# Almost sure convergence by Borel-Cantelli

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?

• 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.