6. KALMAN-BUCY FILTER

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1 6. KALMAN-BUCY FILTER 6.1. Motivation and preliminary. A wa hown in Lecture 2, the optimal control i a function of all coordinate of controlled proce. Very often, it i not impoible to oberve a controlled proce or part of it component. For intance, an information on a controlled traectory i interrupted by a noie. In thi cae the optimal (or reaonable control can not be contructed a a functional of controlled traectorie. It i very natural to etimate a controlled traectory via exiting obervation and then to create a control a a functional of thee etimate. The Kalman-Bucy filter i one of filter which enable u to have on line etimating, or tracking, of unobervable ignal. A tandard filtering problem deal with two random procee, ay X t and Y t, where X t i an unobervable ignal and Y t i an obervable one. The filtering problem conit in the etimating of X t via {Y, t}, that i any filtering etimate X t i a functional of Y t = {Y, t}. The more popular optimal in the mean quare ene filtering etimate i the conditional expectation minimizing the mean quare error X t = E(X t Y t P (t = E(X t X t 2. The conditional expectation E(X t Y t, a a function of Y t, may have too complicated tructure. Therefore, epecially in engineering application, it make ene to ue a linear function of Y t minimizing the mean quare error P (t in a cla of linear filtering etimate. The optimal linear etimate X t i the orthogonal proection on a linear pace generated by Y t1. According to the optimal property of the orthogonal proection it i named the conditional expectation in the wide ene. For Gauian procee the orthogonal proection coincide with the conditional expectation Kalman-Buy model. Let the ignal X t and the obervation Y t are defined by the Itô linear equation with repect to independent Wiener procee W t and W t dx t = a(tx t dt + b(tdw t dy t = A(tX t dt + B(tdW t (6.1 1 thi linear pace i generated by any linear combination of Y t + contant and their cloure in the mean quare ene 1

2 ubect to Gauian initial condition X and Y independent of W t and W t. The determinitic (known function a(t, A(t, b(t, and B(t are aumed to be bounded and and piece-wie continuou for t. The main aumption here i inf t B2 (t c (for ome c >. ( Decription of the Kalman-Bucy filter. The filtering etimate X t and the mean quare filtering error P (t are defined by differential equation (Itô and Riccati ubect to the initial condition: d X t = a(t X t dt + P (ta(t (dy B 2 t A(t (t X t dt P (t = 2a(tP (t + b 2 (t P 2 (ta 2 (t B 2 (t X = EX + cov(x, Y cov(y, Y (Y EY P ( = cov(x, X cov2 (X, Y cov(y, Y. (6.3 ( Derivation of filtering equation. (6.4 i Home work. Since here we deal with Gauian random procee the conditional expectation coincide with the orthogonal proection which i a linear function of obervation. Let u denote X t = L(Y t. To decribe a tructure of L(Y t, introduce partition of the interval [, t: = t n < t n 1 <... < t n n = t o that lim max [t n +1 t n = and find n the optimal in the mean quare filtering etimate of X t given {Y t n, =,..., n}. Since {Y t n, =,..., n} and {Y, Y t n 1, = 1,..., n} contain the ame information, we have L n (Y t = E(X t Y, Y t n 1, = 1,...n and, wherea L n (Y t i a linear function of argument Y, Y t n 1, = 1,...n, we have L n (Y t = G n (t + G n 1(tY + G n (t, t n [Y t n 1, where G n (t, G n 1(t and G n (t, t n are determinitic function. 2

3 We tart with a particular cae EX = and cov(x, Y =. Notice that (6.4 provide G n (, G n 1(. Hence, we have L n (Y t = G n (t, t n [Y t n 1 = G n (t, dy, where G n (t, = G(t, t n, t n < t n +1. For further convenience, we chooe partition {t n 1,..., t n n} {t n+1 1,..., t n+1 n+1}, n 1, o that {Y, Y t n 1, = 1,..., n} {Y =, Y, t}. Then, by the well known property of the conditional expectation ( ( l.i.m. E Y Y X t, Y t n n 1, = 1,..., n = E X t, Y, t, that i E(X t Y t = l.i.m. n Ln (Y t. Our next tak i to how that there exit a function G(t, uch that E(X t Y t = G(t, dy. (6.5 By the Cauchy criteria for the convergence in the mean quare ene we have ( 2 l.i.m. n Ln (Y t exit lim E L n (Y t L m (Y t =. n,m On the other hand, ince L n (Y t i the linear function of Y t, we have and L n (Y t L m (Y t := [G n (t, G m (t, dy ( 2 lim E [G n (t, G m (t, dy =. (6.6 n,m Therefore, taking into account the independence of (X t and (W t and denoting by K(, = E(X X (ee Home work 2., write 3

4 ( 2 E [G n (t, G m (t, dy ( = E [G n (t, G m (t, [A(X d + B(dW ( = E [G n (t, G m (t, A(X d = ( +E + 2 [G n (t, G m (t, B(dW 2 2 [G n (t, G m (t, [G n (t, G m (t, K(, d d [G n (t, G m (t, 2 B 2 (d. Both term in the right ide of thi equality are nonnegative and therefore limit, a n, m, for both are zero. Moreover, ince B 2 ( c >, we obtain lim n,m [G n (t, G m (t, 2 d =. The latter, by the Cauchy criteria provide (6.5 and, moreover, l.i.m. n l.i.m. n Conequently, (6.5 hold true. G n (t, A(X d = G n (t, B(dW = G(t, A(X d G(t, B(dW Wiener-Hopf equation. Now, we how that G(t, olve the Wiener-Hopf integral equation: for every fixed t and t A(K(t, = G(t, ua(k(, ua(udu + G(t, B 2 (, where K(t, i the correlation function of the proce (X t and A(, B( are function involved in (6.1. Proof: Since G(t, dy i the orthogonal proection generated by Y t, the random variable X t G(t, dy i orthogonal to any linear function of Y t. Then, for 4

