# Maximum likelihood estimation when parameters are functions of another data series

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.

-
Are you familiar with a good math programming language, like R? –  jbowman Jan 4 '12 at 19:50
This looks like the discretization of an affine diffusion (SDE) or something close. –  cardinal Jan 4 '12 at 20:10
@jbowman, sadly I don't know R :( –  Grzenio Jan 5 '12 at 9:47
@cardinal, indeed it is –  Grzenio Jan 5 '12 at 9:48

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.
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/… –  Grzenio Jan 5 '12 at 16:08
Not in the first expression. Since $\epsilon_t$-s are independent normally distributed with zero mean and unit variance. –  vinux Jan 5 '12 at 16:53