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Consider the Coin toss problem, i.e. let $Z_{i} : \Omega \rightarrow \{0,1\}$ with $$ Z_{i}\left(\omega\right) = \begin{cases} 1 & \text{if }\omega = H\\ 0 & \text{if } \omega = T \end{cases} $$ be the outcome of the $i$-th coin toss with $\Omega = \{H,T\}$. Assume all the $Z_{i}$ are independent and identical distributed with $$ \mathbb{P}(Z_{i} = 1) = \mathbb{P}(Z_{i} = 0) = \frac{1}{2}. $$ Now define $$ R := \min \{k\geq 1 \mid Z_{k} = 1,Z_{k+1} = 0,Z_{k+2} = 1, Z_{k+3} = 0\}. $$ $R$ is the number of tosses to get the pattern HTHT.

My question ist how to show that $R$ is almost surely finite. Any hints?

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Hints:

  1. Show that $$A_k := \{Z_{4k}=0, Z_{4k+1}=1, Z_{4k+2}=0, Z_{4k+3}=1\}$$ satisfies $\mathbb{P}(A_k) = 1/16$ for each $k \in \mathbb{N}$.
  2. Show that the events $A_k$, $k \geq 1$, are independent.
  3. It follows from Step 1 that $$\sum_{k \geq 1} \mathbb{P}(A_k) = \infty.$$ Apply the Borel Cantelli lemma (using Step 2) to conclude that $$\mathbb{P}(A_k \, \, \text{infinitely often})=1.$$
  4. Conclude that $\mathbb{P}(R<\infty)=1$.
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  • $\begingroup$ Is this correct? $\mathbb{P}(R = \infty) = \mathbb{P}(\lim\limits_{n \to \infty}\inf A_{n}^{c}) = 1 - \mathbb{P}(A_{n}~~ \text{infinitely often}) = 0$ $\endgroup$ – user562724 Nov 7 '18 at 21:05
  • $\begingroup$ I'm sorry. Now see my comment $\endgroup$ – user562724 Nov 7 '18 at 21:10
  • $\begingroup$ @love_math It is correct that $\mathbb{P}(R=\infty)=0$ (simply because $\mathbb{P}(R=\infty) = 1-\mathbb{P}(R<\infty)$) but the other equations are not correct. For instance $\mathbb{P}(R=\infty) = \mathbb{P}(\liminf_n A_n^c)$ does not hold true. $\endgroup$ – saz Nov 7 '18 at 21:13
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    $\begingroup$ @love_math $\mathbb{P}(A_n \, \, \text{infinitely often})=1$ tells us, in particular, that for almost every $\omega$ we can find $n \in \mathbb{N}$ such that $\omega \in A_n$ (in fact there are infinitely many such $n$, but we are fine with one). As $A_n \subseteq \{R<\infty\}$ this implies $\omega \in A_n \subseteq \{R<\infty\}$, i.e. $R(\omega)<\infty$. $\endgroup$ – saz Nov 7 '18 at 21:20
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    $\begingroup$ @love_math We only have $\{R=\infty\} \color{red}{\subseteq} \bigcap_{m \geq 1} A_m^c$ (for instance, if $Z_2(\omega)=1$, $Z_3(\omega)=0$, $Z_4(\omega)=1$, $Z_5(\omega)=0$ for some $\omega$, then $R(\omega)<\infty$ but $\omega \in \bigcap_{m \geq 1} A_m^c$ ... this shows that $\bigcap_m A_m^c$ is stricly larger than $\{R=\infty\}$.) Apart from that, your reasoning is correct, I think. (Note that with the correction which I mentioned you still get the desired result $\mathbb{P}(R=\infty)=0$.) $\endgroup$ – saz Nov 10 '18 at 7:32

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