Recall

In our closed-loop linear predictor we defined the error as $$e_{n}=X_{n}-\sum_{i=1}^{m}a_{i}X_{n-i}$$We then have that $$\begin{align*} E[e_{n}X_{n-j}]&=E\left[ \left( X_{n}-\sum_{i=1}^{m}a_{i}X_{n-i} \right)X_{n-j} \right]\\ &=E[X_{n}X_{n-j}]-\sum_{k=1}^{m}a_{i}E[X_{n-k}X_{n-j}] \end{align*}$$We see that from (*), $a_{1},\dots,a_{m}$ is optimal if and only if $E[e_{n}X_{n-j}]=0$, $j=1,\dots,m$. # Theorem The linear predictor $\hat{X}_{n}=\sum_{k=1}^{m}a_{k}X_{n-k}$ is optimal in the MSE sense if and only if the prediction error is orthogonal to all $X_{n-j}$ i.e.$$\hat{X}_{n}\text{ optimal }\iff(X_{n}-\hat{X}_{n})\perp X_{n-j} \ \ \ j=1,\dots,m$$