1. Introduction
Let X={X(t),0≤t≤1} be a centered and continuous Gaussian process on [0,1] with covariance function
(1)KX(t,s)=EX(t)X(s).
The KarhunenLoève expansion of X is given by the (convergent in mean squares) series
(2)X(t)=∑k=1∞ηkλkfk(t),
where {ηk,k≥1} is a sequence of i.i.d. N(0,1) random variables and {λk,k≥1} is at most the countable set of eigenvalues of Fredholm integral operator
(3)TXf(t)=∫01KX(t,s)f(s)ds{fk(t),k≥1} and forms an orthogonal sequence in L2[0,1] and ∫01KX(t,t)dt<∞.
Deheuvels et al. in [1–4] provided the KarhunenLoève expansions for the processes that are related with Brownian motion. The KarhunenLoève expansion for detrended Brownian motion has been studied by Ai et al. [5]. Note that the detrended Brownian motion in [5] can be viewed as projection to a constant function subspace in L2[0,1]. That is,
(4)∫01W^1(t)2dt=minc1,c2∫01(W(t)c1c2t)2dt.
To generalize the projection idea into nonlinear detrended process, now we consider
(5)minc1,c2,c3∫01(W(t)c1c2tc3t2)2dt
and the optimal constant cj satisfy
(6)∂∂cj∫01(W(t)c1c2tc3t2)2dt=0, j=1,2,3.
It is easy to obtain
(7)c1=9∫01W(s)ds36∫01W(s)s ds+30∫01W(s)s2ds,c2=36∫01W(s)ds+192∫01W(s)s ds180∫01W(s)s2ds,c3=30∫01W(s)ds180∫01W(s)s ds+180∫01W(s)s2ds.
Let
(8)A=(aij)3×3=(936303619218030180180);
we have
(9)cj=∑i=13aij∫01si1W(s)ds, j=1,2,3.
Now we can define the second order detrended process
(10)W^2(t)=W(t)∑j=13cjtj1=W(t)+(9+36t30t2)∫01W(s)ds+(36192t+180t2)∫01W(s)s ds+(30+180t180t2)∫01W(s)s2ds.
2. Main Results
We give the following lemma that provides the explicit covariance function.
Lemma 1.
For convenience, we add KX(s,t) into formula (11), that is
(11)KX(s,t)=E(W^2(t)W^2(s)) =t∧s∑p,q=13apq(tptp+1p(p+1))sq1 ∑i,j=13aij(sisi+1i(i+1))tj1 +∑p,q=13∑i,j=13aijapqp+i+2(p+1)(p+i+1)(i+1)tj1sq1,
where aij,apq,i,j,p,q=1,2,3 is given in (8).
Proof.
Consider
(12)W^2(t)=W(t)∑j=13cjtj1, 0≤t≤1
and W^2(t) is a mean zero Gaussian process; we obtain
(13)E(W^2(t)W^2(s)) =EW^2(t)W^2(s) =E(W(t)∑j=13cjtj1)(W(s)∑q=13cqsq1) =E(W(t)∑i,j=13aij(∫01ui1W(u)du)tj1) ·E(W(s)∑p,q=13apq(∫01vp1W(v)dv)sq1).
We notice that
(14)E(W(t)W(s))=t∧s,(15)E(W(t)∫01vp1W(v)dv) =E(∫01W(t)W(v)vp1dv) =∫01(t∧v)vp1dv =∫0tvpdv+∫t1tvp1dv =tptp+1p(p+1),(16)E(∫01ui1W(u)du)(∫01vp1W(v)dv) =∫01ui1E(W(u)∫01vp1W(v)dv)du =∫01ui1(upup+1p(p+1))du =p+i+2(p+1)(p+i+1)(i+1).