5 f(t, dy with determinitic (bounded function f(t,, it hold ( E X t G(t, dy f(t, dy =. The latter equality implie ( E (X t f(t, dy = E G(t, dy f(t, dy. Write, keeping in mind the independence of (X t and (W t, and E ( EX t f(t, dy = EX t f(t, A(X d + = EX t f(t, A(X d = = f(t, A(E(X t X d f(t, B(dW f(t, A(K(t, d (6.7 ( G(t, dy f(t, dy = E G(t, A(X d + = E (6.7 and (6.8 provide the equality { f(t, A(K(t, = ( f(t, A(X d + +E + G(t, A(X d f(t, B(dW G(t, B(dW f(t, B(dW f(t, A(X d G(t, B(dW f(t, G(t, ua(uk(, ua(dud f(t, G(t, B 2 (d. (6.8 G(t, ua(uk(, ua(du } f(t, G(t, B 2 ( d =. 5

6 So, by an arbitrarine of f(t,, we claim A(K(t, = G(t, ua(k(, ua(udu + G(t, B 2 (. Uniquene of olution of the Wiener-Hopf equation. Aume G i (t,, i = 1, 2 are olution. Set (t, = G 1 (t, G 2 (t,. Then (t, A(K(, ua(ud + (t, ub 2 (u = Multiplying both ide of thi equality by (t, and integrating with repect to d we get (t, A(K(, ua(u (t, udud + 2 (t, ub 2 (udu =. Both part on the left hand ide of above equality equal zero. Since B 2 (u i trictly poitive 2 (t, u =, u t. 1. Show firt that G(t, tb 2 (t = K(t, ta(t Exitence of olution of the Wiener-Hopf equation. = EX 2 t A(t ([ = E X t ([ = E X t ([ = E G(t, t = P (ta(t, (6.9 B 2 (t G(t, ua(tk(t, ua(udu G(t, ua(te(x t X u A(udu G(t, ua(ux u du X t A(t G(t, ua(ux u du X t = E ([X t X t X t A(t = [ + X t X t Xt A(t = P (ta(t + E G(t, udy u X t A(t ([ X t X t Xt A(t [X t X t [X t X t A(t G(t, ub(udw u X t A(t and ince E ([X t X t Xt = we obtain G(t, tb 2 (t = P (ta(t and (6.9. 6

7 2. For < t, we will find olution in a form with mooth in t function ϕ t, ϕ t t = 1. Notice that K(t, i differentiable in t: EX t X, it hold K(t, = E(X X t = EX (X + G(t, = ϕ t G(, (6.1 = EX (X + = K(, + K(t, = a(tk(t, ince K(t, = a(ux u du + a(ux u du a(uk(, udu b(udw u and the reult. Now, ubtituting the right hand ide of (6.1 to the Wiener-Hopf equation, we find A(K(t, ϕ(t, ug(u, ua(k(u, A(udu + ϕ(t, G(, B 2 (. (6.11 and, differentiating both ide of the above identity in t, we get [a(t G(t, ta(ta(k(t, + ϕ(t, u G(u, ua(k(, ua(udu + ϕ(t, G(, B 2 (. Applying the right hand ide of (6.11 in (6.12 intead of A(K(t,, we obtain ( [a(t G(t, ta(t ϕ(t, ug(u, ua(k(, ua(udu + ϕ(t, G(, B 2 ( ϕ(t, u G(u, ua(k(, ua(udu + ϕ(t, G(, B 2 (. (6.12 (6.13 By (6.9 G(t, ta(t = A2 (tp (t and o we claim that the identity (6.13 may hold B 2 (t provided that for ϕ = 1 and for u > dϕ u [ du = a(u P (ua2 (u ϕ u B 2 (u. (6.14 Further, by the olution uniquene of Wiener-Hopf equation we have G(t, = P (ta(t ϕ t B 2 (t. 7

8 The Kalman filter. It i readily to check that that (home work Then X t = G(t, dy = ϕt P (A( dy B 2 ( = ϕ t Itô it hold (home work ϕ t = ϕt. (6.15 ϕ 1 ϕ P (A( dy B 2 (. From thi, by the d X t = a(t X t dt + P (ta(t [ dy B 2 t A(t (t X t dt. (6.16 Since P (t = E(X t X t 2, we derive firt the Itô equation for (t = X t X t. The Itô equation for X t and X t imply (t = ( + a( (d + P (ta(t dw. B(t Further, by the Itô formula we obtain 2 (t = 2 ( a( 2 (d + 2 P (A 2 ( 2 (d B 2 ( P (A( (dw + B( b(dw (b(dw + P 2 (A 2 ( d. B 2 ( P (A 2 ( (d B 2 ( b 2 (d Taking the expectation from both part of the latter equality, we get (home work P (t = P ( + 2 a(p (d + b 2 (d P 2 (A 2 ( d. (6.17 B 2 ( Home work 1. Derive (6.4. Hint: Taking X = c 1 + c 2 Y minimize in c 1, c 2 the mean quare error E(X X Show that under EX = it hold EX t. 3. To prove ( Derive (6.16 and (

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