# A Class of Bridges of Iterated Integrals of Brownian Motion Related to Various Boundary Value Problems Involving the One-Dimensional Polyharmonic Operator

Authors: Aimé Lachal
Journal: International Journal of Stochastic Analysis
 (2011)

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## Abstract

Let(B(t))t∈[0,1] be the linear Brownian motion and (Xn(t))t∈[0,1] the (n−1)-fold integral of Brownian motion, with n being a positive integer: Xn(t)=∫0t((t−s)n−1/(n−1)!)dB(s) for any t∈[0,1]. In this paper we construct several bridges between times 0 and 1 of the process (Xn(t))t∈[0,1] involving conditions on the successive derivatives of Xn at times 0 and 1. For this family of bridges, we make a correspondence with certain boundary value problems related to the one-dimensional polyharmonic operator. We also study the classical problem of prediction. Our results involve various Hermite interpolation polynomials.


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## Body

## 1. Introduction

Throughout the paper, we will denote, for any enough differentiable functionf, its ith derivative by f(i) or dif/dti.Let(B(t))t∈[0,1] be the linear Brownian motion started at 0 and (β(t))t∈[0,1] be the linear Brownian bridge within the time interval [0,1]: (β(t))t∈[0,1]=(B(t)∣B(0)=B(1)=0)t∈[0,1]. These processes are Gaussian processes with covariance functions (1.1)cB(s,t)=s∧t,cβ(s,t)=s∧t-st.
For a given continuous function u, the functions vB and vβ, respectively, defined on [0,1] by (1.2)vB(t)=∫01cB(s,t)u(s)ds,vβ(t)=∫01cβ(s,t)u(s)ds
are the solutions of the respective boundary value problems on [0,1]: (1.3){vB′′=-u,vB(0)=vB′(1)=0,{vβ′′=-u,vβ(0)=vβ(1)=0.
Observe that the differential equations are the same in both cases. Only the boundary conditions differ. They are Dirichlet-type boundary conditions for Brownian bridge while they are Dirichlet/Neumann-type boundary conditions for Brownian motion.These well-known connections can be extended to the polyharmonic operatord2n/dt2n, where n is a positive integer. This latter is associated with the (n-1)-fold integral of Brownian motion (Xn(t))t∈[0,1]: (1.4)Xn(t)=∫0t(t-s)n-1(n-1)!dB(s)foranyt∈[0,1].
(Notice that all of the derivatives at time 0 naturally vanish: Xn(0)=Xn-1(0)=⋯=X2(0)=X1(0)=0.) Indeed, the following facts, for instance, are known (see, e.g., [1, 2]).(i)
The covariance function of the process(Xn(t))t∈[0,1] coincides with the Green function of the boundary value problem
(1.5){v(2n)=(-1)nuon⁡[0,1],v(0)=v′(0)=⋯=v(n-1)(0)=0,v(n)(1)=v(n+1)(1)=⋯=v(2n-1)(1)=0.(ii)
The covariance fonction of the bridge(Xn(t)∣Xn(1)=0)t∈[0,1] coincides with the Green function of the boundary value problem
(1.6){v(2n)=(-1)nuon⁡[0,1],v(0)=v′(0)=⋯=v(n-1)(0)=0,v(n-1)(1)=v(n+1)(1)=⋯=v(2n-1)(1)=0.(iii)
The covariance fonction of the bridge(Xn(t)∣Xn(1)=Xn-1(1)=⋯=X1(1)=0)t∈[0,1] coincides with the Green function of the boundary value problem
(1.7){v(2n)=(-1)nuon⁡[0,1],v(0)=v′(0)=⋯=v(n-1)(0)=0,v(1)=v′(1)=⋯=v(n-1)(1)=0.Observe that the differential equations and the boundary conditions at 0 are the same in all cases. Only the boundary conditions at 1 differ. Other boundary value problems can be found in [3, 4].We refer the reader to [5] for a pioneering work dealing with the connections between general Gaussian processes and Green functions; see also [6]. We also refer to [7–13] and the references therein for various properties, namely, asymptotical study, of the iterated integrals of Brownian motion and more general Gaussian processes as well as to [3, 4, 14, 15] for interesting applications of these processes to statistics.The aim of this work is to examine all the possible conditioned processes of(Xn(t))t∈[0,1] involving different events at time 1: (1.8)(Xn(t)∣Xj1(1)=Xj2(1)=⋯=Xjm(1)=0)t∈[0,1]
for a certain number m of events, 1≤m≤n, and certain indices j1,j2,…,jm such that 1≤j1<j2<⋯<jm≤n, and to make the connection with the boundary value problems:BVP{v(2n)=(-1)nuon⁡[0,1],v(0)=v′(0)=⋯=v(n-1)(0)=0,v(i1)(1)=v(i2)(1)=⋯=v(in)(1)=0
for certain indices i1,i2,…,in such that 0≤i1<i2<⋯<in≤2n-1. Actually, we will see that this connection does not recover all the possible boundary value problems, and we will characterize those sets of indices for which such a connection exists.The paper is organized as follows. In Section2, we exhibit the relationships between general Gaussian processes and Green functions of certain boundary value problems. In Section 3, we consider the iterated integrals of Brownian motion. In Section 4, we construct several bridges associated with the foregoing processes and depict explicitly their connections with the polyharmonic operator together with various boundary conditions. One of the main results is Theorem 4.4. Moreover, we exhibit several interesting properties of the bridges (Theorems 4.1 and 4.3) and solve the prediction problem (Theorem 4.6). In Section 5, we illustrate the previous results on the case n=2 related to integrated Brownian motion. Finally, in Section 6, we give a characterization for the Green function of the boundary value problem (BVP) to be a covariance function. Another one of the main results is Theorem 6.3.


## 2. Gaussian Processes and Green Functions

We consider ann-Markov Gaussian process (X(t))t∈[0,1] evolving on the real line ℝ. By “n-Markov,” it is understood that the trajectory t↦X(t) is n times differentiable and the n-dimensional process (X(t),X′(t),…,X(n-1)(t))t∈[0,1] is a Markov process. Let us introduce the covariance function of (X(t))t∈[0,1]: for s,t∈[0,1], cX(s,t)=𝔼[X(s)X(t)]. It is known (see [6]) that the function cX admits the following representation:(2.1)cX(s,t)=∑k=0n-1φk(s∧t)ψk(s∨t),
where φk,ψk, k∈{0,1,…,n-1}, are certain functions.Let𝒟0,𝒟1 be linear differential operators of order less than p, and let 𝒟 be a linear differential operator of order p defined by (2.2)D=∑i=0pαididti,
where α0,α1,…,αp are continuous functions on [0,1]. More precisely, we have for any p times differentiable function f and any t∈[0,1], (2.3)(Df)(t)=∑i=0pαi(t)f(i)(t).Theorem 2.1.
Assume that the functionsφk,ψk, k∈{0,1,…,n-1}, are p times differentiable and satisfy the following conditions, for a certain constant κ:
(2.4)∑k=0n-1[φkψk(i)-φk(i)ψk]={0if0≤i≤p-2,κifi=p-1,(2.5)Dφk=Dψk=0,(D0φk)(0)=0,(D1ψk)(1)=0.
Then, for any continuous function u on [0,1], the function v defined on [0,1] by
(2.6)v(t)=∫01cX(s,t)u(s)ds
solves the boundary value problem
(2.7){Dv=αpu,(D0v)(0)=(D1v)(1)=0.Remark 2.2.
If the problem (2.7) is determining, that is, if it has a unique solution, then the covariance function cX is exactly the Green function of the boundary value problem (2.7).Proof.
In view of (2.1), the function v can be written as
(2.8)v(t)=∑k=0n-1[ψk(t)∫0tφk(s)u(s)ds+φk(t)∫t1ψk(s)u(s)ds].
The derivative of v is given by
(2.9)v′(t)=∑k=0n-1[ψk′(t)∫0tφk(s)u(s)ds+φk′(t)∫t1ψk(s)u(s)ds]
and its second-order derivative, since ∑k=0n-1[φkψk′-φk′ψk]=0, by
(2.10)v′′(t)=∑k=0n-1[ψk′′(t)∫0tφk(s)u(s)ds+φk′′(t)∫t1ψk(s)u(s)ds+[φk(t)ψ′k(t)-φ′k(t)ψk(t)]u(t)]=∑k=0n-1[ψk′′(t)∫0tφk(s)u(s)ds+φk′′(t)∫t1ψk(s)u(s)ds].
More generally, because of the assumptions (2.4), we easily see that, for i∈{0,1,…,p-1},
(2.11)v(i)(t)=∑k=0n-1[ψk(i)(t)∫0tφk(s)u(s)ds+φk(i)(t)∫t1ψk(s)u(s)ds],
and the pth order derivative of v is given by
(2.12)v(p)(t)=∑k=0n-1[ψk(p)(t)∫0tφk(s)u(s)ds+φk(p)(t)∫t1ψk(s)u(s)ds+[φk(t)ψk(p-1)(t)-φk(p-1)(t)ψ(t)]u(t)∫t1]=∑k=0n-1[ψk(p)(t)∫0tφk(s)u(s)ds+φk(p)(t)∫t1ψk(s)u(s)ds]+κu(t).
Actually, we have proved that, for i∈{0,1,…,p-1},
(2.13)v(i)(t)=∫0t∂icX∂ti(s,t)u(s)ds.
Finally, due to (2.5),
(2.14)Dv(t)=∑k=0n-1[Dψk(t)∫0tφk(s)u(s)ds+Dφk(t)∫t1ψk(s)u(s)ds]+καpu(t)=καpu(t).
Concerning the boundary value conditions, referring to (2.5), we similarly have
(2.15)(D0v)(0)=∑k=0n-1(D0φk)(0)∫01ψk(s)u(s)ds=0,(D1v)(1)=∑k=0n-1(D1ψk)(1)∫01φk(s)u(s)ds=0.
The proof of Theorem 2.1 is finished.In the two next sections, we construct processes connected to the equation𝒟v=u subject to the boundary value conditions at 0: (𝒟0iv)(0)=0 for i∈{0,1,…,n-1} and others at 1 that will be discussed subsequently, where 𝒟 and 𝒟0i are the differential operators (𝒟 being of order p=2n) defined by
(2.16)D=(-1)nd2ndt2n,D0i=didti.


