MEAN-sQUARE FILTERING PROBLEM FOR DIsCRETE VOLTERRA ...

MEAN-SQUARE FILTERING PROBLEM FOR DISCRETE VOLTERRA EQUATIONS A. Bashkov

Moscow Aviation Institute e-mail: [email protected] V. Kolmanovskii

Moscow State Institute of Electronics and Mathematics, Space Research Institute (IKI), Russian Academy of Sciences e-mail: [email protected] X. Mao

University of Strathclyde, Glasgow, UK e-mail: [email protected] A. Matasov

Faculty of Mechanics and Mathematics, M.V. Lomonosov Moscow State University, Space Research Institute (IKI), Russian Academy of Sciences e-mail: [email protected]

Abstract

The mean-square ltering problem for discrete Volterra equations is a nontrivial task due to an enormous amount of operations required for the implementation of optimal lter. A dierence equation of a moderate dimension is chosen as an approximate model for the original system. Then the reduced Kalman lter can be used as an approximate but eÆcient estimator. Using the duality theory of convex variational problems, a level of nonoptimality for the chosen lter is obtained. This level can be eÆciently computed without exactly solving the full ltering problem. Key words:

linear discrete Volterra equations, stochastic mean-square ltering problem, duality theory

1 Introduction Various processes in nature and engineering are described by Volterra integral and integro-dierential equations [1]-[4]. Such type equations appear in viscoelasticity theory [2], in demographers, industrial inventory problems, in a modeling of certain physiological processes and a study of stationary electrode polography [3]. They describe the processes in advertising policies, commodity market, infection diseases, in controlling mechanical systems by proportional-integral and proportional-integral-dierential regulators [4]. These equations appear in dynamics of payload carriers and space vehicles to cover the plants with the chambers partially lled with liquid if the vortical processes are 1

accounted for [5]. The discrete Volterra equations result from appropriate digitization [6]. The discrete Volterra equations are also used for modeling the recurrent events [7]. In this paper, we study classical mean-square ltering problems for discrete Volterra equations, where a system perturbations and measurement noise are zero-mean white noise processes. Clearly, a discrete Volterra equation can be treated as an extended single-step equation of a large order. So, the classical Kalman ltering technique can be used to solve the problem. Nevertheless, if the amount of measurements is large, then we have to operate with matrices of high dimension. Moreover, these matrices would have a lot of zero elements. Recently an optimal ltering algorithm was developed speci cally for discrete Volterra equations [8]. However, in this algorithm we have to compute the matrices that depend on two time-domain arguments and therefore the amount of operations also increases dramatically. Thus the use of the existing algorithms can lead to the accumulation of computational errors, to ineÆcient overload of computer memory, and to the critical retarding of calculations. A way to overcome these unfavorable factors is to construct simpli ed lters of moderate dimension. In the latter case, we need to evaluate the level of nonoptimality for these simpli ed algorithms, remaining in the framework of relatively modest computation. Just this approach is treated in the paper. 1.1

Problem Statement

Consider a discrete linear Volterra equation

x(j + 1) =

j

X

k

=0

A(j; k)x(k) + B (j )u(j );

j = 0; : : : ; N

1;

(1)

where x(j ) = (x1 (j ); : : : ; xn (j ))0 2 Rn is the system state vector; A(j; k) 2 Rnn, B (j ) 2 Rnr , and u(j ) = (u1 (j ); : : : ; ur (j ))0 2 Rr is a perturbation vector. A prime denotes the transposition sign. We assume that u(j ) is a sequence of independent random vectors with zero means and covariance matrices Q(j )  0: E

u(j ) = 0;

E

(u(j )u0(k)) = Q(j )Æjk ;

j; k = 0; : : : ; N

1;

(2)

z (k); %(k) 2 Rm ; H (k) 2 Rnm ;

(3)

where Æjk is the Kronecker delta. Let measurements be made for the state vector

z (k) = H 0 (k)x(k) + %(k);

k = 0; : : : ; N: We assume that the measurement errors %(k) also form a sequence of independent random vectors with zero means and covariance matrices R(k) > 0: %(k) = 0;

E

E

(%(k)%0 (s)) = R(k)Æks;

k; s = 0; : : : ; N:

(4)

Moreover, the initial state vector of the system is assumed to be a zero-mean random variable 0 Ex(0) = 0; E (x(0)x (0)) = P0  0: (5) Stochastic elements x(0), u(
), and %(
) are supposed to be mutually independent. 2

The problem is to evaluate the scalar quantity l = a0 x(N ) (where a 2 Rn is a given vector) with the help of linear functionals N

(i) 2 Rm ; i = 0; : : : ; N;

^l() = X 0 (i)z (i); i=0

 = (0 (0); : : : ; 0 (N ))0 : (6)

The optimal mean-square ltering problem is to nd an estimator 0 that minimizes the mean-square estimation error

d(0 ) = Lemma 1.

inf 2Rm N (

+1)

d();

d() =

n

(^l()

E

o l )2 : 1 2

(7)

The solution of Volterra equation (1) has the form

x(j + 1) = R(j + 1; 0)x(0) +

j

X

k

=0

j  0;

R(j + 1; k + 1)B (k)u(k);

(8)

where the resolvent matrix R(t; s) (t; s  0) is de ned by the following equations

R(j + 1; t) = R(j + 1; t) =

j

X

=

k t

j

X

=

k t

(9)

t  j;

R(j + 1; k + 1)A(k; t);

