Law of iterated expectations with absolute value in front of inner expectation

Question

I have that $Y$ is an $\mathcal{F}$-measurable r.v. with $E{|Y|}<\infty$. I define $$X_{n}=E[Y\mid \mathcal{F_{n}}]$$ I want to compute $E[|X_n|]$. Is it correct that, by applying tower rule, we get the following?

$$E[|X_n|]=E|E[Y\mid \mathcal{F}_n]|=|E[Y]|$$

Or the correct result is the following

$$E[|X_n|]=E|E[Y\mid \mathcal{F}_n]|=E[Y]$$ with $E[Y]$ not in absolute value?

Answer 1

Both are false. You can only say that $E|X_n| \leq E|Y|$.

Tower rule works when you don't have absolute value signs.

Proof of the inequality:

$|Y| -Y \geq 0$ and this implies $E(|Y||\mathcal F_n) \geq E(Y|\mathcal F_n)$. Similarly, $E(|Y||\mathcal F_n) \geq -E(Y|\mathcal F_n)$ and hence $E(|Y||\mathcal F_n) \geq |E(Y|\mathcal F_n)|$. Now take expectation on both sides.

Answer 2

No.

If e.g. $Y$ is $\mathcal F_n$-measurable then $X_n=Y$ a.s. but statements like $\mathbb E|Y|=|\mathbb EY|$ and $\mathbb E|Y|=\mathbb EY$ are definitely not true in general.

Answer 3

I think that now I am able to try to respond to my own question.
$$E\left\{|Y|\right\}=E\left\{E\{|Y||\mathcal{F}_n\}\right\}\geq E\{|E\{Y|\mathcal{F}_n\}|\}=E\left\{|X_n|\right\}$$ where the inequality follows from Jensen's inequality, while the first equality follow from Law of Iterated Expectations and the second equality follows from definition of $X_n$ given in the question.