We have two time series: $X_t$ and $R_t$, and a model saying that $R_{t+1} = (\mu(X_t) - \frac{1}{2}\sigma^2(X_t))\Delta T + \sigma(X_t) \sqrt{\Delta T} \epsilon_t$, where $\Delta T$ is given constant and $\epsilon_t$-s are independent normally distributed with zero mean and unit variance. Further we assume that the functions $\mu(x)$ and $\sigma(x)$ are linear for simplicity. I would like to use some standard method (MLE comes to my mind) to estimate parameters of functions $\mu(x)$ and $\sigma(x)$, but I am not sure how to do this.

I would be grateful for detailed answers, because I am not really experienced with statistics.

  • 1
    $\begingroup$ This looks like the discretization of an affine diffusion (SDE) or something close. $\endgroup$
    – cardinal
    Jan 4, 2012 at 20:10
  • $\begingroup$ @cardinal, indeed it is $\endgroup$
    – Grzenio
    Jan 5, 2012 at 9:48

1 Answer 1


Let $\theta$ be the parameters involved in $\mu(x)$ and $\sigma(x)$.

Your likelihood function will be $$ \mathcal{L}(\theta\,|\,\epsilon_1,\ldots,\epsilon_n) = f(\epsilon_1,\epsilon_2,\ldots,\epsilon_n\;|\;\theta) = \prod_{t=1}^n f(\epsilon_t|\theta)= \prod_{t=1}^{n} \frac{1}{\sqrt{2\pi}\ } \exp\big(-\epsilon_t^2/2\big) \>. $$ You may need to take $t=1$ to $n-1$ (for a large sample it doesn't matter, assuming you have $n$ observations).

Substitute $$ \epsilon_t=\dfrac{R_{t+1} - (\mu(X_t) - \frac{1}{2} \sigma^2(X_t))\Delta T}{\sigma(X_t) \sqrt{\Delta T}} \>. $$

This will be in terms of $\theta$, $R_t$, and $X_t$. MLE estimates are the parameters which optimize the likelihood function found above.

  • $\begingroup$ I think there is $1/\sigma(X_t)$ missing in the likelihood function. In the limiting case when the parameters are constant we simply have a normal distribution for which Wikipedia gives a different anwswer: en.wikipedia.org/wiki/… $\endgroup$
    – Grzenio
    Jan 5, 2012 at 16:08
  • $\begingroup$ Not in the first expression. Since $\epsilon_t$-s are independent normally distributed with zero mean and unit variance. $\endgroup$
    – vinux
    Jan 5, 2012 at 16:53

Your Answer

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

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