= Solution
Write $s_i=p_i+q_i>0$ and $d_i=p_i-q_i$ for $i=0,1$. For the parity correction $h_\theta(k)=k-\theta\mathbf1_{\{k\text{ odd}\}}$, the <conditional expectations> of the increments of $h_\theta(X_n)$ are
$$
\begin{cases}
p_0(1-\theta)+q_0(-1-\theta)=d_0-\theta s_0,&X_n\text{ even},\\
p_1(1+\theta)+q_1(-1+\theta)=d_1+\theta s_1,&X_n\text{ odd}.
\end{cases}
$$
The holding move contributes zero. Thus the <period-two martingale corrector for a birth-death chain> exists precisely when
$$
\boxed{\theta=\frac{d_0}{s_0}=-\frac{d_1}{s_1},\qquad \frac{d_0}{s_0}+\frac{d_1}{s_1}=0.}
$$
This 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 $|X_n|\leq|X_0|+n$; 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 $p_0p_1=q_0q_1$. Positivity makes $|\theta|<1$.
Put $L=-aN$, $U=bN$. The exit time $T_N$ is almost surely finite. Indeed, set $m=U-L$ and $p_* =\min(p_0,p_1)>0$. From any interior point, consecutive right steps reach $U$ in at most $m$ moves, with <probability> at least $p_*^m$. The <Markov property> gives $\mathbb P(T_N>km)\leq(1-p_*^m)^k$. 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 gives
$$
0=\pi_Nh_\theta(L)+(1-\pi_N)h_\theta(U).
$$
With $\varepsilon_a=\mathbf1_{\{aN\text{ odd}\}}$ and $\varepsilon_b=\mathbf1_{\{bN\text{ odd}\}}$,
$$
\pi_N=\frac{bN-\theta\varepsilon_b}{(a+b)N+\theta\varepsilon_a-\theta\varepsilon_b},\qquad
\boxed{\lim_{N\to\infty}\pi_N=\frac{b}{a+b}.}
$$
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> yields
$$
p_k(u(k+1)-u(k))=q_k(u(k)-u(k-1)),\qquad u(L)=0,\quad u(U)=1,
$$
where $p_k,q_k$ 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,
$$
\pi_N=\frac{\sum_{j=1}^{bN}w_j}{\sum_{j=-aN+1}^{bN}w_j},\qquad
w_{j+2}=\rho w_j,\quad \rho=\frac{q_0q_1}{p_0p_1}.
$$
If $\rho<1$, the positive-index sum stays bounded as $N\to\infty$, while the negative-index contribution diverges geometrically, so $\pi_N\to0$. If $\rho>1$, the negative-index sum stays bounded while the positive-index sum diverges, so $\pi_N\to1$. Therefore the complete classification is
$$
\boxed{\lim_{N\to\infty}\pi_N=
\begin{cases}
0,&p_0p_1>q_0q_1,\\
b/(a+b),&p_0p_1=q_0q_1,\\
1,&p_0p_1<q_0q_1.
\end{cases}}
$$
Thus a fixed nonzero imbalance drives the macroscopic exit to the corresponding side, for every positive $a,b$.
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