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Put $h_0=1$ and $h_r=0$. The Markov property and first-step conditioning give $h_i={\sum }_j{P}_{ij}{h}_j$. Consider a finite history ending at state $i$ and having positive conditional probability given $A$. The chance of eventual absorption at zero after that history is $h_i$: the future depends only on the present state, and if zero has already been reached, absorption keeps the chain there. If the next state is $j$, this chance is $h_j$. Bayes' rule therefore givesThis depends on the history only through $i$, proving the conditional Markov property. The conditional transition matrix isIts row sums are one by harmonicity of $h$, and $Q_{ir}=0$. State $r$ is never visited under the conditioning and is naturally removed; any row assigned there on an enlarged formal state space is immaterial, since conditioning starting at $r$ is undefined. State zero stays absorbing. For a random initial law $\nu$, its conditional law is ${\nu }_i{h}_i/{\sum }_j{\nu }_j{h}_j$. This is the Doob h-transform with the absorption probability as harmonic function.

For the simple symmetric walk, $h_i=(r-i)/r$ solves the linear harmonic recurrence and its boundary values. ThusIn particular, the conditioned chain moves downward with probability one from $r-1$.

Let $\tau$ be the first hit of $\{0,r\}$ in the original walk and let $g_i={\mathbb{E}}_i[\tau 1_A]$. Its absorption time has finite expectation: from any transient state a run of at most $r$ downward steps hits zero with probability at least $2^{-r}$, giving a geometric tail bound by blocks. First-step conditioning then givesEquivalently, $g_{i+1}-2g_i+g_{i-1}=-2(r-i)/r$. The cubic polynomialhas exactly these second differences and boundary values. It is the unique solution, since the difference of any two solutions is a linear sequence vanishing at both endpoints. Conditional on $A$, $\tau$ is the first time in zero, so division by $h_i$ provesThis is the duration of gambler's ruin conditioned on a chosen boundary.