Variance of Market-Maker Gains with Symmetric Informed Trades

Summary

The document examines a variance calculation for a market maker’s gain per trade when some trades are informed. The questioner argues that the variance should subtract the squared mean, while the quoted expression appears to omit that term.

The answer models the gain as a random variable that is positive or negative by the jump size for informed trades, with equal probabilities for each direction, and zero for uninformed trades. Under that symmetric outcome distribution, the expected gain is zero, so the variance equals the probability of an informed trade multiplied by the squared jump size. This resolves the discrepancy by making the directional assumption explicit. The result depends on symmetry and zero expected gain; a distribution with an imbalance between upward and downward outcomes would require computing variance from both its first and second moments.

Key ideas

• Represent the per-trade gain as a random outcome that can be positive, negative, or zero.
• The answer assumes informed trades predict jumps in either direction with equal probability.
• Symmetry makes the expected gain zero, so the variance is the second moment.
• If upward and downward outcomes are not balanced, the mean must be included when calculating variance.

Tags

# Is the variance calculation correct in the book?


# Is the variance calculation correct in the book?












I'm reading the book "Financial Markets Under the Microscope" for my market microstructure studies. In the book, the variance of the market maker's gain is calculated as follows:

> Assume that with probability $φ$, an arriving trade is informed, and correctly predicts a price jump of size $±J$. Otherwise (with probability $1 − φ$), the arriving trade is uninformed and does not predict a future price change. The variance $σ^2$ of the market-maker’s gain per trade is then given by $σ^2 = (1 − φ) × 0 + φ × J^2$

But I calculate the variance as $σ^2 = φ × J^2 - φ^2 × J^2$ which is different from the book.

Do I miss something?

## Answer by Pontus Hultkrantz (score 2)

https://quant.stackexchange.com/a/54437

As I see it, a reasonable possibility is that it is a directional bet; money will only be made if the direction of the jump is correct, else the bet is lost.

So if the random outcome is $X \in \{-J, 0, J\}$ with

$\mathbb{P}(+J) = \mathbb{P}(-J) = \varphi /2 \;$ such that $\mathbb{P}(\pm J) = \varphi$,

then it follows that $\mathbb{E}[X] = 0$, and $\mathbb{V}[X] = \varphi \times J^2$. Q.E.D.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.