Assume that I have two sequences $(X_n)_{n \in \mathbb{N}}$ and $(Y_n)_{n \in \mathbb{N}}$ of random variables. Further assume $X_n \overset{\mathcal{D}}{=} Y_n$ and that $$\tag{$*$}\frac{X_n}{n} \to 0 \,almost\,surely.$$

Then, in general, we could not infer that $\frac{Y_n}{n} \to 0 \,almost\,surely$. But if we have derived $(*)$ by the Borel-Cantelli lemma, i.e. we have shown that for any $\varepsilon > 0$ $$\sum_{n=1}^\infty \mathbb{P}(\vert \frac{X_n}{n} \vert > \varepsilon) < \infty,$$

then we could infer that also $\frac{Y_n}{n} \to 0 \,almost\,surely$ holds true, since the above equation would also hold for $Y_n$. Is this approach right?


1 Answer 1

  • If you only assume that $X_n \overset{d}{=} Y_n$ for every $n$, then $X_n/n \to 0$ almost surely does not imply that $Y_n/n \to 0$ almost surely. For example take $X_n = n\, \mathbf{1}_{\{U\leq 1/n\}}$ where $U$ is uniformly distributed on $[0,1]$. Then it should be clear that $X_n/n \to 0$ a.s. Now, if you take $Y_n$ a sequence of independent random variables such that $\mathbb{P}(Y_n = n) = 1/n = 1-\mathbb{P}(Y_n =0)$ then $X_n \overset{d}{=} Y_n$ but $Y_n/n$ does not converge to $0$ a.s. since $$\sum_{n} \mathbb{P}(|Y_n|/n>\epsilon)= \sum_{n} 1/n = +\infty.$$

  • The best you can hope for in general is convergence in probability since $$\mathbb{P}(|Y_n|/n >\epsilon) = \mathbb{P}(|X_n|/n >\epsilon) \to 0.$$

  • If you have the stronger assumption that the whole sequence $(X_n, \, n \geq 0)$ is distributed as the sequence $(Y_n, \, n \geq 0)$ then a.s. convergence for one implies the same for the other.

  • To answer your question, if a.s. convergence for $X_n$ is obtained through Borel-Cantelli as you mentioned then it holds also for $Y_n$ and your argument is absolutely correct.


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