Stochastic differential equations with boundary conditions driven by a Poisson noise

We consider one-dimensional stochastic differential equations with a boundary condition, driven by a Poisson process. We study existence and uniqueness of solutions and the absolute continuity of the law of the solution. In the case when the coefficients are linear, we give an explicit form of the solution and study the reciprocal process property.


Introduction
Stochastic differential equations (s.d.e.) with boundary conditions driven by a Wiener process have been extensively studied in the last fifteen years, both in the ordinary and the partial differential cases.We highlight the papers of Ocone and Pardoux [13], Nualart and Pardoux [10], Donati-Martin [6], Buckdahn and Nualart [4], and Alabert, Ferrante and Nualart [1].These equations arise from the usual ones when we replace the customary initial condition by a functional relation h(X 0 , X 1 ) = 0 between two variables of the solution process X, which is only considered in the bounded time interval [0, 1].Features that have been considered include existence and uniqueness, absolute continuity of the laws, numerical approximations, and Markovian-type properties.
Recently, in our work [2], we have considered boundary value problems where the stochastic integral with respect to the Wiener process is replaced by an additive Poisson perturbation N where the boundary condition is written in a more manageable form.We established an existence and uniqueness result, studied the absolutely continuity of the laws, and characterised several classes of coefficients f which lead to the so-called reciprocal property of the solution.
Let us recall that X = {X t , t ∈ [0, 1]} is a reciprocal process if for all times 0 ≤ s < t ≤ 1, the families or random variables {X u , u ∈ [s, t]} and {X u , u ∈ [0, 1] − [s, t]} are conditionally independent given X s and X t .This property is weaker than the usual Markov property.
Interest in reciprocal processes dates back to Bernstein [3] (they are also called Bernstein processes by physicists) because of their role in the probabilistic interpretation of quantum mechanics.It is by far not true that all s.d.e. with boundary conditions give rise to reciprocal processes and it is also false that a general reciprocal process could be represented as the solution of some sort of first order boundary value problem, no matter which type of driving process is taken.Nevertheless, it is interesting to try to find in which cases the probabilistic dynamic representation given by a first order s.d.e., together with a suitable boundary relation, is indeed able to represent a reciprocal process.
In this paper, we develop the same program as in our previous paper [2], but with a multiplicative Poisson perturbation.Specifically, we consider the equation where f, F : [0, 1]×R → R and ψ : R → R are measurable functions satisfying certain hypotheses, and N = {N t , t ≥ 0} is a Poisson process with intensity 1.
Due to the boundary condition, the solution will anticipate any filtration to which N is adapted, and therefore the stochastic integral appearing in the equation is, strictly speaking, an anticipating integral.However, the bounded variation character of the Poisson process permits to avoid most of the technical difficulties of the anticipating stochastic integrals with respect to the Wiener process.
Equation (1.1) is a "forward equation".One can also consider the "backward equation" (1.2) and the Skorohod-type equation where δ Ñr denotes the Skorohod integral with respect to the compensated Poisson process.
While the stochastic integrals in (1.1) and (1.2) are no more than Stieljes integrals, the Skorohod integral operator is defined by means of the chaos decomposition on the canonical Poisson space.We refer the reader to [8], [12] or [11] for an introduction to the canonical Poisson space, the chaos decomposition and the Skorohod integral.The paper is organised as follows.Section 2 is devoted to the study of the stochastic flow (initial condition problem) associated with s.d.e.(1.1), which will give us the preliminary results needed for the boundary condition case.In Section 3 we study existence, uniqueness, regularity and absolute continuity of the solution to the problem (1.1).In both Sections 2 and 3, the case of linear equations is studied as a special example.In Section 4, we find some sufficient conditions for the solution of the linear equation to enjoy the reciprocal property.In the final Section 5, the relation of the forward equation (1.1) with the backward equation (1.2) and the Skorohod equation (1.3) is established.The linear equation is again considered with particular attention, and the chaos decomposition of the solution is computed in two very simple special cases.
We will use the notation ∂ i g for the derivative of a function g with respect to the i-th coordinate, g(s − ) and g(s + ) for lim t↑s g(t) and lim t↓s g(t) respectively, and the acronym càdlàg for "right continuous with left limits".Throughout the paper, we employ the usual convention that a summation and a product with an empty set of indices are equal to zero and one, respectively.

