Consider a simple symmetric random walk, $\left\{X_n\right\}$, where the following hold;
$$X_n = X_{n-1} + Y_n$$ for $n = 1, 2, 3,\ldots$
$$Y_1, Y_2, Y_3, \ldots \text{iid }, \mathbb{P}(Y_1 = 1) = \mathbb{P}(Y_1 = -1) = 1/2$$
$a$ and $b$ are positive integers such that $-a \leq X_0 \leq b$,
$T$ is the time to absorption for a walk with absorbing barriers at $-a$ and $b$.
We let $\eta_k = \mathbb{E}(T \mid X_0 = k)$ and I wish to show
$$\eta_{k+1} – 2\eta_k – \eta_{k-1} = -2$$
Many thanks.
Best Answer
We can deduce that \begin{align} \eta_k&=E(T\mid X_0=k)=1+\frac 12E(T\mid X_1=k-1)+\frac 12 E(T\mid X_1=k+1)\\ &=1+\frac 12 \eta_{k-1}+\frac 12\eta_{k+1} \end{align} This is the same as your equation, except for a minus sign. I think your equation is wrong, since is should be symmetrical in $\eta_{k-1}$ and $\eta_{k+1}$. (There is no way to distinguish going to the left or right).
To obtain this equation, we just calculate the expected absorbing time for the situation after the next move and add one to that for the current move.