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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 '15 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 '15 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 '15 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 '15 at 6:42
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    $\begingroup$ It's better to call this Jensen's inequality imo. The convex function being $x\rightarrow |x|$ $\endgroup$ – Guillaume Dehaene Nov 4 '15 at 10:04
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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 '15 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? $\endgroup$ – Andrew Charneski Nov 4 '15 at 6:55

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