Erratum to: Fully coupled forward-backward stochastic differential equations on Markov chains

*Correspondence: xinlingxiao@mail.sdu.edu.cn 2School of Mathematical Sciences, Shandong Normal University, Jinan, 250014, P.R. China Full list of author information is available at the end of the article After publication of this work [], it was noted that some information was missed within the acknowledgements section. This work was also supported by the National Natural Science Foundation of China (No. ), and the Programme of Introducing Talents of Discipline to Universities of China (No. B). The correct version of the acknowledgements section can be found below:


Introduction
Since the first introduction by Pardoux and Peng [14] in 1990, the theory of nonlinear backward stochastic differential Equations(BSDEs, for short) driven by a Brownian motion has been intensively researched by many researchers and has been achieved abundant theoretical results. Now the theory of BSDEs has risen to be a powerful part of the stochastic analyst's tool. It also has many important applications, namely in stochastic control, stochastic differential game, finance and the theory of partial differential equations(PDEs, for short).
In the classic BSDEs theory, we consider the Brownian motion as the driver, but Brownian motion is a kind of very idealized stochastic model which restricts greatly the applications of the classic BSDEs. There are many theories of BSDEs with jump processes recently. Tang and Li [17] first discussed BSDEs driven by Brownian motion and Poisson Process; Nualart, Schoutens [12] gave BSDEs driven by Brownian motion and Levy Process. Furthermore there are results on BSDEs that other process instead of the Brown motion in diffusion term, such as Ouknine [13] researched BSDE driven by Poisson random measure. In 2008, Cohen [3] studied BSDE driven by a continuous time, finite state Markov Chain. After then, many results such as comparison theorem about this kind of BSDE, nonlinear expected results( [4], [5]) and so on were achieved.
Along with the rapid development of the BSDE theory, the theory of fully coupled forward backward stochastic differential equations(FBSDEs, for short) which is closely related to BSDEs has been developing very rapidly. Fully coupled FBS-DEs with Brownian motion can be encountered in the optimization problem when applying stochastic maximum principle and mathematical finance considering large investor in security market. As we know now, to get results on the existence and the uniqueness of fully coupled FBSDE's solutions, there are mainly three methods: method of contraction Mapping( [1], [15]), four-step scheme( [11], [10]), method of continuation( [7], [16]). For more contents on fully coupled FBSDEs, the reader is refereed to Yong [21], Wu [20] or Ma, Wu, Zhang and Zhang [9] and the references therein. Wu ([18], [19])also studied FBSDEs driven by Brownian motion and Poisson Process. Li [8] studied FBSDEs driven by Brownian motion and Levy Process.
In this paper, we study fully coupled FBSDEs driven by a Martingale which is generated by a continuous time, finite state Markov Chain when SDE and BSDE have different dimensions. Inspired by Peng and Wu [16], we also introduced a m × n full rank matrix G to overcome the problem caused by the different dimensions of SDE and BSDE. Using the method of continuation, the Itô product rule of Semimartingales and the fixed point principle, with the help of the theory of BSDE driven by a continuous time, finite state Markov Chain, We obtain the existence and uniqueness results of the FBSDEs on Markov Chains. Because of the property of the martingale generated by the finite state Markov Chain is different from the property of the Brownian motion, so the form of the monotone assumptions we got here is different from Peng and Wu [16].
In Section 2, we consider formulation of the problem; in Section 3, we give preliminary concerns; in Section 4 we study the existence and uniqueness results of the fully coupled FBSDEs on Markov chains under some monotone assumptions; in Section 5 we give proofs of Lemmas in Section 4; in Section 6 we discuss the existence and uniqueness results of the fully coupled FBSDEs on Markov chains under another set of monotone assumptions; in Section 7 we give proofs of Lemmas in Section 6. We consider stochastic processes defined on the filtered probability space (Ω, F , F t , P), where {F t } is the completed natural filtration generated by the σ-fields F t = σ({m s , s ≤ t}, F ∈ F T : P(F ) = 0), and F = F T . Observe that m is rightcontinuous, this filtration is right-continuous. If A t denotes the rate matrix for m at time t, then this chain has the following representation:

Formulation of the problem
where M t is a martingale (see [6]). We consider the following forward-backward stochastic differential equations: where X, Y, Z take values in R n , R m , R m×d , T > 0 is an arbitrarily fixed number and b, σ, f, Φ are functions with following appropriate dimensions: We seek a F t -adapted triple (X, Y, Z) such that it satisfies the above forwardbackward stochastic differential equations, on[0, T ] , P-almost surely. That is, our aim is to find the F t -adapted solution of the the above forward-backward stochastic differential equations.
Note that, as it is only the s − lef t limit σ(s−, X s− , Y s− , Z s− ) which enters into Eqs(1), there is no loss of generality to assume that σ(s, X s , Y s , Z s ) is left continuous in s for each w and X, Y. Note also that as M is a semimartingale, X, Y is càdlàg and adapted. We suppose the existence of the left limits of Z. So Z must have at most a countable number of discontinuities, and then it must be left-continuous at each t except possibly on a dt-null set. Hence, if Z s satisfies Eqs(1), then so does Z s− , Writing Z * * t := Z t− we have a left-continuous process Z * * which also satisfy the desired equations, and therefore the writing of the left limits Z t− is unnecessary as we simply assume our solution is left-continuous.
Given there arguments, we rewrite Eqs(1) as As a side note, observe that Eqs(1) is equivalent to Obviously, Eqs(2) is driven by Markov chain m = {m t , t ∈ [0, T ]}.

preliminary concerns (1) preliminary notations
We know that the optional quadratic variation of M t is given by the matrix process Observing that A is the rate matrix of the Markov chain m, the predictable quadratic variation is Then it can be seen through considering Equating these two we can get where L is some martingale. This in turn suggests that We will define the following notations: (, )denotes the usual inner product in R n or R m ; we use the usual Euclidean norm in R n and R m ; and for z ∈ R m×d , we define |z| = {tr(zz * )} For z 1 ∈ R m×d , z 2 ∈ R m×d , We are given an m × n full-rank matrix G.
Furthermore we define

(2) BSDEs on Markov chains
We first discuss the existence and uniqueness of a solution to the following BSDEs on Markov chains. We consider the equations of the form These functions are assumed to be progressively measurable, i.e. F 1 (·, s, Y s , Z s ) and From Cohen [3], we have the following result about martingale representation: can be represented as a stochastic ( in this case Stieltjes) integral with respect to the martingale process M, up to equality P−a.s.. This representation is unique up to a d M, M t ×P−null set. That is where Z s is a predictable R n×d valued matrix process.
Cohen [3] gave the existence and uniqueness of a solution to the above BSDEs.
Theorem 1 Assume Lipschitz continuity on the generators F 2 and F 2 . we shall require there to exist c ∈ R such that for all s ∈ [0, T ] Under the above Lipschitz condition, the above equation has at most one solution up to indistinguishability for Y and equality a.s.d M, M t × P for Y .

Existence and uniqueness for above forwardbackward equations
We give two definitions as following: Definition 2 A triple of processes (X, Y, Z) : Ω×[0, T ] → R n ×R m ×R m×d is called an adapted solution of the Eqs (1) and (2) ,and it satisfies Eqs (1) and (2) P-almost surely.
The solution's adaptedness allows us to put Eqs(1) in a differential form below: Now we give the critical assumptions of our paper: (1)F (t, u) is unif ormly lipschitz with respect to u; (2)H(t, u) is unif ormly lipschitz with respect to u; (3)Φ(x) is unif ormly lipschitz with respect to x.
Where c 2 , c ′ 2 and c 3 are given positive constants.
In this section, we will give the main result of our paper.
Theorem 2 Let Assumption 1 and Assumption 2 hold, then there exists a unique adapted solution (X, Y, Z) for Eqs(1).

From Assumption 1, it follows
Using Stieltjes chain rule for products and hence,taking expectation and evaluating at t = T, by Assumption 2, we can get we get then As c 2 , c ′ 2 and c 3 are given positive constants, then We now consider the following family of FBSDEs parametrized by l ∈ [0, 1] : For proving the existence part of the theorem, we first need the following two lemmas.

Lemma 2
The following equation has a unique solution: where λ is a nonnegative constant.

Now we can give
Proof of Existence From Lemma 2 we see immediately that, when λ = 1 in Eqs(5), eqs(3) for l = 0(that is Eqs(4)) has a unique solution. It then follows from Lemma 3 that there exists a positive constant δ 0 depending on Lipschitz constants, c 2 , c ′ 2 , c 3 and T such that, for each δ ∈ [0, δ 0 ], Eqs(3) for l = l 0 + δ has a unique solution. We can repeat this process for N − times with 1 ≤ Nδ 0 < 1 + δ 0 . It then follows that, in particular, Eqs(3) for l = 1 with ξ ≡ 0 has a unique solution. The proof is complete.