R(t; t) = En ; and En is identity matrix of order n. Proof.

j  t;

A(j; k)R(k; t);

(10)

R(t; s) = 0; s > t;

(11)

For a xed s  0, consider an equation

y (t + 1) =

t X

=

k s

A(t; k)y (k) + B (t)u(t);

y (s) = y0 ;

0  s  t:

(12)

Multiplying equation (12) from the left by a matrix R(j + 1; t + 1), j  t, and summing the corresponding equations over all t, s  t  j , we have j

X

=

t s

R(j + 1; t + 1)y (t + 1) =

j

X

=

t s



R(j + 1; t + 1)

t X

=

k s



A(t; k)y (k) + B (t)u(t) :

Modifying the left-hand side of this equation and interchanging the summation order we obtain the relation

R(j + 1; j + 1)y (j + 1) R(j + 1; s)y0 + =

j

j

XX

R(j + 1; k + 1)A(k; t)y (t) +

= = whence it follows that t sk t

3

j

X

=

k s

j

X

=

t s

R(j + 1; t)y (t)

R(j + 1; k + 1)B (k)u(k)

R(j + 1; j + 1)y (j + 1) +

j  X

=

t s

R(j + 1; t)

= R(j + 1; s)y0 +

j

X

j

X

=

k t



R(j + 1; k + 1)A(k; t) y (t)

R(j + 1; k + 1)B (k)u(k):

= If we de ne the matrix R(j + 1; t) by (10) and (11), then we get that

y (j + 1) = R(j + 1; s)y0 +

j

X

=

k s

k s

R(j + 1; k + 1)B (k)u(k);

0  s  j:

(13)

Thus, taking s = 0 and putting y (j )  x(j ), we obtain the desired relation (8). Now let us derive (9). Consider the homogeneous case: B (k)u(k)  0. Then we have from (13) that y (j + 1) = R(j + 1; s)y0 for any vector y0 : Hence, every column of R(j + 1; t) satis es (12) with B (t)u(t)  0 and therefore (9) holds. 2 By virtue of (1), (3), (6), and (8), the estimation error is given by the following equalities: N ^l l = X 0 (i)z (i) a0 x(N ) (14) i=0 N X

= +

=0

i

1

N X i X i

N X

=

=0

i

1

N X k

!

=0

0 @

0 (i)H 0 (i)R(i; 0)

N X

= +1

i k

N

X a0 R(N; 0) x(0) + 0 (i)%(i)

0 (i)H 0 (i)R(i; k + 1)B (k)u(k) a0

=1 k=0

+

0 (i)H 0 (i)R(i; 0)

1

N X k

!

=0

=0

i

R(N; k + 1)B (k)u(k) N

X a0 R(N; 0) x(0) + 0 (i)%(i)

=0

i

1

0 (i)H 0 (i)R(i; k + 1) a0 R(N; k + 1)A B (k)u(k):

Let us introduce the quantities: (F1) a function   (j ) that, for a given f(j )gN0 , is de ned by the following dierence equation: N X1   (j ) = A0 (k; j ) (k + 1) H (j )(j ); (15) k =j   (N ) = a H (N )(N ); j = 0; : : : ; N 1; (F2) a functional J ( ; ; w): Rn  Rm(N +1)  RrN ! R1 : 

i

1

N

N X

=0

j

X J ( ; ; w) = 0 P0  + 0 (i)R(i)(i) +

=0

where w(j + 1) 2 Rr , w = (w0 (1); : : : ; w0 (N ))0 2 RrN . 4

1

w0 (j + 1)Q(j )w(j + 1)

2

;

(16)

Using (10) and (15) we can easily show (e.g. by direct substitution) that

R0 (N; k)a

N X

=

i k

R0 (i; k)H (i)(i) =   (k);

k = 0; : : : ; N:

(17)

Therefore, by virtue of (14), (17), we have

 0 (0)x(0) +

^l l =

N X

=0

i

0 (i)%(i)

1

N X j

=0

  0 (j + 1)B (j )u(j ):

(18)

Hence, from (2), (4), (5), (7), (16), and (18), we obtain that

d() = J ( (); ; w()); where  () and w() are de ned by the constraints    (0) = 0; w(j + 1) B 0 (j )  (j + 1) = 0;

(19)

j = 0; : : : ; N

1:

(20)

Thus, the optimal mean-square ltering problem reduces to the following linear-quadratic variational problem:

J0 =

inf J ( ; ; w)  2Rn ; 2Rm(N +1) ; w2RrN

(21)

under constraints (20). 1.2

Basic Relations for the Filtering Problem Solution

Let us formulate and prove the basic results for the solution of the ltering problem using the notion of a primal and dual variational problems. 0 0 0 0 Theorem 1. 1 . The solution f ;  ; w g of the linear-quadratic problem (20), (21) is given by the relations 0 =  (0); (22) 0 (i) = R 1 (i)H 0 (i) (i);

i = 0; : : : ; N;

w0 (j + 1) = B 0 (j ) (j + 1); j = 0; : : : ; N 1; where  (j ),  (j ) satisfy the boundary value problem N X1  (j ) = A0 (k; j ) (k + 1) H (j )R 1 (j )H 0 (j ) (j ); k =j  (j + 1) =

j

X

k

=0

A(j; k) (k) + B (j )Q(j )B 0 (j ) (j + 1); j = 0; : : : ; N

1;

with the boundary conditions

 (N ) = a H (N )R 1 (N )H 0 (N ) (N ); 5

 (0) = P0  (0):

