# Conditional Variance of an ARMA(p,q) or even just ARMA(1,1) to illustrate...

I know how to derive the unconditional variance but I want to show that, for example in a standard (i.e. not mixed with ARCH) ARMA(1,1) that the conditional variance at time t, conditional on information to time t-1, is constant. I assume this is the case because otherwise ARCH wouldn't make sense since it's whole purpose is to introduce non-constant conditional variance.

I think I have an answer but it boils down to something I'm not sure on, goes like this.

$Var(X_t|F_{t-1})=Var(\epsilon_t +(\phi+\theta )\sum_{i=1}^\infty \epsilon_{t-i}|F_{t-1})=Var(\epsilon_t|F_{t-1})=E[\epsilon_t^2|F_{t-1}]$

where I believe that the whole 2nd term:

$(\phi+\theta )\sum_{i=1}^\infty \epsilon_{t-i}$

Is a constant, conditional on $F_{t-1}$, so that just leaves the first term. At this point can I just assume that the error term at time "t" is independent of $F_{t-1}$ so that:

$E[\epsilon_t^2|F_{t-1}] =E[\epsilon_t^2]=\sigma^2$

This if for a standard ARMA(1,1) i.e

$X_t=\phi_1X_{t-1}+\theta_1\epsilon_{t-1}+\epsilon_t$ with $\epsilon_t$~$WN(0,\sigma^2)$ (not strict white noise)

Please let me know if this is right or if I've made an error.

In ARMA$(p,q)$ $\epsilon_t$'s are independent of each other and $\epsilon_t$ is independent of $X_{t-i}, i\geq 1$, therefore your assumption is correct i.e. $\epsilon_t$ is independent of $F_{t-1}$.