# Applications of Borel Cantelli lemmas and almost sure bounds for a sequence of random variables

Consider a sequence of probability measure $$(P_{\theta,n})_{n=1}^\infty$$ on $$\mathbb{R}$$, assume that $$X_{n}$$ is distributed according to $$P_{\theta,n}$$ and let $$c_{\theta,n}$$ be a diverging sequence of constants. Here $$\theta$$ can be thought of as a parameter. If $$P_{\theta,n}(X_n/c_{\theta,n}>1)\lesssim n^{-\delta}$$ for $$\delta>1$$, then, denoting by $$P_{\theta,\infty}$$ the law of the sequence $$(X_n)_{n=1}^\infty$$, by Borel-Cantelli lemma $$P_{\theta,\infty}( X_n/c_{\theta,n}>1, \, \text{i.o.})=0$$ where $$\text{i.o.}$$ stands for infinitely often. First question: can we then conclude that for every $$\epsilon>0$$ there exists $$n_{\epsilon, \theta}$$ such that $$P_{\theta,\infty}( X_n/c_{\theta,n}<1+\epsilon, \, \forall n \geq n_{\epsilon,\theta})=1?$$

Assume next that the parameter satisfies $$\theta \in \Theta$$ and that, in fact, $$\sup_{\theta \in \Theta }P_{\theta,n}(X_n/c_{\theta,n}>1)\leq \kappa n^{-\delta}$$ for dome $$\kappa>0$$. Second question: can we then conclude that for every $$\epsilon>0$$ there exists $$n_\epsilon$$ such that $$\inf_{\theta \in \Theta }P_{\theta,\infty}( X_n/c_{\theta,n}<1+\epsilon, \, \forall n \geq n_\epsilon)=1?$$

The answer to the first question is already negative, so $$\theta$$ is irrelevant. The Borel-Cantelli lemma grants a random $$N$$ such that $$P( X_n/c_{\theta,n}\le 1 \, \forall n \geq N)=1.$$ However, this does not yield a non-random $$N$$, even in an "$$\epsilon$$-version".
Precisely, consider the infinite (fair) coin-flipping space, and let $$X_n=n$$ if the number of the first head is $$n$$ and $$0$$ otherwise; $$c_{n}=\sqrt{n}$$. Then, for $$n\ge 2$$, $$P(X_n/c_{n} >1) = P(X_n=n)= 2^{-n}.$$ However, for any $$\epsilon>0$$ and any (fixed, non-random) $$N\ge 1$$ $$P( X_n/c_{n}<1+\epsilon, \, \forall n \ge N)<1.$$
• Yes, in fact this was my doubt. But I was wondering: are there situations in which a random $N$ can be in fact substituted by a large but deterministic $n$ or, in practice, this is never the case? – Jack London Oct 3 at 15:01