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I got the general PDF of the student t distribution, that is:

$\frac{\Gamma[\frac{(\nu+1)}{2}]}{\Gamma(\frac{\nu}{2})}\,\frac{1}{\sqrt{\pi\,\nu}}\,\bigg[1 + \frac{x^2}{\nu}\bigg]^{-(\nu+1)/2}$

Bollerslev (1987) proposed a t distribution for GARCH estimation that looks like this:

$\frac{\Gamma[\frac{(\nu+1)}{2}]}{\Gamma(\frac{\nu}{2})}\,\frac{1}{\sqrt{(\nu-2)\sigma^2}}\,\bigg[1 + \frac{\epsilon_t^2}{(\nu-2)\sigma^2}\bigg]^{-(\nu+1)/2}$

I suppose that omitting $\pi$ is okay, but what I do not understand is how and why it is possible to replace the $\nu$ with $(\nu - 2)\sigma^2$? What am I missing here?

EDIT: In order not to answer my own question, a quick edit. The solution is to consider the variance of the t-distribution, that is, $\sigma^2 = \frac{\nu}{\nu - 2}$ and then substitute (if I got it right).

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  • $\begingroup$ Have you worked out what the variance of this distribution is? Bollerslev claims it is $\sigma^2$ (explicitly assuming $\nu \gt 2$)--and that's something you can check. (Although omitting a multiplicative function of $\pi$ is not OK in this definition, it can be taken as a typographical error; and omitting such a function in using a log likelihood to perform maximum likelihood estimation is usually harmless.) $\endgroup$
    – whuber
    Mar 20, 2016 at 15:18
  • $\begingroup$ No, I was not able to work it out fully. I am not a trained statistician and new in the field of distributions. Nevertheless, I know that $ \sigma^2 = \frac{v}{v-2}$ is the variance of the t-distribution (with $\nu > 2$). So I suppose some algebraic manipulation together with substitutions took place. Actually this is the solution, right? $\endgroup$
    – Taufi
    Mar 20, 2016 at 17:00
  • $\begingroup$ The check can be done with units analysis alone--it doesn't require any training beyond understanding how a probability distribution represents probabilities. BTW, it's fine to answer your own question. $\endgroup$
    – whuber
    Mar 20, 2016 at 21:24

2 Answers 2

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For GARCH modelling with a t-distribution, we want $y_t$ to be t-distributed with mean $\mu$ and variance $\sigma_t^2$. One way to obtain this is to consider

$$ y_t = \mu + \sigma_t \frac{1}{\sqrt{\frac{v}{v-2}}}T $$ where T is t-distributed with $v$ degree of freedoms. Thus,

$$ T = \frac{y_t - \mu}{\sqrt{\frac{v-2}{v}}\sigma_t} $$

Then, replace x with the above expression in the first formula for the density of a t-distribution (put $\mu = 0$).

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Consider a general GARCH model: \begin{align} r_t&=\mu_t+\epsilon_t \\ \epsilon_t&=\sigma_t u_t \end{align} Usually it is assumed that $E(u_t)=0$ and $V(u_t)=1$. However, if $u_t$ follows a "normal" t-distribution with the pdf \begin{align*} f(u_t;\nu)=\frac{\Gamma\left(\frac{\nu+1}{2}\right)}{\sqrt{\nu \pi} \Gamma\left(\frac{\nu}{2}\right)}\left(1+\frac{u_t^2}{\nu}\right)^\left(-\frac{\nu+1}{2}\right) \end{align*} then $E(u_t)=0$ and $V(u_t)=\frac{\nu}{\nu-2}\neq 1$ for a finite $\nu$. So the idea is to transform this density such that the variance becomes one.

To simplify notation, assume $X$ follows a "normal" t-distribution and we consider the random variabe $Y=\sigma\cdot X$ ($\sigma$ is a constant and not the variance of $X$). Then it is obvious that: \begin{align} E(Y)&=E(\sigma X)=\sigma E(X)=0 \\ V(Y)&=V(\sigma X)=\sigma^2V(Y)=\sigma^2\frac{\nu}{\nu-2} \end{align} If we choose $\sigma=\sqrt{\frac{\nu-2}{\nu}}$, we can calculate that: \begin{align} V(Y)=\left(\sqrt{\frac{\nu-2}{\nu}}\right)^2\frac{\nu}{\nu-2}=\frac{(\nu-2)\nu}{\nu(\nu-2)}=1 \end{align} So the random variable $Y=\sqrt{\frac{\nu-2}{\nu}}X$ has an expected value of zero and unit variance, as desired. Now we want to derive the pdf of Y. Since $\sqrt{\frac{\nu-2}{\nu}}$ is strictly increasing, we get: \begin{align} f_Y(y;\nu)&=\frac{1}{\sigma}f_X\left(\frac{y}{\sigma}\right) \\ &=\frac{1}{\sigma}\frac{\Gamma\left(\frac{\nu+1}{2}\right)}{\sqrt{\nu \pi} \Gamma\left(\frac{\nu}{2}\right)}\left(1+\frac{\left(\frac{y}{\sigma}\right)^2}{\nu}\right)^\left(-\frac{\nu+1}{2}\right) \\ &=\frac{1}{\sigma}\frac{\Gamma\left(\frac{\nu+1}{2}\right)}{\sqrt{\nu \pi} \Gamma\left(\frac{\nu}{2}\right)}\left(1+\frac{y^2}{\sigma^2\nu}\right)^\left(-\frac{\nu+1}{2}\right) \\ &=\frac{1}{\sqrt{\frac{\nu-2}{\nu}}}\frac{\Gamma\left(\frac{\nu+1}{2}\right)}{\sqrt{\nu \pi} \Gamma\left(\frac{\nu}{2}\right)}\left(1+\frac{y^2}{\left(\sqrt{\frac{\nu-2}{\nu}}\right)^2\nu}\right)^\left(-\frac{\nu+1}{2}\right) \\ &=\frac{\Gamma\left(\frac{\nu+1}{2}\right)}{\sqrt{\frac{(\nu-2)\nu \pi}{\nu}} \Gamma\left(\frac{\nu}{2}\right)}\left(1+\frac{y^2}{\frac{\nu-2}{\nu}\nu}\right)^\left(-\frac{\nu+1}{2}\right) \\ &=\frac{\Gamma\left(\frac{\nu+1}{2}\right)}{\sqrt{(\nu-2)\pi} \Gamma\left(\frac{\nu}{2}\right)}\left(1+\frac{y^2}{\nu-2}\right)^\left(-\frac{\nu+1}{2}\right) \\ \end{align} This is the pdf of what is called a "standardized t-distribution". Because $\epsilon_t=\sigma_tu_t$, you get for the conditional distribution: \begin{align} f(\epsilon_t\vert {\cal F}_{t-1})=\frac{\Gamma\left(\frac{\nu+1}{2}\right)}{\sqrt{(\nu-2)\pi} \Gamma\left(\frac{\nu}{2}\right)}\frac{1}{\sigma_t}\left(1+\frac{y^2}{(\nu-2)\sigma_t^2}\right)^\left(-\frac{\nu+1}{2}\right) \end{align} However, it seems like $\pi$ is missing in your notation above and without $\pi$ this is not a valid density. Later when calculating the log-likelihood, this factor can be neglected.

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