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).

  • 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
    Commented Feb 15, 2023 at 18:23
  • $\begingroup$ Hint: what's the joint distribution of the increments $X_{t_i}-X_{t_{i-1}}$? $\endgroup$
    – Chris Haug
    Commented Feb 15, 2023 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
    Commented Feb 15, 2023 at 18:53

1 Answer 1


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 $\sigma$.


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