Consider the time series defined by $Y_t = \phi Y_{t-1}+ \epsilon_t + \theta \epsilon_{t-1}$
Why is $E(\epsilon_{t} Y_{t}) = \sigma_{\epsilon}^{2}$?
Consider the time series defined by $Y_t = \phi Y_{t-1}+ \epsilon_t + \theta \epsilon_{t-1}$
Why is $E(\epsilon_{t} Y_{t}) = \sigma_{\epsilon}^{2}$?
Simply work it out. By assumption, the white noise term $\varepsilon_t$ satisfies the following:
$E[\varepsilon_t]=0$
$E[\varepsilon^2_t]=\sigma_\varepsilon^2$
$E[\varepsilon_t\varepsilon_s]=0$, for $t\neq s$
Now multiply $Y_t$ by $\varepsilon_t$ and use linearity of expectation. The white noise terms are all uncorrelated for different times, so their expectations vanish. Explictely, write out $Y_t$ as a geometric series by recursively using the equation for $Y_t$. Otherwise apply induction. The point is that $\varepsilon_t$ and $Y_{s}$ for $s
Use what sam mentioned but it is not necessary to write Yt out as a geometric series. Substitute ϕY$_t$_−$_1$+ϵ$_t$+θϵ$_t$_−$_1$ for Y$_t$ multiply by e$_t$ to get ϕϵ$_t$Y$_t$_−$_1$+ϵ$^2$_t$+θ(ϵ$_t$_−_1$ ϵ$_t$) take expectations.
You get ϕ E(ϵ$_t$Y$_t_−$_1$) + E(ϵ$^2$_t$) + θ E(ϵ$_t$_−$_1$ ϵ$_t$).
Now from condition 3 given by Sam E(ϵ$_t$_−$_1$ ϵ$_t$)=0 and since ϵ$_t$ is independent of Y$_t$_-$_1$,
E(ϵ$_t$Y$_t$_−$_1$)=0. So you are only left with E(ϵ$^2$_t$) which you know is σ$^2$_ϵ$.