Absorbing gambler's-ruin chain

For $0<x<N$, the only row masses are $p$ and $q$, which are nonnegative and [F1, F2] sum to one. At $0$ and $N$ the rows are the corresponding Dirac masses. Thus these rows are exactly [F1] applied to any base kernel having the displayed interior transitions, and [F2] verifies the kernel. When $N=1$ there are no interior rows; when $p=0$ or $p=1$ the interior motion is deterministic.

Define [given] ${\tau }_D=\inf \{n\ge 0:X_n\in \{0,N\}\}\mathrm{.}$ Then $\{{\tau }_D\le n\}={\bigcup }_{k=0}^n\{X_k\in D\}\in {\mathcal{F}}_n$, so it is a hitting time. If $X_0\in D$, then ${\tau }_D=0$; otherwise it may be infinite in the general eventwise formulation.

By [F3], on $\{{\tau }_D<\mathrm{\infty }\}$ the conditional future is the chain started [F3, step 1.1, step 1.2] from $X_{{\tau }_D}$. Step 1.1 gives $P(X_{{\tau }_D},X_{{\tau }_D})=1$, so every subsequent coordinate equals the same boundary state. Constants zero and one give respectively zero and the finite-hit event in the eventwise formula. Choice is used only through [F3].