11footnotetext: Department of Mathematics, University of Tunis El Manar,
Tunis, Tunisia.
Email<EMAIL_ADDRESS>
# Optimal insider control of stochastic Volterra equations

Olfa Draouil1

(26 March 2017 )

###### Abstract

We study the problem of optimal inside control of a stochastic Volterra
equation driven by a Brownian motion and a Poisson random measure. We prove a
sufficient and a necessary maximum principle for the optimal control when the
trader has only partial information available to her decisions and on the
other hand, may have some inside information about the future of the system.
The results are applied to the problem of finding the optimal insider
portfolio in a financial market where the risky asset price is given by a
stochastic Volterra equation.

#### MSC(2010):

60H10, 91A15, 91A23, 91B38, 91B55, 91B70, 93E20

#### Keywords:

Stochastic Volterra equation; optimal control; inside information; Donsker
delta functional; stochastic maximum principle.

##  1 Introduction

In this paper we consider an optimal control problem for a stochastic process
$X(t)=X(t,Z)=X(t,z)|_{z=Z}$ defined as the solution of a stochastic Volterra
(integral) equation (SVIE) given by

$\displaystyle X(t)=X^{u}(t,Z)=\xi(t)+\int_{0}^{t}b(t,s,X(s,Z),u(s,Z))ds$
$\displaystyle+\int_{0}^{t}\sigma(t,s,X(s,Z),u(s,Z))dB(s)+\int_{0}^{t}\int_{\mathbb{R}}\gamma(t,s,X(s,Z),u(s,Z),\zeta)\tilde{N}(ds,d\zeta)$
(1.1)

Here $B(t)$ and $\tilde{N}(dt,d\zeta)$ is a Brownian motion and an independent
compensated Poisson random measure, respectively, jointly defined on a
filtered probability space $(\Omega,\mathbb{F}=\\{\mathcal{F}_{t}\\}_{t\geq
0},P)$ satisfying the usual conditions. The process $u(t)=u(t,Z)=u(t,z)_{z=Z}$
is our insider control process, where $Z$ is a given
$\mathcal{F}_{T_{0}}$-measurable random variable for some $T_{0}>0$ ,
representing the inside information available to the controller. Note that
from (1) we get:

$\displaystyle
dX(t)=\xi^{\prime}(t)dt+b(t,t,X(t,Z),u(t,Z))dt+(\int_{0}^{t}\frac{\partial
b}{\partial t}(t,s,X(s,Z),u(s,Z))ds)dt$
$\displaystyle+\sigma(t,t,X(t,Z),u(t,Z))dB(t)+(\int_{0}^{t}\frac{\partial\sigma}{\partial
t}(t,s,X(s,Z),u(s,Z))dB(s))dt$
$\displaystyle+\int_{\mathbb{R}}\gamma(t,t,X(t,Z),u(t,Z),\zeta)\tilde{N}(dt,d\zeta)+(\int_{0}^{t}\int_{\mathbb{R}}\frac{\partial\gamma}{\partial
t}(t,s,X(s,Z),u(s,Z),\zeta)\tilde{N}(ds,d\zeta))dt.$ (1.2)

We assume that the inside information is of _initial enlargement_ type.
Specifically, we assume that the inside filtration $\mathbb{H}$ has the form

$\mathbb{H}=\\{\mathcal{H}_{t}\\}_{0\leq t\leq T},\text{ where
}\mathcal{H}_{t}=\mathcal{F}_{t}\vee\sigma(Z)$ (1.3)

for all $t$, where $Z$ is a given $\mathcal{F}_{T_{0}}$-measurable random
variable, for some $T_{0}>0$ (constant). Here and in the following we use the
right-continuous version of $\mathbb{H}$, i.e. we put
$\mathcal{H}_{t}=\mathcal{H}_{t^{+}}=\bigcap_{s>t}\mathcal{H}_{s}.$

We also assume that the _Donsker delta functional_ of $Z$ exists (see below).
This assumption implies that the Jacod condition holds, and hence that
$B(\cdot)$ and $N(\cdot,\cdot)$ are semimartingales with respect to
$\mathbb{H}$. See e.g. [DØ2] for details. Let $\mathcal{A}$ be a given family
of admissible controls, required to be
$\mathcal{K}_{t}=\mathcal{G}_{t}\vee\sigma(Z)$-predictable, where
$\mathbb{G}=\\{\mathcal{G}_{t}\\}_{0\leq t\leq T}$ is a given subfiltration of
$\mathbb{F}=\\{\mathcal{F}_{t}\\}_{0\leq t\leq T}$ in the sense that
$\mathcal{G}_{t}\subseteq\mathcal{F}_{t}$ for all $t\in[0,T]$. That is mean
that $\mathcal{K}_{t}\subseteq\mathcal{H}_{t}$ for all $t\in[0,T]$. We assume
that the value at time $t$ of our insider control process $u(t,x)$ is allowed
to depend on both $Z$ and $\mathcal{G}_{t}$. In other words, $u(.,x)$ is
assumed to be $\mathbb{K}$-adapted. Therefore it has the form

$u(t,\omega)=u_{1}(t,Z,\omega)$ (1.4)

for some function
$u_{1}:[0,T]\times\mathbb{R}\times\Omega\rightarrow\mathbb{R}$ such that
$u_{1}(.,z)$ is $\mathbb{G}$-adapted for each $z\in\mathbb{R}$. For simplicity
(albeit with some abuse of notation) we will in the following write $u$
instead of $u_{1}$. In other word, we study the case when the controller has
only partial information available to her decisions and on the other hand, may
have some inside information about the future of the system.

Let $\mathbb{U}$ denote the set of admissible control values. We assume that
the functions

$\displaystyle b(t,s,x,u,z)$
$\displaystyle=b(t,s,x,u,z,\omega):[0,T]\times[0,T]\times\mathbb{R}\times\mathbb{U}\times\mathbb{R}\times\Omega\mapsto\mathbb{R}$
$\displaystyle\sigma(t,s,x,u,z)$
$\displaystyle=\sigma(t,s,x,u,z,\omega):[0,T]\times[0,T]\times\mathbb{R}\times\mathbb{U}\times\mathbb{R}\times\Omega\mapsto\mathbb{R}$
$\displaystyle\gamma(t,s,x,u,z,\zeta)$
$\displaystyle=\gamma(t,s,x,u,z,\zeta,\omega):[0,T]\times[0,T]\times\mathbb{R}\times\mathbb{U}\times\mathbb{R}\times\mathbb{R}\times\Omega\mapsto\mathbb{R}$

are given bounded $C^{1}$ functions with respect to $t,x$ and $u$ and
$\mathbb{F}$-adapted processes in $(s,\omega)$ for each given $t,x,u,z,\zeta$.
The _performance functional_ $J(u)$ of a control process $u\in\mathcal{A}$ is
defined by

$J(u)=\mathbb{E}[\int_{0}^{T}f(t,X(t),u(t,Z),Z)dt+g(X(T),Z)],$ (1.6)

where

$\displaystyle
f(t,x,u,z):[0,T]\times\mathbb{R}\times\mathbb{U}\times\mathbb{R}\mapsto\mathbb{R}$
$\displaystyle g(x,z):\mathbb{R}\times\mathbb{R}\mapsto\mathbb{R}$ (1.7)

are given bounded functions, $C^{1}$ with respect to $x$ and $u$, $f$ is
$\mathbb{F}$-adapted for each $x,u,z$ and $g$ is $\mathcal{F}_{T}$-measurable
for each $x,z$. The functions $f$ and $g$ are called the _profit rate density_
and _terminal payoff density_ , respectively. For completeness of the
presentation we allow these functions to depend explicitly on the future value
$Z$ also, although this would not be the typical case in applications. But it
could be that $f$ and $g$ are influenced by the future value $Z$ directly
through the action of an insider, in addition to being influenced indirectly
through the control process $u$ and the corresponding state process $X$.

###### Problem 1.1

Find $u^{\star}\in\mathcal{A}$ such that

$\sup_{u\in\mathcal{A}}J(u)=J(u^{\star}).$ (1.8)

##  2 The Donsker delta functional

To study this problem we adapt the technique of the paper [DØ1] to the
stochastic Volterra equation (SVE) and we combine this with the method for
optimal control of SVE developed in [AØ]. We first recall briefly the
definition and basic properties of the Donsker delta functional:

###### Definition 2.1

Let $Z:\Omega\rightarrow\mathbb{R}$ be a random variable which also belongs to
$(\mathcal{S})^{\ast}$. Then a continuous functional

$\delta_{Z}(.):\mathbb{R}\rightarrow(\mathcal{S})^{\ast}$ (2.1)

is called a Donsker delta functional of $Z$ if it has the property that

$\int_{\mathbb{R}}g(z)\delta_{Z}(z)dz=g(Z)\quad a.s.$ (2.2)

for all (measurable) $g:\mathbb{R}\rightarrow\mathbb{R}$ such that the
integral converges.