(23)

20 . Boundary value problem (23) has a unique solution. 30 . The dual problem to the primal problem (20), (21) has the form J0 = sup a0 x~(N ; p; ) rN p2R ; 2Rn

(24)

under the constraint (

0 P0 1  +

N X

x~0 (j ; p; )H (j )R 1(j )H 0 (j )~x(j ; p; ) j =0 ) N X1 0 1 + p (j + 1)Q (j )p(j + 1) j =0 where the process x~(j ; p; ) satis es the Volterra equation x~(j + 1; p; ) =

j

X

k

=0

A(j; k)~x(k; p; ) + B (j )p(j + 1);

p(j + 1) 2 Rr ;  2 Rn;

(25) 1 2

 1;

x~(0; p; ) = ;

j = 0; : : : ; N

1:

Moreover,

J0 = J 0 : 40 . The solution for the dual problem (24), (25) is determined by the formulas p0 (j + 1) = J0 1 Q(j )B 0 (j ) (j + 1); 0 = J0 1 P0  (0); j = 0; : : : ; N 1: Note that (22) and (23) imply the equality   (j ) =  (j ); j = 0; : : : ; N:

(26) (27)

(28)

0

Let us introduce the Euclidean space H = Rn  Rm(N +1)  RrN with elements = ( ; ; w) and the inner product

Proof.

h

(1) ; (2) iH = 0 (1)  (2) +

N X

=0

i

0(1) (i)(i) (i) +

N X j

=1

0 (j )w (j ): w(1) (2)

So, J in (16) maps H into R1 . Let us also introduce the Euclidean space H~ = RnRrN with the elements ~ = (g; G) and the inner product 0 g + h ~ (1) ; ~ (2) iH~ = g(1) (2)

N X j

=1

G0(1) (j )G(2) (j );

~ (i) 2 H~ ;

g(i) 2 Rn ; G(i) (j ) 2 Rr ; i = 1; 2: Taking into account formula (17), the constraints (20) can be rewritten in the form R0 (N; 0)a

N X k

=0

R0 (k; 0)H (k)(k)  = 0; 6

(29)



B 0 (j ) R0 (N; j + 1)a



N X

R0 (k; j + 1)H (k)(k)

= +1 j = 0; : : : ; N 1: So, the variational problem (21), (20) takes the form k j

w(j + 1) = 0;

J0 = inf J ( ; ; w) under the constraint ( ; ; w) = 0; H

(30)

where the aÆne operator  is de ned by the left-hand side of (29). The cost functional J ( ; ; w): H ! R1 de ned by (16) is a convex continuous functional, the operator : H ! H~ is a continuous aÆne operator. Let us further assume for simplicity that the matrices P0 and Q(j ) are positive de nite. Note that Theorem 1 remains valid for singular matrices as well. In this case, one should turn to the matrices P0 + In , Q(j ) + Ir and let  ! 0 (here In and Ir are the identity matrices of order n and r). The corresponding calculation is similar to this in [9]. Remark 1.

With nonsingular matrices P0 and Q(j ) one can easily show that J ( ; ; w) is coercive. Hence an optimal solution (0 ; 0 ; w0 ) exists [11]. Now let us construct a dual problem [10],[11]. The perturbation of problem (30) has the form

J0 (Æa; Æw) = inf J ( ; ; w) H under the constraint !

Æa Æw

( ; ; w) =

Æa 2 Rn ; Æw 2 RrN ;

;

(31)

where Æw = (Æw0 (1); : : : ; Æw0 (N ))0 . With Æw = 0 and Æa = 0 we get the initial problem (21), (20). The Lagrange function for problem (21), (29) has the form 

N

N

=0

j

1

1

X X LH( ; ; w; p; ) = 0 P0  + 0 (i)R(i)(i) + w0(j + 1)Q(j )w(j + 1)

i

+

1

N X j

=0

p0 (j + 1)



w(j + 1) B 0 (j )

N X

= +1

k j



+ 0 R0 (N; 0)a

N X

=0

2

(32) 

R0 (k; j + 1)H (k)(k) + B 0 (j )R0 (N; j + 1)a R0 (k; 0)H (k)(k)



 ; =0 where p(j + 1) 2 Rr ,  2 Rn are Lagrange multipliers. It can be easily veri ed that J0 (Æa; Æw) is convex [10],[11]. Obviously, the triple (Æ ; 0; wÆ ), where Æ = R0 (N; 0)a Æa;

k

wÆ = (wÆ 0 (1); : : : ; wÆ 0 (N ))0 ;

wÆ (j + 1) = B 0 (j )R0 (N; j + 1)a Æw(j + 1); 7

j = 0; : : : ; N

1;

satis es (31). Consequently J0 (Æa; Æw)  J (Æ ; 0; wÆ ) and therefore J0 (Æa; Æw) is bounded in a neighborhood of (0; 0). Then, using fundamental theorems of the duality theory [10],[11], we can establish that J0 (Æa; Æw) is continuous at (0; 0) and has a non-empty subdierential. Moreover, the dual problem

J0 =

(

sup

inf  2Rn ;2Rm(N +1) ; w2RrN

2RrN ; 2Rn

p

LH( ; ; w; p; )

)

has a solution (p0 ; 0 ), the Lagrange function has a saddle point: LH(0 ; 0 ; w0; p; )  LH(0 ; 0 ; w0; p0; 0)  LH( ; ; w; p0; 0)  2 Rm(N +1) ; w; p 2 RrN ;

