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Suppose we have $X_n$ a random variable, that can take two values:

$X_n = \begin{cases} 0, & \text{with probability 1 - $\frac{1}{2n}$,} \\ n, & \text{with probability $\frac{1}{2n}$} \end{cases}$

Does $X_n$ converge almost sure to $0$? I don't need a rigorous proof, but I would like to have an intuition for why it doesn't converge almost surely to 0.

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Just use the contrapositive Borel Cantelli Lemma, which says that if your events $A_i$ are independent and $\sum_i P(A_i)=\infty$ then $A_i$ occur infinitely often. In this case, pick $A_i=\{X_i>1\}$. Can you finish it from here?

For "intuition," $X_n$ grows too fast with high probability. If you want, modify $P(X_n=n)=\frac{1}{2n}$ to $P(X_n)=\frac{1}{2n^2}$ [and modify $P(X_n=0)=1-P(X_n)$].. Now apply the usual Borel Cantelli to conclude in this case that $X_n$ converges to $0$ almost surely.

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  • $\begingroup$ Sorry, i'm not familiar with Borel Cantelli, i'm kind of new in Probability Theory. $\endgroup$
    – S. Cow
    May 19, 2016 at 20:24
  • $\begingroup$ The thing is that i want to show that this does not converges to 0 almost surely. $\endgroup$
    – S. Cow
    May 19, 2016 at 20:26
  • $\begingroup$ @S.Cow: How did you happen upon this question then? Borel Cantelli is pretty much the only tool we have for showing something converges almost surely. $\endgroup$
    – Alex R.
    May 19, 2016 at 20:26
  • $\begingroup$ @S.Cow: Yes, that's why you need the contrapositive Borel Cantelli Lemma: en.wikipedia.org/wiki/… $\endgroup$
    – Alex R.
    May 19, 2016 at 20:27
  • $\begingroup$ @AlexR . It might help if you separate out your growth suggestion from your response. I think S.Cow is getting confusing your suggestion with your response in your answer. $\endgroup$ May 19, 2016 at 21:05
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You can't prove it because $X_n$ could either converge to zero almost surely or not. You haven't provided enough information.

As has already been pointed out if the $X_n$ are independent then the fact that $X_n = n$ infinitely often with probability one follows from the second Borel-Cantelli lemma. However, we could let $U \sim$ uniform$(0, 1)$ and set $X_n = n$ if $U < 1 / 2n$. Then our sequence satisfies the conditions of the problem and $X_n \to 0$ almost surely.

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  • $\begingroup$ I asked a question related to this answer here. $\endgroup$ May 16, 2020 at 15:18
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Can i just pick for example a $n=4$ and then i write:

for $i \ge 4$ we have:

$X_i = \begin{cases} 0, & \text{with probability $\frac{7}{8}$,} \\ 4, & \text{with probability $\frac{1}{8}$} \end{cases}$

and then obviusly $\lim_{n \rightarrow \infty} Prob( \lvert x_i\rvert > \varepsilon, \forall i \ge 4) \neq 0 $ ??

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  • $\begingroup$ I don't think this proves it though. How is it obvious that in the limit the probability doesn't approach 0 when $1/(2n)$ does approach 0 as $n$ approaches infinity? $\endgroup$ May 19, 2016 at 21:02

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