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I don't really know how to start proving this question.

Let $\xi$ and $\eta$ be independent, identically distributed random variables with $E(|\xi|)$ finite.

Show that $E(\xi\mid\xi+\eta)=E(\eta\mid\xi+\eta)=\frac{\xi+\eta}{2}$

Does anyone here have any idea for starting this question?

Nate Eldredge
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kkk
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4 Answers4

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There's a subtle point here, which bothered me the first time I saw this problem.

Henry's answer has the essential idea, which is to use symmetry. User Did's comment points out that the symmetry comes from the fact that $(\xi, \eta)$ and $(\eta, \xi)$ are identically distributed. But, straight from the definition of conditional expectation, it isn't clear that symmetry in the joint distributions is enough to get the result. I ended up having to prove the following lemma:

Lemma. Let $X,Y$ be random variables. There is a measurable function $f$ such that $E[X\mid Y] = f(Y)$ a.s. Moreover, if $(X', Y')$ is identically distributed to $(X,Y)$, then $E[X' \mid Y'] = f(Y')$ a.s. for the same function $f$.

Proof. The existence of $f$ is a consequence of the Doob-Dynkin Lemma. For the second part, we use the definition of conditional expectation. $f(Y')$ is clearly $\sigma(Y')$-measurable, so it remains to show that for any $A \in \sigma(Y')$, we have $E[1_A f(Y')] = E[1_A X']$. Since $A \in \sigma(Y')$, $A = (Y')^{-1}(B)$ for some Borel set $B$ (this fact is part of the proof of Doob-Dynkin). But since $(X',Y')$ has the same distribution as $(X,Y)$, we get $$\begin{align*} E[1_A f(Y')] &= E[1_B(Y') f(Y')] \\ &= E[1_B(Y) f(Y)] \\ &= E[1_B(Y) E[X \mid Y]] \\ &= E[1_B(Y) X] && \text{since $1_B(Y)$ is $\sigma(Y)$-measurable}\\ &= E[1_B(Y') X'] \\ &= E[1_A X'] \end{align*}$$ as desired.

It is worth noting that the function $f$ is generally not unique. In particular, we could modify $f$ almost arbitrarily on any set $C \subset \mathbb{R}$ such that $P(Y \in C)=0$.

Also, to address the point in kkk's comment: Just knowing that $\xi, \eta$ are identically distributed is not sufficient. Here is a counterexample. Let $\Omega = \{a,b,c\}$ have three outcomes, each with probability $1/3$ (and $\mathcal{F} = 2^\Omega$). Let $X(a) = 0$, $X(b)=1$, $X(c)=2$; and $Y(a)=1$, $Y(b)=2$, $Y(c)=0$. Thus $X$ is uniformly distributed on $\{0,1,2\}$, and $Y = X + 1 \bmod 2$, so $Y$ is also uniformly distributed on $\{0,1,2\}$.

Now we have $(X+Y)(a) = 1$, $(X+Y)(b)=3$, $(X+Y)(c)=2$. So $X+Y$ is a 1-1 function on $\Omega$ and thus $\sigma(X+Y) = \mathcal{F}$, so both $X,Y$ are $\sigma(X+Y)$-measurable. Thus $E[X\mid X+Y]=X$, $E[Y\mid X+Y]=Y$. However, $X$, $Y$, and $\frac{X+Y}{2}$ are all different.

Community
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Nate Eldredge
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  • I like this lemma, and the example too! –  Nov 03 '11 at 14:34
  • This lemma IS a straight consequence of the definition of conditional expectation (and I remember seeing it in textbooks under headings like Conditional expectations depend only on the distributions). But it is nice to see it stated separately. – Did Nov 03 '11 at 16:47
  • Now things seem to be complicated for me when it comes to measure theory.So Byron, according to that lemma, are you saying, we still could stick to proof u gave before, if there exists such function? – kkk Nov 04 '11 at 06:54
  • Sorry, the proof henry presents – kkk Nov 04 '11 at 06:56
  • @NateEldredge Great answer, but I can't figure out how to prove the lemma in the general (i.e. not necessarily absolutely continuous) case. Any hints on how to procede without using joint and conditional densities, just the definition of conditional expectation? Thanks a lot – Emilio Ferrucci Sep 14 '14 at 21:43
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    @EmilioFerrucci: I added a proof. – Nate Eldredge Sep 15 '14 at 01:13
  • It seems silly to think that identical distribution of $\xi$ and $\eta$ would be of consequence here. It is identical distribution of the _pairs $(\eta,\xi)$ and $(\xi,\eta)$ that matters. $\qquad$ – Michael Hardy Feb 10 '18 at 17:32
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I state in full and prove in details.