## 3. The(n-1)-Fold Integral of Brownian Motion

Let(B(t))t∈[0,1] be the linear Brownian motion limited to the time interval [0,1] and started at 0. We introduce the (n-1)-fold integral of Brownian motion: for any t∈[0,1], (3.1)Xn(t)=∫0t(t-s)n-1(n-1)!dB(s).
In particular, X1=B. The trajectories of (Xn(t))t∈[0,1] are n times differentiable and we have Xn(i)=Xn-i for 0≤i≤n-1. Moreover, we have at time 0 the equalities Xn(0)=Xn-1(0)=⋯=X2(0)=X1(0)=0. The process (Xn(t))t∈[0,1] is an n-Markov Gaussian process since the n-dimensional process (Xn(t),Xn-1(t),…,X1(t))t∈[0,1] is Markovian. The covariance function of the Gaussian process (Xn(t))t∈[0,1] is given by (3.2)cXn(s,t)=∫0s∧t(s-u)n-1(n-1)!(t-u)n-1(n-1)!du.
In order to apply Theorem 2.1, we decompose cXn into the form (2.1). We have for, e.g., s≤t, (3.3)cXn(s,t)=∫0s(s-u)n-1(n-1)!2[∑k=0n-1(n-1k)(-u)n-1-ktk]du=∑k=0n-1(-1)n-1-ktkk!∫0s(s-u)n-1(n-1)!un-1-k(n-1-k)!du=∑k=0n-1(-1)n-1-ks2n-1-k(2n-1-k)!tkk!.
We then obtain the following representation: (3.4)cXn(s,t)=∑k=0n-1φk(s)ψk(t)
with, for any k∈{0,1,…,n-1}, (3.5)φk(s)=(-1)n-1-ks2n-1-k(2n-1-k)!,ψk(t)=tkk!.
We state below a result of [2] that we revisit here by using Theorem 2.1.Theorem 3.1.
Letu be a fixed continuous function on [0,1]. The function v defined on [0,1] by
(3.6)v(t)=∫01cXn(s,t)u(s)ds
is the solution of the boundary value problem
(3.7){v(2n)=(-1)nuon[0,1],v(i)(0)=0fori∈{0,1,…,n-1},v(i)(1)=0fori∈{n,n+1,…,2n-1}.Proof.
Let us check that conditions (2.4) and (2.5) of Theorem 2.1 are fulfilled. First, we have
(3.8)∑k=0n-1[φk(t)ψk(i)(t)-φk(i)(t)ψk(t)]=∑k=0n-1(-1)n-1-k[1{k≥i}t2n-1-k(2n-1-k)!tk-i(k-i)!-1{k≤2n-1-i}t2n-1-i-k(2n-1-i-k)!tkk!]=(-1)n-1t2n-1-i(2n-1-i)![1{i≤n-1}∑k=in-1(-1)k(2n-1-ik-i)-∑k=0(2n-1-i)∧(n-1)(-1)k(2n-1-ik)]=(-1)n-1t2n-1-i(2n-1-i)![1{i≤n-1}(∑k=0n-1-i(-1)i+k(2n-1-ik)-∑k=0n-1(-1)k(2n-1-ik))-1{i≥n}∑k=02n-1-i(-1)k(2n-1-ik)].
Performing the transformation k↦2n-1-i-k in the first sum lying within the last equality, we get
(3.9)∑k=0n-1-i(-1)i+k(2n-1-ik)-∑k=0n-1(-1)k(2n-1-ik)=∑k=n2n-1-i(-1)k(2n-1-i2n-1-i-k)+∑k=0n-1(-1)k(2n-1-ik)=∑k=02n-1-i(-1)k(2n-1-ik)=δi,2n-1,
and then
(3.10)∑k=0n-1[φk(t)ψk(i)(t)-φk(i)(t)ψk(t)]=(-1)nδi,2n-1.
On the other hand, setting
(3.11)D=(-1)nd2ndt2n,D0i=didti,D1i=di+ndti+n,
we clearly see that, for any k∈{0,1,…,n-1},
(3.12)Dφk=Dψk=0,D0iφk(0)=D1iψk(1)=0fori∈{0,1,…,n-1}.
Consequently, by Theorem 2.1, we see that the function v solves the boundary value problem (3.7). The uniqueness part will follow from a more general argument stated in the proof of Theorem 4.4.


## 4. Various Bridges of the(n-1)-Fold Integral of Brownian Motion

In this section, we construct various bridges related to(Xn(t))t∈[0,1]. More precisely, we take (Xn(t))t∈[0,1] conditioned on the event that certain derivatives vanish at time 1. Let us recall that all the derivatives at time 0 naturally vanish: Xn(0)=Xn-1(0)=⋯=X2(0)=X1(0)=0.For anym∈{0,1,…,n}, let J={j1,j2,…,jm} be a subset of {1,2,…,n} with 1≤j1<j2<⋯<jm≤n and the convention that for m=0, J=∅. We see that, for each m∈{0,1,…,n}, we can define (nm) subsets of indices J, and the total number of sets J is then ∑m=0n(nm)=2n. Set for any t∈[0,1](4.1)Y(t)=(Xn(t)∣Xj(1)=0,j∈J)=(Xn(t)∣Xj1(1)=Xj2(1)=⋯=Xjm(1)=0).
In this way, we define 2n processes (Y(t))t∈[0,1] that we will call “bridges’’ of (Xn(t))t∈[0,1]. In particular, (i)
forJ=∅, we simply have (Y(t))t∈[0,1]=(Xn(t))t∈[0,1];(ii)
forJ={1}, the corresponding process is the (n-1)-fold integral of Brownian bridge
(4.2)(Xn(t)∣X1(1)=0)t∈[0,1]=(∫0t(t-s)n-1(n-1)!dβ(s))t∈[0,1];(iii)
forJ={n}, the corresponding process is the “single’’ bridge of (n-1)-fold integral of Brownian motion:
(4.3)(Xn(t)∣Xn(1)=0)t∈[0,1]=(∫0t(t-s)n-1(n-1)!dB(s)∣∫01(1-s)n-1(n-1)!dB(s)=0)t∈[0,1];(iv)
forJ={1,2,…,n}, the corresponding process is
(4.4)(Xn(t)∣Xn(1)=Xn-1(1)=⋯=X1(1)=0)t∈[0,1].
This is the natural bridge related to the n-dimensional Markov process (Xn(t),Xn-1(t),…,X1(t))t∈[0,1].In this section, we exhibit several interesting properties of the various processes(Y(t))t∈[0,1]. One of the main goals is to relate these bridges to additional boundary value conditions at 1. For this, we introduce the following subset I of {0,1,…,2n-1}: (4.5)I=(n-J)∪[{n,n+1,…,2n-1}∖(J+n-1)]
with (4.6)n-J={n-j,j∈J}={n-j1,…,n-jm},J+n-1={j+n-1,j∈J}={j1+n-1,…,jm+n-1}.
The cardinality of I is n. Actually, the set I will be used further for enumerating the boundary value problems which can be related to the bridges labeled by J. Conversely, I yields J through (4.7)J=(n-I)∩{1,2,…,n}.
In Table 1, we give some examples of sets I and J.Table 1I{0,1,…,n-1}{n,n+1,…,2n-1}{n-1,n+1,…,2n-1}{0,n+1,…,2n-2}J{1,2,…,n}∅{1}{n}


### 4.1. Polynomial Drift Description

Below, we provide a representation of(Y(t))t∈[0,1] by means of (Xn(t))t∈[0,1] subject to a random polynomial drift.Theorem 4.1.
One has the distributional identity(4.8)(Y(t))t∈[0,1]=d(Xn(t)-∑j∈JPj(t)Xj(1))t∈[0,1],
where the functions Pj, j∈J, are Hermite interpolation polynomials on [0,1] characterized by
(4.9){Pj(2n)=0,Pj(i)(0)=0fori∈{0,1,…,n-1},Pj(i)(1)=δj,n-ifori∈I.Remark 4.2.
In the case wheren=2, we retrieve a result of [16]. Moreover, conditions (4.9) characterize the polynomials Pj, j∈J. We prove this fact in Lemma A.1 in the Appendix.Proof.
By invoking classical arguments of Gaussian processes theory, we have the distributional identity(4.10)(Y(t))t∈[0,1]=d(Xn(t)-∑j∈JPj(t)Xj(1))t∈[0,1],
where the functions Pj, j∈J, are such that 𝔼[Y(t)Xk(1)]=0 for all k∈J. We get the linear system
(4.11)∑j∈JE[Xj(1)Xk(1)]Pj(t)=E[Xn(t)Xk(1)],k∈J.
We plainly have
(4.12)E[Xj(s)Xk(t)]=∫0s∧t(s-u)j-1(j-1)!(t-u)k-1(k-1)!du.
Then, the system (4.11) writes
(4.13)∑j∈J1j+k-1Pj(t)(j-1)!=∫0t(t-u)n-1(n-1)!(1-u)k-1du.
The matrix of this system (1/(j+k-1))j,k∈J is regular as it can be seen by introducing the related quadratic form which is definite positive. Indeed, for any real numbers xj, j∈J, we have
(4.14)∑j,k∈Jxjxkj+k-1=∫01(∑j,k∈Jxjxkuj+k-2)du=∫01(∑j∈Jxjuj-1)2du≥0,∑j,k∈Jxjxkj+k-1=0iff∀j∈J,xj=0.
Thus, the system (4.11) has a unique solution. As a result, the Pj are linear combinations of the functions t↦∫0t(t-u)n-1(1-u)k-1du which are polynomials of degree less than n+k. Hence, the  Pj are polynomials of degree at most 2n-1.
We now compute the derivatives ofPj at 0 in 1. We have Pj(i)(0)=0 for i∈{0,1,…,n-1} since the functions t↦∫0t(t-u)n-1(1-u)k-1du plainly enjoy this property. For checking that Pj(i)(1)=δj,n-i for i∈I, we successively compute (i)
fori∈{0,1,…,n-1},
(4.15)didtiE[Xn(t)Xk(1)]=E[Xn-i(t)Xk(1)];(ii)
fori=n-1,
(4.16)didtiE[Xn(t)Xk(1)]=E[B(t)Xk(1)]=∫0t(1-u)k-1(k-1)!du;(iii)
fori=n,
(4.17)didtiE[Xn(t)Xk(1)]=(1-t)k-1(k-1)!;(iv)
fori∈{n,n+1,…,n+k-1},
(4.18)didtiE[Xn(t)Xk(1)]=(-1)i+n(1-t)k+n-i-1(k+n-i-1)!;(v)
fori=k+n-1,
(4.19)didtiE[Xn(t)Xk(1)]=(-1)k-1;(vi)
fori≥k+n,
(4.20)didtiE[Xn(t)Xk(1)]=0. Consequently, at time t=1, for i≥n,
(4.21)didtiE[Xn(t)Xk(1)]|t=1=(-1)k-1δi,k+n-1.
Now, by differentiating (4.11), we get for i∈{0,1,…,n-1},
(4.22)∑j∈JE[Xj(1)Xk(1)]Pj(i)(1)=E[Xn-i(1)Xk(1)],k∈J.
In particular, if i∈(n-J), this can be rewritten as
(4.23)∑j∈JE[Xj(1)Xk(1)]Pj(i)(1)=∑j∈Jδj,n-iE[Xj(1)Xk(1)],k∈J,
which by identification yields Pj(i)(1)=δj,n-i. Similarly, for i∈{n,n+1,…,2n-1}, in view of (4.21), we have
(4.24)∑j∈JE[Xj(1)Xk(1)]Pj(i)(1)=(-1)k-1δi,k+n-1,k∈J.
In particular, if i∈{n,n+1,…,2n-1}∖(J+n-1), we have δi,k+n-1=0 for k∈J, and then
(4.25)∑j∈JE[Xj(1)Xk(1)]Pj(i)(1)=0,k∈J,
which by identification yields Pj(i)(1)=0=δj,n-i. The proof of Theorem 4.1 is finished.


