On general filtering problem of stationary processes with fixed transformation

R E S E A R C H

Open Access

On general filtering problem of stationary

processes with fixed transformation

Long suo Li

Correspondence: [email protected] Department of Mathematics, Harbin Institute of Technology, Harbin 150001, P.R.China

Abstract

A fixed transformation are given for one-dimensional stationary processes in this paper. Based on this, we propose a general filtering problem of stationary processes with fixed transformation. Finally, on a stationary processes with no any additional conditions, we get the spectral characteristics ofPHη(t)ξ in the spaceL2(FX(dl)), and

then we calculate the value of the best predict quantityQ of the general filtering problem.

Keywords:stationary processes, fixed transformation, fillering

1. Introduction

The Prediction theory is an important part of stationary processes, also linear filter problems is an important part of Prediction theory. The linear filtering problem of multidimensional stationary sequence and processes for a linear system are firsted stu-died by Rosanov in [1], and then a series of general filter problem of stationary process for a linear system are studied in [2-9]. Theoretically, this problem is a extend of the classic prediction problem. But it has high practical value, it also widely applied in communication, exploration, space technology and automatic control, etc.

2. Propose the problem

LetX(t), tÎRbe (simple) wide stationary process. Let

HX=L{X(t), t∈R}

HX(t) =L{X(s), s≤t, s∈R}

Suppose the complex-value functionb(t) statisfing the following conditions 1)

b(t)∈L[0, +∞)∩L2[0, +∞) (2 1)

2)

b(t) for t<0 (2 2)

Let

B(λ) =

_∞

0

b(t)e−iλtdt (2 3)

then B(l) is boundary values analytic functionB(z) in the low half plane We can obtain

B(λ)= 0, a.e. Leb

Let

E={λ:B(λ) = 0}, E= (−∞, ∞)−E (2 4)

then

L(E) = 0

Let

η(t) =

_∞

0

b(s)·X(t−s)ds=

_∞

0

b(s)U−sX(t)ds (2 5)

whereUis the shift oprator ofX(t) in the spaceHX.

Then, for eachξÎHX

1) Find out the value ofQ,

Q= inf

ξ∈HX(t)

₀∞b(s)U−s_ξds−ξ 2

(2 6)

2) Then, we will prove that

Q= ||PHη(t)ξ−ξ||2

and solve the spectral characteristics ofPHη(t)in the spaceL2(FX(dl)).

3. Main result

Let the random spectral measure of X(t) is FX(dl), and the spectral measure isFX

(dl), the broad spectral measure which also named the spectral measure of absolutely continuous part of Fis fX(l).

The Lebesgue decomposition ofFX(dl) is

FX(dλ) =fX(λ)dλ+δ(dλ) (3 1)

where the Lebesgue measure of δ(dl) is singular, namelyδ(l) = cΔ(l)FX(dl)

⊂(−∞, ∞), L() = 0, = (−∞, ∞)−

Let

At=

h:h=

_∞

0

b(s)U−syds, y ∈HX(t)

(3 2)

Obviously, At is linear set.

_−∞∞f(λ)ϕ(λ)Z(dλ)≤M

_∞

∞ f(λ)Z(dλ)

Proof. According to the nature of the random integral, we have

_−∞∞f(λ)ϕ(λ)Z(dλ)

=

_∞

−∞|f(λ)ϕ(λ)|

2

F(dλ)

1

2

=

_∞

−∞|f(λ)|

2_{· |ϕ}

(λ)|2F(dλ)

12

≤M

∞

−∞|f(λ)|

2_·

F(dλ)

1

2

=M

_∞

−∞f(λ)Z(dλ)

Lemma 2.

L{At}=Hη(t) (3 3)

whereL(At)is the linear closed manifold ofAt.

Proof.∀h(τ0),τ0≤t, Lety=X(τ0), we havey ÎHX(t), and

η(τ0) =

_∞

0

b(s)U−syds=

_∞

0

b(s)U−sX(τ0)ds

namely,h(τ0)ÎAt, then

H_η(t)⊂L(At)

On the other hand, ∀hÎAt, we have

h=

_∞

0

b(s)U−syds

whereyÎ HX(t).

