A market maker's inventory is an integer number of lots from 0 to 3. Each step, from 0 it moves to 1 with probability 1. From 1 or 2 it moves up by 1 lot with probability 2/3 and down by 1 lot with probability 1/3. Reaching 3 stops the process. Starting from 0, what is the expected number of steps to reach 3? Give a fraction a/b in lowest terms.

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Inventory hits the limit

ProbabilityHard

A market maker's inventory is an integer number of lots from $0$ to $3$ . Each step, from $0$ it moves to $1$ with probability $1$ . From $1$ or $2$ it moves up by $1$ lot with probability $\frac{2}{3}$ and down by $1$ lot with probability $\frac{1}{3}$ . Reaching $3$ stops the process. Starting from $0$ , what is the expected number of steps to reach $3$ ? Give a fraction a/b in lowest terms.

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Hint

Let $d$ be the expected steps needed to go up one level for the first time, from each level, and add them.

Solution

Let $h 0, h 1, h 2$ be the expected steps to reach $3$ , with $h 3 = 0$ , and let $d 0 = h 0 - h 1$ , $d 1 = h 1 - h 2$ , $d 2 = h 2$ .
From $0$ : $h 0 = 1 + h 1$ , so $d 0 = 1$ .
From $1$ : $h 1 = 1 + (\frac{2}{3}) h 2 + (\frac{1}{3}) h 0$ , which rearranges to $(\frac{2}{3}) d 1 = 1 + (\frac{1}{3}) d 0$ , so $d 1 = 2$ .
From $2$ : $h 2 = 1 + (\frac{1}{3}) h 1$ , which rearranges to $(\frac{2}{3}) d 2 = 1 + (\frac{1}{3}) d 1$ , so $d 2 = \frac{5}{2}$ .
Then $h 0 = d 0 + d 1 + d 2 = 1 + 2 + \frac{5}{2}$ , which equals $\frac{11}{2}$ .

Answer: $\frac{11}{2}$

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