# Why is the $E(y_{t-1}\epsilon_{t-1})=\sigma^2$?

I am trying to understand how to derive the ACF and PACF of and ARMA(1,1) process. In one of my lectures I have the following ARMA(1,1)$$y_t=\phi y_{t-1}-\theta \epsilon_{t-1}+\epsilon_t$$

I am being told that I have to multiply by $y_{t-k}$ and take the expectation so I end up with: $$E[y_t*y_{t-k}] \ or \ \gamma_k=\phi \gamma_{k-1}-\theta E[\epsilon_{t-1}*y_{t-k}]+E[\epsilon_t *y_{t-k}]$$ and at one point it says that for $k=0$ $E[\epsilon_t*y_t]=\sigma^2$ and $$E[\epsilon_{t-1}*y_t]=E[\epsilon_{t-1}(\phi_1y_{t-1}+\epsilon_t-\theta_1 \epsilon_{t-1})]=\sigma^2(\phi_1-\theta_1) \ [1]$$ I do not understand the last equality [1]. I know that the $Cov(\epsilon_t*\epsilon_{t-1})=0$ and that $E(\epsilon_{t-1}*\epsilon_{t-1})=\sigma^2$ because of the White Noise prpoerties, but why is the $E(y_{t-1}*\epsilon_{t-1})=\sigma^2$???

• It may be helpful to do repetitive susbstitutions of $y_{t-1}$, $y_{t-2}$,... in the model for $y_t$. You will arrive to an expression that depends on some initial values $y_0$ and $\varepsilon_0$, on $\varepsilon_t$ and on lags of $\varepsilon_t$. Taking the expectation of the product of $\varepsilon_t$ (or $\varepsilon_{t-1}$) with this expression will show where the $\sigma^2$ comes from. – javlacalle Apr 26 '17 at 11:02

At $t-1$, the process $\{y_t\}$ is given by $$y_{t-1}=\phi y_{t-2}-\theta \epsilon_{t-2}+\epsilon_{t-1},$$ so that $$E[y_{t-1}\epsilon_{t-1}]=E[(\phi y_{t-2}-\theta \epsilon_{t-2}+\epsilon_{t-1})\epsilon_{t-1}]=\sigma^2$$