# Prove $E[|Z|] < \infty \to E[|Z_n|] < \infty$

Let $Z$ be an integrable random variable on filtered probability space $(\Omega , \mathscr F, (\mathscr {F_n})_{\{n \in \mathbb{N}\}}, \mathbb P)$

Define $Z_{n} := E[Z|\mathscr {F_n}]$. Show that $Z_n$ is integrable i.e. $E[|Z_n|] < \infty$

What I tried:

$E[|Z_n|] = E[|E[Z|\mathscr {F_n}]|] \le E[E[|Z||\mathscr {F_n}]] = E[|Z|] < \infty$ QED

Did I do that inequality* right? Any mistakes?

*It's the Triangle Inequality for Integrals.

• Correct! (Why can't I put just that word?) – Zhanxiong Nov 4 '15 at 3:44
• @Solitary Thanks! How about post as answer including name of inequality? I forgot what it's called. Something about convergence test or triangle inequality hahaha – BCLC Nov 4 '15 at 5:57
• I don't think it has a name. The underlying reason is that if you have $X \leq Y$, then $E(X|\mathscr{F}) \leq E(Y|\mathscr{F})$, which in turn relies on the fact that $X \leq Y$ implies that $\int X dP \leq \int Y dP$. – Zhanxiong Nov 4 '15 at 6:03
• @Solitary That's monotonicity. I didn't simply apply monotonicity on $Z \le |Z|$. LHS of ineq has | | in the middle. – BCLC Nov 4 '15 at 6:42
• It's better to call this Jensen's inequality imo. The convex function being $x\rightarrow |x|$ – Guillaume Dehaene Nov 4 '15 at 10:04

For a random variable $X$ let $X^- = \max(-X, 0)$ and $X^+ = \max(X, 0)$ be its negative and positive parts. By definition $\mathbb E X = \mathbb E X^+ - \mathbb EX^{-}$ whenever at least one of the terms is finite. Thus, $\mathbb E X < \infty$ if and only if $\mathbb E X^+<\infty$ and $\mathbb E X^- < \infty$. Thus, $\mathbb E X <\infty \iff \mathbb E X^+ + \mathbb EX^{-} = \mathbb E \vert X \vert <\infty$. But $\mathbb E Z_n = \mathbb E Z <\infty$ so indeed $\mathbb E \vert Z_n \vert <\infty$, which is what we wanted to prove.
• What do you mean? if we take expectation of $Z_n$, there are no absolute value signs there. – BCLC Nov 4 '15 at 6:47