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    $\begingroup$ @Aksakal, I did not say GARCH is directly equivalent to AR, as it is not. However, it is not a coincidence that it is called GARCH. I explain this in detail here, using an unorthodox representation of ARMA for better comparability with GARCH. I have also edited my answer to avoid confusion. $\endgroup$ Commented Jan 4, 2022 at 20:07
  • $\begingroup$ I'm a bit confused by your response. If h-step-ahead predictions will lag h steps behind in autoregressive models, what's the point of using them? From what I know, that only happens when the autoregressive model can't derive any actual information from your data, and is just replicating the most recent data point. $\endgroup$ Commented Jan 4, 2022 at 21:04
  • $\begingroup$ @VladimirBelik AR models have a stochastic component, e.g. $x_t=\phi_1 x_{t-1}+\varepsilon_t$ - notice the index of the error term, it is not observed yet at $t-1$. the point is that these models describe some processes, hence you can use them to forecast too. the error variance gives you an idea of the least possible amount of uncertainty you can accomplish in forecasting these processes. $\endgroup$ Commented Jan 4, 2022 at 21:25
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    $\begingroup$ @VladimirBelik, I think my last sentence addresses this: in typical applications of GARCH models, conditional variance is often found to be quite close to a random walk. For such a case, last observed value is close to the optimal prediction. Saying that this is not satisfactory indicates your view of the process (you do not like how it behaves) rather than the model that happens to represent the process. GARCH need not be the best model in the most general sense, but just try beating it without use of additional data and see how you fare. People have tried that, often without much success. $\endgroup$ Commented Jan 6, 2022 at 18:16
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    $\begingroup$ @VladimirBelik, also try GARCH vs. $\log(\sigma_t^2)=\log(\sigma_{t-1}^2)+\zeta_t$ with $\mathbb{E}(\zeta_t)=0$ (a multiplicative random walk). You will probably see GARCH doing slightly better. (I suggest multiplicative random walk since additive random work does not work here; variance cannot be negative, while there is nothing preventing additive random walk from going below zero.) $\endgroup$ Commented Jan 6, 2022 at 18:21