8  ;  2 Rn; and the duality relation holds:

J0 = J 0 : Interchanging the sums, we obtain from (32) that

(33)

LH( ; ; w; p; ) = k kH h ; iH + ~0(N )a with



(34) 

= P0  ; fR (i)(i)g0 ; fQ (j )w(j + 1)g0 1 ; 

1 2

1 2

 = P0 ; fR

where

1 2

1 2



~(i) = R(i; 0) +

1

i X j

=0



R(i; j + 1)B (j )p(j + 1) ; i = 1; : : : ; N:

Since

inf LH ( ; ; w; p; ) =  ;; w the dual problem takes the form sup

2RrN ; 2Rn

p

N

1 (i)H 0 (i)~(i)gN0 ; fQ 2 (j )p(j + 1)gN0 1 ;



~(0) = ;

1 2

N



a0 R(N; 0) +

1

N X j

=0

(

(35)

a0 ~(N ) if kkH  1 1 otherwise;

R(N; j + 1)B (j )p(j + 1)



(36)

under the constraint (

0 P0 1  +



N X j



R(j; 0) +

=0 1

jX

1

jX k

=0

0 R(j; k + 1)B (k)p(k + 1) H (j )R 1(j )H 0 (j ) 



1

N X

(37)

)1

p0 (j + 1)Q 1 (j )p(j + 1)

 1: =0 k =0 Obviously, taking into account formula (8), the problem (36), (37) can be rewritten in a more compact form (24), (25).  R(j; 0) +

R(j; k + 1)B (k)p(k + 1) +

8

j

2

The Euclidean space RrN  Rn is nite-dimensional, the cost function a0 ~(N ) is continuous (linear), and constraint (37) de nes a compact set. Hence we see again that a solution (p0 ; 0 ) for (36), (37) exists. Using the saddle point property, from (34) we nd that, for the minimum point 0 , k 0kH h 0; 0 iH  k kH h ; 0iH; 8 2 H; where 0 = (p0 ; 0 ) is determined by a solution of the dual problem. Consequently k 0 kH = h 0 ; 0 iH and therefore 0 = k0 for a k 2 R1 . In other words, 0 = kP0 1 0 ; (38) 0 (i) = kR 1 (i)H 0 (i)~ (i) p=p ; = ; i = 0; : : : ; N; w0 (j + 1) = kQ 1 (j )p0 (j + 1); j = 0; : : : ; N 1; for a k 2 R1 . Let us now nd the solution (p0 ; 0 ) of dual problem (36), (37). Without loss of generality, the exponent 21 in condition (37) can be omitted. In accordance with the Kuhn-Tucker theorem [10],[11], the corresponding Lagrange function achieves its minimum at the minimum point (p0 ; 0 ) of (36), (37). The Lagrange function has the form    N X1 L0(p; ; 0; ) = 0 a0 R(N; 0) + R(N; j + 1)B (j )p(j + 1) (39) j =0  0 jX 1 N  X 0 1 +   P0  + R(j; 0) + R(j; k + 1)B (k)p(k + 1) H (j )R 1(j )H 0 (j ) j =0 k =0    jX 1 N X1  R(j; 0) + R(j; k + 1)B (k)p(k + 1) + p0(j + 1)Q 1(j )p(j + 1) ; j =0 k =0 where 0 ;  are non-negative and are not both zero. First we show that one can put 0 = 1,  > 0. Obviously, the Slater condition holds, i.e. there exists a pair (p; ) = (0; 0) such that inequality (37) is strict. Hence, we can put 0 = 1. Suppose  = 0; then   N X1 a0 R(N; 0)0 + R(N; j + 1)B (j )p0 (j + 1) (40) j =0   N X1 0  a R(N; 0) + R(N; j + 1)B (j )p(j + 1) 8 (p; ): j =0 So, inequality (40) implies 0

0

a0 R(N; 0) = 0 and a0 R(N; j + 1)B (j ) = 0;

j = 0; : : : ; N

1:

(41)

By virtue of (8) equalities (41) mean that a0 x(N ) = 0 for all x(0) and u(j ). Clearly, this is a very degenerate case. Moreover, it follows from (29) that  = 0,  = 0, w = 0 are admissible elements for the variational problem. Consequently, the optimal estimator is 9

zero and the minimax problem is resolved with J0 = 0. In the sequel, we omit this trivial case. Then it follows from (40) that  > 0. Note that if, nevertheless, (41) holds, then it can be shown that assertions 10 -30 of Theorem 1 are still valid. The function L0 is linear-quadratic with positive de nite quadratic form. Therefore 0 L is convex and its minimum is given by a stationary point [10],[11] @ L0 @ L0 = 0; = 0; j = 0; : : : ; N 1: @p(j + 1) @ Direct calculation of partial derivatives in these relations lead to the following equations for (p0 ; 0 ): B 0 (j )R0 (N; j + 1)a + 2Q 1 (j )p(j + 1) + 2B 0 (j ) and

N X

= +1

k j

R0 (k; j + 1)H (k)R 1 (k)H 0(k)~ (k) = 0;

R0 (N; 0)a + 2P0 1 + 2

N X

1;

R0 (k; 0)H (k)R 1(k)H 0 (k)~(k) = 0;

=0 where ~(k) is de ned by (35). Thus we get p0 (j + 1) = Q(j )B 0 (j )~(j + 1); 0 = P0 ~(0); k

j = 0; : : : ; N

j = 0; : : : ; N

1;

