Fully coupled drift-less forward and backward stochastic differential equations in a degenerate case

Abstract

Existence and uniqueness results of fully coupled forward stochastic differential equations without drifts and backward stochastic differential equations in a degenerate case are obtained for an arbitrarily large time duration.

Appendix

In this section, for a given $\sigma$, and for any $(t, x) \in [0,T] \times {\mathbb {R}}^{l}$, let $X^{t, x} = \left\{ X^{t, x} (r) \right\} _{r \in [t,T]}$ be a solution to the equation,

$$\begin{aligned} X (r) = x + \int _{t}^{r} \sigma _{*} \left( s, X (s)\right) \textrm{dW}(s), \quad t \le r \le T. \end{aligned}$$

(5)

Lemma 5.1

(Key lemma) Suppose that $\sigma$ satisfy (A.1). Let $X^{t, x} = \left\{ X^{t, x} (r) \right\} _{r \in [t,T]}$ be a solution to the equation (5). Then, for all $0 \le t \le r$ and $x, h \in {\mathbb {R}}^{l}$,

$$\begin{aligned} {\mathbb {E}} \left[ \left| X^{t, x+ h } (r) - X^{t, x} (r) \right| \right] \le \sqrt{l} \left| h\right| . \end{aligned}$$

Proof

For any fixed $(t, x) \in [0,T] \times {\mathbb {R}}^{l}$, we write for $j=1, 2, \dots , l$, $e_{(j)} = (0, \dots , 0, \overbrace{1}^{j}, 0, \dots , 0)$ and we set for any fixed $\epsilon \in (-1,1) {\setminus } \{0\}$, $\delta \ge 0$ and $j=1, 2, \dots , l$,

$$\begin{aligned} M^{\epsilon ^{-1} \delta e_{(j)} } (r) \equiv \epsilon ^{-1} \left( X^{t, x+ \delta e_{(j)} } (r) - X^{t, x} (r) \right) , \quad r \in [t,T]. \end{aligned}$$

Thus, we have

$$\begin{aligned} \begin{aligned}&M^{\epsilon ^{-1} \delta e_{(j)} } (r) = \epsilon ^{-1}\delta e_{(j)} \\ {}&+ \sum _{k=1}^{d}\int _{t}^{r} \varSigma _{k, j}\left( s, X^{t, x+ \delta e_{(j)} } (s), X^{t, x} (s) \right) M^{\epsilon ^{-1} \delta e_{(j)} } \textrm{d}W_{k} (s) \end{aligned} \end{aligned}$$

where $\varSigma _{k, j}$ is a function such that for $k=1, \dots , d$ and $j=1, \dots , l$

$$\begin{aligned} \varSigma _{k, j} (s, a, b) = \int _{0}^{1} (\nabla _x \sigma _{*})_k \left( s, (1-\theta ) a + \theta b \right) \textrm{d} \theta , \quad (s, a, b) \in [0,T] \times {\mathbb {R}}^{2l}, \end{aligned}$$

where $(\nabla _x \sigma _{*})_k$ stands for the k-component of the space derivative $\sigma _{*}$. We note that

$$\begin{aligned} \left\{ \left( X^{t, x+ \delta e_{(j)} } (r), X^{t, x} (r), M^{\epsilon ^{-1} \delta e_{(j)} } (r) \right) \right\} _{r \in [t,T]} \end{aligned}$$

is a solution to the SDE with Lipschitz continuous coefficient. In particular, when $x=\delta =0$ it has the trivial solution, $0 \in {\mathbb {R}}^{3l}$. It follows from the pathwise uniqueness that the $l$-dimensional process $M^{\epsilon ^{-1} \delta e_{(j)} }$ satisfies

$$\begin{aligned} M^{\epsilon ^{-1} \delta e_{(j)} } = (0, \dots , 0,\epsilon ^{-1} \delta {\mathcal {E}}(X, j), 0, \dots , 0), \end{aligned}$$

where we denote

$$\begin{aligned} {\mathcal {E}}(X, j) = \exp \left[ -\frac{1}{2} \int _{t}^{r} \langle \varSigma _{\cdot , j}(s), \varSigma _{\cdot , j}(s) \rangle _{{\mathbb {R}}^{d}}\textrm{d}s+ \sum _{k=1}^{d}\int _{t}^{r} \varSigma _{k, j}(s) \textrm{d}W_{k} (s)\right] . \end{aligned}$$

