Deriving a Market Maker’s Gain Variance from Price Jumps

Summary

The answer explains a variance formula for a market maker’s gain per trade using a simple jump model. A price jump has equal probability of being positive or negative, with each direction occurring with probability equal to half the total jump probability. With the remaining probability, no jump occurs. The expected jump amount is therefore zero because the upward and downward outcomes offset, while the no-jump outcome contributes nothing.

Variance is the expected squared deviation from the mean. Since the mean is zero in this setup, it reduces to the expected squared gain: each nonzero outcome contributes the square of the jump size, and the two directional probabilities sum to the overall jump probability. The result relies on symmetric jumps and a zero outcome otherwise; asymmetric probabilities, a nonzero mean, or a different gain definition would require recalculating the expectation and variance.

Key ideas

• The model assigns equal probability to positive and negative jumps and allows no jump otherwise.
• Symmetric jump outcomes make the expected jump amount zero.
• When the mean is zero, variance is the expected squared gain.
• The two nonzero outcomes yield variance equal to jump probability times squared jump size.
• Changing jump probabilities or assumptions can change the variance calculation.

Tags

# Market-maker's gain variance


# Market-maker's gain variance












I am reading the book "Trades, Quotes and Prices" by JEAN-PHILIPPE BOUCHAUD and have stuck in the very beginning with understanding the formula of variance of MM's gain per trade (see picture). How is this formula derived since it is not like a standard variance formula with expected value and mean in it? It is also strange to me that we literally calculate variance for a single variable. Would be very greatful for your answers

## Answer by mark leeds (score 6, accepted)

https://quant.stackexchange.com/a/65929

The probablility of a jump of $J = \phi$. ( in either direction so I'll assume $\frac{\phi}{2} = $ probability of J and $\frac{\phi}{2} = $ probability of -J ). The probability of a jump of $0 = (1-\phi)$.

So, the expectation of the of jump amount, MM,

$ = E(MM) = \frac{\phi}{2} \times J + \frac{\phi}{2} \times -J + (1-\phi) \times 0 = 0$

The variance, $\sigma^2_{MM}$ of the jump amount = $E( MM - 0)^2 = E(MM)^2$.

So, $\sigma^2_{mm}$ becomes $ \frac{\phi}{2} \times J^2 + \frac{\phi}{2} \times (-J)^2 + (1-\phi) \times 0 ^2 = \phi J^2$.

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