Let $(a_n)$ be a sequence of positive real numbers such that $\sum_{n \geqslant 1} a_n < \infty $ and $$\sum_{n \geqslant 1} \Pr\left(\left|X_{n+1} - X_n\right| > a_n\right) < \infty $$

Why does $(X_n)$ converge almost surely?

  • 4
    $\begingroup$ Use the Borel-Cantelli lemma. $\endgroup$ – Ewan Delanoy Mar 20 '13 at 16:36
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    $\begingroup$ This is not stochastic-analysis. Corrected the tags. $\endgroup$ – Did Mar 20 '13 at 18:18
  • $\begingroup$ But the Borel-Cantelli Lemma doesn't say anything about the almost-surely convergency, right? $\endgroup$ – JohnD Mar 20 '13 at 20:38

As in the comment: Since $\sum_n {\Bbb P}(|X_{n+1} - X_n| > a_n)<\infty$, the Borel-Cantelli lemma implies that, with probability $1$, there are only finitely many $n$s such that $|X_{n+1}-X_n|>a_n$. Since $\sum_n a_n$ is finite, this means that $\sum_n |X_{n+1}-X_n|$ is finite with probability $1$. So, $\lim_{N\to\infty} \sum_{n\ge N} |X_n-X_{n+1}|=0$ a.s., which, by the triangle inequality, means that $(X_n)$ is a.s. Cauchy. Therefore, $(X_n)$ almost surely converges.


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