In short, we conclude that for any $x \in {\mathbb {R}}^{l}$ and $\delta \ge 0$, it holds

$$\begin{aligned} {\mathbb {E}}[| X^{t, x+ \delta e_{(j)} } (r) - X^{t, x} (r) |] = {\mathbb {E}}[| \delta {\mathcal {E}}(X, j) |] = \left| \delta \right| , \quad j=1,2, \dots , l. \end{aligned}$$

Moreover, for any $h=(h_{1}, h_{2}, \dots , h_{l}) \in {\mathbb {R}}^{l}$, it holds that

$$\begin{aligned} X^{x+h}(r) -X^{x}(r) = (h_{1} {\mathcal {E}}(X, 1), h_{2} {\mathcal {E}}(X, 2), \dots , h_{l} {\mathcal {E}}(X, l)). \end{aligned}$$

It leads that for any $(t, x) \in [0,T] \times {\mathbb {R}}^{l}$, we have

$$\begin{aligned} \begin{aligned}&{\mathbb {E}} \left[ \left| X^{t, x+h}(r) -X^{t,x}(r) \right| \right] = {\mathbb {E}} \left[ \left\{ \sum _{j=1}^{l}|h_{j}|^2 |{\mathcal {E}}(X, j) (r)|^2 \right\} ^{\frac{1}{2}} \right] \\ {}&\le {\mathbb {E}} \left[ \sum _{j=1}^{l} | h_j | | {\mathcal {E}}(X, j)(r)| \right] = \sum _{j=1}^{l} | h_j | {\mathbb {E}} [ |{\mathcal {E}}(X, j)(r)|] =\sum _{j=1}^{l} | h_j |. \end{aligned} \end{aligned}$$

The desired result is followed from the inequality,

$$\begin{aligned} \begin{aligned} \sum _{j=1}^{l} | h_j | \le \sqrt{l} \left( \sum _{j=1}^{l} | h_j |^2 \right) ^{1/2}. \end{aligned} \end{aligned}$$

$\square$

Remark 5.2

This estimate holds for the $L^1$ but not $L^2$. In fact, we consider an exponential martingale;

$$\begin{aligned} X(r) = x + \int _{t}^{r} L_{\sigma , x}\, X(s)\textrm{dW}(s) = x \exp \left( {L_{\sigma , x}} W(r) -\frac{1}{2}{L_{\sigma , x}^2} r \right) . \end{aligned}$$

Thus we have

$$\begin{aligned} {\mathbb {E}}[|X^{t, x+h}(r) - X^{t, x}(r)|^p] =h^p \exp \left( \frac{1}{2}\left( p^2 - 1\right) {L_{\sigma , x}^2} r \right) , \quad p>0. \end{aligned}$$

From the observation, the above inequality is interesting itself.

Remark 5.3

(Calderón Zygmund lemma) A key of the estimation is that of heat kernel, thus, which can be proven by a fundamental analytical approach known as Calderón Zygmund lemma. Moreover, the transition function of SDE, has a uniformly equicontinuous for the time and space if the coefficients satisfies a bounded condition see the book of [15, Theorem 7.2.4]. For one dimensional unbounded setting, a simple proof also given by [16].

1.1 Dominated property and decoupling FBSDEs

We note a relation between the decoupling expression and the non-linear Feynman-Kac formula. According to the result was given [10], the decoupling FBSDEs;

$$\begin{aligned} \left\{ \begin{array}{l} X(r) = X (t) + \int _{t}^{r} b \left( s, X(s) \right) \textrm{d}s+ \int _{t}^{r} \sigma \left( s, X(s) \right) \textrm{dW}(s) \\ \\ Y(r) = \varphi \left( X(T) \right) + \int _{r}^{T} f \left( s, X(s), Y(s), Z(s) \right) \textrm{d}s - \int _{r}^{T} Z (s) \textrm{dW}(s), \quad r \in [t,T]. \end{array}\right. \end{aligned}$$

gives a expression such that there exists a pair of functions, (u, v) satisfies

$$\begin{aligned} Y(s) = u(s, X(s)), \quad Z(s) = \nabla _x u (s, X(s)) \cdot \sigma (s, X(s) ), \quad s \in [0,T]. \end{aligned}$$

