An application of the Borel-Cantelli lemma

Question

Problem. Consider the independent standard Cauchy variables $ X_1, X_2, X_3, \ldots $, i.e. their probability density function is of form $f(x) = \dfrac{C}{1+x^2}$ (for some $C\in\mathbb{R}$). If $ Z = \limsup\limits_{n\to \infty} \dfrac{\ln |X_n|}{\ln n} $, then show that $Z$ is constant (eventually $\infty$) almost surely. Calculate the value of $Z$!

I strongly believe that I should use the Borel-Cantelli lemma to prove this.

By definition, $$\limsup_{n \to \infty} \dfrac{\ln |X_n|}{\ln n} = \lim_{n \to \infty} \sup_{m \geq n} \dfrac{\ln |X_m|}{\ln m}. $$

So I thought about looking at the event $\left\{ \dfrac{\ln |X_m|}{\ln m} < c \right\}$ for some constant $c\in \mathbb{R}$. It is easy to find out that the left hand side is less then $c$ if and only if $|X_m| < m^c $. Now, $$ \mathbb{P}( |X_m| < m^c ) = \int\limits_{-m^c}^{m^c} f_{X_m}(x) \, dx = C_m\left(\arctan(m^c) - \arctan(-m^c)\right) = 2C_m \arctan(m^c) ,$$ which means that for $c>0$, we have $$ \sum_{m=1}^\infty \mathbb{P}( |X_m| < m^c ) = \infty,$$ since $\lim\limits_{m \to \infty} \arctan(m^c) = \dfrac{\pi}{2} \neq 0 $. So now we can use the (second) Borel-Cantelli lemma, which states that in our situation, $$ \mathbb{P} \left(\limsup_{m \to \infty} \{ \omega \in \Omega \colon |X_m(\omega)| < m^c \}\right) = 1 $$ or with other words, the event $\{|X_m| < m^c\}$ occurs infinitely often. Now I should conclude something on $ Z $, but I'm not sure if what I've done helps...

Any advice would be appreciated!

Answer 1

No need to apply Kolmogorov's $0$-$1$ law. As you computed, for every $c>0$ we have $$\mathbb P(|X_n|\ge n^c)=1-\frac2\pi\,\arctan(n^c)\sim\frac1{n^c}$$ as $n\to\infty$. In particular, $$\sum_{n\ge1}\mathbb P(|X_n|\ge n)=\infty\qquad\text{and}\qquad \forall c>1,\enspace\sum_{n\ge1}\mathbb P(|X_n|\ge n^c)<\infty.$$ By the Borel-Cantelli lemmas, this implies that $$\limsup_{n\to\infty}\:\frac{\ln{|X_n|}}{\ln n}\ge1,\enspace \text{a.s.,}\qquad\text{and}\qquad\forall c>1,\enspace\limsup_{n\to\infty}\:\frac{\ln{|X_n|}}{\ln n}\le c,\enspace\text{a.s.}$$ As we can take $c\downarrow1$ along a countable set we easily conclude that $$\limsup_{n\to\infty}\:\frac{\ln{|X_n|}}{\ln n}=1,\enspace\text{a.s.}$$

Answer 2

That $Z$ is a constant (a.e.) random variable follows from the Kolmogorov Zero-One Law. Let $\mathcal{F}_{n}=\sigma(X_{n},X_{n+1},\ldots)$ and $\mathcal{F}=\cap_{n}\mathcal{F}_{n}$. $\mathcal{F}$ is known as the tail $\sigma$-algebra. Kolmogorov Zero-One Law says that for any $A\in\mathcal{F}$, $P(A)=0$ or $P(A)=1$. Therefore, if we can prove that $Z$ is $\mathcal{F}$-measurable, then $Z$ must be constant a.e..

To prove that $Z$ is $\mathcal{F}$-measurable: Let $n\in\mathbb{N}$ be arbitrary. For each $m$, let $Y_{m}=\sup_{k\geq m}\frac{\ln|X_{k}|}{\ln k}$. Note that $Y_{1}\geq Y_{2}\geq\ldots$. Therefore, we have that \begin{eqnarray*} Z & = & \inf_{m\geq1}Y_{m}\\ & = & \inf_{m\geq n}Y_{m}. \end{eqnarray*} Observe that for each $m$, $Y_{m}$ is $\mathcal{F}_{m}$-measurable. Hence, $\inf_{m\geq n}Y_{m}$ is $\mathcal{F}_{n}$-measurable because $\mathcal{F}_{n}\supseteq\mathcal{F}_{m}$ for any $m\geq n$. Since $n\in\mathbb{N}$ is arbitrary, it follows that $Z$ is $\mathcal{F}$-measurable.