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) $N\ge 1$ $$ P( X_n/c_{n}<1+\epsilon, \, \forall n \ge N)<1. $$

  • $\begingroup$ 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? $\endgroup$ – Jack London Oct 3 at 15:01
  • $\begingroup$ I would say it's not the case almost surely :) $\endgroup$ – zhoraster Oct 3 at 18:55

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