Sunday, September 6, 2015

Comparison of Brownian Motion Expected Drawdown and simulated results


Can anyone tell me whether results as predicted by Brownian Motion for a given mean and std, match what you get by measuring actual drawdown from simulated results over a number of iterations?



Answer



Its very simple,


One of Brownian Motion (a.k.a. Wiener process in Mathematics) properties is that each increment from s->t is normally distributed with mean = 0 and sd = t-s.


So, if the process that drives your simulated results is ~N(0, t-s) distributed for each increment s->t with 0<=s<=t then yes, your simulated draw downs should match the ones predicted by a Wiener process (one which is driven by a Brownian Motion). Otherwise, it is not.


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