Proposition: Let $(\Omega,\mathcal{F},P$) be a probability space. Let $X,Y$ be i.i.d. random variables with $E\left[|X|\right]<\infty$. Let $\mathcal{G}=\sigma(X+Y)$. Then $\operatorname{E} \left[X \mid \mathcal{G}\right] = \operatorname{E} \left[Y \mid \mathcal{G}\right]=\frac{1}{2}(X+Y)$.

Proof: Let $\mu_{XY}$ be the joint distribution measure on $\mathbb{R}^{2}$ induced by $(X,Y)$. That is, $\mu_{XY}(B)=P\left(\left\{ \omega \mid (X(\omega), Y(\omega)) \in B\right\} \right)$. Let $\mu_X$ and $\mu_Y$ be the distribution measures on $\mathbb{R}$ induced by $X$ and $Y$ respectively. Since $X$ and $Y$ are independent, we have $\mu_{XY}=\mu_X\times\mu_Y$. Moreover, since $X$ and $Y$ are identically distributed, $\mu_X=\mu_Y$. We denote $\mu=\mu_X=\mu_Y$.

Let $A\in\mathcal{G}$ be arbitrary. There exists a Borel set $B\subseteq\mathbb{R}$ such that $A=(X+Y)^{-1}(B)$. Hence $1_{A}(\omega)=1_{B}(X(\omega)+Y(\omega))$ for any $\omega\in\Omega$.

We have \begin{align} & \int_A \operatorname{E}\left[X\mid\mathcal{G}\right]\,dP = \int_A X\,dP=\int 1_B(X+Y)X \, dP = \int 1_B(x+y)x\,d\mu_{XY}(x,y) \\[10pt] = {} & \iint1_{B}(x+y)x\,d\mu_{X}(x) \, d\mu_Y(y) = \iint 1_B(x+y)x \, d\mu(x) \, d\mu(y). \end{align} By the same argument, $$ \int_A \operatorname{E}\left[Y\mid\mathcal{G}\right]\,dP=\iint1_{B}(x+y)y \, d\mu(x) \, d\mu(y). $$ Now it is clear that $$ \int_A \operatorname{E}\left[X\mid\mathcal{G}\right]\,dP=\int_A \operatorname{E} \left[Y\mid\mathcal{G}\right] \,dP $$ and hence $\operatorname{E} \left[X \mid \mathcal{G}\right] = \operatorname{E}\left[Y \mid \mathcal{G}\right]$. Lastly, $\operatorname{E}\left[X+Y\mid\mathcal{G}\right]=X+Y$. It follows that $\operatorname{E}\left[X\mid\mathcal{G}\right]=\operatorname{E} \left[Y \mid \mathcal{G} \right]=\frac 1 2 (X+Y)$.

Michael Hardy
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Danny Pak-Keung Chan
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  • I forget to point out that all equalities involving conditional expectation should be interpreted as $P$-.a.s., for example, I write $E[X\mid\mathcal{G}]=E[Y\mid\mathcal{G}]$. It actually means that they are equal a.s.. – Danny Pak-Keung Chan Aug 07 '17 at 18:46
  • may you explain why the line $\int 1_B(X+Y)X , dP = \int 1_B(x+y)x , d \mu_{X,Y}$ holds? – Bryan Shih Nov 12 '17 at 14:43
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    @CWL: It is from the following fact: For any Borel function $f:\mathbb{R}^2\rightarrow \mathbb{R}$ that is $\mu_{X,Y}$-integrable, we have $\int f(X,Y),dP = \int f(x,y) d\mu_{X,Y}(x,y)$. This fact can be proved in the standard way: Firstly it is true for all indicator functions of Borel subsets of $\mathbb{R}^2$. Then, by linearity, it is true for all simple functions. Then, by Monotone Convergence Theorem, it is true for all non-negative Borel functions. Lastly, by decomposing into positive part and negative part... – Danny Pak-Keung Chan Nov 13 '17 at 17:16
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$E(\xi\mid \xi+\eta)=E(\eta\mid \xi+\eta)$ since $\xi$ and $\eta$ are exchangeable, i.e. $(\xi,\eta)$ and $(\eta,\xi)$ are identically distributed. (Independent does not matter here.)

