| I don't really get it either so here's an attempt to justify it with gambler's ruin: http://mathworld.wolfram.com/GamblersRuin.html > Let two players each have a finite number of pennies (say, n_1 for player one and n_2 for player two). Now, flip one of the pennies (from either player), with each player having 50% probability of winning, and transfer a penny from the loser to the winner. Now repeat the process until one player has all the pennies. > The chances P_1 and P_2 that players one and two, respectively, will be rendered penniless are > P_1 = n_2 / (n_1 + n_2) > P_2 = n_1 / (n_1 + n_2) So if the current order book looks like: - Sell 5 for $99.00 (best offer) - Buy 10 for $98.75 (best bid) And you chomp up orders randomly, flipping a completed order to the opposite position: - the probability that the price is above $99.00 (all 5 sell got filled first) is 10 / (5 + 10) - the probability that the price is below $98.75 (all 10 buy got filled first) is 5 / (5 + 10) So expected value is (99.00 * 10 + 98.75 * 5) / (10 + 5) = 98.9167. I am not happy with this explanation. It seems like after you execute an buy for $98.75 you need to immediately put a sell for it at slightly above $98.75 (and vice versa) to fit the random walk model described above. And then I took the expected value using the price at two different points in time. Overall I am still confused myself. |