Let zl= ml

k=1

al_kX(tl_k),tl_k≤t,k= 1, 2,· · ·,ml,al

k, is complex number. Let

||y−zl|| →0 (l → ∞)

while

_∞

0

b(s)U−szlds= ml

k=1

al_k

_∞

0

b(s)U−sX(tl_k)ds∈H_η(t), l = 1, 2, · · ·

Let, the spectral characteristics of yand ztare ψy(l) andψzl(λ)in the space L2(FX

|B(λ)| =

_∞

0

b(s)e−iλsds≤

_∞

0 |

b(s)|ds∧=M

whereM >0 is constant. According to the lemma 1, we get

h−

_∞

0

b(s)U−szlds

=

_∞

0

b(s)U−s(y−zl)ds = _∞ 0 b(s) _∞ −∞e −isλ_(ψ

y(λ)−ψzl(λ))X(dλ)ds

=

_∞

−∞(ψy(λ)−ψzl(λ))

_∞

0

b(s)e−isλdsX(dλ)

=

_∞

−∞(ψy(λ)−ψzl(λ))B(λ)X(dλ)

≤ M

_∞

−∞(ψy(λ)−ψzl(λ))X(dλ)

= M

_∞

−∞|ψy(λ)−ψzl(λ)|

2_F(dλ)

1

2

= M||y−zl|| →0 (l → ∞)

so, hÎ Hh(t), namely

L{At} ⊂Hη(t)

Then, it shows the equation (3-3) is correct. According to the lemma 2, we have

Q= inf

ξ∈HX(t)

₀∞b(s)U−sξds−ξ

2

= inf

h∈At||h−ξ|| 2

= inf

h∈L{At}||h−ξ||

2

= inf

h∈Hη(t)||h−ξ|| 2

=||PHη(t)ξ−ξ||2

(3 4)

According to the equation (2-5) and (2-1), we get

η(t) =

_∞

0

b(s)X(t−s)ds=

_∞

0

b(s)

_∞

−∞e

i(t−s)λ X(dλ)ds

=

_∞

−∞e

itλ ∞

0

b(s)e−isλds

X(dλ) =

_∞

−∞e

itλ_B(λ) X(dλ)

on the other hand

η(t) =

_∞

−∞e

itλ η(dλ)

η(dλ) =B(λ)X(dλ) (3 5)

Fη(dλ) = |B(λ)|2FX(dλ) (3 6)

f_η(dλ) = |B(λ)|2fX(λ) (3 7)

where F_h(dl), Fh(dl), fh(l) representative the random spectral measure, spectral measure and broad spectral measure ofh(t).

Lemma 3. LetξÎHX,ξ =ξ+ξ, whereξ ∈Hη,ξ⊥Hη, then

Q= ||ξ||2+||PHη(t)ξ−ξ||2 (3 8)

If, the wold decomposition of h(t)is

η(t) =ηr(t) +ηs(t)

and

ξ =ξr+ξs

where hr is regular process, hs(t) is singular process, and _ξr_⊥S

η· ξs∈Sη,

S_η=

t

H_η(t)_{, then}

Q= ||ξ||2+||PHηr(t)ξr−ξr||2 (3 9)

Proof. According to_ξ=ξ+ξ,_{ξ ∈}H_η,_ξ⊥H_η, So

PHη(t)ξ=PHη(t)ξ+PHη(t)ξ =PHη(t)ξ

According to the equation (3-4),

Q = ||PHη(t)ξ−ξ||2= ||PHη(t)ξ −(ξ+ξ)||2

= ||(PHη(t)ξ−ξ)−ξ||2

also, according toξ⊥(PH_η(t)ξ−ξ), So

Q= ||ξ||2+||PHη(t)ξ−ξ||2

namely, the equation (3-8) is correct.

When h(t) have the wold decomposition, notice _ξS_=P

Hηξ, Hη(t) =Hηr(t)⊕S_η

we get

PHη(t)ξs =PHη(t)(PSηξ) =PSηξ =ξ

s _{(3 10)}

PHη(t)ξr=PHηr(t)ξr+PSηξ

r_=P

Hηr(t)ξr (3 11)

So

||PHη(t)ξ−ξ||2

= ||(PHη(t)ξr+PHη(t)ξs−(ξr+ξs)||2

Then, according to the equation (3-8), we get that the equation (3-9) is correct. Lemma 4. LetξÎHX, ξ=ξ +ξ, whereξ∈Hη,ξ⊥Hη,ψ(l) is spectral characteristics of ξin the space ofL2(FX(dl)),ψ(λ)is spectral characteristics ofξ in the space ofL2

(Fh(dl)), then 1) when_λ∈E,

ψ(λ) =ψ(λ)B(λ), a.e. FX(dλ) (3 12)

2)

||ξ||2₌

E|ψ

(λ)|2FX(dλ) (3 13)

Proof. 1) According to the given conditions, we have

ξ =

_∞

−∞ψ(λ)X(dλ)

