Journal
of Applied Mathematics
and Stochastic Analysis, 12:1
(1999), 85-90.
LINEAR FILTERING WITH FRACTIONAL
BROWNIAN MOTION IN THE SIGNAL
AND OBSERVATION PROCESSES
M.L. KLEPTSYNA
Russian Academy of Sciences
Institute of Information Transmission Problems
Moscow 1017, Russia
marina @sci. lpi. ac. ru
P.E. KLOEDEN
Johan Wolfgang Goethe Universitiit
Fachbereich Mathematik
D-6005 Frankfurt am Main Germany
kloeden @math. uni-frankfurt, de
V.V. ANH
Queensland University of Technology
School
of Mathematical Sciences
Brisbane 001, Australia
v.anh@fsc.qut.edu.au
(Received December, 1997; Revised July, 1998)
Integral equations for the mean-square estimate are obtained for the linear
filtering problem, in which the noise generating the signal is a fractional
Brownian motion with Hurst index h E (3/4, 1) and the noise in the observation process includes a fractional Brownian motion as well as a Wiener
process.
Key words: Linear Filtering, Fractional Brownian Motion, LongRange Dependence, Optimal Mean-Square Filter.
AMS subject classifications: 93Ell, 60G20, 60G35.
1. Introduction
We consider the linear problem with the signal 0 and the observation
the linear equations
Ot--
J
a(s)Os ds+Bht, t
0
Printed in the U.S.A.
()1999 by North
/ A( )Osds+Wt+Bh
t,
t defined by
(1)
0
Atlantic Science Publishing Company
85
86
M.L. KLEPTSYNA, P.E. KLOEDEN and V.V. ANH
where the noise generating the signal is a fractional Brownian motion (fBm) Bth with
Hurst index h E (3/4, 1) and the noise disturbing the observation of the signal consists
of both a standard Wiener process W and the fractional Brownian motion Bth. The
coefficients a(t) and A(t) are bounded measurable functions and the noise processes
Bth and W are independent.
Fractional Brownian motion Bth with Hurst index h E (1/2,1)is often used to
model the long-range dependence in random data commonly encountered in many
financial and environmental applications [7, 9]. It is a zero mean Gaussian process
having the correlation function
1/2 (t
Fh(t,s)
2h
+ s 2h- It--sigh), 1/2<h<1.
(2)
It is known that Bth is not a semimartingale (see e.g. [4, 6]), so neither is the signal
process t nor the observation process t, and the martingale approach to filtering
expounded in [6] is not applicable here. In particular, as shown in [8], we cannot
uniquely determine an innovation process corresponding to t" Nevertheless, we can
derive an explicit expression for the conditional expectation of the signal
a E(e
o <_ _< t),
using a theorem on normal correlation in [5] provided we restrict the Hurst index h to
the interval (3/4, 1). We formulate this results as a theorem in the next section and
present its proof in Section 3. Finally, a simple example is provided in Section 4 to
illustrate the result.
2. The Optimal Filter
Let
.
t be the a-algebra r(s, 0 _s _< t)and note that
g(t, s) A_ E(0t0s), (t, s) A_ E(0tBsh).
Then it follows directly from the first equation of
the system of integral equations
(1)
that
K(t,s) and K(t,s) satisfy
a(l)K(t, l)dl + K (t, s),
K(t, s)
Define
(3)
0
"
/ a(l)( (1, s)ds + rh(t, s).
(t, s)
(4)
0
With these we can obtain an explicit closed-form representation of the optimal meansquare filter for system (1).
Theorem 2.1: There exists a unique deterministic function ( L2([0, T]2,)
satisfying
,(t,s)
/ O(t, )[h(2h
0
1) Is
-12- u
(5)
87
Linear Filtering with Fractional Brownian Motion
Oh" s)
+ A(s)-v (s, 7)]dv
+ A(s)A(r)K(v, s)+ A(v)--(7,
OK s)
+ A(s)K(t, s) + -f-(t,
such that the optimal mean-square filtering estimate 0
of the
linear system
(1)
satis-
fies
(6)
((t,s)ds
0
0
for t E [0, T],
where the integral is understood in the mean-square sense.
a solution. This solution is in fact unique.
Theorem 2.2: The system of integral equations (3)-(5) has a unique solution.
