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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)?

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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 $\delta$.

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  • $\begingroup$ Your explanation makes sense. To make sure I understand it well, I will specify another question. If the initial value was set to the unconditional variance of the sample, would the estimated (fitted) sigmas all be constant and equal to this same unconditional variance? I suspect that this should be the case. This could be derived by hand by solving the maximum likelihood problem, but I am not very confident of my algebra skills... :) $\endgroup$ – Richard Hardy Aug 26 '14 at 19:46
  • $\begingroup$ The MLE starts with a log-likelihood function to be minimized based on parameters. To calculate the log-likelihood function in each step of the minimization requires updating the conditional variance, which depends on the selection of the initial variance and the parameters in the minimization. The first time you run the MLE (given your stated choice of the initial variance), it may be that the conditional variance is not homoskedastic. However, it might be that when the MLE is done, then $\delta$ goes near zero (b/c it doesn't matter much) and $w$ goes to the unconditional variance. $\endgroup$ – John Aug 26 '14 at 19:55
  • $\begingroup$ In that case you might be right, but I'm not sure if I can prove it. $\endgroup$ – John Aug 26 '14 at 19:55
  • $\begingroup$ Thanks! Your post "The MLE starts with..." is quite enlightening! I now recognize that GARCH model will be estimated without computer having solved it algebraically. Perhaps algebraic solution yields the outcome I am proposing but the computer will not follow this logic - it will rather engage in some sort of optimization instead. $\endgroup$ – Richard Hardy Aug 26 '14 at 20:25
  • $\begingroup$ I now recognize that the answer to my original question depends on algebraic derivation rather than discussion of how a computer program will solve this kind of problem. Whatever answer a computer will arrive to for a given time series, it still might not make sense if one could algebraically prove that the MLE of GARCH(p,0) model yields all sigmas to be equal to the same constant (which means conditional homoskedasticity). Perhaps I should try my luck in solving this algebraic problem after all... :) $\endgroup$ – Richard Hardy Aug 26 '14 at 20:30
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GARCH($p$,0) can generate the following patterns of the conditional variance $\sigma_t^2$:

  1. If $\omega=0$ and $\delta>1$, $\sigma_t^2$ explodes exponentially (a bit slower).
  2. If $\omega>0$ and $\delta>1$, $\sigma_t^2$ explodes exponentially (a bit faster).
  3. If $\omega=0$ and $\delta=1$, $\sigma_t^2$ is constant and equals the unconditional variance $\sigma^2$, $\sigma_t^2 \equiv \sigma^2$.
    This is the case where the GARCH($p$,0) model is redundant.
  4. If $\omega>0$ and $\delta=1$, $\sigma_t^2$ grows linearly at the speed of $\omega$.
  5. If $\omega=0$ and $\delta<1$, $\sigma_t^2$ converges to zero.
  6. If $\omega>0$ and $\delta<1$, $\sigma_t^2$ converges to $\frac{\omega}{1-\delta}$.

enter image description here

Whether any of the patterns makes sense depends on the applications.
Nevertheless, there seems to be nothing wrong with the model definition per se, except perhaps for case 3. where the conditional variance equation is redundant.

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If GARCH(p, 0) is "redundant", then why did the authors of GARCH put together such a queer model statement? Apparently, the GARCH part is "redundant" only when both p and q are equal to zero. That's a common sense explanation.

The problem is you are confusing the conditional and unconditional variances. Any GARCH(p, q) process is a stationary process with a constant unconditional variance that can be computed similarly to how you got the limit for $ \sigma^2_t$. The value of that limit answers a question: if one selects $y_t$ at random, what is its variance going to be?

Conditional variance answers a question: if one knows the entire process history upto $(t-1)$, inclusive, then what is the variance of $y_t$ going to be? Apparently, given the relationship:

$$ \sigma^2_t = \omega + \delta \sigma^2_{t-1} $$

you will have a much better estimate for the variance of $y_t$ in the second case. On the other hand, if it were a “regular” ARMA(p,q) process, the answer would be exactly the same in both cases.

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  • $\begingroup$ I think I follow the overall logic of your post. However, I am confused in some details. Note that conditional variance is never observed. Therefore it is not straightforward to obtain $\sigma^2_{t-1}$ so as to be able to forecast $\sigma^2_t$ using the equation above. In practice, all sigmas have to be estimated via maximum likelihood. I am trying to see whether the exact maximum likelihood estimation (where initial value is treated as a parameter) would yield $\sigma^2_t$ to be constant for all t. $\endgroup$ – Richard Hardy Aug 26 '14 at 19:33
  • $\begingroup$ Regarding authors putting together a queer model: actually, I have never seen a GARCH(p,0) model in the literature (neither in textbooks nor in research articles). Therefore I am asking the question - I want to clarify whether such a model makes sense at all. $\endgroup$ – Richard Hardy Aug 26 '14 at 19:38
  • $\begingroup$ Of course it does, just like AR(p) does. It may be not very practical, but that's beside the point. $\endgroup$ – James Aug 26 '14 at 19:47
  • $\begingroup$ The conditional variance always depends on t. If it doesn't, it's GARCH(0, 0). $\endgroup$ – James Aug 26 '14 at 19:54
  • $\begingroup$ You might be missing one important point: unlike ARMA model, GARCH model does not have its own error term. That is, according to GARCH model, conditional variance of today is precisely (!) determined by conditional variance of yesterday (and innovations of the past - if we include ARCH terms in the GARCH equation, as is common). Meanwhile, in AR(p) models, value of today is imprecisely determined by value of yesterday due to an extra error term in AR(p) model. Thus AR(p) making sense does not lead to GARCH(p,0) making sense immediately. $\endgroup$ – Richard Hardy Aug 26 '14 at 20:16

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