Dynamic Robust Duality in Utility Maximization

Abstract

A celebrated financial application of convex duality theory gives an explicit relation between the following two quantities:

• (i) The optimal terminal wealth $X^*(T) : = X_{\varphi ^*}(T)$ of the problem to maximize the expected U-utility of the terminal wealth $X_{\varphi }(T)$ generated by admissible portfolios $\varphi (t); 0 \le t \le T$ in a market with the risky asset price process modeled as a semimartingale;

• (ii) The optimal scenario $\frac{dQ^*}{dP}$ of the dual problem to minimize the expected V-value of $\frac{dQ}{dP}$ over a family of equivalent local martingale measures Q, where V is the convex conjugate function of the concave function U.

In this paper we consider markets modeled by Itô-Lévy processes. In the first part we use the maximum principle in stochastic control theory to extend the above relation to a dynamic relation, valid for all $t \in [0,T]$. We prove in particular that the optimal adjoint process for the primal problem coincides with the optimal density process, and that the optimal adjoint process for the dual problem coincides with the optimal wealth process; $0 \le t \le T$. In the terminal time case $t=T$ we recover the classical duality connection above. We get moreover an explicit relation between the optimal portfolio $\varphi ^*$ and the optimal measure $Q^*$. We also obtain that the existence of an optimal scenario is equivalent to the replicability of a related T-claim. In the second part we present robust (model uncertainty) versions of the optimization problems in (i) and (ii), and we prove a similar dynamic relation between them. In particular, we show how to get from the solution of one of the problems to the other. We illustrate the results with explicit examples.

Appendix: Maximum Principles for Optimal Control

Consider the following controlled stochastic differential equation

$$\begin{aligned} dX(t)&= b(t, X(t), u(t), \omega ) dt + \sigma (t, X(t), u(t), \omega ) dB(t) \nonumber \\&\quad + \int _\mathbb {R}\gamma (t, X(t), u(t), \omega , \zeta ) \tilde{N}(dt, d\zeta ) \; ; \; 0 \le t \le T \; ; \; X(0) = x \in \mathbb {R}. \end{aligned}$$

(4.1)

The performance functional is given by

$$\begin{aligned} J(u) = E \left[ \int _0^T f(t, X(t), u(t), \omega ) dt + \phi (X(T), \omega ) \right] \end{aligned}$$

(4.2)

where $T>0$ and u is in a given family $\mathcal {A}$ of admissible $\mathcal {F}$-predictable controls. For $u \in \mathcal {A}$ we let $X^u(t)$ be the solution of (4.1). We assume this solution exists, is unique and satisfies

$$\begin{aligned} E\left[ \int _0^T |X^u(t) |^{2 } dt\right] < \infty . \end{aligned}$$

(4.3)

We want to find $u^* \in \mathcal {A}$ such that

$$\begin{aligned} \sup _{u \in \mathcal {A}} J(u) = J(u^*). \end{aligned}$$

(4.4)

We make the following assumptions

$$\begin{aligned} f \in C^1 \text{ and } E\left[ \int _0^T | \nabla f |^2(t) dt\right] < \infty , \end{aligned}$$

(4.5)

$$\begin{aligned} b, \sigma , \gamma \in C^1 \text{ and } E\left[ \int _0^T \big ( | \nabla b |^2 + | \nabla \sigma |^2 + \Vert \nabla \gamma \Vert ^2\big ) (t) dt\right] < \infty , \end{aligned}$$

(4.6)

$$\begin{aligned} \text { where } \Vert \nabla \gamma (t, \cdot ) \Vert ^2 := \int _\mathbb {R}\gamma ^2(t, \zeta ) \nu (d \zeta ) \nonumber \\ \phi \in C^1 \text{ and } \text { for all } u \in \mathcal {A}, \; E\left[ \phi {^\prime }(X(T))^{2 }\right] < \infty . \end{aligned}$$

(4.7)

Let $\mathbb {U}$ be a convex closed set containing all possible control values $u(t); t \in [0,T]$.

