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

Question

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.

Answer 1

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.