# Characterization of degenerate scaling

This post is a follow-up to my post on 12/29/2023, where I posed the following question:

Let $(X_n)$ be an i.i.d. sequence of $\mathbb{R}$-valued random variables. Let $(a_n)$ be a positive sequence such that $\lim a_n=\infty.$ Under what condition of $(a_n)$ can one conclude that $\lim\frac{X_n}{a_n}=0$ almost surely?

I have found a full characterization of the sequence $(a_n).$ After all, it is an interesting exercise.

Proposition: Let $(X_n)$ be an i.i.d. sequence of $\mathbb{R}$-valued random variables Let $(a_n)$ be a positive sequence such that $\lim a_n=\infty.$ If $\sum P\left(|X_1|>\frac{a_n}{c}\right)<\infty$ for any constant $c>0,$ then $\lim\frac{X_n}{a_n}=0$ almost surely. Otherwise, almost surely $\frac{X_n}{a_n}$ does not converge.

As an application, if $(X_n)$ is an i.i.d. sequence of mean-one exponentially distributed random variables, then almost surely the sequence of $\frac{X_n}{\ln n}$ does not converge.

Proof: First, suppose that there exists $c>0$ such that $\sum P\left(|X_1|>\frac{a_n}{c}\right)=\infty.$ Then the independent events $E_n=[|X_n|>\frac{a_n}{c}]$ satisfies

$\displaystyle \sum \mathbb{P}(E_n)=\infty.$

By Borel-Cantelli’s Lemma, almost surely the events $E_n$ occurs for infinitely many $n.$ Therefore, with probability 1, the sequence $(X_n/a_n)$ does not converge to 0. Now fix $b\neq 0.$ We show that almost surely $X_n/a_n\not\to b$. Consider the independent events $E'_n=\left[|X_n|<\frac{|b|a_n}{2}\right].$ One has

$\displaystyle \sum\mathbb{P}(E'_n)=\infty$

By Borel-Cantelli’s Lemma, almost surely the events $E'_n$ occurs for infinitely many $n.$ Therefore, with probability 1, the sequence $(X_n/a_n)$ does not converge to $b.$

Next, suppose that for every $c>0$, one has $\sum P\left(|X_1|>\frac{a_n}{c}\right)<\infty.$ For a fixed constant $c>0$, the event $A_c=\left[\limsup\frac{|X_n|}{a_n}\le \frac{1}{c}\right]$ is a tail event. Thus, $\mathbb{P}(A_c)=0$ or $\mathbb{P}(A_c)=1$ according to Kolmogorov’s 0-1 Law. Note that $\cap_{n=1}^\infty \left[|X_n|\le \frac{a_n}{c}\right]\subset A_c$ and

$\displaystyle \mathbb{P}\left(\bigcap\limits_{n=1}^\infty \left[|X_n|\le \frac{a_n}{c}\right]\right)=\prod_{n=1}^\infty \left(1-P\left(|X_1|>\frac{a_n}{c}\right)\right)$

which is positive because $\sum P\left(|X_1|>\frac{a_n}{c}\right)<\infty.$ Thus, $\mathbb{P}(A_c)>0$ and therefore, $\mathbb{P}(A_c)=1.$ Now note that

$\displaystyle \mathbb{P}\left(\lim\frac{X_n}{a_n}=0\right)= \mathbb{P}\left(\limsup\frac{|X_n|}{a_n}=0\right)=\lim_{k\to\infty} \mathbb{P}(A_{k})=1.$

This completes the proof.