# Estimate parameters in Brownian Motion with drift, $dX_t = \mu dt + \sigma dW_t$

Consider a Brownian Motion with drift, $$X$$, on the interval $$[0; T]$$ given by $$dX_t = \mu dt + \sigma dW_t.$$

Suppose that the interval is split into $$n$$ pieces of equal size to define $$\Delta:=T/n$$ at time points $$t_i, \; i=0,...,n$$. Assume we have observed $$X$$ at these points. That is, we have a sequence of observed values $$(x_i)_{i=0,...,n}$$.

I want to estimate $$\mu$$ and $$\sigma$$ from the data. How can I do this?

My first approach is to look at the distribution of $$X_t$$. It is $$X_t \sim N(\mu t, \sigma^2 t)$$ as I use that $$W_t \sim N(0, t)$$. However, this does not seem to help me estimate $$\mu$$ and $$\sigma$$ - as I would use Maximum Likelihood Estimation (MLE) to find the mean and variance of $$X_t$$ (but these are not even identically distributed... they are time-dependtent).

• MLE doesn't only apply to i.i.d. data, as long as you can formulate the likelihood function, then you can always use it. Feb 15 at 18:23
• Hint: what's the joint distribution of the increments $X_{t_i}-X_{t_{i-1}}$? Feb 15 at 18:31
• It is normal, with constant coefficients. And then I could use MLE to get estimates of those - but these are not the same af $\mu$ and $\sigma$. How can proceed from here? Feb 15 at 18:53

It is well known that (note that $$\{W_t\}$$ by definition is a Gaussian process) for $$0 < t_1 < \cdots < t_k$$, the joint density of $$(W_{t_1}, \ldots, W_{t_k})$$ is (where $$t_0 = w_0 = 0$$) \begin{align} f_{t_1\cdots t_k}(w_1, \ldots, w_k) = \prod_{i = 1}^k\frac{1}{\sqrt{2\pi(t_i - t_{i - 1})}} \exp\left[-\frac{(w_i - w_{i - 1})^2}{2(t_i - t_{i - 1})}\right]. \end{align}
Since the transformation $$\mathbf{X} = \mu\mathbf{t} + \sigma\mathbf{W}$$ is affine (where $$\mathbf{W} = (W_{t_1}, \ldots, W_{t_k})$$, $$\mathbf{X} = (X_{t_1}, \ldots, X_{t_k})$$, $$\mathbf{t} = (t_1, \ldots, t_k)$$), the joint density of $$(X_{t_1}, \ldots, X_{t_k})$$ is then given by (where $$t_0 = x_0 = 0$$): \begin{align} & g_{t_1\cdots t_k}(x_1, \ldots, x_k) \\ =& \frac{1}{\sigma^k} \prod_{i = 1}^k\frac{1}{\sqrt{2\pi(t_i - t_{i - 1})}} \exp\left[-\frac{((\sigma^{-1}(x_i - \mu t_i) - \sigma^{-1}(x_{i - 1} - \mu t_{i - 1}))^2}{2(t_i - t_{i - 1})}\right] \\ =& \frac{1}{\sigma^k} \prod_{i = 1}^k\frac{1}{\sqrt{2\pi(t_i - t_{i - 1})}} \exp\left[-\frac{(x_i - x_{i - 1} - \mu(t_i - t_{i - 1}))^2}{2\sigma^2(t_i - t_{i - 1})}\right]. \end{align}
This means that given data $$x_1, \ldots, x_k$$ observed at $$0 < t_1 < \cdots < t_k$$, the log-likelihood function of $$(\mu, \sigma)$$ is \begin{align} -k\log\sigma - \frac{1}{2}\sum_{i = 1}^k\log(2\pi(t_i - t_{i - 1})) - \frac{1}{2\sigma^2}\sum_{i = 1}^k((x_i - x_{i - 1} - \mu(t_i - t_{i - 1}))^2. \tag{1} \end{align}
From $$(1)$$ it is easy to determine the MLE of $$\mu$$ and $$\sigma$$.