_ξ =

∞

−∞

ψ(λ)η(dλ) =

∞

−∞

ψ(λ)B(λ)X(dλ)

η(t) =

_∞

−∞e

itλ η(dλ) =

_∞

−∞e

itλ_B(λ) X(dλ)

So

(ξ, η(−t)) =

_∞

−∞e

itλ_ψ(λ)B(λ)F X(dλ)

on the other hand

(ξ, η(−t)) = (ξ+ξ, η(−t)) = (ξ, η(−t))

=

_∞

−∞e

itλ_{ψ(λ)|B(λ)|}2

FX(dλ)

According to the Fourier transformation, we have

ψ(λ)B(λ) =ψ(λ)|B(λ)|2, a.e. FX(dλ)

So, when_λ∈E, we have

ψ(λ) =ψ(λ)B(λ), a.e. FX(dλ)

2) According to _ξ=ξ +ξ, and_ξ⊥ξ, Thus

||ξ||2_{= ||ξ+ξ||}2_{= ||ξ||}2_+||ξ||2

||ξ||2_{= ||ξ||}2_{− ||ξ||}2

= _∞

−∞|ψ(λ)|

2_F X(dλ)−

_∞

−∞|

ψ(λ)|2Fη(dλ)

= _∞

−∞|ψ(λ)|

2_F X(dλ)−

_∞

−∞|

ψ(λ)| · |B(λ)|2FX(dλ)

= ∞

−∞χE(λ)|ψ(λ)|

2_F X(dλ) +

∞

−∞χE(λ)|ψ(λ)|

2_F X(dλ)−

∞

−∞χE(λ)|

ψ(λ)|2_{· |B(λ)|}2_F X(dλ)

= ∞

−∞χE(λ)|ψ(λ)|

2_F X(dλ) =

E

namely, the equation (3-13) is correct.

Theorem. Let ξÎ HX,ξ =ξ+ξ, whereξ∈Hη,ξ⊥Hη,ψ(l) representative the spec-tral characteristics of ξ in the space ofL2(FX(dl)),ψ(λ)representative the spectral

characteristics of _ξ in the space ofL2(Fh(dl)). Then

1) WhenlogfX(λ)

1+λ2 ∈/ L1(−∞, ∞)

Q=

E|ψ

(λ)|2FX(dλ) (3 14)

now the spectral characteristics of PHη(t)ξin the space ofL2(FX(dl)) is

ψ(λ) =

⎧ ⎨ ⎩

0, λ∈E,

a.e. FX(dλ) ψ(λ),λ∈E.

(3 15)

2) WhenlogfX(λ)

1+λ2 ∈L1(−∞, ∞)

Q=

E

|ψ(λ)|2FX(dλ) +

_∞

t

|ϕ(−s)|2ds (3 16)

now the spectral characteristics of PHη(t)ξin the space ofL2(FX(dl)) is

ψ(λ) =

⎧ ⎪ ⎨ ⎪ ⎩

0, λ∈E,

ψ(λ), λ∈E,

B(λ)₋∞_te−isλ_ϕ(s)ds

(λ) ,λ∈E

(3 17)

whereb(t), B(l),h(t), E andΔis decided by the equation 1), 2), 3), 5), (2-4) and (3-1) respectively. Where Γ(l) is the maximal factor boundary values of the spectral density fh(l),(l) is the Fourier transformation ofψ(λ)(λ), where_ψ(λ)is determined by equation (3-12) and_ψ(λ) =ψ(λ)/B(λ), a.e.L.

Proof. 1) When log₁₊fX_λ(2λ) ∈/ L1(−∞, ∞). According to

f_η(λ) = |B(λ)|2fX(λ) (3 18)

log|B(λ)| ∈L1(−∞, ∞) (3 19)

we know that

logf_η(λ) 1 +λ2 ∈/ L

1_{(−∞, ∞)}

Thus, h(t) is singular process. So

H_η =S_η

PH_η(t)ξ=PH_ηξ =PS_ηξ =ξ

According to (3-8)

Q= ||ξ||2=

E

On the other hand

PHη(t)ξ=PHη(ξ+ξ) =PHηξ =PSηξ =ξ

The spectral characteristics ofPHη(t)ξin the spaceL2(FX(dl)) is

ψ(λ) =ψ(λ)B(λ), a.e. FX(dλ)

According to the equation (3-12), we get

ψ(λ) =

⎧ ⎨ ⎩

0, λ∈E,

a.e. FX(dλ) ψ(λ),λ∈E,

2) When log₁₊fX_λ(2λ) ∈L1(−∞, ∞), according to (3-18) and (3-19), we

getlogfη(λ)

1+λ2 ∈L1(−∞, ∞).

