DISCRETIZING A BACKWARD STOCHASTIC DIFFERENTIAL EQUATION

where (Yt,Zt) are unknown predictable processes. We will assume that f is a Lipschitz function with respect to its arguments throughout this paper. Since this equation has its important applications into control theory and mathematical finance, many mathematicians are not satisfied merely by descriptive existence theorems. They are also interested in constructing the numerical solutions. In order to make real construction, Antonelli [1] solved in short time the coupled forward-backward stochastic differential equations, in which it is assumed that ξ = g(VT ) where {Vt}t is the solution to a forward stochastic differential equation


Introduction.
Given a probability space (Ω,F,P).Let W t be a standard Brownian motion with (F t ) ⊂ F as its natural filtration.Given any positive constant T < ∞ and a random variable ξ ∈ F T .A backward stochastic differential equation is the equation [9] where (Y t ,Z t ) are unknown predictable processes.We will assume that f is a Lipschitz function with respect to its arguments throughout this paper.Since this equation has its important applications into control theory and mathematical finance, many mathematicians are not satisfied merely by descriptive existence theorems.They are also interested in constructing the numerical solutions.In order to make real construction, Antonelli [1] solved in short time the coupled forward-backward stochastic differential equations, in which it is assumed that ξ = g(V T ) where {V t } t is the solution to a forward stochastic differential equation Later, Hu and Peng [4] and Yong [11] proved the long time existence of the coupled forward-backward stochastic differential equations.Moreover, a four step scheme was suggested by Ma et al. in [7] to solve (1.1) and (1.2) jointly.However, their scheme is related to solving a high dimensional semi-linear partial differential system, which is nontrivial and numerically difficult as noticed by Zhang [12].Bally [2] has a random time discretization scheme, which requires to approximate integrals of dimension as high as the partition size.Chevance [3] has a scheme under higher regularity condition (C 4 ).Ma et al. have a quite general numerical method which converge weakly to the true solution [6].
Zhang [12] studied a numerical scheme to solve a coupled forward-backward stochastic differential equations, which converge strongly to the real solution.His approach is still quite different than ours.The main difference is that Zhang's finite difference equation is a little less natural than ours (3.4), but certainly his condition is more general.
The purpose of this paper is to develop the idea appeared in [7] and to give a simpler numerical scheme to solve (1.1) completely.Our method also gives a numerical probability scheme to solve a semi-linear partial differential equation related to a backward stochastic differential equation [10] where we assume that h is Lipschitz with respect to its arguments, φ ∈ H 2+α and φ(x) → 0 (|x| → ∞).It is well known [5, page 306] that (1.3) has a unique solution u ∈ H 2+α and it is easy to see (e.g., using Feyman-Kac formula) that u(x, t) → 0 (|x| → ∞).
For simplifying our notation, we only consider the one-dimensional case.However, it is easy to see that our argument is still valid for multidimensional case as well.

Discretization.
In this section, we consider the problem of discretization.From the above discussion, it is sufficient to discuss the case where ξ = g(W T ).The more general case that ξ = g(W t 1 ,...,W T ) follows from the fact that we can patch the pieces corresponding to different intervals (t i ,t i+1 ) together as in Section 2. Since there is no closed form solution to (1.3), our first goal is to solve it numerically by a discrete probabilistic approach.
Denote by P t the standard semigroup of Gaussian operators, that is, Given an integer N > 0. We first consider the following backward equation on discrete time {kT /N} k=0,1,...,N .Set X (N) T

= g(W T ) and define X (N)
kT /N inductively by the following equation: where we used the fact that X (N) (k+1)T /N is a function and still denoted as defined as the result of P t operating on X (N) (k+1)T /N .The same type of notation will be carried out through this paper.Since f is Lipschitz, by implicit function theorem, when N is large enough, (3.2) has a unique solution Next, define Finally, we consider the forward equation 0 .Consider several useful facts.We may also get X (N) through the following equation: with the initial condition X (N) T = g(W T ).In fact, since (∂/∂s)P T −s = −(1/2)∆P T −s , we easily have, from Ito's formula, that ) (for all s ≥ (N − 1)T /N) and take conditional expectation with respect to W (N−1)T /N on both sides of (3.5), where we used the exchangeability between P t−s and ∂/∂x.Then we get the solution to (3.5) for k = N − 1. Repeating this procedure, we get the solution to (3.5) for all k by mathematical induction.It is also easy to check that We will show in Section 5 the rate of convergence of X (N) towards Y and in Section 6 the rate of convergence of Q (N) towards Z.Moreover, we will show in Section 6 that Z s is Hölder continuous.Thus we may compare two forward stochastic differential equations, (3.4) and (3.5), and easily see that

Stability of difference equation.
Let X iT /N and XiT/N be two solutions to (3.5), that is, with the initial condition that X T = g(W T ) and XT = g(W T ).Then we have the following stability result, which is the discretized version of Pardoux-Peng's remarkable result [9].
Proof.We have 3) The two parts on the right-hand side are orthogonal, and the expectation of the last term vanishes, we get By (4.4) and (4.6), However, (4.9) For a > 0, b > 0, and λ > 0 we have ab where (4.13) Since (1 − κ/N) −N → e −κ , we get the conclusion. (N) to Y .We are going to estimate the error between X (N) and Y in this section.

Theorem 5.1. There is a constant
Proof.We introduce for each n ≤ N a discrete time process {X (n,N) iT /N } i=0,1,...,N as follows: define X (n,N) iT /N = Y iT /N (for all i ≥ n) and for each i ≤ n − 1 we define the process repeatedly by (5. 2) It is easy to see that that is, (5.4) Since the two terms on the left-hand side of (5.4) are orthogonal, we get where we used Lipschitz condition on f .We use C to denote an arbitrary constant, whose value may change according to the context but not depending on the given variables.
We estimate the last two terms separately (5.6) In the deduction (5.2), (5.3), (5.4), (5.5), and (5.6), we used repeatedly the fact that u ∈ H 2+α so that u and its derivatives up to the second order are bounded.
For the last term of (5.5), we have (5.7) Since u xx is bounded, it is easy to see from the property of Gaussian kernel that |P T /N u x (T − (n + 1)T /N, x) − u x (T − (n + 1)T /N, x)| 2 converge to 0 uniformly in the order of 1/N when N → ∞.Therefore where we used the fact that u x is Hölder continuous in t with exponential greater than 1/2 and Lipschitz in x (see [5, page 46] for the definition of H 2+α ).Finally, by (5.5), (5.6), and (5.8), we get (5.9) Applying Theorem 4.1, we deduce that It is easy to check that the last term has a bound which is independent on n.Summing up the above inequalities over n, we deduce that 6. Convergence of Q (N) to Z.We can also prove the convergence of Q (N) to Z as follows: according to the discussion in Section 2, we may assume that Z t = u x (T − t, W t ), where u ∈ H 2+α , for simplicity.Since u x is Hölder continuous with exponent greater than 1/2 in t and with bounded derivative in x, it is easy to see that Z t is Hölder continuous with exponent greater than 1/2 in L 2 (Ω).
For any pair 0 where we used the fact that s |W iT /N ] when s is greater than iT /N and the Hölder L 2 -continuity of Z s .