Gambler's ruin: The fair game is the longest

Question

Consider a gambler who, in every trial of a game, wins or loses a dollar with probability $p\in\left( 0,1\right) $ and $q=1-p$, respectively. Let his initial capital be $z>0$ and let him play against an adversary with the same capital $z>0$. The game continues until one of the players is ruined.

In the language of random variables, this amounts to consider a sequence $X_{1}^{\left( p\right) },X_{2}^{\left( p\right) },...\ $of random variables (on a sample space $\Omega$) taking on two values $+1$ and $-1$ with probabilities $\Pr\left[ X_{n}^{\left( p\right) }=+1\right] =p$ and $\Pr\left[ X_{n}^{\left( p\right) }=-1\right] =q$. In particular, $X_{n}^{\left( p\right) }$ describes the gambler's gain on the $n$th trial and $$ S_{n}^{\left( p\right) }\left( \omega\right) =X_{1}^{\left( p\right) }\left( \omega\right) +\cdots+X_{n}^{\left( p\right) }\left( \omega\right) ,\quad S_{0}^{\left( p\right) }\left( \omega\right) \equiv0\qquad\forall \,\omega\in\Omega $$ describes his net cumulated gain after $n$ trials. The duration of the game is the random number of trials before he is either ruined or wins the game (and his adversary wins or is ruined) $$ D^{\left( p\right) }\left( \omega\right) =\min\left\{ t\in\mathbb{N}% :\left\vert S_{t}^{\left( p\right) }\left( \omega\right) \right\vert =z\right\} \qquad\forall \,\omega\in\Omega. $$

Q. I need a reference or a simple proof of the folk result according to which, for each $p\in\left( 0,1\right) $, $$ \Pr\left[ D^{\left( p\right) }\leq s\,\right] \geq\Pr\left[ D^{\left( 1/2\right) }\leq s\,\right] \qquad\forall\, s=1,2,... $$

Answer 1

Let us show a bit more, that $F_p(s):=P(D^{(p)}\le s)$ is nondecreasing in $|p-1/2|$, for any real $s$. That is, take any $p$ and $p_1$ in $(0,1)$ such that $|p_1-1/2|>|p-1/2|$, and let, for brevity, $F:=F_p$ and $G:=F_{p_1}$. We shall show that then $F\le G$.

Indeed, by the formula (10), referenced by Carlo Beenakker, \begin{equation*} f_p(n):=P(D^{(p)}=n)=a_{n,z} b_{p,z}(pq)^{n/2} \end{equation*} for natural $n$, where $a_{n,z}$ depends only on $n$ and $z$; $b_{p,z}:=(p/q)^{z/2}+(q/p)^{z/2}$; and $q:=1-p$. Since $pq$ decreases in $|p-1/2|$, the family $(f_p)$ of pmf's has the monotone likelihood ratio (MLR) property, meaning that \begin{equation*} r(n):=\frac{f(n)}{g(n)} \end{equation*} increases in $n\in N$, where $N:=N_z:=\{n\colon a_{n,z}\ne0\}$, $f:=f_p$, $g:=f_{p_1}$, and $p$ and $p_1$ are as before. Therefore, for any given real $s$, \begin{align*} F(s)-G(s)&=F(s)[1-G(s)]-G(s)[1-F(s)] \\ &=\sum_{m,n\colon\, N\ni m\le s<n\in N}[f(m)g(n)-g(m)f(n)] \\ &=\sum_{m,n\colon\, N\ni m\le s<n\in N}[r(m)-r(n)]g(n)g(m)\le0, \end{align*} since $r$ is increasing. Thus, indeed $F\le G$.