Therefore, we obtain

Lemma 5.4

Suppose that (A.1) holds. Then, there exists a series $(u_n, v_n)$ satisfying the equation (3) and it is uniquely determined for all $n \in {\mathbb {N}}$.

Proof

Firstly, for $(t, x) \in [0,T]\times {\mathbb {R}}^{l}$, we denote $X_{0}^{t, x} (r) \equiv x$ for all $t \le r \le T$. Thus, the (A.1) implies that $(Y_0, Z_0) \in {\mathscr {S}}^2 \times {\mathscr {H}}^2$ be a unique solution of BSDE such that

$$\begin{aligned} Y_{0} (r) = \varphi \left( X_{0} (T) \right) + \int _{r}^{T} f \left( s, X_{0}(s), Y_{0}(s), Z_{0}(s) \right) \textrm{d}s- \int _{r}^{T} Z_{0}(s) \textrm{d}s, \quad t \le r \le T. \end{aligned}$$

Moreover, we have for all $(t, x) \in [0,T] \times {\mathbb {R}}^{l}$,

$$\begin{aligned} Y_0 (r) = \varphi ( x ), \quad Z_0 (r)=0, \quad t \le r \le T. \end{aligned}$$

Denoting $u_{0}(t, x)= \varphi ( x )$ and $v_{0}(t, x)=0$ for all $(t, x) \in [0,T] \times {\mathbb {R}}^{l}$, it follows from (A.1) again that there exists a unique strong solution to the following SDE,

$$\begin{aligned} X_{1} (r) = x + \int _{t}^{r} \sigma \left( s, X_{1} (s), u_0 (s, X_{1} (s)), v_0 (s, X_{1} (s) ) \right) \textrm{dW}(s), \quad t \le r \le T. \end{aligned}$$

For $k \in {\mathbb {N}}$, suppose that there exists a series of a pair of smooth functions $(u_{k-1}, v_{k-1})$ such that

$$\begin{aligned} v_{k-1} (t, x) = \left( \nabla _x u_{k-1}\right) (t, x) \cdot \sigma (t, x, u_{k-1}(t, x), v_{k-1}(t, x)), \quad (t, x) \in [0,T]\times {\mathbb {R}}^{l}. \end{aligned}$$

It induces the following decoupling FBSDE’s and it follows from [9] such that the solution $\varTheta _{k}^{t, x}$ exists uniquely;

$$\begin{aligned} \left\{ \begin{array}{l} X_{k} (r) = x + \int _{t}^{r} \sigma \left( s, X_{k}(s), u_{k-1} (s, X^{t, x}_{k} (s) ), v_{k-1} (s, X^{t, x}_{k} (s) ) \right) \textrm{dW}(s) \\ \\ Y_{k} (r) = \varphi \left( X_{k} (T) \right) + \int _{r}^{T} f \left( s, \varTheta _{k}^{t, x}(s) \right) \textrm{d}s - \int _{r}^{T} Z_{k}(s) \textrm{dW}(s), \end{array}\right. \end{aligned}$$

for any $(t, x) \in [0,T] \times {\mathbb {R}}^{l}$. Thanks to [10], there exists a semi-linear parabolic PDE solution $u_{k}$ such that it satisfies for all $t \le r \le T$,

$$\begin{aligned} Y_{k}(r) = u_{k} (r, X_{k} (r) ), \ Z_{k}(r) = \left( \nabla _x u_{k} \right) (r, X_{k} (r) ) \cdot \sigma \left( \nabla _x u_{k} \right) (r, \varTheta _{k}^{t, x}(r) ). \end{aligned}$$

