I'm having issues with understanding how Borel-Cantelli lemma applies to the following exercise:

If a coin is tossed infinitely many times (and the tosses are independent), prove that the probability of getting 2 consecutive heads (or tails) infinite times is 1.

Could someone please show me?

For reference: https://en.wikipedia.org/wiki/Borel%E2%80%93Cantelli_lemma#Converse_result


The coin tosses can be modelled as a sequence of independent $\text{Ber}\left(\frac{1}{2}\right)$ random variables $(X_n)_{n\in\mathbb{N}}$

Letting $A_n=\{X_n=1\wedge X_{n+1}=1\}$, you want to know $P(\lim\sup_n A_n)$

Consider $\lim\sup_nA_{2n}\subseteq\lim\sup_nA_n$. Notice that $(A_{2n})_{n\in\mathbb{N}}$ are independent.

Since $\sum_nP(A_{2n})=\sum_n\frac{1}{4}=+\infty$, second BC lemma yields $P(\lim\sup_nA_{2n})=1$. Therefore $P(\lim\sup_nA_n)=1$. QED

  • $\begingroup$ Thank you very much for the answer, I'll need to read a bit of Bernoulli random variables but that just for the little paper I have to present. Thanks again. $\endgroup$ – Cesar Carlos Molina Apr 26 '15 at 15:40
  • $\begingroup$ @G.Sassatelli, sorry to bother you again but I was wondering if P(limsupnA2n)=1 therefore P(limsupnAn)=1 comes from limsupnA2n ⊆ limsupnAn or other results and why is A2n independent while An isn't. $\endgroup$ – Cesar Carlos Molina Apr 27 '15 at 1:18
  • $\begingroup$ @CesarCarlosMolina It comes from $\lim\sup_nA_{2n}\subseteq\lim\sup_nA_n$: if $A\subseteq B$, then $P(A)\leq P(B)$. $\endgroup$ – user228113 Apr 27 '15 at 1:26
  • $\begingroup$ @G.Sassatelli Got it, and about the independence of A2n? $\endgroup$ – Cesar Carlos Molina Apr 27 '15 at 2:34
  • $\begingroup$ It should be possible to verify it directly. The way I see it comes after a result that should be in any probability book:$$$$ Let $(X_n)_{n\in\mathbb{N}}$ be independent random variables $\Omega\rightarrow \mathbb{R}$. Let $(G_n)_{n\in\mathbb{N}}$ random variables $\Omega\rightarrow\mathbb{R}$ such that $\exists (K_n)_{n\in\mathbb{N}}$ pair-wise disjoint subsets of $\mathbb{N}$,$\exists (g_n)_{n\in\mathbb{N}}$ measurable functions $g_n:\mathbb{N}^{K_n}\rightarrow \mathbb{R}$ such that $G_n=g_n((X_i)_{i\in K_n})$. Then, $G_n$ are independent. $$$$ @CesarCarlosMolina $\endgroup$ – user228113 Apr 27 '15 at 6:31

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