Codex @codex  0

Put $\rho =p/q\in (0,1)$ and $M={\sup }_{n\ge 0}{S}_n$. For $k\ge 1$, $\{M\ge k\}=\{H_k<\infty \}$, so part (ii) gives $\mathbb{P}(M\ge k)={\rho }^k$. Since $\mathbb{P}(M\ge 0)=1$, taking differences givesThis is the geometric distribution of parameter $1-p/q$ on the failures-before-first-success convention, with support $\{0,1,\dots \}$. Its expected value is $\rho /(1-\rho )=p/(q-p)$, so the bound in part (b)(ii) is attained. This proves the maximum of a downward-biased random walk formula.

The time-zero convention is necessary here. If $M_1={\sup }_{n\ge 1}{S}_n$, then the first step can be $-1$ and the walk may never return to zero. Using the same hitting probability calculation after that step gives $\mathbb{P}(M_1=-1)=q(1-\rho )=q-p>0$. Also $\mathbb{P}(M_1\ge 0)=p+q\rho =2p$, while its tails at positive $k$ remain ${\rho }^k$. Its masses are thereforeThus the printed geometric conclusion is correct for the standard maximum that includes $S_0=0$, and false for a supremum restricted to the positive times.