I wonder if a GARCH model with only "autoregressive" terms and no lagged innovations makes sense. I have never seen examples of GARCH($p$,0) in the literature. Should the model be discarded altogether?
E.g. GARCH(1,0):
$$ \sigma^2_t = \omega + \delta \sigma^2_{t-1}. $$
From the above expression one can derive (by repeated substitution) that
$$ \sigma^2_t \rightarrow \frac{ \omega }{ 1-\delta } $$
for all $t$, if an infinite past of the process is assumed. In other words, GARCH(1,0) implies homoskedasticity and thus the "autoregressive" term, and indeed the whole model, becomes redundant.
Edit:
My argumentation in the paragraph above was imprecise and likely misleading. The point I was trying to make (and John's answer below helped me realize and formulate it better) is that whatever the initial conditional variance is, after a long enough time the conditional variance will stabilize around the level $\frac{ \omega }{ 1-\delta }$. However, it will at the same time obey the law of motion $\sigma^2_t = \omega + \delta \sigma_{t-1}^2$. The two can only be reconciled with $\omega=0$ and $\delta=1$. The latter implies constant conditional variance. Hence, GARCH(1,0) only makes sense when $\omega=0$ and $\delta=1$, which means the whole GARCH model is redundant as the conditional variance is constant.
(End of edit)
Of course, when estimating models in practice, we do not have infinite past; but for long enough time series this approximation should be reasonably representative.
Is this right? Should we never use GARCH($p$,0)?
Best Answer
Why bother with GARCH(1,0)? The $q$ term is easier to estimate than the $p$ term (i.e. you can estimate ARCH($q$) with OLS) anyway.
Nevertheless, my understanding of the way MLE GARCH programs work is they will set the initial GARCH variance equal to either the sample variance or the expected value (that you derive for this case). Without any ARCH terms, the sample variance version would converge to the long-term one (depending on the size of $\delta$). I don't think there would be any change for the expected variance version. So, I'm not sure if you could say it is homoskedastic no matter what (it depends on how you choose the initial variance), but it likely would converge quickly to the expected value for common values of $\delta$.