Bernoulli String of Luck Question

Question

Question: Let $X_k \sim Bern(p)$ and for $n,l \in \mathbb N$: $$ A_n^l = \{X_n = X_{n+1} = .... = X_{n+l-1} = 1 \} $$ which is the event that a run of luck of at least length l starts at time n, further, let: $$ A^l = \limsup_{n \to \infty} A^l_n $$ Show that $P(\bigcap_{l = 1}^{\infty} A^l)=1$ That is, with probability 1 there are infinitely many runs of luck of arbitrary length.

Attempt: Im really confused with how to interpret $A^l$. For example, when $l=1$, then: \begin{align*} A^1 &= \limsup_{n \to \infty} A^1_n\\ &=\bigcap^\infty_{j=1} \left ( \bigcup^{\infty}_{n=j} A_n^1\right)\\ &=(A_1^1 \cup A_2^1 \cup A_3^1 \cup...) \cap (A_2^1 \cup A_3^1 \cup ...)\cap ( A_3^1 \cup A_4^1 \cup ...)\\ \end{align*}

thanks to @GenericNickname, we can see that $\mathbb P(A^1) = 1$. By the same reasoning for $l=2$: $$ \mathbb P(A^2) = \lim _{n \to \infty} (1 -\mathbb P \text{(two successes in a row occurs at least once for $m\ge n$)} ) $$

Intuitively this will be equal to 1 as well. Not as intuitive when we take $l=100$ for example. The next hint from @GenericNickname is to use the Borel-Cantelli lemma to prove this, which states for a sequence $S_n$:

$$ \sum_{n=1}^{\infty} \mathbb P(S_n) < \infty \implies \mathbb P(\limsup_{n \to \infty} S_n) = 0 $$

Still confused but working on it:

If we take $S_n = A^l_n$ above. Then: $$ \sum_{n=1}^{\infty} \mathbb P(A_n^l)=p^l + p^l+p^l+.... \to \infty $$ So, the lemma doesn't hold here

Answer 1

You seem to be using a wrong notion of $\limsup$. Usually one has $$\limsup_{n \to \infty} A^l_n = \bigcap_{n = 1}^{\infty} \bigcup_{m=n}^{\infty} A^l_m$$ According to this definition, $\limsup A^l_n$ is the set of elementary events that occur in infinitely many $A^l_n$. In this exercise it would be the set of runs that contain infinitely many strings of luck of length $l$.

Hence, $$A^1 = \{ X_n = 1 \text{ for infinitely many } n \in \mathbb{N}\}$$ and $$\mathbb{P}(A^1) = 1.$$

EDIT: Maybe I should explain in detail how one can compute $\mathbb{P}(A^1)=1$. First off, I use continuity from above of $\mathbb{P}$, i.e. for sets $B_n$ such that $B_{n+1}\subseteq B_n$ we have $$\mathbb{P}(\bigcap_{n=1}^{\infty} B_n) = \lim_{n \to \infty} \mathbb{P}(B_n).$$ If we set $B_n:= \bigcup_{m=n}^{\infty} A_m^1$ we get $B_n = (A_n^1 \cup B_{n+1}) \supseteq B_{n+1}$ and hence $$\mathbb{P}(A^1) = \mathbb{P}( \bigcap_{n=1}^{\infty} B_n ) = \lim_{n \to \infty} \mathbb{P}(B_n).$$

Now lets look at $\mathbb{P}(B_n)$. We have $A_m^1 = \{X_m = 1\}$ and \begin{align*} \mathbb{P}(B_n) &= \mathbb{P}(\bigcup_{m=n}^{\infty} A_m^1 ) = \mathbb{P}(\{ X_k = 1 \text{ for some } m \geq n\})\\ & = 1 - \mathbb{P}(\{X_k = 0 \text{ for all } m \geq n\}). \end{align*}

We can use either intuition or a similiar argument to before to see that $$\mathbb{P}(\{X_k = 0 \text{ for all } k \geq m\}) = 0$$ for all $n$. But this means that $\mathbb{P}(B_n) =1$ for all $n$ and hence $$\mathbb{P}(A^1) = \lim_{n \to \infty} \mathbb{P}(B_n) = 1.$$ I hope this makes it a little bit clearer for you.

EDIT #2: My hint regarding the Borel-Cantelli Lemma was aimed at a statement that is often regarded as the lemma's 2nd statement:

If $(E_n)$ is a sequence of pairwise independent events such that $\sum_{n=1}^{\infty} \mathbb{P}(E_n) = \infty$, then $$\mathbb{P}(\limsup_n E_n) = 1$$

Answer 2

I preassume that the $X_n$ are independent (this should be mentioned in your question) and step in where it has been proved allready that $\mathbb P(A^1)=1$.

Define $Y_k:=X_{kl}\times\cdots\times X_{kl+l-1}$. Then the $Y_k$ are iid with Bernouilli-distribution having parameter $p^l$ and $A_{kl}^l=\{Y_k=1\}$.

With the same method used to prove that $\mathbb P(A^1)=1$ it can be proved that $P\{Y_k=1\mid\text{ for infinitely many }k\}=1$ and this implies that $P(A^l)=1$.