It shows that h(t) is non-singular process, soh(t) has regular singular decomposition, and it consistent with Lebesgue-Gramer decomposition. Let the decomposition equa-tion is

η(t) =ηr(t) +ηs(t)

wherehr(t) is regular process,hs(t) is singular process ηr_{(t) =}

_∞

−∞e

itλ ηr(dλ) =

_∞

−∞e

itλ_χ

(λ)η(dλ)

ηs_{(t) =}

_∞

−∞e

itλ ηs(dλ) =

_∞

−∞e

itλ_χ

(λ)η(dλ)

Let V(ds) is basic cross stochastic measure, namely

V(1) =

_∞

−∞

eiλt2−_eiλt1 iλ (dλ)

whereΔ1= (t1, t2], stochastic measure(dλ) = (1λ)ηr(dλ), Thus ηr_{(t) =}

t

−∞C(t−s)V(ds)

whenC(s) = ₂1_π−∞∞ eisλ_{(λ)dλ, andΓ(l) satisfate the follow conditions}

f(λ) = 1 2π|(λ)|

2

alsoΓ(l) is the boundary values of the lower half plane maximum analytic functions

Γ(z), notice _ξr₌

_∞

−∞

ψ(λ)χ(λ)η(dλ) =

_∞

−∞

ψ(λ)ηr(dλ)

=

_∞

−∞

ψ(λ)(λ)(dλ)

_ξs₌

_∞

−∞

ψ(λ)χ(λ)_η(dλ) =

_∞

−∞

Let (s) is the Fourier transforation of_ψ(λ)(λ), namely ϕ(s) = 1

2π

_∞

−∞e

isλ_{ψ(λ)(λ)dλ}

According to the Fourier transformation of random measure

_ξr₌

_∞

−∞

ψ(λ)(λ)(dλ) =

_∞

−∞

_∞

−∞e

−iλs_ϕ(s)ds(dλ)

=

∞

−∞

∞

−∞e

iλs_{ϕ(−s)ds(dλ) =} ∞

−∞ϕ(−s)V(ds)

Thus

PHηr(t)ξˆr= t

−∞ϕ(−s)V(ds) ||PH_ηr(t)ξˆr− ˆξr||2= _t∞ϕ(−s)V(ds)2=

_∞

t |ϕ(−s)| 2_ds

According to the equation (3-9), we get

Q=

E

|ψ(λ)|2FX(dλ) +

_∞

t

|ϕ(−s)|2ds

According to the Lemma 3

PHη(t)ξ=PHη(t)ξ =PHη(t)ξr+PHη(t)ξs=PHηr(t)ξr+ξs

notice

PHηr(t)ξr=

t

−∞ϕ(−s)V(ds) =

_∞

−∞

t

−∞e

iλs_{ϕ(−s)ds(dλ)}

=

_∞

−∞

∞

−t

e−iλsϕ(s)ds

1

(λ)ηr(dλ)

=

_∞

−∞

∞

−t

e−iλsϕ(s)ds

1

(λ)χ(λ)B(λ)X(dλ)

_ξs₌

_∞

−∞

ψ(λ)χ(λ)_η(dλ) =

_∞

−∞

ψ(λ)χ(λ)B(λ)X(dλ)

Thus

PHη(t)ξ=

_∞

−∞

_∞

−t

e−iλsϕ(s)dsB(λ)

(λ)χ(λ) +ψ(λ)B(λ)χ(λ)

X(dλ)

So, the spectral characteristics ofPHη(t)ξ in the space ofL2(FX(dl)) is

ψ(λ) =

⎧ ⎪ ⎨ ⎪ ⎩

0, λ∈E

ψ(λ), λ∈E

B(λ)₋∞_te−iλs_ϕ(s)ds

(λ) ,λ∈E

4. Conclusions

A fixed transform is given which based on one-dimensional a stationary processes in this paper. Also, we propose a general filtering problem. Then, in the space of L2(FX

Acknowledgements

This work was supported by the Natural Science Foundation of China (Grant no. 10771047). Authors’contributions

The studies and manuscript of this paper was written by Longsuo Li independently. Competing interests

The author declares that they have no competing interests.

Received: 5 May 2011 Accepted: 23 September 2011 Published: 23 September 2011

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doi:10.1186/1029-242X-2011-68

Cite this article as:Li:On general filtering problem of stationary processes with fixed transformation.Journal of Inequalities and Applications20112011:68.

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