The Hamiltonian associated to the problem (4.4) is defined by

$$\begin{aligned} H: [0,T] \times \mathbb {R}\times \mathbb {U}\times \mathbb {R}\times \mathbb {R}\times \mathcal{R } \times \Omega \mapsto \mathbb {R}\end{aligned}$$

$$\begin{aligned} H(t,x, u,p,q,r, \omega )= & {} f(t,x,u, \omega ) + b (t,x,u, \omega ) p + \sigma (t,x,u, \omega ) q\\&+ \displaystyle \int _\mathbb {R}\gamma (t,x,u, \zeta , \omega ) r(t, \zeta ) \nu (d \zeta ). \end{aligned}$$

For simplicity of notation the dependence on $\omega$ is suppressed in the following. We assume that H is Fréchet differentiable in the variables x, u. We let m denote the Lebesgue measure on [0, T].

The associated BSDE for the adjoint processes (p, q, r) is

(4.8)

Here and in the following we are using the abbreviated notation

$$\begin{aligned} \frac{\partial H}{\partial x} (t) = \frac{\partial H}{\partial x} (t, X(t), u(t)) \text { etc } \end{aligned}$$

We first formulate a sufficient maximum principle.

Theorem 4.1

(Sufficient maximum principle)Let $\hat{u}\in \mathcal {A}$ with corresponding solutions $\hat{X}$, $\hat{p}, \hat{q}, \hat{r}$ of equations (4.1)–(4.8). Assume the following:

• The function $x \mapsto \phi (x)$ is concave

• (The Arrow condition) The function

  $$\begin{aligned} \mathcal{H }(x) := \sup _{v \in \mathbb {U}} H(t, x, v, \hat{p}(t),\hat{q}(t),\hat{r}(t, \cdot )) \end{aligned}$$

  (4.9)

  is concave for all $t \in [0,T]$.

• $$\begin{aligned} \sup _{v \in \mathbb {U}} H(t, \hat{X}(t), v, \hat{p}(t),\hat{q}(t),\hat{r}(t, \cdot )) = H(t, \hat{X}(t), \hat{u}(t), \hat{p}(t),\hat{q}(t),\hat{r}(t, \cdot )) ; \; t \in [0,T]. \end{aligned}$$

  (4.10)

Then $\hat{u}$ is an optimal control for the problem (4.4).

Next, we state a necessary maximum principle. For this, we need the following assumptions:

• For all $t_0 \in [0, T]$ and all bounded $\mathcal {F}_{t_0}$-measurable random variables $\alpha (\omega )$ the control

  $$\begin{aligned} \theta (t) := \chi _{[t_0, T]}(t) \alpha (\omega ) \end{aligned}$$

  (4.11)

  belongs to $\mathcal {A}$.

• For all $u, \beta \in \mathcal {A}$ with $\beta$ bounded, there exists $\delta >0$ such that the control

  $$\begin{aligned} \tilde{u}(t) := u(t) + a \beta (t) ; \; t \in [0,T] \end{aligned}$$

  belongs to $\mathcal {A}$ for all $a \in ( - \delta , \delta )$.

• The derivative process

  $$\begin{aligned} x(t) := \frac{d}{da} X^{u + a \beta } (t) \mid _{a =0}, \end{aligned}$$

  exists and belongs to $L^2(dm \times dP)$, and

  $$\begin{aligned} \left\{ \begin{array}{ll} dx(t) = \{ \frac{\partial b}{\partial x} (t) x(t) + \frac{\partial b}{\partial u} (t) \beta (t)\} dt + \{ \frac{\partial \sigma }{\partial x} (t) x(t) + \frac{\partial \sigma }{\partial u} (t) \beta (t) \} dB(t) \\ \qquad \displaystyle + \displaystyle \int _\mathbb {R}\{ \frac{\partial \gamma }{\partial x} (t, \zeta ) x(t) + \frac{\partial \gamma }{\partial u} (t, \zeta ) \beta (t) \} \tilde{N}(dt, d \zeta ) \\ x(0) =0 \end{array}\right. \end{aligned}$$

  (4.12)

Theorem 4.2

(Necessary maximum principle) The following are equivalent

$$\begin{aligned} \bullet&\;\; \frac{d}{da} J(u + a \beta ) \mid _{a=0} = 0 \text { for all bounded } \beta \in \mathcal {A}\\ \bullet&\;\; \frac{\partial H}{\partial u}(t) = 0 \text { for all } t \in [0, T]. \end{aligned}$$

For detailed proofs of Theorems 4.1 and 4.2 we refer to proofs of Theorem 2.1 and 2.2 of [15]. We give below the ideas of the proofs. Sketch of proof of Theorem 4.2: By introducing a suitable family of stopping times as in the proof of Theorem 2.1 in [15], we may assume that all the local martingales below are martingales and hence have expectation 0.

