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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.

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    $\begingroup$ 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

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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.

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  • $\begingroup$ 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

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