(42)

where, for (p; ) = (p0 ; 0 ),

~(j ) =

R0 (N; j )a ; R0 (k; j )H (k)R 1(k)H 0 (k)~ (k) + 2 k =j N X

(43)

jX 1 ~ ~(j ) = R(j; 0)P0  (0) + R(j; k + 1)B (k)Q(k)B 0 (k)~(k + 1); k =0 j = 0; : : : ; N (remind that ~(0) def = P0 ~(0)): Similar to (15), (17), from (9), (10), and (43) we nd that the functions ~(j ) and ~(j ) satisfy the boundary value problem N X1 ~(j ) = A0 (k; j )~(k + 1) H (j )R 1 (j )H 0 (j )~(j ); (44) k =j

~(j + 1) =

j

X

k

=0

A(j; k)~ (k) + B (j )Q(j )B 0 (j )~(j + 1); j = 0; : : : ; N

1;

with the boundary conditions a H (N )R 1 (N )H 0 (N )~ (N ); ~(0) = P0 ~(0): ~(N ) = 2 Problem (44) has a unique solution. In fact, in the contrary case there exist nonzero functions Æ (j ) and Æ (j ) that satisfy (44) with the boundary conditions Æ (N ) = H (N )R 1 (N )H 0 (N )Æ (N ); Æ (0) = P0 Æ (0): (45) 10

Multiplying the rst equation by Æ 0 (j ), the second one by Æ 0 (j + 1), we obtain N X1 Æ 0 (j )Æ (j ) = Æ 0 (j ) A0 (k; j )Æ (k + 1) Æ 0 (j )H (j )R 1 (j )H 0 (j )Æ (j ); (46) k =j

Æ 0 (j + 1)Æ (j + 1) = Æ 0 (j + 1)

j

X

A(j; k)Æ (k) + Æ 0 (j + 1)B (j )Q(j )B 0 (j )Æ (j + 1):

=0 Subtracting the rst equality of (46) from the second one, we have k

Æ 0 (j + 1)Æ (j + 1) Æ 0 (j )Æ (j )

(47)

= Æ 0 (j + 1)B (j )Q(j )B 0 (j )Æ (j + 1) + Æ 0 (j )H (j )R 1 (j )H 0(j )Æ (j ) j N X X1 + Æ 0(j + 1) A(j; k)Æ (k) Æ 0(k + 1) A(k; j )Æ (j ); k =0 k =j j = 0; : : : ; N 1: Summing up the equalities (47) with boundary conditions (45) and interchanging the sums we get N X1 Æ 0 (0)P0 Æ (0) + Æ 0 (j + 1)B (j )Q(j )B 0 (j )Æ (j + 1) j =0 +

N X j

whence, 1 2

P0 Æ (0) = 0;

=0

Æ 0(j )H (j )R 1 (j )H 0 (j )Æ (j ) = 0;

Q (k)B 0 (k)Æ (k + 1) = 0; 1 2

R (j )H 0(j )Æ (j ) = 0; 1 2

(48)

k = 0; : : : ; N 1; j = 0; : : : ; N: If (48) holds, then we have from the boundary value problem (44) and boundary conditions (45) that N X1 Æ (j ) = A0 (k; j )Æ (k + 1); Æ (N ) = 0; k =j Æ (j + 1) =

j

X

k

=0

A(j; k)Æ (k);

Æ (0) = P0 Æ (0);

j = 0; : : : ; N 1: Consequently, Æ (j )  Æ (j )  0 and the solution of (44) is unique. The existence of the solution for (44) is guaranteed by the existence of the solution (p0 ; 0 ) for the dual problem. Then the solution of (23) also exists and is unique and, obviously,  (j ) = 2~(j );  (j ) = 2~(j ): (49) Substituting (38) into (29), by virtue of (43), we obtain !

1

k R0 (N; 0)a = 0; 2 11

(50)

!

k 1 B 0 (j )R0 (N; j + 1)a = 0; j = 0; : : : ; N 1: 2 Since trivial case (41) is excluded, (50) implies k = 2. Then from (38), (42), (49) it follows that 0 =  (0); (51) 0 (i) = R 1 (i)H 0 (i) (i);

w0 (j + 1) = B 0 (j ) (j + 1);

i = 0; : : : ; N; j = 0; : : : ; N

1:

With  > 0 in (39), the complementary slackness condition of the Kuhn-Tucker theorem [10],[11] gives that at the maximum point inequality (37) turns into the equality. Thus we have from (37), (42), (43), and (49) that 

N

X (42 ) 1  0(0)P0  (0) +  0 (j )H (j )R 1(j )H 0 (j ) (j ) (52) j =0  N X1 0 0 +  (j + 1)B (j )Q(j )B (j ) (j + 1) = 1: j =0 Condition (52) with (51) takes the form   N N X X1 (42 ) 1 0 0 P0 0 + 0 0 (j )R(j )0 (j ) + w0 0 (j + 1)Q(j )w0 (j + 1) = 1: j =0 j =0

Hence, taking into account formula (16) and duality relation (33), we get that 2 = J0 = J 0 : Moreover, it follows from (42), (49), and (53) that p0 (j + 1) = J0 1 Q(j )B 0 (j ) (j + 1); 0 = J0 1 P0  (0); Theorem 1 is proved.

j = 0; : : : ; N

(53) 1:

It follows from Theorem 1 that the nding of the optimal estimator reduces to the solving of the boundary value problem (23). Unfortunately, the solving of (23) can involve serious numerical diÆculties. Therefore we propose the following approach. Using natural simpli cation of the original ltering problem, we search a simpli ed estimator ' instead of 0 . We de ne the approximation quality by the ratio
=

d(') : d(0 )

(54)

Obviously,
 1. Since 0 is unknown,
is unknown. Then our aim is to construct an upper bound
0 for
that can be calculated without exactly solving the problem (21), (20). If
0 is close to unity, then the use of the estimator ' will be justi ed. The structure of the optimal estimator 0 given by Theorem 1 turns out to be very helpful to obtain the desired upper bound
0 . 12

2 Kalman Filter for Reduced System A natural way to simplify our Volterra equation is to replace it by the following reduced model in which we discount the \tails" y (j s 1); : : : ; y (0):

y (j + 1) =

j

X

=

k j s

A(j; k)y (k) + B (j )u(j );

j = 0; : : : ; N

1;

(55)

y (0) = x(0); A(j; k) = 0 for k < 0: The noise u(j ) 2 Rr is a zero-mean white noise sequence with a covariance matrix Q (j ): Eu (j ) = 0; E ( u(j )u0 (k)) = Q (j )Æjk ; j; k = 0; : : : ; N 1: (56) Let the model for the measurements be de ned by the equation

z (k) = H 0 (k)y (k) + %(k);

z (k); %(k) 2 Rm ;

k = 0; : : : ; N:

(57)

The measurement noise %(k) is a zero-mean white noise sequence with a covariance matrix R (k): E% (k) = 0; E ( %(k)%0 (s)) = R (k)Æks; k; s = 0; : : : ; N: (58) Obviously, the model equation (55) has the same coeÆcients as equation (1) but with

A(j; k) = 0; In particular, we can put Q (j ) = 1 Q(j );

j

R (k) = 2 R(k);

k > s:

(59)

1 ; 2 2 R1 ; 1 ; 2 > 0;

(60)

j = 0; : : : ; N 1; k = 0; : : : ; N; where parameters 1 , 2 are set by a designer. The model system (55){(58) can be represented in a single-step form by introducing the augmented vector state X (j ) = (y0(j ); y0(j 1); : : : ; y0(j s))0; X (j ) 2 Rn(s+1): Then the system (55){(58) turns into the system

X (j + 1) = A(j )X (j ) + B(j )u(j ); j = 0; : : : ; N 1; (61) z (k) = H0 (k)X (k) + %(k); k = 0; : : : ; N; (62) 0 EX (0) = 0; E (X (0)X (0)) = P0 and a scalar l = a~0 X (N ) is to be estimated. The augmented matrices A(j )  fn(s +1) n(s + 1)g, X (0)  fn(s + 1)  1g, B(j )  fn(s + 1)  rg, H(k)  fn(s + 1)  mg, a~  fn(s +1)  1g, P0  fn(s +1)  n(s +1)g are de ned by the following equalities: 0

A(j ) =

B B B B @

A(j; j ) A(j; j 1) : : : A(j; j s + 1) A(j; j s) En 0 ::: 0 0 .. .. .. .. . . . . 0 0 ::: En 0 13

1 C C C C A

;

0

1

0

x(0) B C y B ( 1) C B X (0) = B .. CC; @ . A y ( s) 0

1

B (j ) B C B 0 B B(j ) = B .. CCC; @ . A 0 1

a B C B 0 C B a~ = B .. C C; @ . A 0

0

0

1

H (k) B 0 C B B H(k) = B .. CCC; @ . A 0 1

P0 0 : : : 0 B En : : : 0 C B 0 C B P0 = B .. .. : .. C @ . A . . C 0 0 : : : En

(63)

The quantities y ( 1); : : : ; y ( s), which enter into X (j ) for j = 0; : : : ; s 1, are multiplied by zero matrices A(j; k), k < 0. Therefore y ( 1); : : : ; y ( s) may be assigned arbitrarily. In particular, if we need the matrix P0 be positive de nite, then we can put Ey (k)y 0(k) = En for k = 1; : : : ; s as this is done in (63). In case the system (1) is stationary, i.e. A(j; k) = A(j k), it can be convenient to keep this feature for k < 0 as well. Then y ( 1); : : : ; y ( s) must be put equal to zero. System (61), (62) has a standard Kalman form. Therefore the suboptimal linear mean-square estimate ^l' is yielded by the discrete Kalman lter [12]-[14] Remark 2.

^l' = a~0 X^ (N jN );

X^(j jj ) = X^(j jj 1) + K(j )(z(j ) H0(j )X^(j jj 1)); j = 0; : : : ; N; X^ (j jj 1) = A(j 1)X^ (j 1jj 1); j = 1; : : : ; N; X^(0j 1) = 0; K(j ) = P (j jj 1)H(j )(H0 (j )P (j jj 1)H(j ) + R (j )) 1 = P (j jj )H(j )R 1 (j ); P (j jj ) = (E K(j )H0 (j ))P (j jj 1); j = 0; : : : ; N; P (0j 1) = P0 ; P (j jj 1) = A(j 1)P (j 1jj 1)A0(j 1) + B(j 1)Q (j 1)B0 (j 1);

(64)