### 4.2.  Covariance Function

LetcY be the covariance function of (Y(t))t∈[0,1]: cY(s,t)=𝔼[Y(s)Y(t)]. In the next theorem, we supply a representation of cY of the form (2.1).Theorem 4.3.
The covariance function of(Y(t))t∈[0,1] admits the following representation: for s,t∈[0,1],
(4.26)cY(s,t)=∑k=0n-1φk(s∧t)ψ̃k(s∨t)
with, for any k∈{0,1,…,n-1},
(4.27)ψ̃k(t)=ψk(t)-∑j∈Jψk(n-j)(1)Pj(t).
Moreover, the functions ψ̃k, k∈{0,1,…,n-1}, are Hermite interpolation polynomials such that
(4.28){ψ̃k(2n)=0,ψ̃k(i)(0)=δi,kfori∈{0,1,…,n-1},ψ̃k(i)(1)=0fori∈I.Proof.
We decomposeY(t) into the difference
(4.29)Y(t)=Xn(t)-Z(t)withZ(t)=∑j∈JPj(t)Xj(1).
We have
(4.30)cY(s,t)=E[Xn(s)Xn(t)]+E[Z(s)Z(t)]-E[Xn(s)Z(t)]-E[Z(s)Xn(t)]=cXn(s,t)+∑j,k∈JE[Xj(1)Xk(1)]Pj(s)Pk(t)-∑j∈JE[Xn(t)Xj(1)]Pj(s)-∑j∈JE[Xn(s)Xj(1)]Pj(t).
By definition (4.11) of the Pj’s, we observe that
(4.31)∑j,k∈JE[Xj(1)Xk(1)]Pj(s)Pk(t)-∑k∈JE[Xn(s)Xk(1)]Pk(t)=∑k∈J(∑j∈JE[Xj(1)Xk(1)]Pj(s)-E[Xn(s)Xk(1)])Pk(t)=0.
Then, we can simplify cY(s,t) into
(4.32)cY(s,t)=cXn(s,t)-∑j∈JE[Xn(t)Xj(1)]Pj(s).
Since the covariance functions cY and cXn are symmetric, we also have
(4.33)cY(s,t)=cXn(s,t)-∑j∈JE[Xn(s)Xj(1)]Pj(t).
Let us introduce the symmetric polynomial
(4.34)Q(s,t)=∑j∈JE[Xn(s)Xj(1)]Pj(t)=∑j∈JE[Xn(t)Xj(1)]Pj(s).
It can be expressed by means of the functions φk,ψk’s as follows:
(4.35)Q(s,t)=∑j∈J∑k=0n-1φk(s∧t)ψk(n-j)(1)Pj(s∨t).
We can rewrite cY(s,t) as
(4.36)cY(s,t)=cXn(s,t)-Q(s,t)=∑k=0n-1φk(s∧t)[ψk(s∨t)-∑j∈Jψk(n-j)(1)Pj(s∨t)],
and then, for s,t∈[0,1],
(4.37)cY(s,t)=∑k=0n-1φk(s∧t)ψ̃k(s∨t)withψ̃k(t)=ψk(t)-∑j∈Jψk(n-j)(1)Pj(t).
We immediately see that ψ̃k is a polynomial of degree less than 2n such that ψ̃k(i)(0)=δi,k for i∈{0,1,…,n-1} and, since Pj(i)(1)=δj,n-i,
(4.38)ψ̃k(i)(1)=ψk(i)(1)-∑j∈Jψk(n-j)(1)Pj(i)(1)=(1-1{i∈n-J})ψk(i)(1).
We deduce that
(4.39){ψ̃k(i)(1)=0ifi∈(n-J),ψ̃k(i)(1)=ψk(i)(1)=0ifi∈{n,n+1,…,2n-1}∖(J+n-1).
Then ψ̃k(i)(1)=0 for any i∈I. This ends up the proof of Theorem 4.3.


### 4.3.  Boundary Value Problem

In this section, we write out the natural boundary value problem which is associated with the process(Y(t))t∈[0,1]. The following statement is the main connection between the different boundary value conditions associated with the operator d2n/dt2n and the different bridges of the process (Xn)t∈[0,1] introduced in this work.Theorem 4.4.
Letu be a fixed continuous function on [0,1]. The function v defined on [0,1] by
(4.40)v(t)=∫01cY(s,t)u(s)ds
is the solution of the boundary value problem
(4.41){v(2n)=(-1)nuon[0,1],v(i)(0)=0fori∈{0,1,…,n-1},v(i)(1)=0fori∈I.Proof.
Step 1.
Recall thatcY(s,t)=cXn(s,t)-Q(s,t). We decompose the function v into the difference w-z, where, for t∈[0,1],
(4.42)w(t)=∫01cXn(s,t)u(s)ds,z(t)=∫01Q(s,t)u(s)ds.
We know from Theorem 3.1 that w(2n)=(-1)nu, w(i)(0)=(∂iQ/∂ti)(s,0)=0 for i∈{0,1,…,n-1} and w(i)(1)=0 for i∈{n,n+1,…,2n-1}. Moreover, with the function t↦Q(s,t) being a polynomial of degree less than 2n, the function z is also a polynomial of degree less than 2n. Then z(2n)=0, z(i)(0)=0 for i∈{0,1,…,n-1}, and
(4.43)v(2n)=w(2n)-z(2n)=(-1)nu,v(i)(0)=w(i)(0)-z(i)(0)=0fori∈{0,1,…,n-1}.
On the other hand, we learn from (2.13) that, for i∈{0,1,…,2n-1},
(4.44)w(i)(1)=∫01∂icXn∂ti(s,1)u(s)ds,
and then, for i∈{0,1,…,n-1},
(4.45)w(i)(1)=∫01E[Xn(s)Xn-i(1)]u(s)ds.
We also have, for any i∈{0,1,…,2n-1},
(4.46)∂iQ∂ti(s,1)=∑j∈JE[Xn(s)Xj(1)]Pj(i)(1)=1{i∈(n-J)}E[Xn(s)Xn-i(1)].
As a result, we see that
(4.47)z(i)(1)=∫01∂iQ∂ti(s,1)u(s)ds=1{i∈(n-J)}w(i)(1).
This implies that, for i∈(n-J), z(i)(1)=w(i)(1) and for i∈{n,n+1,…,2n-1}∖(J+n-1), z(i)(1)=0=w(i)(1). Then z(i)(1)=w(i)(1) for any i∈I, that is, v(i)(1)=0. The function v is a solution of (4.41).Step 2.
We now check the uniqueness of the solution of (4.41). Let v1 and v2 be two solutions of 𝒟v=u. Then the function w=v1-v2 satisfies 𝒟w=0, w(i)(0)=0 for i∈{0,1,…,n-1} and w(i)(1)=0 for i∈I. We compute the following “energy” integral:
(4.48)∫01w(n)(t)2dt=(-1)n+1[∑i=0n-1(-1)iw(i)(t)w(2n-1-i)(t)]01+(-1)n∫01w(t)w(2n)(t)dt=(-1)n+1∑i=0n-1(-1)iw(i)(1)w(2n-1-i)(1).
We have constructed the set I in order to have w(i)(1)w(2n-1-i)(1)=0 for any i∈{0,1,…,n-1}: when we pick an index i∈{0,1,…,n-1}, either i∈I or 2n-1-i∈I. Indeed, (i)
ifi∈I∩{0,1,…,n-1}, w(i)=0;(ii)
ifi∈{0,1,…,n-1}∖I, by observing that {0,1,…,n-1}∖I={0,1,…,n-1}∖(n-J), we have 2n-1-i∈{n,n+1,…,2n-1}∖(J+n-1). Since {n,n+1,…,2n-1}∖(J+n-1)⊂I, we see that 2n-1-i∈I and then w(2n-1-i)=0.Next, ∫01w(n)(t)2dt=0 which entails w(n)=0, that is, w is a polynomial of degree less than n. Moreover, with the boundary value conditions at 0, we obtain w=0 or v1=v2.
The proof of Theorem4.4 is finished.Remark 4.5.
We have seen in the previous proof that uniqueness is assured as soon as the boundary value conditions at 1 satisfyw(i)(1)w(2n-1-i)(1)=0 for any i∈{0,1,…,n-1}. These conditions are fulfilled when the set I={i1,i2,…,in}⊂{0,1,…,2n-1} is such that i1∈{0,2n-1}, i2∈{1,2n-2}, i3∈{2,2n-3},…,in∈{n-1,n}. This is equivalent to say that I and (2n-1-I) make up a partition of {0,1,…,2n-1}, or
(4.49)2n-1-I={0,1,…,2n-1}∖I.
In this manner, we get 2n different boundary value problems which correspond to the 2n different bridges we have constructed. We will see in Section 6 that the above identity concerning the differentiating set I characterizes the possibility for the Green function of the boundary value problem (4.41) to be the covariance of a Gaussian process.


