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We define the conditional expectation of a random variable $X$ on a given probability space $(\Omega,\mathscr F, P) $ w.r.t a sub $\sigma$-algebra $\mathscr G$ is a random variable denoted by $E[X|\mathscr G]$ defined as random variable that satisfies the following condition

$\int_{A}E[X|\mathscr{G}] \hspace{0.1 cm}dP = \int_A X\hspace{0.1 cm}dP \hspace{0.3 cm}\text{for all } A \in \mathscr{G}$

But we know that if $X_1$ and $X_2$ are two random variables satisfying the condition that

$\int_{A}X_1 \hspace{0.1 cm}dP = \int_A X_2\hspace{0.1 cm}dP \hspace{0.3 cm}\text{for all } A \in \mathscr{G}$, then $X_1 = X_2$ a.s.

Therefore for the conditonal expectation defined as above, we conclude that $E[X|\mathscr G] = X$ a.s. Then what is the major difference between $E[X|\mathscr G]$ and $X$?

Uday Kumar
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  • $\int_A X_1\ \mathsf d\mathbb P=\int_A X_2\ \mathsf d\mathbb P$ for all $A\in\mathcal G$ implies that $\mathbb E[X_1\mid\mathcal G]=\mathbb E[X_2\mid\mathcal G]$ a.s., not that $X_1=X_2$ a.s. – Math1000 Apr 04 '16 at 07:51
  • $\int_{A}X_1 \hspace{0.1 cm}dP = \int_A X_2\hspace{0.1 cm}dP \hspace{0.3 cm}$ for all $ A \in \mathscr{G}$, then $X_1 = X_2$ a.s. is TRUE I believe strongly. Do you agree that if $\int_A Y \hspace{0.1 cm} dP=0 $ for all $A \in \mathscr{G}$, then $Y=0$ a.s. Do you agree this? If you don;t agree this..pls provide a counter example. – Uday Kumar Apr 04 '16 at 08:45
  • Counterexample - $\mathcal G={\varnothing, \Omega}$. – Math1000 Apr 04 '16 at 16:26
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    This is not the correct counter example. Only constant functions(or random variables) are measurable with respect to the trivial sigma algebra $\mathcal G = { \Phi,\Omega}$ – Uday Kumar Apr 05 '16 at 06:03
  • Fair enough, I was being a bit glib :) – Math1000 Apr 05 '16 at 06:07
  • Actually, @Math1000's counterexample is spot on: for integrable random variables $X_1$ and $X_2$, the fact that the condition [$E(X_1;A)=E(X_2;A)$ for every $A$ in $\mathcal G$] implies [$X_1=X_2$ almost surely] holds only if $X_1-X_2$ is $\mathcal G$-measurable. Here you are trying to apply this true fact to $$X_1=X\qquad X_2=E(X\mid\mathcal G)$$ For this choice, $X_2$ is $\mathcal G$-measurable by definition hence $X_1-X_2$ is $\mathcal G$-measurable if and only if $X$ is, and then indeed $E(X\mid\mathcal G)=X$ almost surely. But not otherwise... – Did Feb 06 '17 at 10:30

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Let $$\mathbb P(X=1) = \mathbb P(X=-1)=\frac12, $$ $Y=2X$, and $$\mathbb P(W=1) = \mathbb P(W=0)=\frac12, $$ with $W$ and $X$ independent. Then $$\mathbb E[X\mid\sigma(W)]=\mathbb E[X] = 0,\quad E[Y\mid\sigma(W)]=\mathbb E[Y] = 0\ a.s. $$ so for any $A\in\sigma(W)$, $$\mathbb E[\mathbb E[X\mathsf \mid\sigma(W)]1_A] = \mathbb E[\mathbb E[X]\mathsf 1_A]=\mathbb E[X]\mathbb P(A)=0$$ and similarly for $Y$. But $\mathbb P(X\ne Y)=1$, so clearly $X\ne Y$ a.s.

Math1000
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