Almost sure convergence is usually proved by Borel-Cantelli lemma. If the condition of Borel-Cantelli lemma doesn't hold, does the almost sure convergence still hold? If so, how to construct a sequence of identically distributed random variables $\{X_n,n\geq1\}$ and a sequence of real numbers $0\leq a_n\rightarrow\infty$ such that $$\sum_{n=1}P[|X_n|>a_n]=\infty,\mbox{ and } \ X_n/a_n\rightarrow 0\mbox{ a.s.} $$
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Just consider $X_n=X$ where $X$ is not integrable. Then for any sequence $\left(a_n\right)_{n\geqslant 1}$ such that $a_n\to +\infty$, $X_n/a_n=X/a_n$ goes to $0$ as $n$ goes to infinity. Now, for the choice $a_n=n$ or something which growth slower, we have $\sum_{n=1}^{+\infty}\Pr\left\{X_n> a_n\right\}= +\infty$.
answered Nov 4, 2016 at 12:13
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$\begingroup$ For any non-integrable random variable $X$ and sequence $(a_n)_{n\geq1}$ goes to infiity, why is $X/a_n$ goes to 0 as $n$ goes to infinity? $\endgroup$– SHANNov 4, 2016 at 12:37
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$\begingroup$ If you fix $\omega$, the sequence $\left(X(\omega) /a_n\right)_n$ goes to $0$ as $n$ goes to infinity. $\endgroup$ Nov 4, 2016 at 16:13
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$\begingroup$ why is it important for $X$ to not be integrable? $\endgroup$– AlexJul 6, 2020 at 12:08