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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 !

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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.

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  • $\begingroup$ 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

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