Parameters for numerically fitting t-distribution to log-returns

Question

I am fitting the t-distribution to log-returns numerically (not using R, MATLAB, Stata, etc.), but rather using general programming. Assuming the log-return values are $r_t$, and the $t$-variates are $x_t$, then given $r_t=ax_t + b$, and knowing the variance of t-distribution, $\nu/(\nu-2)$, the variance of $r_t$ becomes $\sigma^2(r_t)=a^2(\nu/(\nu-2))$. Would the parameters to fit for the $r_t$ then be $b$, the location, and $a=\sqrt{\sigma^2(r_t)/(\nu/(\nu-2))}$, where $\sigma(r_t)$ is the measured standard deviation of the observed log-return?

The solution would be to first initialize $a=1$ and $b=0$ and then calculate

$$ \hat{r_t}=x_t=\frac{r_t-b}{a}=\frac{r_t-b}{\sqrt{\frac{\sigma^2(r_t)}{\frac{\nu}{\nu-2}}}} $$

with minimization of $MSE=\frac{1}{T}\sum_t (r_t-\hat{r}_t)^2$ ?

Answer 1

It could be much more simple: if you use the method of moments (MM) then you estimate the mean and the variance and for example the kurtosis of your sample. Then you fit the parameters to these statistics. Alternatively you use maximum-likelihood (MLE).

For MM: from wikipedia you get the mean and the variance. In your notation you can fit $b = \bar{r}$ so $b$ equals the empirical mean. The excess kurtosis is defined here and you can solve for the parameter $\nu$ from the estimated kurtosis of $r_t$ without estimating its volatility (if I am not mistaken this cancels out due to the definition). Then you go back and estimate your $a$ from the sample variance and your fitted $\nu$.

For the MLE you would need to write down the density of $r_t$.

Another comment to your proposed method. It is not that far from what I propose. But you don't use higher moments in order to estimate $\nu$. Using the density (MLE) or the kurtosis (MM) you go beyond modelling mean and variance only.