Expectation of folded Student's-t

[AlexanderMcFarlane]

The EGARCH model is given by,

A detailed summary can be found in the Matlab documentation here.

If the innovation distribution is normal then,

I've figured out that this is true because the expectation is that of the folded normal distribution which is described on the wikipedia page here (just set the mean and standard deviation to zero and it simplifies).

I'm therefore assuming that if the EGARCH innovation distribution is student's-t, then the following relation,

must be the expectation of the folded Student's-t distribution.

I am wondering how you would go about deriving this assuming the standardised Student's-t distribution is given by:

From what I can guess, this is two parts:

1. Calculate analytic form of folded Student's-t
2. Calculate (\int_{-{\infty}}^{+{\infty}} xf(x) dx) for the expectation value

and I am stumped by the first part.

 

[ferdia]

I don't think you need the analytic form of the folded Student's t-distribution, similarly to how you don't need the analytic form of the folded normal distribution to calculate the expected absolute value. If $X\sim f$ and $g$ is some function (which is sufficiently nice, etc.) then the expectation of $g(X)$ is given by the law of the unconscious statistician:
$\mathbf E[g(X)]=\int\limits_{-\infty}^\infty g(x)f(x)\,\mathrm dx$
It is not too difficult to calculate the pdf of the folded distribution though: the distribution is symmetric, so the pdf of the folded distribution is just twice the pdf of the original one for nonnegative arguments and $0$ for negative arguments. More generally the pdf of a folded distribution will be $f+f\circ P$ for positive arguments and $0$ for negative arguments, with $f$ the original pdf and $P$ parity change (multiplication by $-1$), i.e.
$f_\text{folded}(x)=\begin{cases}0\quad &\text{for}\quad& x<0\\f(x)+f(-x) \quad&\text{for}\quad & x\geqslant 0\end{cases}$
You may see this intuitively or with a proof: Let $X\sim f$ with $f$ a continuous pdf and cdf $F$. Then we have that the cdf of $|X|$ is given by
$F_\text{folded}(x)=\mathbf P[|X|\leqslant x]=\mathbf P[-x\leqslant X\leqslant x]=\begin{cases}0\quad&\text{for}\quad&x<0\\F(x)-F(-x)\quad&\text{for}\quad& x\geqslant 0 \end{cases}$
Differentiating we get that the pdf of $|X|$, the folded pdf, is given by
$f_\text{folded}(x)=\begin{cases}0\quad &\text{for}\quad&x<0\\F'(x)--F'(-x)=f(x)+f(-x)\quad &\text{for}\quad&x\geqslant 0\end{cases}$

If $z_{t-j}$ is distributed according to the standardised Student's t-distribution and $f_\nu$ denotes the pdf of the standardised Student's t-distribution with $\nu$ degrees of freedom, the expectation you are looking for is simply given by

 

[AlexanderMcFarlane]

ok thanks, I've never done much math with the gamma function but according to this article, the recursion relationship with the gamma function is:

$\Gamma\big(n+\frac{1}{2}\big) = \big(n-\frac{1}{2}\big)\Gamma\big(n-\frac{1}{2}\big)$​

Therefore, from your second to last step you would get:

$... = \frac{2}{(v-1)}\sqrt{\frac{\nu}{\pi}}\frac{\Gamma\big(\frac{\nu+1}{2}\big)}{\Gamma\big(\frac{\nu}{2}\big)}$

$= \frac{2}{(v-1)}\frac{(v-1)}{2}\sqrt{\frac{\nu}{\pi}}\frac{\Gamma\big(\frac{\nu-1}{2}\big)}{\Gamma\big(\frac{\nu}{2}\big)}$

$= \sqrt{\frac{\nu}{\pi}}\frac{\Gamma\big(\frac{\nu-1}{2}\big)}{\Gamma\big(\frac{\nu}{2}\big)}$​

The standardised t distribution I posted has a factor of $\frac{1}{\sqrt{(\nu-2)\pi}}$ instead of $\frac{1}{\sqrt{\pi\nu}}$ at the front and $\sqrt{\nu-2}$ instead of $\nu$ inside the integral...
I think from looking at it the $\nu$ that popped out half way through when you did a change of variables will instead pop out as $\nu-2$ and give the following:

$... = \frac{2}{(v-1)}\frac{1}{\sqrt{\pi(\nu-2)}}\frac{\Gamma\big(\frac{\nu+1}{2}\big)}{\Gamma\big(\frac{\nu}{2}\big)}$

$= \sqrt{\frac{(\nu-2)}{\pi}}\frac{\Gamma\big(\frac{\nu-1}{2}\big)}{\Gamma\big(\frac{\nu}{2}\big)}$
​

which equals the equation given by the Matlab lot!

Thanks for the heads up on the folding of the p.d.f.!

 

[ferdia]

Indeed you are right about the recurrence relation for $\Gamma$ and that if you take the pdf as you said, it works out right. Also I see now that you are right in using the pdf that you used, as it is, like you said, standardised, i.e. expectation $0$, variance $1$, which is what one naturally demands of the innovations.

 

[AlexanderMcFarlane]

when using the student's-t distribution in volatility models you need to modify it to include a scale parameter, $\gamma$:

\(f(x|\nu,\sigma) = \frac{\Gamma\large(\frac{\nu+1}{2}\right)}{\sqrt{\nu\pi\gamma^2}\Gamma\large(\frac{\nu}{2}\right)}\large(1+\frac{x^2}{\nu\gamma^2}\right)^{-\frac{\nu+1}{2}}\)​

the scale parameter is related to the variance by:

$\sigma^2=\gamma^2\frac{v}{v-2}\ for\ v>2$
​

Then if you sub for the variance, $\sigma^2$:

\(f(x|\nu,\sigma) = \frac{\Gamma\large(\frac{\nu+1}{2}\right)}{\sigma\sqrt{\pi(\nu-2)}\Gamma\large(\frac{\nu}{2}\right)}\large(1+\frac{x^2}{(\nu-2)\sigma^2}\right)^{-\frac{\nu+1}{2}}\)​

the innovations should have unit standard deviation because you specify the volatility in the volatility equation so just eliminate the standard deviation by setting it equal to 1.

(I got this all from wikipedia and a book by Carol Alexander "Practical Financial Econometrics", part II of the "Market Risk Analysis" series)

 

[ferdia]

Yes thanks for the effort but in the meanwhile I had realised that as well. Good luck with your program