= Maximum of a downward-biased random walk
{title2=$\mathbb P(M=j)=(1-p/q)(p/q)^j,\quad j\geq0$}
For a nearest-neighbour <random walk> with upward probability $0<p<q=1-p$, start at $S_0=0$ and put $M=\sup_{n\geq0}S_n$. 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, $Z_{n\wedge H_k}\leq(q/p)^k$; 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 $\mathbb P(M\geq k)=(p/q)^k$ 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>.
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