# A Distributional Identity for the Bivariate Brownian Bridge: A Nontensor Gaussian Field

Authors: Xiaohui Ai
Journal: Mathematical Problems in Engineering
 (2018)

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## Abstract

The bivariate Brownian bridge, a nontensor Gaussian Field, is defined byB(t1,t2)=W(t1,t2)W(1,1)=0=W(t1,t2)-t1t2W(1,1), where t1,t2∈I=[0,1] and W(t1,t2) is a Brownian sheet. We obtain a distributional identity, a consequence of the Karhunen-Loève expansion for the bivariate Brownian bridge by Fredholm integral equation and Laplace transform approach.


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## Body

## 1. Introduction

LetX=Xt,t∈0,12 be a mean zero Gaussian process on C(0,12) with covariance function KX(t,s)=EX(t)X(s),t,s∈0,12. Then the well-known Karhunen-Loève (KL) expansion is (1)Xt=∑k≥1,j≥1ηk,jλk,jfk,jt,where ηk,j,k≥1,j≥1 is a sequence of 2-index i.i.d. N(0,1) random variables, fk,jt,k≥1,j≥1 forms an orthogonal sequence in L2(0,12), and λk,j,k≥1,j≥1 is the set of eigenvalues of the integral operator TXf(t)=∫0,12KX(t,s)f(s)ds. For the random variables X and Y, X=lawY means that X and Y have the same law; then a natural consequence of the KL expansions is the distributional identity (2)∫0,12X2tdt=law∑k,j=1∞λk,jηk,j2.As we know, a tied-down Brownian bridge (see [1]) is defined by (3)B∗t1,t2=Wt1,t2-t1W1,t2-t2Wt1,1+t1t2W1,1, whose covariance (s1∧t1-s1t1)(s2∧t2-s2t2),s1,s2,t1,t2∈I=[0,1] is tensor (separate) product and its Karhunen-Loève expansion is well known; see [2–4]. A bivariate Brownian bridge is defined by (4)Bt1,t2=Wt1,t2W1,1=0=Wt1,t2-t1t2W1,1, whose covariance (t1∧s1)(t2∧s2)-t1t2s1s2,s1,s2,t1,t2∈I=[0,1], a nontensor (nonseparate) product, is mentioned in the literature, such as [3, 5–10] and references therein. In this paper, we provide the Karhunen-Loève expansion for the bivariate Brownian bridge, which is our main goal and the highlight of our work.In particular, Deheuvels et al. in [6] give the explicit Laplace transform of the bivariate Brownian bridge; however, there are few references for the explicit eigenvalues and the associated simple eigenfunctions expression of the nontensor Gaussian process, which is one of our motivations. Somayasa in [9] studied the approximation methods for computing the quantiles about the bivariate Brownian bridge in our paper, which is considered as the residual partial sums limit process associated with a constant model; if we know the eigenvalues of the process, then we can compute the explicit quantiles analytically. Meanwhile, the explicit and simple eigenvalues can be used in other aspects of statistics, which is our another motivation.The rest of the paper is organized as follows. In Section2, we give the first part eigenvalues in (23) of the bivariate Brownian motion through solving the differential equation converted from the associated Fredholm integral equation of the covariance function of the process. The other part eigenvalues in (23) can be found via Laplace transform.


