Arch Model and $\sigma$

Question

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

Answer 1

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.

Answer 2

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.