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For a nearest-neighbour random walk with upward probability $0<p<q=1-p$, start at $S_0=0$ and put $M={\sup }_{n\ge 0}{S}_n$. Its probability distribution is geometric on the nonnegative integers, with the displayed mass and expected value $p/(q-p)$. To prove this, the exponential martingale of a biased simple random walk $Z_n=(q/p{)}^{S_n}$ has expectation one. Stop at the first hit $H_k$ of $k$. Before that hit, $Z_{n\wedge H_k}\le (q/p{)}^k$; the strong law of large numbers gives $S_n\to -\infty$ on paths without a hit. Dominated convergence theorem therefore gives $1=(q/p{)}^k\mathbb{P}(H_k<\infty )$. Taking differences of $\mathbb{P}(M\ge k)=(p/q{)}^k$ yields the mass. If time zero is excluded, the maximum can equal $-1$ with probability $q-p$ and does not have this nonnegative geometric distribution.

Assume $0<p<1$, so the ratio defining $Z_n$ is meaningful. Extend the random walk by $S_0=0$, the filtration by a trivial ${\mathcal{F}}_0$, and the process by $Z_0=1$. Write $r=q/p$. Each $Z_n=r^{S_n}$ is ${\mathcal{F}}_n$-measurable and integrable, since $|S_n|\le n$. The next increment is independent of ${\mathcal{F}}_n$, soTherefore $\boxed{(Z_n{)}_{n\ge 0}\text{ is a nonnegative martingale},\mathbb{E}{Z}_n=1}$. This is the exponential martingale of a biased simple random walk, with exponent $\log (q/p)$. At $p=q=1/2$, it is the constant process one. The endpoint cases $p=0$ or $p=1$ require separate definitions and are not covered by the printed ratio.

The moment-generating function of one increment isThus withindependence givesThis is the exponential martingale of a biased simple random walk.