Stochastic flows induced by Poisson equations
Let N = {N t , t ≥ 0} be a standard Poisson process with intensity 1 defined on some probability space (Ω, F, P ); that means, N has independent increments, N t − N s has a Poisson law with parameter t − s, N 0 ≡ 0, and all its paths are integer-valued, non-decreasing, càdlàg, with jumps of size 1.
Throughout the paper, S n will denote the jump times of N : The sequence S n is strictly increasing to infinity, and Let us consider the pathwise equation where x ∈ R, and assume that f, F : [0, 1]× R → R are measurable functions such that f satisfies For every x ∈ R, denote by Φ(s, t; x) the solution to the deterministic equation All conclusions of the following lemma are well known or easy to show: Lemma 2.1 Under hypotheses (H 1 ) and (H 2 ), there exists a unique solution Φ(s, t; x) of equation (2.2).Moreover: 1) For every 0 ≤ s ≤ t ≤ 1, and every x ∈ R, |Φ(s, t; x)| ≤ (|x| + M 1 )e K 1 (t−s) .
3) For every 0 ≤ s ≤ t ≤ 1, and every In particular, for every s, t, the function x → Φ(s, t; x) is a homeomorphism from R into R.Then, for each x ∈ R, there exists a unique process ϕ(x) = {ϕ st (x), 0 ≤ s ≤ t ≤ 1} that solves (2.1).Moreover:
The existence and regularity of the function ϕ st (s 1 , . . ., s n ; x) of ( 3) are also clear from (2.3).We compute now its derivative with respect to s j .For n = 1, we get Suppose that (3) holds for n = k.Then, for n = k + 1 and j = 1, . . ., k, Taking into account that we obtain, for j = k + 1, In the next proposition we find that under the regularity hypotheses of Proposition 2.4 and an additional condition relating f and F , the law of ϕ t (x) is a weighted sum of a Dirac-δ and an absolutely continuous probability.
Let ϕ(x) = {ϕ t (x), t ∈ [0, 1]} be the solution to (2.1) for s = 0.Then, for all t > 0, the distribution function L of ϕ t (x) can be written as and ) and condition (2.7), we obtain |∂ n G| > 0. It is known that, conditionally to {N t = n}, (S 1 , . . ., S n ) follows the uniform distribution on and therefore ϕ t (x) is absolutely continuous on {N t = n}, for every n ≥ 1, with conditional density and the result follows.
which is the hypothesis used by Carlen and Pardoux in [5] (Theorem 4.3) to prove that, in the autonomous case, the law of ϕ 1 (x) is absolutely continuous on the set {N 1 ≥ 1}.

Equations with boundary conditions
In this section we establish first an easy existence and uniqueness theorem, based on Proposition 2.2 above, when the initial condition is replaced by a boundary condition.Then we prove in this situation the analogue of Propositions 2.4(3) and 2.5 on the differentiability with respect to the jump times and the absolute continuity of the laws (Proposition 3.4 and Theorem 3.5 below, respectively).
and (H 2 ) of Section 2, with constants K 1 and M 1 respectively, and there exists a constant ψ is a continuous and non-increasing function. Then admits a unique solution X, which is a càdlàg process.
Remark 3.2 In general, the condition k 2 ≥ −1 cannot be relaxed.For instance, the problem has no solutions.Indeed, the first equality implies X t = X 0 e t (−1) Nt (see Example 2.3), which gives X 1 = −X 0 e for N 1 ∈ {1, 3, 5, . . .}, and this is incompatible with the boundary condition.
On the other hand, if we change the boundary condition to X 0 = −1 e X 1 the new problem has an infinite number of solutions: where x(ω) is an arbitrary real number.
Notice that the purpose of (H 3 ) is to ensure that x → ψ(ϕ 1 (x)) has a unique fix point.Alternative hypotheses that lead to the same consequence may be used instead for particular cases.See for instance the comments at the end of Example 3.3.
Proof: Since X 0 = ψ(ϕ 1 (X 0 )), we have On the other hand, for X t = ϕ t (X 0 ), The following theorem is the counterpart of Proposition 2.5 for the case of boundary conditions.The proof follows the same lines but using at the end the decomposition Assume in addition that ψ ′ < 0 and that condition (2.7) holds.Let x * be the unique solution to x = ψ(Φ(0, 1; x)), and X the solution to (3.1).Then, the distribution function of and where h 0n is the density of X t conditioned to N t = 0, N 1 = n, and h n is the density of X t conditioned to N t = n.For t = 0, the formula is also valid taking h n ≡ 0.