For example, consider the special case when $Z$ is a first order chaos random
variable of the form

$Z=Z(T_{0});\text{ where
}Z(t)=\int_{0}^{t}\beta(s)dB(s)+\int_{0}^{t}\int_{\mathbb{R}}\psi(s,\zeta)\tilde{N}(ds,d\zeta),\mbox{
for }t\in[0,T_{0}]$ (2.3)

for some deterministic functions $\beta\neq 0,\psi$ such that

$\int_{0}^{T_{0}}\\{\beta^{2}(t)+\int_{\mathbb{R}}\psi^{2}(t,\zeta)\nu(d\zeta)\\}dt<\infty\text{
a.s. }$ (2.4)

and for every $\epsilon>0$ there exists $\rho>0$ such that

$\int_{\mathbb{R}\setminus(-\epsilon,\epsilon)}e^{\rho\zeta}\nu(d\zeta)<\infty.\\\
$

This condition implies that the polynomials are dense in $L^{2}(\mu)$, where
$d\mu(\zeta)=\zeta^{2}d\nu(\zeta)$. It also guarantees that the measure $\nu$
integrates all polynomials of degree $\geq 2$.
In this case it is well known (see e.g. [MØP], [DiØ1], Theorem 3.5, and
[DØP],[DiØ2]) that the Donsker delta functional exists in
$(\mathcal{S})^{\ast}$ and is given by

$\displaystyle\delta_{Z}(z)$ $\displaystyle=$
$\displaystyle\frac{1}{2\pi}\int_{\mathbb{R}}\exp^{\diamond}\big{[}\int_{0}^{T_{0}}\int_{\mathbb{R}}(e^{ix\psi(s,\zeta)}-1)\tilde{N}(ds,d\zeta)+\int_{0}^{T_{0}}ix\beta(s)dB(s)$
(2.5) $\displaystyle+$
$\displaystyle\int_{0}^{T_{0}}\\{\int_{\mathbb{R}}(e^{ix\psi(s,\zeta)}-1-ix\psi(s,\zeta))\nu(d\zeta)-\frac{1}{2}x^{2}\beta^{2}(s)\\}ds-
ixz\big{]}dx,$

where $\exp^{\diamond}$ denotes the Wick exponential. Moreover, we have for
$t<T_{0}$

$\displaystyle\mathbb{E}[\delta_{Z}(z)|\mathcal{F}_{t}]$ $\displaystyle=$
$\displaystyle\frac{1}{2\pi}\int_{\mathbb{R}}\exp\big{[}\int_{0}^{t}\int_{\mathbb{R}}ix\psi(s,\zeta)\tilde{N}(ds,d\zeta)+\int_{0}^{t}ix\beta(s)dB(s)$
(2.6)
$\displaystyle+\int_{t}^{T_{0}}\int_{\mathbb{R}}(e^{ix\psi(s,\zeta)}-1-ix\psi(s,\zeta))\nu(d\zeta)ds-\int_{t}^{T_{0}}\frac{1}{2}x^{2}\beta^{2}(s)ds-
ixz\big{]}dx.$ (2.7)

If $D_{t}$ and $D_{t,\zeta}$ denotes the _Hida-Malliavin derivative_ at $t$
and $t,\zeta$ with respect to $B$ and $\tilde{N}$, respectively, we have

$\displaystyle\mathbb{E}[D_{t}\delta_{Z}(z)|\mathcal{F}_{t}]=$
$\displaystyle\frac{1}{2\pi}\int_{\mathbb{R}}\exp\big{[}\int_{0}^{t}\int_{\mathbb{R}}ix\psi(s,\zeta)\tilde{N}(ds,d\zeta)+\int_{0}^{t}ix\beta(s)dB(s)$
$\displaystyle+\int_{t}^{T_{0}}\int_{\mathbb{R}}(e^{ix\psi(s,\zeta)}-1-ix\psi(s,\zeta))\nu(d\zeta)ds-\int_{t}^{T_{0}}\frac{1}{2}x^{2}\beta^{2}(s)ds-
ixz\big{]}ix\beta(t)dx$ (2.8)

and

$\displaystyle\mathbb{E}[D_{t,z}\delta_{Z}(z)|\mathcal{F}_{t}]=$
$\displaystyle\frac{1}{2\pi}\int_{\mathbb{R}}\exp\big{[}\int_{0}^{t}\int_{\mathbb{R}}ix\psi(s,\zeta)\tilde{N}(ds,d\zeta)+\int_{0}^{t}ix\beta(s)dB(s)$
$\displaystyle+\int_{t}^{T_{0}}\int_{\mathbb{R}}(e^{ix\psi(s,\zeta)}-1-ix\psi(s,\zeta))\nu(d\zeta)ds-\int_{t}^{T_{0}}\frac{1}{2}x^{2}\beta^{2}(s)ds-
ixz\big{]}(e^{ix\psi(t,z)}-1)dx.$ (2.9)

For more information about the Donsker delta functional, Hida-Malliavin
calculus and their properties, see [DØ1].

From now on we assume that $Z$ is a given random variable which also belongs
to $(\mathcal{S})^{\ast}$, with a Donsker delta functional
$\delta_{Z}(z)\in(\mathcal{S})^{\ast}$ satisfying

$\mathbb{E}[\delta_{Z}(z)|\mathcal{F}_{T}]\in\mathbf{L}^{2}(\mathcal{F}_{T},P)$
(2.10)

and

$\mathbb{E}[\int_{0}^{T}(\mathbb{E}[D_{t}\delta_{Z}(z)|\mathcal{F}_{t}])^{2}dt]<\infty,\text{
for all }z.$ (2.11)

##  3 Transforming the insider control problem to a related parameterized non-
insider problem

Since $X(t)$ is $\mathbb{H}$-adapted, we get by using the definition of the
Donsker delta functional $\delta_{Z}(z)$ of $Z$ that

$X(t)=X(t,Z)=X(t,z)_{z=Z}=\int_{\mathbb{R}}X(t,z)\delta_{Z}(z)dz$ (3.1)

for some $z$-parameterized process $X(t,z)$ which is $\mathbb{F}$-adapted for
each $z$. Then, again by the definition of the Donsker delta functional we can
write

$\displaystyle X(t)=\xi(t,Z)+\int_{0}^{t}b(t,s,X(s,Z),u(s,Z))ds$
$\displaystyle+\int_{0}^{t}\sigma(t,s,X(s,Z),u(s,Z))dB(s)+\int_{0}^{t}\int_{\mathbb{R}}\gamma(t,s,X(s,Z),u(s,Z),\zeta)\tilde{N}(ds,d\zeta)$
$\displaystyle=\xi(t,z)_{z=Z}+\int_{0}^{t}b(t,s,X(s,z),u(s,z))_{z=Z}ds$
$\displaystyle+\int_{0}^{t}\sigma(t,s,X(s,z),u(s,z))_{z=Z}dB(s)$
$\displaystyle+\int_{0}^{t}\int_{\mathbb{R}}\gamma(t,s,X(s,z),u(s,z),\zeta)_{z=Z}\tilde{N}(ds,d\zeta)$
$\displaystyle=\int_{\mathbb{R}}\xi(t,z)\delta_{Z}(z)dz+\int_{0}^{t}\int_{\mathbb{R}}b(t,s,X(s,z),u(s,z))\delta_{Z}(z)dzds$
$\displaystyle+\int_{0}^{t}\int_{\mathbb{R}}\sigma(t,s,X(s,z),u(s,z))\delta_{Z}(z)dzdB(s)$
$\displaystyle+\int_{0}^{t}\int_{\mathbb{R}}\int_{\mathbb{R}}\gamma(t,s,X(s,z),u(s,z),\zeta)\delta_{Z}(z)dz\tilde{N}(ds,d\zeta)$
$\displaystyle=\int_{\mathbb{R}}\\{\xi(t,z)+\int_{0}^{t}b(t,s,X(s,z),u(s,z))ds+\int_{0}^{t}\sigma(t,s,X(s,z),u(s,z))dB(s)$
$\displaystyle+\int_{0}^{t}\int_{\mathbb{R}}\gamma(t,s,X(s,z),u(s,z),\zeta)\tilde{N}(ds,d\zeta)\\}\delta_{Z}(z)dz.$
(3.2)

Comparing (3.1) and (3) we see that (3.1) holds if we for each $z$ choose
$X(t,z)$ as the solution of the classical (but parameterized ) SVIE

$\displaystyle
dX(t,z)=\xi^{\prime}(t,z)dt+b(t,t,X(t,z),u(t,z))dt+(\int_{0}^{t}\frac{\partial
b}{\partial t}(t,s,X(s,z),u(s,z))ds)dt$
$\displaystyle+\sigma(t,t,X(t,z),u(t,z))dB(t)+(\int_{0}^{t}\frac{\partial\sigma}{\partial
t}(t,s,X(s,z),u(s,z))dB(s))dt$
$\displaystyle+\int_{\mathbb{R}}\gamma(t,t,X(t,z),u(t,z),\zeta)\tilde{N}(dt,d\zeta)+(\int_{0}^{t}\int_{\mathbb{R}}\frac{\partial\gamma}{\partial
t}(t,s,X(s,z),u(s,z),\zeta)\tilde{N}(ds,d\zeta))dt.$ (3.3)

As before let $\mathcal{A}$ be the given family of admissible
$\mathbb{K}-$adapted controls $u$. Then in terms of $X(t,z)$ the performance
functional $J(u)$ of a control process $u\in\mathcal{A}$ defined in (1.6) gets
the form

$\displaystyle J(u)$
$\displaystyle=\mathbb{E}[\int_{0}^{T}f(t,X(t,Z),u(t,Z),Z)dt+g(X(T,Z),Z)]$
$\displaystyle=\mathbb{E}[\int_{\mathbb{R}}\Big{\\{}\int_{0}^{T}f(t,X(t,z),u(t,z),z)\mathbb{E}[\delta_{Z}(z)|\mathcal{F}_{t}]dt$
$\displaystyle+g(X(T,z),z)\mathbb{E}[\delta_{Z}(z)|\mathcal{F}_{T}]\Big{\\}}dz]$
$\displaystyle=\int_{\mathbb{R}}j(u)(z)dz,$ (3.4)

where

$\displaystyle j(u)(z)$
$\displaystyle:=\mathbb{E}[\int_{0}^{T}f(t,X(t,z),u(t,z),z)\mathbb{E}[\delta_{Z}(z)|\mathcal{F}_{t}]dt$
$\displaystyle+g(X(T,z),z)\mathbb{E}[\delta_{Z}(z)|\mathcal{F}_{T}].$ (3.5)

Thus we see that to maximize $J(u)$ it suffices to maximize $j(u)(z)$ for each
value of the parameter $z\in\mathbb{R}$. Therefore Problem 1.1 is transformed
into the problem

###### Problem 3.1

For each given $z\in\mathbb{R}$ find $u^{\star}=u^{\star}(t,z)\in\mathcal{A}$
such that

$\sup_{u\in\mathcal{A}}j(u)(z)=j(u^{\star})(z).$ (3.6)