Proof of Lemma 2 and Lemma 3
Proof of Lemma 2: We observe that the matrix G is of full rank. The proof of the existence for equation (5) will be divided into two cases: n ≤ m and n > m.
For the first case, the matrix G * G is of full rank. We set Multiplying G * on both sides of the BSDE for (Y, Z) yields Similarly, multiplying (I m − G(G * G) −1 G * ) on both sides of the same equation Obviously the pair (Y ′′ , Z ′′ ) is uniquely determined (by [cohen]). The uniqueness of (X ′ , Y ′ , Z ′ ) follows from Theorem 2. In order to solve Eqs(6), we introduce the following n × n-symmetric matrix-valued ODE, known as the matrix-Riccati equation: It is well known that this equation has a unique nonnegative solution K(·) ∈ C 1 ([0, T ); S n ). Where S n stands for the space of all n × n-symmetric matrices. We then consider the solution (p, q) ∈ M 2 (0, T ; R n+n×d) of the following linear simple BSDE: We now let X ′ t be the solution of the SDE Then it is easy to check that ( is the solution of equation (6). Once (X ′ , Y ′ , Z ′ ) and (Y ′′ , Z ′′ ) are resolved, then the triple (X, Y, Z) is uniquely obtained by For the second case, the matrix GG * is of full rank. We set ′′ is the unique solution of the following linear SDE: The triple (X ′ , Y ′ , Z ′ ) solves the FBSDE: To solve this equation, we introduce the following m×m-symmetric matrix-valued matrix-Raccati equation: It is well known that this equation has a unique nonnegative solution K(·) ∈ C 1 ([0, T ); S m ). Where S m stands for the space of all m × m-symmetric matrices. We then consider the solution (p, q) ∈ M 2 (0, T ; R m+m×d) of the following linear simple BSDE: We now let X ′ t be the solution of the SDE: Then it is easy to check that( The proof is complete. Proof of Lemma 3: Since for each φ ∈ M 2 (0, T ; R n ), γ ∈ M 2 (0, T ; R m ), ψ ∈ M 2 (0, T ; R n×d ), ξ ∈ L 2 (Ω, F T , P), x ∈ R n , l 0 ∈ [0, 1), there exists a unique solution of (3), therefore, for each U s = (X s , Y s , Z s ) ∈ M 2 (0, T ; R n+m+m×d ) there exists a unique triple u s = (x s , y s , z s ) ∈ M 2 (0, T ; R n+m+m×d ) satisfying the following FB-SDE: We want to prove that the mapping defined by Using Stieltjes chain rule for product (Gx s ,ŷ s ), and hence, taking expectations and evaluating at t = T yields From Assumption 1 and Assumption 2, c 2 > 0, c 3 > 0, we obtain For the difference of the solutions (ŷ,ẑ) = (y − y ′ , z − z ′ ), we apply the usual technique to the BSDE, we can get Similarly, for the difference of the solutionx = x−x ′ , we apply the usual technique to the forward part and get Here the constant c 4 depends on the Lipschitz constants as well as G, c 2 , c ′ 2 and T.
Combing the above estimates, it is clear that, we always have we can get where the constant c 5 depending on c 4 , G, c 2 and c 3 . If we choose δ 0 = 1 2c 5 , then it is clear that, for each fixed δ ∈ [0, δ 0 ], the mapping I α+δ is contract in the sense that It indicates that this mapping has a unique fixed point (u l 0 +δ ) = (x l 0 +δ , y l 0 +δ , z l 0 +δ ), which is the solution of equation (3) for l = l 0 + δ. The proof is complete.

Another existence and uniqueness theorem
Next, we give the other existence and uniqueness theorem. First, we give another assumption.
Assumption 3 There exists a constant c 2 > 0, c Where c 2 , c ′ 2 and c 3 are given positive constants.
Theorem 3 Let Assumption 1 and Assumption 3 hold, then there exists a unique adapted solution (X, Y, Z) for Eqs (1).
Proof of Uniqueness: Using the same procedure as the proof of uniqueness of Theorem 2 and by Assumption 3 we can get We get then As c 2 , c 7 Proof of Lemma 4 and Lemma 5 Proof of Lemma 4: We observe that the matrix G is of full rank. The proof of the existence for equation (10) will be divided into two cases: n ≤ m and n > m.
For the first case, the matrix G * G is strictly positive. We set Multiplying G * on both sides of the BSDE for (Y, Z) yields Similarly, multiplying (I m − G(G * G) −1 G * ) on both sides of the same equation Obviously the pair (Y ′′ , Z ′′ ) is uniquely determined. The uniqueness of (X ′ , Y ′ , Z ′ ) follows from Theorem 7.1. In order to solve Eqs(11), we introduce the following n×nsymmetric matrix-valued ODE, known as the matrix-Raccati equation: It is well known that this equation has a unique nonnegative solution K(·) ∈ C 1 ([0, T ); S n ). Where S n stands for the space of all n × n-symmetric matrices. We then consider the solution (p, q) ∈ M 2 (0, T ; R n+n×d) of the following linear simple BSDE: We now let X It indicates that this mapping has a unique fixed point (u l 0 +δ ) = (x l 0 +δ , y l 0 +δ , z l 0 +δ ), which is the solution of equation (8) for l = l 0 + δ.
The proof is complete.