(65)

j = 1; : : : ; N: It follows directly from (64), (65) that N

^l' = X '0 (i)z (i); '(i) = K0 (i)0 (N; i)~a; i = 0; : : : ; N; (66) i=0 N Y1
(N 1)
: : :
(i + 1)
(i);
E (N; i)= (j ) = (N; N ) = n(s+1) ; j =i (i) = (En(s+1) K(i + 1)H0 (i + 1))A(i); i = 0; : : : ; N 1: Thus for moderate s the estimator '(i) can be computed by a simple and reliable algorithm. 14

3 Level of Nonoptimality for Simpli ed Filter We shall use the estimator ' obtained from the solution of the reduced system (55){(58) instead of the optimal one. Theorem 3. Let ' be a simpli ed estimator de ned by (66). Then the level of nonoptimality (54) satis es the following inequalities:

J '
k'
0 = 0 ' ja x~ (N )j ;

1 

0 ;

(67)

where n

J ' =  ' 0 (0)P0  '(0) +

N X

=0

i

'0 (i)R(i)'(i) +

n

k' =  '0 (0)P0  ' (0) + +

N X

=0

i

1

N X j

=0

o1

 ' 0 (j +1)B (j )Q(j )B 0 (j ) '(j +1) ; (68) 2

x~' 0 (i)H (i)R 1 (i)H 0 (i)~x' (i) 1

N X

(69) o1

 ' 0 (j + 1)B (j )Q(j )B 0 (j ) ' (j + 1) ; 2

=0 the function  (j ), j = 0; : : : ; N , is given by (15), and the auxiliary process x~' (j ), j = 0; : : : ; N , is de ned by the Volterra equation j

'

x~ (j +1) = '

j

X

k

=0

A(j; k)~x' (k) + B (j )Q(j )B 0 (j ) ' (j +1)

(70)

with the initial value x~' (0) = P0  ' (0).

It follows from (19){(21) and (54) that the level of nonoptimality
can be rewritten in the form J ( ('); '; w(')) J ' = : (71) 1
= J0 J0 In order to get the upper bound for the level of nonoptimality (71) we should nd a lower bound for the optimal estimation error J0 . Let us denote by S the set of points (p; ) that belong to (25). Due to the assertion 30 of Theorem 1 J0 = J 0  a0 x~(N ; p; ) for any (p; ) 2 S : (72) Proof.

Let us construct a successful pair (p; ) for (72) basing on the simpli ed model introduced above. The reduced model (55){(58) approximates the initial Volterra system. Hence the estimator ' from (66) approximates the unknown optimal estimator 0 . Consequently, by virtue of (28), the function  '(j ) approximates the function  (j ). Therefore the pair (J0 1 p ; J0 1  ), where

p (j + 1) = Q(j )B 0 (j ) '(j + 1);  = P0  ' (0); j = 0; : : : ; N 1; (73) gives us an idea about the unknown optimal pair (p0 ; 0 ) determined by (27). In particular, the direction of the dual vector (J0 1 p ; J0 1  ) can be exploited to nd a successful 15

approximating pair (p; ) in (72). Basing on this reasoning, we shall search the lower bound in (72) on a linear subspace

M = (p; ) 2 S (p; ) =  (p ; );  2 R1 ;

n

o

where (p ;  ) is given by (73). Since M and S are balanced, M  S , and x~(N ; p; ) is linear in (p; ), we get from (70), (72), and (73) that, for all admissible  , J0 = J 0  ja0 x~(N ; p ;  )j = j j ja0 x~(N ; p ;  )j = j j ja0 x~' (N )j: Therefore, by virtue of (71), 1


J' j j ja0x~' (N )j :

(74)

Obviously, the minimum in (74) is attained with maximal admissible j j, that is for  such that the left-hand side of (25) equals 1. Since x~(j ; p ;  ) =  x~(j ; p ;  ) =  x~' (j ), the constraint (25) with strict equality takes the form 

j j  '0(0)P0 '(0) + +

N X

=0

i

x~' 0 (i)H (i)R 1 (i)H 0 (i)~x' (i) 1

N X j

=0

 0 (j + 1)B (j )Q(j )B 0 (j ) '(j + 1) '

(75) 1 2

= 1:

Then the desired estimate (67) follows from (74) and (75) with j j 1 = k' .

2

The algorithm for calculating
0 consists of the following steps.  The calculation of the Kalman gains fK(j )gN0 1 de ned by (65).  The calculation of the simpli ed estimator f'(i)gN0 by formulas (66).  The calculation of the approximate function f '(j )gN0 by dierence equation (15).  The calculation of the auxiliary process fx~'(j )gN0 , in accordance with Volterra equation (70).  The calculation of the quadratures J ' and k' by (68) and (69).  The calculation of
0 by formula (67). Note that all these operations can be easily implemented.

4 Numerical Examples Let us study two simple examples of discrete Volterra equations. Example 1. First consider a stationary scalar equation of the form

x(j + 1) =

j

X

k

=0

j k+1x(k) + u(j ); 16

j = 0; : : : ; N

1;

(76)

Table 1: Example 1: Values of
0 ; lter order s = 8

N 120 160 200 240 280 320 360
0 1.22 1.34 1.48 1.64 1.82 2.06 2.24
0opt 1.21 1.33 1.47 1.62 1.79 1.97 2.16 1 0.8 0 .7 0.7 0.7 0.7 0.7 0.7 2 1.0 1.0 1.0 1.0 1.0 1.0 1.0