It follows from the proof of Theorem 1 that system (5) has
3. Proof of Theorem 1
We note that the joint distribution of (s, Ot) for all 0 < s,t
Theorem 13.1 of [5] on normal correlation holds here.
t be the dyadic partition of [0, t], that is, with t(n)=
J
2
and denote the ,-algebra r[,
,, -, ,...,, -t(t)_l )
t)=
for all t E
for all t E
<T
is Gaussian, so
-et 0. t(0n)< tn)< <
n,
Ant ,for j 0,1,...,2
,n
_
by
St
Then
[0, T]. Furthermore,
(7)
[0, T]. Hence using Theorem 13.1 of [5]
2n-1
we obtain
(Otl ,n)_ (Otl )+ n (t,tn,) (t(n)
(s)
[0, T], where (I)n:[0, T]2 is a deterministic function. Denote (I)(t, s)
n(t,t.n)) for n) _<s< tn)+ 1" Then we can rewrite (8)as
for all t
t
(9)
0
But the processes W and
(Bht,Ot) are independent, so
f
0
(I).(t,s)-am(t,s)12ds
M.L. KLEPTSYNA, P.E. KLOEDEN and V.V. ANIt
88
+z
On(t
s)(dBhs + A(s)0sds
m(t
0
s)(dBhs + A(s)Osds
0
(The integration of a deterministic function with respect to an fBm here is understood in the mean-square sense, cf. [4]). Applying (7) we obtain that
n,
/
limm_o
( (t’ s)- (m(t, s) 12ds
O,
0
L2[0, t].
so the sequence {(I)n} is a Cauchy sequence in
(I) E
t] such that
L2[O,
lirn
/
Hence there exist a function
O,(t, s)- O(t, s) 12ds O.
0
It then follows from [1] that
(I)n(t
nlirn:
(I)(t, s)dBh2
s)dBhs
0
0
and
lim IF
(b(t, s)A(s)Osds
On(t s)A(s)Osds
0
0
so we obtain
((t,s)ds.
0
0
We shall now show that (I) satisfies equation
jointly measurable function, so the integral
It:-
(5). Let f: [0, T]2---,R
be a bounded and
f f(t,s)d
0
is well-defined and the process
Ot)It)
I,
O. Consequently
is
t-measurable
f(t,s)d s
O(t,s)d s
0
with
0
-(It2) < oo
and
-((e t-
Linear Filtering with Fractional Brownian Motion
89
02 Fh(7 s)dT"ds
/ep(t,s)f(t )ds+/ i((t,’)f(t, sJOsOr
,s
0
0
+
0
i i O(t,v)f(t,s)A(v)A(s)K(’,s)drds
0
0
+
ep(t, 7)f(t,
s)A(s)-(s, 7)dvds
0
0
+
ep(t, 7)f(t, s)A(r
0
7,
s)dTds
0
and
f(t, s)---(t, s)ds +
f(t, s)ds
0
0
Now, f
so
0
f(t, s)A(s)K(t, s)ds.
0
02 rh(t s)
is otherwise arbitrary and the function 0-’
h(2h 1) It
s eh-
\OtOs
0
0
Bht
is equivalent in an innovation sense [8] to W t.
which means the process W +
Hence equation (5) is valid. Uniqueness follows from the linearity of the equations
under consideration.
This completes the proof of Theorem 2.1.
4. Example
Consider the case
a(t)
0 and
A(t)
O,
so system
(1) reduces to
(10)
By Theorem 2.1, the filtering estimate
equation
@(t, s)
f
0
0- f toeP(t,s)d
h(2h l),I,(t-)
s
s
-I
-
and (I) satisfies the integral
+1/2-s {t2h +s- it-sigh},
d(11)
which is a Fredholm integral equation with a singular kernel. However, for h > 3/4
this kernel is square integrable and equation (11) has a unique solution. In this case,
M.L. KLEPTSYNA, P.E. KLOEDEN and V.V. ANH
90
the filtering error
_
"),t-z([Ot-’tl2)is given by
%t 2h
(I
((l, s)ds
0
_2h_
/ O(t,s)
O(,s)+
h(2h-1)q(,’)ls-rl 2h-2dr ds
0
0
using
(11).
Acknowledgement
Partially supported by the Australian Research Council Grants A89601825 and
C1960019. The authors wish to thank the referee for his many constructive comments.
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[6]
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[7]
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[9]
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