By Randall L. Eubank
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This is true for our our setting is that C2 /C1 2 ≤ F 2 /(1 + F 2 )2 . 30) R(∞) and, hence, Cov(x(t), ε(t)) ≈ Cov(x(t), ε(j)) ≈ F R(∞) − W0 , H t−j R(∞) − W0 , H for j ≤ t − 1 and Cov(x(t), ε(j)) ≈ F j−t R(∞) − W0 H W0 j−t R(∞) for j ≥ t + 1. Recall from Chapter 1 that the R(t) have interpretations as the elements of the diagonal matrix R in the Cholesky decomposition Var(y) = LRLT for the response variance-covariance matrix. Consequently, characterization of R(∞) provides us with information about how the elements of R (and, hence, the variances of the ε(t)) will behave for large t.
One way to evaluate the entirety of ΣXε in a single recursion is to work forward from the upper left hand corner of the matrix in an L-shaped pattern. 2. Then the above diagonal blocks for the (t + 1)st column block are evaluated. 4. The above diagonal blocks for row block 1 consist of the matrices σXε (1, 2) = S(1|0)M T (1)H T (2) and σXε (1, j) = S(1|0)M T (1) · · · M T (j − 1)H T (j) for j = 2, . , n. , followed by post-multiplication by the relevant H T (j) matrices to obtain the actual block element of ΣXε .
The first row and second column block) of the matrix. 2, we see that O(n2 ) flops will also be needed to evaluate the upper diagonal blocks of ΣXε . Accordingly, the entire matrix can be obtained in order n2 operations. 4. 9) and (F2). 2) to see that x(t) = F (t − 1) · · · F (j)x(j) + Z(t) with Z(t) depending only on u(t − 1), . , u(j). 15). 18) is somewhat more difficult to establish. 18) for j = t + 1. By exactly the same process we used for j = t + 1 we find that Cov(x(t), ε(t + 2)) has the form Cov(x(t), x(t + 1) − x(t + 1|t + 1))F T (t + 1)H T (t + 2).
A Kalman Filter Primer by Randall L. Eubank