Let $(X_{n})$ be a sequence of nonnegative uniformly integrable random variables. Is it true that $\limsup X_{n} \in L_{1}$? Thanks!
limit superior of uniformly integrable random variables is integrable?
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probability-theory
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0@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
1 Answers
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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.