# Macroscopic Market Making

Ivan Guo  Ivan Guo, Centre for Quantitative Finance and Investment Strategies,
School of Mathematics, Monash University<EMAIL_ADDRESS>,  Shijia Jin
Shijia Jin, School of Mathematics, Monash University<EMAIL_ADDRESS>and  Kihun Nam  Kihun Nam, Centre for Quantitative Finance and Investment
Strategies, School of Mathematics, Monash University<EMAIL_ADDRESS>
###### Abstract.

We propose a macroscopic market making model à la Avellaneda-Stoikov, using
continuous processes for orders instead of discrete point processes. The model
intends to bridge the gap between market making and optimal execution
problems, while shedding light on the influence of order flows on the optimal
strategies. We demonstrate our model through three problems. The study
provides a comprehensive analysis from Markovian to non-Markovian noises and
from linear to non-linear intensity functions, encompassing both bounded and
unbounded coefficients. Mathematically, the contribution lies in the existence
and uniqueness of the optimal control, guaranteed by the well-posedness of the
strong solution to the Hamilton-Jacobi-Bellman equation and the
(non-)Lipschitz forward-backward stochastic differential equation.

Keywords: Market making, Stochastic optimal control, Forward-backward
stochastic differential equation

###### Contents

  1. 1 Introduction
  2. 2 Markovian Order Flows with Linear Intensity Function
  3. 3 Non-Markovian Order Flows: General Intensity Functions
  4. 4 Non-Markovian Order Flows: Unbounded Coefficients
  5. 5 Appendix: Stochastic Maximum Principle

##  1\. Introduction

In this paper, we focus on the strategic behaviour of the market maker, which
is defined as a liquidity provider in the financial market. More specifically,
the market maker provides bid and ask prices on one or several assets, and
makes profit by the bid–ask spread (the price difference between buy and sell
orders). Pioneered by the work of [13] and subsequently [3], market making—as
a stochastic control problem—has been the subject of extensive literature in
market microstructure. We propose a macroscopic model à la Avellaneda-Stoikov
[3] in view of two major motivations.

Firstly, while the market making problem focuses on the provision of
liquidity, the other side of the coin studies how liquidity can be consumed in
an optimal fashion. This is known as the optimal execution problem, set in
motion by [6] and [2]. We refer to [11] and [8] for excellent overviews of
both topics. Although market making and optimal execution problems are highly
correlated, they have been studied separately in two streams of literature.
One of the main reasons is the difference in the underlying order
characteristics: the total volume of (market) orders in the optimal execution
is absolutely continuous with respect to time, allowing for the consideration
of trading rates; However, in the market making problem, orders follow
prescribed point processes. To bridge the gap between these two topics, we
study a market making problem similar to the framework in [3], but replace
discrete point processes by continuous processes, aligning with the order rate
concept prevalent in the optimal execution literature. In practical terms, we
expand the time horizon of the market making from seconds, as traditionally
approached, to minutes or hours (see the introduction section in [12]). This
expanded horizon enables us to safely model orders using continuous processes,
resulting in what we refer to as a macroscopic model.

Secondly, our macroscopic model is expected to address several deficiencies of
the original model, while offering new insights simultaneously. Pointed out by
[14], in certain liquid stocks, the execution probability for limit orders
located more than one tick away from the best quote is significantly low in
short time scales. Under the expanded time horizon, we anticipate a higher and
hence more meaningful execution probability. The assumption that all market
orders are of the same size is no longer needed in the macroscopic model, and
our order dynamics are allowed to be non-Markovian. Furthermore, akin to the
fact that investments are made based on the belief of future prices, the
market maker should also keep an eye on the order dynamics to devise
appropriate strategies. This is made possible in our setting. Additionally,
the solution of the optimal control problem enables the observation of quotes
price changes corresponding to the order flows. This can offer a market-making
explanation to the concave price impact—see [16] for a microstructural
explanation.

The goal of this paper is to propose the macroscopic market making problems as
stochastic control problems, and to study the associated mathematical
issues—the existence and uniqueness of the optimal control. The study on each
problem is conducted in three steps: we introduce the market making setup,
providing a clear description of the stochastic control problem; the
stochastic control problem is then transformed to either a partial
differential equation (PDE) or a forward-backward stochastic differential
equation (FBSDE), depending on the specific context; we finally establish the
well-posedness of the equation, hence ensuring the existence and uniqueness of
the optimal control.

We start with the case when noises are Markovian, and the intensity
function—which is used to model the execution probability—is linear. Applying
the dynamic programming principle, the resulting PDE exhibits similarities to
the main PDEs studied in [4], [9] and [19], all of which investigate the
optimal execution problems with stochastic price impact. While [4], [9] study
the approximation solutions and [19] deduces the well-posedness of viscosity
solutions, our contribution lies in the existence and uniqueness of classical
solutions. Besides the connection with optimal execution problem, we also
examine the case of linear intensity function due to the applicability of the
technique employed in proving the well-posedness, which will be extended to
the next problem. We conclude the first problem by presenting a simple example
with an explicit solution, demonstrating how order flows are incorporated into
the market maker’s strategy.

Since order flows are known to be non-Markovian, the second problem adopts
non-Markovian noises and also more general non-linear intensity functions
introduced in [10]. By utilizing a truncation function and a version of the
stochastic maximum principle, we obtain a Lipschitz FBSDE, the well-posedness
of which is proved via the decoupling method pioneered by [15]. Furthermore,
the solution of the Lipschitz FBSDE is used to verify the well-posedness of a
non-Lipschitz FBSDE that corresponds to the case when the truncation is not
applied.

Moving beyond bounded noises and coefficients considered in the previous two
problems, the third and final problem explores unbounded coefficients, along
with non-Markovian noises and the linear intensity function. By leveraging a
convex-analytic approach, we obtain a non-Lipschitz FBSDE and establish its
well-posedness by constructing the solution using a monotonic sequence. By
organizing the problems in this manner, we gradually extend the analysis from
Markovian to non-Markovian noises, and from linear to non-linear intensity
functions, encompassing situations with both bounded and unbounded
coefficients.

The organization of the article aligns with the three problems outlined in
preceding paragraphs: Section 2 delves into the first problem, Section 3
explores the second problem, and Section 4 focuses on the third problem.
Appendix 5 encompasses a version of the stochastic maximum principle.

Notation: Throughout the present work, we fix $T>0$ to represent our finite
trading horizon. We denote by
$(\Omega,\mathcal{F},\mathbb{F}=(\mathcal{F}_{t})_{0\leq t\leq T},\mathbb{P})$
a complete filtered probability space, with $\mathcal{F}_{T}=\mathcal{F}$. An
$m$-dimensional Brownian motion $W=(W^{1},\dots,W^{m})$ is defined on such
space, for a fixed positive integer $m$, and the filtration $\mathbb{F}$ is
generated by $W$ and augmented. Let $\mathcal{G}$ represents an arbitrary
$\sigma$-algebra contained in $\mathcal{F}$ and consider the following spaces:

$\displaystyle L^{p}(\Omega,\mathcal{G}):=\big{\\{}X:X\text{ is
}\mathcal{G}\text{-measurable and }\mathbb{E}|X|^{p}<\infty\big{\\}};$
$\displaystyle\mathbb{H}^{p}:=\Big{\\{}X:X\text{ is
}\mathbb{F}\text{-progressively measurable and
}\mathbb{E}\Big{[}\big{(}\int_{0}^{T}|X_{t}|^{2}\,dt\big{)}^{p/2}\Big{]}<\infty\,\Big{\\}};$
$\displaystyle\mathbb{S}^{p}:=\Big{\\{}X\in\mathbb{H}^{p}:\mathbb{E}\big{[}\sup_{0\leq
t\leq T}|X_{t}|^{p}\big{]}<\infty\Big{\\}};$
$\displaystyle\mathcal{M}:=\big{\\{}M:M_{t}\in
L^{2}(\Omega,\mathcal{F}_{t})\text{ for a.e. }t\in[0,T]\text{ and
}\\{M_{t},\mathcal{F}_{t}\\}_{0\leq t\leq T}\text{ is a martingale}\big{\\}}.$

##  2\. Markovian Order Flows with Linear Intensity Function

This section is devoted to introducing the macroscopic market making as a
stochastic control problem. We will begin with the linear intensity function
and Markovian order flows, solving the problem by the dynamic programming
principle. The reason is twofold: it reveals several interesting connections
with optimal execution problems under stochastic price impact; the same
technique used in the problem will be also applied in the next section, where
we will look at the non-Markovian context.

Let $S=(S_{t})_{0\leq t\leq T}$ be the mid-price of the asset being
continuously traded throughout the time horizon $T$. We follow the framework
in [5] and consider the following stochastic differential equation for
$L:=(L_{t})_{t\in[0,T]}$ in $\mathbb{R}^{d}$:

$\displaystyle L_{0}$ $\displaystyle=l_{0}\in\mathbb{R}^{d},$ (1)
$\displaystyle dL_{t}$
$\displaystyle=\Gamma(t,L_{t})\,dt+\sum_{j=1}^{m}\Sigma_{j}(t,L_{t})\,dW_{t}^{j},\hskip
28.45274ptt\in[0,T],$

for continuous functions $\Gamma:[0,T]\times\mathbb{R}^{d}\to\mathbb{R}^{d}$
and $\Sigma_{j}:[0,T]\times\mathbb{R}^{d}\to\mathbb{R}^{d}$, $j=1,\dots,m$.
Define the matrix-valued function
$\Sigma:[0,T]\times\mathbb{R}^{d}\to\mathbb{R}^{d\times m}$ by
$\Sigma^{i,j}:=(\Sigma_{j})^{i}$. The process $L$ serves as the underlying
noise. We introduce the following assumptions.

######  Assumption 2.1.

(1) The mid-price process $S$ belongs to $\mathcal{M}$. Hence, for any
$K\in\mathbb{H}^{2}$, it holds

$\mathbb{E}\int_{0}^{T}K_{t}\,dS_{t}=0.$ (2)

(2) For $G\in\\{\Gamma,\Sigma_{1},\dots,\Sigma_{m}\\}$, there exists a
constant $C_{L}>0$ such that

$|G(t,x)-G(t,y)|\leq C_{L}|x-y|,$

for all $t\in[0,T]$ and $x,y\in\mathbb{R}^{d}$. Further,
$\det(\Sigma(t,l)\,\Sigma^{tr}(t,l))\neq 0$ for all
$(t,l)\in[0,T]\times\mathbb{R}^{d}$.

######  Remark 2.2.

The first assumption is common in optimal execution and market making
literatures, where the mid-price $S$ usually has no effect on the optimization
problem. In this case, the emphasis is placed on how the market maker deals
with the order flows. The last condition on $\Sigma$ is added for technical
reasons.

A market maker keeps trading such an asset to profit from the bid-ask spread
by posting bid and ask quotes during the time interval $[0,T]$. More
precisely, she posts the ask orders and bid orders at the price levels

$S_{t}^{a}:=S_{t}+\delta_{t}^{a}\text{\quad and
\quad}S_{t}^{b}:=S_{t}-\delta_{t}^{b}$ (3)

respectively, where
$\boldsymbol{\delta}:=(\delta^{a},\delta^{b})\in\mathbb{H}^{2}\times\mathbb{H}^{2}$
represents the control. While market orders are described by point processes
in the discrete setting, here we regard them as flows—continuous processes
adapted to the Brownian filtration $\mathbb{F}$. The market order flows at the
ask and bid side are modelled by

$a_{t}:=a(L_{t})\text{\quad and \quad}b_{t}:=b(L_{t})$ (4)

accordingly, for some bounded positive functions
$a,b:\mathbb{R}^{d}\to\mathbb{R}_{+}$. Denote by $\bar{a},\bar{b}>0$ the upper
bounds for functions $a$ and $b$ accordingly. On a macroscopic scale, the
execution probability in the original Avellaneda-Stoikov framework is now
interpreted as the portion of order flows captured by the agent. More
specifically, suppose that execution probability is $p\in(0,1)$ in the
discrete setting, implying the agent has a probability of $p$ to execute each
market order. Then, given a sufficient amount of orders, one expects that $p$
portion of total orders is matched by the agent on average. Since such
probability is determined by the intensity function $\Lambda$, we first look
at a linear function

$\Lambda(\delta)=\zeta-\gamma\,\delta,$ (5)

where $\zeta,\gamma>0$ represents the competition level among market makers
and we also assume the bid-ask symmetry. Intuitively, as the agent posts
further away from the mid-price, the chance of executions decreases in a
linear fashion. While the action $\delta$ makes the most sense in
$[0,\zeta/\gamma]$, we do not restrict the action to this closed interval,
which just requires a small modification overall. Consequently, the inventory
and cash processes of the market maker now follow

$\displaystyle
Q_{t}=q_{0}-\int_{0}^{t}a_{u}\cdot(\zeta-\gamma\delta^{a}_{u})\,du+\int_{0}^{t}b_{u}\cdot(\zeta-\gamma\delta^{b}_{u})\,du,$
$\displaystyle
X_{t}=x_{0}+\int_{0}^{t}a_{u}\cdot(\zeta-\gamma\delta^{a}_{u})\,(S_{u}+\delta_{u}^{a})\,du-\int_{0}^{t}b_{u}\cdot(\zeta-\gamma\delta^{b}_{u})\,(S_{u}-\delta_{u}^{b})\,du.$

The term $a_{u}(\zeta-\gamma\delta^{a}_{u})$ shows the (passive) selling rate
of the agent, and $S_{u}+\delta_{t}^{u}$ represents the selling price at time
$u$. The agent then aims at maximizing the following objective functional

$\mathbb{E}\big{[}X_{T}+S_{T}\,Q_{T}-\int_{0}^{T}\phi_{t}\,(Q_{t})^{2}\,dt-A\,(Q_{T})^{2}\big{]}$
(6)

by controlling $\boldsymbol{\delta}\in\mathbb{H}^{2}\times\mathbb{H}^{2}$,
where $A:=A(L_{T})$ and $\phi_{t}:=\phi(L_{t})$ for some non-negative bounded
functions $A,\phi:\mathbb{R}^{d}\to\mathbb{R}_{+}$ with the upper bounds
$\bar{A},\bar{\phi}>0$, accordingly.

######  Assumption 2.3.

Function A is a continuous non-negative function with the upper bound
$\bar{A}$. Functions $a,b$ are positive functions bounded above by
$\bar{a},\bar{b}$ accordingly; function $\phi$ is a non-negative function
bounded above by $\bar{\phi}$. Moreover, functions $a,b,\phi$ are all
Lipschitz continuous in any bounded domain of $\mathbb{R}^{d}$.

######  Remark 2.4.

In the functional (6), the random variable $A$ stands for the terminal
penalty. It can be interpreted as the overnight risk as in [1] or the terminal
clearance cost as in the optimal execution literatures. The integral term is
the well-known reduced-form model of risk and process $\phi$ could be
understood as the product of a risk parameter and the stochastic volatility of
the mid-price process (see [19] for an example).