Choose $u \in \mathcal {A}$ and consider

$$\begin{aligned} J(u) - J(\hat{u}) = J_1 + J_2 , \end{aligned}$$

where

$$\begin{aligned} J_1 = E \left[ \int _0^T \{ f(t) - \hat{f}(t)\} dt \right] ,\; J_2 = E [ \phi (X(T)) - \phi (\hat{X}(T))], \end{aligned}$$

where $f(t) = f(t, X(t), u(t))$, with $X(t) = X^u(t)$ etc.

By the definition of H we have

$$\begin{aligned} J_1&= E \left[ \int _0^T \{ H(t) - \hat{H}(t) - \hat{p}(t) (b(t) - \hat{b}(t)) \right. \nonumber \\&\left. \left. \quad - \hat{q}(t) (\sigma (t) - \hat{\sigma }(t)) - \int _\mathbb {R}\hat{r}(t,\zeta )(\gamma (t,\zeta ) - \hat{\gamma }(t,\zeta )) \nu (d\zeta ) \right\} dt \right] . \end{aligned}$$

(4.13)

By concavity of $\phi$ and the Itô formula,

$$\begin{aligned} J_2&\le E[ \phi {^\prime }(\hat{X}(T))(X(T) - \hat{X}(T))] \nonumber \\&= E [ \hat{p}(T) (X(T) - \hat{X}(T))] - E [ \hat{\lambda }(T) h'(\hat{X}(T))(X(T) - \hat{X}(T))] \nonumber \\&= E \left[ \int _0^{T} \hat{p}(t^-) (dX(t) - d \hat{X}(t)) + \int _0^{T} (X(t^-) - \hat{X}(t^-)) d \hat{p}(t) \right. \nonumber \\&\left. \quad + \int _0^{T} \hat{q}(t) (\sigma (t) - \hat{\sigma }(t))dt + \int _0^{T} \int _\mathbb {R}\hat{r}(t,\zeta )(\gamma (t,\zeta ) - \hat{\gamma }(t,\zeta )) \nu (d \zeta )\right] \nonumber \\&= E \left[ \int _0^T \hat{p}(t) (b(t) - \hat{b}(t))dt + \int _0^T (X(t) - \hat{X}(t)) \left( - \frac{\partial \hat{H}}{\partial x}(t)\right) dt \right. \nonumber \\&\left. \quad \quad + \int _0^T \hat{q}(t)(\sigma (t) - \hat{\sigma }(t))dt + \int _0^T \int _\mathbb {R}\hat{r}(t,\zeta ) (\gamma (t,\zeta ) - \hat{\gamma }(t,\zeta )) \nu (d\zeta ) dt \right] . \end{aligned}$$

(4.14)

Adding (4.13) and (4.14) we get

$$\begin{aligned} J(u)&- J(\hat{u}) = J_1 +J_2 \nonumber \\&\le E \left[ \int _0^T\left\{ H(t) - \hat{H}(t) - \frac{\partial \hat{H}}{\partial x} (X(t) - \hat{X}(t)) \right\} dt \right] . \end{aligned}$$

(4.15)

By a separating hyperplane argument (see e.g. [19], , Chapt. 5, Sec. 23) we get that

$$\begin{aligned} J(u) - J(\hat{u})&\le \hat{\mathcal H}(X(t)) - \hat{\mathcal H}(\hat{X}(t), ) - \frac{\partial \hat{\mathcal H}}{\partial x} (\hat{X}(t))(X(t) - \hat{X}(t)) \\&\le 0, \text { by concavity of } \mathcal H. \end{aligned}$$

$\square$

Sketch of proof of Theorem 4.2 : By introducing a suitable sequence of stopping times as in the proof of Theorem 2.2 in [15], we may assume that all the local martingales below are martingales and hence have expectation 0.