##  4 A sufficient-type maximum principle

In this section we will establish a sufficient maximum principle for Problem
3.1.
Problem 3.1 is a stochastic control problem with a standard (albeit
parameterized) stochastic Volterra equation (3) for the state process
$X(t,z)$, but with a non-standard performance functional given by (3). We can
solve this problem by a modified maximum principle approach, as follows:
Let $L$ and $L_{\zeta}$ be the set of all stochastic processes with parameter
space $[0,T]$ and $[0,T]\times\mathbb{R}_{0}$, respectively, where
$\mathbb{R}_{0}=\mathbb{R}\setminus\\{0\\}$. Define the Hamiltonian
functionals:

$H_{0}:[0,T]\times\mathbb{R}\times\mathbb{U}\times\mathbb{R}\times\mathbb{R}\times\mathbb{R}\rightarrow\mathbb{R}$
(4.1)

and

$H_{1}:[0,T]\times\mathbb{R}\times\mathbb{U}\times\mathbb{R}\times L\times
L_{\zeta}\rightarrow\mathbb{R}$ (4.2)

by

$\displaystyle
H_{0}(t,x,z,u,p,q,r):=f(t,x,z,u)E[\delta_{Z}(z)|\mathcal{F}_{t}]+b(t,t,x,z,u)p+\sigma(t,t,x,z,u)q$
$\displaystyle+\int_{\mathbb{R}}\gamma(t,t,x,z,u,\zeta)r(\zeta)\nu(d\zeta)$
(4.3)

and

$\displaystyle
H_{1}(t,x,z,u,p,D_{t}p(.),D_{t,\zeta}p(.)):=\int_{t}^{T}\frac{\partial
b}{\partial s}(s,t,x,z,u)p(s,z)ds+\int_{t}^{T}\frac{\partial\sigma}{\partial
s}(s,t,x,z,u)E[D_{t}p(s,z)|\mathcal{F}_{t}]ds$
$\displaystyle+\int_{t}^{T}\int_{\mathbb{R}}\frac{\partial\gamma}{\partial
s}(s,t,x,z,u,\zeta)E[D_{t,\zeta}p(s,z)|\mathcal{F}_{t}]\nu(d\zeta)ds.$ (4.4)

Here $\mathcal{R}$ denotes the set of all functions
$r(\cdot):\mathbb{R}\rightarrow\mathbb{R}$ such that the last integral above
converges.The quantities $p,q,r(\cdot)$ are called the _adjoint variables_.
Define

$H(t,x,z,u,p(.),q(.),r(.)):=H_{0}(t,x,z,u,p,q,r)+H_{1}(t,x,u,p(.),D_{t}p(.),D_{t,\zeta}p(.)).$
(4.5)

The _adjoint processes_ $p(t,z),q(t,z),r(t,z,\zeta)$ are defined as the
solution of the $z$-parameterized backward stochastic differential equation
(BSDE)

$\left\\{\begin{array}[]{l}dp(t,z)=-\frac{\partial H}{\partial
x}(t,z)dt+q(t,z)dB(t)+\int_{\mathbb{R}}r(t,z,\zeta)\tilde{N}(dt,d\zeta);\\\
\quad(t,z)\in(0,T)\times\mathbb{R}\\\ p(T,z)=\frac{\partial g}{\partial
x}(X(T,z),z)\mathbb{E}[\delta_{Z}(z)|\mathcal{F}_{T}];\quad
z\in\mathbb{R}\end{array}\right.$ (4.6)

where

$\frac{\partial H}{\partial x}(t,z)=\frac{\partial H}{\partial
x}(t,X(t,z),u(t,z),z,p(t,z),q(t,z),r(t,z,.)).$ (4.7)

We can now state the first maximum principle for our problem (3.6):

###### Theorem 4.1

[Sufficient-type maximum principle]
Let $\hat{u}\in\mathcal{A}$, and denote the associated solution of (3) and
(4.6) by $\hat{X}(t,z)$ and
$(\hat{p}(t,z),\hat{q}(t,z),\hat{r}(t,z,\zeta))$, respectively. Assume that
the following hold:

  1. 1.

$x\rightarrow g(x,z)$ is concave for all $z$

  2. 2.

$(x,u)\rightarrow
H(t,x,u,z,\widehat{p}(t,z),\widehat{q}(t,z),\hat{r}(t,z,\zeta))$ is concave
for all $t,z,\zeta$

  3. 3.

$\sup_{w\in\mathbb{U}}E[H\big{(}t,\hat{X}(t,z),w,\widehat{p}(t,z),\widehat{q}(t,z),\hat{r}(t,z,\zeta)\big{)}|\mathcal{G}_{t}]$
$=E[H\big{(}t,\widehat{X}(t,z),\widehat{u}(t,z),\widehat{p}(t,z),\widehat{q}(t,z),\hat{r}(t,z,\zeta)\big{)}|\mathcal{G}_{t}]$
for all $t,z,\zeta.$

Then $\widehat{u}(\cdot,z)$ is an optimal insider control for Problem 3.1.

Proof. By considering an increasing sequence of stopping times $\tau_{n}$
converging to $T$, we may assume that all local integrals appearing in the
computations below are martingales and hence have expectation 0. See [ØS2]. We
omit the details.
Choose arbitrary $u(.,z)\in\mathcal{A}$, and let the corresponding solution of
(3) and (4.6) be $X(t,z)$, $p(t,z)$, $q(t,z)$, $r(t,z,\zeta)$. For simplicity
of notation we write
$f=f(t,X(t,z),u(t,z),z)$,
$\widehat{f}=f(t,\widehat{X}(t,z),\widehat{u}(t,z),z)$ and similarly with $b$,
$\widehat{b}$, $\sigma$, $\widehat{\sigma}$ and so on.
Moreover put

$\hat{H}(t)=H(t,\widehat{X}(t,z),\widehat{u}(t,z),\widehat{p}(t,z),\widehat{q}(t,z),\widehat{r}(t,z,.))$
(4.8)

and

$H(t)=H(t,X(t,z),u(t,z),\widehat{p}(t,z),\widehat{q}(t,z),\widehat{r}(t,z,.))$
(4.9)

and similarly with $\hat{H}_{0}(t)$ and $H_{0}(t)$. In the following we write
$\widetilde{f}=f-\widehat{f}$, $\widetilde{b}=b-\widehat{b}$,
$\widetilde{X}=X-\widehat{X}$.
Consider

$j(u(.,z))-j(\widehat{u}(.,z))=I_{1}+I_{2},$

where

$I_{1}=\mathbb{E}[\int_{0}^{T}\\{f(t)-\widehat{f}(t)\\}\mathbb{E}[\delta_{Z}(z)|\mathcal{F}_{t}]dt],\quad
I_{2}=\mathbb{E}[\\{g(x)-\hat{g}(x)\\}\mathbb{E}[\delta_{Z}(z)|\mathcal{F}_{T}]].$
(4.10)

By the definition of $H$ we have

$\displaystyle I_{1}$
$\displaystyle=\mathbb{E}[\int_{0}^{T}\\{H_{0}(t)-\widehat{H_{0}}(t)-\widehat{p}(t)\widetilde{b}(t)-\widehat{q}(t)\widetilde{\sigma}(t)$
$\displaystyle-\int_{\mathbb{R}}\hat{r}(t,\zeta)\tilde{\gamma}(t,\zeta)\nu(d\zeta)\\}dt].$
(4.11)

Since $g$ is concave with respect to $x$ we have

$\displaystyle(g(X(T,z),z)-g(\hat{X}(T,z),z))\mathbb{E}[\delta_{Z}(z)|\mathcal{F}_{T}]$
$\displaystyle\leq\frac{\partial g}{\partial
x}(\hat{X}(T,z),z)\mathbb{E}[\delta_{Z}(z)|\mathcal{F}_{T}](X(T,z)-\hat{X}(T,z)),$
(4.12)

and hence

$\displaystyle I_{2}$ $\displaystyle\leq\mathbb{E}\Big{[}\frac{\partial
g}{\partial
x}(\widehat{X}(T,z))\mathbb{E}[\delta_{Z}(z)|\mathcal{F}_{T}]\tilde{X}(T,z)\Big{]}=\mathbb{E}[\widehat{p}(T,z)\widetilde{X}(T,z)]$
(4.13)
$\displaystyle=E\Big{[}\int_{0}^{T}\Big{\\{}\hat{p}(t,z)\Big{(}b(t,z)-\hat{b}(t,z)+\int_{0}^{t}(\frac{\partial
b}{\partial t}(t,s)-\frac{\partial\hat{b}}{\partial t}(t,s))ds$
$\displaystyle+\int_{0}^{t}(\frac{\partial\sigma}{\partial
t}(t,s)-\frac{\partial\hat{\sigma}}{\partial t}(t,s))dB(s)$
$\displaystyle+\int_{0}^{t}\int_{\mathbb{R}}(\frac{\partial\gamma}{\partial
t}(t,s,\zeta)-\frac{\partial\hat{\gamma}}{\partial
t}(t,s,\zeta))\tilde{N}(ds,d\zeta)\Big{)}$
$\displaystyle-\frac{\partial\hat{H}}{\partial x}(t)(X(t,z)-\hat{X}(t,z))$
$\displaystyle+\hat{q}(t,z)[\sigma(t,z)-\hat{\sigma}(t,z)]+\int_{\mathbb{R}}\hat{r}(t,z,\zeta)[\gamma(t,z,\zeta)-\hat{\gamma}(t,z,\zeta)]\nu(d\zeta)\Big{\\}}dt\Big{]}$

