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I have a time series of daily data that I want to calibrate GBM parameters $\mu$ and $\sigma$ to. Using the discretized solution

$$ S_{t_{i+1}} = S_{t_i}\exp\left(\left(\mu - \frac{\sigma^2}{2}\right)\Delta t + \sigma \sqrt{\Delta{}}Z_{i+1}\right), $$ calibrating the parameters $\mu$ and $\sigma$ to a given time series with $n$ values turns out to be simply computing

$$ \sigma = \frac{std(R)}{\sqrt{\Delta t}}, \qquad \mu = \frac{\mathbb{E}[R]}{t} + \frac{\sigma^2}{2}, $$

where $R$ is a vector of log returns with components $R_{i+1} = \log S_{t_{i+1}} / S_{t_i}$, $1 \leq i \leq n-1$. The term $std(R)$ denotes the standard deviation of $R$.

Now, the time step $\Delta t = t_{i+1} - t_i$ is supposed to be the length of time between values in the series. Recall the closed-form solution to a GBM evaluated at "final" time $T$ is $$ S_T = S_0\exp\left(\left(\mu - \frac{\sigma^2}{2}\right)T + \sigma W(T)\right). $$ So, if I have a time series history of daily prices spanning exactly one year (say 28 Oct 2013 - 28 Oct 2014), what should $T$ and $\Delta t$ be? In addition, $n=253$ in my series, even though the dates cover 365 days.

Some results: using natural gas futures prices with dates given above.

$T = 1$ and $\Delta t = 1/365$, I get $\sigma = 0.32$ and $\mu = 0.07$.

$T = 1$ and $\Delta t = 1/253$, I get $\sigma = 0.27$ and $\mu = 0.05$.

$T = 365$ and $\Delta t = 1$, I get $\sigma = 0.02$ and $\mu = 0.0002$.

$T = 253$ and $\Delta t = 1$, I get $\sigma = 0.02$ and $\mu = 0.0002$ (same as before).

The first two seem more reasonable for my time series. Any thoughts?

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3 Answers 3

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The second one will be the best estimate. Also, a smaller timestep usually corresponds to a smaller bias. But I agree, the answer is not obvious.

You should be careful about increasing $T$ though, because for negative drifts there is a threshold value ($2\mu + \sigma^2 < 0$) beyond which the variance of the price process stops increasing. It's an interesting proof to go over.

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Time is expressed in fractions of year in the GBM formula. Therefore, $T=1$ year and $\Delta t = 1/m$. Considered that you have $253$ observations, I would use $m = 253$, so the second option as Drew suggested.

In general, using 253 or 365 days in a year depends on how you consider reality: do you think that when markets are closed (i.e. weekends) the price evolves? In general, it does: there may be an event when markets are closed that may change the price $S$. In practice, using the number of business days ($253$ in your case) or $365$ does not change much in most of the applications.

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In my case (and I work mostly with natural gas) what I do in the calibration is to use the real value of $\Delta t$ from the historical data, and measure the time in days. In this way, $\Delta t=1$ in most cases, and $\Delta t=1$ in the weekends, so that you take into account the invisible changes in the markets during the weekends.

It also help after the calibration, because counting days when doing simulations get easier, because you can simulate for every day, even if the market is closed, using always $\Delta t=1$. Or, if for example, you need to calculate the time to maturity of an option, you can just subtract the present date from the maturity date.

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