In particular, putting $v_{k}(t, x) \triangleq Z^{t, x}_{k} (t)$ we have

$$\begin{aligned} v_{k}(t, x) = \left( \nabla _x u_{k} \right) (t, x) \cdot \sigma \left( t, x, u_{k}(t, x), v_{k}(t, x) \right) , \quad (t, x) \in [0,T] \times {\mathbb {R}}^{l}. \end{aligned}$$

The smooth condition of the diffusion coefficient $\sigma (s, x, u_{k} (s, x), v_{k} (s, x))$ implies that there exists a strong unique solution in ${\mathscr {S}}^2$ such that

$$\begin{aligned} X_{k+1} (r) = x + \int _{t}^{r} \sigma \left( s, X_{k+1} (s), u_{k}(s, X_{k+1} (s) ), v_{k}(s, X_{k+1} (s) ) \right) \textrm{dW}(s). \end{aligned}$$

Therefore, we obtain the desired result. $\square$

Remark 5.5

The series of functions $\{ u_n \}$ and $\{ v_n \}$ is Lipschitz continuous for all $(t, n) \in [0,T] \times {\mathbb {N}}$:

$$\begin{aligned} \begin{aligned}&\left| u_{n}(t, x+ h) - u_{n}(t, x) \right| \le L_{u_n, x} |h|, \\ {}&\left| v_{n}(t, x+ h) - v_{n}(t, x) \right| \le L_{v_n, x} |h|, \quad (x, h) \in {\mathbb {R}}^{l} \times {\mathbb {R}}^{l}. \end{aligned} \end{aligned}$$

It stands for the construction of the scheme is easy but the problem is to select the convergence sequence if it exists.

The following dominated property is convenient.

Lemma 5.6

(${\overline{Y}}$ dominated by ${\overline{X}}$) Suppose that $(X_i, Y_i, Z_i) \in {\mathscr {S}}^2 \times {\mathscr {S}}^2 \times {\mathscr {H}}^2$ satisfies

$$\begin{aligned} \begin{aligned} Y_i (r)=\varphi (X_i (T))&+ \int _{r}^{T} f \left( s, X_i (s), Y_i (s), Z_i (s) \right) \textrm{d}s \\ {}&- \int _{r}^{T} Z_i (s) \textrm{dW}(s), \quad r \in [t,T], \ i=1,2. \end{aligned} \end{aligned}$$

Denote and write

$$\begin{aligned} ({\overline{X}}, {\overline{Y}}, {\overline{Z}}) \triangleq (X_1 - X_2, Y_1 - Y_2, Z_1 - Z_2 ). \end{aligned}$$

For any $\epsilon \in (0,1)$ we denote $L = (1+ 2 L_{f, y} +\epsilon ^{-1} ) + (1-\epsilon )$. Then, it holds that

$$\begin{aligned} \begin{aligned}&{\mathbb {E}} \left[ \left| {\overline{Y}} (t) \right| ^2 e^{-Lt} \right] +(1-\epsilon )\int _t^T {\mathbb {E}} \left[ \left( \left| {\overline{Y}} (s) \right| ^2+ \left| {\overline{Z}} (s) \right| ^2 \right) e^{-Ls}\right] \textrm{d}s \\ {}&\le {\mathbb {E}} \left[ \left| \delta _{X} \varphi (T)\right| ^2 e^{-LT}\right] + \int _{t}^{T} {\mathbb {E}} \left[ \left| \delta _{X} f (s) \right| ^2 e^{-Ls} \right] \textrm{d}s. \end{aligned} \end{aligned}$$

where we define $\delta _{X} f (s) = f (s, X_1(s), Y_1 (s), Z_1 (s) ) - f (s, X_2 (s), Y_1 (s), Z_1 (s) )$ and $\delta _{X} \varphi (T)\triangleq \varphi \left( X_1 (T) \right) -\varphi \left( X_2 (T) \right)$.

