A gambler has a starting fortune of $M$. He will repeatedly play a game the game and his bets are such that he gains/loses $s$ with probability $p(s)$ (gains if $s\geq 0$, loses if $s<0$), where $p(s)$ is the same every game. If his fortune ever goes below 0 or above $T>M$ then he stops playing. Additionally, the expected return $\int_{-\infty}^\infty sp(s)ds=0$.
I am only interested in the duration of the game. I believe that the game will last the longest if $M=\frac{T}{2}$, i.e. the game starts in the middle. I am looking for help with the proof or a counterexample.
It maybe useful to consider that if $D_M$ is the expected duration of the game when the game starts at $M$, then we have the constraints
$$D_M=1+\int_{-\infty}^\infty p(s)D_{M+s}ds$$ $$D_0=D_T=0$$
For the standard Gambler's ruin with $p(+1)=p(-1)=\frac{1}{2}$ and $p(s)=0$ for $s\neq\pm 1$, the statement is true since the expected duration is then given by $D_M=M(T-M)$, which is maximal for $M=\frac{T}{2}$. I do not know however how to prove this for the general game.
Examples that do not appear to be counterexamples:
- $p(2)=\frac{1}{2}$, $p(-1)=p(-3)=\frac{1}{4}$
- $p(-\alpha)=p$, $p(\beta)=(1-p)$ with $p=\frac{\beta}{\alpha+\beta}$, $\alpha,\beta\geq 0$
Let $T = 8$, and let $p(s)$ be given by
\begin{align*} p(-2) &= 1/8\\[6pt] p(-1) &= 1/2\\[6pt] p(2) &= 3/8 \end{align*}
For $M$ from $1$ to $7$, let $E[M]$ denote the expected number of rounds until the player's remaining wealth is is no longer in the open interval $(0,T)$, given a starting wealth of $M$.
Then, assuming my Maple program is correct,
\begin{align*} E[1] &= 164984/46004 \approx 3.586296844\\[6pt] E[2] &= 252681/46004 \approx 5.492587601\\[6pt] E[3] &= 317280/46004 \approx 6.896791583\\[6pt] E[4] &= 331160/46004 \approx 7.198504478\\[6pt] E[5] &= 331500/46004 \approx 7.205895140\\[6pt] E[6] &= 253149/46004 \approx 5.502760630\\[6pt] E[7] &= 214016/46004 \approx 4.652117207 \end{align*}
In particular, since $E[5] > E[4]$, the expected duration is not maximized at $M=T/2$.