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Write $s_i=p_i+q_i>0$ and $d_i=p_i-q_i$ for $i=0,1$. For the parity correction $h_{\theta }(k)=k-\theta 1_{\{k\text{ odd}\}}$, the conditional expectations of the increments of $h_{\theta }(X_n)$ areThe holding move contributes zero. Thus the period-two martingale corrector for a birth-death chain exists precisely whenThis is the condition that $h_{\theta }$ be a harmonic function for a Markov chain. The process has integrable values from a fixed initial state because $|X_n|\le |X_0|+n$; the Markov property therefore turns the zero conditional increments into the required martingale property. Conversely, both parities are reached with positive probability because all right and left probabilities are positive, so the two conditional equations are necessary. Expanding their numerator also gives the equivalent balance condition $p_0{p}_1=q_0{q}_1$. Positivity makes $|\theta |<1$.

Put $L=-aN$, $U=bN$. The exit time $T_N$ is almost surely finite. Indeed, set $m=U-L$ and $p_{\ast }=\min (p_0,p_1)>0$. From any interior point, consecutive right steps reach $U$ in at most $m$ moves, with probability at least $p_{\ast }^m$. The Markov property gives $\mathbb{P}(T_N>km)\le (1-p_{\ast }^m{)}^k$. This is the finite-interval exit times have geometric tails argument. The walk cannot skip a boundary because its jumps have size at most one.

Under the balance condition, the stopped martingale $h_{\theta }(X_{n\wedge T_N})$ is bounded, because the stopped position stays in $[L,U]$. Apply the optional stopping theorem at $n\wedge T_N$, then dominated convergence as $n\to \infty$. Starting from zero givesWith ${\epsilon }_a=1_{\{aN\text{ odd}\}}$ and ${\epsilon }_b=1_{\{bN\text{ odd}\}}$,The bounded parity corrections disappear after division by $N$; the holding probability of a Markov chain may be positive or zero and may differ between parities.

For failure of balance, use scale increments for a birth-death chain rather than the nonexistent bounded correction. Let $u(k)$ be the upper exit probability from $k$ for boundaries $L,U$. The Markov property yieldswhere $p_k,q_k$ depend only on parity. Choose positive numbers $w_j$ with $w_{j+1}=(q_j/p_j)w_j$. Then $u(k)={\sum }_{j=L+1}^k{w}_j/{\sum }_{j=L+1}^U{w}_j$ satisfies these equations and boundary values. Stopping this bounded harmonic function proves that it is the exit probability. In particular,If $\rho <1$, the positive-index sum stays bounded as $N\to \infty$, while the negative-index contribution diverges geometrically, so ${\pi }_N\to 0$. If $\rho >1$, the negative-index sum stays bounded while the positive-index sum diverges, so ${\pi }_N\to 1$. Therefore the complete classification isThus a fixed nonzero imbalance drives the macroscopic exit to the corresponding side, for every positive $a,b$.