How to estimate the parameters of a geometric Brownian motion (GBM)? It seems rather simple but actually took me quite some time to solve it. The most intuitive way is by using the method of moments.

# Estimation of ABM

First let us consider a simpler case, an arithmetic Brownian motion (ABM). The evolution is given by

$dS = \mu dt + \sigma dW.$

By integrating both sides over $(t,t+T]$ we have

$\Delta \equiv S(t+T) - S(t) = \left(\mu - \frac{\sigma^2}{2}\right) T + \sigma W(T)$

which follows a normal distribution with mean $(\mu - \sigma^2/2)T$ and variance $\sigma^2 T$. That is to say, given $T$ and i.i.d. observations $\\{\Delta_1,\Delta_2,\ldots,\Delta_n\\}$ for different $t$ values, with sample mean

$\hat{\mu}_{\Delta} = \frac{\sum_{i=1}^n\Delta_i}{n}\overset{p}{\to}\left(\mu - \frac{\sigma^2}{2}\right)T$

and modified sample variance

$\hat{\sigma}_{\Delta}^2 = \frac{\sum_{i=1}^n (\Delta_i - \hat{\mu}_{\Delta})^2}{n-1} \overset{p}{\to} \sigma^2 T,$

we have unbiased estimator for $\mu$

$\hat{\mu} = \frac{2\hat{\mu}_{\Delta} + \hat{\sigma}_{\Delta}^2}{2T}$

and for $\sigma^2$ we have

$\hat{\sigma}^2 = \frac{\hat{\sigma}_{\Delta}^2}{T}.$

Now we prove the consistency. First we consider the variance of $\hat{\mu}_{\Delta}$

$Var(\hat{\mu}_{\Delta}) = \frac{Var(\Delta_1)}{n} = \frac{\sigma^2 T}{n}$

and the variance of $\hat{\sigma}_{\Delta}^2$

$Var(\hat{\sigma}_{\Delta}^2) = E(\hat{\sigma}_{\Delta}^4) - E(\hat{\sigma}_{\Delta}^2)^2 = \frac{n E[(\Delta_1-\hat{\mu}_{\Delta})^4] + n(n-1) E[(\Delta_1-\hat{\mu}_{\Delta})^2]^2}{(n-1)^2} - \sigma^4T^2 = \frac{2\sigma^4T^2}{n}.$

The variance of $\hat{\mu}$ is therefore given by

$Var(\hat{\mu}) = \frac{4Var(\hat{\mu}_{\Delta}) + Var(\hat{\sigma}_{\Delta}^2)}{4T^2} = \frac{\sigma^2 (2 + \sigma^2T)}{2nT}$

and the variance of $\hat{\sigma}^2$ is given by

$Var(\hat{\sigma}^2) = \frac{Var(\hat{\sigma}_{\Delta}^2)}{T^2} = \frac{2\sigma^4}{n}.$

So the two estimators are also both consistent. It should be noticed that there exists certain "trade-off" between the efficiency of $\hat{\mu}_{\Delta}$ and $\hat{\sigma}_{\Delta}^2$ by varying the value of $T$.

# Estimation of GBM

For a general GBM with drift $\mu$ and diffusion $\sigma$, we have PDE

$\frac{dS}{S} = \mu dt + \sigma dW,$

so we can integrate the both sides within $(t,t+T]$ for any $t$ and get

$\Delta \equiv \ln S(t+T) - \ln S(t) = \left(\mu - \frac{\sigma^2}{2}\right) T + \sigma W(T).$

The rest derivation is exactly the same.

# Validation

Now we numerically validate this against monte Carlo simulation.

Statistics monte Carlo Method of moment P Value
E(mu_hat) 1.994533e-03 2.000000e-03 0.222191
Var(mu_hat) 4.010866e-07 3.924000e-07 -
E(sigma2_hat) 3.596733e-03 3.600000e-03 0.201573
Var(sigma2_hat) 1.308537e-07 1.296000e-07 -

Now we may safely apply this estimation in application.

1. 1.To assure that $\{\Delta_i\}$ are i.i.d., we need time slots $(t,t+T]$ consecutive and have no overlay. Furthermore, in order to achieve the most efficient estimators for a given $T$, it is clear that we opt for end-to-end slots over the total timespan. ↩︎
2. 2.Although the LHS looks related to $\ln(S)$, it's actually not. We need to use Itō calculus to derive this stochastic integral. ↩︎