By Itô’s formula, we compute that

$\displaystyle X_{T}+Q_{T}S_{T}$
$\displaystyle=x_{0}+\int_{0}^{T}(S_{t}+\delta_{t}^{a})\,a_{t}(\zeta-\gamma\delta_{t}^{a})\,dt-\int_{0}^{T}(S_{t}-\delta_{t}^{b})\,b_{t}(\zeta-\gamma\delta_{t}^{b})\,dt$
$\displaystyle\hskip
41.25641pt+q_{0}s_{0}+\int_{0}^{T}S_{t}\,b_{t}(\zeta-\gamma\delta_{t}^{b})\,dt-\int_{0}^{T}S_{t}\,a_{t}(\zeta-\gamma\delta_{t}^{a})\,dt+\int_{0}^{T}Q_{t}\,dS_{t}$
$\displaystyle=x_{0}+q_{0}s_{0}+\int_{0}^{T}\delta_{t}^{a}\,a_{t}(\zeta-\gamma\delta_{t}^{a})\,dt+\int_{0}^{T}\delta_{t}^{b}\,b_{t}(\zeta-\gamma\delta_{t}^{b})\,dt+\int_{0}^{T}Q_{t}\,dS_{t}.$

While the last term will vanish after taking the expectation due to (2), we
can then rewrite the functional (6) as

$\mathbb{E}\big{[}\int_{0}^{T}\delta_{t}^{a}\,a_{t}(\zeta-\gamma\delta_{t}^{a})\,dt+\int_{0}^{T}\delta_{t}^{b}\,b_{t}(\zeta-\gamma\delta_{t}^{b})\,dt-\int_{0}^{T}\phi_{t}\,(Q_{t})^{2}\,dt-A\,(Q_{T})^{2}\big{]},$
(7)

where the mid-price and cash no longer play a role. Define

$\displaystyle
H(t,l,q):=\sup_{\delta^{a},\,\delta^{b}\in\mathbb{H}^{2}}\mathbb{E}\big{[}\int_{t}^{T}\delta_{u}^{a}\,a_{u}^{t,l}(\zeta-\gamma\delta_{u}^{a})$
$\displaystyle
du+\int_{t}^{T}\delta_{u}^{b}\,b_{u}^{t,l}(\zeta-\gamma\delta_{u}^{b})\,du$
(8)
$\displaystyle-\int_{t}^{T}\phi_{u}^{t,l}\,(Q_{u}^{t,l,q,\boldsymbol{\delta}})^{2}\,du-A\,(Q_{T}^{t,l,q,\boldsymbol{\delta}})^{2}\big{]}.$

We use the superscript $(t,l)$ to show the initial condition of the underlying
Markov diffusions, and $(t,l,q,\boldsymbol{\delta})$ to display the additional
dependence on the control. Let $\mathcal{L}$ be the generator of $L$ as

$(\mathcal{L}f)(t,l,q):=\sum_{i=1}^{d}\Gamma^{i}(t,l)\,\frac{\partial
f}{\partial
l^{i}}(t,l,q)+\frac{1}{2}\sum_{i,j=1}^{d}(\Sigma\Sigma^{tr})^{i,j}(t,l)\,\frac{\partial^{2}f}{\partial
l^{i}\partial l^{j}}(t,l,q)$

for some sufficiently smooth function
$f:[0,T]\times\mathbb{R}^{d}\times\mathbb{R}\to\mathbb{R}$, then the dynamic
programming principle suggests the following partial differential equation
(PDE)

$\displaystyle\frac{\partial H}{\partial t}+\mathcal{L}H(t,l,q)$
$\displaystyle+\sup_{\delta^{a}\in\mathbb{R}}\big{\\{}(\delta^{a}-\frac{\partial
H}{\partial q})\,a(l)(\zeta-\gamma\delta^{a})\big{\\}}$ (9)
$\displaystyle+\sup_{\delta^{b}\in\mathbb{R}}\big{\\{}(\delta^{b}+\frac{\partial
H}{\partial q})\,b(l)(\zeta-\gamma\delta^{b})\big{\\}}-\phi(l)\,(q)^{2}=0$

for $(t,l,q)\in[0,T]\times\mathbb{R}^{d}\times\mathbb{R}$, subjecting to the
boundary condition $H(T,l,q)=-A(l)\,(q)^{2}$. The optimal control in feedback
form reads

$\delta_{t}^{b,*}=\frac{\zeta}{2\gamma}-\frac{1}{2}\frac{\partial H}{\partial
q},\hskip
28.45274pt\delta_{t}^{a,*}=\frac{\zeta}{2\gamma}+\frac{1}{2}\frac{\partial
H}{\partial q}.$

Combined with the affine ansatz widely used in optimal execution literatures
(thanks to the linear structure of our problem):

$H(t,l,q)=h_{0}(t,l)+h_{1}(t,l)\,q+h_{2}(t,l)\,(q)^{2},$

we plug them back to (9) to obtain, for $(t,l)\in[0,T]\times\mathbb{R}^{d}$,
that

$\frac{\partial h_{2}}{\partial
t}+\mathcal{L}h_{2}(t,l)+\gamma\big{(}a(l)+b(l)\big{)}\,h_{2}(t,l)^{2}-\phi(l)=0,$
(10) $\frac{\partial h_{1}}{\partial
t}+\mathcal{L}h_{1}(t,l)+\zeta\big{(}b(l)-a(l)\big{)}\,h_{2}(t,l)+\gamma\big{(}a(l)+b(l)\big{)}\,h_{2}(t,l)\,h_{1}(t,l)=0,$
(11) $\frac{\partial h_{0}}{\partial
t}+\mathcal{L}h_{0}(t,l)+\zeta\big{(}b(l)-a(l)\big{)}\,h_{1}(t,l)+\frac{\gamma\,b(l)}{4}\big{(}\frac{\zeta}{\gamma}-h_{1}(t,l)\big{)}^{2}+\frac{\gamma\,a(l)}{4}\big{(}\frac{\zeta}{\gamma}+h_{1}(t,l)\big{)}^{2}=0,$
(12)

subjecting to the boundaries $h_{2}(T,l)=-A(l)$, $h_{1}(T,l)=0$ and
$h_{0}(T,l)=0$. One can see that the main challenge lies in the nonlinear PDE
(10), while the others are all linear and the well-posedness is direct given
$h_{2}$ with sufficient regularity. We summarize this result in the next
theorem with the verification step. Hence, the remaining of this section is
devoted to (10).

######  Remark 2.5.

We compare our PDE system with the ones from related literatures:
(1) Regarding the optimal execution problems that take the order flow into
account (see [8] section 7.3), in the PDE for $h_{2}$, the quadratic term of
$h_{2}$ has a constant coefficient, that is,

$\frac{\partial h_{2}}{\partial
t}+\mathcal{L}h_{2}(t,l)-\phi+\frac{1}{k}\,h_{2}(t,l)^{2}=0,$

where $k,\phi>0$ represent the permanent price impact and risk parameter
accordingly. As there are no source terms $a$ and $b$, they deduce that
$h_{2}$ is independent of the order flow and solve a Riccati equation.
However, there is a source term of order flows in our case.

(2) With respect to execution problems with stochastic parameters ([4], [9]
and [19]), they all consider PDEs for $h_{2}$ that are similar to ours. While
[4] and [9] study the approximation solutions, the existence and uniqueness of
the viscosity solution is proven by [19]. Our contribution lies in the well-
posedness of the classical solution. Let us look at the equation from [4]
(equation (2.17) with no permanent price impact):

$\frac{\partial h_{2}}{\partial
t}+\mathcal{L}h_{2}(t,l)-\varphi+\frac{1}{f(l)}\,h_{2}(t,l)^{2}=0,$

where $\varphi>0$ stands for the risk and $f(l)$ is the stochastic impact
parameter. Comparing $1/f(l)$ with $\gamma(a(l)+b(l))$ of our equation, it is
interesting to see that the lower friction in optimal executions is analogous
to the higher market activity in market makings. Furthermore, our method can
be applied to prove the well-posedness of (2.17) by a change of variable.

######  Theorem 2.6.

Suppose (10) admits a bounded classical solution denoted by $h_{2}$, it holds:

(1) Both (11) and (12) also possess bounded classical solutions, written as
$h_{1}$ and $h_{0}$ accordingly;

(2) Define $\tilde{H}:[0,T]\times\mathbb{R}^{d}\times\mathbb{R}\to\mathbb{R}$
as

$\tilde{H}(t,l,q)=h_{0}(t,l)+h_{1}(t,l)\,q+h_{2}(t,l)\,(q)^{2},$

then $H=\tilde{H}$ on $[0,T]\times\mathbb{R}^{d}\times\mathbb{R}$;

(3) Let $\boldsymbol{\delta}^{*}=(\delta^{a,*},\delta^{b,*})$ be defined by

$\delta_{t}^{b,*}=\frac{\zeta}{2\gamma}-\frac{1}{2}\frac{\partial\tilde{H}}{\partial
q}\text{ \; and \;
}\delta_{t}^{a,*}=\frac{\zeta}{2\gamma}+\frac{1}{2}\frac{\partial\tilde{H}}{\partial
q},$ (13)

then $\boldsymbol{\delta}^{*}$ is the unique optimal control.

###### PROOF.

(1) The fact that both (11) and (12) accepts classical solutions follows from
[5]. It suffices to apply the Feynman–Kac formula to derive the boundedness.

(2) It is straightforward to see that $\tilde{H}$ is sufficiently regular to
apply the Itô’s formula and satisfies the quadratic growth condition in $q$:

$|\tilde{H}(t,l,q)|\leq C\,(1+q^{2}),$

for a constant $C>0$. We follow the arguments in [18]. First, we will show
that $\tilde{H}\geq H$. For all
$(t,l,q)\in[0,T)\times\mathbb{R}^{d}\times\mathbb{R}$,
$\boldsymbol{\alpha}\in\mathbb{H}^{2}\times\mathbb{H}^{2}$, and any stopping
time $\tau$ valued in $[t,\infty)$, it holds that

$\displaystyle\tilde{H}(s\wedge\tau,L_{s\wedge\tau}^{t,l},Q_{s\wedge\tau}^{t,l,q,\boldsymbol{\alpha}})=\tilde{H}$
$\displaystyle(t,l,q)+\int_{t}^{s\wedge\tau}\frac{\partial\tilde{H}}{\partial
l}(u,L_{u}^{t,l},Q_{u}^{t,l,q,\boldsymbol{\alpha}})^{tr}\,\Sigma(u,L_{u}^{t,l})\,dW_{u}$
$\displaystyle+\int_{t}^{s\wedge\tau}\big{[}\frac{\partial\tilde{H}}{\partial
t}(u,L_{u}^{t,l},Q_{u}^{t,l,q,\boldsymbol{\alpha}})+\mathcal{L}\tilde{H}(u,L_{u}^{t,l},Q_{u}^{t,l,q,\boldsymbol{\alpha}})$
$\displaystyle\hskip 51.21504pt-\frac{\partial\tilde{H}}{\partial
q}\,a(L_{u}^{t,l})\,(\zeta-\gamma\,\alpha_{u}^{a})$ $\displaystyle\hskip
79.6678pt+\frac{\partial\tilde{H}}{\partial
q}\,b(L_{u}^{t,l})\,(\zeta-\gamma\,\alpha_{u}^{b})\big{]}\,du.$

If we choose

$\tau=\tau_{n}=\inf\Big{\\{}r\geq
t:\;\int_{t}^{r}\big{|}\frac{\partial\tilde{H}}{\partial
l}(u,L_{u}^{t,l},Q_{u}^{t,l,q,\boldsymbol{\alpha}})^{tr}\,\Sigma(u,L_{u}^{t,l})\big{|}^{2}\,du\geq
n\Big{\\}},$

the stopped stochastic integral is then a martingale and hence

$\displaystyle\mathbb{E}\big{[}\tilde{H}(s\wedge\tau_{n},L_{s\wedge\tau_{n}}^{t,l},Q_{s\wedge\tau_{n}}^{t,l,q,\boldsymbol{\alpha}})\big{]}$
$\displaystyle=\tilde{H}(t,l,q)+\mathbb{E}\int_{t}^{s\wedge\tau_{n}}\big{[}\frac{\partial\tilde{H}}{\partial
t}(u,L_{u}^{t,l},Q_{u}^{t,l,q,\boldsymbol{\alpha}})+\mathcal{L}\tilde{H}(u,L_{u}^{t,l},Q_{u}^{t,l,q,\boldsymbol{\alpha}})$
$\displaystyle\hskip 56.9055pt-\frac{\partial\tilde{H}}{\partial
q}\,a(L_{u}^{t,l})\,(\zeta-\gamma\,\alpha_{u}^{a})+\frac{\partial\tilde{H}}{\partial
q}\,b(L_{u}^{t,l})\,(\zeta-\gamma\,\alpha_{u}^{b})\big{]}\,du$
$\displaystyle\leq\tilde{H}(t,l,q)+\mathbb{E}\int_{t}^{s\wedge\tau_{n}}\big{[}\frac{\partial\tilde{H}}{\partial
t}(u,L_{u}^{t,l},Q_{u}^{t,l,q,\boldsymbol{\alpha}})+\mathcal{L}\tilde{H}(u,L_{u}^{t,l},Q_{u}^{t,l,q,\boldsymbol{\alpha}})$
$\displaystyle\hskip
56.9055pt+\sup_{\delta^{a}\in\mathbb{R}}\big{\\{}(\delta^{a}-\frac{\partial\tilde{H}}{\partial
q})\,a(L_{u}^{t,l})\,(\zeta-\gamma\,\delta^{a})\big{\\}}$ $\displaystyle\hskip
56.9055pt+\sup_{\delta^{b}\in\mathbb{R}}\big{\\{}(\delta^{b}+\frac{\partial\tilde{H}}{\partial
q})\,b(L_{u}^{t,l})\,(\zeta-\gamma\,\delta^{b})\big{\\}}$ $\displaystyle\hskip
56.9055pt-\alpha_{u}^{a}\,a(L_{u}^{t,l})\,(\zeta-\gamma\,\alpha_{u}^{a})-\alpha_{u}^{b}\,b(L_{u}^{t,l})\,(\zeta-\gamma\,\alpha_{u}^{b})\big{]}\,du$
$\displaystyle=\tilde{H}(t,l,q)+\mathbb{E}\int_{t}^{s\wedge\tau_{n}}\Big{[}\phi(L_{u}^{t,l})\,(Q_{u}^{t,l,q,\boldsymbol{\alpha}})^{2}$
$\displaystyle\hskip
56.9055pt-\alpha_{u}^{a}\,a(L_{u}^{t,l})\,(\zeta-\gamma\,\alpha_{u}^{a})-\alpha_{u}^{b}\,b(L_{u}^{t,l})\,(\zeta-\gamma\,\alpha_{u}^{b})\big{]}\,du.$

Note that

$\displaystyle\big{|}\int_{t}^{s\wedge\tau_{n}}\Big{[}\phi(L_{u}^{t,l})\,(Q_{u}^{t,l,q,\boldsymbol{\alpha}})^{2}-\alpha_{u}^{a}\,a(L_{u}^{t,l})\,(\zeta-\gamma\,\alpha_{u}^{a})-\alpha_{u}^{b}\,b(L_{u}^{t,l})\,(\zeta-\gamma\,\alpha_{u}^{b})\big{]}\,du\,\big{|}$
$\displaystyle\hskip
56.9055pt\leq\int_{t}^{T}\big{|}\phi(L_{u}^{t,l})\,(Q_{u}^{t,l,q,\boldsymbol{\alpha}})^{2}-\alpha_{u}^{a}\,a(L_{u}^{t,l})\,(\zeta-\gamma\,\alpha_{u}^{a})$
$\displaystyle\hskip
85.35826pt-\alpha_{u}^{b}\,b(L_{u}^{t,l})\,(\zeta-\gamma\,\alpha_{u}^{b})\big{|}\,du$

and

$\tilde{H}(s\wedge\tau,L_{s\wedge\tau_{n}}^{t,l},Q_{s\wedge\tau_{n}}^{t,l,q,\boldsymbol{\alpha}})\leq
C\,(1+\sup_{u\in[t,T]}|Q_{u}^{t,l,q,\boldsymbol{\alpha}}|^{2}).$

The right hand sides of both above terms are integrable and we apply the
dominated convergence theorem twice to see