## 2. Karhunen-Loève Expansion

There are some lemmas needed to prove Theorem4. Before we prove the lemmas, we introduce some notations.Set(5)Im=km,l,jm,l:2km,l+12jm,l+1=2m+1,km,1<⋯<km,l<⋯<km,Im,1≤l≤Im,m,l∈N, and let I be the disjoint union of Im; that is, (6)I=k,j:k≥0,j≥0=⋃m=1∞Im, and then set (7)Im×Im=⋃m=1∞km,l,jm,l,km,l′,jm,l′:2km,l+12jm,l+1=2m+1,2km,l′+12jm,l′+1=2m+1,km,1<⋯<km,l<km,l+1<⋯<km,l′-1<km,l′<⋯<km,Im,1≤l<l′≤Im,m,l,l′∈N.It is proved in Lemma 1 that Im-1 is the number of pairs of (k,j), which is also the dimension number of the space spanned by the eigenfunctions associated with eigenvalues (41).We use(k,j) to denote (km,l,jm,l) and utilize (p,q) to represent (km,l′,jm,l′) for convenience. Then we can simplify the eigenfunctions to the following: (8)fm,km,l,km,l′t1,t2=2sin⁡2km,l+12πt1sin⁡2jm,l+12πt2-sin2km,l′+12πt1sin2jm,l′+12πt2.We note that ifl=l′, then f(t1,t2)=0, and thus l≠l′. It is easy to check that the eigenfunctions are orthogonal between different m, so we only need to find the multiplicity of the eigenfunctions for the same m. In Lemma 1, we use the formula (9)dimspanfm,2,…,fm,3,…,fm,Im=Im-1 instead of (10)dimspanfm,km,1,km,2,…,fm,km,1,km,l,…,fm,km,1,km,Im=Im-1 for convenience.Now we are in the position to provide the useful lemmas.Lemma 1.
Let(11)2km,i+12jm,i+1=2p+12q+1=2m+1 and the eigenfunctions be defined by (12)fm,i=2sin⁡2km,1+12πt1sin⁡2jm,1+12πt2-sin⁡2km,i+12πt1sin⁡2jm,i+12πt2with the associated eigenvalues 16/2m+12π4; then, for i=1,2,…,Im, m=1,2,…, (13)dimspanfm,1,…,fm,2,…,fm,Im-1=Im-1.Proof.
Assume∑i=1Im-1cifm,i=0; then we only need to check ci=0, i=0,1,…,Im-1.
In fact(14)c1fm,1+c2fm,2+⋯+cIm-1fm,Im-1=0 implies (15)∑i=1Im-1cisin⁡2km,1+12πt1sin⁡2jm,1+12πt2-∑i=2Imci-1sin⁡2km,i+12πt1sin⁡2jm,i+12πt2=0.
By multiplyingsin⁡2km,l+1/2πt1, 2≤l≤Im on both sides of (15) and integrate from 0 to 1 with respect to t1, we obtain (16)∑i=1Im-1cisin⁡2jm,1+12πt2∫01sin⁡2km,l+12πt1sin⁡2km,1+12πt1dt1-∑i=2Imci-1sin⁡2jm,i+12πt2∫01sin⁡2km,l+12πt1sin⁡2km,i+12πt1dt1=0,which gives c1=c2=⋯=cIm-1=0. Thus we claim the conclusion.The following lemma is from [11].Lemma 2.
Fors>1, (17)ζs=∏p1-p-s-1, where the above product takes over all prime numbers p and ζ(s)=∑n=1∞n-s.Lemma 3.
Letd(n) be the number of divisors of positive integer n; then (18)∑m=0∞d2m+12m+12=π426.Proof.
By the representation of Dirichlet series∑n=1∞d(n)/n2 and prime number decomposition, we obtain (19)∑n=1∞dnn2=∏p∑k=0∞dpkp2k,and the product of (19) takes over all primes (20)∏p∑k=0∞dpkp2k=∏p∑k=0∞k+1p-2k=∏p1-p-2-2=ζ22,and the last equal sign is given by Lemma 2.
On the other hand, the integern in the left hand side of (19) can be expressed by the product form of 2 and all the other odds; thus we get(21)∑n=1∞dnn2=∑k=0∞∑m=0∞k+1d2m+122k2m+12=∑k=0∞k+122k∑m=0∞d2m+12m+12=169∑m=0∞d2m+12m+12.
Notice thatζ(2)=∑n=1∞n-2=π2/6; by (20) and (21), the lemma is proved.Some important eigenvalues can be provided by the following analysis from Fredholm integral equation, but the entire eigenvalues can be found by the Laplace transform in Theorem4.We use Mercer’s theorem to compute the following integral equation of the covariance(22)KBt1,t2;s1,s2=t1∧s1t2∧s2-t1t2s1s2,s1,s2,t1,t2∈I=0,12, of the bivariate Brownian bridge; that is, (23)TBft1,t2=∫I2KBt1,t2;s1,s2fs1,s2ds1ds2=λft1,t2which is simplified to be (24)λft1,t2=1-t1t2∫0t2∫0t1s1s2fs1,s2ds1ds2+t1∫0t2∫t11s21-t2s1fs1,s2ds1ds2+t2∫t21∫0t1s11-t1s2fs1,s2ds1ds2+t1t2∫t21∫t111-s1s2fs1,s2ds1ds2.Taking derivative of (24) with respect to t1, we obtain (25)λ∂∂t1ft1,t2=-t2∫0t2∫0t1s1s2fs1,s2ds1ds2+∫0t2∫t11s21-t2s1fs1,s2ds1ds2+t2∫t21∫0t1s1-s2fs1,s2ds1ds2+t2∫t21∫t111-s1s2fs1,s2ds1ds2.Taking derivative of (25) with respect to t1 again, we obtain (26)-∫0t2s2ft1,s2ds2-t2∫t21ft1,s2ds2=λ∂2∂t12ft1,t2.By differentiating both sides of (26) with respect to t2, we have (27)-∫t21ft1,s2ds2=λ∂3∂t12t2ft1,t2.By differentiating both sides of (27) with respect to t2 again, we get (28)ft1,t2=λ∂4∂t12t22ft1,t2.From (24) to (27), we know the following facts:(29)ft1,0=0,f0,t2=0,(30)f1,1=0,(31)∂3∂t22∂t1f1,t2=0,∂3∂t12∂t2ft1,1=0.Define the following functions:(32)Fnt1=c1nsin⁡ant1+c1n′cos⁡ant1,Gnt2=c2nsin⁡t2λan+c2n′cos⁡t2λan, where cin, cin′, i=1,2 and, for n∈N, an>0 are constants (depending on λ) to be determined by (29) to (31). Then it is easy to see that Fn(t1)Gn(t2) are solutions of (28) and thus (33)ft1,t2=∑n≥1Fnt1Gnt2 are a set of solutions of (28). It is easy to check that when there is only one item in the series in (33) there is the trivial solution of (28), and when the items are more than two, the set of eigenfunctions is the same as the case of the two items. So the eigenfunction question is reduced to consider the sum of two items F1(t1)G1(t2) and F2(t1)G2(t2) in (33) (Fn(t1)=0 for n≥3), where the four functions F1(t1), F2(t1), G1(t2), and G2(t2) satisfy(34)d2F1t1dt12=-a1F1t1,d2G1t2dt22=F2t2-λa1,d2F2t1dt12=-a2G1t1,d2G2t2dt22=G2t2-λa2,for a1≠a2, a1>0, a2>0. Actually, if a1=a2, then F1(t1)=F2(t1), G1(t2)=G2(t2), and thus f(t1,t2)=2F1(t1)G1(t2), which is not the solution of the original integral equation (23). Hence a1≠a2.Then the eigenfunction can be expressed by(35)ft1,t2=F1t1G1t2+F2t1G2t2=c1nsin⁡a1t1+c2cos⁡a1t1c12sin⁡1λa1t2+c12′cos⁡1λa1t2+c21sin⁡a2t1+c21′cos⁡a2t1c22sin⁡1λa2t2+c22′cos⁡1λa2t2.Firstly, we deal with (35). Using (29), for any t1∈[0,1] and t2∈[0,1], we have(36)c12′c1nsin⁡a1t1+c11′cos⁡a1t1+c22′c21sin⁡a2t1+c21′cos⁡a2t1=0,c11′c12sin⁡1λa1t2+c12′cos⁡1λa1t2+c21′c22sin⁡1λa2t2+c22′cos⁡1λa2t2=0,which give (37)c12′=0,c22′=0,c11′=0,c21′=0.Then we can rewrite (35) as (38)ft1,t2=F1t1G1t2+F2t1G2t2=c11c12sin⁡a1t1sin⁡1λa1t2+c21c22sin⁡a2t1sin⁡1λa2t2.Making use of (31), for any t1∈[0,1] and t2∈[0,1], we obtain (39)a1λc11c12sin⁡a1t1cos⁡1λa1+c21c22a2λsin⁡a2t1cos⁡1λa2=0c11c121λa12a1cos⁡a1sin⁡1λa1t2+c21c221λa22a2cos⁡a2sin⁡1λa2t2=0,which give (40)a1=k+12π,a2=p+12π,1λa1=j+12π,1λa2=q+12π.Thus the eigenvalues are(41)λk,j=162k+122j+12π4,k,j∈N×N∖0,0.Based on fact (30), the condition (2k+1)(2j+1)=(2p+1)(2q+1), and the normalization condition, we can obtain c11c12=-c21c22=2 and then the simple eigenfunctions are, for k,j,p,q∈N×N∖{0,0},(42)ft1,t2=2sin⁡k+12πt1sin⁡j+12πt2-sin⁡p+12πt1sin⁡q+12πt2with associated eigenvalues (41). The multiplicity of the eigenfunctions (42) associated with (41) is Im-1, which has been shown by Lemma 1.Althoughsin⁡(1/2πt1)sin⁡(1/2πt2) is the solution of (28), it does not satisfy (23), so the index becomes k,j∈N×N∖{0,0} (or only p,q∈N×N∖{0,0}). Hence we have to search the solution form of (23). We find the interesting fact that 2sin⁡k+1/2πt1sin⁡j+1/2πt2 and sin⁡(p+1/2)πt1sin⁡(q+1/2)πt2 satisfy differential equation (28), but they are not the solutions of original integral equation (23). A