We have an AR(1) process,
$X_t=alpha X_{t-1}+varepsilon_t$ with $varepsilonsim(0,sigma^2)$, $X_0=0$ and $|phi|<1$.
We have the conditional expected a value with respect to $X_{t-1}$:
$E(X_t|X_{t-1}) = E(alpha+X_{t-1}+varepsilon_t|X_{t-1})= E(alpha X_{t-1}|X_{t-1})+E(varepsilon_t|X_{t-1})= alpha E(X_{t-1}|X_{t-1})+E(varepsilon_t|X_{t-1})=
alpha X_{t-1}+E(varepsilon_t)=alpha X_{t-1}.$
The conditional variance,
$V(X_t|X_{t-1})= E[(X_t - E[X_t|X_{t-1}])^2|X_{t-1}]=E[(alpha X_{t-1}+varepsilon_t-E[X_t|X_{t-1}])^2|X_{t-1}]=E[(alpha X_{t-1}+varepsilon_t-[alpha X_{t-1}])^2|X_{t-1}]=E[(alpha^2 X_{t-1}+varepsilon^2+alpha^2 X_{t-1}+2 alpha X_{t-1}varepsilon_t-2 varepsilon_t alpha X_{t-1}-2 alpha^2 X_{t-1}^2)|X_{t-1}]=E(varepsilon^2|X_{t-1})=E(varepsilon^2)=sigma^2.$
If I show it in this way, is it tautological?
$V(X_t|X_{t-1})=V(alpha X_{t-1}+varepsilon_t|X_{t-1})= V(alpha X_{t-1}|X_{t-1})+V(varepsilon|X_{t-1})+2Cov(alpha X_{t-1},varepsilon_t|X_{t-1})=alpha^2Var(X_{t-1}|X_{t-1})+V(varepsilon_t|X_{t-1})+2Cov(alpha X_{t-1},varepsilon_t|X_{t-1})=V(varepsilon_t|X_{t-1})=V(varepsilon_t)=sigma^2.$
Because $X_{t-1}$ is independent from $varepsilon_t$ so $Cov(cdot)=0$ and $V(X_{t-1}|X_{t-1})=0$.
Am I right?