0
$\begingroup$

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).

$\endgroup$
3
  • 1
    $\begingroup$ 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. $\endgroup$
    – Zhanxiong
    Feb 15 at 18:23
  • $\begingroup$ Hint: what's the joint distribution of the increments $X_{t_i}-X_{t_{i-1}}$? $\endgroup$
    – Chris Haug
    Feb 15 at 18:31
  • $\begingroup$ 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? $\endgroup$
    – Landscape
    Feb 15 at 18:53

1 Answer 1

3
$\begingroup$

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 $\sigma$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.