# How to calculate mean and volatility parameters for Geometric Brownian motion?

Say I have a time series $$S_K$$ for monthly asset prices for the last 30 years. I want to run a monte carlo simulation using geometric brownian motion

$$S_t = S_0\exp\left(\left(\mu - \frac{\sigma^2}{2}\right)t + \sigma W_t\right)$$

In my monte carlo simulation, I plan to use a time increment $$dt=\frac{1}{12}$$ to simulate 1 month increments.

What is the mean $$\mu$$ and volatility $$\sigma$$ that should be used in the calculation? Intuitively, using the long term (30 year) mean and standard deviation seem incorrect as the simulation will have 1 month time steps, so I'm unsure what values to use.

If you want to rely on historical values at all (as opposed to a forward curve and implied volatilities), then $$\mu$$ would be the annualized exponential growth rate measured over a period T, calculated as $$\mu=\frac{ln(S_{T}/S_{0})}{T}$$ (where T is measured in years), and $$\sigma$$ would be the annualized volatility, determined as the variance of log-returns over a period of N days, annualized with a factor of $$\sqrt{N_{trade}/N}$$, where $$N_{trade}$$ is the number of trading days per year (frequently taken as 252):
$$\sigma=\frac{\sqrt{N_{trade}/N}}{\sqrt{n-1}}\sqrt{\sum_{i=1}^{n}ln^{2}(\frac{S_{i+N}}{S_{i}})}$$
Perhaps worth mentioning the reason for omission of the average $$\mu$$ of log returns in above formula - $$\mu$$ is typically much smaller than the standard deviation $$\sigma$$ of log returns. Using these formulae, you don't have to worry about the length of the observation interval, as long as you set $$T$$ and $$N$$ correctly.
• Thanks for the explanation. What is $n$ in the volatility term? – PyRsquared Jan 31 '19 at 17:13