# Parameter Estimation of Brownian Motions by Method of Moments

### 2018-05-05

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 $$t$$ values1, 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 integrate2 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.

import numpy as np
import pandas as pd
from scipy.stats import ttest_1samp as ttest

np.random.seed(1)

timespan = 10000
mu = 0.002
sigma = 0.06
T = 50
S0 = 10
n_sim = 20000
sigma2 = sigma**2
mu_hat_list = []
sigma2_hat_list = []

for i in range(n_sim):
# simulate log_S
log_S0 = np.log(S0)
log_St = mu - sigma2 / 2 + sigma * np.random.normal(size=timespan)
log_S = np.cumsum(np.append(log_S0, log_St))
# calculate delta
delta = log_S[T::T] - log_S[:-T:T]
n = len(delta)
# estimate mu and sigma2
mu_delta = delta.mean()
sigma2_delta = ((delta - mu_delta)**2).sum() / (n - 1)
mu_hat = (2 * mu_delta + sigma2_delta) / T / 2
sigma2_hat = sigma2_delta / T
mu_hat_list.append(mu_hat)
sigma2_hat_list.append(sigma2_hat)

E_mu_hat_mc = np.mean(mu_hat_list)
Var_mu_hat_mc = np.var(mu_hat_list)
Var_mu_hat_mm = sigma2 * (2 + sigma2 * T) / n / T / 2
E_sigma2_hat_mc = np.mean(sigma2_hat_list)
Var_sigma2_hat_mc = np.var(sigma2_hat_list)
Var_sigma2_hat_mm = 2 * sigma2**2 / n

result = pd.DataFrame([
[E_mu_hat_mc, mu, ttest(mu_hat_list, mu)],
[Var_mu_hat_mc, Var_mu_hat_mm, '-'],
[E_sigma2_hat_mc, sigma2, ttest(sigma2_hat_list, sigma2)],
[Var_sigma2_hat_mc, Var_sigma2_hat_mm, '-']
], columns = ['monte Carlo', 'Method of moment', 'P Value'],
index = ['E(mu_hat)', 'Var(mu_hat)', 'E(sigma2_hat)', 'Var(sigma2_hat)'])
result

Statistics monte Carlo Method of moment P Value
E($$\hat{\mu}$$) 1.994533e-03 2.000000e-03 0.222191
Var($$\hat{\mu}$$) 4.010866e-07 3.924000e-07 -
E($$\hat{\sigma}^2$$) 3.596733e-03 3.600000e-03 0.201573
Var($$\hat{\sigma}^2$$) 1.308537e-07 1.296000e-07 -

Now we may safely apply this estimation in application.

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. Although the LHS looks related to $$\ln(S)$$, it’s actually not. We need to use Itō calculus to derive this stochastic integral. ↩︎