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I know the following formulation of the Fatou's lemma:

From Jacod-Protter (2004)

Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space and $(X_n)_n$ be a sequence of random variables defined on it.
If the random variables $X_n$ satisfy $X_n\geq Y$ a.s. $(Y\in\mathcal{L}^1)$, all $n$, we have $$E\{\liminf_{n\to\infty}X_n\}\leq\liminf_{n\to\infty}E\{X_n\}\tag{1}$$In particular, if $X_n\geq 0$ a.s. all $n$, then $$E\{\liminf\limits_{n\to\infty}X_n\}\leq\liminf\limits_{n\to\infty}E\{X_n\}\tag{2}$$

Given the above result, how is that possible to show that:

$$\displaystyle\mathbb{E}( {\limsup_{n \mathop \to \infty} X_n}) \ge \limsup_{n \mathop \to \infty} \mathbb{E}({X_n})\tag{3}$$

? If so, how?

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  • $\begingroup$ Try Fatou with $Y-X_n$ $\endgroup$ – Meowdog Oct 5 '20 at 20:45
  • $\begingroup$ But how to pass from expectation of limit to probability of limit? @Shaqinho $\endgroup$ – Strictly_increasing Oct 5 '20 at 20:55
  • $\begingroup$ At the risk of asking something stupid, what is $E_n$? $\endgroup$ – zugzug Oct 5 '20 at 21:26
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    $\begingroup$ Is there missing information? In the edit, I see that $E_n$ was changed to $X_n$. But $\limsup X_n$ is not an event so taking the probability doesn't make sense. $\endgroup$ – zugzug Oct 5 '20 at 21:45
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    $\begingroup$ Oh, I think I understand. The probabilities should change to expected values. $\endgroup$ – zugzug Oct 5 '20 at 21:47
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The idea is to use the fact that $\liminf (-X_n)=-\limsup X_n.$ So then \begin{align} &\phantom{\implies} E(\liminf (-X_n))=E(-\limsup(X_n))=-E(\limsup(X_n))\\ &\leq \liminf E(-X_n)=\liminf(-E(X_n))=-\limsup(E(X_n)). \end{align} Then just flip the inequalities. Typically the assumption to justify this is that $|X_n|\leq Y$. Then $-Y\leq X_n \leq Y$. In particular, $-X_n\geq -Y$, so Fatou's is justified applied to $-X_n$.

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