Simplified Solution to Closed-loop Predictor Coefficients Problem

For WSS processes an optimal $\mathbf{a}=(a_{1},\dots,a_{m})^{T}$ must satisfy $$\mathbf{R}_{m}\mathbf{a}=\mathbf{v}_{m}$$where $$\mathbf{R}_{m}=\begin{bmatrix}r_{0} & r_{1} & \dots & r_{m-1} \\ r_{1} & r_{0} & \dots & r_{m-2} \\ \vdots & \vdots & \ddots & \vdots \\ r_{m-1} & r_{m-2} & \dots & r_{0}\end{bmatrix}\quad \mathbf{v}_{m}=\begin{bmatrix}r_{1} \\ r_{2} \\ \vdots \\ r_{m}\end{bmatrix}$$Note how the symmetric nature of these processes (the second property) greatly simplifies the predictor coefficients problem.

If $\mathbf{R}_{m}$ invertible $$\mathbf{a}=\mathbf{R}_{m}^{-1}\mathbf{v}_{m}$$ ## Note $\mathbf{R}_{m}$ and the optimal $\mathbf{a}$ depend only on the prediction order $m$, not the time index $n$.

Results

Now let $$\mathbf{c}=(c_{0},c_{1},\dots,c_{m})^{T}=(1,-a_{1},-a_{2},\dots,-a_{m})^{T}=(1,-\mathbf{a})^{T}$$then $$\begin{align*} E[e_{n}^{2}]&=E\left[ \left( X_{n}-\sum_{i=1}^{m}a_{i}X_{n-i} \right)^{2} \right]\\ &=E\left[ \left( \sum_{i=0}^{m}c_{i}X_{n-i} \right)^{2} \right]\\ &=E\left[ \sum_{k=0}^{m}\sum_{j=0}^{m}c_{k}c_{j}X_{n-k}X_{n-j} \right]\\ &=\sum_{k=0}^{m}\sum_{j=0}^{m}c_{j}r_{j-k}c_{k}\\ &=\mathbf{c}^{T}\mathbf{R}_{m+1}\mathbf{c} \end{align*}$$This gives us the following results 1. For an optimal $\mathbf{c}$ we have $\mathbf{c}^{T}\mathbf{R}_{m+1}\mathbf{c}=0$ iff $\mathbf{R}_{m+1}$ is singular. Hence $$E[e_{n}^{2}]=0\iff \mathbf{R}_{m+1}\text{ is singular}$$ 2. If $\mathbf{R}_{m}$ is nonsingular $\forall m\ge 1$, the WSS process is called nondeterministic 3. If $\mathbf{a}$ is optimal, $\mathbf{v}_{m}=\mathbf{R}_{m}^{-1}\mathbf{a}$, so $$E[e_{n}^{2}]=r_{0}-\mathbf{a}^{T}\mathbf{v}_{m}=r_{0}-\sum_{j=1}^{m}a_{j}r_{j}$$