A Kalman Filter Primer by Randall L. Eubank

By Randall L. Eubank

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The problem with this is that M (t) = F (t) − F (t)S(t|t − 1)H T (t)R −1 (t)H(t) so that all the M (t), t = 2, . , n, will not be available unless we have already evaluated S(t|t−1), t = 1, . , n. Consequently, if we want to compute the S(t|t − 1) and R(t) in tandem with evaluation of ΣXε we need a slightly more subtle strategy. Now, in general, for the tth row block the above diagonal blocks appear like σXε (t, j) = S(t|t − 1)M © 2006 by Taylor & Francis Group, LLC T (t) · · · M T (j − 1)H T (j) A Kalman Filter Primer 42 for j = t+ 1, .

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). 15) we can express x(t + 1) − x(t + 1|t + 1) as t x(t + 1) − Cov(x(t + 1), ε(j))R −1 (j)ε(j) j=1 −Cov(x(t + 1), ε(t + 1))R −1 (t + 1)ε(t + 1) = F (t)[x(t) − x(t|t)] −S(t + 1|t)H T (t + 1)R −1 (t + 1)ε(t + 1) + u(t). 8), the definition of M (t) and our previous result for j = t + 1 we see that the covariance © 2006 by Taylor & Francis Group, LLC The Fundamental Covariance Structure 39 between x(t) and x(t + 1) − x(t + 1|t + 1) is Cov(x(t), x(t) − x(t|t))F T −Cov(x(t), ε(t + 1))R = S(t|t − 1)M −H T T (t + 1)R (t) −1 (t + 1)H(t + 1)S(t + 1|t) (t)[I −1 (t + 1)H(t + 1)S(t + 1|t)].

If we now let C1 = F 2 W0 + H 2 Q0 + W0 and C2 = 2 , it then follows that F 2 W0 R(t) = C1 − C2 R(t − 1) C2 = C1 − C1 − C2 R(t − 2) C2 = C1 − C2 C1 − C2 C1 − C1 − C2 R(t − 3) which reveals a continued fraction representation for R(t) as described, for example, in Khinchin (1997). © 2006 by Taylor & Francis Group, LLC The Fundamental Covariance Structure 49 General results for convergence of continued fractions can be found in Chapter 3 of Wall (1948). 1) for 2 ≤ 1/4. This is true for our our setting is that C2 /C1 2 ≤ F 2 /(1 + F 2 )2 .

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