$\displaystyle\mathbb{E}\big{[}\tilde{H}(s,$ $\displaystyle
L_{s}^{t,l},Q_{s}^{t,l,q,\boldsymbol{\alpha}})\big{]}\leq\tilde{H}(t,l,q)$
$\displaystyle+\mathbb{E}\int_{t}^{s}\Big{[}\phi(L_{u}^{t,l})\,(Q_{u}^{t,l,q,\boldsymbol{\alpha}})^{2}-\alpha_{u}^{a}\,a(L_{u}^{t,l})\,(\zeta-\gamma\,\alpha_{u}^{a})-\alpha_{u}^{b}\,b(L_{u}^{t,l})\,(\zeta-\gamma\,\alpha_{u}^{b})\big{]}\,du,$

and further

$\displaystyle\mathbb{E}\big{[}-A(L_{T}^{t,l})\,(Q_{T}^{t,l,q,\boldsymbol{\alpha}})^{2}\big{]}$
$\displaystyle\leq\tilde{H}(t,l,q)$
$\displaystyle+\mathbb{E}\int_{t}^{T}\Big{[}\phi(L_{u}^{t,l})\,(Q_{u}^{t,l,q,\boldsymbol{\alpha}})^{2}-\alpha_{u}^{a}\,a(L_{u}^{t,l})\,(\zeta-\gamma\,\alpha_{u}^{a})$
$\displaystyle\hskip
99.58464pt-\alpha_{u}^{b}\,b(L_{u}^{t,l})\,(\zeta-\gamma\,\alpha_{u}^{b})\big{]}\,du.$

Therefore, the arbitrariness of $\boldsymbol{\alpha}$ implies

$H(t,l,q)\leq\tilde{H}(t,l,q).$

To see the reversed inequality, let $Q^{t,l,q,\boldsymbol{\delta}^{*}}$ be the
associated inventory of the control $\boldsymbol{\delta}$ defined in (13) and
apply the Itô’s formula to
$\tilde{H}(u,L_{u}^{t,l},Q_{u}^{t,l,q,\boldsymbol{\delta}^{*}})$ between
$t\in[0,T)$ and $s\in[t,T)$. Through similar localizing and limiting
procedures, one can obtain

$\displaystyle-A(L_{T}^{t,l})\,(Q_{T}^{t,l,q,\boldsymbol{\delta}^{*}})^{2}$
$\displaystyle=\tilde{H}(t,l,q)+\mathbb{E}\int_{t}^{T}\big{[}\frac{\partial\tilde{H}}{\partial
t}(u,L_{u}^{t,l},Q_{u}^{t,l,q,\boldsymbol{\delta}^{*}})+\mathcal{L}\tilde{H}(u,L_{u}^{t,l},Q_{u}^{t,l,q,\boldsymbol{\delta}^{*}})$
$\displaystyle\hskip 85.35826pt-\frac{\partial\tilde{H}}{\partial
q}\,a(L_{u}^{t,l})\,(\zeta-\gamma\,\delta_{u}^{a,*})$ $\displaystyle\hskip
108.12054pt+\frac{\partial\tilde{H}}{\partial
q}\,b(L_{u}^{t,l})\,(\zeta-\gamma\,\delta_{u}^{b,*})\big{]}\,du$
$\displaystyle=\tilde{H}(t,l,q)+\mathbb{E}\int_{t}^{T}\Big{[}\phi(L_{u}^{t,l})\,(Q_{u}^{t,l,q,\boldsymbol{\delta}^{*}})^{2}$
$\displaystyle\hskip
56.9055pt-\alpha_{u}^{a}\,a(L_{u}^{t,l})\,(\zeta-\gamma\,\delta_{u}^{a,*})-\alpha_{u}^{b}\,b(L_{u}^{t,l})\,(\zeta-\gamma\,\delta_{u}^{b,*})\big{]}\,du.$

To finally conclude that

$H(t,l,q)=\tilde{H}(t,l,q),$

it suffices to show that $\boldsymbol{\delta}^{*}\in\mathbb{H}^{2}$. Since the
inventory satisfies the linear random ODE

$\displaystyle dQ_{u}^{t,l,q,\boldsymbol{\delta}^{*}}$
$\displaystyle=-a_{u}\,(\zeta-\gamma\,\delta_{u}^{a,*})\,du+b_{u}\,(\zeta-\gamma\,\delta_{u}^{b,*})\,du$
$\displaystyle=\frac{1}{2}\,[\zeta\,(b_{u}-a_{u})+\gamma\,h_{1}(u,l)\,(a_{u}-b_{u})]\,du+\gamma\,(a_{u}+b_{u})\,h_{2}(u,l)\,Q_{u}^{t,l,q,\boldsymbol{\delta}^{*}}\,du,$

the process $Q^{t,l,q,\boldsymbol{\delta}^{*}}$ turns out to be bounded due to
the boundedness of $a,b,h_{1},$ and $h_{2}$. Consequently, the control
$\boldsymbol{\delta}^{*}$ is also bounded and thus admissible.

(3) It suffices to prove the uniqueness of the optimal control. The statement
is true if the objective functional (7) is strictly concave. The strict
concavity of (7) can be deduced from the strict concavity of
$\delta_{u}^{a}\,a_{u}(\zeta-\gamma\delta_{u}^{a})+\delta_{u}^{b}\,b_{u}(\zeta-\gamma\delta_{u}^{b})$
and the convexity of $Q_{t}^{2}$ with respect to the control. The strict
concavity of
$\delta_{u}^{a}\,a_{u}(\zeta-\gamma\delta_{u}^{a})+\delta_{u}^{b}\,b_{u}(\zeta-\gamma\delta_{u}^{b})$
is clear because they are second order polynomials with negative signs for the
leading terms. Let
$\boldsymbol{\delta},\boldsymbol{\beta}\in\mathbb{H}^{2}\times\mathbb{H}^{2}$
be two generic controls, and set $\lambda\in[0,1]$. We check the convexity of
the quadratic inventory through

$\displaystyle(Q_{u}^{\lambda\,\boldsymbol{\delta}+(1-\lambda)\,\boldsymbol{\beta}})^{2}$
$\displaystyle-\lambda\,(Q_{u}^{\boldsymbol{\delta}})^{2}-(1-\lambda)\,(Q_{u}^{\boldsymbol{\beta}})^{2}$
$\displaystyle=\big{(}\lambda\,Q_{u}^{\boldsymbol{\delta}}+(1-\lambda)\,Q_{u}^{\boldsymbol{\beta}}\big{)}^{2}-\lambda\,(Q_{u}^{\boldsymbol{\delta}})^{2}-(1-\lambda)\,(Q_{u}^{\boldsymbol{\beta}})^{2}$
$\displaystyle\leq 0,$

where we have applied the linearity of the inventory with respect to the
control. ∎

We intend to prove the existence and uniqueness of the classical solution to
(10) on the basis of [5], the result of which cannot be applied directly
because of the quadratic term $h_{2}(t,l)^{2}$. To begin with, we introduce
two auxiliary independent linear PDEs

$\displaystyle\frac{\partial h}{\partial
t}+\mathcal{L}h(t,l)+\gamma\big{(}a(l)+b(l)\big{)}\,g(t,l)\,h(t,l)-\phi(l)=0,\quad\text{such
that}\;\;h(T,l)=-A;$ $\displaystyle\frac{\partial h}{\partial
t}+\mathcal{L}h(t,l)+\gamma\big{(}a(l)+b(l)\big{)}\,\big{[}\psi\big{(}g(t,l)\big{)}\big{]}^{2}-\phi(l)=0,\quad\text{such
that}\;\;h(T,l)=-A,$

where $\psi:\mathbb{R}\to\mathbb{R}$ is a truncation function with the
parameter $\xi>0$ defined by

$\psi(x)=\begin{cases}-\xi,\quad&\text{if }x<-\xi\\\ x,\quad&\text{if
}-\xi\leq x\leq\xi\\\ \xi,\quad&\text{if }x>\xi,\end{cases}$

and $g$ is a bounded continuous function to be specified later. If the
function $g$ and the solution of the linear PDE are considered as input and
output accordingly, we can define two mappings $F_{1}$ and $F_{2}$ for the two
equations by the Feynman-Kac formula:

$\displaystyle(F_{1}\,g)(t,l)=\mathbb{E}\Big{[}-A\,e^{\int_{t}^{T}\gamma(a_{s}+b_{s})\,g(s,L_{s}^{t,l})\,ds}-\int_{t}^{T}\phi_{s}\,e^{\int_{t}^{s}\gamma(a_{s}+b_{s})\,g(u,L_{u}^{t,l})\,du}\,ds\Big{]},$
$\displaystyle(F_{2}\,g)(t,l)=\mathbb{E}\Big{[}-A+\int_{t}^{T}\gamma(a_{s}+b_{s})\,\big{(}\psi(g(s,L_{s}^{t,l}))\big{)}^{2}\,ds-\int_{t}^{T}\phi_{s}\,ds\Big{]}.$

It turns out that $F_{1}$ and $F_{2}$ are contraction mappings in the space of
bounded functions.

######  Lemma 2.7.

Denote by $C_{b}^{-}$ the set of bounded, non-positive and continuous
functions on $[0,T]\times\mathbb{R}^{d}$, then the following statements hold:
(1) $F_{1}$ defines a contraction mapping on $C_{b}^{-}$ with respect to the
norm

$\|v\|_{\zeta_{1}}:=\sup_{(t,x)\in[0,T]\times\mathbb{R}^{d}}e^{-\zeta_{1}\,(T-t)}\cdot|v(t,x)|,$

for some $\zeta_{1}\in\mathbb{R}_{+}$ large enough.
(2) $F_{2}$ defines a contraction mapping on $C_{b}$ with respect to the norm

$\|v\|_{\zeta_{2}}:=\sup_{(t,x)\in[0,T]\times\mathbb{R}^{d}}e^{-\zeta_{2}\,(T-t)}\cdot|v(t,x)|,$

for some $\zeta_{2}\in\mathbb{R}_{+}$ large enough.

###### PROOF.

We first introduce an inequality that will be used for both statements. For
$\zeta_{1}>0$ and some bounded functions $v$ and $w$, one can see

$\displaystyle\int_{t}^{T}|v(s,L_{s})-w(s,L_{s})|\,ds$
$\displaystyle=\int_{t}^{T}|v(s,L_{s})-w(s,L_{s})|e^{-\zeta_{1}\,(T-s)}\,e^{\zeta_{1}\,(T-s)}\,ds$
(14)
$\displaystyle\leq\|v-w\|_{\zeta_{1}}\int_{t}^{T}e^{\zeta_{1}\,(T-s)}\,ds$
$\displaystyle=\frac{e^{\zeta_{1}\,(T-t)}-1}{\zeta_{1}}\cdot\|v-w\|_{\zeta_{1}}.$

For any $v\in C_{b}^{-}$, it is clear that $F_{1}v$ is also a bounded negative
function. Since the term inside the conditional expectation is continuous and
uniformly bounded by $\bar{A}+\bar{\phi}\,T$, the continuity of $F_{1}v$
follows from the dominated convergence theorem. Consider any $v$, $w\in
C_{b}^{-}$ and it follows that, for all $(t,l)\in[0,T]\times\mathbb{R}^{d}$,

$\displaystyle e^{-\zeta_{1}\,(T-t)}$
$\displaystyle|(F_{1}\,v)(t,l)-(F_{1}\,w)(t,l)|$ $\displaystyle\leq
e^{-\zeta_{1}\,(T-t)}\,\mathbb{E}\Big{[}A\,|e^{\int_{t}^{T}\gamma(a_{s}+b_{s})\,v(s,L_{s}^{t,l})\,ds}-e^{\int_{t}^{T}\gamma(a_{s}+b_{s})\,w(s,L_{s}^{t,l})\,ds}|$
$\displaystyle\hskip
71.13188pt+\bar{\phi}\int_{t}^{T}|e^{\int_{t}^{s}\gamma(a_{u}+b_{u})\,v(u,L_{u}^{t,l})\,du}-e^{\int_{t}^{s}\gamma(a_{u}+b_{u})\,w(u,L_{u}^{t,l})\,du}|\,ds\Big{]}$
$\displaystyle\leq
e^{-\zeta_{1}\,(T-t)}\,\mathbb{E}\Big{[}A\,\big{|}\int_{t}^{T}\gamma(a_{s}+b_{s})\,v(s,L_{s}^{t,l})\,ds-\int_{t}^{T}\gamma(a_{s}+b_{s})\,w(s,L_{s}^{t,l})\,ds\big{|}$
$\displaystyle\hskip
71.13188pt+\bar{\phi}\int_{t}^{T}\big{|}\int_{t}^{s}\gamma(a_{u}+b_{u})\,v(u,L_{u}^{t,l})\,du$
$\displaystyle\hskip
128.0374pt-\int_{t}^{s}\gamma(a_{u}+b_{u})\,v(u,L_{u}^{t,l})\,du\big{|}\,ds\Big{]}$
$\displaystyle\leq
e^{-\zeta_{1}\,(T-t)}\cdot\gamma\,(\bar{a}+\bar{b})\,(\bar{A}+\bar{\phi}\,T)\cdot\mathbb{E}\Big{[}\int_{t}^{T}|v(s,L_{s}^{t,l})-w(s,L_{s}^{t,l})|\,ds\Big{]}$
$\displaystyle\leq\frac{\gamma\,(\bar{a}+\bar{b})\,(\bar{A}+\bar{\phi}\,T)}{\zeta_{1}}\cdot\|v-w\|_{\xi_{1}}$

where the positiveness of $a,b$ and negativeness of $v$, $w$ are used to
derive the second inequality, and the proposed inequality (14) is applied in
the last line. For a contraction mapping, it suffices to pick
$\zeta_{1}\in\mathbb{R}_{+}$ such that

$\zeta_{1}>\gamma\,(\bar{a}+\bar{b})\,(\bar{A}+\bar{\phi}\,T).$

Similarly, we look at the second statement and compute that

$\displaystyle e^{-\zeta_{2}\,(T-t)}\,|(F_{2}\,v)(t,l)$
$\displaystyle-(F_{2}\,w)(t,l)|$
$\displaystyle=e^{-\zeta_{2}\,(T-t)}\,\mathbb{E}\big{|}\int_{t}^{T}\gamma(a_{s}+b_{s})\,\big{[}\psi(v(s,L_{s}^{t,l}))^{2}-\psi(w(s,L_{s}^{t,l}))^{2}\big{]}\,ds\,\big{|}$
$\displaystyle\leq
2\xi\,e^{-\zeta_{2}\,(T-t)}\,\mathbb{E}\int_{t}^{T}\big{|}\gamma(a_{s}+b_{s})\,\big{[}\psi(v(s,L_{s}^{t,l}))-\psi(w(s,L_{s}^{t,l}))\big{]}\,\big{|}\,ds$
$\displaystyle\leq\frac{2\xi\,\gamma\,(\bar{a}+\bar{b})}{\zeta_{2}}\cdot\|v-w\|_{\xi_{2}}.$

Again, it suffices to pick $\zeta_{2}$ large enough such that
$\zeta_{2}>2\xi\,\gamma\,(\bar{a}+\bar{b})$ for a contraction mapping. The
boundedness and continuity of $F_{2}v$ follows from the uniform boundedness
and continuity inside the conditional expectation. ∎

Due to the Banach fixed-point theorem, both $F_{1}$ and $F_{2}$ have their own
fixed points. One reason why we consider both two mappings is that they
actually share the same fixed point if the constant $\xi$ is sufficiently
large.

######  Lemma 2.8.

Let $v\in C_{b}^{-}$ be the fixed point of the $F_{1}$, then it is also the
one of $F_{2}$, provided that the constant $\xi$ of the truncation $\psi$ is
large enough.

