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Let $(X_{n})$ be a sequence of nonnegative uniformly integrable random variables. Is it true that $\limsup X_{n} \in L_{1}$? Thanks!

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    First, the question should be rephrased to avoid giving orders. Second, the result is false.2011-03-12
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    Can you give a counterexample?2011-03-12
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    Yes, I can. But did you read the first part of my comment?2011-03-12
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    Now that you modified your question, let us turn to the second part of my comment, saying that the result is false. Hint 1: any i.i.d. sequence of integrable random variables is uniformly integrable. Hint 2: what is the limsup of such a sequence?2011-03-12
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    what is the limsup of such a sequence?2011-03-13
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    @Didier, I think you need another condition on the sequence to get your hint to pan out. But, maybe that was intentionally left out.2011-03-13
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    @cardinal Such as *distribution of unbounded support*, maybe? And, yes that was intentional... :-)2011-03-13
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    @user7762 Consider for example $(X_n)$ i.i.d. exponential with parameter $1$ and $Y=\limsup X_n$. Can you write the event $[Y\le x]$ thanks to events like $[X_n\le x]$? You could then evaluate $P(Y\le x)$--and this would be a big step towards an answer to the question you asked.2011-03-13
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    @Didier, I figured. For a sequence of i.i.d random variables with arbitrary distribution $F$ of unbounded support, my first thought was to use the second Borel--Cantelli lemma. Is there an "easier" way? (That's pretty easy, I think.)2011-03-13
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    @cardinal (Sorry I missed your last comment.) BC2 is allright to me here. The alternatives I know more or less amount to rediscovering BC2 in a simple case, so they are not really worth the trouble.2011-03-25

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Take $\{X_n\}$ a sequence of i.i.d. , with a density of unbounded support, and integrable.

  • The supremum in the definition of uniform integrability $\sup_{n\in\Bbb N}\int_{\{|X_n|\geq R}|X_n|dP$ is actually $\int_{\{|X_1|\geq R\}}|X_1|dP$ which converges to $0$ when $R\to +\infty$, using monotone convergence for example.
  • Let $j$ a fixed integer, and $E_{j,n}:=\{X_n\geq j\}$. Since the series $\sum_{n\geq 1}P(E_{j,n})$ is divergent and the events $\{E_{j,n}\}$ are independent, by Borel-Cantelli lemma, $P(\limsup_n E_{j,n})=1$. Hence for almost all $\omega$, there exist $S_{\omega,j}\subset \Bbb N$ infinite such that for each element $n$ of this set $X_n(\omega)\geq j$. Hence $\limsup_{n\to +\infty}X_n(\omega)\geq j$. Since it's true for almost all $\omega$, we conclude that $\limsup_{n\to +\infty}X_n$ is not integrable.