Proof

Applying the Itô formula to $\left| {\overline{Y}} (t) \right| ^2 e^{- L t }$ for some $L>0$, we have

$$\begin{aligned} \begin{aligned}&\left| {\overline{Y}} (t) \right| ^2 e^{-Lt} + L \int _t^T \left| {\overline{Y}} (s) \right| ^2 e^{-Ls} \textrm{d}s + \int _t^T \left| {\overline{Z}} (s) \right| ^2 e^{-Ls} \textrm{d}s \\ {}&= \left| \delta _{X} \varphi (T) \right| ^2 e^{-LT} + 2 \int _{t}^{T} \langle {\overline{Y}}(s), {\overline{f}} (s) \rangle e^{-Ls} \textrm{d}s - 2 \int _{t}^{T} e^{-Ls} \langle {\overline{Y}}(s), {\overline{Z}}(s) \textrm{d}W(s) \rangle . \end{aligned} \end{aligned}$$

where we define ${\overline{f}} (s) = f (s, X_1(s), Y_1(s), Z_1 (s) ) - f (s, X_2 (s), Y_2 (s), Z_2 (s) )$. For arbitrary $\epsilon >0$, we have $2 ab \le {\epsilon a}^2 + {\epsilon ^{-1} b^2}$ for all $a, b \ge 0$ holds, we have

$$\begin{aligned} \begin{aligned}&2\left| \langle {\overline{Y}}(s), {\overline{f}} (s) \rangle \right| \\ {}&\le 2 \left| {\overline{Y}}(s) \right| \left( |\delta _X f (s)| + L_{f, y}|{\overline{Y}}(s)| + L_{f, z} |{\overline{Z}}(s)| \right) \\ {}&\le |\delta _{X} f (s)|^2 + (1+ 2L_{f, y} +\epsilon ^{-1} )|{\overline{Y}}(s)| ^2 + \epsilon L_{f, z}^2 |{\overline{Z}}(s)| ^2. \end{aligned} \end{aligned}$$

Since ${\overline{Y}}, {\overline{Z}}$ is squared integrable, $\left\{ \int _{r}^{T} \langle {\overline{Y}}(s), {\overline{Z}}(s) \textrm{d}W(s) \rangle \right\} _{r \in [t,T] }$ is martingale vanishing at T. Therefore, we obtain

$$\begin{aligned} \begin{aligned}&{\mathbb {E}} \left[ \left| {\overline{Y}} (t) \right| ^2 e^{-Lt}\right] + L \int _t^T {\mathbb {E}} \left[ \left| {\overline{Y}} (s) \right| ^2 e^{-Ls}\right] \textrm{d}s + \int _t^T {\mathbb {E}} \left[ \left| {\overline{Z}} (s) \right| ^2 e^{-Ls}\right] \textrm{d}s \\ {}&\le {\mathbb {E}} \left[ \left| \delta _{X} \varphi \right| ^2 e^{-LT}\right] \\ {}&+ \int _{t}^{T} {\mathbb {E}} \left[ \left| \delta _{X} f (s) \right| ^2 + (1+ 2 L_{f, y} +\epsilon ^{-1} ) L_{f, y}^2 \left| {\overline{Y}} (s) \right| ^2 + \epsilon L_{f, z}^2 |{\overline{Z}}(s)| ^2 \right] e^{-Ls} \textrm{d}s. \end{aligned} \end{aligned}$$

where we write $\delta _{X} \varphi (T)\triangleq \varphi \left( X_1 (T) \right) -\varphi \left( X_2 (T) \right)$. Now, putting $L = (1+ 2 L_{f, y} +\epsilon ^{-1} ) + (1-\epsilon )$, we obtain

$$\begin{aligned} \begin{aligned}&{\mathbb {E}} \left[ \left| {\overline{Y}} (t) \right| ^2 e^{-Lt} \right] +(1-\epsilon )\int _t^T {\mathbb {E}} \left[ \left( \left| {\overline{Y}} (s) \right| ^2+ \left| {\overline{Z}} (s) \right| ^2 \right) e^{-Ls}\right] \textrm{d}s \\ {}&\le {\mathbb {E}} \left[ \left| \delta _{X} \varphi (T)\right| ^2 e^{-LT}\right] + \int _{t}^{T} {\mathbb {E}} \left[ \left| \delta _{X} f (s) \right| ^2 e^{-Ls} \right] \textrm{d}s. \end{aligned} \end{aligned}$$

$\square$