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 !
Best Answer
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.