Independent sequence of random variables such that each random variable has the Bernoulli distribution.

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I'm quite confused with this question. I understand the first half of question itself but I'm unsure about what $A_n^m$ is describing - is it just the sum of the random variables is equal to 1?
I'm also confused about how I'd show that the probability is equal to 1.
Any hints or similar examples would be greatly appreciated.
Thanks in advance.

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The event $A_n^m$ is the event that each variable in the block of $m$ variables starting at $X_n$ is observed to have value $1$.

Hints:

  1. To show that $P(\cap_m A^m)=1$, it is enough to show $P(A^m)=1$ for each $m$. (Why?)

  2. To show $P(A^m)=1$ for a fixed $m$ it looks like you could try the second Borel-Cantelli lemma, since $A^m$ is by definition the limsup of a sequence of events $\{A_n^m, n=1,2,\ldots\}$, and your aim is to show the limsup of that sequence has probability one. But to apply the second Borel-Cantelli lemma we require the events to be independent, and the $A_n^m$ here are not (why not?). But there exists a subcollection $\{B_n\}$ of $\{A_n^m\}$ such that the $B_n$'s are independent and still $P(\limsup_n B_n)=1$. (How would you prove this, and how does this imply the desired result?)