Good morning everyone, I am new to this website so I hope everything is going to be ok.

Here is a question for my homework :

Let $(X_n)_{n \geq 1}$ be a sequence of independent random variables. For every $k$ greater than or equal to $1$, $X_k$ has a density $f_X(t) = \frac{1}{2} \exp(-|t|) $ with respect to the Lebesgue measure on $\mathbb{R}$. Show that : $$S_n = \sum_{k=1}^n \frac{X_k}{k^2}$$ converges almost surely. We can calculate $\mathbb{P}(|X_k| \geq 2 \log(k))$ for every $k \geq 1$.

I tried to use Borel-Cantelli lemmas but I don't know the limit of this series. I also tried to use the law of large numbers, without success. All I did is :

$$\mathbb{P}(|X_k| \geq 2 \log(k)) = \mathbb{P}(|X_1| \geq 2 \log(k)) = \frac{1}{k^2}$$

Hence :

$$S_n = \sum_{k=1}^n \frac{X_k}{k^2} = \sum_{k=1}^n X_k \cdot \mathbb{P}(|X_1| \geq 2 \log(k))$$

What can I do now ? Thank you ! (PS : I'm not english so I apologize for the mistakes...).

  • 1
    $\begingroup$ Your first question is excellent, keep up the same template and you will get attention on your questions as well as good answers. $\endgroup$ Apr 18, 2020 at 13:34
  • $\begingroup$ Oh great, thank you ! $\endgroup$
    – NaNoS
    Apr 18, 2020 at 13:43
  • 2
    $\begingroup$ You can get the proper font and spacing for $\exp$ and $\log$ using \exp and \log. For operators that don't have a command of their own, you can use \operatorname{name}. $\endgroup$
    – joriki
    Apr 18, 2020 at 13:49
  • $\begingroup$ Thanks for the tip, that’s really kind ! For the problem, I think I found something. If we use Borel-Cantelli, we can see that there is almost surely a finite set of indices such that $|X_k| \geq 2 \log(k)$. We can say now that the series converges (absolutely) almost surely by comparison. Can someone confirm ? $\endgroup$
    – NaNoS
    Apr 18, 2020 at 13:51

2 Answers 2


$$ \mathbb{P}(|X_k|\geqslant 2\log k)=\int_{2\log k}^{+\infty}e^{-t}dt=\frac{1}{k^2} $$ Thus the series $\sum\mathbb{P}(|X_k|\geqslant 2\log k)$ converges, by Borel-Cantelli lemma we have $$ \mathbb{P}\left(\bigcap_{n\geqslant 1}\bigcup_{k\geqslant n}\{|X_k|\geqslant 2\log k\}\right)=0 $$ Because of what said above, there exists a finite number of $k$ such that $|X_k|\geqslant 2\log k$ almost surely. Thus for $k\gg1$, $|X_k|<2\log k$ almost surely so that $$ \frac{X_k}{k^2}=\mathcal{O}\left(\frac{\log k} {k^2}\right)=\mathcal{O}\left(\frac{1}{k^{3/2}}\right) $$ and thus the series $\sum\frac{X_k}{k^2}$ converges almost surely.

  • 1
    $\begingroup$ Wow, thank you a lot ! What a great forum. Have a nice day ;) $\endgroup$
    – NaNoS
    Apr 18, 2020 at 14:17

The a.s convergence of $S_n$ is implied by the a.s convergence of $$\sum_{k = 1}^\infty \frac{|X_k|}{k^2}$$ But $$E \left( \sum_{k = 1}^\infty \frac{|X_k|}{k^2}\right) = \sum_{k = 1}^\infty \frac{E(|X_k|)}{k^2} = \frac{\pi^2}{6}E(|X_1|) < \infty$$ so the result follows from the fact that a random variable with finite expectation is a.s. finite.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .