Why does it hold $\operatorname{E}[Y\mid\mathcal{F}]=\operatorname{E}[Y\mid Y]=Y$, if $Y$ is $\mathcal{F}$-measurable? Let


*

*$(\Omega,\mathcal{A},\operatorname{P})$ be a probability space

*$\mathcal{F}\subseteq\mathcal{A}$ be a $\sigma$-algebra on $\Omega$

*$Y\in\mathcal{L}^1(\Omega,\mathcal{A},\operatorname{P})$ be measurable wrt $\mathcal{F}$


Then, $$\operatorname{E}[Y\mid\mathcal{F}]=\operatorname{E}[Y\mid Y]=Y$$

From the definition of conditional expectation (see below) it's easy to see that we've got $$\operatorname{E}[Z]=\operatorname{E}[Z'],$$ where


*

*$Z:=\operatorname{E}[Y\mid\mathcal{F}]$

*$Z':=\operatorname{E}[Y\mid\sigma(Y)]=\operatorname{E}[Y\mid Y]$


However, I don't see how we can conclude $Z\equiv Z'$ (almost surely) and $\operatorname{E}[Z']=Y$.

Please note:
A random variable $Z$ is called conditional expectation of $X$ given $\mathcal{F}$ $:\Leftrightarrow$


*

*$Z$ is $\mathcal{F}$-measurable

*$\operatorname{E}[1_AX]=\operatorname{E}[1_AZ]$ for all $A\in\mathcal{F}$


We write $\operatorname{E}[X\mid\mathcal{F}]=:Z$. The notation $\operatorname{E}[X\mid X']$, with $X'$ being another random variable, is a shorthand for $\operatorname{E}[X\mid\sigma(X')]$
 A: First we prove that $E[Y\mid\mathscr{F}]=Y$ whenever $Y$ is $\mathscr{F}$-measurable. For this, you only need to demonstrate that $Y$ satisfies the two given requirements: 


*

*$Y$ is $\mathscr{F}$-measurable: this is given;

*$E[1_AY]=E[1_AY]$ for all $A\in\mathscr{F}$: this is trivial.


Next, consider $E[Y\mid Y]$. Let $\mathscr{G}=\sigma(Y)$, the $\sigma$-algebra generated by $Y$. Of course, $Y$ is $\mathscr{G}$-measurable, so by the result just proved, we have $E[Y\mid Y]=E[Y|\mathscr{G}]=Y$. We conclude
$$
E[Y\mid\mathscr{F}]=Y=E[Y\mid Y]\quad\text{whenever }Y\text{ is }\mathscr{F}\text{-measurable}.
$$
A: Suppose $Z$ and $Z'$ are two versions of $E(Y|\mathcal F)$. Look at $X:=Z-Z'$. By definition of conditional expectation, $E(X1_A)=0$ for every $A\in\mathcal F$. Apply this to the set $X>\varepsilon$:
$$0 = E(X1_{\{X>\varepsilon\}})\ge\varepsilon P(X>\varepsilon)$$
Conclude that $P(X>\varepsilon) =0$ for every $\varepsilon>0$. Finally use the fact that $\{X>\frac1n\}\uparrow\{X>0\}$ to conclude $P(X>0)=0$. Similarly $P(X<0)=0$.
