 Maximum of a downward-biased random walk

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.