###### PROOF.

Since $v$ is the fixed point of $F_{1}$, for any
$(t,l)\in[0,T]\times\mathbb{R}^{d}$ it holds

$v(t,l)=\mathbb{E}\Big{[}-A\,e^{\int_{t}^{T}\gamma\,(a_{s}+b_{s})\,v(s,L_{s}^{t,l})\,ds}-\int_{t}^{T}\phi_{s}\,e^{\int_{t}^{s}\gamma\,(a_{u}+b_{u})\,v(u,L_{u}^{t,l})\,du}\,ds\Big{]}.$

We take the derivative of
$\exp(\int_{t}^{T}\gamma(a_{s}+b_{s})\,v(s,L_{s}^{t,l})\,ds)$ to obtain

$\displaystyle de^{\int_{r}^{T}\gamma(a_{s}+b_{s})\,v(s,L_{s}^{t,l})\,ds}$
$\displaystyle=-\gamma(a_{r}+b_{r})\,v(r,L_{r}^{t,l})\,e^{\int_{r}^{T}\gamma(a_{s}+b_{s})\,v(s,L_{s}^{t,l})\,ds}\,dr,$
$\displaystyle A-A\,e^{\int_{t}^{T}\gamma(a_{s}+b_{s})\,v(s,L_{s}^{t,l})\,ds}$
$\displaystyle=-A\int_{t}^{T}\gamma(a_{s}+b_{s})\,v(s,L_{s}^{t,l})\,e^{\int_{s}^{T}\gamma(a_{u}+b_{u})\,v(u,L_{u}^{t,l})\,du}\,ds,$

and then take the expectation to see

$\displaystyle\mathbb{E}\big{[}A-A\,e^{\int_{t}^{T}\gamma(a_{s}+b_{s})\,v(s,L_{s}^{t,l})\,ds}\big{]}$
$\displaystyle=\mathbb{E}\int_{t}^{T}\gamma(a_{s}+b_{s})\,v(s,L_{s}^{t,l})$
(15) $\displaystyle\hskip
14.22636pt\cdot\mathbb{E}_{s}\big{[}(-A)\,e^{\int_{s}^{T}\gamma(a_{u}+b_{u})\,v(u,L_{u}^{t,l})\,du}\big{]}\,ds,$
$\displaystyle
v(t,l)+\mathbb{E}\Big{[}\int_{t}^{T}\phi_{s}\,e^{\int_{t}^{s}\gamma\,(a_{u}+b_{u})\,v(u,L_{u}^{t,l})\,du}\,ds\Big{]}$
$\displaystyle=\mathbb{E}[-A]+\mathbb{E}\int_{t}^{T}\gamma(a_{s}+b_{s})\,v(s,L_{s}^{t,l})$
$\displaystyle\hskip
14.22636pt\cdot\mathbb{E}_{s}\big{[}(-A)\,e^{\int_{s}^{T}\gamma(a_{u}+b_{u})\,v(u,L_{u}^{t,l})\,du}\big{]}\,ds.$

If one notes that

$\displaystyle\mathbb{E}\int_{t}^{T}\gamma(a_{s}+b_{s})\,v(s,L_{s}^{t,l})\,\mathbb{E}_{s}\big{[}(-A)\,e^{\int_{s}^{T}\gamma(a_{u}+b_{u})\,v(u,L_{u}^{t,l})\,du}\big{]}\,ds$
(16)
$\displaystyle=\mathbb{E}\int_{t}^{T}\gamma(a_{s}+b_{s})\,v(s,L_{s}^{t,l})\,\big{(}v(s,L_{s}^{t,l})+\mathbb{E}_{s}\int_{s}^{T}\phi_{r}\,e^{\int_{s}^{r}\gamma(a_{u}+b_{u})v(u,L_{u}^{s,L_{s}^{t,l}})\,du}\,dr\big{)}\,ds,$

which follows from the fixed point property, and also

$\displaystyle\mathbb{E}\Big{[}\int_{t}^{T}$
$\displaystyle\gamma\,(a_{s}+b_{s})\,v(s,L_{s}^{t,l})\,\mathbb{E}_{s}\big{(}\int_{s}^{T}\phi_{r}\,e^{\int_{s}^{r}\gamma\,(a_{u}+b_{u})\,v(u,L_{u}^{s,L_{s}^{t,l}})\,du}\,dr\big{)}\,ds\Big{]}$
(17)
$\displaystyle=\mathbb{E}\Big{[}\int_{t}^{T}\phi_{r}\,\big{(}\int_{t}^{r}\gamma\,(a_{s}+b_{s})\,v(s,L_{s}^{t,l})\,e^{\int_{s}^{r}\gamma\,(a_{u}+b_{u})\,v(u,L_{u}^{t,l})\,du}\,ds\big{)}\,dr\Big{]}$
$\displaystyle=\mathbb{E}\Big{[}\int_{t}^{T}\phi_{r}\,\big{(}e^{\int_{t}^{r}\gamma\,(a_{u}+b_{u})\,v(u,L_{u}^{t,l})\,du}-1\big{)}dr\Big{]},$

one can observe the following result via (15)–(17):

$v(t,l)=\mathbb{E}\Big{[}-A+\int_{t}^{T}\gamma\,(a_{s}+b_{s})\,v(s,L_{s}^{t,l})^{2}\,ds-\int_{t}^{T}\phi_{s}\,ds\Big{]}.$

Since $v$ is the fixed point of $F_{1}$, it’s clear that $v$ is uniformly
bounded by $\bar{A}+\bar{\phi}\,T$ and, if we pick
$\xi\geq\bar{A}+\bar{\phi}\,T$, the equation above is equivalent to

$v(t,l)=\mathbb{E}\Big{[}-A+\int_{t}^{T}\gamma\,(a_{s}+b_{s})\,\psi(v(s,L_{s}^{t,l}))^{2}\,ds-\int_{t}^{T}\phi_{s}\,ds\Big{]},$

showing that $v$ is also the fixed point of $F_{2}$. ∎

The other reason why we consider both two mappings is that the mapping $F_{2}$
has been studied by [5] under some relative general assumptions. We can now
extend its result to our equation (10).

######  Theorem 2.9.

The equation (10) accepts a unique classical solution in $C_{b}^{1,2}$. More
specifically, the solution takes value in $[-\bar{A}-\bar{\phi}\,T,0)$.

###### PROOF.

We start to look at the PDE (10) with a regularization:

$\frac{\partial h_{2}}{\partial
t}+\mathcal{L}h_{2}(t,l)+\gamma\big{(}a(l)+b(l)\big{)}\,\psi(h_{2}(t,l))^{2}-\phi(l)=0,\quad\text{such
that}\;\;h_{2}(T,l)=-A(l).$ (18)

In order to apply the proposition 2.3 from [5], for convenience we set
$\iota(t,l,v)=\gamma\,(a(l)+b(l))\,\psi(v)^{2}-\phi(l)$ and check the
following set of conditions:

  * 1.

functions A and $\iota$ are continuous and bounded, with $\iota$ being
Lipschitz in $v$ uniformly in $t$ and $l$;

  * 2.

there exists a sequence $(D_{n})_{n\in\mathbb{N}}$ of bounded domains with
closure $\bar{D}_{n}\subseteq\mathbb{R}^{d}$ such that
$\bigcup^{\infty}_{n=1}D_{n}=\mathbb{R}^{d}$ and each $D_{n}$ has a
$C^{2}$-boundary;

  * 3.

the functions $\Gamma$ and $\Sigma\,\Sigma^{tr}$ are uniformly Lipschitz-
continuous on $[0,T]\times\bar{D}_{n}$;

  * 4.

$\det(\Sigma(t,l)\,\Sigma^{tr}(t,l))\neq 0$ for all
$(t,l)\in[0,T]\times\mathbb{R}^{d}$;

  * 5.

$(t,l,v)\mapsto\iota(t,l,v)$ is uniformly Hölder-continuous on
$[0,T]\times\bar{D}_{n}\times\mathbb{R}$.

The first two conditions can be verified directly based on the Assumption 2.3,
while the third and fourth ones follow from the Assumption 2.1. For any
$(t,l,v),(t^{\prime},l^{\prime},v^{\prime})\in[0,T]\times\bar{D}_{n}\times\mathbb{R}$,
note that

$\displaystyle|\iota(t^{\prime},l^{\prime},v^{\prime})-\iota(t,l,v)|$
$\displaystyle\leq\gamma\,(a(l^{\prime})+b(l^{\prime}))\,|\psi(v^{\prime})^{2}-\psi(v)^{2}|$
(19) $\displaystyle\hskip
51.21504pt+\gamma\,\psi(v)^{2}\,|a(l^{\prime})+b(l^{\prime})-a(l)-b(l)|+|\phi(l^{\prime})-\phi(l)|$
$\displaystyle\leq
2\,\gamma\,\xi\,(\bar{a}+\bar{b})\,|v^{\prime}-v|+2\,\gamma\,\xi^{2}\,C_{n}\,|l^{\prime}-l|+C_{n}\,|l^{\prime}-l|,$

where $C_{n}$ denotes the Lipschitz coefficient of $a$, $b$ and $\phi$ in the
domain $\bar{D}_{n}$ using the Assumption 2.3. Equation (19) shows that
$\iota$ is Lipschitz-continuous on $[0,T]\times\bar{D_{n}}\times\mathbb{R}$,
and hence uniformly Hölder-continuous due to the boundedness. Consequently,
the Proposition 2.3 from [5] tells us that the ‘regular’ PDE (18) has a unique
classical solution in $C_{b}^{1,2}$, which is given by the fixed point
$\hat{v}$ of the mapping $F_{2}$. Since $F_{1}$ and $F_{2}$ share the same
fixed point, $\hat{v}$ is also the fixed point of $F_{1}$ and we already know
that $\hat{v}$ is bounded by $\bar{A}+\bar{\phi}\,T$. Therefore, if $\xi$ is
chosen to be larger than $\bar{A}+\bar{\phi}\,T$, it follows that

$\displaystyle\frac{\partial\hat{v}}{\partial t}+\mathcal{L}\hat{v}(t,l)$
$\displaystyle+\gamma\big{(}a(l)+b(l)\big{)}\,\psi(\hat{v}(t,l))^{2}-\phi(l)$
$\displaystyle=\frac{\partial\hat{v}}{\partial
t}+\mathcal{L}\hat{v}(t,l)+\gamma\big{(}a(l)+b(l)\big{)}\,\hat{v}(t,l)^{2}-\phi(l)=0,$

establishing the existence of solutions.

For the uniqueness, suppose that $u\in C_{b}^{1,2}$ is another solution to the
PDE (10). Consider another truncation function $\tilde{\psi}$ with the
parameter $\tilde{\xi}$ satisfying

$\tilde{\xi}\;\geq\;(\bar{A}+\bar{\phi}\,T)\;\vee\sup_{(t,l)\in[0,T]\times\mathbb{R}^{d}}|u(t,l)|.$

Note that both $\hat{v}$ and $u$ solve the following ‘regular’ PDE:

$\frac{\partial h}{\partial
t}+\mathcal{L}h(t,l)+\gamma\big{(}a(l)+b(l)\big{)}\,\tilde{\psi}(h(t,l))^{2}-\phi(l)=0,\quad\text{such
that}\;\;h(T,l)=-A(l).$

which has a unique solution according to previous discussions. Hence,
$\hat{v}\equiv u$ and the uniqueness is obtained. ∎

######  Remark 2.10.

Our method can be further generalized to unbounded functions $a$ and $b$. If
$a,b$ satisfy proper growth conditions, consider equations

$\frac{\partial h_{n}}{\partial
t}+\mathcal{L}h_{n}(t,l)+\gamma\big{(}a(l)\wedge n+b(l)\wedge
n\big{)}\,h_{n}(t,l)^{2}-\phi(l)=0,\quad\text{such that}\;\;h_{n}(T,l)=-A(l),$

for $n\in\mathbb{N}$. A sequence of solutions $(h_{n})_{n\in\mathbb{N}}$,
being uniformly bounded, is guaranteed by the Theorem 2.9 and is non-
decreasing by the comparison result. If we denote by $g$ as the limit of the
sequence, it turns out that $g$ satisfies the conditional representation

$g(t,l)=\mathbb{E}\Big{[}-A+\int_{t}^{T}\gamma(a_{s}+b_{s})\,\big{(}\psi(g(s,L_{s}^{t,l}))\big{)}^{2}\,ds-\int_{t}^{T}\phi_{s}\,ds\Big{]}$

for a truncation function $\psi$ with the parameter $\bar{A}+\bar{\phi}\,T$.
Then, one can use the proof of [5] to show that $g\in C^{1,2}$ solves the PDE.
We do not cover it here, because the boundedness of order flows guarantees the
square integrability of the inventory (and thus (2)). But, the idea will be
illustrated in the later section through some other equations.

######  Example 2.11.

We conclude this section with a particular example of (10) that has an
explicit solution. Some economic intuition is also helpful in a later proof.
First, the boundedness of $\partial h_{2}/\partial l$ needs to be derived. For
this purpose, several additional conditions are imposed:

  * •

each element in the matrix-valued function $\Sigma(t,l)$ is a bounded
function;

  * •

$\phi\equiv 0$, i.e., the agent is risk neutral;

  * •

functions $a,b$, and $A$ are all uniformly Lipschitz, the Lipschitz
coefficient of which is denoted by $C_{\text{Lip}}$.

Define the function $g:[0,T]\times\mathbb{R}^{d}\to\mathbb{R}$ as

$g(t,l)=\frac{h_{2}(t,l+\epsilon\,e^{i})-h_{2}(t,l)}{\epsilon},$

where $\epsilon>0$ is a constant and $e^{i}$ is the unit vector of the
$i$-coordinate in $\mathbb{R}^{d}$ for $i\in\\{1,\dots,d\\}$. Applying a
similar transform to the equation (10), one can observe that $g$ satisfies the
following linear PDE:

$\displaystyle\frac{\partial g}{\partial t}+\mathcal{L}g(t,l)$
$\displaystyle+\gamma\big{(}a(l+\epsilon\,e^{i})+b(l+\epsilon\,e^{i})\big{)}\,\big{(}h_{2}(t,l+\epsilon\,e^{i})+h_{2}(t,l)\big{)}\,g(t,l)$
(20) $\displaystyle+\big{(}\tilde{a}(l)+\tilde{b}(l)\big{)}\,h_{2}(t,l)=0,$

such that $g(T,l)=-\tilde{A}(l)$, where we have defined

$\tilde{\theta}(l)=\frac{\theta(t,l+\epsilon\,e^{i})-\theta(t,l)}{\epsilon}$

for $\theta\in\\{a,b,A\\}$. To verify the term
$(a(l+\epsilon\,e^{i})+b(l+\epsilon\,e^{i}))\,(h_{2}(t,l+\epsilon\,e^{i})+h_{2}(t,l))$
is uniformly Hölder-continuous, it suffices to notice that:

  * •

functions $a$ and $b$ are uniformly Lipschitz;

  * •

function $h_{2}(t,l)$ is uniformly Lipschitz on $[0,T]\times D$ for any
compact set $D\subset\mathbb{R}^{d}$, since partial derivatives of $h_{2}$ are
continuous;

  * •

functions $a,b$, and $h_{2}$ are uniformly bounded.

