# Thread: limit of random variables, expectation

1. ## limit of random variables, expectation

Hallo!

Is the following equality true?
$\lim\limits_{h \to 0}\mathbb{E}[X_h]=\mathbb{E}[X]$

$X$ is a random variable $X>0$
$X_h$ is a sequence of random variables, $1>h>0, X_h>0$

with
$\mathbb{E}[X_h] \leq \mathbb{E}[X]$
$p\lim\limits_{h \to 0}X_h=X,$ where $p$ denotes convergence in probability

Now, I think it follows with Fatou's Lemma for non-negative random variables, that
$\liminf\limits_{h \to 0}\mathbb{E}[X_h]\geq \mathbb{E}[\liminf\limits_{h \to 0}X_h]$.

Can I now say, that
$\lim\limits_{h \to 0}\mathbb{E}[X_h]=\mathbb{E}[X]$
is true?

3. OK. Thanks!

4. The issue I have with your logic is the invocation of Fatou's Lemma. You should need almost sure convergence for that. Everything else goes through fine I think (if you have Fatou's Lemma then you can get $\limsup \mathbb E X_h \le \mathbb E X \le \liminf \mathbb E X_h$).

5. Okay, Fatou's Lemma looks fine. Apparently all you need is convergence in probability for that. Sorry for any confusion my ignorance may have caused

6. So, you think under the given assumptions
$\lim\limits_{h \to 0}EX_h=EX$ is correct?

But do you have any idea how to show it?

7. Okay, this question is starting to make me quite nervous. The first inequality I posted should come from the fact that $\mathbb E X_h \le \mathbb E X$ and the second inequality is the version of Fatou's Lemma I was able to find that concerned convergence in probability - that is, $\mathbb E X \le \liminf \mathbb E X_h$ provided we have convergence in probability, i.e. you can get rid of the liminf.

8. This question is making me very nervous.
But thank you so much for helping me,

Julia