L^2$, this is a generalization.

[2] It suffice to prove that the joint cdf of $(X_1, X_1 + X_2)$ and $(X_2, X_1 + X_2)$ are the same. As the OP @LeBtz pointed out in the comments, the correct argument is as follows: Since $(X_1, X_2)$ and $(X_2, X_1)$ have the same distribution, it follows that $f(X_1, X_2)$ and $f(X_2, X_1)$ also have the same distribution. Applying it to the function $f(x,y) = (x, x+y)$ we prove the result.

Ant
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    Sorry, but i think this doesn't really answer my question. You saysing that $E[X_2Z] = E[X_1Z]$ (which is the key part in the proof) because $X_1,X_2$ are i.i.d is exactly what i meant by people just stating the property I am interested in as though it was obvious. – Lukas Betz Sep 28 '16 at 20:55
  • @LeBtz Oh, I see, I'm sorry :) See my edit and tell me if something it's still missing ;-) – Ant Sep 28 '16 at 21:11
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    First: Thanks for your time! However I don't see how the first equality is justified. If we set $x=y=0$ for example on the right side $X_1,X_2$ both have to be negative. On the left side however $X_2$ can be positive if $X_1$ is negative. – Lukas Betz Sep 28 '16 at 21:25
  • @LeBtz Indeed, you're right. I will think about it – Ant Sep 28 '16 at 21:52
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    I think I found a way now: Since $(X_1,X_2)$ and $(X_2,X_1)$ have the same distribution, so have $f(X_1,X_2)$ and $f(X_2,X_1)$ where $f:\mathbb R^2\to\mathbb R^2, (x,y)\mapsto (x,x+y)$. Should work, right? – Lukas Betz Sep 28 '16 at 22:15
  • I managed to show the mentioned theorem on conditional expectations in general now as well as [2] from you. Thanks again to you two, this was just the help I needed. – Lukas Betz Sep 28 '16 at 22:38
  • @LeBtz You're welcome; yes I think it works! :D – Ant Sep 28 '16 at 22:50
  • @LeBtz This is the argument. Well done. – Did Sep 29 '16 at 13:10
  • So the first equation in [2] is not correct? – peer Oct 01 '16 at 11:28
  • @peer Indeed. Let me remove it – Ant Oct 01 '16 at 11:32