Fixed order-recursive prediction

• Recall that the least-squares solution for $A$ is obtained from the matrix equation:
  

  $$A R = H T^{*}$$ (12)

  
  where $R = TT^{*}$.

• Using autocorrelation windowing, we verify $R_{A}$ (which is Hermitian by construction) has also the block Toeplitz property. Consider the $(i,j)$ block entry of dimension $m$-by-$m$, where $1\!<\!i\!<\!p$, $1\!<\!j\!<\!p$:
  

  $$\left[R_{A}\right]_{(i,j)} = \sum_{n} y_{n+1-i} y^{*}_{n+1-j}$$

  
  With $y_{1-p} \ldots y_{0} = 0$, $y_{N+1} \ldots y_{N+p} = 0$, we restrict the summation index to satisfy $n>\mathrm{max}(i,j)-1$ and $n<N + \mathrm{min}(i,j)$. Defining $n' = n + 1 - j$, the sum is rewritten:
  

  $$\left[R_{A}\right]_{(i,j)} = \sum_{\mathrm{max}(i,j)-j+1}^{\mathrm{min}(i,j)-j+N} y_{n-(i-j)}y^{*}_{n}$$

  $$= \sum_{\mathrm{max}(i-j,0)+1}^{\mathrm{min}(i-j,0)+N} y_{n-(i-j)}y^{*}_{n}$$

  
  which depends only on $i-j$.

• Thus, $R_{A}$ for the order $p$ is parameterized by $p$ block entries:
  

  $$R_{A,p} = \left[\begin{array}{cccc} R_{0} & R^{*}_{1} & \ldots & R^{*}_{p-1... ...ts & \ddots & \vdots \\ R_{p-1} & R_{p-2} & \ldots & R_{0} \end{array} \right]$$

  
  

• Furthermore, it is easily verified:
  

  $$\rho_{p} \stackrel{\Delta}{=}H_{A,p} T^{*}_{A,p} = [R_{1} \ldots R_{p}]$$

  
  

• Therefore, we see there are two ways to partition $R_{A,p}$, such that $R_{A,p-1}$ appears as a submatrix in the lower left or upper right corners. This motivates a recursive algorithm. The ``hard work'' in $\ref{eq:lssol}$ is that of inverting $R_{A,p}$: if we know already the inverse of $R_{A,p-1}$, perhaps it is not so hard to get the inverse of $R_{A,p}$.

• We choose the following partition. Defining ``#'' as the operation which reverses the block elements of a vector, we verify:
  

  $$R_{A,p} = \left[\begin{array}{cc} R_{A,p-1} & \rho^{*\char93 }_{p-1} \\ \rho^{\char93 }_{p-1} & R(0) \end{array}\right]$$ (13)