### 4.4. Prediction

Now, we tackle the problem of the prediction for the process(Y(t))t∈[0,1].Theorem 4.6.
Fixt0∈[0,1]. The shifted process (Y(t+t0))t∈[0,1-t0] admits the following representation:
(4.50)(Y(t+t0))t∈[0,1-t0]=(Ỹt0(t)+∑i=0n-1Qi,t0(t)Y(i)(t0))t∈[0,1-t0],
where
(4.51)Ỹt0(t)=X̃n(t)-∑j∈JP̃j,t0(t)X̃j(1-t0).
The process (X̃n(t))t∈[0,1-t0] is a copy of (Xn(t))t∈[0,1-t0] which is independent of (Xn(t))t∈[0,t0].
The functionsP̃j,t0, j∈J, and Qi,t0, i∈{0,1,…,n-1}, are Hermite interpolation polynomials on [0,1-t0] characterized by
(4.52){P̃j,t0(2n)=0,P̃j,t0(ι)(0)=0forι∈{0,1,…,n-1},P̃j,t0(ι)(1-t0)=δι,n-jforι∈I,{Qi,t0(2n)=0,Qi,t0(ι)(0)=δι,iforι∈{0,1,…,n-1},Qi,t0(ι)(1-t0)=0forι∈I.
Actually, these functions can be expressed by means of the functions Pj, j∈J, and ψ̃i, i∈{0,1,…,n-1}, as follows:
(4.53)P̃j,t0(t)=(1-t0)n-jPj(t1-t0),Qi,t0=(1-t0)iψ̃i(t1-t0).
In other words, the process (Ỹt0(t))t∈[0,1-t0] is a bridge of length (1-t0) which is independent of (Y(t))t∈[0,t0], that is,
(4.54)(Ỹt0(t))t∈[0,1-t0]=d(X̃n(t)∣X̃j(1-t0)=0,j∈J)t∈[0,1-t0].Proof.
Step 1.
Fixt0∈[0,1]. We have the well-known decomposition, based on the classical prediction property of Brownian motion stipulating that (B(t+t0))t∈[0,1-t0]=(B̃t0(t)+B(t0))t∈[0,1-t0], where (B̃t0(t))t∈[0,1-t0] is a Brownian motion independent of (B(t))t∈[0,t0],
(4.55)(Xn(t+t0))t∈[0,1-t0]=(X̃n(t)+∑k=0n-1tkk!Xn-k(t0))t∈[0,1-t0]
with X̃n(t)=∫01((t-s)n-1/(n-1)!)dB̃(s). Differentiating this equality (n-j) times, j∈{1,2,…,n}, we obtain
(4.56)Xj(t+t0)=X̃j(t)+∑k=n-jn-1tj+k-n(j+k-n)!Xn-k(t0).
Therefore,
(4.57)Y(t+t0)=Xn(t+t0)-∑j∈JPj(t+t0)Xj(1)=X̃n(t)+∑k=0n-1tkk!Xn-k(t0)-∑j∈JPj(t+t0)[X̃j(1-t0)+∑k=n-jn-1(1-t0)j+k-n(j+k-n)!Xn-k(t0)]=[X̃n(t)-∑j∈JPj(t+t0)X̃j(1-t0)]+∑k=0n-1[tkk!-∑j∈J:j≥n-k(1-t0)j+k-n(j+k-n)!Pj(t+t0)]Xn-k(t0).
We are going to express the Xn-k(t0), k∈{1,…,n}, by means of the X̃j(1-t0), j∈J. We have, by differentiating (4.57) i times, for i∈{0,1,…,n-1},
(4.58)Y(i)(t+t0)=[X̃n-i(t)-∑j∈JPj(i)(t+t0)X̃j(1-t0)]+∑k=0n-1[tk-i(k-i)!1{k≥i}-∑j∈J:j≥n-k(1-t0)j+k-n(j+k-n)!Pj(i)(t+t0)]Xn-k(t0).
For t=0, this yields
(4.59)Y(i)(t0)=-∑j∈JPj(i)(t0)X̃j(1-t0)+∑k=0n-1[δi,k-∑j∈J:j≥n-k(1-t0)j+k-n(j+k-n)!Pj(i)(t0)]Xn-k(t0).
Set for i,k∈{0,1,…,n-1}(4.60)aik=δi,k-∑j∈J:j≥n-k(1-t0)j+k-n(j+k-n)!Pj(i)(t0),
and let us introduce the matrix
(4.61)A=(aik)0≤i,k≤n-1
together with its inverse matrix
(4.62)B=A-1=(bik)0≤i,k≤n-1.
Equalities (4.59) for i∈{0,1,…,n-1} read as a linear system of (n-1) equations and (n-1) unknowns:
(4.63)∑k=0n-1aikXn-k(t0)=Y(i)(t0)+∑j∈JPj(i)(t0)X̃j(1-t0),i∈{0,1,…,n-1},
which can be rewritten into a matrix form as
(4.64)A(Xn(t0)Xn-1(t0)⋮X2(t0)X1(t0))=(Y(t0)Y′(t0)⋮Y(n-2)(t0)Y(n-1)(t0))+∑j∈JX̃j(1-t0)(Pj(t0)Pj′(t0)⋮Pj(n-2)(t0)Pj(n-1)(t0)).
The solution is given by
(4.65)(Xn(t0)Xn-1(t0)⋮X2(t0)X1(t0))=B(Y(t0)Y′(t0)⋮Y(n-2)(t0)Y(n-1)(t0))+∑j∈JX̃j(1-t0)B(Pj(t0)Pj′(t0)⋮Pj(n-2)(t0)Pj(n-1)(t0)),
and we see that Xn-k(t0), k∈{0,1,…,n-1}, is of the form
(4.66)Xn-k(t0)=∑i=0n-1bkiY(i)(t0)+∑j∈J[∑i=0n-1bkiPj(i)(t0)]X̃j(1-t0).
Therefore, by plugging (4.66) into (4.57), we obtain
(4.67)Y(t+t0)=X̃n(t)-∑j∈JPj(t+t0)X̃j(1-t0)+∑k=0n-1[tkk!-∑j∈J:j≥n-k(1-t0)j+k-n(j+k-n)!Pj(t+t0)]×[∑i=0n-1bkiY(i)(t0)+∑j∈J(∑i=0n-1bkiPj(i)(t0))X̃j(1-t0)]=X̃n(t)-∑j∈J[Pj(t+t0)-∑0≤i,k≤n-1bkiPj(i)(t0)(tkk!-∑j∈J:j≥n-k(1-t0)j+k-n(j+k-n)!Pj(t+t0))]X̃j(1-t0)+∑0≤i,k≤n-1bki[tkk!-∑j∈J:j≥n-k(1-t0)j+k-n(j+k-n)!Pj(t+t0)]Y(i)(t0).
Finally, Y(t+t0) can be written as
(4.68)Y(t+t0)=Ỹt0(t)+∑i=0n-1Qi,t0(t)Y(i)(t0),
where
(4.69)Qi,t0(t)=∑k=0n-1bki[tkk!-∑j∈J:j≥n-k(1-t0)j+k-n(j+k-n)!Pj(t+t0)],Ỹt0(t)=X̃n(t)-∑j∈JP̃j,t0(t)X̃j(1-t0)
with
(4.70)P̃j,t0(t)=Pj(t+t0)-∑0≤i,k≤n-1bkiPj(i)(t0)[tkk!-∑j∈J:j≥n-k(1-t0)j+k-n(j+k-n)!Pj(t+t0)]=Pj(t+t0)-∑i=0n-1Pj(i)(t0)Qi,t0(t).Step 2.
We easily see that the functionsP̃j,t0 and Qi,t0 are polynomials of degree less than 2n. Let us compute now their derivatives at 0 and t0. First, concerning Qi,t0 we have
(4.71)Qi,t0(ι)(t)=∑k=0n-1bki[tk-ι(k-ι)!1{k≥ι}-∑j∈J:j≥n-k(1-t0)j+k-n(j+k-n)!Pj(ι)(t+t0)].
Choosing t=0 and recalling the definition of aιk and the fact that the matrices (aik)0≤i,k≤n and (bik)0≤i,k≤n are inverse, this gives, for ι∈{0,1,…,n-1},
(4.72)Qi,t0(ι)(0)=∑k=0n-1bki[δι,k-∑j∈J:j≥n-k(1-t0)j+k-n(j+k-n)!Pj(ι)(t0)]=∑k=0n-1aιkbki=δι,i.
Choosing t=1-t0, we have for ι∈I(4.73)Qi,t0(ι)(1-t0)=∑k=0n-1bki[(1-t0)k-ι(k-ι)!1{k≥ι}-∑j∈J:j≥n-k(1-t0)j+k-n(j+k-n)!Pj(ι)(1)].
By Theorem 4.1, we know that Pj(ι)(1)=δj,n-ι for ι∈I; then
(4.74)∑j∈J:j≥n-k(1-t0)j+k-n(j+k-n)!Pj(ι)(1)=(1-t0)k-ι(k-ι)!1{k≥ι}1{ι∈(n-J)}.
Observing that, if ι≤k≤n-1, the conditions ι∈(n-J) and ι∈I are equivalent, we simply have
(4.75)∑j∈J:j≥n-k(1-t0)j+k-n(j+k-n)!Pj(ι)(1)=(1-t0)k-ι(k-ι)!1{k≥ι}
which immediately entails, by (4.73),
(4.76)Qi,t0(ι)(1-t0)=0forι∈I.
Next, concerning P̃j,t0(ι), we have
(4.77)P̃j,t0(ι)(t)=Pj(ι)(t+t0)-∑i=0n-1Pj(i)(t0)Qi,t0(ι)(t).
Choosing t=0, this gives, for ι∈{0,1,…,n-1}, since Qi,t0(ι)(0)=δι,i,
(4.78)P̃j,t0(ι)(0)=Pj(ι)(t0)-∑i=0n-1Pj(i)(t0)Qi,t0(ι)(0)=0.
Choosing t=1-t0, we have, for ι∈I, since Pj(ι)(1)=δι,n-j and Qi,t0(ι)(1-t0)=0,
(4.79)P̃j,t0(ι)(1-t0)=Pj(ι)(1)-∑i=0n-1Pj(i)(t0)Qi,t0(ι)(1-t0)=δι,n-j.
The polynomials P̃j,t0 (resp., Qi,t0) enjoy the same properties than the Pj’s (resp., the ψ̃i’s), regarding the successive derivatives, they can be deduced from these latter by a rescaling according as
(4.80)P̃j,t0(t)=(1-t0)n-jPj(t1-t0)(resp.Qi,t0=(1-t0)iψ̃i(t1-t0)).
It is then easy to extract the identity in distribution below, by using the property of Gaussian conditioning:
(4.81)(X̃n(t)-∑j∈JP̃j,t0(t)X̃j(1-t0))t∈[0,1-t0]=d(X̃n(t)∣X̃j(1-t0)=0,j∈J)t∈[0,1-t0].
Theorem 4.6 is established.


