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For random variables $X,Y\in L^1$, if $E[X|\sigma(Y)]=Y$ and $E[Y|\sigma(X)]=X$, prove that $X=Y$ a.s.

My attempt:

  1. Note that $E[X-Y|\sigma(Y)]=E[X|\sigma(Y)]-E[Y|\sigma(Y)]=Y-Y=0$. Similarly, $E[X-Y|\sigma(X)]=0$.

  2. By 1, if I could prove the following lemma, then $E[X-Y|\sigma(X,Y)]=X-Y=0$ as desired.

Lemma (not necessarily true). If $E[Z|\mathcal A]=E[Z|\mathcal B]=0$ where $\mathcal A$ and $\mathcal B$ are two $\sigma$-fields, then $E[Z|\sigma(\mathcal A,\mathcal B)]=0$.

Questions:

  1. Is this "lemma" correct? Is there a counterexample?

  2. If not, how should I solve the original question?

trisct
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  • I remember a theorem that if $E(Z_1 1_B)=E(Z_2 1_B)$ so almost sure $Z_1=Z_2$ (with a condition like $Z_1$ and $Z_2$ are measurable with same sigma field, but i do not remember it exactly). In here $\forall B\in \sigma(X)$ $E(X1_B)=E(E(Y|X)1B)\overset{\star}{=}E(Y1_B)$ and $\forall B\in \sigma(Y)$ $E(Y1_B)=E(E(X|Y)1B)\overset{\star}{=}E(X1_B)$. ($\star$ is by definition of conditional expectation).Perhaps it will help – Masoud May 13 '20 at 13:16
  • Related: https://math.stackexchange.com/q/666843/321264 – StubbornAtom May 13 '20 at 15:00

2 Answers2

1

For the lemma, it may unfortunately not work. Let $\varepsilon$ and $\varepsilon'$ be two i.i.d. random variables such that $\varepsilon$ takes the values $-1$ and $1$ with probability $1/2$. Let $Z=\varepsilon\cdot \varepsilon'$ and $\mathcal A=\sigma(\varepsilon)$, $\mathcal B=\sigma(\varepsilon')$. The random variable $Z$ is $(\mathcal A\vee\mathcal B)$-measurable and since $\varepsilon$ is $\mathcal A$-measurable, it follows that $$ \mathbb E\left[Z\mid\mathcal A\right]=\varepsilon\mathbb E\left[\varepsilon'\mid\sigma(\varepsilon)\right]=0 $$ and similarly, $ \mathbb E\left[Z\mid\mathcal A\right]=0$.

For the second question, it was solved here.

Davide Giraudo
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0

I don't know your lemma is true or not but for the original question, here is my idea

$E[(X-Y)Y]=E[E(X-Y)Y|\sigma(Y)]]=E[YE[X-Y|\sigma(Y)]]=0$

Similarly, $E(X-Y)X]=0$, combine these two equations, we get $E[(X-Y)^2]=0$, which gives the desire result.

N.Quy
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