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I've seen several examples when a sequence of r.v. converge in probability but not almost surely, yet none of them had the sequence to be independent. Would additional conditions of independence and convergence to a constant be sufficient to ensure almost surely convergence?

Leo
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  • What do you mean "converge in probability" ? I am curious about it – Cardinal Jul 29 '15 at 22:32
  • @Cardinal: https://en.wikipedia.org/wiki/Convergence_of_random_variables#Convergence_in_probability – Leo Jul 29 '15 at 22:33
  • Do you mean that each element of the sequence be independent, or it is a series of independent terms, which is converging? – gt6989b Jul 29 '15 at 22:52
  • @gt6989b: I mean it like the first "i" in i.i.d. – Leo Jul 29 '15 at 22:55
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    A related fact that may interest you: if a series $\sum_n X_n$ of independent random variables $X_n$ converges in probability, then it converges almost surely. See http://math.stackexchange.com/questions/90116/how-to-show-convergence-in-probability-imply-convergence-a-s-in-this-case – Nate Eldredge Jul 29 '15 at 23:54

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Let $A_n$ be independent events with $\mathbb{P}(A_n)=1/n$, and define $X_n=1_{A_n}$. Then $X_n\to0$ in probability, but $X_n$ does not converge almost everywhere.

Apply the second Borel-Cantelli lemma twice; once to the sequence $A_n$ and also to the sequence $A_n^c$, to conclude that
$$P([X_n=1\mbox{ infinitely often}] \cap [X_n=0\mbox{ infinitely often}])=1.$$

  • Could you give an explicit example if such a sequence $(A_n)$? – Leo Jul 30 '15 at 00:02
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    @Leo If $(U_n)$ are i.i.d. uniform(0,1) random variables, then set $A_n=(U_n <1/n)$. –  Jul 30 '15 at 00:41