The reciprocal property
Let (Ω, F, P ) be a probability space and let A 1 , A 2 and B be sub-σ-fields of F such that Then the σ-fields A 1 and A 2 are said to be conditionally independent with respect to B. One can show that if X is a Markov process then X is reciprocal, and that the converse is not true.For a proof of this fact, we refer the reader to Alabert and Marmolejo [2] (Proposition 4.2), where we also established the next lemma (Lemma 4.6 of [2]).
In our previous work [2], we obtained several sufficient conditions on f for the solution to enjoy the reciprocal property when the Poisson noise appears additively, namely in The main classes of functions f leading to this property are those of the form f (t, x) = f 1 (t) + f 2 (t)x and those which are 1-periodic in the second variable, f (t, x) = f (t, x + 1).But we showed with examples that there are many more; we also obtained conditions on f ensuring that the solution will not be reciprocal.In contrast, for equations driven by the Wiener process, conditions which are at the same time necessary and sufficient have been obtained in a wide variety of settings, even with multiplicative noise.
With a multiplicative Poisson noise, the techniques currently known do not seem to allow a general analysis.We will restrict ourselves to linear equations.The main result contained in the next theorem is that if both coefficients are truly linear in the second variable (i.e.f (t, x) = f 2 (t)x, and F (t, x) = F 2 (t)x) then the solution is reciprocal.We have not been able to obtain necessary conditions, even when considering only the class of linear equations.Thus, we have to leave open the study of the general linear case, which for white noise driven equations (with boundary conditions also linear) was studied thoroughly in the seminal paper of Ocone and Pardoux [13].1], and ψ : R → R a continuous and non-increasing function.Let X = {X t , t ∈ [0, 1]} be the solution of In each of the following cases, X is a reciprocal process: (1) ψ is constant.
Proof: (1) reduces to the case of initial condition, while in (2) the solution is identically zero.Thus in both situations we obtain a Markov process.
(3) In this case the solution is where A(t) = exp{ t 0 f 2 (r) dr} as before, and X 0 solves ), where g is a Borel function and Since ξ has independents increments, Lemma 4.2 implies that Y , and therefore X, are reciprocal processes.
(5) In this situation, we obtain (4.4) where x * solves x = ψ(xA(1)).Process (4.4) can be thought as the solution to the initial value problem where η is the random variable Although η anticipates the Poisson process, X not only has the reciprocal property, but it is in fact a Markov process.Indeed, it is immediate to check that Y t := X t /A(t) is a Markov chain taking at most three values.

