# limit of random variables, expectation

• Mar 25th 2011, 07:27 AM
Juju
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?

• Mar 26th 2011, 08:33 AM
theodds
• Mar 26th 2011, 08:40 AM
Juju
OK. Thanks!
• Mar 26th 2011, 09:19 AM
theodds
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 $\limsup \mathbb E X_h \le \mathbb E X \le \liminf \mathbb E X_h$).
• Mar 26th 2011, 09:23 AM
theodds
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 :)
• Mar 26th 2011, 09:28 AM
Juju
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?
• Mar 26th 2011, 10:47 AM
theodds
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.
• Mar 26th 2011, 10:52 AM
Juju
This question is making me very nervous.
But thank you so much for helping me,

Julia