natural idea is to combine them to relation (42), and then we should check whether (42) are the solutions of both (28) and (23) or not. We realize that (42) satisfy (28), while we do not make sure it is also doable for (23). By routine calculation, if (42) satisfy original integral equation (23), we must make (2k+1)(2j+1)=(2p+1)(2q+1). Therefore the index for the eigenfunction (42) should also be k,j∈N×N∖{0,0} (or only p,q∈N×N∖{0,0}).The well-known trace-variance formula gives(43)∫01∫01KBt1,t2;t1,t2dt1dt2=∫01∫01t1t2-t1t22dt1dt2=536.However, on the other hand, with the help of Lemmas1 and 3 and the fact that ∑n=1∞1/n2=π2/6, we have the following comparison relation: (44)∑m=1∞162m+12π4Im-1=16π4∑m=1∞Im2m+12-∑m=1∞12m+12=16π4π464-1-π26-π224-1=14-2π2<536.Therefore, there are still other eigenvalues lost so far. Fortunately, Laplace transform of the bivariate Brownian bridge provided them.Based on the analysis of the above, now we give the main result, namely, the following Theorem4 in this paper.Theorem 4.
The spectrum of the KL expansion for the bivariate Brownian bridgeBt1,t2,t1,t2∈0,12 is given by (41) and (52). In particular, (45)∫01∫01Bt1,t22dt1dt2=law∫01∫01Wt1,t2-∫01∫01Ws1,s2ds1ds22dt1dt2=law∑k,j≥1162k+122j+12π4ηk,j2+∑k,j≥1λk,j∗η∗k,j2,where {λk,j∗,k≥1,j≥1} are the reciprocal of the roots of (52) and λk,j∗-1∈[uk,j-,uk,j+], uk,j±=1/162j+12k±1π22, and {ηk,j, k≥1,j≥1} and {η∗k,j, k≥1,j≥1} are two independent sequences of 2-index i.i.d. N(0,1) random variables. Furthermore, for k≥k0, j≥j0, k0 and j0 are large enough, and we have (46)λk,j∗-1=2j+1kπ222+Ok-2j-2,(47)∑k≥k0,j≥j02j+1kπ2/22λk,j∗-1-1<∞.Here the first identity in Theorem4 can be found in Corollary 3.1 in [6].Proof.
Based on the Laplace transform of the bivariate Brownian bridgeB(s,t) (see Proposition 4.1.(i) of page 521 in [6])(48)Eexp⁡-a22∫0,12Bs,t2dsdt=D-a2-1/2=∏j=0∞cosh⁡2a2j+1π4u∑j=0∞tanh⁡2a2j+1π2j+1π-1-1/2and setting -a2=u in (48), we get the Fredholm determinant of the covariance function of the process (49)Du=∏j=0∞cos⁡2u1/22j+1π×4u-1/2∑j=0∞12j+1πtan⁡2u1/22j+1π,And according to (23) and formula (5.10) of page 146 and Theorems 5.1, 5.2, and 5.3 from page 148 to 150 in [12], the entire eigenvalues λk,j of the integral operator (50)∫0,12s1∧t1s2∧t2-s1s2t1t2fs1,s2ds1ds2are the reciprocal of the roots of D(u) in (49), which satisfy (51)cos⁡2u1/22j+1π=0 or (52)∑j=0∞12j+1πtan⁡2u1/22j+1π=0.Equation (51) also gives the eigenvalues of the form (53)λk,j=162k+122j+12π4,k,j∈N×N∖0,0,which are obtained in (41).
LetF(u) stand for (52); then by routine calculation, we know that F(u) is an increasing function on the interval uk,j-,uk,j+, uk,j±=1/162j+12k±1π22 and Fuk,j±=±∞ by the following fact: (54)0<F′u=∑j=0∞12j+12π21ucos2⁡2u1/2/2j+1π<∞.Therefore, there is only one positive zero in [uk,j-,uk,j+], uk,j±=1/162j+12k±1π22, which is denoted by ξk,j. Let ξk,j=λk,j∗-1 denote the solution of (52); then {λk,j∗,k,j=1,2,…} are the other eigenvalues of the integral operator TB.
Fork>k0, j>j0, k0 and j0 are large enough. The inequality(55)ξk,j-2j+1kπ222≤uk,j+-uk,j-=2k2j+1π222implies relation (46). It is also easy to check that (47) holds.Remark 5.
Although the entire eigenvalues can not be solved by the Fredholm integral equation, it gives the important information, which is stated in Theorem4.



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