While the other conditions can be checked in a similar way, one can again
learn from [5] that the PDE (20) accepts a unique solution $g\in C_{b}^{1,2}$.
The solution $g$ possesses the representation

$\displaystyle
g(t,l)=\mathbb{E}\Big{[}-\tilde{A}(L_{T}^{t,l})\,e^{\int_{t}^{T}\gamma(a(L_{s}^{t,l}+\epsilon\,e^{i})+b(L_{s}^{t,l}+\epsilon\,e^{i}))\,(h_{2}(s,L_{s}^{t,l}+\epsilon\,e^{i})+h_{2}(s,L_{s}^{t,l}))\,ds}$
$\displaystyle\hskip
28.45274pt+\int_{t}^{T}(\tilde{a}(L_{s}^{t,l})+\tilde{b}(L_{s}^{t,l}))\,h_{2}(s,L_{s}^{t,l})$
$\displaystyle\hskip 113.81102pt\cdot
e^{\int_{t}^{s}\gamma(a(L_{u}^{t,l}+\epsilon\,e^{i})+b(L_{u}^{t,l}+\epsilon\,e^{i}))\,(h_{2}(u,L_{u}^{t,l}+\epsilon\,e^{i})+h_{2}(u,L_{u}^{t,l}))\,du}\Big{]},$

which further implies

$|g(t,l)|\leq C_{\text{Lip}}\,e^{2\gamma
T(\bar{a}+\bar{b})(\bar{A}+\bar{\phi}\,T)}+2\,T\,C_{\text{Lip}}(\bar{A}+\bar{\phi}\,T)\,e^{2\gamma
T(\bar{a}+\bar{b})(\bar{A}+\bar{\phi}\,T)}$ (21)

for all $(t,l)\in[0,T]\times\mathbb{R}^{d}$. We emphasize that the right hand
side of (21) is independent of $\epsilon$. It follows $\partial h_{2}/\partial
l^{i}$ is uniformly bounded and the same is true for all elements in the
vector $\partial h_{2}/\partial l$.

Let $Y_{t}:=h_{2}(t,L_{t})$, $Z_{t}:=(\partial h_{2}/\partial
l)(t,L_{t})\,\Sigma(t,L_{t})$, and then apply Itô’s formula to obtain

$Y_{t}=-A+\int_{t}^{T}\gamma\,(a_{s}+b_{s})\,Y_{s}^{2}\,ds-\int_{t}^{T}Z_{s}\,dW_{s}.$

Because $Y_{t}<0$, to observe the following we use Itô’s formula again:

$\displaystyle d(Y_{t}^{-1})$
$\displaystyle=-A^{-1}-\int_{t}^{T}\gamma\,(a_{s}+b_{s})\,ds-\int_{t}^{T}\frac{Z_{s}}{Y_{s}^{2}}\big{(}dW_{s}-\frac{Z_{s}}{Y_{s}}\,ds\big{)}$
(22)
$\displaystyle=-A^{-1}-\int_{t}^{T}\gamma\,(a_{s}+b_{s})\,ds-\int_{t}^{T}\frac{Z_{s}}{Y_{s}^{2}}\,d\tilde{W}_{s},$

where $\tilde{W}$ is a standard Brownian motion with respect to some
probability measure $\mathbb{Q}$, guaranteed by the Girsanov theorem. Indeed,
the Girsanov transform can be applied since $Z_{t}$ is bounded and
$Y_{t}<-A\,\exp{(-\gamma\,(\bar{a}+\bar{b})\,A\,T)}$ by the mapping $F_{1}$.
An explicit form of $h_{2}$ is obtained via taking the conditional expectation
of (22):

$h_{2}(t,L_{t})=Y_{t}=\bigg{\\{}\,\mathbb{E}_{t}^{\mathbb{Q}}\Big{[}-A^{-1}-\int_{t}^{T}\gamma\,(a_{s}+b_{s})\,ds\Big{]}\,\bigg{\\}}^{-1}.$

The first-order approximation, mentioned in [4] and [9], turns out to be the
rectangle approximation of the integral $\int_{t}^{T}(a_{s}+b_{s})\,ds$. In
our context, the integral $\int_{t}^{T}(a_{s}+b_{s})\,ds$ stands for the total
volume of future order flows. While $h_{2}$ is the coefficient of the second-
order term in $H$, the coefficient of the first-order term $h_{1}$—by the
Feynman–Kac Formula—can be represented by

$h_{1}(t,l)=\zeta\,\mathbb{E}_{t,l}\Big{[}\int_{t}^{T}(b_{s}-a_{s})\cdot
h_{2}(s,L_{s})\,e^{\gamma\int_{t}^{s}(a_{u}+b_{u})\,h_{2}(u,L_{u})\,du}\,ds\Big{]},$

which estimates a weighted sum of future order imbalances.

##  3\. Non-Markovian Order Flows: General Intensity Functions

In this section, we extend the market making problem in two directions. First,
the order flows and penalty parameters are not necessarily Markovian.

######  Assumption 3.1.

Let $a$, $b\in\mathbb{H}^{2}$ be positive processes and
$\phi\in\mathbb{H}^{2}$ be a non-negative process, such that they are bounded
by constants $\bar{a}$, $\bar{b},$ $\bar{\phi}>0$ respectively. Let $A\in
L^{2}(\Omega,\mathcal{F}_{T})$ be a non-negative random variable, bounded by
$\bar{A}>0$.

On the other hand, more general intensity functions are considered. We borrow
the following notations, definitions, and associated results from [10].

######  Assumption 3.2 ([10]).

A function $\Lambda:\mathbb{R}\to\mathbb{R_{+}}$ belongs to the class of
intensity functions $\boldsymbol{\Lambda}$ if:

  * 1.

$\Lambda$ is twice continuously differentiable;

  * 2.

$\Lambda$ is strictly decreasing and hence $\Lambda^{\prime}(x)<0$ for any
$x\in\mathbb{R}$;

  * 3.

$\lim_{x\to\infty}\Lambda(x)=0\,$ and
$\;\sup_{x\in\mathbb{R}}\frac{\Lambda(x)\,\Lambda^{\prime\prime}(x)}{(\Lambda^{\prime}(x))^{2}}<2$.

######  Lemma 3.3 ([10]).

For any $\Lambda\in\boldsymbol{\Lambda}$, define the function
$\mathcal{W}:\mathbb{R}\to\mathbb{R}$ as
$\mathcal{W}(p)=\sup_{\delta\in\mathbb{R}}\Lambda(\delta)\,(\delta-p)$. Then,
the following holds:

  * 1.

$\mathcal{W}$ is a decreasing function of class $C^{2}$;

  * 2.

The supremum in the definition of $\mathcal{W}$ is attained at a unique
$\delta^{*}(p)$ characterized by

$\delta^{*}(p)=\Lambda^{-1}\big{(}-\mathcal{W}^{\prime}(p)\big{)},$

where $\Lambda^{-1}$ denotes the inverse function of $\Lambda$;

  * 3.

The function $p\mapsto\delta^{*}(p)$ belongs to $C^{1}$ and is increasing. Its
derivative reads

$(\delta^{*})^{\prime}(p)=\Big{[}2-\frac{\Lambda(\delta^{*}(p))\,\Lambda^{\prime\prime}(\delta^{*}(p))}{\Lambda^{\prime}(\delta^{*}(p))^{2}}\Big{]}^{-1}>0.$

Denote by
$(\Lambda^{a},\Lambda^{b})\in\boldsymbol{\Lambda}\times\boldsymbol{\Lambda}$
the intensity functions of the ask and bid sides. Additionally, a strategy
$\boldsymbol{\delta}:=(\delta^{a},\delta^{b})$ is admissible if both two
entries are bounded by $\xi$ for some large $\xi>0$, i.e., the admissible
space is defined as

$\mathbb{A}:=\\{\delta\in\mathbb{H}^{2}:|\delta_{t}|\leq\xi\text{ for all
}t\in[0,T]\,\\}.$

The inventory and cash processes of the market maker are then given by

$\displaystyle
Q_{t}=q_{0}-\int_{0}^{t}a_{u}\Lambda^{a}(\delta_{u}^{a})\,du+\int_{0}^{t}b_{u}\Lambda^{b}(\delta_{u}^{b})\,du,$
$\displaystyle
X_{t}=x_{0}+\int_{0}^{t}a_{u}\Lambda^{a}(\delta_{u}^{a})\,(S_{u}+\delta_{u}^{a})\,du-\int_{0}^{t}b_{u}\Lambda^{b}(\delta_{u}^{b})\,(S_{u}-\delta_{u}^{b})\,du.$

The agent intends to maximise the same objective functional:

$\displaystyle\mathbb{E}\big{[}X_{T}+S_{T}\,Q_{T}-\int_{0}^{T}\phi_{t}\,(Q_{t})^{2}\,dt-A\,(Q_{T})^{2}\big{]}$
$\displaystyle=\mathbb{E}\Big{[}\int_{0}^{T}\delta_{t}^{a}\,a_{t}\,\Lambda^{a}(\delta_{t}^{a})\,dt+\int_{0}^{T}\delta_{t}^{b}\,b_{t}\,\Lambda^{b}(\delta_{t}^{b})\,dt-\int_{0}^{T}\phi_{t}\,(Q_{t})^{2}\,dt-A\,(Q_{T})^{2}\Big{]}$

by controlling $\boldsymbol{\delta}\in\mathbb{A}\times\mathbb{A}$. To apply
the Pontryagin maximum principle, the Hamiltonian of the agent with respect to
the objective functional reads

$\mathscr{H}(t,Q_{t},Y_{t},\boldsymbol{\delta}_{t})=\big{[}b_{t}\,\Lambda^{b}(\delta_{t}^{b})-a_{t}\,\Lambda^{a}(\delta_{t}^{a})\big{]}\,Y_{t}+b_{t}\,\delta_{t}^{b}\Lambda^{b}(\delta_{t}^{b})+a_{t}\,\delta_{t}^{a}\Lambda^{a}(\delta_{t}^{a})-\phi\,Q_{t}^{2}.$

While the function $\mathscr{H}$ is concave in the state variable $Q$, the
concavity with respect to the control $\boldsymbol{\delta}$ is not guaranteed,
which is required by the usual stochastic maximum principle (for example [7]).
However, the separation between the state variable and control enables us to
still apply this principle, the proof of which is placed in the appendix.
Denote by $(\delta^{a*},\delta^{b*})$ the maximisers associated with the
intensity functions $(\Lambda^{a},\Lambda^{b})$ according to Lemma 3.3, and
further define

$\tilde{\delta}^{i*}(p):=\delta^{i*}(p)\wedge\xi\vee(-\xi)$

for $i\in\\{a,b\\}$. Then the optimal feedback controls on the ask and bid
sides are given by

$\tilde{\delta}^{a*}(Y_{t})\quad\text{and}\quad\tilde{\delta}^{b*}(-Y_{t}).$

Indeed, on the ask side, we look for the maximizer of

$\sup_{|\delta_{t}^{a}|\leq\xi}g(\delta_{t}^{a}):=\sup_{|\delta_{t}^{a}|\leq\xi}\Lambda^{a}(\delta_{t}^{a})\cdot(\delta_{t}^{a}-Y_{t}).$

Without the constraint $|\delta_{t}^{a}|\leq\xi$, the maximiser simply reads
$\delta^{a*}(Y_{t})$ according to Lemma 3.3. However, being aware of

$g^{\prime}(\delta_{t}^{a})=\Lambda^{a^{\prime}}(\delta_{t}^{a})\,\big{[}\delta_{t}^{a}+\frac{\Lambda^{a}(\delta_{t}^{a})}{\Lambda^{a^{\prime}}(\delta_{t}^{a})}-Y_{t}\big{]}$

and the fact that
$\delta_{t}^{a}+\Lambda^{a}(\delta_{t}^{a})/\Lambda^{a^{\prime}}(\delta_{t}^{a})$
is an increasing function because

$\big{[}\delta_{t}^{a}+\frac{\Lambda^{a}(\delta_{t}^{a})}{\Lambda^{a^{\prime}}(\delta_{t}^{a})}\big{]}^{\prime}=2-\frac{\Lambda^{a}(\delta_{t}^{a})\,\Lambda^{a^{\prime\prime}}(\delta_{t}^{a})}{\Lambda^{a^{\prime}}(\delta_{t}^{a})^{2}}>0,$

one can see that the function $g$ first increases then decreases. Hence, the
maximiser is $\tilde{\delta}^{a*}(Y_{t})$ under the constraint stated in
$\mathbb{A}$. The bid side can be derived in the same way. Combined with the
stochastic maximum principle, the above discussion yields the next result.

######  Theorem 3.4.

Let $\boldsymbol{\delta}\in\mathbb{A}\times\mathbb{A}$ be an admissible
control, $Q=Q^{\boldsymbol{\delta}}$ be the corresponding controlled
inventory, and $(Y,Z)$ be the adjoint processes. Then $\boldsymbol{\delta}$ is
an optimal control if and only if it holds $\mathbb{P}$-a.s. that

$\mathscr{H}(t,Q_{t},Y_{t},\boldsymbol{\delta}_{t})=\sup_{\boldsymbol{\alpha}\in[-\xi,\xi]^{2}}\mathscr{H}(t,Q_{t},Y_{t},\boldsymbol{\alpha}),\qquad\text{a.e.
in }t\in[0,T],$

or equivalently

$\delta^{a}_{t}=\tilde{\delta}^{a*}(Y_{t}),\quad\text{and}\quad\delta^{b}_{t}=\tilde{\delta}^{b*}(-Y_{t}),\qquad\text{a.e.
in }t\in[0,T].$

Further, the optimal inventory $Q$ together with the adjoint processes $(Y,Z)$
solves the FBSDE

$\left\\{\begin{aligned}
\;&dQ_{t}=-a_{t}\,\Lambda^{a}\big{(}\tilde{\delta}^{a*}(Y_{t})\big{)}dt+b_{t}\,\Lambda^{b}\big{(}\tilde{\delta}^{b*}(-Y_{t})\big{)}dt,\\\
&dY_{t}=2\phi_{t}\,Q_{t}\,dt+Z_{t}\,dW_{t},\\\ &Q_{0}=q_{0},\quad
Y_{T}=-2A\,Q_{T}.\end{aligned}\right.$ (23)

###### PROOF.

See the Appendix for the stochastic maximum principle. ∎

Equation (23) is a degenerate FBSDE with Lipschitz coefficients. To see this,
note that $(\Lambda^{a},\Lambda^{b})$ have bounded derivatives in the
admissible action space, and the same is true for $(\delta^{a*},\delta^{b*})$
since

$0<(\delta^{i*})^{\prime}(p)=\Big{[}2-\frac{\Lambda^{i}(\delta^{i*}(p))\,\Lambda^{i^{\prime\prime}}(\delta^{i*}(p))}{\Lambda^{i^{\prime}}(\delta^{i*}(p))^{2}}\Big{]}^{-1}\leq\Big{[}2-\sup_{u\in\mathbb{R}}\frac{\Lambda^{i}(u)\,\Lambda^{i^{\prime\prime}}(u)}{\Lambda^{i^{\prime}}(u)^{2}}\Big{]}^{-1}<\infty,$

for $i\in\\{a,b\\}$. The Lipschitz property of the forward equation of (23)
then follows from the boundedness of these two derivatives and flows $a$, $b$.
The remaining of section is devoted to the (global) well-posedness of the
FBSDE (23).

######  Remark 3.5.

(1) The constant $\xi$ in the definition of $\mathbb{A}$ can be also
interpreted as a regularizer for the equation (23) to make it Lipschitz.
Later, we will see how it can be removed.
(2) If one additionally imposes that $a$, $b$ and $\phi$ are bounded away from
$0$, the well-posedness of (23) can be verified by the method of continuation
(see [17]).

We first introduce an auxiliary result on a linear degenerate FBSDE, the proof
of which is based on the same technique used for (10).

######  Lemma 3.6.

Let $(\mu_{t})_{0\leq t\leq T}\in\mathbb{H}^{2}$ be a non-negative bounded
process with the upper bound $\bar{\mu}>0$, and $\nu_{T}\in
L^{2}(\Omega,\mathcal{F}_{T})$ be a non-negative bounded random variable
bounded by $\bar{\nu}>0$. The FBSDE

$\left\\{\begin{aligned} \;&dQ_{t}=\mu_{t}\,Y_{t}\,dt,\\\
&dY_{t}=2\phi_{t}\,Q_{t}\,dt+Z_{t}\,dW_{t},\\\ &Q_{0}=q_{0},\quad
Y_{T}=-\nu_{T}\,Q_{T}\end{aligned}\right.$ (24)

admits a unique solution in
$\mathbb{S}^{2}\times\mathbb{S}^{2}\times\mathbb{H}^{2}$. In particular, the
solution accepts the representation

$Y_{t}=P_{t}\cdot Q_{t}\text{\quad and
\quad}Q_{t}=q_{0}\,\exp\big{(}\int_{0}^{t}\mu_{s}\,P_{s}\,ds\big{)},$

with $(P_{t})_{0\leq t\leq T}\in\mathbb{H}^{2}$ being non-positive and bounded
by $\bar{\nu}+2\bar{\phi}\,T$.