By the Fubini Theorem, we get

$\int_{0}^{T}(\int_{0}^{t}\frac{\partial b}{\partial
t}(t,s)ds)\hat{p}(t,z)dt=\int_{0}^{T}(\int_{s}^{T}\frac{\partial b}{\partial
t}(t,s)\hat{p}(t,y)dt)ds=\int_{0}^{T}(\int_{t}^{T}\frac{\partial b}{\partial
s}(s,t)\hat{p}(s,z)ds)dt,$ (4.14)

and similarly, by the duality theorems,

$\displaystyle E[\int_{0}^{T}(\int_{0}^{t}\frac{\partial\sigma}{\partial
t}(t,s)dB(s))\hat{p}(t,z)dt]=\int_{0}^{T}E[\int_{0}^{t}\frac{\partial\sigma}{\partial
t}(t,s)dB(s)\hat{p}(t,z)]dt$
$\displaystyle=\int_{0}^{T}E[\int_{0}^{t}\frac{\partial\sigma}{\partial
t}(t,s)E[D_{s}\hat{p}(t,z)|\mathcal{F}_{s}]ds]dt=\int_{0}^{T}E[\int_{s}^{T}\frac{\partial\sigma}{\partial
t}(t,s)E[D_{s}\hat{p}(t,z)|\mathcal{F}_{s}]dt]ds$
$\displaystyle=E[\int_{0}^{T}\int_{t}^{T}\frac{\partial\sigma}{\partial
s}(s,t)E[D_{t}\hat{p}(s,z)|\mathcal{F}_{t}]dsdt]$ (4.15)

and

$\displaystyle
E[\int_{0}^{T}(\int_{0}^{t}\int_{\mathbb{R}}\frac{\partial\gamma}{\partial
t}(t,s,\zeta)\tilde{N}(ds,d\zeta))\hat{p}(t,z)dt]=\int_{0}^{T}E[\int_{0}^{t}\int{\mathbb{R}}\frac{\partial\gamma}{\partial
t}(t,s,\zeta)\tilde{N}(ds,d\zeta))\hat{p}(t,z)]dt$
$\displaystyle=\int_{0}^{T}E[\int_{0}^{t}\int_{\mathbb{R}}\frac{\partial\gamma}{\partial
t}(t,s,\zeta)E[D_{s,\zeta}\hat{p}(t,z)|\mathcal{F}_{s}]\nu(d\zeta)ds]dt$
$\displaystyle=\int_{0}^{T}E[\int_{s}^{T}\int_{\mathbb{R}}\frac{\partial\gamma}{\partial
t}(t,s,\zeta)E[D_{s,\zeta}\hat{p}(t,z)|\mathcal{F}_{s}]\nu(d\zeta)dt]ds$
$\displaystyle=E[\int_{0}^{T}\int_{t}^{T}\int_{\mathbb{R}}\frac{\partial\gamma}{\partial
s}(s,t,\zeta)E[D_{t,\zeta}\hat{p}(s,z)|\mathcal{F}_{t}]\nu(d\zeta)dsdt].$
(4.16)

Substituting (4.14), (4) and (4) into (4.13), we get

$\displaystyle I_{2}\leq
E\Big{[}\int_{0}^{T}\Big{(}\hat{p}(t,z)[b(t)-\hat{b}(t)]+\hat{p}(t,z)\int_{t}^{T}(\frac{\partial
b}{\partial s}(s,t)-\frac{\partial\hat{b}}{\partial s}(s,t))ds$
$\displaystyle+\int_{t}^{T}(\frac{\partial\sigma}{\partial
s}(s,t)-\frac{\partial\hat{\sigma}}{\partial
s}(s,t))E[D_{t}\hat{p}(s,z)|\mathcal{F}_{t}]ds$
$\displaystyle+\int_{t}^{T}\int_{\mathbb{R}}(\frac{\partial\gamma}{\partial
s}(s,t,\zeta)-\frac{\partial\hat{\gamma}}{\partial
s}(s,t,\zeta))E[D_{t,\zeta}\hat{p}(s)|\mathcal{F}_{t}]ds\nu(d\zeta)$
$\displaystyle-\frac{\partial\hat{H}}{\partial x}(t)(X(t,z)-\hat{X}(t,z))$
$\displaystyle+\hat{q}(t,z)[\sigma(t,z)-\hat{\sigma}(t,z)]+\int_{\mathbb{R}}\hat{r}(t,\zeta)[\gamma(t,z,\zeta)-\hat{\gamma}(t,z,\zeta)]\nu(d\zeta)\Big{)}dt\Big{]}.$
(4.17)

Adding (4) and (4), we get

$\displaystyle J(u)-J(\hat{u})=I_{1}+I_{2}$ $\displaystyle\leq
E\Big{[}\int_{0}^{T}\Big{\\{}H(t)-\hat{H}(t)-\frac{\partial{H}}{\partial
x}(t)(X(t,z)-\hat{X}(t,z))\Big{\\}}dt\Big{]}.$ (4.18)

By the concavity assumption of $H$ in $(x,u)$ we have:

$H(t)-\hat{H}(t)\leq\frac{\partial\widehat{H}}{\partial
x}(t)(X-\hat{X})(t,z)+\frac{\partial\widehat{H}}{\partial
u}(t)(u(t,z)-\hat{u}(t,z)),$ (4.19)

Then equation (4) becomes

$\displaystyle J(u)-J(\hat{u})$ $\displaystyle\leq
E\Big{[}\int_{0}^{T}\frac{\partial\widehat{H}}{\partial
u}(t)(u(t,z)-\hat{u}(t,z))dt\Big{]}$
$\displaystyle=\int_{0}^{T}E\Big{[}\frac{\partial\widehat{H}}{\partial
u}(t)(u(t,z)-\hat{u}(t,z))\Big{]}dt$
$\displaystyle=\int_{0}^{T}E\Big{[}E[\frac{\partial\widehat{H}}{\partial
u}(t)(u(t,z)-\hat{u}(t,z))|\mathcal{G}_{t}]\Big{]}dt$ (4.20)

and the maximum condition implies that

$E[\frac{\partial\widehat{H}}{\partial
u}(t,z)(u(t,z)-\hat{u}(t,z))|\mathcal{G}_{t}]\leq 0.$ (4.21)

Hence by (4) we get $j(u)\leq j(\hat{u})$. Since $u\in\mathcal{A}$ was
arbitrary, this shows that $\hat{u}$ is optimal. $\square$

##  5 A necessary-type maximum principle

We proceed to establish a corresponding necessary maximum principle. For this,
we do not need concavity conditions, but instead we need the following
assumptions about the set of admissible control processes:

  * •

$A_{1}$. For all $t_{0}\in[0,T]$ and all bounded
$\mathcal{G}_{t_{0}}$-measurable random variables $\alpha(z,\omega)$, the
control $\theta(t,z,\omega):=\mathbf{1}_{[t_{0},T]}(t)\alpha(z,\omega)$
belongs to $\mathcal{A}$.

  * •

$A_{2}$. For all $u,\beta_{0}\in\mathcal{A}$ with $\beta_{0}(t,z)\leq
K<\infty$ for all $t,z$ define

$\delta(t,z)=\frac{1}{2K}dist(u(t,z),\partial\mathbb{U})\wedge 1>0$ (5.1)

and put

$\beta(t,z)=\delta(t,z)\beta_{0}(t,z).$ (5.2)

Then the control

$\widetilde{u}(t,x,z)=u(t,z)+a\beta(t,z);\quad t\in[0,T]$

belongs to $\mathcal{A}$ for all $a\in(-1,1)$.

  * •

$A3$. For all $\beta$ as in (5.2) the derivative process

$\chi(t,z):=\frac{d}{da}X^{u+a\beta}(t,z)|_{a=0}$ (5.3)

exists, and belong to $\mathbf{L}^{2}(\lambda\times\mathbf{P})$ and

$\displaystyle\chi(t,z)=\int^{t}_{0}\Big{(}\frac{\partial b}{\partial
x}(t,s)\chi(s,z)+\frac{\partial b}{\partial u}(t,s)\beta(s,z)\Big{)}ds$
$\displaystyle+\int^{t}_{0}\Big{(}\frac{\partial\sigma}{\partial
x}(t,s)\chi(s,z)+\frac{\partial\sigma}{\partial u}(t,s)\beta(s,z)\Big{)}dB(s)$
$\displaystyle+\int^{t}_{0}\int_{\mathbb{R}}\Big{(}\frac{\partial\gamma}{\partial
x}(t,s,\zeta)\chi(s,z)+\frac{\partial\gamma}{\partial
u}(t,s,\zeta)\beta(s,z)\Big{)}\tilde{N}(ds,d\zeta),$ (5.4)

and hence

$\displaystyle d\chi(t,z)=\Big{[}\frac{\partial b}{\partial
x}(t,t)\chi(t,z)+\frac{\partial b}{\partial
u}(t,t)\beta(t,z)+\int_{0}^{t}\Big{(}\frac{\partial^{2}b}{\partial t\partial
x}(t,s)\chi(s,z)+\frac{\partial^{2}b}{\partial t\partial
u}(t,s)\beta(s,z)\Big{)}ds$
$\displaystyle+\int_{0}^{t}\Big{(}\frac{\partial^{2}\sigma}{\partial t\partial
x}(t,s)\chi(s,z)+\frac{\partial^{2}\sigma}{\partial t\partial
u}(t,s)\beta(s,z)\Big{)}dB(s)$
$\displaystyle+\int_{0}^{t}\int_{\mathbb{R}}\Big{(}\frac{\partial^{2}\gamma}{\partial
t\partial x}(t,s,\zeta)\chi(s,z)+\frac{\partial^{2}\gamma}{\partial t\partial
u}(t,s,\zeta)\beta(s,z)\Big{)}\tilde{N}(ds,d\zeta)\Big{]}dt$
$\displaystyle+\Big{(}\frac{\partial\sigma}{\partial
x}(t,t)\chi(t,z)+\frac{\partial\sigma}{\partial u}(t,t)\beta(t,z)\Big{)}dB(t)$
$\displaystyle+\int_{\mathbb{R}}\Big{(}\frac{\partial\gamma}{\partial
x}(t,t)\chi(t,z)+\frac{\partial\gamma}{\partial
u}(t,t)\beta(t,z)\Big{)}\tilde{N}(dt,d\zeta)$ (5.5)

###### Theorem 5.1

[Necessary-type maximum principle]
Let $\hat{u}\in\mathcal{A}$ and $z\in\mathbb{R}$. Then the following are
equivalent:

  1. 1.

$\frac{d}{da}j(\hat{u}+a\beta)(z)|_{a=0}=0$ for all bounded
$\beta\in\mathcal{A}$ of the form (5.2).

  2. 2.

