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(NOTE: also under discussion in sos Math Cyberboard)

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- Aug 15th 2011, 12:33 PMkingwinnerTime Series: ARCH model properties
https://sites.google.com/site/asdfasdf23135/stat9.JPG

(NOTE: also under discussion in sos Math Cyberboard) - Aug 16th 2011, 12:00 AMMooRe: Time Series: ARCH model properties
Hello,

First equality : because $\displaystyle \sigma_t$ depends on $\displaystyle Z_{t-1}$, so $\displaystyle Z_t$ is independent from $\displaystyle \sigma_t$, since the Zt are iid, and hence the equality (that's a common property).

Second equality : same stuff, Xt is independent of all these Zt's, because it depends on Zt.

Third equality : ditto - Aug 16th 2011, 07:02 AMkingwinnerRe: Time Series: ARCH model properties
I think you're trying to say that Z_t and X_{t-1} are independent since X_{t-1} is a function of only Z_{t-1} , Z_{t-2}, ..., etc. and Z_i s are iid.

But now my question is: Why is X_{t-1} is function of ONLY Z_{t-1}, Z_{t-2},... ? How can we**prove**this?

The trouble is I think X_{t-1} a function of Z_{t-1}, Z_{t-2},...__AND__some σ_j.

Thanks! - Aug 16th 2011, 01:30 PMMooRe: Time Series: ARCH model properties
Yeah sorry I probably got messed up with the variables' names, well it seems like you understand what I meant (Itwasntme)

Well to see it, you can say that $\displaystyle X_t$ is $\displaystyle \sigma(\sigma_t,Z_{t-1})$-measurable (the first sigma is "sigma-algebra generated by...")

$\displaystyle \sigma_t$ is $\displaystyle \sigma(X_{t-1})=\sigma(\sigma_{t-1},Z_{t-2})$-measurable, etc... So in the end, by independence of the Zt's, $\displaystyle X_t\perp Z_t$ (independence).

If you want to prove it, you can just use induction, but it's not necessary in your exercise :)