###### PROOF.

Motivated by the linearity of (24), we adopt the affine ansatz
$Y_{t}=P_{t}\cdot Q_{t}$ for some $P:=(P_{t})_{0\leq t\leq T}$ to be
determined. Through applying Itô’s formula and matching the coefficients, one
can see that $P$ satisfies the BSDE

$dP_{t}=\big{(}-\mu_{t}\,(P_{t})^{2}+2\phi_{t}\big{)}\,dt+\tilde{Z}_{t}\,dW_{t},\quad\text{such
that}\;\;P_{T}=-\nu_{T}.$ (25)

To study this non-Lipschitz BSDE, we similarly look at two auxiliary linear
BSDEs:

$\displaystyle
dP_{t}=\big{(}-\mu_{t}\,(G_{t}\wedge(-\tilde{\xi})\vee\tilde{\xi})^{2}+2\phi_{t}\big{)}\,dt+Z_{t}^{1}\,dW_{t},\quad\text{such
that}\;\;P_{T}=-\nu_{T};$ $\displaystyle
dP_{t}=\big{(}-\mu_{t}\,G_{t}P_{t}+2\phi_{t}\big{)}\,dt+Z_{t}^{2}\,dW_{t},\quad\text{such
that}\;\;P_{T}=-\nu_{T},$

and their associated mappings

$\displaystyle(F_{3}\,G)_{t}=\mathbb{E}_{t}\Big{[}-\nu_{T}+\int_{t}^{T}\mu_{s}\,(G_{s}\wedge(-\tilde{\xi})\vee\tilde{\xi})^{2}\,ds-\int_{t}^{T}2\phi_{s}\,ds\Big{]},$
$\displaystyle(F_{4}\,G)_{t}=\mathbb{E}_{t}\Big{[}-\nu_{T}\,e^{\int_{t}^{T}\mu_{u}\,G_{u}\,du}-\int_{t}^{T}2\phi_{s}\,e^{\int_{t}^{s}\mu_{u}\,G_{u}\,du}\,ds\Big{]},$

for some constant $\tilde{\xi}>0$ and process $G\in\mathbb{S}^{2}$ to be
specified. As before, $F_{3}$ is a contraction mappings on the space
$\mathbb{S}^{2}$ with respect to the norm

$\|G\|_{\zeta_{3}}:=\mathbb{E}\big{[}\sup_{0\leq t\leq
T}e^{-2\zeta_{3}\,(T-t)}\cdot|G_{t}|^{2}\,\big{]}^{1/2},$

and $F_{4}$ is a contraction mappings on the space of non-positive processes
in $\mathbb{S}^{2}$ with respect to the norm

$\|G\|_{\zeta_{4}}:=\mathbb{E}\big{[}\sup_{0\leq t\leq
T}e^{-2\zeta_{4}\,(T-t)}\cdot|G_{t}|^{2}\,\big{]}^{1/2},$

for some $\zeta_{3},\zeta_{4}>0$ large enough. We learn from the Banach fixed
point theorem that both $F_{3}$ and $F_{4}$ have their own unique fixed
points. Given $\tilde{\xi}>\bar{\nu}+2\bar{\phi}\,T$, they further share the
same fixed point. Finally, the common fixed point solves the equation (25) and
guarantees the existence of the solution. To see that (25) accepts a unique
bounded solution, let $P^{1}$, $P^{2}\in\mathbb{S}^{2}$ be two solutions with
bounds $\bar{P}^{1}$, $\bar{P}^{2}>0$ respectively. If we let
$\tilde{\xi}\geq\bar{P}^{1}\vee\bar{P}^{2}$, both $P^{1}$ and $P^{2}$ are then
fixed points of the mapping $F_{3}$, implying $P^{1}=P^{2}$ due the uniqueness
of the fixed point.

The uniqueness of (24) follows from the method of continuation as in [17].
Given two solutions $(Q,Y,Z)$ and $(\tilde{Q},\tilde{Y},\tilde{Z})$ in
$\mathbb{S}^{2}\times\mathbb{S}^{2}\times\mathbb{H}^{2}$, then
$(\mathcal{Q},\mathcal{Y},\mathcal{Z}):=(\tilde{Q}-Q,\tilde{Y}-Y,\tilde{Z}-Z)$
solves the FBSDE

$\left\\{\begin{aligned} \;&d\mathcal{Q}_{t}=\mu_{t}\,\mathcal{Y}_{t}dt,\\\
&d\mathcal{Y}_{t}=2\phi_{t}\,\mathcal{Q}_{t}\,dt+\mathcal{Z}_{t}\,dW_{t},\\\
&\mathcal{Q}_{0}=0,\quad\mathcal{Y}_{T}=-\nu_{T}\,\mathcal{Q}_{T}.\end{aligned}\right.$

While Itô’s formula implies

$d(\mathcal{Q}_{t}\,\mathcal{Y}_{t})=\big{[}\mu_{t}(\mathcal{Y}_{t})^{2}+2\phi_{t}(\mathcal{Q}_{t})^{2}\big{]}\,dt+\mathcal{Q}_{t}\,\mathcal{Z}_{t}\,dW_{t},$

through observing

$0\geq\mathbb{E}[-\nu_{T}\,(\mathcal{Q}_{T})^{2}]=\mathbb{E}\int_{0}^{T}\big{[}\mu_{t}\,(\mathcal{Y}_{t})^{2}+2\phi_{t}\,(\mathcal{Q}_{t})^{2}\big{]}\,dt\geq\mathbb{E}\int_{0}^{T}\mu_{t}\,(\mathcal{Y}_{t})^{2}\,dt\geq
0$

we can see $\mu_{t}\,\mathcal{Y}_{t}=0$ a.e. in $t$ and thus $\mathcal{Q}=0$.
The fact $\mathcal{Y}=\mathcal{Z}=0$ then follows immediately. ∎

######  Remark 3.7.

The arguments of the continuation method in proving the uniqueness of the
solution will be applied to some other equations later. We emphasize that such
arguments do not need the boundedness of $\mu$, $\phi$ and $\nu_{T}$.

Since the FBSDE (23) is Lipschitz, denote by $\iota>0$ the Lipschitz constant
for the coefficient functions of forward and backward equations of
$\eqref{Lip_den_FBSDE}$ (excluding the terminal condition), i.e., it holds
$\mathbb{P}$-a.s. for any $y,\tilde{y},q,\tilde{q}$ that

$\displaystyle\Big{|}-a_{t}\,\Lambda^{a}\big{(}\tilde{\delta}^{a*}(y)\big{)}+b_{t}\,\Lambda^{b}\big{(}\tilde{\delta}^{b*}(-y)\big{)}+a_{t}\,\Lambda^{a}\big{(}\tilde{\delta}^{a*}(\tilde{y})\big{)}-b_{t}\,\Lambda^{b}\big{(}\tilde{\delta}^{b*}(-\tilde{y})\big{)}\Big{|}$
$\displaystyle\leq\iota\,|\tilde{y}-y|,$
$\displaystyle\big{|}2\phi_{t}\,q-2\phi_{t}\,\tilde{q}\big{|}$
$\displaystyle\leq\iota\,|\tilde{q}-q|,$
$\displaystyle\big{|}2A\,q-2A\,\tilde{q}\big{|}$
$\displaystyle\leq\iota\,|\tilde{q}-q|.$

To prove the well-posedness, we apply the decoupling approach introduced by
[15]. Actually, Lemma 3.6 studies the growth of the Lipschitz coefficient—a
critical step for such technique.

######  Theorem 3.8.

The FBSDE (23) accepts a unique solution in
$\mathbb{S}^{2}\times\mathbb{S}^{2}\times\mathbb{H}^{2}$. Further, the optimal
control is monotonic with respect to the initial (inventory) condition, i.e.,
if $q_{1}>q_{2}$, then it holds that

$\hat{\delta}_{1,t}^{a}\leq\hat{\delta}_{2,t}^{a}\quad\text{and}\quad\hat{\delta}_{1,t}^{b}\geq\hat{\delta}_{2,t}^{b}$

for $t\in[0,T]$ $\mathbb{P}$-a.s., where
$(\hat{\delta}_{i,t}^{a},\hat{\delta}_{i,t}^{b})_{0\leq t\leq T}$ represents
the optimal control associated with the initial condition $q_{i}$ for
$i\in\\{1,2\\}$.

###### PROOF.

Short time analysis: Since the FBSDE is of Lipschitz type, we start with the
short time analysis. Fix any $q_{0}\in\mathbb{R}$, if $Q=(Q_{t})_{0\leq t\leq
T}\in\mathbb{S}^{2}$ is given such that $Q_{0}=q_{0}$, denote by
$(Y,Z)\in\mathbb{S}^{2}\times\mathbb{H}^{2}$ the unique solution of the BSDE:

$dY_{t}=2\phi_{t}\,Q_{t}\,dt+Z_{t}\,dW_{t},\text{\; such that
\;}Y_{T}=-2A\,Q_{T}.$

Subsequently, given $(Y,Z)$, we let $Q^{\prime}_{t}$ be the uniqueness
solution of the SDE

$dQ^{\prime}_{t}=-a_{t}\,\Lambda^{a}\big{(}\tilde{\delta}^{a*}(Y_{t})\big{)}dt+b_{t}\,\Lambda^{b}\big{(}\tilde{\delta}^{b*}(-Y_{t})\big{)}dt,\text{\;
such that \;}Q^{\prime}_{0}=q_{0}.$

In such way, we have defined a mapping

$\mathbb{S}^{2}\ni Q\hookrightarrow\Phi(Q)=Q^{\prime}\in\mathbb{S}^{2}$

and proceed to show that $\Phi$ is a contraction mapping when $T$ is small
enough. For any $Q$, $\tilde{Q}\in\mathbb{S}^{2}$, write $Q^{\prime}=\Phi(Q)$,
$\tilde{Q}^{\prime}=\Phi(\tilde{Q})$ and note that

$\displaystyle|\tilde{Q}^{\prime}_{t}-Q^{\prime}_{t}|$
$\displaystyle\leq\iota\,T\sup_{0\leq s\leq T}|\tilde{Y}_{s}-Y_{s}|$ (26)
$\displaystyle\mathbb{E}\sup_{0\leq s\leq
T}|\tilde{Q}^{\prime}_{s}-Q^{\prime}_{s}|^{2}$
$\displaystyle\leq(\iota\,T)^{2}\,\mathbb{E}\sup_{0\leq s\leq
T}|\tilde{Y}_{s}-Y_{s}|^{2}.$

To obtain the stability for the $\tilde{Y}$ and $Y$, we further compute

$\displaystyle|\tilde{Y}_{t}-Y_{t}|^{2}$ $\displaystyle\leq(2\bar{A}+2\iota
T)^{2}\,\mathbb{E}_{t}\big{[}\sup_{0\leq s\leq
T}|\tilde{Q}_{s}-Q_{s}|\big{]}^{2},$ (27) $\displaystyle\mathbb{E}\sup_{0\leq
s\leq T}|\tilde{Y}_{s}-Y_{s}|^{2}$ $\displaystyle\leq(2\bar{A}+2\iota
T)^{2}\,\mathbb{E}\Big{[}\sup_{0\leq t\leq T}\mathbb{E}_{t}\big{[}\sup_{0\leq
s\leq T}|\tilde{Q}_{s}-Q_{s}|\big{]}^{2}\Big{]},$ $\displaystyle\leq
4\,(2\bar{A}+2\iota T)^{2}\,\mathbb{E}\sup_{0\leq s\leq
T}|\tilde{Q}_{s}-Q_{s}|^{2},$

where the Doob’s $L^{p}$ inequality is applied in the last line. Combining
(26) and (27), we find that

$\mathbb{E}\sup_{0\leq s\leq T}|\tilde{Q}^{\prime}_{s}-Q^{\prime}_{s}|^{2}\leq
4\,(\iota T)^{2}\,(2\bar{A}+2\iota T)^{2}\,\mathbb{E}\sup_{0\leq s\leq
T}|\tilde{Q}_{s}-Q_{s}|^{2}$

and it suffices to pick $T$ satisfying

$4\,(\iota T)^{2}\,(2\bar{A}+2\iota T)^{2}<1$

for a contraction mapping. The FBSDE (23) accepts a unique solution in
$\mathbb{S}^{2}\times\mathbb{S}^{2}\times\mathbb{H}^{2}$ with respect to the
such $T$.

Decoupling approach: we then extend the local result to any finite time
horizon $T$. Define time step $\Delta>0$ as

$\Delta=\frac{1}{\sqrt{2}}\cdot\big{[}4\iota^{2}\,(2\bar{A}+2\bar{\phi}T+2\iota
T)^{2}\big{]}^{-1/2}.$ (28)

Note that

$4\,(\iota\Delta)^{2}\,(2\bar{A}+2\iota\Delta)^{2}\leq
4\,(\iota\Delta)^{2}\,(2\bar{A}+2\bar{\phi}T+2\iota T)^{2}=\frac{1}{2}$

and thus the following FBSDE is well-posed by the previous discussion:

$\left\\{\begin{aligned}
\;&dQ_{t}=-a_{t}\,\Lambda^{a}\big{(}\tilde{\delta}^{a*}(Y_{t})\big{)}dt+b_{t}\,\Lambda^{b}\big{(}\tilde{\delta}^{b*}(-Y_{t})\big{)}dt,\\\
&dY_{t}=2\phi_{t}\,Q_{t}\,dt+Z_{t}\,dW_{t},\\\
&Q_{T-\Delta}=q_{T-\Delta},\quad Y_{T}=-2A\,Q_{T}.\end{aligned}\right.$ (29)

We can then define the decoupling field
$u:[T-\Delta,T]\times\Omega\times\mathbb{R}\to\mathbb{R}$ by

$u(t,q):=Y_{t}^{t,q},$ (30)

where the superscript $(t,q)$ denotes the initial time and condition of (29).
Let $q_{1}$ and $q_{2}$ be two initial conditions for the FBSDE (29), and—due
to the well-posedness—denote by $(Q^{1},Y^{1},Z^{1})$ and
$(Q^{2},Y^{2},Z^{2})$ the solutions corresponding to these initial conditions.
For $t$ such that $Y_{t}^{1}\neq Y_{t}^{2}$, let

$\displaystyle d(Q_{t}^{1}-Q_{t}^{2})/dt$
$\displaystyle=-a_{t}\big{[}\Lambda^{a}\big{(}\tilde{\delta}^{a*}(Y_{t}^{1})\big{)}-\Lambda^{a}\big{(}\tilde{\delta}^{a*}(Y_{t}^{2})\big{)}\big{]}+b_{t}\,\big{[}\Lambda^{b}\big{(}\tilde{\delta}^{b*}(-Y_{t}^{1})\big{)}-\Lambda^{b}\big{(}\tilde{\delta}^{b*}(-Y_{t}^{2})\big{)}\big{]}$
$\displaystyle=:\mathcal{U}_{t}\cdot(Y_{t}^{1}-Y_{t}^{2}).$

Naturally, for those $t$ such that $Y_{t}^{1}=Y_{t}^{2}$, we let
$\mathcal{U}_{t}=0$. Note that $\mathcal{U}\in\mathbb{H}^{2}$ is bounded by
$\iota$ due to the Lipschitz continuity. It is further non-negative since the
expression

