# Convergence in probability implies Fatou's lemma?

Let $(\Omega, \mathcal{F},P)$ be a probability space and $(X_n)$ be a positive-valued sequence of random variables on $\Omega$. We assume that $(X_n)$ converges in probability to the random variable $X$. Is it true that $$E(X)\leq \liminf_{n\to+\infty}E(X_n)$$

Thanks a lot for the help in advance!

• Fatou's lemma does not require the convergence of the sequenze as $\liminf X_n$ converges a.s. to some r.v. Y. A fortiori $X=Y$ a.S.. – Kolmo Jan 29 '16 at 14:39

## 1 Answer

Yes, it's true. By the very definition of $\liminf$, there exists a subsequence such that

$$\liminf_{n \to \infty} \mathbb{E}(X_n ) = \lim_{k \to \infty} \mathbb{E}(X_{n_k}). \tag{1}$$

Since, by assumption, $X_{n} \to X$ in probability (hence in particular $X_{n_k} \to X$ in probability), we can choose a further subsequence of $(X_{n_k})_{k \in \mathbb{N}}$ which converges almost surely to $X$. Now the claim follows directly from Fatou's lemma and $(1)$.

• Great! Passing from convergence in probability to (a subsequential) a.e. convergence is a part I missed. Thanks a lot! – Nikolaos Skout Jan 29 '16 at 14:01
• I think in (1) you wanted to write somewhere liminf or not? And did you use the positivity of the $X_n$'s? Thank you. – Jimmy R. Jan 29 '16 at 14:07
• I am sure on (1) saz meant $\varlimsup$ instead of $\lim$ on the left side of the equality. By the way, the positivity of the $X_n$'s is used, when applying Fatou's lemma. Thanks for pointing them out Jimmy. – Nikolaos Skout Jan 29 '16 at 14:16
• @JimmyR. Yeah, exactly, thanks... – saz Jan 29 '16 at 15:24