400 2.41 2.36 0.7 1.0

Table 2: Example 1: Values of
0 ; lter order s = 9

N 120 160 200 240 280 320 360
0 1.07 1.11 1.16 1.23 1.30 1.39 1.48
0opt 1.06 1.11 1.16 1.22 1.29 1.38 1.47 1 0.9 0 .9 0.8 0.8 0.8 0.7 0.7 2 1.0 1.0 1.0 1.0 1.0 1.0 1.0

400 1.58 1.56 0.7 1.0

with measurements

z (k) = x(k) + %(k);

k = 0; : : : ; N;

where  is a speci ed scalar, x(0) is a given zero-mean random variable with a variance  2 , the sequences u(j ) and %(k) are scalar zero-mean white noise sequences with constant variances Q and R, respectively. The quantity x(N ) is to be evaluated. Obviously, this is a special case of system (1), (3) with n = r = m =1, A(j; k) = j k+1; B (j ) = 1; H (k) = 1; a = 1: We put

 = 0:5;  2 = 100:0; Q = 1:0; R = 1:0: The values of
0 for various N at two values of the lter order s are shown in Table 1 and Table 2. The quantities
0 for 1 = 2 = 1:0 are presented in the rst row. The quantities
0 that were optimized in scales 1 ; 2 are cited in the second row. The corresponding optimal values of 1 ; 2 are indicated in the third and fourth row. Note that in spite of the modest value of  = 0:5, the estimated process (76) does not decay: Ex(N ) = 0, Ex2 (N )  N for large N . Example 2.

Now consider a two-dimensional Volterra equation

x(j + 1) = x(k) = (x1 (k); x2 (k))0 2 R2 ;

j

X

k

=0

!



j k

+1 w 1 x(k) + u(j ); 0 1

u(j ) = (u1 (j ); u2 (j ))0 2 R2 ; 17

(77)

j = 0; : : : ; N

1;

Table 3: Example 2: Values of
0 for N = 100 and w = 0:8

s
0
0opt 1 2

8 9.74 3.03 42.0 1.0

9 4.50 2.41 22.0 1.0

10 2.33 1.77 9.0 1.0

11 1.43 1.33 4.0 1.0

12 1.12 1.11 2.0 1.0

13 1.03 1.03 1.2 1.0

Table 4: Example 2: Values of
0 for N = 100 and w = 1:0

s
0
0opt 1 2

14 9.43 6.38 70.0 1.0

15 4.66 3.73 21.0 1.0

16 2.44 2.22 7.0 1.0

17 1.48 1.45 2.5 1.0

18 1.14 1.13 1.4 1.0

19 1.03 1.03 1.1 1.0

and scalar measurements

z (k) = x1 (k) + %(k);

k = 0; : : : ; N:

The aim is to estimate the second component x2 (N ). Here  is a speci ed scalar, x(0) is a given zero-mean random vector with a covariance matrix P0 , the sequences u(j ) and %(k) are zero-mean white noise sequences with a covariance matrix Q and a variance R. This is a special case of system (1), (3) with n = r = 2, m = 1, !

A(j; k) = 

j k

+1 w 1 ; 0 1

!

B (j ) = 10 01 ;

!

H (k) = 10 ;

!

a = 01 :

We set !

!

 = 0:5; P0 = 1000 :0 1000 :0 ; Q = 10:0 10:0 ; R = 1:0: The quantities of
0 for N = 100 and various values of lter order s with: (a) w = 0:8 and (b) w = 1:0 are shown in Table 3 and Table 4, respectively. The zero-mean solutions of (77) are also unbounded: for w = 0:8 Ex21 (N )  N ; for w = 1:0 Ex21 (N )  N 3 . It follows from Tables 1-4 that the reduced lters of moderate orders can be successfully used for ltering in Volterra equations. Moreover, in some cases, the scales 1 ; 2 can be adjusted to improve the nonoptimality level. In Table 5 the results for large values of N are presented ( 1 = 2 = 1:0). For each example, we indicate the following quantities: (a) the reduced lter minimal order s such that
0  1:20; (b) the nonoptimality level
0 that corresponds to the s; (c) a rough 18

Table 5: The case of large N

N = 1000 N = 2000

Example 1 s
0 to =ts 12 1.14 15 14 1.07 40

Example 2, w = 0:8 s
0 to =ts 16 1.15 20 17 1.19 70

Example 2, w = 1:0 s
0 to =ts 28 1.18 4 32 1.07 10

estimate for a ratio of the calculation time t0 of the optimal algorithm from [8] to the calculation time ts of our simpli ed algorithm. We see from Table 5 that, for large N , the calculation time of the full-dimensional optimal algorithm from [8] is, as a rule, one order greater than the calculation time in our approach. Note that even for N  100 our algorithm reduces the calculation expenditure hundred of times relative to the direct Kalman lter. The lters presented in the paper seem to be the most natural ones to start the design. They were chosen to emphasize the main idea. One can propose more sophisticated numerically eective schemes.

5 Conclusion EÆcient estimation algorithms for linear discrete Volterra equations are proposed. On the one hand, the ltering algorithms are quite easy for implementation. On the other hand, the levels of nonoptimality for these algorithms are constructed. It is important that the levels of nonoptimality can be easily computed without solving the original full-dimensional ltering problem. If the amount of steps is large, then the calculation time of our algorithm can be considerably less compared with the optimal lter from [8]. Thus our approach can be helpful in various applications. Similar approach was used earlier for many of estimation and control problems, in particular, for continuous dynamic systems with delay [9], [15]-[18]. The work was supported by the grant of the Royal Society (UK) and by the RFBR grant no. 03-01-00010.

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