## 4.1. Polynomial Drift Description

Below, we provide a representation of(Y(t))t∈[0,1] by means of (Xn(t))t∈[0,1] subject to a random polynomial drift.Theorem 4.1.
One has the distributional identity(4.8)(Y(t))t∈[0,1]=d(Xn(t)-∑j∈JPj(t)Xj(1))t∈[0,1],
where the functions Pj, j∈J, are Hermite interpolation polynomials on [0,1] characterized by
(4.9){Pj(2n)=0,Pj(i)(0)=0fori∈{0,1,…,n-1},Pj(i)(1)=δj,n-ifori∈I.Remark 4.2.
In the case wheren=2, we retrieve a result of [16]. Moreover, conditions (4.9) characterize the polynomials Pj, j∈J. We prove this fact in Lemma A.1 in the Appendix.Proof.
By invoking classical arguments of Gaussian processes theory, we have the distributional identity(4.10)(Y(t))t∈[0,1]=d(Xn(t)-∑j∈JPj(t)Xj(1))t∈[0,1],
where the functions Pj, j∈J, are such that 𝔼[Y(t)Xk(1)]=0 for all k∈J. We get the linear system
(4.11)∑j∈JE[Xj(1)Xk(1)]Pj(t)=E[Xn(t)Xk(1)],k∈J.
We plainly have
(4.12)E[Xj(s)Xk(t)]=∫0s∧t(s-u)j-1(j-1)!(t-u)k-1(k-1)!du.
Then, the system (4.11) writes
(4.13)∑j∈J1j+k-1Pj(t)(j-1)!=∫0t(t-u)n-1(n-1)!(1-u)k-1du.
The matrix of this system (1/(j+k-1))j,k∈J is regular as it can be seen by introducing the related quadratic form which is definite positive. Indeed, for any real numbers xj, j∈J, we have
(4.14)∑j,k∈Jxjxkj+k-1=∫01(∑j,k∈Jxjxkuj+k-2)du=∫01(∑j∈Jxjuj-1)2du≥0,∑j,k∈Jxjxkj+k-1=0iff∀j∈J,xj=0.
Thus, the system (4.11) has a unique solution. As a result, the Pj are linear combinations of the functions t↦∫0t(t-u)n-1(1-u)k-1du which are polynomials of degree less than n+k. Hence, the  Pj are polynomials of degree at most 2n-1.
We now compute the derivatives ofPj at 0 in 1. We have Pj(i)(0)=0 for i∈{0,1,…,n-1} since the functions t↦∫0t(t-u)n-1(1-u)k-1du plainly enjoy this property. For checking that Pj(i)(1)=δj,n-i for i∈I, we successively compute (i)
fori∈{0,1,…,n-1},
(4.15)didtiE[Xn(t)Xk(1)]=E[Xn-i(t)Xk(1)];(ii)
fori=n-1,
(4.16)didtiE[Xn(t)Xk(1)]=E[B(t)Xk(1)]=∫0t(1-u)k-1(k-1)!du;(iii)
fori=n,
(4.17)didtiE[Xn(t)Xk(1)]=(1-t)k-1(k-1)!;(iv)
fori∈{n,n+1,…,n+k-1},
(4.18)didtiE[Xn(t)Xk(1)]=(-1)i+n(1-t)k+n-i-1(k+n-i-1)!;(v)
fori=k+n-1,
(4.19)didtiE[Xn(t)Xk(1)]=(-1)k-1;(vi)
fori≥k+n,
(4.20)didtiE[Xn(t)Xk(1)]=0. Consequently, at time t=1, for i≥n,
(4.21)didtiE[Xn(t)Xk(1)]|t=1=(-1)k-1δi,k+n-1.
Now, by differentiating (4.11), we get for i∈{0,1,…,n-1},
(4.22)∑j∈JE[Xj(1)Xk(1)]Pj(i)(1)=E[Xn-i(1)Xk(1)],k∈J.
In particular, if i∈(n-J), this can be rewritten as
(4.23)∑j∈JE[Xj(1)Xk(1)]Pj(i)(1)=∑j∈Jδj,n-iE[Xj(1)Xk(1)],k∈J,
which by identification yields Pj(i)(1)=δj,n-i. Similarly, for i∈{n,n+1,…,2n-1}, in view of (4.21), we have
(4.24)∑j∈JE[Xj(1)Xk(1)]Pj(i)(1)=(-1)k-1δi,k+n-1,k∈J.
In particular, if i∈{n,n+1,…,2n-1}∖(J+n-1), we have δi,k+n-1=0 for k∈J, and then
(4.25)∑j∈JE[Xj(1)Xk(1)]Pj(i)(1)=0,k∈J,
which by identification yields Pj(i)(1)=0=δj,n-i. The proof of Theorem 4.1 is finished.


## 4.2.  Covariance Function

LetcY be the covariance function of (Y(t))t∈[0,1]: cY(s,t)=𝔼[Y(s)Y(t)]. In the next theorem, we supply a representation of cY of the form (2.1).Theorem 4.3.
The covariance function of(Y(t))t∈[0,1] admits the following representation: for s,t∈[0,1],
(4.26)cY(s,t)=∑k=0n-1φk(s∧t)ψ̃k(s∨t)
with, for any k∈{0,1,…,n-1},
(4.27)ψ̃k(t)=ψk(t)-∑j∈Jψk(n-j)(1)Pj(t).
Moreover, the functions ψ̃k, k∈{0,1,…,n-1}, are Hermite interpolation polynomials such that
(4.28){ψ̃k(2n)=0,ψ̃k(i)(0)=δi,kfori∈{0,1,…,n-1},ψ̃k(i)(1)=0fori∈I.Proof.
We decomposeY(t) into the difference
(4.29)Y(t)=Xn(t)-Z(t)withZ(t)=∑j∈JPj(t)Xj(1).
We have
(4.30)cY(s,t)=E[Xn(s)Xn(t)]+E[Z(s)Z(t)]-E[Xn(s)Z(t)]-E[Z(s)Xn(t)]=cXn(s,t)+∑j,k∈JE[Xj(1)Xk(1)]Pj(s)Pk(t)-∑j∈JE[Xn(t)Xj(1)]Pj(s)-∑j∈JE[Xn(s)Xj(1)]Pj(t).
By definition (4.11) of the Pj’s, we observe that
(4.31)∑j,k∈JE[Xj(1)Xk(1)]Pj(s)Pk(t)-∑k∈JE[Xn(s)Xk(1)]Pk(t)=∑k∈J(∑j∈JE[Xj(1)Xk(1)]Pj(s)-E[Xn(s)Xk(1)])Pk(t)=0.
Then, we can simplify cY(s,t) into
(4.32)cY(s,t)=cXn(s,t)-∑j∈JE[Xn(t)Xj(1)]Pj(s).
Since the covariance functions cY and cXn are symmetric, we also have
(4.33)cY(s,t)=cXn(s,t)-∑j∈JE[Xn(s)Xj(1)]Pj(t).
Let us introduce the symmetric polynomial
(4.34)Q(s,t)=∑j∈JE[Xn(s)Xj(1)]Pj(t)=∑j∈JE[Xn(t)Xj(1)]Pj(s).
It can be expressed by means of the functions φk,ψk’s as follows:
(4.35)Q(s,t)=∑j∈J∑k=0n-1φk(s∧t)ψk(n-j)(1)Pj(s∨t).
We can rewrite cY(s,t) as
(4.36)cY(s,t)=cXn(s,t)-Q(s,t)=∑k=0n-1φk(s∧t)[ψk(s∨t)-∑j∈Jψk(n-j)(1)Pj(s∨t)],
and then, for s,t∈[0,1],
(4.37)cY(s,t)=∑k=0n-1φk(s∧t)ψ̃k(s∨t)withψ̃k(t)=ψk(t)-∑j∈Jψk(n-j)(1)Pj(t).
We immediately see that ψ̃k is a polynomial of degree less than 2n such that ψ̃k(i)(0)=δi,k for i∈{0,1,…,n-1} and, since Pj(i)(1)=δj,n-i,
(4.38)ψ̃k(i)(1)=ψk(i)(1)-∑j∈Jψk(n-j)(1)Pj(i)(1)=(1-1{i∈n-J})ψk(i)(1).
We deduce that
(4.39){ψ̃k(i)(1)=0ifi∈(n-J),ψ̃k(i)(1)=ψk(i)(1)=0ifi∈{n,n+1,…,2n-1}∖(J+n-1).
Then ψ̃k(i)(1)=0 for any i∈I. This ends up the proof of Theorem 4.3.


## 4.3.  Boundary Value Problem

In this section, we write out the natural boundary value problem which is associated with the process(Y(t))t∈[0,1]. The following statement is the main connection between the different boundary value conditions associated with the operator d2n/dt2n and the different bridges of the process (Xn)t∈[0,1] introduced in this work.Theorem 4.4.
Letu be a fixed continuous function on [0,1]. The function v defined on [0,1] by
(4.40)v(t)=∫01cY(s,t)u(s)ds
is the solution of the boundary value problem
(4.41){v(2n)=(-1)nuon[0,1],v(i)(0)=0fori∈{0,1,…,n-1},v(i)(1)=0fori∈I.Proof.
Step 1.
Recall thatcY(s,t)=cXn(s,t)-Q(s,t). We decompose the function v into the difference w-z, where, for t∈[0,1],
(4.42)w(t)=∫01cXn(s,t)u(s)ds,z(t)=∫01Q(s,t)u(s)ds.
We know from Theorem 3.1 that w(2n)=(-1)nu, w(i)(0)=(∂iQ/∂ti)(s,0)=0 for i∈{0,1,…,n-1} and w(i)(1)=0 for i∈{n,n+1,…,2n-1}. Moreover, with the function t↦Q(s,t) being a polynomial of degree less than 2n, the function z is also a polynomial of degree less than 2n. Then z(2n)=0, z(i)(0)=0 for i∈{0,1,…,n-1}, and
(4.43)v(2n)=w(2n)-z(2n)=(-1)nu,v(i)(0)=w(i)(0)-z(i)(0)=0fori∈{0,1,…,n-1}.
On the other hand, we learn from (2.13) that, for i∈{0,1,…,2n-1},
(4.44)w(i)(1)=∫01∂icXn∂ti(s,1)u(s)ds,
and then, for i∈{0,1,…,n-1},
(4.45)w(i)(1)=∫01E[Xn(s)Xn-i(1)]u(s)ds.
We also have, for any i∈{0,1,…,2n-1},
(4.46)∂iQ∂ti(s,1)=∑j∈JE[Xn(s)Xj(1)]Pj(i)(1)=1{i∈(n-J)}E[Xn(s)Xn-i(1)].
As a result, we see that
(4.47)z(i)(1)=∫01∂iQ∂ti(s,1)u(s)ds=1{i∈(n-J)}w(i)(1).
This implies that, for i∈(n-J), z(i)(1)=w(i)(1) and for i∈{n,n+1,…,2n-1}∖(J+n-1), z(i)(1)=0=w(i)(1). Then z(i)(1)=w(i)(1) for any i∈I, that is, v(i)(1)=0. The function v is a solution of (4.41).Step 2.
We now check the uniqueness of the solution of (4.41). Let v1 and v2 be two solutions of 𝒟v=u. Then the function w=v1-v2 satisfies 𝒟w=0, w(i)(0)=0 for i∈{0,1,…,n-1} and w(i)(1)=0 for i∈I. We compute the following “energy” integral:
(4.48)∫01w(n)(t)2dt=(-1)n+1[∑i=0n-1(-1)iw(i)(t)w(2n-1-i)(t)]01+(-1)n∫01w(t)w(2n)(t)dt=(-1)n+1∑i=0n-1(-1)iw(i)(1)w(2n-1-i)(1).
We have constructed the set I in order to have w(i)(1)w(2n-1-i)(1)=0 for any i∈{0,1,…,n-1}: when we pick an index i∈{0,1,…,n-1}, either i∈I or 2n-1-i∈I. Indeed, (i)
ifi∈I∩{0,1,…,n-1}, w(i)=0;(ii)
ifi∈{0,1,…,n-1}∖I, by observing that {0,1,…,n-1}∖I={0,1,…,n-1}∖(n-J), we have 2n-1-i∈{n,n+1,…,2n-1}∖(J+n-1). Since {n,n+1,…,2n-1}∖(J+n-1)⊂I, we see that 2n-1-i∈I and then w(2n-1-i)=0.Next, ∫01w(n)(t)2dt=0 which entails w(n)=0, that is, w is a polynomial of degree less than n. Moreover, with the boundary value conditions at 0, we obtain w=0 or v1=v2.
The proof of Theorem4.4 is finished.Remark 4.5.
We have seen in the previous proof that uniqueness is assured as soon as the boundary value conditions at 1 satisfyw(i)(1)w(2n-1-i)(1)=0 for any i∈{0,1,…,n-1}. These conditions are fulfilled when the set I={i1,i2,…,in}⊂{0,1,…,2n-1} is such that i1∈{0,2n-1}, i2∈{1,2n-2}, i3∈{2,2n-3},…,in∈{n-1,n}. This is equivalent to say that I and (2n-1-I) make up a partition of {0,1,…,2n-1}, or
(4.49)2n-1-I={0,1,…,2n-1}∖I.
In this manner, we get 2n different boundary value problems which correspond to the 2n different bridges we have constructed. We will see in Section 6 that the above identity concerning the differentiating set I characterizes the possibility for the Green function of the boundary value problem (4.41) to be the covariance of a Gaussian process.


