$v$ should be the total number of parameters (constants + AR + MA + GARCH + ARCH).
I disagree with @kiwiakos, the student t(df) distribution is used because we are using standard errors which are estimates of standard deviations (and not true standard deviations) to compute the statistic. That is the reason why we use student t test eventhose the asymptotically parameter distribution is Gaussian.
The formula for the p-value is then :
Example :
Coeff = 0.15
Std Error = 0.064 (or robust std errors)
The t-statistic is : 0.15/ 0.064 = 2.3438 (not p-statistic because we use standard error!)
t-prob = p-value = 2*(1-tcdf( | 2.3438 | ,n-v)) ( p-value with a student dist and not p-value with a normal dist !)
EDIT
This is not specific to GARCH parameters but to the theory of tests statistics.
The main idea :
We should use the z-test only if there is no uncertainty regarding the
population variance. However this is rarely the case so the p-value are obtained using the student t distribution. If the sample size is large enough the normal distribution (i.e the z test) can be used.
Let's X to be a gaussian random variable, and let $\bar{X}$ to be the sample mean.
The standardized statistic z is normally distributed ( see the central limit theorem) and is given by:
$z = (\bar{x} - m_{0} )/ \sigma_{\bar{X}} $
where $\sigma_{\bar{X}} $ is the standard deviation of the sample mean. We can compute this standard deviation based on the population variance $\sigma_{X}^{2}$ and it is given by :
\begin{equation*}\begin{aligned}
\sigma_{\bar{X}}^{2} &=& var\{\dfrac{1}{n}(X_{1}+\ldots + X_{n})\}\\
&=&\dfrac{1}{n^2} var\{( X_{1}+\ldots + X_{n} )\}\\
&=&\dfrac{1}{n^2} (\sigma_{X}^{2}+ \ldots+ \sigma_{X}^{2})\\
&=&\dfrac{1}{n^2} (n\sigma_{X}^{2})\\
&=& \frac{\sigma_{X}^{2}}{n}
\end{aligned}\end{equation*}
However since we don’t know the population variance $\sigma^{2}_{X} $ we need to employ an estimation for it and this introduce some uncertainty. To approximate the population variance $\sigma_{X}^{2}$, (generally) we employ the sample variance given by :
\begin{equation*}
s_{X}^{2}=\dfrac{1}{n-1}\sum_{i=1}^{n}(X_{i}-\bar{X})^{2}
\end{equation*}
Then the variance of the sample mean is approximated as follow :
$\widehat{ \sigma^{2}_{\bar{X}}} = \frac{s_{X}^{2} }{ n}$
(instead of $ \frac{\sigma_{X}^{2}}{n}$ )
So the standards errors (which takes into account this uncertainty) are :
$ se = \sqrt{\widehat{ \sigma^{2}_{\bar{X}}} } = \frac{s_{X} }{\sqrt{n}}$
the z test is modified in the following way:
$z = \frac{ \bar{x} - m_{0} }{ \sigma_{\bar{X}} } $
becomes :
$t = \frac{ \bar{x} - m_{0} }{ se } $
and $t$ will be student t distributed with (n-number of parameters) degrees of freedom (see here ).
The fact that we employ the standards errors instead of the standard deviation make the Z test to be a T test . When we use the MLE method, it is the same , we can replace X as the estimated parameters (wich are asymptotically gaussian random variables). We don't employ their true standards deviations but we use an estimation of it. The Hessian gives us the standards errors , not the standards deviations (because there is uncertainty - we don't observe the population but a sample). So we should employ the $t$ test.
Note that, when $n$ (the number of observation ) increases the student t distribution becomes closer to the normal distribution and then p values obtained with the normal distribution or student t distribution will becomes the same ( cf @John comment). This is logical because the uncertainty decreases as the number of observations increases.