Substituting (16), (17), and (19) into (15), we derive
(17)E(W^2(t)W^2(s)) =t∧s∑p,q=13apq(tptp+1p(p+1))sq1 ∑i,j=13aij(sisi+1i(i+1))tj1 +∑p,q=13∑i,j=13aijapqp+i+2(p+1)(p+i+1)(i+1)tj1sq1.
Lemma 2 (see [<xref reftype="bibr" rid="B3">3</xref>]).
If t∈[0,1], ςj(t)=∑k=1∞ωkλk,jek,j(t), j=1,2,…, then the condition
(18)∫[0,1]ς12(t)dt=law∫[0,1]ς22(t)dt
is equivalent to the identity
(19)λk,1=λk,2 ∀k≥1.
In the following, we will give some preliminaries, notions, and facts that are needed in Theorem 3. For v>1, Jv(·) is Bessel function [6] with index v and the positive zeros of Jv(·) are infinite sequence 0<zv,1<zv,2<⋯. When v=3/2, v=5/2, the positive zeros of J3/2,k, J5/2,k are z3/2,k, z5/2,k, k=1,2,…, and they are in such a way that
(20)0<z3/2,1<z5/2,1<z3/2,2<⋯.
Now we can state one of the main results of this paper.
Theorem 3.
For the second order detrended Brownian motion W^2(t) and a generalized Brownian bridge B2(t) with n=2 in [7],
(21)B2(t)=B(t)136t(60t2+18t67)B(1)t(60t296t+11)∫01B(s)ds+10t(12t218t+1)∫01B(s)s ds.
One has the distribution identities
(22)∫01W^2(t)2dt=law∫01B2(t)2dt=law∑k≥1ηk24z3/2,k2+∑k≥1ηk*24z5/2,k2,
where {ηk,k≥1} and {ηk*,k≥1} denote two independent sequences of independently and identically distributed N(0,1) random variables.
Proof.
By straightforward induction based on the equation and splitting the integration range from t, we get
(23)λf(t)=∫0tsf(s)ds+t∫t1f(s)ds∑p,q=13apq(tptp+1p(p+1))∫01sq1f(s)ds∑i,j=13aijtj1∫01(sisi+1i(i+1))f(s)ds+∑p,q=13∑i,j=13aijapqp+i+2(p+1)(p+i+1)(i+1)tj1×∫01sq1f(s)ds.
By differentiation of both sides of (23) with respect to t, we have
(24)λf′(t)=∫t1f(s)ds∑p,q=13apq1tpp∫01sq1f(s)ds∑i=1,j=23(j1)aijtj2∫01(sisi+1i(i+1))f(s)ds+∑p,q=13∑i=1,j=23aijapq(p+i+2)(j1)(p+1)(p+i+1)(i+1)tj2×∫01sq1f(s)ds.
By differentiation of both sides of (24) with respect to t, we have
(25)λf′′(t)+f(t) =∑i,p,q=13ai3apq2(p+i+2)(p+1)(p+i+1)(i+1)∫01sq1f(s)ds 2∑i=13ai3∫01(sisi+1i(i+1))f(s)ds +∑q=13a1q∫01sq1f(s)ds +(∑q=13a2q∫01sq1f(s)ds)t +(∑q=13a3q∫01sq1f(s)ds)t2.
We can simplify this equation to
(26)λf′′(t)+f(t)+b1+b2t+b3t2=0,
where
(27)b1=∑i,p,q=13ai3apq2(p+i+2)(p+1)(p+i+1)(i+1)∫01sq1f(s)ds+2∑i=13ai3∫01(sisi+1i(i+1))f(s)ds∑q=13a1q∫01sq1f(s)ds,(28)b2=∑q=13a2q∫01sq1f(s)ds,(29)b3=∑q=13a3q∫01sq1f(s)ds.
We solve the inhomogeneous second differential equation to obtain