## 4.4. Prediction

Now, we tackle the problem of the prediction for the process(Y(t))t∈[0,1].Theorem 4.6.
Fixt0∈[0,1]. The shifted process (Y(t+t0))t∈[0,1-t0] admits the following representation:
(4.50)(Y(t+t0))t∈[0,1-t0]=(Ỹt0(t)+∑i=0n-1Qi,t0(t)Y(i)(t0))t∈[0,1-t0],
where
(4.51)Ỹt0(t)=X̃n(t)-∑j∈JP̃j,t0(t)X̃j(1-t0).
The process (X̃n(t))t∈[0,1-t0] is a copy of (Xn(t))t∈[0,1-t0] which is independent of (Xn(t))t∈[0,t0].
The functionsP̃j,t0, j∈J, and Qi,t0, i∈{0,1,…,n-1}, are Hermite interpolation polynomials on [0,1-t0] characterized by
(4.52){P̃j,t0(2n)=0,P̃j,t0(ι)(0)=0forι∈{0,1,…,n-1},P̃j,t0(ι)(1-t0)=δι,n-jforι∈I,{Qi,t0(2n)=0,Qi,t0(ι)(0)=δι,iforι∈{0,1,…,n-1},Qi,t0(ι)(1-t0)=0forι∈I.
Actually, these functions can be expressed by means of the functions Pj, j∈J, and ψ̃i, i∈{0,1,…,n-1}, as follows:
(4.53)P̃j,t0(t)=(1-t0)n-jPj(t1-t0),Qi,t0=(1-t0)iψ̃i(t1-t0).
In other words, the process (Ỹt0(t))t∈[0,1-t0] is a bridge of length (1-t0) which is independent of (Y(t))t∈[0,t0], that is,
(4.54)(Ỹt0(t))t∈[0,1-t0]=d(X̃n(t)∣X̃j(1-t0)=0,j∈J)t∈[0,1-t0].Proof.
Step 1.
Fixt0∈[0,1]. We have the well-known decomposition, based on the classical prediction property of Brownian motion stipulating that (B(t+t0))t∈[0,1-t0]=(B̃t0(t)+B(t0))t∈[0,1-t0], where (B̃t0(t))t∈[0,1-t0] is a Brownian motion independent of (B(t))t∈[0,t0],
(4.55)(Xn(t+t0))t∈[0,1-t0]=(X̃n(t)+∑k=0n-1tkk!Xn-k(t0))t∈[0,1-t0]
with X̃n(t)=∫01((t-s)n-1/(n-1)!)dB̃(s). Differentiating this equality (n-j) times, j∈{1,2,…,n}, we obtain
(4.56)Xj(t+t0)=X̃j(t)+∑k=n-jn-1tj+k-n(j+k-n)!Xn-k(t0).
Therefore,
(4.57)Y(t+t0)=Xn(t+t0)-∑j∈JPj(t+t0)Xj(1)=X̃n(t)+∑k=0n-1tkk!Xn-k(t0)-∑j∈JPj(t+t0)[X̃j(1-t0)+∑k=n-jn-1(1-t0)j+k-n(j+k-n)!Xn-k(t0)]=[X̃n(t)-∑j∈JPj(t+t0)X̃j(1-t0)]+∑k=0n-1[tkk!-∑j∈J:j≥n-k(1-t0)j+k-n(j+k-n)!Pj(t+t0)]Xn-k(t0).
We are going to express the Xn-k(t0), k∈{1,…,n}, by means of the X̃j(1-t0), j∈J. We have, by differentiating (4.57) i times, for i∈{0,1,…,n-1},
(4.58)Y(i)(t+t0)=[X̃n-i(t)-∑j∈JPj(i)(t+t0)X̃j(1-t0)]+∑k=0n-1[tk-i(k-i)!1{k≥i}-∑j∈J:j≥n-k(1-t0)j+k-n(j+k-n)!Pj(i)(t+t0)]Xn-k(t0).
For t=0, this yields
(4.59)Y(i)(t0)=-∑j∈JPj(i)(t0)X̃j(1-t0)+∑k=0n-1[δi,k-∑j∈J:j≥n-k(1-t0)j+k-n(j+k-n)!Pj(i)(t0)]Xn-k(t0).
Set for i,k∈{0,1,…,n-1}(4.60)aik=δi,k-∑j∈J:j≥n-k(1-t0)j+k-n(j+k-n)!Pj(i)(t0),
and let us introduce the matrix
(4.61)A=(aik)0≤i,k≤n-1
together with its inverse matrix
(4.62)B=A-1=(bik)0≤i,k≤n-1.
Equalities (4.59) for i∈{0,1,…,n-1} read as a linear system of (n-1) equations and (n-1) unknowns:
(4.63)∑k=0n-1aikXn-k(t0)=Y(i)(t0)+∑j∈JPj(i)(t0)X̃j(1-t0),i∈{0,1,…,n-1},
which can be rewritten into a matrix form as
(4.64)A(Xn(t0)Xn-1(t0)⋮X2(t0)X1(t0))=(Y(t0)Y′(t0)⋮Y(n-2)(t0)Y(n-1)(t0))+∑j∈JX̃j(1-t0)(Pj(t0)Pj′(t0)⋮Pj(n-2)(t0)Pj(n-1)(t0)).
The solution is given by
(4.65)(Xn(t0)Xn-1(t0)⋮X2(t0)X1(t0))=B(Y(t0)Y′(t0)⋮Y(n-2)(t0)Y(n-1)(t0))+∑j∈JX̃j(1-t0)B(Pj(t0)Pj′(t0)⋮Pj(n-2)(t0)Pj(n-1)(t0)),
and we see that Xn-k(t0), k∈{0,1,…,n-1}, is of the form
(4.66)Xn-k(t0)=∑i=0n-1bkiY(i)(t0)+∑j∈J[∑i=0n-1bkiPj(i)(t0)]X̃j(1-t0).
Therefore, by plugging (4.66) into (4.57), we obtain
(4.67)Y(t+t0)=X̃n(t)-∑j∈JPj(t+t0)X̃j(1-t0)+∑k=0n-1[tkk!-∑j∈J:j≥n-k(1-t0)j+k-n(j+k-n)!Pj(t+t0)]×[∑i=0n-1bkiY(i)(t0)+∑j∈J(∑i=0n-1bkiPj(i)(t0))X̃j(1-t0)]=X̃n(t)-∑j∈J[Pj(t+t0)-∑0≤i,k≤n-1bkiPj(i)(t0)(tkk!-∑j∈J:j≥n-k(1-t0)j+k-n(j+k-n)!Pj(t+t0))]X̃j(1-t0)+∑0≤i,k≤n-1bki[tkk!-∑j∈J:j≥n-k(1-t0)j+k-n(j+k-n)!Pj(t+t0)]Y(i)(t0).
Finally, Y(t+t0) can be written as
(4.68)Y(t+t0)=Ỹt0(t)+∑i=0n-1Qi,t0(t)Y(i)(t0),
where
(4.69)Qi,t0(t)=∑k=0n-1bki[tkk!-∑j∈J:j≥n-k(1-t0)j+k-n(j+k-n)!Pj(t+t0)],Ỹt0(t)=X̃n(t)-∑j∈JP̃j,t0(t)X̃j(1-t0)
with
(4.70)P̃j,t0(t)=Pj(t+t0)-∑0≤i,k≤n-1bkiPj(i)(t0)[tkk!-∑j∈J:j≥n-k(1-t0)j+k-n(j+k-n)!Pj(t+t0)]=Pj(t+t0)-∑i=0n-1Pj(i)(t0)Qi,t0(t).Step 2.
We easily see that the functionsP̃j,t0 and Qi,t0 are polynomials of degree less than 2n. Let us compute now their derivatives at 0 and t0. First, concerning Qi,t0 we have
(4.71)Qi,t0(ι)(t)=∑k=0n-1bki[tk-ι(k-ι)!1{k≥ι}-∑j∈J:j≥n-k(1-t0)j+k-n(j+k-n)!Pj(ι)(t+t0)].
Choosing t=0 and recalling the definition of aιk and the fact that the matrices (aik)0≤i,k≤n and (bik)0≤i,k≤n are inverse, this gives, for ι∈{0,1,…,n-1},
(4.72)Qi,t0(ι)(0)=∑k=0n-1bki[δι,k-∑j∈J:j≥n-k(1-t0)j+k-n(j+k-n)!Pj(ι)(t0)]=∑k=0n-1aιkbki=δι,i.
Choosing t=1-t0, we have for ι∈I(4.73)Qi,t0(ι)(1-t0)=∑k=0n-1bki[(1-t0)k-ι(k-ι)!1{k≥ι}-∑j∈J:j≥n-k(1-t0)j+k-n(j+k-n)!Pj(ι)(1)].
By Theorem 4.1, we know that Pj(ι)(1)=δj,n-ι for ι∈I; then
(4.74)∑j∈J:j≥n-k(1-t0)j+k-n(j+k-n)!Pj(ι)(1)=(1-t0)k-ι(k-ι)!1{k≥ι}1{ι∈(n-J)}.
Observing that, if ι≤k≤n-1, the conditions ι∈(n-J) and ι∈I are equivalent, we simply have
(4.75)∑j∈J:j≥n-k(1-t0)j+k-n(j+k-n)!Pj(ι)(1)=(1-t0)k-ι(k-ι)!1{k≥ι}
which immediately entails, by (4.73),
(4.76)Qi,t0(ι)(1-t0)=0forι∈I.
Next, concerning P̃j,t0(ι), we have
(4.77)P̃j,t0(ι)(t)=Pj(ι)(t+t0)-∑i=0n-1Pj(i)(t0)Qi,t0(ι)(t).
Choosing t=0, this gives, for ι∈{0,1,…,n-1}, since Qi,t0(ι)(0)=δι,i,
(4.78)P̃j,t0(ι)(0)=Pj(ι)(t0)-∑i=0n-1Pj(i)(t0)Qi,t0(ι)(0)=0.
Choosing t=1-t0, we have, for ι∈I, since Pj(ι)(1)=δι,n-j and Qi,t0(ι)(1-t0)=0,
(4.79)P̃j,t0(ι)(1-t0)=Pj(ι)(1)-∑i=0n-1Pj(i)(t0)Qi,t0(ι)(1-t0)=δι,n-j.
The polynomials P̃j,t0 (resp., Qi,t0) enjoy the same properties than the Pj’s (resp., the ψ̃i’s), regarding the successive derivatives, they can be deduced from these latter by a rescaling according as
(4.80)P̃j,t0(t)=(1-t0)n-jPj(t1-t0)(resp.Qi,t0=(1-t0)iψ̃i(t1-t0)).
It is then easy to extract the identity in distribution below, by using the property of Gaussian conditioning:
(4.81)(X̃n(t)-∑j∈JP̃j,t0(t)X̃j(1-t0))t∈[0,1-t0]=d(X̃n(t)∣X̃j(1-t0)=0,j∈J)t∈[0,1-t0].
Theorem 4.6 is established.


