# What's the point of (G)ARCH when you can square the residual and use ARMA?

I'm taught that

\begin{aligned} X_t \sim \text{ARCH}(p) & \rightarrow X_t^2 \sim \text{AR}(p) \\ X_t \sim \text{GARCH}(p, q) & \rightarrow X_t^2 \sim \text{ARMA}(\max(p, q), p) \\ \end{aligned}

When I got a time series whose residual appears to be white noise, instead of fitting GARCH on it, I can simply

1. Square the residual.
2. Fit ARMA on the squared residual.
3. Do some prediction, or inference, or whatever with ARMA.
4. "Restore" the residual by taking the square root.

Why do I need a (G)ARCH model then?

## 2 Answers

I see two reasons for preferring the GARCH representation to its equivalent ARMA representation.

1. You typically want to model the conditional distribution or certain distributional characteristics (such as the conditional mean $$\mu_t$$ and the conditional variance $$\sigma^2_t$$) of $$X_t$$ rather than $$X^2_t$$. A GARCH model does that directly by specifying a conditional mean equation for $$\mu_t$$, a conditional variance equation for $$\sigma^2_t$$, and the distribution of the standardized innovations. If your central interest is in the conditional variance of the process, it is more convenient to use the standard GARCH representation.
From GARCH, you get a specific equation for the conditional variance $$\sigma^2_t$$ directly (both the point estimates and the standard errors), while from the equivalent ARMA representation you have to back it up to get it expressed in terms of the objects of central interest ($$\sigma^2_t$$). It should be easy to back up the point estimates, but obtaining the relevant standard errors might be more challenging.
(Thanks to Johan Stax Jakobsen for some elaboration on the topic in the comments to his own answer.)
2. If your process has a nonconstant conditional mean in addition to a nonconstant GARCH-type conditional variance, then you can estimate the model (both the conditional mean part and the conditional variance part) more efficiently if you do it simultaneously rather than doing it in a stepwise fashion (estimating the conditional mean model first, obtaining residuals, and then estimating the conditional variance model).
• Thanks, but I'm a little confused. Are you referring to the "ARMA-GARCH model" by "GARCH model"? As far as I know, GARCH only models the conditional variance of the residual (which is also the conditional variance of $X_t$ since all other terms are deterministic), but it tells us nothing about the conditional mean of $X_t$. I think you're comparing "ARMA-GARCH" with "ARMA-ARMA", where the latter ARMA in ARMA-ARMA models residual squared. – nalzok Jun 11 '19 at 1:57
• @nalzok, when I say A GARCH model does that directly by specifying a conditional mean equation for $\mu_t$, a conditional variance equation for $\sigma^2_t$, and the distribution of the standardized innovations, I am talking about any conditional mean model (constant, ARIMA, any other) plus a GARCH-type equation for the conditional variance plus a distribution of the standardized innovations. A GARCH model must have some equation for the conditional mean because it specifies the whole conditional distribution of $X_t$, including location. – Richard Hardy Jun 11 '19 at 5:46
• @nalzok, When I mention alternative representations, by GARCH I mean the equation for $\sigma_t^2$ while by ARMA I mean the equation for squared raw residuals of the conditional mean model (whatever the conditional mean model is). – Richard Hardy Jun 11 '19 at 5:50

It is correct that, we can obtain an AR(p) representation for $$X_t^2$$ if $$X_t$$ follows an ARCH(p) process and an ARMA(max(p,q),p) representation for $$X_t^2$$ if $$X_t$$ follows a GARCH(p,q) process (see this questions - the generalization to the GARCH case is obvious).

The reason why we are using a GARCH type of model is that we want to model the conditional volatility and not interested in modelling the squared observations. We take the squared observations to be a "realized measure" or signal of volatility.

Assume that we are interested in forecasting, then for the AR(1) model, we have

$$\sigma_{t+1}^2 = E_t[x_{t+1}^2] = w + \alpha E_t[x_t^2] + E_t[x_{t+1}^2 - \sigma^2_{t+1}] = w + \alpha x_t^2$$

Thus, the forecasting equation corresponds to ARCH recursion.

Typically, one sets up the (quasi)-loglikelihood of the GARCH model and maximizes it to get parameter estimates. Your "strategy" suggests a potential way of estimating ARCH type models described in Chapter 6 of Francq and Zakoian's book "GARCH Models: Structure, Statistical Inference and Financial Applications" (2010).

To sum up. The simple answer to your question is that we are using GARCH type models because we want to model the conditional volatility!

• I think this is an interesting question, but I do not quite understand the essence of your answer. If GARCH and the corresponding ARMA are equivalent representations of the same process (are they?) and if there is a one-to-one relationship between the parameters of the two models (is there?), it should not matter much which representation we use for estimation, because we can back up the other representation without a problem. Does this sound logical? – Richard Hardy May 29 '19 at 13:35
• The problems with the OLS approach are that OLS is inefficient (due to the heteroskedasticity in the error term) and that it requires the existence of order 8 for the observed procees. The properties of the standard MLE are often better. Given parameter estimates, the reason why one wants the GARCH representation is that we want to make inference and forecast the conditional variance and not the squared errors. Does this make sense? – Johan Stax Jakobsen May 29 '19 at 19:58
• Sorry, I still do not understand. First, estimation method for a given representation does not seem to be the core issue; I would not use OLS for either GARCH or ARMA since it is not a feasible estimation method in either case (e.g. in ARMA, error terms are unobserved, so you cannot construct the design matrix $X$ which is needed for $\hat\beta=(X^\top X)^{-1}X^\top y$). Second and more importantly, my main question is, if ARMA and GARCH are equivalent representations of the same process, why cannot I make inference and forecast the conditional variance from ARMA if I can do it from GARCH? – Richard Hardy May 30 '19 at 6:34
• The latter seems to go against the notion of equivalent representations. So once again: If GARCH and the corresponding ARMA are equivalent representations of the same process (are they?) and if there is a one-to-one relationship between the parameters of the two models (is there?), it should not matter much which representation we use for estimation, because we can back up the other representation without a problem. – Richard Hardy May 30 '19 at 6:36
• My point is simply that it is natural to focus on the representation that describes the conditional variance directly because we want to model it and not the squared errors. The forecast of $x_t^2$ will be equal to the forecast of $\sigma_t^2$ as far as I can see. – Johan Stax Jakobsen May 30 '19 at 7:26