# Showing a result starting from Fatou's lemma

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?

• Try Fatou with $Y-X_n$ – Meowdog Oct 5 '20 at 20:45
• But how to pass from expectation of limit to probability of limit? @Shaqinho – Strictly_increasing Oct 5 '20 at 20:55
• At the risk of asking something stupid, what is $E_n$? – zugzug Oct 5 '20 at 21:26
• 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. – zugzug Oct 5 '20 at 21:45
• Oh, I think I understand. The probabilities should change to expected values. – zugzug Oct 5 '20 at 21:47

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 $$-X_n$$.