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In my problem set about ARCH models I'm given that

$\epsilon^2_{t}=\alpha\epsilon^2_{t-1}+v_{t}$

But then I'm asked to calculate $E(\sigma^2_{t+n}|I_{t-1})$.

So is the same to calculate $E(\epsilon^2_{t+n}|I_{t-1})$, isn't it?

Thanks in advance

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Yes they are equivalent it is just different notation.

For example, in Hamilton Time Series analysis the ARCH is given on pp 658 as:

$$E(u^2_t | u^2_{t-1} ... u^2_{t-m}) = \zeta + \alpha_1 u^2_{t-1}+ ...+ \alpha_m u^2_{t-m}$$

which for only one lag collapses to:

$$E(u^2_t | u^2_{t-1} ) = \zeta + \alpha_1 u^2_{t-1}$$

Just because $u$ is used instead of $\epsilon$ or $\sigma$ that does not make it different model (in addition here also the information set is just described directly instead of putting $I$ there but it is still an ARCH model).

However, this being said usually $\sigma$ is reserved to be the parameter of interest in ARCH model. For example, following Verbeek (2008) a guide to modern econometrics 4th ed pp 325 by definition:

$$\sigma^2_t \equiv E \{ \epsilon_t^2 | \mathcal{I}_{t-1} \}= \omega +\alpha \epsilon^2_{t-1}.$$

It is useful to use notation which does not use the same letter for expected variance as in GARCH model you would also want to include lags of expected/predicted variance (e.g. $\sigma^2_t = \omega +\alpha \epsilon^2_{t-1} + \beta \sigma^2_{t-1}$), but since in question you say this is for ARCH I don't think it makes any difference.

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    $\begingroup$ $\sigma_t^2$ is not the dependent variable but a parameter of interest. The dependent variable is $x_t$ modelled as $x_t=\mu_t+u_t$ where $\mu_t$ is the conditional mean and $u_t$ is the residual. This might be relevant. $\endgroup$ – Richard Hardy Dec 6 '20 at 12:21
  • $\begingroup$ @RichardHardy thank you for pointing that out I edited my answer $\endgroup$ – 1muflon1 Dec 6 '20 at 12:25
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It is the same in practice, they are both variances some people will write it as $\sigma^2$ some as $\epsilon^2$. Variance in error tells you what is the variance in the variable you predict.

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  • $\begingroup$ This is a theoretical exercise so, they are the same in practice but also in theory? $\endgroup$ – GregorSilvei Nov 29 '20 at 20:07
  • $\begingroup$ they are the same also in theory $\endgroup$ – csilvia Nov 30 '20 at 1:15

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