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I read this definition of an ARCH(1) model:

$$r_{ŧ}=\sigma_{t|t-1}\epsilon_{t}$$ $$\sigma^{2}_{t|t-1} = \omega + \alpha r_{t-1}^{2}$$

However, when it comes to forecasting the h-step-ahead variance, I don't understand why is defined in this way. This is the h-step-ahead conditional variance:

$$\sigma^{2}_{t+h|t}=E(r^{2}_{ŧ+h}|r_{t},r_{t-1},...)$$

How can I see that the equation above is correct?

Another question. I'd like to know what is going on in the first steps of the derivation of the recursive conditional variance:

$$\sigma^{2}_{t+h|t} = E(r^{2}_{t+h}|r_{t},r_{t-1},...) = E[E(\sigma^{2}_{t+h|t+h-1}\epsilon^{2}_{t+h}|r_{t+h-1},r_{t+h-2},...)|r_{t},r_{t-1},...] = \omega + \alpha \sigma^{2}_{t+h-1|t}$$

I skipped the final steps because they are easier to follow. I guess this is an instance of $E[E(X|Y)]=E(X)$ but I don't know why those conditional variables are chosen in this case.

UPDATE

A couple of additional questions.

  1. Based on the expression for $E(r^{2}_{t+h}|r_{t}, r_{t-1},...)$, does it mean that, for instance, $E(r^{2}_{t+h+1}|r_{t},r_{t-1}) = \sigma^{2}_{t+h+1|t}$ but since $\sigma^{2}_{t+h+1|t} = \omega + \alpha r_{t}$, then any other expectation for additional steps $t+ h + 2$, $t+h+3$, etc are going to produce the same result?

  2. When we take the expectation $E(r^{2}_{t+h}|r_{t}, r_{t-1},...)$, why don't we modify the conditional subindex from $t$ to $t+h-1$ in $\sigma^{2}_{t+h|t}$ and we do change it when we write:

$$E(r^{2}_{t+h}|r_{t},r_{t-1},...) = E[E(\sigma^{2}_{t+h|t+h-1}\epsilon^{2}_{t+h}|r_{t+h-1},r_{t+h-2},...)|r_{t},r_{t-1},...]$$

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$E(r_t^2|r_{t-j},j=1,2,...)=E(\sigma^2_{t|t-1}\epsilon_t^2|r_{t-j},j=1,2,...)$. Now given $r_{t-j},j=1,2,...$, you know the values of $\sigma_{t|t-1}$ because $\sigma^2_{t|t-1}=\omega+\alpha r_{t-1}^2$. So you can take it out of expectation to have: $E(r_t^2|r_{t-j},j=1,2,...)=\sigma^2_{t|t-1}E(\epsilon_t^2|r_{t-j},j=1,2,...)$. Also $\epsilon_t$ is independent of $r_{t-j},j=1,2,...$, so (roughly speaking) any function of that is also independent of $r_{t-j},j=1,2,...$. In particular, $f(x)=x^2$. So, $\epsilon_t^2$ is independent of $r_{t-j},j=1,2,...$. Hence, $E(r_t^2|r_{t-j},j=1,2,...)=\sigma^2_{t|t-1}E(\epsilon_t^2)$. Finally note that $\epsilon_t\sim N(0,1)$. So $Var(\epsilon_t)=E(\epsilon_t^2)=1$. Therefore, $E(r_t^2|r_{t-j},j=1,2,...)=\sigma^2_{t|t-1}.$ Now we can change $t$ to $t+h$ to get $h$-step ahead forecasts, i.e. $E(r_{t+h}^2|r_{t-j},j=1,2,...)=\sigma^2_{t+h|t-1}.$ I think you can now answer your 2nd question. Hint: you condition on previous returns $r_t$ that makes your expectation easy to compute.

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  • $\begingroup$ 1+. Thanks! Please, see my update to the question. I have a couple of questions I hope you can clarify. $\endgroup$ – Robert Smith Oct 22 '13 at 19:21

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