$E[\frac{\partial H}{\partial u}(t,z)|\mathcal{G}_{t}]_{u=\hat{u}}=0$ for all
$t\in[0,T].$

Proof. For simplicity of notation we write $u$ instead of $\hat{u}$ in the
following.
By considering an increasing sequence of stopping times $\tau_{n}$ converging
to $T$, we may assume that all local integrals appearing in the computations
below are martingales and have expectation 0. See [ØS2]. We omit the details.
We can write

$\frac{d}{da}j((u+a\beta)(z))|_{a=0}=I_{1}+I_{2}$

where

$I_{1}=\frac{d}{da}\mathbb{E}[\int_{0}^{T}f(t,X^{u+a\beta}(t,z),u(t,z)+a\beta(t,z),z)\mathbb{E}[\delta_{Z}(z)|\mathcal{F}_{t}]dt]|_{a=0}$

and

$I_{2}=\frac{d}{da}\mathbb{E}[g(X^{u+a\beta}(T,z),z)\mathbb{E}[\delta_{Z}(z)|\mathcal{F}_{T}]]|_{a=0}.$

By our assumptions on $f$ and $g$ and by (5.3) we have

$I_{1}=\mathbb{E}[\int_{0}^{T}\\{\frac{\partial f}{\partial
x}(t,z)\chi(t,z)+\frac{\partial f}{\partial
u}(t,z)\beta(t,z)\\}\mathbb{E}[\delta_{Z}(z)|\mathcal{F}_{t}]dt],$ (5.6)
$I_{2}=\mathbb{E}[\frac{\partial g}{\partial
x}(X(T,z),z)\chi(T,z)\mathbb{E}[\delta_{Z}(z)|\mathcal{F}_{T}]]=\mathbb{E}[p(T,z)\chi(T,z)].$
(5.7)

We have

$\displaystyle I_{1}=\mathbb{E}[\int_{0}^{T}\Big{(}\\{\frac{\partial
H_{0}}{\partial x}(t)-\frac{\partial b}{\partial
x}(t,t)p(t,z)-\frac{\partial\sigma}{\partial x}(t,t)q(t,z)$
$\displaystyle-\int_{\mathbb{R}}\frac{\partial\gamma}{\partial
x}(t,t,\zeta)r(t,z,\zeta)\nu(d\zeta)\\}X(t,z)+\\{\frac{\partial
H_{0}}{\partial u}-\frac{\partial b}{\partial
u}(t,t)p(t,z)-\frac{\partial\sigma}{\partial u}(t,t)q(t,z)$
$\displaystyle-\int_{\mathbb{R}}\frac{\partial\gamma}{\partial
u}(t,t,\zeta)r(t,z,\zeta)\nu(d\zeta)\\}\beta(t,z)\Big{)}dt]$ (5.8)

Since $H_{0}=H-H_{1}$ then

$\displaystyle I_{1}=\mathbb{E}[\int_{0}^{T}\Big{(}\\{\frac{\partial
H}{\partial x}(t)-\frac{\partial H_{1}}{\partial x}(t)-\frac{\partial
b}{\partial x}(t,t)p(t,z)-\frac{\partial\sigma}{\partial x}(t,t)q(t,z)$
$\displaystyle-\int_{\mathbb{R}}\frac{\partial\gamma}{\partial
x}(t,t,\zeta)r(t,z,\zeta)\nu(d\zeta)\\}\chi(t,z)+\\{\frac{\partial H}{\partial
u}(t)-\frac{\partial H_{1}}{\partial u}(t)-\frac{\partial b}{\partial
u}(t,t)p(t,z)-\frac{\partial\sigma}{\partial u}(t,t)q(t,z)$
$\displaystyle-\int_{\mathbb{R}}\frac{\partial\gamma}{\partial
u}(t,t,\zeta)r(t,z,\zeta)\nu(d\zeta)\\}\beta(t,z)\Big{)}dt]$
$\displaystyle=\mathbb{E}[\int_{0}^{T}\Big{(}\\{\frac{\partial H}{\partial
x}-\int_{t}^{T}\frac{\partial^{2}b}{\partial s\partial
x}(s,t)p(s,z)ds-\int_{t}^{T}\frac{\partial^{2}\sigma}{\partial s\partial
x}(s,t)E[D_{t}p(s,z)|\mathcal{F}_{t}]ds$
$\displaystyle-\int_{t}^{T}\int_{\mathbb{R}}\frac{\partial^{2}\gamma}{\partial
s\partial
x}(s,t)E[D_{t,\zeta}p(s,z)|\mathcal{F}_{t}]\nu(d\zeta)ds-\frac{\partial
b}{\partial x}(t,t)p(t,z)-\frac{\partial\sigma}{\partial x}(t,t)q(t,z)$
$\displaystyle-\int_{\mathbb{R}}\frac{\partial\gamma}{\partial
x}(t,t,\zeta)r(t,z,\zeta)\nu(d\zeta)\\}\chi(t,z)$
$\displaystyle+\\{\frac{\partial H}{\partial
u}(t)-\int_{t}^{T}\frac{\partial^{2}b}{\partial s\partial
u}(s,t)p(s,z)ds-\int_{t}^{T}\frac{\partial^{2}\sigma}{\partial s\partial
u}(s,t)E[D_{t}p(s,z)|\mathcal{F}_{t}]ds$
$\displaystyle-\int_{t}^{T}\int_{\mathbb{R}}\frac{\partial^{2}\gamma}{\partial
s\partial
u}(s,t)E[D_{t,\zeta}p(s,z)|\mathcal{F}_{t}]\nu(d\zeta)ds-\frac{\partial
b}{\partial u}(t,t)p(t,z)$ $\displaystyle-\frac{\partial\sigma}{\partial
u}(t,t)q(t,z)-\int_{\mathbb{R}}\frac{\partial\gamma}{\partial
u}(t,t,\zeta)r(t,z,\zeta)\nu(d\zeta)\\}\beta(t,z)\Big{)}dt]$ (5.9)

By the Itô formula

$\displaystyle I_{2}$
$\displaystyle=\mathbb{E}[p(T,z)\chi(T,z)]=\mathbb{E}[\int_{0}^{T}p(t,z)d\chi(t,z)+\int_{0}^{T}\chi(t,z)dp(t,z)$
$\displaystyle+\int_{0}^{T}d[\chi,p](t,z)]$
$\displaystyle=E\Big{[}\int_{0}^{T}p(t)(\frac{\partial b}{\partial
x}(t,t)\chi(t,z)+\frac{\partial b}{\partial u}(t,t)\beta(t,z))dt$
$\displaystyle+\int^{T}_{0}p(t)\Big{\\{}\int^{t}_{0}(\frac{\partial^{2}b}{\partial
t\partial x}(t,s)\chi(s,z)+\frac{\partial^{2}b}{\partial t\partial
u}(t,s)\beta(s,z))ds\Big{\\}}dt$
$\displaystyle+\int^{T}_{0}p(t)\Big{\\{}\int^{t}_{0}(\frac{\partial^{2}\sigma}{\partial
t\partial x}(t,s)\chi(s,z)+\frac{\partial^{2}\sigma}{\partial t\partial
u}(t,s)\beta(s,z))dB(s)\Big{\\}}dt$
$\displaystyle+\int^{T}_{0}p(t)\Big{\\{}\int^{t}_{0}(\frac{\partial^{2}\gamma}{\partial
t\partial x}(t,s,\zeta)\chi(s,z)+\frac{\partial^{2}\gamma}{\partial t\partial
x}(t,s,\zeta)\beta(s,z))\tilde{N}(ds,d\zeta)\Big{\\}}dt$
$\displaystyle-\int^{T}_{0}\chi(t,z)\frac{\partial H}{\partial
x}(t)dt+\int_{0}^{T}q(t,z)(\frac{\partial\sigma}{\partial
x}(t,t)\chi(t,z)+\frac{\partial\sigma}{\partial u}(t,t)\beta(t,z))dt$
$\displaystyle+\int^{T}_{0}\int_{\mathbb{R}}r(t,\zeta)(\frac{\partial\gamma}{\partial
x}(t,t,\zeta)\chi(t,z)+\frac{\partial\gamma}{\partial
u}(t,t,\zeta)\beta(t,z))\nu(d\zeta)dt\Big{]}.$ (5.10)

From (4.14), (4) and (4), we have

$\displaystyle E[p(T)Y(T)]$
$\displaystyle=E\Big{[}\int_{0}^{T}\Big{\\{}\frac{\partial b}{\partial
x}(t,t)p(t,z)+\int_{t}^{T}\Big{(}\frac{\partial^{2}b}{\partial s\partial
x}(s,t)p(s,z)+\frac{\partial^{2}\sigma}{\partial s\partial
x}(s,t)E[D_{t}p(s,z)|\mathcal{F}_{t}]$
$\displaystyle+\int_{\mathbb{R}}\frac{\partial^{2}\gamma}{\partial s\partial
x}(s,t,\zeta)E[D_{t,\zeta}p(s)|\mathcal{F}_{t}]\nu(d\zeta)\Big{)}ds\Big{\\}}\chi(t,z)dt$
$\displaystyle+\int_{0}^{T}\Big{\\{}\frac{\partial b}{\partial
u}(t,t)p(t,z)+\int_{t}^{T}\Big{(}\frac{\partial^{2}b}{\partial s\partial
u}(s,t)p(s,z)+\frac{\partial^{2}\sigma}{\partial s\partial
u}(s,t)E[D_{t}p(s,z)|\mathcal{F}_{t}]$
$\displaystyle+\int_{\mathbb{R}}\frac{\partial^{2}\gamma}{\partial s\partial
u}(s,t,\zeta)E[D_{t,\zeta}p(s)|\mathcal{F}_{t}]\nu(d\zeta)\Big{)}ds\Big{\\}}\beta(t,z)dt$
$\displaystyle-\int_{0}^{T}\frac{\partial H}{\partial
x}(t)X(t,z)dt+\int_{0}^{T}(\frac{\partial\sigma}{\partial
x}(t,t)\chi(t,z)+\frac{\partial\sigma}{\partial u}(t,t)\beta(t))q(t,z)dt$
$\displaystyle+\int_{0}^{T}\int_{\mathbb{R}}(\frac{\partial\gamma}{\partial
x}(t,t,\zeta)\chi(t,z)+\frac{\partial\sigma}{\partial
u}(t,t,\zeta)\beta(t,z))r(t,\zeta)\nu(d\zeta)dt\Big{]}.$ (5.11)

Summing (5) and (5) we get

$\frac{d}{da}j((u+a\beta)(.,z))|_{a=0}=I_{1}+I_{2}=\mathbb{E}[\int_{0}^{T}\frac{\partial
H}{\partial u}(t,z)\beta(t,z)dt].$

We conclude that

$\frac{d}{da}j(u+a\beta)(z))|_{a=0}=0$

if and only if

$\mathbb{E}[\int_{0}^{T}\frac{\partial H}{\partial u}(t,z)\beta(t,z)dt]=0,$
(5.12)

for all bounded $\beta\in\mathcal{A}$ of the form (5.2).