Backward and Skorohod equations
In this Section we consider the backward and Skorohod versions of our boundary value problems.There are very simple cases where the backward equation, even in the initial condition situation, does not possess a solution.For example, for k ∈ R, the equation In general, for the existence of a solution of (5.1) it is necessary that the mapping A r (y) := y − F (r, y) be invertible for each r.Assume now that f : [0, 1] × R → R satisfies the hypotheses (H 1 ) and (H 2 ) of Section 2, and that either Then there exists a unique process ϕ(x) = {ϕ t (x), t ∈ [0, 1]} that satisfies the backward equation (5.1).This follows from Theorem 5.1 of León, Solé and Vives [8], where it is shown that ϕ is a solution to (5.1) if and only if ϕ is a solution to the forward equation The existence of A −1 r is assured by (5.2) or (5.3).We consider now the backward equation with boundary condition Theorem 5.1 Assume that f satisfies hypotheses (H 1 ) and (H 2 ) of Section 2 and that for some functions α and β such that α − 1 ≤ β ≤ α < 1. Assume moreover that ψ satisfies hypothesis (H 3 ) of Theorem 3.1.Then (5.4) admits a unique solution X = {X t , t ∈ [0, 1]}, which is a càdlàg process.
Proof: By the relation between the forward and the backward equation with initial condition given above, the solution to (5.4) coincides with the solution to the forward equation with boundary condition provided it exists.By Theorem 3.1, it is enough to show there exists a constant .
We find and the conclusion follows.
The study of the properties of backward equations can thus be reduced to the case of the forward equations when the above condition (5.5) on F holds true.In particular, when F (t, x) = F 1 (t) + F 2 (t)x, we obtain x , and condition (5.5) reduces to F 2 < 1.
Example 5.2 (Linear backward equation).Now consider the problem where f 1 , f 2 , F 1 , F 2 : [0, 1] → R are continuous functions with F 2 < 1 and ψ : R → R is a continuous and non-increasing function.By Theorem 5.1, this problem has unique solution, given by (see Example 3.3) where and X 0 solves We turn now to the Skorohod equation with boundary condition (5.7) We place ourselves in the canonical Poisson space (Ω, F, P ) (see e.g.[8], [12] or [11] for a more detailed introduction to the analysis in this space).The elements of Ω are sequences ω = (s 1 , . . ., s n ), n ≥ 1, with s j ∈ [0, 1], together with a special point a.The canonical Poisson process is defined in (Ω, F, P ) as the measure-valued process where δ s i means the Dirac measure on s i .Any square integrable random variable H in this space can be decomposed in Poisson-Itô chaos H = ∞ n=0 I n (h n ), where for almost all t ∈ [0, 1], Nualart and Vives [12] define its Skorohod integral as δ(u) := 1 0 u s δ Ñs := ∞ n=0 I n+1 (ũ n ), where ũn is the symmetrization of u n with respect to its n + 1 variables, provided u ∈ Dom δ, that means, if ∞ n=0 (n + 1)! ũn For a process u with integrable paths, define the random variable where ωj means (s 1 , . . ., s j−1 , s j+1 , . . ., s n ).One can also consider, for any random variable H and for almost all t ∈ [0, 1], the random variable The following Lemma is shown in Nualart and Vives [12].Lemma 5.3 With the notations introduced above, we have Two concepts of solution for initial value Skorohod equations were introduced in [7] by León, Ruiz de Chávez and Tudor, which they called "strong solution" and "φ-solution".In the latter, the process F (r, X r ) is only required to have square integrable paths, and its integral is interpreted as φ(F (r, X r )).If F (r, X r ) belongs to Dom δ, Lemma 5.3 (a) ensures that both concepts coincide.We only need here a version of the first notion, which we will call simply "solution".We supplement the definition in [7] with the requirement of càdlàg paths, for the boundary condition to be meaningful.Definition 5.4 A measurable process X is a solution of (5.7), if Denoting by X n (ω, x) the corresponding solutions starting at ζ ≡ x ∈ R, it is easy to show that for any These inequalities and (H ′′ 3 ) imply that there exists a unique point x * such that (5.12) x * = ψ(X 0 1 (a, x * )) , which we define as X 0 (a).Then, X 0 (a, X 0 (a)) satisfies (5.9) with ζ(a) = X 0 (a) and the boundary condition of (5.7).
In general, given ω = (s 1 , . . ., s n ), once we know X n−1 , and using equation (5.11), one shows that there exists a unique point x * such that which we define as X 0 (ω).We have then that X n (ω, X 0 (ω)) satisfies (5.11) with ζ(ω) = X 0 (ω) and the boundary condition of (5.7).Since ψ is bounded, X 0 is a bounded random variable.The process thus constructed clearly satisfies (5.8) together with the boundary condition, and the theorem is proved.
Remark 5.6 Skorohod equations can be converted to forward ones in special situations: When F (t, x) ≡ F (t) or when ψ ≡ x 0 ∈ R, the solution of (5.7) coincides with the solution of The results of Sections 3 and 4 are automatically translated to Skorohod equations in the situations of the previous remark.In other cases, the inductive construction of the solution X, in which the value of X t (ω) (with ω ∈ [0, 1] n ) depends on the values X t (ω) (with ω ∈ [0, 1] n−1 ), does not allow the equivalence.
Therefore X t belongs the first order chaos and

Proposition 2 . 5
Let f satisfy hypotheses (H ′ 1 ) and (H ′ 2 ) of Proposition 2.4, and assume that F , ∂ 1 F and ∂ 2 F are continuous functions.Assume moreover that and ψ : R → R satisfy the hypotheses of Proposition 3.4.
dr , where h n is the density function of the law of ϕ t (x) conditioned to N t = n.