# An application of the Borel-Cantelli lemma

borel-cantelli-lemmasprobability distributionsprobability theoryrandom variables

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…

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 $$\limsup_{n\to\infty}\:\frac{\ln{|X_n|}}{\ln n}=1,\enspace\text{a.s.}$$