In particular, applying this to $\beta(t,z)=\theta(t,z)$ as in $A1$, we get
that this is again equivalent to

$\mathbb{E}[\alpha(z,\omega)\int_{t_{0}}^{T}\frac{\partial H}{\partial
u}(t,z)dt]=0,$ (5.13)

Differentiating the right-hand side of (5.13), we get

$\mathbb{E}[\alpha(z,\omega)\frac{\partial H}{\partial u}(t_{0},z)]=0$ (5.14)

Since this holds for all bounded $\mathcal{G}_{t_{0}}$-measurable $\alpha$, so
we deduce that

$\mathbb{E}[\frac{\partial H}{\partial
u}(t_{0},z)|\mathcal{F}_{t_{0}}]=0,\forall t_{0}\in[0,T]$ (5.15)

$\square$

##  6 Applications

###  6.1 The case when the coefficients do not depend on $x$

Consider the case when the coefficients do not depend on $x$, i.e., the system
has the form:

$\displaystyle
X(t,z)=\xi(t,z)+\int_{0}^{t}b(t,s,u(s,z),z)ds+\int_{0}^{t}\sigma(t,s,u(s,z),z)dB(s)+\int_{0}^{t}\int_{\mathbb{R}}\gamma(t,s,u(s,z),\zeta)\tilde{N}(ds,d\zeta)$
(6.1)

with performance functional

$J(u)=E[\int_{0}^{T}f(t,u(t))E[\delta_{Z}(z)|\mathcal{F}_{t}]dt+g(X(T))E[\delta_{Z}(z)|\mathcal{F}_{T}]].$
(6.2)

In this case the Hamiltonian $H$ given in (4.5) takes the form

$\displaystyle
H(t,z,u,p,q,r):=f(t,z,u)E[\delta_{Z}(z)|\mathcal{F}_{t}]+b(t,t,z,u)p+\sigma(t,t,z,u)q+\int_{\mathbb{R}}\gamma(t,t,z,u,\zeta)r(\zeta)\nu(d\zeta)$
$\displaystyle\int_{t}^{T}\frac{\partial b}{\partial
s}(s,t,z,u)p(s)ds+\int_{t}^{T}\frac{\partial\sigma}{\partial
s}(s,t,z,u)E[D_{t}p(s)|\mathcal{F}_{t}]ds+\int_{t}^{T}\int_{\mathbb{R}}\frac{\partial\gamma}{\partial
s}(s,t,z,u,\zeta)E[D_{t,\zeta}p(s)|\mathcal{F}_{t}]\nu(d\zeta)ds$ (6.3)

The BSDE (4.6) for the adjoint variables $p,q,r$ gets the form

$\begin{cases}dp(t,z)=q(t,z)dB(t)+\int_{\mathbb{R}}r(t,z,\zeta)\tilde{N}(dt,d\zeta),0\leq
t\leq T\\\ p(T,z)=g^{\prime}(X(T))E[\delta_{Z}(z)|\mathcal{F}_{T}]\end{cases}$
(6.4)

which has the solution

$\displaystyle
p(t,z)=E[g^{\prime}(X(T,z))E[\delta_{Z}(z)|\mathcal{F}_{T}]|\mathcal{F}_{t}]$
(6.5) $\displaystyle
q(t,z)=D_{t}p(t,z)=E[D_{t}(g^{\prime}(X(T,z))E[\delta_{Z}(z)|\mathcal{F}_{T}])|\mathcal{F}_{t}]$
(6.6) $\displaystyle
r(t,z,\zeta)=D_{t,\zeta}p(t,z)=E[D_{t,\zeta}(g^{\prime}(X(T,z))E[\delta_{Z}(z)|\mathcal{F}_{T}])|\mathcal{F}_{t}].$
(6.7)

Substituting (6.5)-(6.7) into (6.1) we get

$E[H(t,z,u,p(.),q(.),r(.))|\mathcal{F}_{t}]=E[H_{0}(t,z,u,p,q,r)|\mathcal{F}_{t}],$
(6.8)

where

$\displaystyle
H_{0}(t,z,u,p,q,r):=f(t,z,u)E[\delta_{Z}(z)|\mathcal{F}_{t}]+b(t,t,z,u)E[g^{\prime}(X(T,z))E[\delta_{Z}(z)|\mathcal{F}_{T}]|\mathcal{F}_{t}]$
$\displaystyle+\sigma(t,t,z,u)E[D_{t}(g^{\prime}(X(T,z))E[\delta_{Z}(z)|\mathcal{F}_{T}])|\mathcal{F}_{t}]$
$\displaystyle+\int_{\mathbb{R}}\gamma(t,t,z,u,\zeta)E[D_{t,\zeta}(g^{\prime}(X(T,z))E[\delta_{Z}(z)|\mathcal{F}_{T}])|\mathcal{F}_{t}]\nu(d\zeta)$
$\displaystyle+\int_{t}^{T}\frac{\partial b}{\partial
s}(s,t,z,u)g^{\prime}(X(T,z))E[\delta_{Z}(z)|\mathcal{F}_{T}]ds$
$\displaystyle+\int_{t}^{T}\frac{\partial\sigma}{\partial
s}(s,t,z,u)E[D_{t}\\{g^{\prime}(X(T,z))E[\delta_{Z}(z)|\mathcal{F}_{T}]\\}|\mathcal{F}_{t}]ds$
$\displaystyle+\int_{t}^{T}\int_{\mathbb{R}}\frac{\partial\gamma}{\partial
s}(s,t,z,u,\zeta)E[D_{t,\zeta}\\{g^{\prime}(X(T,z))E[\delta_{Z}(z)|\mathcal{F}_{T}]\\}|\mathcal{F}_{t}]\nu(d\zeta)ds$
(6.9)

Since we have

$\displaystyle\int_{t}^{T}\frac{\partial b}{\partial
s}(s,t,z,u)g^{\prime}(X(T,z))E[\delta_{Z}(z)|\mathcal{F}_{T}]ds$
$\displaystyle=\int_{t}^{T}\frac{\partial b}{\partial
s}(s,t,z,u)dsg^{\prime}(X(T,z))E[\delta_{Z}(z)|\mathcal{F}_{T}]$
$\displaystyle=(b(T,t,z,u)-b(t,t,z,u))g^{\prime}(X(T,z))E[\delta_{Z}(z)|\mathcal{F}_{T}],$
(6.10)

$\displaystyle\int_{t}^{T}\frac{\partial\sigma}{\partial
s}(s,t,z,u)E[D_{t}\\{g^{\prime}(X(T,z))E[\delta_{Z}(z)|\mathcal{F}_{T}]\\}|\mathcal{F}_{t}]ds$
$\displaystyle=\int_{t}^{T}\frac{\partial\sigma}{\partial
s}(s,t,z,u)dsE[D_{t}\\{g^{\prime}(X(T,z))E[\delta_{Z}(z)|\mathcal{F}_{T}]\\}|\mathcal{F}_{t}]$
$\displaystyle=(\sigma(T,t,z,u)-\sigma(T,t,z,u))E[D_{t}\\{g^{\prime}(X(T,z))E[\delta_{Z}(z)|\mathcal{F}_{T}]\\}$
(6.11)

and

$\displaystyle\int_{t}^{T}\int_{\mathbb{R}}\frac{\partial\gamma}{\partial
s}(s,t,z,u,\zeta)E[D_{t,\zeta}\\{g^{\prime}(X(T,z))E[\delta_{Z}(z)|\mathcal{F}_{T}]\\}|\mathcal{F}_{t}]\nu(d\zeta)ds$
$\displaystyle=\int_{\mathbb{R}}(\int_{t}^{T}\frac{\partial\gamma}{\partial
s}(s,t,z,u,\zeta)ds)E[D_{t,\zeta}\\{g^{\prime}(X(T,z))E[\delta_{Z}(z)|\mathcal{F}_{T}]\\}|\mathcal{F}_{t}]\nu(d\zeta)$
$\displaystyle=\int_{\mathbb{R}}(\gamma(T,t,z,u,\zeta)-\gamma(t,t,z,u,\zeta))E[D_{t,\zeta}\\{g^{\prime}(X(T,z))E[\delta_{Z}(z)|\mathcal{F}_{T}]\\}|\mathcal{F}_{t}]\nu(d\zeta).$
(6.12)

Then $H_{0}(t,v,p,q,r)$ is reduced to

$\displaystyle\mathbb{H}_{0}(t,z,u,X(T,z))=f(t,z,u)E[\delta_{Z}(z)|\mathcal{F}_{t}]+b(T,t,z,u)E[g^{\prime}(X(T,z))E[\delta_{Z}(z)|\mathcal{F}_{T}]|\mathcal{F}_{t}]$
$\displaystyle+\sigma(T,t,z,u)E[D_{t}(g^{\prime}(X(T,z))E[\delta_{Z}(z)|\mathcal{F}_{T}])|\mathcal{F}_{t}]$
$\displaystyle+\int_{\mathbb{R}}\gamma(T,t,z,u,\zeta)E[D_{t,\zeta}(g^{\prime}(X(T,z))E[\delta_{Z}(z)|\mathcal{F}_{T}])|\mathcal{F}_{t}]\nu(d\zeta)$
(6.13)