(30)f(t)=c1costλ+c2sintλ+2λb3b1b2tb3t2.
We substitute f(t) into (28) and (29) to obtain
(31)(λsin1λ+6λcos1λ12λλsin1λ+6λ)c1 +(λcos1λ+6λsin1λiiiiiiiiiiiiii+12λλcos1λ12λλ+λ)c2 =0,(2λsin1λ14λcos1λ+30λλsin1λ16λ)c1 +(2λcos1λ14λsin1λiiiiiiiiiiiiii30λλcos1λ+30λλ3λ)c2 =0.
In order that there are nonzero choices for c1,c2, the determinant of the above two equations has to be zero, which can be written as
(32)D11D22D12D21=0,
where
(33)D11=λsin1λ+6λcos1λ12λλsin1λ+6λ,D12=λcos1λ+6λsin1λ+12λλcos1λ12λλ+λ,D21=2λsin1λ14λcos1λ+30λλsin1λ16λ,D22=2λcos1λ14λsin1λ30λλcos1λ+30λλ3λ.
We obtain, after some simplification,
(34)24λ2λ+4λλ =(24λ2λ)sin1λ+(24λ2λ8λλ)cos1λ.
Then λ≠0 is an eigenvalue if and only if (34) holds. We therefore obtain
(35)D(λ)=720((24λ7/2λ5/2)sinλ1/2iiiiiiiiiiii+(24λ48λ3)cosλ1/224λ44λ3),
with D(0)=1.
According to the trigonometric function formula
(36)sin1λ=2sin12λcos12λ,cos1λ=2 cos212λ1=12sin212λ,
we can observe that
(37)D11D22D12D21=12πλJ3/2(12λ)J5/2(12λ)=0,
where J3/2(z), J5/2(z) are Bessel functions as follows:
(38)J3/2(z)=2π·zπ(sinzz2coszz),J5/2(z)=2π·zπ((1z+3z3)sinz3z2cosz),
which gives two sequences of eigenvalues of (37), namely, (2z3/2,k)2 and (2z5/2,k)2.
Similarly, we can obtain the two eigenvalues (2z3/2,k)2, (2z5/2,k)2 corresponding to those of integral operator of a generalized Brownian bridge B2(t). Note that the integral operator is
(39)∫01K2(s,t)f(s)ds.
Actually, in Lemma 2, we have the distribution identities
(40)∫01W^2(t)2dt=law∫01B2(t)2dt=law∑k≥1ηk24z3/2,k2+∑k≥1ηk*24z5/2,k2.
Remark 4.
From (11) and (22), we derive that
(41)∫01KX(t,t)dt=∫01E(W^2(t)2)dt=E∫01W^2(t)2dt=∑k≥114z3/2,k2+∑k≥114z5/2,k2=140+156=3140
by using the Rayleigh’s formula, for v=3/2 and v=5/2 (see, e.g., [3, (1.91), page 77] and [6, page 502]).
To check (41), from (11), we infer that
(42)∫01KX(t,t)dt =∫01[t∑p,q=13apq(tqptp+qp(p+1))iiiiiiiiiiiiiiii∑i,j=13aij(tjiti+ji(i+1))iiiiiiiiiiiiiiii+∑p,q=13 ∑i,j=13aijapqiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii∑p,q=13apq×p+i+2(p+1)(p+i+1)(i+1)tj+q]dt =3140
which is in agreement with (41).
3. Applications
In this section, the relevant applications of KarhunenLoève expansion are given.
Proposition 5.
For each θ∈R, one has
(43)Eexp(θ22∫01W^2(t)2dt) ={720((24θ7θ5)sinθ+(24θ8+8θ6)cosθiiiiiiiiiiiiiiiiiiiiiii+24θ8+4θ6)}1/2.
Proof.
(44)
E
exp
(

