My understanding of ARMA-GARCH models for a variable $X$ is as follows: I estimate a conditional mean of a variable $X$ by use of the ARMA part of the model. I estimate the conditional variance of variable $X$ by use of the GARCH part of the model.

And as far as I understand those models, this means that the variance of $X$ is simply interpreted as the squared residual of the mean model at a specific point in time.

Is my understanding correct?


1 Answer 1


Not quite. The conditional variance of $X_t$, conditional on the information up to and including time $t-1$, equals the conditional variance of the squared error: $$ \text{Var}(X_t|I_{t-1})=\mathbb{E}(u_t^2|I_{t-1}) $$ where $u_t$ is the raw error. The squared residual $\hat u_t^2$ is a rather noisy proxy for it / estimate thereof.

The unconditional / long-run variance of $X$ depends on the ARMA-GARCH model but in any case is not equal to the square residual at any period, except when/if it is numerically equal, which is by chance.

  • $\begingroup$ Before accepting this as an answer: (1) Why is the conditional variance of the squared error simply the expected value of the squared error? (2) Why is the squared residual a noisy proxy for it? Does this "noisiness" still remain, even when assuming my conditional mean model true? Thanks in advance. $\endgroup$
    – shenflow
    Dec 14, 2020 at 14:16
  • 1
    $\begingroup$ @shenflow, (1) is by definition of $u_t$; $X_t=\mu_t+u_t$ where $\mu_t:=\mathbb{E}(X_t\mid I_{t-1})$ and so $\mathbb{E}(u_t\mid I_{t-1})=0$. (2) It just is, even for correctly specified conditional mean models; see Andersen & Bollerslev "Answering the Skeptics: Yes, Standard Volatility Models do Provide Accurate Forecasts" (1998) p. 889 and onwards. The magnitude of correlation between squared errors and true underlying volatilities may be pretty low (say, 0.5, 0.4, 0.3 or even 0.2) for plausible data generating processes. $\endgroup$ Dec 14, 2020 at 19:29

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