The method of moments estimator for AR processes can be had with the Yule-Walker equations. But how is it derived?
The equation for AR(1):
$$Y_t =aY_{t-1}+\epsilon_t$$
Where $\epsilon $ ~ $N(0,\sigma^2 )$.
So the moment conditions are $E(\epsilon)=0, E(\epsilon^2)=\sigma^2$
But the "standard" solution can't be used as $E(\epsilon)=E(Y_t-aY_{t-1})=0$ is true for any $a$.
I figured the answer:
Multiply the equation by $Y_{t-1}$
$Y_tY_{t-1}=aY_{t-1}Y_{t-1}+\epsilon_tY_{t-1}$
$E(Y_tY_{t-1})=E(aY_{t-1}Y_{t-1}+\epsilon_tY_{t-1})$
$E(Y_tY_{t-1})=aE(Y_{t-1}^2)$
$E(Y_tY_{t-1})/E(Y_{t-1})=a$
$E(Y_tY_{t-1})/E(Y_{t})=a$
And for $\sigma^2$:
$E((Y_t+aY_{t-1})^2)=\sigma^2$
$E(Y_t^2+a^2Y_{t-1}^2-2aY_tY_{t-1})=\sigma^2$
$E(Y_t^2+a^2Y_{t-1}^2-2aY_tY_{t-1})=\sigma^2$
$E(Y_t^2+a^2Y_{t-1}^2-2aY_{t-1}(aY_{t-1}+\epsilon_t))=\sigma^2$
$E(Y_t^2+a^2Y_{t-1}^2-2a^2Y_{t-1}^2)=\sigma^2$
$E(Y_t^2-a^2Y_{t-1}^2)=\sigma^2$
$E(Y_t^2)-a^2E(Y_{t-1}^2)=\sigma^2$
$(1-a^2)E(Y_{t}^2)=\sigma^2$