θ
2
2
∫
0
1
W
^
2
(
t
)
2
d
t
)
=
E
exp
(

θ
2
2
∑
k
=
1
∞
λ
k
ξ
k
2
)
=
∏
k
=
1
∞
(
1
+
λ
k
θ
2
)

1
/
2
=
(
D
(

θ
2
)
)

1
/
2
=
{

720
(
(
24
θ

7

θ

5
)
sin
θ
+
(

24
θ

8
+
8
θ

6
)
cos
θ
iiiiiiiiiiiiiiiiiiiiiii
+
24
θ

8
+
4
θ

6
)
}

1
/
2
,
where λ1>λ2>⋯>0 and ∑k=1∞λk<∞.
Proposition 6.
If x>0, then
(45)P(∫01W^22(t)dt>x) =1π∑k=1∞(1)k+1 ×∫γ2k1γ2k(eux/224u44u3)1/2)1iiiiiiiiiiiiiiiiiiiiii×(24u44u3))1/2)1u(24u44u3))1/2720((24u7/2u5/2)sinu1/224u44u3)1/2iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii+(24u48u3)cosu1/2iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii24u44u3))1/2)1)du,
where γk=λk1, k=1,2,….
Proof.
It can be proved by the Smirnov formula [8, 9], formula (23), and the definition of the Fredholm determinant. Similar proof method can be found from Proposition 3.3 in [10].
Next, we give the large deviation and small deviation probabilities of the second order detrended Brownian motion with respect to the norm in the Hilbert Space L2[0,1].
Proposition 7.
Consider x→∞,
(46)P(∫01W^2(t)2dt>x) =(1+o(1))(2π)1/2(2z3/2,1)2x1/2exp(2z3/2,12x) ·{720((324z3/2,19+1328z3/2,17)sin2z3/2,1iiiiiiiiiiiiiiiiiiiii+(325z3/2,110+926z3/2,18127z3/2,16)iiiiiiiiiiiiiiiiiiiii×cos2z3/2,1+325z3/2,110+326z3/2,18)}1/2.
Proof.
By Deheuvels [2] and Martynov [8], we have for all x>0(47)P(∫01W^2(t)2dt>x) =(1+o(1))(2π)1/2γ11(D′(γ1))1/2 ×x1/2exp(γ1x2);
we take D(λ) and γ1=(2z3/2,1)2 into (47), and then the proof is completed.
Proposition 8.
There exists a constant c>0 such that
(48)P(∫01W^2(t)2dt≤ε) =(c+o(1))ε2exp(18ε), as ε⟶0.
Proof.
We start with proving (48) by recalling Li, 1992 [11, 12].
Given two sequences ak>0 and bk>0 with
(49)∑k≥1ak<∞, ∑k≥1bk<∞, ∑k≥11akbk<∞,
we have, as ε→0,
(50)P(∑k≥1akξk2≤ε) =(1+o(1))(∏k≥1bkak)1/2P(∑k≥1bkξk2≤ε).
By the asymptotic formula for zeros of Bessel function
(51)z3/2,k=(k+12)π+O(k1), k⟶∞,z5/2,k=(k+1)π+O(k1), k⟶∞,
then ak=λk, b2k1=((2k+1)π)2, and b2k=((2k+2)π)2, k∈N, which satisfy (49) and by the distribution identity ∫01W^22(t)dt=∑k≥1λ2k1ηk2+∑k≥1λ2kη2k*2 and (50), there exists a constant c1, such that
(52)P(∫01W^2(t)2dt≤ε) =P(∑k≥1λ2k1ηk2+∑k≥1λ2kηk*2≤ε) =(1+o(1))∏k≥1(bkak)1/2P(∑k≥1bkξk2≤ε) =(1+o(1))c1 ×P(∑k≥1ξ2k12((2k+1)π)2+∑k≥1ξ2k2((2k+2)π)2≤ε) =(1+o(1))c1P(∑k≥1(k+2)2ξk2≤επ2), as ε⟶0.
Also, for all d>1, there exists a constant c2>0, such that, as ε→0,
(53)P(∑k≥1(k+d)2ξk2≤επ2)=(1+o(1))c2εdexp(18ε).
Connecting (52) with (53), we can obtain the proposition.