Independent, Identically Distributed Random Variables

Question

Let $(X_n)_{n∈\mathbb{N}}$ a sequence of i.i.d. random variables uniformly distributed on the interval $[0, 1]$. Show that $$\limsup_{n\to+\infty} \frac{X_{2n}}{X_{2n+1}}=+\infty$$ a.s.

I tried something that first I thought it was correct, but late I realized that maybe it is not. I tried to think about the random variable $\frac{X_{2n}}{X_{2n+1}}$ and use Borel-Cantelli to prove that the probability of this fraction being limited is $0$. The problem was that I don't know if this is a random variable.

Answer 1

Let $$E_n = \left\{\frac{X_{2n}}{X_{2n+1}} > n\right\}. $$ Then $$\mathbb P(E_n) = \int_n^{\infty} \frac1{2z^2}\ \mathsf dz =\frac1{2n}. $$ Since the $E_n$ are independent and $$\sum_{n=1}^\infty \mathbb P(E_n) = \frac1{2n}=\infty, $$ so by the second Borel-Cantelli lemma we conclude that $$\mathbb P\left(\limsup_{n\to\infty} E_n\right)=1.$$ It follows that there is a subsequence $n_k$ such that $$\frac{X_{2n_k}}{X_{2n_k+1}}>n_k $$ and hence that $$\limsup_{n\to\infty}\frac{X_{2n_k}}{X_{2n_k+1}}=\infty. $$

Answer 2

You can indeed say that $Y_n=X_{2n}/X_{2n+1}$ is a random variable, since functions of random variables are also random variables. So if viewing $Y_n$ as random variables gives you a proof, that could be correct. Your approach sounds good to me.