We conclude that, in this case, we have the following maximum principles:

###### Theorem 6.1

(Sufficient maximum principle)
Suppose that the coefficients $f(t,z,u),b(t,s,z,u),\sigma(t,s,z,u)$ and
$\gamma(t,s,z,u,\zeta)$ of the stochastic control system (6.1)-(6.2) do not
depend on $x$. Let $\hat{u}\in\mathcal{A}$ with associated solution $\hat{X}$
of (6.1). Suppose that the functions

$x\rightarrow g(x)$ (6.14)

and

$u\rightarrow\mathbb{H}_{0}(t,z,u,\hat{X}(T,z))$ (6.15)

are concave and that, for all $t$,

$\max_{u\in\mathbb{U}}E[\mathbb{H}_{0}(t,z,u,\hat{X}(T,z))|\mathcal{G}_{t}]=E[\mathbb{H}_{0}(t,\hat{u}(t,z),\hat{X}(T,z))|\mathcal{G}_{t}].$
(6.16)

Then, $\hat{u}$ is an optimal control, i.e.,

$\sup_{u\in A}J(u)=J(\hat{u}).$ (6.17)

###### Theorem 6.2

(Necessary maximum principle) Let $X(t,z)$ and $J(u)$ be as in Theorem 6.1.
Let $\hat{u}\in\mathcal{A}$ with associated solution $\hat{X}$ of (6.1). Then
the following, (i) and (ii), are equivalent:
(i)$\frac{d}{da}J(\hat{u}+a\beta)|_{a=0}=0$ for all processes
$\beta\in\mathcal{A}$ of the form (5.2).
(ii)$E[\frac{\partial\mathbb{H}_{0}}{\partial
u}(t,u,\hat{X}(T,z))|\mathcal{G}_{t}]|_{u=\hat{u}(t,z)}=0$

##  7 Example: Optimal insider portfolio in a financial market modeled by a
Volterra equation

In this example, we choose $\mathbb{G}=\mathbb{F}$. Consider a financial
market where the unit price $S_{0}(t)$ of the risk free asset is

$S_{0}(t)=1,t\in[0,T]$ (7.1)

and the unit price process $S(t)$ of the risky asset has no jumps and is given
by

$\begin{cases}dS(t)=b_{0}(t,t,Z)S(t)dt+\sigma_{0}(t,t,Z)S(t)dB(t)\\\
+(\int_{0}^{t}\frac{\partial b_{0}}{\partial
t}(t,s)S(s)ds+\int_{0}^{t}\frac{\partial\sigma_{0}}{\partial
t}(t,s)S(s)dB(s))dt;t\in[0,T]\\\ S(0)>0\end{cases}$ (7.2)

Then the wealth process $X(t)=X^{\Pi}(t)$ associated to a portfolio
$u(t)=\Pi(t)$, interpreted as the fraction of the wealth invested in the risky
asset at time $t$, is described by the linear stochastic Volterra equation

$X(t)=x_{0}+\int_{0}^{t}b_{0}(t,s,Z)\Pi(s)X(s)ds+\int_{0}^{t}\sigma_{0}(t,s,Z)\Pi(s)X(s)dB(s);t\geq
0$ (7.3)

or, in differential form

$\begin{cases}dX(t)=b_{0}(t,t,Z)\Pi(t)X(t)dt+\sigma_{0}(t,t,Z)\Pi(t)X(t)dB(t)\\\
+(\int_{0}^{t}\frac{\partial b_{0}}{\partial
t}(t,s)\Pi(s)X(s)ds+\int_{0}^{t}\frac{\partial\sigma_{0}}{\partial
t}(t,s)\Pi(s)X(s)dB(s))dt;t\in[0,T]\\\ X(0)=x_{0}>0\end{cases}$ (7.4)

Let $U$ be a given utility function. We want to find $\Pi\in\mathcal{A}$ such
that

$J(\Pi^{*})=\sup_{\Pi\in\mathcal{A}}J(\Pi),$ (7.5)

where

$J(\Pi):=E[U(X^{\Pi}(T))].$ (7.6)

Note that, in terms of our process $X(t,z)$ we have

$\begin{cases}dX(t,z)=b_{0}(t,t,z)\Pi(t)X(t,z)dt+\sigma_{0}(t,t,z)\pi(t,z)X(t,z)dB(t)\\\
+(\int_{0}^{t}\frac{\partial b_{0}}{\partial
t}(t,s,z)\pi(s,z)X(s,z)ds+\int_{0}^{t}\frac{\partial\sigma_{0}}{\partial
t}(t,s,z)\pi(s,z)X(s,z)dB(s))dt;t\in[0,T]\\\ X(0,z)=x_{0}>0\end{cases}$ (7.7)

or in the integral form:

$X(t,z)=x_{0}+\int_{0}^{t}b_{0}(t,s,z)\pi(s,z)X(s,z)ds+\int_{0}^{t}\sigma_{0}(t,s,z)\pi(s,z)X(s,z)dB(s);t\in[0,T]$
(7.8)

We assume that $b_{0}(t,s,z)=b_{0}(t,s,z,\omega)$ and
$\sigma_{0}(t,s,z)=\sigma_{0}(t,s,z,\omega)$ are given bounded processes, and
that $b_{0}(t,s)$ and $\sigma_{0}(t,s)$ are $\mathcal{F}_{s}$-measurable for
all $s,t$ and $C^{1}$ with respect to $t$ for all $s$, a.s. We also assume
that

$\sigma_{0}(t,s,z)\geq c_{0}\text{ a.s. for all t, }s\in[0,T]\text{ for some
constant }c_{0}>0.$ (7.9)

We assume that $x_{0}>0$. If $\pi\in\mathcal{A}$, then it follows that
$X(t,z)>0$ for all $t\in[0,T]$. To see this, note that from (7.7) we get

$\displaystyle
X(t,z)=x_{0}\exp(\int_{0}^{t}\sigma_{0}(s,s,z)\pi(s,z)dB(s)+\int_{0}^{t}b_{0}(s,s,z)\pi(s,z)-\frac{1}{2}\sigma_{0}^{2}(s,s,z)\pi^{2}(s,z)+\alpha(s,z)ds)>0,$
(7.10)

where

$\displaystyle\alpha(s,z):=\int_{0}^{s}\frac{\partial b_{0}}{\partial
s}(s,r,z)\pi(r,z)X(r,z)dr+\int_{0}^{s}\frac{\partial\sigma_{0}}{\partial
s}(s,r,z)\pi(r,z)X(r,z)dB(r).$ (7.11)

The performance functional gets the form

$J(\pi)=E[U(X(T,z))E[\delta_{Z}(z)|\mathcal{F}_{T}]]$ (7.12)

This is a problem of the type investigated in the previous sections (in the
special case with no jumps) and we can apply the results there to solve it, as
follows: The Hamiltonian gets the form, with $u=\pi$,

$\displaystyle H(t,x,z,\pi,p,q)=b_{0}(t,t,z)\pi xp+\sigma_{0}(t,t,z)\pi xq$
$\displaystyle+\int_{t}^{T}\frac{\partial b_{0}}{\partial s}(s,t,z)\pi
xp(s,z)ds+\int_{t}^{T}\frac{\partial\sigma_{0}}{\partial s}(s,t,z)\pi
xE[D_{t}p(s,z)|\mathcal{F}_{t}]ds$ (7.13)

while the BSDE of the adjoint processes becomes

$\begin{cases}dp(t,z)=[b_{0}(t,t,z)\pi(t,z)p(t,z)+\sigma_{0}(t,t)\pi(t,z)q(t,z)\\\
+\int_{t}^{T}\frac{\partial b_{0}}{\partial
s}(s,t,z)\pi(s,z)p(s,z)ds+\int_{t}^{T}\frac{\partial\sigma_{0}}{\partial
s}(s,t,z)\pi(s,z)E[D_{t}p(s)|\mathcal{F}_{t}]ds]dt\\\
+q(t,z)dB(t);t\in[0,T]\\\
p(T,z)=U^{\prime}(X(T,z))E[\delta_{Z}(z)|\mathcal{F}_{T}]\end{cases}$ (7.14)

Suppose there exists an optimal control $\pi\in\mathcal{A}$ for (5.18) with
corresponding $\hat{X},\hat{p},\hat{q}$. Then,

$E[\frac{\partial
H}{\partial\pi}(t,\hat{X}(t,z),\pi,\hat{p},\hat{q})|\mathcal{F}_{t}]_{\pi=\hat{\pi}(t)}=0$
(7.15)

i.e.,

$\displaystyle
E[b_{0}(t,t,z)\hat{X}(t,z)\hat{p}(t,z)+\sigma_{0}(t,t,z)\hat{X}(t)\hat{q}(t,z)$
$\displaystyle+\int_{t}^{T}\frac{\partial b_{0}}{\partial
s}(s,t,z)\hat{X}(t,z)\hat{p}(s,z)ds+\int_{t}^{T}\frac{\partial\sigma_{0}}{\partial
s}(s,t,z)\hat{X}(t,z)E[D_{t}\hat{p}(s,z)|\mathcal{F}_{t}]ds|\mathcal{F}_{t}]=0.$
(7.16)

Since $\hat{X}(t)>0$, this is equivalent to

$\displaystyle b_{0}(t,t,z)\hat{p}(t)+\sigma_{0}(t,t,z)\hat{q}(t)$
$\displaystyle+E[\int_{t}^{T}\frac{\partial b_{0}}{\partial
s}(s,t,z)\hat{p}(s)ds+\int_{t}^{T}\frac{\partial\sigma_{0}}{\partial
s}(s,t,z)E[D_{t}\hat{p}(s)|\mathcal{F}_{t}]ds|\mathcal{F}_{t}]=0.$ (7.17)

