Interpretation of Drift

Question

Consider the common model of stock prices given by a geometric Brownian motion (GBM), which follows the SDE $$ dS(t) = \mu S(t) dt + \sigma S(t) dW(t). $$

Below is a plot of a simulation of such a GBM using $N = 1000$ points starting at $S(0) = 100$ and volatility $\sigma = 0.2$. What would you think the drift $\mu$ was set to?

If you were to do an MLE estimate of $\mu$ and $\sigma$ on this time series you would get $\mu = 0.5054$ and $\sigma = 0.1966$. In fact, this simulation was run with $\mu = 0$.

It's clear to me that I don't have a good intuition for $\mu$, as I would be tempted to agree with the MLE estimates. How should I be thinking of $\mu$?

I think this has important implications for understanding calibration in practice. I've done an MLE estimate to a GBM because I believe the process I'm trying to model follows a GBM to some extent - in this case, it IS a GBM, so there is no "modelling error". My "best guess" of its parameters are $\mu = 0.5054$ and $\sigma = 0.1966$. But, in fact, $\mu = 0$. That is, my GBM model for a GBM with drift $0$ has drift $0.5054$. What.

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Some requested details: Set $R_i = \log \left( \frac{S(t_i)}{S(t_{i-1})} \right)$, $i = 1,\ldots,N$. The likelihood function $L$ for the data $\{R_i\}$ is \begin{align*} L(\mu, \sigma) = \prod_{i=1}^N \phi(R_i; \left(\mu - \frac{\sigma^2}{2}\right)\Delta t, \sigma \sqrt{\Delta t}) \end{align*} where $\phi(x; m,s)$ is the pdf for a $\mathcal{N}(m,s^2)$-distributed random variable.

The MATLAB code I'm using:

% Simulates a GBM(mu,sigma) and plots the path.
% Here, GBM(mu,sigma) ~ log N((mu - sigma^2/2)*t, sigma^2*t)

% ----------------- Simulation (Generation of Data) ---------------------

%------------ set up parameters ----------------
N=1000; % number of points in path
mu=0;  % drift
sigma=0.2;  % vol
S0=100;  % initial value of process
T=1;  % final time
dt=T/N;  % time step

%------------ allocation & initialization ----------------
t=linspace(0,T,N+1);  % time axis values
S=zeros(1,N+1);  % allocate vector of process values
S(1)=S0;  % assign initial value to process' first value
Z=normrnd(0,1,1,N);  % std normals

%------------ compute paths ----------------
for i=2:N+1
    S(i)=S(i-1)*exp((mu - sigma^2/2)*dt + sigma*sqrt(dt)*Z(i-1));
end

% create plot of process
plot(t,S);



% ----------------- Parameter Estimation ----------------------
% create vector of log returns
R=zeros(1,N);
for i=2:N+1
  R(i-1) = log(S(i)/S(i-1));  
end

%------------ MLE Estimates ----------------
mle_est = mle(R,'distribution','normal');
theta(2) = sqrt(mle_est(2)^2/dt);  % sigma
theta(1) = mle_est(1)/dt + 0.5*theta(2)^2;  % mu

disp('MLE Estimates:')
disp(sprintf('mu = %f, sigma = %g',theta(1), theta(2)));

Answer 1

You generated only one realization of the GBM. The variance of a GBM increases with time and so you must generate more realizations to get accurate estimates.

Here see a sample of Brownian simluations: https://tex.stackexchange.com/questions/59926/how-to-draw-brownian-motions-in-tikz-pgf

Answer 2

This is related to a common misconception called the Time Diversification fallacy. Returns do not average out over longer periods of time. On the contrary, as emcor points out, the variance increases.

You might find this article named Risk and Time by John Norstad interesting.