## 5. Example: Bridges of Integrated Brownian Motion (n=2)

Here, we have a look on the particular case wheren=2 for which the corresponding process (Xn(t))t∈[0,1] is nothing but integrated Brownian motion (the so-called Langevin process): (5.1)X2(t)=∫0tB(s)ds.
The underlying Markov process is the so-called Kolmogorov diffusion (X2(t),X1(t))t∈[0,1]. All the associated conditioned processes that will be constructed are related to the equation v(4)(t)=u(t) with boundary value conditions at time 0: v(0)=v′(0)=0. There are four such processes: (i)
(X2(t))t∈[0,1] (integrated Brownian motion);(ii)
(X2(t)∣X1(1)=0)t∈[0,1] (integrated Brownian bridge);(iii)
(X2(t)∣X2(1)=0)t∈[0,1] (bridge of integrated Brownian motion);(iv)
(X2(t)∣X1(1)=X2(1)=0)t∈[0,1] (another bridge of integrated Brownian motion).On the other hand, when adding two boundary value conditions at time 1 to the foregoing equation, we find six boundary value problems:v(1)=v′(1)=0, v(1)=v′′(1)=0, v(1)=v′′′(1)=0, v′(1)=v′′(1)=0, v′(1)=v′′′(1)=0, v′′(1)=v′′′(1)=0. Actually, only four of them can be related to some Gaussian processes—the previously listed processes—in the sense of our work whereas two others cannot be.For each process, we provide the covariance function, the representation by means of integrated Brownian motion subject to a random polynomial drift, the related boundary value conditions at 1, and the decomposition related to the prediction problem. Since the computations are straightforward, we will omit them and we only report here the results.For an account on integrated Brownian motion in relation with the present work, we refer the reader to, for example, [16] and references therein.


### 5.1. Integrated Brownian Motion

The process corresponding to the setI=∅ is nothing but integrated Brownian motion: (5.2)(X2(t))t∈[0,1]=(∫0tB(s)ds)t∈[0,1].
The covariance function is explicitly given by (5.3)c(s,t)=16[s∧t]2[3(s∨t)-s∧t].
This process is related to the boundary value conditions at 1 (J={2,3}): v′′(1)=v′′′(1)=0. The prediction property can be stated as follows: (5.4)(X2(t+t0))t∈[0,1-t0]=(X̃2(t)+X2(t0)+tX1(t0))t∈[0,1-t0].


### 5.2.  Integrated Brownian Bridge

The process corresponding to the setI={1} is integrated Brownian bridge: (5.5)(Y(t))t∈[0,1]=(∫0tB(s)ds∣B(1)=0)t∈[0,1]=(∫0tβ(s)ds)t∈[0,1]=(X2(t)∣X1(1)=0)t∈[0,1].
This process can be represented as (5.6)(Y(t))t∈[0,1]=(X2(t)-12t2X1(1))t∈[0,1].
The covariance function is explicitly given by (5.7)c(s,t)=16[s∧t]2[3(s∨t)-s∧t]-14s2t2.
The process (Y(t))t∈[0,1] is related to the boundary value conditions at 1 (J={1,3}): v′(1)=v′′′(1)=0. The prediction property says that (5.8)(Y(t+t0))t∈[0,1-t0]=(Ỹt0(t)+Y(t0)+(t-t22(1-t0))Y′(t0))t∈[0,1-t0].


### 5.3.  Bridge of Integrated Brownian Motion

The process corresponding to the setI={2} is the bridge of integrated Brownian motion: (5.9)(Y(t))t∈[0,1]=(∫0tB(s)ds∣∫01B(s)ds=0)t∈[0,1]=(X2(t)∣X2(1)=0)t∈[0,1].
The bridge is understood as the process is pinned at its extremities: Y(0)=Y(1)=0. This process can be represented as (5.10)(Y(t))t∈[0,1]=(X2(t)-12t2(3-t)X2(1))t∈[0,1].
The covariance function is explicitly given by (5.11)c(s,t)=16[s∧t]2[3(s∨t)-s∧t]-112s2t2(3-s)(3-t).
The process (Y(t))t∈[0,1] is related to the boundary value conditions at 1 (J={0,2}): v(1)=v′′(1)=0. The prediction property says that (5.12)(Y(t+t0))t∈[0,1-t0]=(Ỹt0(t)+t3-3(1-t0)t2+2(1-t0)32(1-t0)3Y(t0)+t3-3(1-t0)t2+2(1-t0)2t2(1-t0)2Y′(t0))t∈[0,1-t0].


### 5.4.  Other Bridge of Integrated Brownian Motion

The process corresponding to the setI={1,2} is another bridge of integrated Brownian motion (actually of the two-dimensional Kolmogorov diffusion): (5.13)(Y(t))t∈[0,1]=(∫0tB(s)ds∣∫01B(s)ds=B(1)=0)t∈[0,1]=(X2(t)∣X2(1)=X1(1)=0)t∈[0,1].
The bridge here is understood as the process is pinned at its extremities together with its derivatives: Y(0)=Y′(0)=Y(1)=Y′(1)=0. This process can be represented as (5.14)(Y(t))t∈[0,1]=(X2(t)-t2(t-1)X1(1)-t2(3-2t)X2(1))t∈[0,1].
The covariance function is explicitly given by (5.15)c(s,t)=16[s∧t]2[1-s∨t]2[3(s∨t)-s∧t-2st].
The process (Y(t))t∈[0,1] is related to the boundary value conditions at 1 (J={0,1}): v(1)=v′(1)=0. The prediction property says that (5.16)(Y(t+t0))t∈[0,1-t0]=(Ỹt0(t)+t2(t+t0-1)(1-t0)3Y(t0)+t2(3-3t0-2t)(1-t0)2Y′(t0))t∈[0,1-t0].


### 5.5.  Two Counterexamples

(i) The solution of the problem associated with the boundary value conditionsv(1)=v′′′(1)=0 (which corresponds to the set I1={0,3}) is given by
(5.17)v(t)=∫01G1(s,t)u(s)ds,t∈[0,1],
where (5.18)G1(s,t)=16[s∧t]2[3(s∨t)-s∧t]+16s2t2(s-3).(ii) The solution of the problem associated with the boundary value conditionsv′(1)=v′′(1)=0 (which corresponds to the set I2={1,2}) is given by (5.19)v(t)=∫01G2(s,t)u(s)ds,t∈[0,1],
where (5.20)G2(s,t)=16[s∧t]2[3(s∨t)-s∧t]+16s2t2(t-3).
We can observe the relationships G1(s,t)=G2(t,s) and I2={0,1,2,3}∖(3-I1). The Green functions G1 and G2 are not symmetric, so they cannot be viewed as the covariance functions of any Gaussian process. In the next section, we give an explanation of these observations.


## 5.1. Integrated Brownian Motion

The process corresponding to the setI=∅ is nothing but integrated Brownian motion: (5.2)(X2(t))t∈[0,1]=(∫0tB(s)ds)t∈[0,1].
The covariance function is explicitly given by (5.3)c(s,t)=16[s∧t]2[3(s∨t)-s∧t].
This process is related to the boundary value conditions at 1 (J={2,3}): v′′(1)=v′′′(1)=0. The prediction property can be stated as follows: (5.4)(X2(t+t0))t∈[0,1-t0]=(X̃2(t)+X2(t0)+tX1(t0))t∈[0,1-t0].


## 5.2.  Integrated Brownian Bridge

The process corresponding to the setI={1} is integrated Brownian bridge: (5.5)(Y(t))t∈[0,1]=(∫0tB(s)ds∣B(1)=0)t∈[0,1]=(∫0tβ(s)ds)t∈[0,1]=(X2(t)∣X1(1)=0)t∈[0,1].
This process can be represented as (5.6)(Y(t))t∈[0,1]=(X2(t)-12t2X1(1))t∈[0,1].
The covariance function is explicitly given by (5.7)c(s,t)=16[s∧t]2[3(s∨t)-s∧t]-14s2t2.
The process (Y(t))t∈[0,1] is related to the boundary value conditions at 1 (J={1,3}): v′(1)=v′′′(1)=0. The prediction property says that (5.8)(Y(t+t0))t∈[0,1-t0]=(Ỹt0(t)+Y(t0)+(t-t22(1-t0))Y′(t0))t∈[0,1-t0].


## 5.3.  Bridge of Integrated Brownian Motion

The process corresponding to the setI={2} is the bridge of integrated Brownian motion: (5.9)(Y(t))t∈[0,1]=(∫0tB(s)ds∣∫01B(s)ds=0)t∈[0,1]=(X2(t)∣X2(1)=0)t∈[0,1].
The bridge is understood as the process is pinned at its extremities: Y(0)=Y(1)=0. This process can be represented as (5.10)(Y(t))t∈[0,1]=(X2(t)-12t2(3-t)X2(1))t∈[0,1].
The covariance function is explicitly given by (5.11)c(s,t)=16[s∧t]2[3(s∨t)-s∧t]-112s2t2(3-s)(3-t).
The process (Y(t))t∈[0,1] is related to the boundary value conditions at 1 (J={0,2}): v(1)=v′′(1)=0. The prediction property says that (5.12)(Y(t+t0))t∈[0,1-t0]=(Ỹt0(t)+t3-3(1-t0)t2+2(1-t0)32(1-t0)3Y(t0)+t3-3(1-t0)t2+2(1-t0)2t2(1-t0)2Y′(t0))t∈[0,1-t0].


