# Question regarding Borel-Cantelli lemma

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?

• I don't think Borel-Cantelli is helpful in this case. What if you compute the probability of never getting another 1 after index $n$? – Michael Jul 8 '18 at 5:16
• @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 $)$ ? – student_R123 Jul 8 '18 at 13:26
• @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 ? – student_R123 Jul 8 '18 at 15:18
• I meant to use $$\{\mbox{finite number of 1s}\} = \cup_{n=1}^{\infty}\{\mbox{there are no 1s after index n}\}$$ – Michael Jul 9 '18 at 3:54
• 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 $$P[\mbox{finite number of 1s}] \leq \sum_{n=1}^{\infty}P[\mbox{there are no 1s after index n}]$$ – Michael Jul 9 '18 at 16:19

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 $\mathsf{P}\left(\bigcap_{n\ge m} A_n^c\right)=0$.