I'm studying convergence in my probability class and I'm asked to show if there exists any convergence for the following sequence of random variables:

$$\left\{\frac{W_n}{ln(n)}\right\}_{n\geq1} \ s.t. W_n\sim exp(1)$$

I have been able to show that this sequence converges to $0$ in probability by Markov inequality, but I'm struggling to prove if there is almost sure convergence to $0$ in this case. I know I'm assumed fo use Borel Cantelli lemma, and my specific doubt is if it is right to assume that the right set for which to apply the lemma is:


If so, I guess $\sum P(A_n)=0$ because $W_n$ has probability zero of assuming a point as it is a continuous distribution. I thought this is right but I'm only suspicious I did something wrong with this logic because $\lim \frac{W_n}{ln(n)}$ could become a different random variable that has probability 1 of assuming the value 0, but I dont know how to show that if it is the case.

If anyone could help me with this, I'd be very grateful.

  • $\begingroup$ If the $X_n$ are independent you want the Borel-Cantelli lemma: en.wikipedia.org/wiki/… $\endgroup$ Commented Jun 23, 2020 at 3:44
  • $\begingroup$ Okay, I understand why I should use the Borel Cantelli lemma, I'll edit to be more specific $\endgroup$ Commented Jun 23, 2020 at 3:46
  • $\begingroup$ @ThomasLumley Now I explained my doubts with a little bit more details $\endgroup$ Commented Jun 23, 2020 at 3:55

1 Answer 1


Fix $\epsilon>0$ and consider the event $A_n = \{X_n > \epsilon\log n\}$. If we write $Z_n$ for your sequence $X_n/\log n$, then $A_n$ is the event $\{Z_n>\epsilon\}$.

So when $P(\textrm{$A_n$ infinitely often})=0$ for all $\epsilon>0$ you have $$Z_n\stackrel{a.s.}{\to} 0$$ Conversely, if there is an $\epsilon>0$ such that $P(\textrm{$A_n$ infinitely often})\neq 0$ then you don't have almost sure convergence.

And that's where you can use the Borel-Cantelli Lemma, in either direction


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