Consider a sequence of Bernoulli trials $X_1 , X_2 , X_3 , \ldots$, where $X_n = 1$ or $0$.
If we suppose that $$\Pr\{X_n=1\mid X_1,X_2,\ldots,X_{n-1}\} \geq \alpha > 0, \ n=1,2,\ldots $$
Could someone help me to prove that:
$$\Pr\{X_n=1 \text{ for some } n \}= 1 \text{ and } \Pr\{X_n=1 \text{ infinitely often } n \}= 1 $$
A Bernoulli trial results only in 1 or 0. It takes the trials of only the happening or non happening of any event.
I've been super stuck on this question for a while now. Would really appreciate it if someone could break down the solution for me or give me hints.
Thanks for your hints, help and time.
\begin{align} \Pr(X_{k+1}=0,\ldots,X_{k+n}=0) &= \Pr(\Pr(X_{k+1}=0,\ldots,X_{k+n}=0\mid X_1,\ldots,X_k))\\ &=\Pr(\Pr(X_{k+n}=0\mid X_1,\ldots,X_k, X_{k+1}=0,\ldots,X_{k+n-1}=0)\cdot \Pr(X_{k+1}=0,\ldots,X_{k+n-1}=0\mid X_1,\ldots,X_k))\\ &\le (1-\alpha)\Pr(\Pr(X_{k+1}=0,\ldots,X_{k+n-1}=0\mid X_1,\ldots,X_k))\\ \end{align} and continuing by induction $\Pr(X_{k+1}=0,\ldots,X_{k+n}=0)\le (1-\alpha)^n$ for any $k$. Letting $n\to\infty,$ $\Pr(X_n=0:n\ge k)=0$ for any $k$, so $$ \Pr(X=0 \text{ eventually})\le \sum_{k=1}^\infty\Pr(X_n=0:n\ge k)\le 0, $$ so $X_n$ is $1$ infinitely often.