Let $X_1,...X_n$ be a sequence of random variables such that $X_n=1 $ or $0$ and $P(X_1=1) \geq \alpha$ and $P(X_n=1|X_1,...X_{n-1}) \geq \alpha$ for $n=2,3,...$ where $\alpha >0$

I need to show $P(X_n=1$ infinitely often )$ =1$

Using Borel-Cantelli lemma,

I can show that $\sum P(X_n=1) =\infty $ because

$\sum P(X_n=1) = P(X_1=1) + P(X_2 =1| X1) + P(X_3|X_1,X2) + .... > \alpha + \alpha + .... =\infty $

But to apply the Borel-Cantelli lemma, the sequence should be independent . But in this case it is not.

Can anyone help me to figure out how to find an independent sub sequence ? Also is there any better approach than Borel-Cantelli lemma for this question?

  • $\begingroup$ I don't think Borel-Cantelli is helpful in this case. What if you compute the probability of never getting another 1 after index $n$? $\endgroup$
    – Michael
    Jul 8, 2018 at 5:16
  • $\begingroup$ @Michael . Did you mean to calculate $P(X_n =0 $ I.O $) $ and take the complement or calculate $p(X_n =1 $ for some n $)$ ? $\endgroup$ Jul 8, 2018 at 13:26
  • $\begingroup$ @Michael I tried to calculate $P(X_n =0 $ i.o $)$ . Then i need to calculate $\sum P(X_n=0)$ isn't it ? . But then i am getting $\sum P(X_n=0) = \infty $ because if $\alpha >0 $ , then $ 1- \alpha > 0 $ too. so the sum goes to $ \infty $ . Am i correct or did i do something wrong ? $\endgroup$ Jul 8, 2018 at 15:18
  • $\begingroup$ I meant to use $$\{\mbox{finite number of 1s}\} = \cup_{n=1}^{\infty}\{\mbox{there are no 1s after index $n$}\} $$ $\endgroup$
    – Michael
    Jul 9, 2018 at 3:54
  • 1
    $\begingroup$ Yes, except the last step should be $\leq$ because $P[\cap_{m=n}^k \{X_m=0\}]\leq (1-\alpha)^{k-n+1}$. Similar to how you are using the union bound, you could also say $$ P[\mbox{finite number of 1s}] \leq \sum_{n=1}^{\infty}P[\mbox{there are no 1s after index $n$}]$$ $\endgroup$
    – Michael
    Jul 9, 2018 at 16:19

1 Answer 1


Let $\mathcal{F}_n=\sigma(X_1,\ldots,X_n)$ and $\mathcal{F}_0=\{0,\emptyset\}$. Then by the generalized Borell-Cantelli lemma, $$ \{X_n=1\text{ i.o.}\}=\left\{\sum_{n\ge 1 }\mathsf{P}(X_n=1\mid\mathcal{F}_{n-1})=\infty\right\}, $$ and the probability of the RHS is $1$.

Instead you may adjust the proof of the unconditional B-C lemma. Specifically, let $A_n=\{X_n=1\}$. Indpendence is used to show that $$ \mathsf{P}\left(\bigcap_{n\ge m} A_n^c\right)=\prod_{n\ge m}\mathsf{P}(A_n^c)=0. $$ In your case, \begin{align} \mathsf{P}\left(\bigcap_{m\le n\le r} A_n^c\right)&=\mathsf{E}\left[\mathsf{E}\left[1_{A_r^c}\mid \mathcal{F}_{r-1}\right]1_{\bigcap_{m\le n\le r-1} A_n^c}\right]\le (1-\alpha)\mathsf{P}\left(\bigcap_{m\le n\le r-1} A_n^c\right) \\ &\le(1-\alpha)^2\mathsf{P}\left(\bigcap_{m\le n\le r-2} A_n^c\right)\le\cdots\le(1-\alpha)^{r-m+1}. \end{align} Taking $r\to\infty$ implies that $\mathsf{P}\left(\bigcap_{n\ge m} A_n^c\right)=0$.

  • 2
    $\begingroup$ True, but this seems to be a homework problem and usually it is not a good idea to solve homeworks by throwing a more sophisticated theorem at a problem. If the asker had access to this generalization, likely the question would not be asked. Though I give +1 since it is useful to learn of existence of this generalization. $\endgroup$
    – Michael
    Jul 8, 2018 at 6:03
  • $\begingroup$ @Michael It seems that the question fits this result perfectly... $\endgroup$
    – user140541
    Jul 8, 2018 at 6:04
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    $\begingroup$ It does fit it perfectly but (according to my best guess) this homework problem expects students to use basic principles. $\endgroup$
    – Michael
    Jul 8, 2018 at 6:15

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