= Minimum of an upward-biased random walk
{title2=$\mathbb P(-\min_{n\geq0}S_n=m)=(1-q/p)(q/p)^m$}
Let $S_0=0$ and let the <independent and identically distributed random variables> $S_n-S_{n-1}$ be $+1$ with <probability> $p>1/2$ and $-1$ with <probability> $q=1-p>0$. The <strong law of large numbers> gives $S_n\to+\infty$ <almost surely>, so the global <minimum> $W$ is finite and attained. For each <nonnegative integer> $m$, the unit-step property and translation of the starting point give
$$
\mathbb P(-W\geq m)=\mathbb P_0(T_{-m}<\infty)
=\mathbb P_m(T_0<\infty)=(q/p)^m
$$
by the <biased random walk hitting probability>. Subtracting consecutive tails gives the displayed zero-based <geometric distribution>, with success parameter $1-q/p$. Inclusion of time zero is important: it ensures $-W\geq0$. Reflecting the sign of the path yields the corresponding <maximum of a downward-biased random walk>.
Back to article page
