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Let $X_i$ be a sequence of iid Rademacher random variables (i.e. $\mathbf{P}(X_i = 1) = 1/2 = \mathbf{P}(X_i=-1)$, and let $S_n=\sum_{i=1}^n X_i$. Let $\tau = \inf \{n \geq 5 : S_n = S_{n-5}+5\}$ be the first time the random walk takes $5$ positive steps in a row.

I am interested in computing $\mathbf{E}\tau$.

As a first step, we know that $\tau=n$ is equivalent to the last $5$ steps being positive and (if $n\geq 5+1$), the $6th$ last step being negative, and not encountering 5 positive steps in a row at times $\leq n-6$. Thus, we have $$\mathbf{P}(\tau=n) = \frac{1}{64}\mathbf{P}(\tau\not<n-5)\\ = \frac{1}{64}(1- \mathbf{P}(\tau \geq n-5)).$$

I have been struggling with how to manipulate this expression to get a closed form expression either of $\mathbf{P}(\tau=n)$ or $\mathbf{P}(\tau \geq n)$. Any help is appreciated.

RobPratt
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Brian
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1 Answers1

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Let random variable $Y_i$ denote the number of steps to get $5$ positive steps in a row, starting from $i$ positive steps in a row, and let $m_i=\mathbb{E}[Y_i]$. Then $m_5=0$, and we want to compute $m_0$. By conditioning on the first step out of state $i$, we obtain for $i<5$ that $$m_i = 1 + \frac{1}{2}m_{i+1} + \frac{1}{2}m_0.$$ Solving this linear system of equations yields $$m_0=\color{red}{62},m_1=60,m_2=56,m_3=48,m_4=32,m_5=0.$$

More generally, for $k$ positive steps in a row, $m_0=2^{k+1}-2$.

RobPratt
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