Gamblers ruin- Expected number of returns

Question

Consider a symmetric gamblers ruin problem on $\{0,1,\ldots,n\}$. Suppose that the initial fortune is $k$ dollars, I want to calculate the expected number of visits to $j$ dollars when (i) $j=k$ and (ii) $k<j<n$. I think I have to find the probability $p$ that the chain returns to state k before getting absorbed and then the expected number of visits can be calculated as the mean of Geometric random variable with parameter $1-p$. But I am not sure how to calculate this $p$. Any help will be appreciated.

Answer 1

In (i), with probability $\frac12$ you move to $j-1$ and with probability $\frac12$ you move to $j+1$. In the first case, the probability that you return to $j$ before hitting $0$ is $\frac{j-1}j$. In the second case, the probability that you return to $j$ before hitting $n$ is $\frac{n-j-1}{n-j}$. Thus the probability to return to $j$ before absorption is

\begin{eqnarray*} p &=& \frac12\cdot\frac{j-1}j+\frac12\cdot\frac{n-j-1}{n-j} \\ &=& 1-\frac1{2j}-\frac1{2(n-j)}\;, \end{eqnarray*}

and the expected number of returns is

\begin{eqnarray*} \frac p{1-p} &=& \frac{1-\frac1{2j}-\frac1{2(n-j)}}{\frac1{2j}+\frac1{2(n-j)}} \\ &=& \frac1{\frac1{2j}+\frac1{2(n-j)}}-1 \\ &=& \frac{2j(n-j)}n-1\;. \end{eqnarray*}

In (ii), you get to $j$ before hitting $0$ with probability $\frac kj$, which yields one “return” plus the expected number of returns calculated in (i), so the expected total is

$$ \frac kj\cdot\frac{2j(n-j)}n=\frac{2k(n-j)}n\;. $$

Note that you can consider (i) as a special case of (ii) if you count starting at $k=j$ as a first “return”.