We deduce that the corresponding BSDE (7.14) reduces to

$\begin{cases}d\hat{p}(t,z)=\hat{q}(t,z)dB(t);0\leq t\leq T\\\
\hat{p}(T,z)=U^{\prime}(\hat{X}(T,z))E[\delta_{Z}(z)|\mathcal{F}_{T}],\end{cases}$
(7.18)

which has the unique solution

$\hat{p}(t,z)=E[U^{\prime}(\hat{X}(T,z))E[\delta_{Z}(z)|\mathcal{F}_{T}]|\mathcal{F}_{t}],\quad\hat{q}(t,z)=D_{t}\hat{p}(t,z).$
(7.19)

Substituted (7.19) into (7), this gives the equation

$\displaystyle
E[b_{0}(t,t,z)U^{\prime}(\hat{X}(T,z))E[\delta_{Z}(z)|\mathcal{F}_{T}]+\sigma_{0}(t,t,z)D_{t}\\{U^{\prime}(\hat{X}(T,z))E[\delta_{Z}(z)|\mathcal{F}_{T}]\\}$
$\displaystyle+\int_{t}^{T}\frac{\partial b_{0}}{\partial
s}(s,t,z)E[U^{\prime}(\hat{X}(T,z))E[\delta_{Z}(z)|\mathcal{F}_{T}]|\mathcal{F}_{s}]ds$
$\displaystyle+\int_{t}^{T}\frac{\partial\sigma_{0}}{\partial
s}(s,t,z)E[D_{t}\\{U^{\prime}(\hat{X}(T,z))E[\delta_{Z}(z)|\mathcal{F}_{T}]\\}|\mathcal{F}_{s}]ds|\mathcal{F}_{t}]=0,$
(7.20)

where we have used that

$\displaystyle
D_{t}E[U^{\prime}(\hat{X}(T,z))E[\delta_{Z}(z)|\mathcal{F}_{T}]|\mathcal{F}_{t}]=E[D_{t}\\{U^{\prime}(\hat{X}(T,z))E[\delta_{Z}(z)|\mathcal{F}_{T}]\\}|\mathcal{F}_{t}],$
(7.21)

Equation (7) can be simplified to

$\displaystyle
b_{0}(t,t,z)E[U^{\prime}(\hat{X}(T,z))E[\delta_{Z}(z)|\mathcal{F}_{T}]|\mathcal{F}_{t}]+\sigma_{0}(t,t,z)E[D_{t}\\{U^{\prime}(\hat{X}(T,z))E[\delta_{Z}(z)|\mathcal{F}_{T}]\\}|\mathcal{F}_{t}]$
$\displaystyle+E[\int_{t}^{T}\frac{\partial b_{0}}{\partial
s}(s,t,z)U^{\prime}(\hat{X}(T,z))E[\delta_{Z}(z)|\mathcal{F}_{T}]ds|\mathcal{F}_{t}]$
$\displaystyle+E[\int_{t}^{T}\frac{\partial\sigma_{0}}{\partial
s}(s,t,z)D_{t}\\{U^{\prime}(\hat{X}(T,z))E[\delta_{Z}(z)|\mathcal{F}_{T}]\\}ds|\mathcal{F}t]=0,$
(7.22)

or

$\sigma_{0}(T,t,z)D_{t}E[U^{\prime}(\hat{X}(T,z))E[\delta_{Z}(z)|\mathcal{F}_{T}]|\mathcal{F}_{t}]+b_{0}(T,t,z)E[U^{\prime}(\hat{X}(T,z))E[\delta_{Z}(z)|\mathcal{F}_{T}]|\mathcal{F}_{t}]=0.$
(7.23)

By (7.9) we see that (7.23) can be written

$\frac{D_{t}Y(t)}{Y(t)}=-\frac{b_{0}(T,t,z)}{\sigma_{0}(T,t,z)},$ (7.24)

where

$Y(t,z)=E[U^{\prime}(\hat{X}(T,z))E[\delta_{Z}(z)|\mathcal{F}_{T}]|\mathcal{F}_{t}].$
(7.25)

By the chain rule for Malliavin derivatives, we deduce from that

$D_{t}(lnY(t,z))=-\frac{b_{0}(T,t,z)}{\sigma_{0}(T,t,z)}.$ (7.26)

On the other hand, since $Y(t,z)$ is a positive martingale, there exists an
adapted process $\theta_{0}(t,z)$ such that

$dY(t,z)=\theta_{0}(t,z)Y(t,z)dB(t)$ (7.27)

i.e.,

$Y(t,z)=Y(0,z)\exp(\int_{0}^{t}\theta_{0}(s,z)dB(s)-\frac{1}{2}\int_{0}^{t}\theta_{0}^{2}(s,z)ds).:$
(7.28)

From (7.28) we get

$D_{t}(\ln
Y(t,z))=D_{t}(\int_{0}^{t}\theta_{0}(s,z)dB(s)-\frac{1}{2}\int_{0}^{t}\theta_{0}^{2}(s,z)ds)=\theta_{0}(t,z),$
(7.29)

Comparing (7.26) and (7.29) we conclude that

$\theta_{0}(t,z)=-\frac{b_{0}(T,t,z)}{\sigma_{0}(T,t,z)}$ (7.30)

and hence, by (7.25),

$\displaystyle
E[U^{\prime}(\hat{X}(T,z))E[\delta_{Z}(z)|\mathcal{F}_{T}]|\mathcal{F}_{t}]=Y(t,z)$
$\displaystyle=E[U^{\prime}(\hat{X}(T,z))E[\delta_{Z}(z)|\mathcal{F}_{T}]]\exp(\int_{0}^{t}\theta_{0}(s,z)dB(s)-\frac{1}{2}\int_{0}^{t}\theta_{0}^{2}(s,z)ds).$
(7.31)

It remains to find the constant

$c=E[U^{\prime}(\hat{X}(T,z))E[\delta_{Z}(z)|\mathcal{F}_{T}]].$ (7.32)

From (7) with $t=T$ we get

$U^{\prime}(\hat{X}(T,z))E[\delta_{Z}(z)|\mathcal{F}_{T}]=c\exp(\int_{0}^{T}\theta_{0}(s,z)dB(s)-\frac{1}{2}\int_{0}^{T}\theta_{0}^{2}(s,z)ds).$
(7.33)

To make this more explicit, we proceed as follows: Define

$M(t,z):=E[\delta_{Z}(z)|\mathcal{F}_{t}]$ (7.34)

Then by the generalized Clark-Ocone theorem

$\begin{cases}dM(t,z)=E[D_{t}\delta_{Z}(z)|\mathcal{F}_{t}]dB(t)=\Phi(t,z)M(t,z)dB(t)\\\
M(0,z)=1\end{cases}$ (7.35)

where

$\displaystyle\Phi(t,z)=\frac{E[D_{t}\delta_{Z}(z)|\mathcal{F}_{t}]}{E[\delta_{Z}(z)|\mathcal{F}_{t}]}.$
(7.36)

Solving this SDE for $M(t,z)$ we get

$M(t,z)=\exp(\int_{0}^{t}\Phi(s,z)dB(s)-\frac{1}{2}\int_{0}^{t}\Phi^{2}(s,z)ds).$
(7.37)

Substituting this into (7.33) we get

$U^{\prime}(\hat{X}(T,z))=c\exp(\int_{0}^{T}(\theta_{0}(s,z)-\Phi(s,z))dB(s)-\frac{1}{2}\int_{0}^{T}(\theta_{0}^{2}(s,z)-\Phi^{2}(s,z))ds).$
(7.38)

Then we can deduce that

$\hat{X}(T,z)=(U^{\prime})^{-1}(c\exp(\int_{0}^{T}(\theta_{0}(s,z)-\Phi(s,z))dB(s)-\frac{1}{2}\int_{0}^{T}(\theta_{0}^{2}(s,z)-\Phi^{2}(s,z))ds))=F(c).$
(7.39)

On the other hand, if we define

$\hat{K}(t,s,z):=\sigma_{0}(t,s,z)\hat{\pi}(s,z)\hat{X}(s,z),$ (7.40)

then by (7.8), the pair $(\hat{X},\hat{K})$ solves the following (Yong type)
backward stochastic Volterra integral equation (BSVIE)

$\hat{X}(t,z)=F(c)-\int_{t}^{T}\frac{b_{0}(t,s,z)}{\sigma_{0}(t,s,z)}\hat{K}(t,s,z)ds-\int_{t}^{T}\hat{K}(t,s,z)dB(s);0\leq
s\leq T.$ (7.41)

By Theorem 3.2 in [20] the solution of this equation is unique. Putting $t=0$
and taking expectation in (7.41), we get

$x_{0}=E[F(c)]-\int_{0}^{T}E[\frac{b_{0}(t,s,z)}{\sigma_{0}(t,s,z)}\hat{K}(t,s,z)]ds.$
(7.42)

This equation determines implicitly the value of $c$. Hence by (7.39) we have
found the optimal terminal wealth $\hat{X}(T,z)$. Then, finally we obtain the
optimal portfolio $\hat{\pi}$ by (7.40). Conversely, since the functions
$x\rightarrow U(x)$ and $(x,\pi)\rightarrow H(t,x,\pi,\hat{p},\hat{q})$ are
concave, we see that $\hat{\pi}$ found above satisfies the conditions of
Theorem 4.1, and hence $\hat{\pi}$ is indeed optimal. We summarize what we
have proved as follows:

###### Theorem 7.1

Assume that $\sigma_{0}(t,s)>0$ is bounded away from 0, for $s,t\in[0,T]$.
Then, the optimal portfolio $\hat{\Pi}(t)$ for the problem (7.6) is

$\hat{\Pi}(t,Z)=\hat{\pi}(t,z)_{z=Z}=\int_{\mathbb{R}}\pi(z)\delta_{Z}(z)dz$
(7.43)

where

$\hat{\pi}(s,z)=\frac{\hat{K}(t,s,z)}{\sigma_{0}(t,s,z)\hat{X}(s,z)};s\in[0,T],$
(7.44)

with $(\hat{X},\hat{K})$ is the unique solution of the BSVIE (7.41) with $F$
defined by (7.39), and the constant $c$ is the solution of (7.42).

Acknowledgment
We are grateful to Bernt Øksendal for helpful comments.

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