So $2E(\xi\mid \xi+\eta)=2E(\eta\mid \xi+\eta) = E(\xi\mid \xi+\eta)+E(\eta\mid \xi+\eta) =E(\xi+\eta\mid \xi+\eta) = \xi+\eta$ since the sum $\xi+\eta$ is fixed.

Now divide by two.

Henry
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    Qualification: $E(\xi\mid\xi+\eta)=E(\eta\mid\xi+\eta)$ because $(\xi,\eta)$ and $(\eta,\xi)$ are identically distributed. The condition that $\xi$ and $\eta$ are identically distributed is not enough. – Did Nov 03 '11 at 11:36
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    Thank you so much, i was thinking in a more complicated way.But Didier Might be right, Even if $\xi$ and $\eta$ are identically distirbuted,would that be enough to show the conditional expecation. are the same? – kkk Nov 03 '11 at 11:37
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    Might be right is absolutely sweet. – Did Nov 03 '11 at 12:03
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    $\mathbb P(\xi=1)=\mathbb P(\xi=2)=\mathbb P(\xi=3)=\frac13$, $\eta=\xi+1 \mod 3$. Then $\mathbb E(\xi|\xi+\eta=3)=1$ while $\mathbb E(\eta|\xi+\eta=3)=2$ and so on. – NCh Jun 03 '17 at 04:31
  • @NCh: I think that was Didier's point: I implicitly used $(\xi,\eta)$ and $(\eta,\xi)$ being identically distributed, which is not satisfied by your example – Henry Jun 03 '17 at 10:02
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    Sure. And therefore the independence does matter here. – NCh Jun 04 '17 at 01:37
  • Why not correct the post? – Did Aug 07 '17 at 20:43
  • @Did: I did it by stating the proposition in full and proving it in details. – Danny Pak-Keung Chan Aug 08 '17 at 00:33
  • @DannyPak-KeungChan Believe it or not, my comment was not referring to your answer (whose necessity, by the way, nearly 6 years after Nate provided a full answer, I fail to be convinced about). – Did Aug 08 '17 at 06:44
  • @Henry this proof seem so require the statement: if $X_1, X_2\sim X$ are iid, $Y_1, Y_2\sim Y$ are iid and $X_1+X_2=Y_1+Y_2$ then $X\sim Y$. – Andrew Aug 30 '23 at 19:38
  • @Andrew - I think that was related to the point Didier Pau made 12 years ago and Nate Eldredge made in another answer. My answer relied on $(ξ,η)$ and $(η,ξ)$ being identically distributed which is stronger than $ξ$ and $η$ being identically distributed but weaker than them being iid. – Henry Aug 30 '23 at 19:54
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The assumption of independence can be weakened. You say $\zeta,\eta$ are i.i.d. A consequence is:

$$ \text{ The pairs } (\zeta,\eta) \text{ and }(\eta,\zeta) \text{ both have the same distribution.} \tag 1 $$

Therefore $$\operatorname E(\zeta\mid \zeta+\eta) = \operatorname E(\eta\mid \zeta+\eta).$$ Next, observe that $$\operatorname E(\zeta\mid\zeta+\eta) + \operatorname E(\eta\mid\zeta+\eta) = \operatorname E(\zeta+\eta \mid \zeta+\eta) = \zeta+\eta.$$

Thus we have $$ \operatorname E(\zeta\mid\zeta+\eta) = \operatorname E(\eta\mid \zeta+\eta) = \frac{\zeta+\eta} 2. $$ The statement $(1)$ above is weaker than "i.i.d." but stronger than identical distribution.

Michael Hardy
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