A market maker receives 8 orders in sequence. Each order is independently a buy with probability 1/4 and a sell with probability 3/4. Count the times that consecutive orders are on opposite sides. What is the expected value of this count? Give a fraction a/b in lowest terms.

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A market maker receives $8$ orders in sequence. Each order is independently a buy with probability $\frac{1}{4}$ and a sell with probability $\frac{3}{4}$ . Count the times that consecutive orders are on opposite sides. What is the expected value of this count? Give a fraction a/b in lowest terms.

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Hint

Compute the chance that one consecutive pair differs, then sum over all pairs.

Solution

There are $7$ consecutive pairs of orders.
A pair differs if it is buy then sell or sell then buy, with probability $\frac{1}{4} \times \frac{3}{4} + \frac{3}{4} \times \frac{1}{4} = \frac{3}{8}$ .
Each pair has an indicator with expectation $\frac{3}{8}$ , and linearity does not need independence between pairs.
The total is $7 \times \frac{3}{8} = \frac{21}{8}$ .

Answer: $\frac{21}{8}$

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