(NOTE: also under discussion in sos Math Cyberboard)
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
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!
Yeah sorry I probably got messed up with the variables' names, well it seems like you understand what I meant
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