I have recently been working through problems on the convergence of random variables. In the following exercise, I believe I have shown that the following random variable does not converge almost surely. However, the question asks me to show that it does. I am unsure about what the mistake is in my process and would be grateful for any help.
Consider the random variable $X$ following the continuous uniform distribution on $(-1,1)$, and define the sequence of random variables $Y_n= \frac{X}{n}$. Show that $Y_n$ converges almost surely.
I have been able to show that $Y_n$ converges in distribution and converges in probability to $Y=0$ so far.
My attempt to show almost sure convergence was to show the following:
$$\sum_{n=1}^{\infty}P(|Y_n-0| \ge\epsilon) < \infty$$
I believe that this should be sufficient to prove almost sure convergence.
I have computed the CDF of $Y_n$ as $F_{Y_n}(y)= \frac{ny+1}{2}$ for $y \in (-\frac{1}{n}, \frac{1}{n})$
Therefore, if I have computed this correctly, the sum should simplify (for $\epsilon \in (-\frac{1}{n}, \frac{1}{n})$), we should have:
$$\sum_{n=1}^{\infty} n \epsilon $$
However, this appears to diverge, as if we try to bound the sum by replacing $\epsilon$ with $\frac{1}{n}$, then we end up with the infinite sum of a constant – which clearly diverges rather than converges (as we need to show).
I would be grateful for any help resolving my problem.
Best Answer
We shall weaken the assumptions and prove $P(|Y_n|>\varepsilon\textrm{ i.o.})=0$ with the Borel-Cantelli lemma.
Let $|X|\leq M$ a.s., that is $P(|X|\leq M)=1$. Then $P(|Y_n|>\varepsilon\textrm{ i.o.})=0$. To see this: $$P(|Y_n|> \varepsilon)=P(|X|> n\varepsilon)=1-P(-n\varepsilon \leq X\leq n\varepsilon)$$ For any $\varepsilon>0$, there exists $N_\varepsilon$ s.t. $\forall n > N_\varepsilon$, $P(-n\varepsilon \leq X\leq n\varepsilon)=1$. Indeed, just choose $N_\varepsilon=M/\varepsilon$. Thus $\sum_{n}P(|Y_n|>\varepsilon)=\sum_{n\leq N_\varepsilon}P(|Y_n|>\varepsilon)<\infty$. By Borel-Cantelli, $P(|Y_n|>\varepsilon\textrm{ i.o.})=0$.