# Time Series: ARCH model properties

1. Aug 15, 2011

### kingwinner

Consider an ARCH(1) model:
Xt = σtZt, where Zt~ i.i.d. N(0,1)
σt2 = w0 + w1 Xt-12
Find (i) E(Xt)
and (ii) the autocovariance function γX(h) for h=0,1,2,3,..., assuming the process is second-order stationary.

Solution:
(i) E(Xt) = E[E(Xtt2)] =E[E(σtZtt2)]
=E[σtE(Ztt2)] = E[σtE(Zt)] = E[σt * 0] = 0
(here I don't understand why E(Ztt2)=E(Zt).)

(ii) γX(0)=E(Xt2)
=E(σt2 Zt2)=E(σt2)E(Zt2)
=E(w0 + w1 Xt-12) * 1
= w0 + w1 γX(0)
Solve for γX(0) => γX(0) = w0/(1-w1)

γX(h)=E(XtXt+h)
=E[E(XtXt+h|Zt+h-1,...,Zt+1)] = E[XtE(Xt+h|Zt+h-1,...,Zt+1)]
(here I don't understand why we can pull the Xt out of the expectation.)
=E[XtE(σt+hZt+h|Zt+h-1,...,Zt+1)] = E[Xtσt+hE(Zt+h|Zt+h-1,...,Zt+1)]
(here I don't understand why we can pull the σt+h out of the expectation.)
=E[Xtσt+hE(Zt+h)] = E[Xtσt+h * 0] = 0 for all h>0.

I'm cannot follow the reasoning of the three equalities labelled in red above. Can someone explain why they are true?
Any help would be much appreciated! :)

2. Aug 16, 2011

### Stephen Tashi

$Z_t$ is independent of $\sigma_t^2$. I think this step is an example of the fact that if X and Y are independent random variables then E(X|Y) = E(X).

The conditional expectation with respect to the sequence of values $Z_{h+t-1}...Z_{t+1}$ amounts to an integration with respect to the joint density of a sequence of independent random variables. These are events that happen after time $t$ and $X_t$ doesn't depend on them. I think this step is an example of the idea that $\int g(x_3) f(x_1,x_2)) dx_1 dx_2 = g( x_3) \int f(x_1,x_2) dx_1 dx_2$

I don't understand that step yet. I wonder if the phrase "assuming the process is second-order stationary" tells us anything important. The definition of the process looks very specific, so it isn't clear to me what that phase adds to it.

3. Aug 17, 2011

### Stephen Tashi

I could understand the above if it said:

$E(X_t E(\sigma_{t+h} X_{t+h}| Z_{t+h-1}...Z_{t+1} ) = E(X_t E(\sigma_{t+h}|Z_{t+h-1}...Z_{t+1}) E(Z_{t+h}|Z_{t+h-1}...Z_{t+1}))$

since $Z_{t+h}$ is independent of $\sigma_{t+h}$. The result will follow because the factor of zero still appears.