probability | Tuan Pham

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.

There are well-known results about the scaling and centering laws of a sequence of $\mathbb{R}$ -valued random variables. That is the problem of finding the right coefficients $a_n$ and $b_n$ such that the sequence $\frac{X_n-b_n}{a_n}$ converges in law. See, for example, Section 14.8 of [1]. Normally, one tries to avoid the degenerate type, which is when the limit distribution is a Dirac mass at 0. I recently ran into a problem where the degenerate type is desirable. More specifically:

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?

This problem seems to be quite classic. If $X_1$ has a bounded range, i.e. $\mathbb{P}(X_1\in[a,b])=1$ for some numbers $a$ and $b$ , then no additional condition of $(a_n)$ is needed. How about the case $X_1$ has an unbounded range? This is essentially Problem 51 on page 263 of [1]. The textbook does not ask for a full characterization of the sequence $a_n$ . I find the problem quite interesting.

The first observation is that the event $A=\left[\lim\frac{X_n}{a_n}=0\right]$ is a tail event. By Kolmogorov’s 0-1 Law, either $\mathbb{P}(A)=0$ or $\mathbb{P}(A)=1$ . We just need to decide between the two. One may be tempted to think that $\frac{X_n}{a_n}$ is essentially the same as $\frac{X_1}{a_n}$ , so it should almost surely converge 0 as $n\to\infty$ . This argument is not correct because $\frac{X_n}{a_n}$ only has the same distribution as $\frac{X_1}{a_n}.$ If the range of values of $X_1$ is unbounded, then so is the range of values of $\frac{X_1}{a_n}$ . Let us consider two following cases, one of which gives $\mathbb{P}(A)=0$ and the other gives $\mathbb{P}(A)=1.$ Let us denote the cumulative distribution function of $|X_1|$ by $F(x).$

Case 1: there exists a positive sequence $(c_n)$ such that

$\displaystyle \lim\frac{c_n}{a_n}=0$ and $\displaystyle\sum(1-F(c_n))<\infty.$

Let $E_n=[|X_n|\le c_n]$ and $E=\cap_{n=1}^\infty E_n.$ Then $\mathbb{P}(E_n)=F(c_n)$ and

$\displaystyle \mathbb{P}(E)=\prod_{n=1}^\infty F(c_n)=\prod_{n=1}^\infty (1-(1-F(c_n))$

which has a finite positive value because $\sum(1-F(c_n))<\infty.$ It is easy to see that $E\subset A$ and thus $\mathbb{P}(A)=1.$

Case 2: there exists a positive sequence $(c_n)$ such that

$\displaystyle \lim\frac{c_n}{a_n}>0$ and $\displaystyle\sum(1-F(c_n))=\infty.$

Let $E_n=[|X_n|> c_n]$ and $E=\limsup E_n.$ The events $E_n$ are independent and

$\displaystyle \sum \mathbb{P}(E_n)=\sum(1-F(c_n))=\infty.$

By Borel-Cantelli’s Lemma, $\mathbb{P}(E)=1$ . It is easy to see that $E\subset A^c.$ Thus, $\mathbb{P}(A)=1-\mathbb{P}(A^c)=0.$

Final thoughts: Cases 1 and 2 do not cover all possibilities. One can see this through the simple case $F(x)=1-e^{-x}$ (the case when the random variables $X_n$ are i.i.d. exponentials of mean one) and $a_n=2\ln n.$ It would be interesting to see a full characterization of the sequence $(a_n)$ for which one has $\mathbb{P}(A)=1.$ I think this should be known somewhere in the literature.

Update 01/01/2024: I have resolved the problem.

References:

“A modern approach to Probability Theory” (1997) by Friestedt and Gray.

Tuan Pham

Department of Mathematics

Tag Archives: probability

Characterization of degenerate scaling

Degenerate scaling of random variables

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