= Solution
A <martingale> is an adapted integrable process $M_n$ with $\mathbb E(M_{n+1}\mid\mathcal F_n)=M_n$. A <stopping time> $T$ satisfies $\{T\leq n\}\in\mathcal F_n$. The bounded <optional sampling theorem> states that for bounded <stopping times> $S\leq T$, $\mathbb E(M_T\mid\mathcal F_S)=M_S$.
To prove it, let $A\in\mathcal F_S$ and take a common deterministic bound $N$. Then
$$
\mathbb E[\mathbf1_A(M_T-M_S)]=\sum_{j=0}^{N-1}\mathbb E[\mathbf1_{A\cap\{S\leq j<T\}}(M_{j+1}-M_j)]=0,
$$
since each indicator is $\mathcal F_j$-measurable. This proves the conditional identity. The theorem extends to almost surely finite <stopping times> for a uniformly integrable <martingale> by truncating the times and passing in $L^1$; without such a hypothesis unrestricted optional sampling is false.
For the proposed exponential process, conditioning on one increment gives the <martingale> condition $z(pe^\theta+qe^{-\theta})=1$. Thus, when $q>0$, the complete set is
$$
\boxed{\theta_\pm=\log\left(\frac{1\pm\sqrt{1-4pqz^2}}{2pz}\right).}
$$
These are the two real logarithms: $\theta_-<0<\theta_+$. If $q=0$, the sole finite value is $\theta=-\log z$. Stop the $\theta_+$ <martingale> at $n\wedge\tau_k$. Before the first hit, $S_n<k$, so the stopped process is bounded above by $e^{k\theta_+}$. On failure to hit, its value tends to zero because of $z^n$. Bounded convergence after bounded optional sampling gives $1=e^{k\theta_+}\mathbb E[z^{\tau_k};\tau_k<\infty]$, hence
$$
\boxed{G_k(z)=\left(\frac{2pz}{1+\sqrt{1-4pqz^2}}\right)^k.}
$$
Its limit at $z=1$ is one, proving almost-sure hitting as well. Differentiating at one gives
$$
\boxed{\mathbb E\tau_k=\frac{k}{p-q}.}
$$
For instance implicit differentiation of the <martingale> equation gives $\theta_+'(1)=-1/(p-q)$.
For the first subsequent return $\tau'_k$, the walk may first rise above $k$, so the uniform bound used in the stopping-limit argument fails. In fact after reaching $k$ it returns with probability $q+p(q/p)=2q<1$: a downward first step is followed by an almost-sure upward hit, while an upward first step hits the level below only with probability $q/p$. Thus $\tau'_k=\infty$ with positive probability and its unconditional mean is infinite. This pinpoints why repeating the earlier differentiation argument is invalid; bounded-time optional sampling itself remains valid.
Back to article page
