Let $(X_n)$ be a sequence of random variables. I want to show that if $E[X_n] \rightarrow C$ and $Var(X_n) \leq \frac{C}{n^2}$, where $C$ is some constant, then $X_n$ converge almost surely to $C$.

I use Borel-Cantelli Lemma, i.e. $$P\left[|X_n - E[X_n]| > \frac{1}{\sqrt[4]{n}}\right] \leq \sqrt{n} Var(X_n) \leq \frac{C}{n^{\frac{3}{2}}}$$ Hence by Borel-Cantelli lemma $$P\left[|X_n-E[X_n]| > \frac{1}{\sqrt[4]{n}} i.o.\right] = 0$$ On the other hand, $|E[X_n]-C| < \frac{1}{\sqrt[4]{n}}$ for some large $n$. And my claim will follow?

  • $\begingroup$ You mean... "then $X_n\to E(X)$ almost surely", not $X_n\to X$ almost surely which would be absurd? $\endgroup$
    – Did
    Nov 25, 2014 at 10:45
  • $\begingroup$ Yes, I edited it, instead of X, I have a constant C now. $\endgroup$
    – Meryl
    Nov 25, 2014 at 10:52
  • $\begingroup$ Borelwith only one l. $\endgroup$
    – Did
    Nov 26, 2014 at 14:28

1 Answer 1


Your proof is correct. More generally:

If $E(X_n)\to C$ and the series $\sum\limits_n\mathrm{var}(X_n)$ converges, then $X_n\to C$ almost surely.

Hint: There exists some positive sequence $(\alpha_n)$ such that $\alpha_n\to0$ and $\sum\limits_n\frac1{\alpha_n^2}\mathrm{var}(X_n)$ converges. Consider the events $A_n=[|X_n-E(X_n)|\geqslant\alpha_n]$.

  • $\begingroup$ Is there a name for the “more general” result that you state? Or is it just a consequence of Borel-Cantelli? Also, if we want to show that $X_n \to X$ a.s., is that the same as showing $(X_n - X) \to 0$ a.s.? Or is that not equivalent for some reason? $\endgroup$
    – Joe
    Apr 17, 2021 at 15:31

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