# If $Y\in\mathcal{L}^1$ and $(\mathcal{F}_t,t\in I)$ is a filtration, then $\text{E}[\text{E}[|Y|\mid\mathcal{F}_t]]=\text{E}[|Y|]$

Let

• $(\Omega,\mathcal{A},\operatorname{P})$ be a probability space
• $I\subseteq\mathbb{R}$
• $(\mathcal{F}_t,t\in I)$ be a filtration on $(\Omega,\mathcal{A})$
• $Y\in\mathcal{L}^1(\operatorname{P})$ and $$X_t:=\operatorname{E}\left[Y\mid\mathcal{F}_t\right]\;\;\;\text{for }t\in I$$

I don't understand why it holds $$\operatorname{E}\left[\operatorname{E}\left[\left|Y\right|\mid\mathcal{F}_t\right]\right]=\operatorname{E}\left[|Y|\right]\tag{1}$$

$(1)$ would imply $$\operatorname{E}\left[\left|Y\right|\mid\mathcal{F}_t\right]=|Y|\;\;\;\color{blue}{\operatorname{P}\text{-almost surely}}$$ So, $Y$ would need to be $\mathcal{F}_t$-measurable. I really don't understand why this should hold.

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

• $Z$ is $\mathcal{F}$-measurable
• $\operatorname{E}[1_AX]=\operatorname{E}[1_AZ]$ for all $A\in\mathcal{F}$

We write $\operatorname{E}[X\mid\mathcal{F}]=:Z$.

• No (1) does not imply that. Likewise E(U)=E(V) does not imply U=V P-almost surely.
– Did
Commented Apr 2, 2015 at 20:20
• @Did You're right. But we've got $\operatorname{E}[X]=\operatorname{E}[Y]$ $\Rightarrow$ $X=Y$ almost surely, if $X\ge Y$ almost surely. I'd consufed that. Commented Apr 2, 2015 at 20:26
• It is not true that $E[|Y||\mathcal{F}_t]\leq |Y|$ almost surely, nor is the reverse true. Commented Apr 2, 2015 at 20:28
• No, @Shalop's point is different and, should I admit, slightly subtler than mine (and I see nothing wrong with the notations in this comment). Say, 0/oxbadfo/00/od, what are you playing at?
– Did
Commented Apr 2, 2015 at 20:32
• @Did He has edited his comment. Commented Apr 2, 2015 at 20:46

If $X$ is any random variable, and $\mathcal{H} \subset \mathcal{A}$ is any sub-sigma-algebra, then by the definition of conditional expectation, it is always true that $$E[E[X|\mathcal{H}]] = \int_{\Omega} E[X|\mathcal{H}] dP = \int_{\Omega} X dP = E(X)$$ Now just apply this when $X=|Y|$ and $\mathcal{H}=\mathcal{F}_t$.