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Mathematical Problems in Engineering
Volume 2018, Article ID 9687039, 6 pages
https://doi.org/10.1155/2018/9687039
Research Article

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

1Mathematics School, Jilin University, Changchun 130012, China
2Department of Mathematics, Northeast Forestry University, Harbin 150040, China

Correspondence should be addressed to Xiaohui Ai; moc.361@628_hxa

Received 16 November 2017; Revised 11 March 2018; Accepted 26 March 2018; Published 6 May 2018

Academic Editor: Kishin Sadarangani

Copyright © 2018 Xiaohui Ai. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

The bivariate Brownian bridge, a nontensor Gaussian Field, is defined by where and 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.

1. Introduction

Let be a mean zero Gaussian process on with covariance function Then the well-known Karhunen-Loève (KL) expansion is where is a sequence of 2-index i.i.d. random variables, forms an orthogonal sequence in , and is the set of eigenvalues of the integral operator . For the random variables and , means that and have the same law; then a natural consequence of the KL expansions is the distributional identity

As we know, a tied-down Brownian bridge (see [1]) is defined by whose covariance is tensor (separate) product and its Karhunen-Loève expansion is well known; see [24]. A bivariate Brownian bridge is defined by whose covariance , a nontensor (nonseparate) product, is mentioned in the literature, such as [3, 510] 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 Section 2, 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 Theorem 4. Before we prove the lemmas, we introduce some notations.

Set and let be the disjoint union of ; that is, and then set It is proved in Lemma 1 that is the number of pairs of , which is also the dimension number of the space spanned by the eigenfunctions associated with eigenvalues (41).

We use to denote and utilize to represent for convenience. Then we can simplify the eigenfunctions to the following:

We note that if , then , and thus . It is easy to check that the eigenfunctions are orthogonal between different , so we only need to find the multiplicity of the eigenfunctions for the same . In Lemma 1, we use the formula instead of for convenience.

Now we are in the position to provide the useful lemmas.

Lemma 1. Let and the eigenfunctions be defined by with the associated eigenvalues ; then, for , ,

Proof. Assume ; then we only need to check ,
In fact implies By multiplying , on both sides of (15) and integrate from 0 to 1 with respect to , we obtain which gives . Thus we claim the conclusion.

The following lemma is from [11].

Lemma 2. For , where the above product takes over all prime numbers and .

Lemma 3. Let be the number of divisors of positive integer ; then

Proof. By the representation of Dirichlet series and prime number decomposition, we obtain and the product of (19) takes over all primes and the last equal sign is given by Lemma 2.
On the other hand, the integer in the left hand side of (19) can be expressed by the product form of 2 and all the other odds; thus we getNotice that ; 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 Theorem 4.

We use Mercer’s theorem to compute the following integral equation of the covariance of the bivariate Brownian bridge; that is, which is simplified to be

Taking derivative of (24) with respect to , we obtain

Taking derivative of (25) with respect to again, we obtain

By differentiating both sides of (26) with respect to , we have

By differentiating both sides of (27) with respect to again, we get

From (24) to (27), we know the following facts:

Define the following functions: where , , and, for , are constants (depending on ) to be determined by (29) to (31). Then it is easy to see that are solutions of (28) and thus 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 and in (33) ( for ), where the four functions , , , and satisfyfor , , Actually, if , then , , and thus , which is not the solution of the original integral equation (23). Hence .

Then the eigenfunction can be expressed by Firstly, we deal with (35). Using (29), for any and , we havewhich give

Then we can rewrite (35) as

Making use of (31), for any and , we obtain which give

Thus the eigenvalues are

Based on fact (30), the condition , and the normalization condition, we can obtain and then the simple eigenfunctions are, for ,with associated eigenvalues (41). The multiplicity of the eigenfunctions (42) associated with (41) is , which has been shown by Lemma 1.

Although is the solution of (28), it does not satisfy (23), so the index becomes (or only ). Hence we have to search the solution form of (23). We find the interesting fact that and 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 . Therefore the index for the eigenfunction (42) should also be (or only ).

The well-known trace-variance formula gives

However, on the other hand, with the help of Lemmas 1 and 3 and the fact that , we have the following comparison relation:

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 Theorem 4 in this paper.

Theorem 4. The spectrum of the KL expansion for the bivariate Brownian bridge is given by (41) and (52). In particular, where are the reciprocal of the roots of (52) and , , and , and , are two independent sequences of 2-index i.i.d. random variables. Furthermore, for , , and are large enough, and we have

Here the first identity in Theorem 4 can be found in Corollary 3.1 in [6].

Proof. Based on the Laplace transform of the bivariate Brownian bridge (see Proposition 4.1.(i) of page 521 in [6])and setting in (48), we get the Fredholm determinant of the covariance function of the process 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 of the integral operator are the reciprocal of the roots of in (49), which satisfy or Equation (51) also gives the eigenvalues of the form which are obtained in (41).
Let stand for (52); then by routine calculation, we know that is an increasing function on the interval , and by the following fact: Therefore, there is only one positive zero in , , which is denoted by . Let denote the solution of (52); then are the other eigenvalues of the integral operator .
For , , and are large enough. The inequalityimplies 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 Theorem 4.

Conflicts of Interest

The author declares that there are no conflicts of interest regarding the publication of this paper.

Acknowledgments

The author is supported by the NSFC Grant (no. 11401085) and the Fundamental Research Funds for the Central Universities (no. 2572017CB25, no. 2572015BB14).

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