We can write $\displaystyle \frac{d}{da} J(u + a \beta ) \mid _{a=0} = I_1 + I_2$, where

$$\begin{aligned} I_1&= \frac{d}{da} E \left[ \int _0^T f(t,X^{u+a \beta }(t), u(t)) + a \beta (t))dt\right] _{a=0} \\ I_2&= \frac{d}{da} E[ \phi (X^{u+a \beta }(T))]_{a=0}. \end{aligned}$$

By our assumptions on f and $\phi$ we have

$$\begin{aligned} I_1&= \left[ \int _0^T \left\{ \frac{\partial f}{\partial x}(t) x(t) + \frac{\partial f}{\partial u}(t) \beta (t)\right\} dt\right] \nonumber \\ I_2&= E[ \phi {^\prime } (X(T)) x(T)] = E[p(T) x(T)]. \end{aligned}$$

(4.16)

By the Itô formula

$$\begin{aligned} I_2&= E[p(T) x(T)] \nonumber \\&=E \left[ \int _0^{T} p(t) dx(t) + \int _0^{T} x(t) dp(t) + \int _0^{T} d[p,x](t) \right] \nonumber \\&= E \left[ \int _0^{T} p(t) \left\{ \frac{\partial b}{\partial x}(t) x(t) + \frac{\partial b}{\partial u}(t) \beta (t) \right\} dt + \int _0^{T}x(t) \left( - \frac{\partial H}{\partial x}(t)\right) dt \right. \nonumber \\&= + \int _0^{T} q(t) \left\{ \frac{\partial \sigma }{\partial x}(t) x(t) + \frac{\partial \sigma }{\partial u}(t) \beta (t) \right\} dt \nonumber \\&\left. \quad + \int _0^{T}\int _\mathbb {R}r(t,\zeta ) \left\{ \frac{\partial \gamma }{\partial x}(t,\zeta )x(t) + \frac{\partial \gamma }{\partial u}(t,\zeta )\beta (t)\right\} \nu (d\zeta ) dt \right] \nonumber \\&= E \left[ \int _0^{T} x(t) \left\{ \frac{\partial b}{\partial x}(t) p(t) + \frac{\partial \sigma }{\partial x}(t) q(t) + \int _\mathbb {R}\frac{\partial \gamma }{\partial x}(t,\zeta ) r(t,\zeta ) \nu (d\zeta ) - \frac{\partial H}{\partial x}(t) \right\} dt\right. \nonumber \\&= E \left[ \int _0^{T} x(t) \left\{ - \frac{\partial f}{\partial x}(t) \right\} dt \right. \nonumber \\&\quad \left. + \int _0^{T} \beta (t) \left\{ \frac{\partial H}{\partial u}(t)- \frac{\partial f}{\partial u}(t) \right\} dt \right] \nonumber \\&= - I_1 + E \left[ \int _0^T \frac{\partial H}{\partial u} (t) \beta (t) dt\right] . \end{aligned}$$

(4.17)

Summing (4.16) and (4.17) we get

$$\begin{aligned} \frac{d}{da} J(u + a \beta ) \mid _{a=0} = I_1 + I_2 = E \left[ \int _0^T \frac{\partial H}{\partial u}(t) \beta (t)dt \right] . \end{aligned}$$

We conclude that

$$\begin{aligned} \frac{d}{da} J(u + a \beta ) \mid _{a=0} = 0 \end{aligned}$$

if and only if

$$\begin{aligned} E \left[ \int _0^T \frac{\partial H}{\partial u}(t) \beta (t) dt \right] = 0 \; ; \; \text { for all bounded } \beta \in \mathcal {A}. \end{aligned}$$

In particular, applying this to $\beta (t) = \theta (t)$ as in (4.11), we get that this is again equivalent to

$$\begin{aligned} E \left[ \frac{\partial H}{\partial u}(t) \mid \mathcal {E}_t\right] = 0 \text { for all } t \in [0,T]. \end{aligned}$$

$\square$