A strategy's P&L starts at 0. Each step it gains 1 with probability 2/3 and loses 1 with probability 1/3, independently. It stops at +2 or at −1, whichever comes first. What is the expected number of steps? Give a fraction in lowest terms.

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A strategy's P&L starts at $0$ . Each step it gains $1$ with probability $\frac{2}{3}$ and loses $1$ with probability $\frac{1}{3}$ , independently. It stops at $+ 2$ or at $- 1$ , whichever comes first. What is the expected number of steps? Give a fraction in lowest terms.

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Hint

Use $(\frac{1}{2})^{S_{t}}$ for the exit odds, then $S_{t} - \frac{t}{3}$ for the time.

Solution

With $r = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2}$ , optional stopping on $r^{S_{t}}$ gives $P \times (\frac{1}{4}) + (1 - P) \times 2 = 1$ , so $P = \frac{2 - 1}{2 - \frac{1}{4}} = \frac{4}{7}$ .
The expected end point is $(\frac{4}{7}) \times 2 - (\frac{3}{7}) \times 1 = \frac{5}{7}$ .
Each step has mean $\frac{2}{3} - \frac{1}{3} = \frac{1}{3}$ , and $S_{t} - \frac{t}{3}$ is a martingale, so the expected time is $\frac{\frac{5}{7}}{\frac{1}{3}} = \frac{15}{7}$ .

Answer: $\frac{15}{7}$

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