## 5.4.  Other Bridge of Integrated Brownian Motion

The process corresponding to the setI={1,2} is another bridge of integrated Brownian motion (actually of the two-dimensional Kolmogorov diffusion): (5.13)(Y(t))t∈[0,1]=(∫0tB(s)ds∣∫01B(s)ds=B(1)=0)t∈[0,1]=(X2(t)∣X2(1)=X1(1)=0)t∈[0,1].
The bridge here is understood as the process is pinned at its extremities together with its derivatives: Y(0)=Y′(0)=Y(1)=Y′(1)=0. This process can be represented as (5.14)(Y(t))t∈[0,1]=(X2(t)-t2(t-1)X1(1)-t2(3-2t)X2(1))t∈[0,1].
The covariance function is explicitly given by (5.15)c(s,t)=16[s∧t]2[1-s∨t]2[3(s∨t)-s∧t-2st].
The process (Y(t))t∈[0,1] is related to the boundary value conditions at 1 (J={0,1}): v(1)=v′(1)=0. The prediction property says that (5.16)(Y(t+t0))t∈[0,1-t0]=(Ỹt0(t)+t2(t+t0-1)(1-t0)3Y(t0)+t2(3-3t0-2t)(1-t0)2Y′(t0))t∈[0,1-t0].


## 5.5.  Two Counterexamples

(i) The solution of the problem associated with the boundary value conditionsv(1)=v′′′(1)=0 (which corresponds to the set I1={0,3}) is given by
(5.17)v(t)=∫01G1(s,t)u(s)ds,t∈[0,1],
where (5.18)G1(s,t)=16[s∧t]2[3(s∨t)-s∧t]+16s2t2(s-3).(ii) The solution of the problem associated with the boundary value conditionsv′(1)=v′′(1)=0 (which corresponds to the set I2={1,2}) is given by (5.19)v(t)=∫01G2(s,t)u(s)ds,t∈[0,1],
where (5.20)G2(s,t)=16[s∧t]2[3(s∨t)-s∧t]+16s2t2(t-3).
We can observe the relationships G1(s,t)=G2(t,s) and I2={0,1,2,3}∖(3-I1). The Green functions G1 and G2 are not symmetric, so they cannot be viewed as the covariance functions of any Gaussian process. In the next section, we give an explanation of these observations.


## 6. General Boundary Value Conditions

In this last part, we address the problem of relating the general boundary value problem(6.1){v(2n)=(-1)nuon⁡[0,1],v(0)=v′(0)=⋯=v(n-1)(0)=0,v(i1)(1)=v(i2)(1)=⋯=v(in)(1)=0,
for any indices i1,i2,…,in such that 0≤i1<i2<⋯<in≤2n-1, to some possible Gaussian process. Set I={i1,i2,…,in}. We have noticed in Theorem 4.4 and Remark 4.5 that, when I satisfies the relationship 2n-1-I={0,1,…,2n-1}∖I, the system (6.1) admits a unique solution. We proved this fact by computing an energy integral. Actually, this fact holds for any set of indices I; see Lemma A.1.Our aim is to characterize the set of indicesI for which the Green function of (6.1) can be viewed as the covariance function of a Gaussian process. A necessary condition for a function of two variables to be the covariance function of a Gaussian process is that it must be symmetric. So, we characterize the set of indices I for which the Green function of (6.1) is symmetric and we will see that in this case this function is a covariance function. For this, we make use of general results concerning the relationships between boundary value problems and symmetric linear operators in Hilbert spaces. Since our results below are direct applications of Theorem 2 in Section 18.2 of [17, Chapter V], which describes the general form of boundary conditions corresponding to the self-adjoint extensions of standard symmetric operator, we only state them without any proof. We refer the reader to a draft [18] for more details and self-contained proofs.


### 6.1.  Representation of the Solution

We first write out a representation for the Green function of (6.1).Theorem 6.1.
The boundary value problem (6.1) has a unique solution. The corresponding Green function admits the following representation, for s,t∈[0,1]:
(6.2)GI(s,t)=(-1)n1{s≤t}(t-s)2n-1(2n-1)!-(-1)n∑ι∈I(1-s)2n-1-ι(2n-1-ι)!RI,ι(t),
where the RI,ι, ι∈I, are Hermite interpolation polynomials satisfying
(6.3){RI,ι(2n)=0,RI,ι(i)(0)=0fori∈{0,1,…,n-1},RI,ι(i)(1)=δι,ifori∈I.Remark 6.2.
Conditions (6.3) completely characterize the polynomials RI,ι, ι∈I (see Lemma A.1 in the appendix).A necessary condition forGI to be the covariance function of a Gaussian process is that it must be symmetric: GI(s,t)=GI(t,s) for any s,t∈[0,1]. The theorem below asserts that if the set of indices I is not of the form displayed in the preamble of Section 4, that is, I≠{0,1,…,2n-1}∖(2n-1-I), the Green function GI is not symmetric and consequently this function cannot be viewed as a covariance function, that is, we cannot relate the boundary value problem (6.1) to any Gaussian process.Theorem 6.3.
The Green functionGI is symmetric (and it corresponds to a covariance function) if and only if the set of indices I satisfies 2n-1-I={0,1,…,2n-1}∖I.


### 6.2. Example: Bridges of Twice Integrated Brownian Motion (n=3)

Here, we have a look on the particular case wheren=3 for which the corresponding process (Xn(t))t∈[0,1] is the twice integrated Brownian motion: (6.4)X3(t)=∫0t(t-s)B(s)ds=∫0t(∫0s2B(s1)ds1)ds2.
All the associated conditioned processes that can be constructed are related to the equation v(6)(t)=-u(t) with boundary value conditions at time 0: v(0)=v′(0)=v′′(0)=0. There are 23=8 such processes. Since the computations are tedious and the explicit results are cumbersome, we only report the correspondence between bridges and boundary value conditions at time 1 through the sets of indices I and J. These are written in Table 2.Table 2Conditioning setJ∅{1}{2}{3}{1,2}{1,3}{2,3}{1,2,3}Differentiating setI{3,4,5}{2,4,5}{1,3,5}{0,3,4}{1,2,5}{0,2,4}{0,1,3}{0,1,2}The Green functions related to the other sets cannot be related to some Gaussian processes. The sets are written in Table3 with the correspondence I2={0,1,2,3,4,5}∖(5-I1).Table 3Differentiating setI1{0,1,4}{0,1,5}{0,2,5}{0,3,5}{0,4,5}{1,4,5}Differentiating setI2{0,2,3}{1,2,3}{1,2,4}{1,3,4}{2,3,4}{2,3,5}


### 6.3.  Example: Bridges of Thrice Integrated Brownian Motion (n=4)

Forn=4, only the 24=16 following differentiating sets can be related to bridges: (6.5){0,1,2,3},{0,1,2,4},{0,1,3,5},{0,1,4,5},{0,2,3,6},{0,2,4,6},{0,3,5,6},{0,4,5,6},{1,2,3,7},{1,2,4,7},{1,3,5,7},{1,4,5,7},{2,3,6,7},{2,4,6,7},{3,5,6,7},{4,5,6,7}.


## 6.1.  Representation of the Solution

We first write out a representation for the Green function of (6.1).Theorem 6.1.
The boundary value problem (6.1) has a unique solution. The corresponding Green function admits the following representation, for s,t∈[0,1]:
(6.2)GI(s,t)=(-1)n1{s≤t}(t-s)2n-1(2n-1)!-(-1)n∑ι∈I(1-s)2n-1-ι(2n-1-ι)!RI,ι(t),
where the RI,ι, ι∈I, are Hermite interpolation polynomials satisfying
(6.3){RI,ι(2n)=0,RI,ι(i)(0)=0fori∈{0,1,…,n-1},RI,ι(i)(1)=δι,ifori∈I.Remark 6.2.
Conditions (6.3) completely characterize the polynomials RI,ι, ι∈I (see Lemma A.1 in the appendix).A necessary condition forGI to be the covariance function of a Gaussian process is that it must be symmetric: GI(s,t)=GI(t,s) for any s,t∈[0,1]. The theorem below asserts that if the set of indices I is not of the form displayed in the preamble of Section 4, that is, I≠{0,1,…,2n-1}∖(2n-1-I), the Green function GI is not symmetric and consequently this function cannot be viewed as a covariance function, that is, we cannot relate the boundary value problem (6.1) to any Gaussian process.Theorem 6.3.
The Green functionGI is symmetric (and it corresponds to a covariance function) if and only if the set of indices I satisfies 2n-1-I={0,1,…,2n-1}∖I.


## 6.2. Example: Bridges of Twice Integrated Brownian Motion (n=3)

Here, we have a look on the particular case wheren=3 for which the corresponding process (Xn(t))t∈[0,1] is the twice integrated Brownian motion: (6.4)X3(t)=∫0t(t-s)B(s)ds=∫0t(∫0s2B(s1)ds1)ds2.
All the associated conditioned processes that can be constructed are related to the equation v(6)(t)=-u(t) with boundary value conditions at time 0: v(0)=v′(0)=v′′(0)=0. There are 23=8 such processes. Since the computations are tedious and the explicit results are cumbersome, we only report the correspondence between bridges and boundary value conditions at time 1 through the sets of indices I and J. These are written in Table 2.Table 2Conditioning setJ∅{1}{2}{3}{1,2}{1,3}{2,3}{1,2,3}Differentiating setI{3,4,5}{2,4,5}{1,3,5}{0,3,4}{1,2,5}{0,2,4}{0,1,3}{0,1,2}The Green functions related to the other sets cannot be related to some Gaussian processes. The sets are written in Table3 with the correspondence I2={0,1,2,3,4,5}∖(5-I1).Table 3Differentiating setI1{0,1,4}{0,1,5}{0,2,5}{0,3,5}{0,4,5}{1,4,5}Differentiating setI2{0,2,3}{1,2,3}{1,2,4}{1,3,4}{2,3,4}{2,3,5}


## 6.3.  Example: Bridges of Thrice Integrated Brownian Motion (n=4)

Forn=4, only the 24=16 following differentiating sets can be related to bridges: (6.5){0,1,2,3},{0,1,2,4},{0,1,3,5},{0,1,4,5},{0,2,3,6},{0,2,4,6},{0,3,5,6},{0,4,5,6},{1,2,3,7},{1,2,4,7},{1,3,5,7},{1,4,5,7},{2,3,6,7},{2,4,6,7},{3,5,6,7},{4,5,6,7}.



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*Source: 762486-2011-12-13.xml*