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