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Prove that if random variables $X_n$ are dominated by an integrable random variable then $E[X_n] \to E[X]$ follows if $X_n$ converges to $X$ in probability.

Hint: Use the following theorem :

A necessary and sufficient condition for $X_n \to _p X$ is that each subsequence $\{X_{n_k}\}$ contain a further subsequence $\{X_{n_{k_j}}\}$ such that $\{X_{n_{k_j}}\}\to X$ with probability 1 as $j, n \to \infty$

The problem is that I have a.s. convergence for a subsequence one level lower for which I have to prove convergence in mean. So, stuck.

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aaaaaa
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1 Answers1

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The key point is the following lemma which is known as subsequence principle:

Lemma Let $(a_n)_{n \in \mathbb{N}}$ be a sequence. If for any subsequence $(a_{n'})_n$ of $(a_n)_n$ there exists a convergent subsequence $(a_{n(k)'})_k$ and the limit does not depend on the subsequence, then $(a_n)_n$ is convergent.

See this question for a proof.

Now let $a_n := \mathbb{E}(X_n)$ and $(a_{n'})_n$ an arbitrary subsequence. By the mentioned theorem, we can choose a subsequence $(X_{n(k)'})_k$ such that $X_{n(k)'} \to X$ almost surely. From the dominated convergence theorem, we see that $$a_{n(k)'} = \mathbb{E}(X_{n(k)'}) \to \mathbb{E}X.$$ Applying the lemma finishes the proof.

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saz
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    @aaaaaa No, the point is that any subsequence (of the original sequence) has a convergent subsequence - as stated in the lemma. And this implies convergence of the "original" sequence. – saz Apr 14 '14 at 07:50
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    @aaaaaa See this question: http://math.stackexchange.com/q/173590/. (And please do not delete your comments.) – saz Apr 14 '14 at 09:01
  • I understood the proof before you made the previous comment, that's why deleted. Thanks for the solution. – aaaaaa Apr 14 '14 at 09:05