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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.

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    $\begingroup$ Correct! (Why can't I put just that word?) $\endgroup$
    – Zhanxiong
    Nov 4, 2015 at 3:44
  • $\begingroup$ @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 $\endgroup$
    – BCLC
    Nov 4, 2015 at 5:57
  • $\begingroup$ 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 $\int X dP \leq \int Y dP$. $\endgroup$
    – Zhanxiong
    Nov 4, 2015 at 6:03
  • $\begingroup$ @Solitary That's monotonicity. I didn't simply apply monotonicity on $Z \le |Z|$. LHS of ineq has | | in the middle. $\endgroup$
    – BCLC
    Nov 4, 2015 at 6:42
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    $\begingroup$ It's better to call this Jensen's inequality imo. The convex function being $x\rightarrow |x|$ $\endgroup$ Nov 4, 2015 at 10:04

2 Answers 2

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You could invoke Jensen's inequality for conditional expectation. Here's an easier way to prove it, in the sense that it requires less background machinery.

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.

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The math you stated looks right. I'm not aware of any name for the inequality, but just from the title it is fairly intuitive. If you had any (finite) set of points, any non-finite value would cause the expectation to be non-finite.

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  • $\begingroup$ What do you mean? if we take expectation of $Z_n$, there are no absolute value signs there. $\endgroup$
    – BCLC
    Nov 4, 2015 at 6:47
  • $\begingroup$ Negative/Positive is somewhat orthogonal to Finite/Nonfinite. The absolute value operation doesn't really do much. Am I missing something? $\endgroup$ Nov 4, 2015 at 6:55

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