Arrow Sufficient Conditions for Optimality of Fully Coupled Forward–Backward Stochastic Differential Equations with Applications to Finance

Abstract

This paper is concerned with optimal control problems of fully coupled forward–backward stochastic differential equations on finite horizon and infinite horizon with partial information. Two sufficient conditions for optimality are established for the above problems. We demonstrate their applications by four illustrative examples in the framework of cash management, risk minimizing, and linear-quadratic optimal control problems. These examples are explicitly solved based on the sufficient conditions and the optimal filtering of forward–backward stochastic differential equations derived in this paper.

Appendix

Proof of Theorem 2.1 For any $v(\cdot )\in \mathcal U_{ad}$, we consider

$$\begin{aligned} J(v(\cdot ))-J(u(\cdot ))=I_1+I_2+I_3 \end{aligned}$$

(23)

with

$$\begin{aligned} I_1:=&\ \phi (y^v(0))-\phi (y(0)), \quad I_2:=\text {l}\!\text {E}[\psi (x^v(T))-\psi (x(T))],\\ I_3:=&\ \text {l}\!\text {E}\int _0^T\left( l(t, v(t))-l(t)\right) \mathrm{{d}}t. \end{aligned}$$

Recall that $\phi$ is convex in $y$ and $p(0)=-\phi _y(y(0))$. Applying Itô’s formula to $\langle p(\cdot ), y^v(\cdot )-y(\cdot )\rangle$, we get

$$\begin{aligned} I_1&\ge -p^\top (0)(y^v(0)-y(0))\nonumber \\&= -\text {l}\!\text {E}[p^\top (T)A(x^v(T)-x(T))]-\text {l}\!\text {E}\int _0^T\langle H_y(t), y^v(t)-y(t)\rangle \mathrm{{d}}t\nonumber \\&-\,\text {l}\!\text {E}\int _0^T\langle H_z(t), z^v(t)-z(t)\rangle \mathrm{{d}}t-\text {l}\!\text {E}\int _0^T\langle p(t), f(t, v(t))-f(t)\rangle \mathrm{{d}}t. \end{aligned}$$

(24)

Similarly,

$$\begin{aligned} I_2\ge \text {l}\!\text {E}[(q^\top (T)+p^\top (T)A)(x^v(T)-x(T))] \end{aligned}$$

(25)

with

$$\begin{aligned}&\text {l}\!\text {E}[q^\top (T)(x^v(T)-x(T))]\\&= \text {l}\!\text {E}\int _0^T\langle q(t), b(t, v(t))-b(t)\rangle \mathrm{{d}}t+\text {l}\!\text {E}\int _0^T\langle k(t), \sigma (t, v(t))-\sigma (t)\rangle \mathrm{{d}}t\\&-\,\text {l}\!\text {E}\int _0^T\langle H_x(t), x^v(t)-x(t)\rangle \mathrm{{d}}t. \end{aligned}$$

From (23)–(25) and $I_3$, it is easy to see that

$$\begin{aligned} J(v(\cdot ))-J(u(\cdot ))&\ge \text {l}\!\text {E}\int _0^T \left( H(t, v(t))-H(t)\right) \mathrm{{d}}t-\text {l}\!\text {E}\int _0^T\langle H_x(t), x^v(t)-x(t)\rangle \mathrm{{d}}t\\&-\,\text {l}\!\text {E}\int _0^T\langle H_y(t), y^v(t)-y(t)\rangle \mathrm{{d}}t-\text {l}\!\text {E}\int _0^T\langle H_z(t), z^v(t)-z(t)\rangle \mathrm{{d}}t. \end{aligned}$$

Now we affirm that $J(v(\cdot ))-J(u(\cdot ))\ge 0$ holds for any $v(\cdot )\in \mathcal U_{ad}$.

The detailed proof of the affirmance is as follows. According to the minimum assumption, for any $(t, x, y, z)$,

$$\begin{aligned} \tilde{H}(t, x, y, z)-\tilde{H}(t, x(t), y(t), z(t))\le H(t, x, y, z, v, p(t), q(t), k(t))-H(t).\qquad \end{aligned}$$

(26)

Hence, it suffices to show that

$$\begin{aligned}&\tilde{H}(t, x, y, z)-\tilde{H}(t, x(t), y(t), z(t))\nonumber \\&\ge \langle H_x(t), x-x(t)\rangle +\langle H_y(t), y-y(t)\rangle +\langle H_z(t), z-z(t)\rangle . \end{aligned}$$

(27)

Fix $t\in [0, T]$. Since $\tilde{H}(t, x, y, z)$ is convex in $(x, y, z)$, there exist $a(t)$, $b(t)\in \text {l}\!\text {R}^n$ and $c(t)\in \text {l}\!\text {R}^{n\times m}$ such that

$$\begin{aligned} \tilde{H}(t, x, y, z)-\tilde{H}(t, x(t), y(t), z(t))&\ge \langle a(t), x-x(t)\rangle +\langle b(t), y-y(t)\rangle \nonumber \\&+\,\langle c(t), z-z(t)\rangle . \end{aligned}$$

