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

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1 Answer 1


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$.

  • $\begingroup$ Thank you so much for this step by step explanation, post factum looks quite easy and logical, now everything is clear! $\endgroup$
    – Artemy
    Jul 9, 2021 at 14:31
  • $\begingroup$ Hi: I'm glad it helped. They didn't state the assumption regarding $\frac{\phi}{2}$ which would have made the question clearer but I figure that must be the assumption. $\endgroup$
    – mark leeds
    Jul 10, 2021 at 15:19

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