Consider a sequence of independant random variables such that $X_{n} \in\{0,1\}$ almost surely, $$ \mathbb{P}\left(X_{n}=1\right)=\frac{1}{n}, \quad \text { and } \quad \mathbb{P}\left(X_{n}=0\right)=1-\frac{1}{n} \quad \forall n \geq 1 $$ the sequence $X_{n}$ converges in probability towards $0,$ but not almost surely.

My attempt :

$\mathbb{P}(\lim_{k} X_k = 0) = \mathbb{P}(X_n = 0) < 1$

my reasoning is that only $0$ can converge point-wise to zero, $1$ can't.

is this method correct ?

also I was told that one can use the Borel-Cantelli lemma, how so ? I'm only used to proving a.s comvergence using that, not disproving it.

thanks !


A sequence of $0$'s and $1$'s only converges to $0$ if it is eventually ALWAYS $0$.

So, almost sure convergence to $0$ is equivalent to saying that almost surely, the sequence contains only finitely many $1$'s.

So, to show that the sequence does not almost surely converge to $0$, it is sufficient to show that there is a non-zero probability that the sequence contains infinitely many $1$'s. You can prove this using the second Borel-Cantelli lemma.

  • $\begingroup$ just making sure I get it, $\sum \mathbb{P}(X_n = 1) = +\infty$, therefore by the second Borel-Cantelli lemma, $\mathbb{P}(\limsup X_n = 1) = 1$, and so there is infinitely many one's. $\endgroup$ Feb 21 '20 at 18:33
  • $\begingroup$ also, can you please tell me what's wrong with my attempt ? $\endgroup$ Feb 21 '20 at 18:36
  • $\begingroup$ That's right, yes. As for your attempt: I can't quite make sense of it. Saying that $P(\lim_k X_k=0)=P(X_n=0)$ is not only false... it doesn't actually make sense. (For which $n$?) $\endgroup$ Feb 21 '20 at 19:13
  • $\begingroup$ I get it now, I was confusing between the indices and the $\omega$'s $\endgroup$ Feb 21 '20 at 19:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.