(28)

Define

$$\begin{aligned} \varphi (t, x, y, z)&:= H(t, x, y, z, u(t), p(t), q(t), k(t))-H(t)\\&-\,\langle a(t), x-x(t)\rangle +\langle b(t), y-y(t)\rangle +\langle c(t), z-z(t)\rangle . \end{aligned}$$

Then, it follows from (26) and (28) that $\varphi (t, x, y, z)\ge 0$ for all $(t, x, y, z)$. Moreover, $\varphi (t, x(t), y(t), z(t))=0$. It implies that $(x(\cdot ), y(\cdot ), z(\cdot ))$ is a minimum point of $\varphi$. Because $\varphi$ is differentiable with respect to $(x, y, z)$, we have the partial derivatives

$$\begin{aligned} \varphi _x(t, x(t), y(t), z(t))=0, \quad \varphi _y(t, x(t), y(t), z(t))=0, \quad \varphi _z(t, x(t), y(t), z(t))=0, \end{aligned}$$

i.e., $H_x(t)=a(t), H_y(t)=b(t), H_z(t)=c(t).$ Inserting them into (28), we derive (27), i.e., $u(\cdot )$ is an optimal control. Then, the proof is complete. $\Box$

Proof of Theorem 3.1 For any $v(\cdot )\in \mathcal V_{ad}[0, \infty [$, we write

$$\begin{aligned} J(v(\cdot ))-J(u(\cdot ))=I_1+I_2 \end{aligned}$$

(29)

with

$$\begin{aligned} I_1:=\phi (y^v(0))-\phi (y(0)),\quad I_2:=\text {l}\!\text {E}\int _0^\infty \left( l(t, v(t))-l(t)\right) \mathrm{{d}}t. \end{aligned}$$

Noting the convexity of $\phi$ and applying Itô’s formula to $\langle p(\cdot ), y^v(\cdot )-y(\cdot )\rangle$, we have

$$\begin{aligned} I_1&\ge -p^\top (0)(y^v(0)-y(0))\nonumber \\&= -\text {l}\!\text {E}\int _0^\infty \langle H_y(t), y^v(t)-y(t)\rangle \mathrm{{d}}t-\text {l}\!\text {E}\int _0^\infty \langle H_z(t), z^v(t)-z(t)\rangle \mathrm{{d}}t\nonumber \\&-\,\text {l}\!\text {E}\int _0^\infty \langle p(t), f(t, v(t))-f(t))\rangle \mathrm{{d}}t. \end{aligned}$$

(30)

Similarly, it follows from the transversality condition that

$$\begin{aligned} 0&\ge \liminf _{T\uparrow \infty }\text {l}\!\text {E}[q^\top (T)(x^v(T)-x(T))]\nonumber \\&= \text {l}\!\text {E}\int _0^\infty \langle q(t), b(t, v(t))-b(t)\rangle \mathrm{{d}}t+\text {l}\!\text {E}\int _0^\infty \langle k(t), \sigma (t, v(t))-\sigma (t)\rangle \mathrm{{d}}t\nonumber \\&-\,\text {l}\!\text {E}\int _0^\infty \langle H_x(t), x^v(t)-x(t)\rangle \mathrm{{d}}t. \end{aligned}$$

(31)

Similar to the proof of Theorem 2.1, it is easy to see from (29), (30), (31), and $I_2$ that

$$\begin{aligned} J(v(\cdot ))-J(u(\cdot ))&\ge \text {l}\!\text {E}\int _0^\infty \left( H(t, v(t))-H(t)\right) \mathrm{{d}}t-\text {l}\!\text {E}\int _0^\infty \langle H_x(t), x^v(t)-x(t)\rangle \mathrm{{d}}t\\&-\,\text {l}\!\text {E}\int _0^\infty \langle H_y(t), y^v(t)\!-\!y(t)\rangle \mathrm{{d}}t-\!\text {l}\!\text {E}\int _0^\infty \langle H_z(t), z^v(t)\!-\!z(t)\rangle \mathrm{{d}}t\\&\ge \text {l}\!\text {E}\int _0^\infty \langle H_v(t), v(t)-u(t)\rangle \mathrm{{d}}t. \end{aligned}$$

Furthermore, using the minimum condition in Theorem 3.1, we get

$$\begin{aligned} 0&\le \ \Big (\frac{\partial }{\partial v}\text {l}\!\text {E}[H\Big (t, x(t), y(t), z(t), u(t), p(t), q(t), k(t)\Big )|\mathcal G_t]\Big )^\top (v-u(t))\\&= \ \text {l}\!\text {E}[H_v(t)^\top (v-u(t))|\mathcal G_t]. \end{aligned}$$

This implies that $J(u(\cdot ))=\min _{v(\cdot )\in \mathcal V_{ad}[0, \infty [}J(v(\cdot ))$. Then, the proof is complete. $\Box$