$-a\,\Lambda^{a}\big{(}\tilde{\delta}^{a*}(y)\big{)}+b\,\Lambda^{b}\big{(}\tilde{\delta}^{b*}(-y)\big{)}$

is non-decreasing with respect to $y$, which can be deduced from the
increasing property of $(\delta^{a*},\delta^{b*})$ and decreasing property of
$(\Lambda^{a},\Lambda^{b})$. One can then observe that
$(Q^{1}-Q^{2},Y^{1}-Y^{2},Z^{1}-Z^{2})$ satisfies the following FBSDE

$\left\\{\begin{aligned}
\;&d\mathcal{Q}_{t}=\mathcal{U}_{t}\,\mathcal{Y}_{t}dt,\\\
&d\mathcal{Y}_{t}=2\phi_{t}\,\mathcal{Q}_{t}\,dt+\mathcal{Z}_{t}\,dW_{t},\\\
&\mathcal{Q}_{T-\Delta}=q_{1}-q_{2},\quad\mathcal{Y}_{T}=-2A\,\mathcal{Q}_{T},\end{aligned}\right.$
(31)

for $t\in[T-\Delta,T]$, which is called the variational FBSDE associated with
the original (23). While (31) is a special case of the result in Lemma 3.6,
equation (31) admits a unique solution

$(\mathcal{Q}_{t},\mathcal{Y}_{t},\mathcal{Z}_{t})=\big{(}Q_{t}^{1}-Q_{t}^{2},Y_{t}^{1}-Y_{t}^{2},Z_{t}^{1}-Z_{t}^{2}\big{)}$

and, more importantly, process $\mathcal{Y}$ is bounded by
$|q_{1}-q_{2}|\cdot(2\bar{A}+2\bar{\phi}\,\Delta)$. According to the
definition of the decoupling field, we deduce that $u$ is uniformly Lipschitz
continuous $\mathbb{P}$-a.s. with respect to variable $q$ with Lipschitz
coefficient $2\bar{A}+2\bar{\phi}\,\Delta$. Consider the FBSDE in the next
interval

$\left\\{\begin{aligned}
\;&dQ_{t}=-a_{t}\,\Lambda^{a}\big{(}\tilde{\delta}^{a*}(Y_{t})\big{)}dt+b_{t}\,\Lambda^{b}\big{(}\tilde{\delta}^{b*}(-Y_{t})\big{)}dt,\\\
&dY_{t}=2\phi_{t}\,Q_{t}\,dt+Z_{t}\,dW_{t},\\\
&Q_{T-2\Delta}=q_{T-2\Delta},\quad
Y_{T-\Delta}=u(T-\Delta,Q_{T-\Delta}).\end{aligned}\right.$ (32)

With the same short time analysis, it suffices to make sure

$4\,(\iota\Delta)^{2}\,(2\bar{A}+2\bar{\phi}\Delta+2\iota\Delta)^{2}<1,$

and our choice (28) is clearly qualified. Referring to Lemma 3.6, we study the
variational FBSDE and observe the linear growth of the Lipschitz coefficient
of the decoupling field. Because the potential Lipschitz coefficient of the
decoupling field can not exceed $2\bar{A}+2\bar{\phi}T$, the time step
$\Delta$ will always qualify, and hence the field $u$ can be extended to the
whole time horizon $[0,T]$ by a finite number of iterations. The global
existence of the solution then can be inferred from the (global) decoupling
field $u$. Because the uniqueness of solution can be proved in the same
fashion as Lemma 3.6, the well-posedness of (23) is established.

Monotonicity: to verify the monotonicity property, we let $q_{1}>q_{2}$ be two
initial conditions, denoting by $(Q^{1},Y^{1},Z^{1})$ and
$(Q^{2},Y^{2},Z^{2})$ the associated solutions. Since
$(Q^{1}-Q^{2},Y^{1}-Y^{2},Z^{1}-Z^{2})$ solves the variational FBSDE (31) with
initial condition $q_{1}-q_{2}>0$, again by Lemma 3.6 we see $Y^{1}-Y^{2}$ is
negative. Recalling that the optimal controls read

$\hat{\delta}_{t}^{a}=\tilde{\delta}^{a*}(Y_{t})\quad\text{and}\quad\hat{\delta}_{t}^{b}=\tilde{\delta}^{b*}(-Y_{t}),$

the smaller value of $Y_{t}$ results in a smaller $\hat{\delta}_{t}^{a}$ but a
larger $\hat{\delta}_{t}^{b}$. ∎

We finish this section by removing the constant $\xi$. It turns out that,
provided $\xi$ is picked large enough, the corresponding constraint will have
no impact on the control. That is to say, we are able to find a (unique)
solution to the equation

$\left\\{\begin{aligned}
\;&dQ_{t}=-a_{t}\,\Lambda^{a}\big{(}\delta^{a*}(Y_{t})\big{)}dt+b_{t}\,\Lambda^{b}\big{(}\delta^{b*}(-Y_{t})\big{)}dt,\\\
&dY_{t}=2\phi_{t}\,Q_{t}\,dt+Z_{t}\,dW_{t},\\\ &Q_{0}=q_{0},\quad
Y_{T}=-2A\,Q_{T}.\end{aligned}\right.$ (33)

Note that this FBSDE is non-Lipschitz because the derivatives of
$(\Lambda^{a},\Lambda^{b})$ are not necessarily bounded.

######  Proposition 3.9.

The FBSDE (33) accepts a unique solution in
$\mathbb{S}^{2}\times\mathbb{S}^{2}\times\mathbb{H}^{2}$. Further, the optimal
control is monotonic with respect to the initial (inventory) condition, i.e.,
if $q_{1}>q_{2}$, then it holds that

$\hat{\delta}_{1,t}^{a}<\hat{\delta}_{2,t}^{a}\quad\text{and}\quad\hat{\delta}_{1,t}^{b}>\hat{\delta}_{2,t}^{b}$

for $t\in[0,T]$ $\mathbb{P}$-a.s., where
$(\hat{\delta}_{i,t}^{a},\hat{\delta}_{i,t}^{b})_{0\leq t\leq T}$ represents
the optimal control associated with the initial condition $q_{i}$ for
$i\in\\{1,2\\}$.

###### PROOF.

We intend to construct a solution to (33) via the ‘truncated’ version:

$\left\\{\begin{aligned}
\;&dQ_{t}=-a_{t}\,\Lambda^{a}\big{(}\tilde{\delta}^{a*}(Y_{t})\big{)}dt+b_{t}\,\Lambda^{b}\big{(}\tilde{\delta}^{b*}(-Y_{t})\big{)}dt,\\\
&dY_{t}=2\phi_{t}\,Q_{t}\,dt+Z_{t}\,dW_{t},\\\ &Q_{0}=q_{0},\quad
Y_{T}=-2A\,Q_{T}.\end{aligned}\right.$ (34)

The idea comes from the solution of the linear case. Recall that the value
function $H$ is composed of three terms with different order (see Theorem
2.6): the one of second order is induced from the terminal inventory penalty
and is uniformly bounded by the penalty parameters; the one of first order is
the estimation of the integral of future order imbalance. We are then
motivated to look at two extreme unbalanced situations: the case when the buy
market order dominates and the case when the dominance is achieved by the sell
market order. Thus, we look at two associate FBSDEs:

$\left\\{\begin{aligned}
\;&dQ_{t}=b_{t}\,\Lambda^{b}\big{(}\tilde{\delta}^{b*}(-Y_{t})\big{)}dt,\\\
&dY_{t}=2\phi_{t}\,Q_{t}\,dt+Z_{t}\,dW_{t},\\\ &Q_{0}=q_{0},\quad
Y_{T}=-2A\,Q_{T};\end{aligned}\right.$ (35) $\left\\{\begin{aligned}
\;&dQ_{t}=-a_{t}\,\Lambda^{a}\big{(}\tilde{\delta}^{a*}(Y_{t})\big{)}dt,\\\
&dY_{t}=2\phi_{t}\,Q_{t}\,dt+Z_{t}\,dW_{t},\\\ &Q_{0}=q_{0},\quad
Y_{T}=-2A\,Q_{T}.\end{aligned}\right.$ (36)

We hope the solutions of these two equations would reveal certain properties
of (34), so that the parameter $\xi$ can be removed safely. Denote by
$(\tilde{Q},\tilde{Y},\tilde{Z})$ and $(\hat{Q},\hat{Y},\hat{Z})$ the (unique)
solutions of (35) and (36) accordingly. By definition, process $\tilde{Y}$ has
the representation

$-\tilde{Y}_{t}=\mathbb{E}_{t}\Big{[}2A\,\tilde{Q}_{T}+2\int_{t}^{T}\phi_{s}\,\tilde{Q}_{s}\,ds\Big{]}\geq\mathbb{E}_{t}\Big{[}2A\,\tilde{Q}_{0}+2\int_{t}^{T}\phi_{s}\,\tilde{Q}_{0}\,ds\Big{]}\geq(-2)\,|q_{0}|\,(\bar{A}+\bar{\phi}\,T)$

for all $t\in[0,T]$, where we have used the fact that $\tilde{Q}$ is non-
decreasing. Consequently, a uniform upper bound for $\tilde{Q}$ can be derived
by

$\displaystyle q_{0}\leq\tilde{Q}_{t}$
$\displaystyle=q_{0}+\int_{0}^{t}b_{s}\,\Lambda^{b}\big{(}\tilde{\delta}^{b*}(-\tilde{Y}_{s})\big{)}\,ds$
$\displaystyle\leq
q_{0}+\bar{b}\int_{0}^{T}\Lambda^{b}\Big{(}\delta^{b*}\big{(}(-2)\,|q_{0}|\,(\bar{A}+\bar{\phi}\,T)\big{)}\Big{)}\,ds$
$\displaystyle=q_{0}+\bar{b}\,T\,\Lambda^{b}\Big{(}\delta^{b*}\big{(}(-2)\,|q_{0}|\,(\bar{A}+\bar{\phi}\,T)\big{)}\Big{)}.$

In light of this, the lower bound for $\tilde{Y}_{t}$ can also be found by

$-\tilde{Y}_{t}=\mathbb{E}_{t}\Big{[}2A\,\tilde{Q}_{T}+2\int_{t}^{T}\phi_{s}\,\tilde{Q}_{s}\,ds\Big{]}\leq
2\,(\bar{A}+\bar{\phi}\,T)\cdot\bigg{[}|q_{0}|+\bar{b}\,T\,\Lambda^{b}\Big{(}\delta^{b*}\big{(}(-2)\,|q_{0}|\,(A+\bar{\phi}\,T)\big{)}\Big{)}\bigg{]}.$
(37)

We remark that these bounds for $(\tilde{Q},\tilde{Y})$ are independent of the
parameter $\xi$, implying that we can already find a solution for this ‘one-
sided’ non-Lipschitz FBSDE when $\xi$ is large enough. In a similar fashion,
the corresponding bounds for $(\hat{Q},\hat{Y})$ read

$\displaystyle\hat{Y}_{t}=\mathbb{E}_{t}\Big{[}-2A\,\hat{Q}_{T}-2\int_{t}^{T}\phi_{s}\,\hat{Q}_{s}\,ds\Big{]}\geq(-2)\,|q_{0}|\,(\bar{A}+\bar{\phi}\,T),$
$\displaystyle
q_{0}\geq\hat{Q}_{t}=q_{0}-\int_{0}^{t}a_{s}\,\Lambda^{a}\big{(}\tilde{\delta}^{a*}(\hat{Y}_{s})\big{)}\,ds\geq
q_{0}-\bar{a}\,T\,\Lambda^{a}\Big{(}\delta^{a*}\big{(}(-2)\,|q_{0}|\,(\bar{A}+\bar{\phi}\,T)\big{)}\Big{)},$
(38) $\displaystyle\hat{Y}_{t}\leq
2\,(\bar{A}+\bar{\phi}\,T)\cdot\bigg{[}|q_{0}|+\bar{a}\,T\,\Lambda^{a}\Big{(}\delta^{a*}\big{(}(-2)\,|q_{0}|\,(\bar{A}+\bar{\phi}\,T)\big{)}\Big{)}\bigg{]},$
(39)

holding for all $t\in[0,T]$. Again, such bounds are independent of the
parameter $\xi$.

Set $(Q,Y,M)$ as the solution to the original FBSDE (34), and we then compute
the difference of two equations (34) and (35). Note that

$\displaystyle d(\tilde{Q}_{t}-Q_{t})$
$\displaystyle=b_{t}\,\Big{[}\Lambda^{b}\big{(}\tilde{\delta}^{b*}(-\tilde{Y}_{t})\big{)}-\Lambda^{b}\big{(}\tilde{\delta}^{b*}(-Y_{t})\big{)}\Big{]}\,dt+a_{t}\,\Lambda^{a}\big{(}\tilde{\delta}^{a*}(Y_{t})\big{)}\,dt$
$\displaystyle=b_{t}\,\Big{[}\Lambda^{b}\big{(}\tilde{\delta}^{b*}(-\tilde{Y}_{t})\big{)}-\Lambda^{b}\big{(}\tilde{\delta}^{b*}(-Y_{t})\big{)}\Big{]}\,dt$
$\displaystyle\hskip
56.9055pt+a_{t}\,\Big{[}\Lambda^{a}\big{(}\tilde{\delta}^{a*}(Y_{t})\big{)}-\Lambda^{a}\big{(}\tilde{\delta}^{a*}(\tilde{Y}_{t})\big{)}\Big{]}\,dt+a_{t}\,\Lambda^{a}\big{(}\tilde{\delta}^{a*}(\tilde{Y}_{t})\big{)}\,dt$
$\displaystyle=\pi_{t}\,(\tilde{Y}_{t}-Y_{t})\,dt+a_{t}\,\Lambda^{a}\big{(}\tilde{\delta}^{a*}(\tilde{Y}_{t})\big{)}\,dt,$

where we let

$\pi_{t}=\frac{b_{t}\,\Big{[}\Lambda^{b}\big{(}\tilde{\delta}^{b*}(-\tilde{Y}_{t})\big{)}-\Lambda^{b}\big{(}\tilde{\delta}^{b*}(-Y_{t})\big{)}\Big{]}+a_{t}\,\Big{[}\Lambda^{a}\big{(}\tilde{\delta}^{a*}(Y_{t})\big{)}-\Lambda^{a}\big{(}\tilde{\delta}^{a*}(\tilde{Y}_{t})\big{)}\Big{]}}{\tilde{Y}_{t}-Y_{t}}$

when $\tilde{Y}_{t}-Y_{t}\neq 0$, and $\pi_{t}=0$ in the case of
$\tilde{Y}_{t}-Y_{t}=0$. Due to the monotonicity properties of
$(\Lambda^{a},\Lambda^{b})$ and $(\delta^{a*},\delta^{b*})$, we know that
$\pi\in\mathbb{H}^{2}$ is a bounded non-negative process. Afterwards, the
triplet $(\tilde{Q}-Q,\tilde{Y}-Y,\tilde{Z}-Z)$ solves the following linear
FBSDE:

$\left\\{\begin{aligned}
\;&d\mathcal{Q}_{t}=\pi_{t}\,\mathcal{Y}_{t}dt+a_{t}\,\Lambda^{a}\big{(}\tilde{\delta}^{a*}(\tilde{Y}_{t})\big{)}\,dt,\\\
&d\mathcal{Y}_{t}=2\phi_{t}\,\mathcal{Q}_{t}\,dt+\mathcal{Z}_{t}\,dW_{t},\\\
&\mathcal{Q}_{0}=0,\quad\mathcal{Y}_{T}=-2A\,\mathcal{Q}_{T},\end{aligned}\right.$

which accepts a unique solution as a simple variation of Lemma 3.6. To solve
this equation, its linear structure suggests the affine ansatz
$\mathcal{Y}_{t}=\mathscr{A}_{t}\,\mathcal{Q}_{t}+\mathscr{B}_{t}$, where
processes $\mathscr{A}$ and $\mathscr{B}$ satisfy two coupled BSDEs:

$\displaystyle
d\mathscr{A}_{t}=\big{(}-\pi_{t}\,(\mathscr{A}_{t})^{2}+2\phi_{t}\big{)}\,dt+\mathscr{Z}^{\mathscr{A}}_{t}\,dW_{t},\quad\text{with}\quad\mathscr{A}_{T}=-2A;$
$\displaystyle
d\mathscr{B}_{t}=-\Big{[}\pi_{t}\,\mathscr{A}_{t}\,\mathscr{B}_{t}+\mathscr{A}_{t}\,a_{t}\,\Lambda^{a}\big{(}\tilde{\delta}^{a*}(\tilde{Y}_{t})\big{)}\Big{]}\,dt+\mathscr{Z}^{\mathscr{B}}_{t}\,dW_{t},\quad\text{with}\quad\mathscr{B}_{T}=0.$

We already know the first BSDE accepts a solution $\mathscr{A}$ that is
negative and bounded by $2(\bar{A}+\bar{\phi}\,T)$. In turn, the solution of
the second linear BSDE is given explicitly by

$\mathscr{B}_{t}=\mathbb{E}_{t}\Big{[}\int_{t}^{T}\mathscr{A}_{s}\,a_{s}\,\Lambda^{a}\big{(}\tilde{\delta}^{a*}(\tilde{Y}_{s})\big{)}\,e^{\int_{t}^{s}\pi_{u}\,\mathscr{A}_{u}\,du}\,ds\Big{]}\leq
0.$

Note that the expression inside the conditional expectation is uniformly
bounded on $[0,T]$ by some constant independent of $\xi$. Since
$\mathcal{Q}_{t}$ satisfies the simple ODE

$d\mathcal{Q}_{t}=\pi_{t}\,(\mathscr{A}_{t}\,\mathcal{Q}_{t}+\mathscr{B}_{t})\,dt+a_{t}\,\Lambda^{a}\big{(}\tilde{\delta}^{a*}(\tilde{Y}_{t})\big{)}\,dt,$

the solution can be also given precisely by

$\mathcal{Q}_{t}=\int_{0}^{t}\Big{[}\pi_{s}\,\mathscr{B}_{s}+a_{s}\,\Lambda^{a}\big{(}\tilde{\delta}^{a*}(\tilde{Y}_{s})\big{)}\Big{]}\,e^{\int_{s}^{t}\pi_{u}\,\mathscr{A}_{u}\,du}\,ds.$

By definition, we further deduce

$\displaystyle\tilde{Q}_{t}-Q_{t}$
$\displaystyle=\int_{0}^{t}\Big{[}\pi_{s}\,\mathscr{B}_{s}+a_{s}\,\Lambda^{a}\big{(}\tilde{\delta}^{a*}(\tilde{Y}_{s})\big{)}\Big{]}\,e^{\int_{s}^{t}\pi_{u}\,\mathscr{A}_{u}\,du}\,ds$
$\displaystyle\leq\int_{0}^{t}a_{s}\,\Lambda^{a}\big{(}\tilde{\delta}^{a*}(\tilde{Y}_{s})\big{)}\,ds,$
$\displaystyle Q_{t}$
$\displaystyle\geq\tilde{Q}_{t}-\int_{0}^{t}a_{s}\,\Lambda^{a}\big{(}\tilde{\delta}^{a*}(\tilde{Y}_{s})\big{)}\,ds.$

The uniform boundedness of $(\tilde{Q},\tilde{Y})$ implies that $Q$ is bounded
below by a constant that is independent of $\xi$. On the other hand, to search
for an upper bound for $Q$, we study the relation between $(Q,Y,Z)$ and
$(\hat{Q},\hat{Y},\hat{Z})$. Taking the difference of (34) and (36), we apply
a similar transform to see

$\displaystyle d(\hat{Q}_{t}-Q_{t})$
$\displaystyle=-b_{t}\,\Lambda^{b}\big{(}\tilde{\delta}^{b*}(-Y_{t})\big{)}\,dt-
a_{t}\,\Big{[}\Lambda^{a}\big{(}\tilde{\delta}^{a*}(\hat{Y}_{t})\big{)}-\Lambda^{a}\big{(}\tilde{\delta}^{a*}(Y_{t})\big{)}\Big{]}\,dt$
$\displaystyle=b_{t}\,\Big{[}\Lambda^{b}\big{(}\tilde{\delta}^{b*}(-\hat{Y}_{t})\big{)}-\Lambda^{b}\big{(}\tilde{\delta}^{b*}(-Y_{t})\big{)}\Big{]}\,dt$
$\displaystyle\hskip
56.9055pt+a_{t}\,\Big{[}\Lambda^{a}\big{(}\tilde{\delta}^{a*}(Y_{t})\big{)}-\Lambda^{a}\big{(}\tilde{\delta}^{a*}(\hat{Y}_{t})\big{)}\Big{]}\,dt-
b_{t}\,\Lambda^{b}\big{(}\tilde{\delta}^{b*}(\hat{Y}_{t})\big{)}\,dt$
$\displaystyle=\kappa_{t}\,(\hat{Y}_{t}-Y_{t})\,dt-
b_{t}\,\Lambda^{b}\big{(}\tilde{\delta}^{b*}(\hat{Y}_{t})\big{)}\,dt,$

where we let

$\kappa_{t}=\frac{b_{t}\,\Big{[}\Lambda^{b}\big{(}\tilde{\delta}^{b*}(-\hat{Y}_{t})\big{)}-\Lambda^{b}\big{(}\tilde{\delta}^{b*}(-Y_{t})\big{)}\Big{]}+a_{t}\,\Big{[}\Lambda^{a}\big{(}\tilde{\delta}^{a*}(Y_{t})\big{)}-\Lambda^{a}\big{(}\tilde{\delta}^{a*}(\hat{Y}_{t})\big{)}\Big{]}}{\hat{Y}_{t}-Y_{t}}$

when $\hat{Y}_{t}-Y_{t}\neq 0$, and $\kappa_{t}=0$ in the case of
$\hat{Y}_{t}-Y_{t}=0$. One can see that $\kappa\in\mathbb{H}^{2}$ is non-
negative and bounded as well. Similarly, the triplet
$(\hat{Q}_{t}-Q_{t},\hat{Y}_{t}-Y_{t},\hat{Z}_{t}-Z_{t})$ solves the following
(well-posed) linear FBSDE:

$\left\\{\begin{aligned} \;&d\mathcal{Q}_{t}=\kappa_{t}\,\mathcal{Y}_{t}dt-
b_{t}\,\Lambda^{b}\big{(}\tilde{\delta}^{b*}(\hat{Y}_{t})\big{)}\,dt,\\\
&d\mathcal{Y}_{t}=2\phi_{t}\,\mathcal{Q}_{t}\,dt+\mathcal{Z}_{t}\,dW_{t},\\\
&\mathcal{Q}_{0}=0,\quad\mathcal{Y}_{T}=-2A\,\mathcal{Q}_{T}.\end{aligned}\right.$

The linear structure again advocates the affine ansatz
$\mathcal{Y}_{t}=\mathscr{P}_{t}\,\mathcal{Q}_{t}+\mathscr{H}_{t}$, with
processes $\mathscr{P}$ and $\mathscr{H}$ solving two coupled BSDEs

$\displaystyle
d\mathscr{P}_{t}=(-\kappa_{t}\,\mathscr{P}_{t}^{2}+2\phi_{t})\,dt+\mathscr{Z}^{\mathscr{P}}_{t}\,dW_{t},\quad\text{with}\quad\mathscr{P}_{T}=-2A;$
$\displaystyle
d\mathscr{H}_{t}=-\Big{[}\kappa_{t}\,\mathscr{P}_{t}\,\mathscr{H}_{t}-\mathscr{P}_{t}\,b_{t}\,\Lambda^{b}\big{(}\tilde{\delta}^{b*}(\hat{Y}_{t})\big{)}\Big{]}\,dt+\mathscr{Z}^{\mathscr{H}}_{t}\,dW_{t},\quad\text{with}\quad\mathscr{H}_{T}=0.$

We know the first BSDE accepts a solution $\mathscr{P}\in\mathbb{H}^{2}$ that
is negative and bounded. In turn, the solution of the second linear BSDE is
given explicitly by

$\mathscr{H}_{t}=-\,\mathbb{E}_{t}\Big{[}\int_{t}^{T}\mathscr{P}_{s}\,b_{s}\,\Lambda^{b}\big{(}\tilde{\delta}^{b*}(\hat{Y}_{s})\big{)}\,e^{\int_{t}^{s}\kappa_{u}\,\mathscr{P}_{u}\,du}\,ds\Big{]}\geq
0.$

Note that $\mathscr{H}$ is also uniformly bounded by some constant being
independent of $\xi$. As $\mathcal{Q}$ can be represented by

$\mathcal{Q}_{t}=\int_{0}^{t}\Big{[}\kappa_{s}\,\mathscr{H}_{s}-b_{s}\,\Lambda^{b}\big{(}\tilde{\delta}^{b*}(\hat{Y}_{s})\big{)}\Big{]}\,e^{\int_{s}^{t}\kappa_{u}\,\mathscr{P}_{u}\,du}\,ds,$

by definition we observe

$\displaystyle\hat{Q}_{t}-Q_{t}$
$\displaystyle=\int_{0}^{t}\Big{[}\kappa_{s}\,\mathscr{H}_{s}-b_{s}\,\Lambda^{b}\big{(}\tilde{\delta}^{b*}(\hat{Y}_{s})\big{)}\Big{]}\,e^{\int_{s}^{t}\kappa_{u}\,\mathscr{P}_{u}\,du}\,ds$
$\displaystyle\geq-\int_{0}^{t}b_{s}\,\Lambda^{b}\big{(}\tilde{\delta}^{b*}(\hat{Y}_{s})\big{)}\,ds,$
$\displaystyle Q_{t}$
$\displaystyle\leq\hat{Q}_{t}+\int_{0}^{t}b_{s}\,\Lambda^{b}\big{(}\tilde{\delta}^{b*}(\hat{Y}_{s})\big{)}\,ds.$

The uniform boundedness of $(\hat{Q},\hat{Y})$ suggests that $Q$ is bounded
above by a constant that is independent of $\xi$. While now we know $Q$ is
uniformly lower bounded, it leads to the uniform boundedness of $Y$ since

$Y_{t}=\,\mathbb{E}_{t}\Big{[}-2A\,Q_{T}-2\int_{t}^{T}\phi_{s}\,Q_{s}\,ds\Big{]}.$

In consequence, by picking $\xi$ large enough, the solution $(Q,Y,Z)$ to the
Lipschitz FBSDE (34) also solves the non-Lipschitz (33). The uniqueness again
follows from the continuation method in Lemma 3.6.

With respect to the monotonicity, it suffices to apply Theorem 3.8 and the
fact that the truncation has been removed. ∎

##  4\. Non-Markovian Order Flows: Unbounded Coefficients

In the previous two cases, the coefficients—order flows $(a,b)$ and
penalisation parameters $(\phi,A)$—are all bounded. Since we have mentioned
the possibility of unbounded order flows in Remark 2.10, this section is
devoted to such situations. For convenience of notation, we will present the
result in the case of linear intensity function, i.e.,
$\Lambda(\delta)=\zeta-\gamma\,\delta$ for some constants $\zeta,\gamma>0$.
Regarding the general intensity function, it suffices to add a polynomial
growth condition in Assumption 3.2, while the exponential function will be
excluded.

We start by introducing proper spaces for the order flows $(a,b)$ and
penalisation parameters $(\phi,A)$.

######  Assumption 4.1.

Let $A\in L^{8\mathscr{D}}(\mathcal{F}_{T},\mathbb{R}_{+})$, $a$,
$b\in\mathbb{S}^{8\mathscr{D}}$, and $\phi\in\mathbb{H}^{8\mathscr{D}}$, where
$\mathscr{D}\in\mathbb{N}$ will be specified later.

Since the intensity function is $\Lambda(\delta)=\zeta-\gamma\,\delta$, we
recall the inventory and cash processes of the market maker as follows:

$\displaystyle
Q_{t}=q_{0}-\int_{0}^{t}a_{u}(\zeta-\gamma\delta^{a}_{u})\,du+\int_{0}^{t}b_{u}(\zeta-\gamma\delta^{b}_{u})\,du,$
$\displaystyle
X_{t}=x_{0}+\int_{0}^{t}a_{u}(\zeta-\gamma\delta^{a}_{u})\,(S_{u}+\delta_{u}^{a})\,du-\int_{0}^{t}b_{u}(\zeta-\gamma\delta^{b}_{u})\,(S_{u}-\delta_{u}^{b})\,du,$

where $\delta^{a},\delta^{b}\in\mathbb{H}^{8}$ represent the control. Note
that here we seek strategies in the space $\mathbb{H}^{8}$ to make sure the
associated inventory $Q$ is still in $\mathbb{H}^{4}$ (and indeed in
$\mathbb{S}^{4}$). Hence, the following objective functional is well-defined
and can be simplified as before:

$\displaystyle J(\boldsymbol{\delta}):$
$\displaystyle=\mathbb{E}\big{[}X_{T}+S_{T}\,Q_{T}-\int_{0}^{T}\phi_{t}\,(Q_{t})^{2}\,dt-A\,(Q_{T})^{2}\big{]}$
$\displaystyle=\mathbb{E}\Big{[}\int_{0}^{T}\delta_{t}^{a}\,a_{t}(\zeta-\gamma\delta_{t}^{a})\,dt+\int_{0}^{T}\delta_{t}^{b}\,b_{t}(\zeta-\gamma\delta_{t}^{b})\,dt-\int_{0}^{T}\phi_{t}\,(Q_{t})^{2}\,dt-A\,(Q_{T})^{2}\Big{]},$

by controlling $\boldsymbol{\delta}\in\mathbb{H}^{8}\times\mathbb{H}^{8}$.
Indeed, it suffices to observe

$\displaystyle\big{|}\int_{0}^{T}\delta_{t}^{a}\,a_{t}(\zeta-\gamma\delta_{t}^{a})\,dt\big{|}$
$\displaystyle\leq C\,\sup_{0\leq t\leq
T}|a_{t}|^{2}+C\,\big{|}\int_{0}^{T}(\delta_{t}^{a})^{2}\,dt\big{|}+C\,\big{|}\int_{0}^{T}(\delta_{t}^{a})^{2}\,dt\big{|}^{2},$
$\displaystyle\big{|}\int_{0}^{T}\phi_{t}\,(Q_{t})^{2}\,dt\big{|}$
$\displaystyle\leq C\,\sup_{0\leq t\leq
T}|Q_{t}|^{4}+C\,\big{|}\int_{0}^{T}(\phi_{t})^{2}\,dt\big{|}$

to justify the integrability. Since the convex nature of the objective
functional $J$ is revealed in the proof of Theorem 2.6, the convex-analytic
method is used to characterize the (unique) optimal control as the solution of
a FBSDE.

######  Theorem 4.2.

The functional $\boldsymbol{\delta}\hookrightarrow J(\boldsymbol{\delta})$ is
Gâteaux differentiable and the derivative in the direction of
$\boldsymbol{\beta}:=(\beta^{a},\beta^{b})\in\mathbb{H}^{8}\times\mathbb{H}^{8}$
reads

$\displaystyle\frac{d}{d\epsilon}$ $\displaystyle
J(\boldsymbol{\delta}+\epsilon\boldsymbol{\beta})\big{|}_{\epsilon=0}:=\lim_{\epsilon\searrow
0}\frac{1}{\epsilon}\,\big{[}J(\boldsymbol{\delta}+\epsilon\boldsymbol{\beta})-J(\boldsymbol{\delta})\big{]}$
(40)
$\displaystyle=\mathbb{E}\Big{[}\int_{0}^{T}\big{(}a_{t}\,\beta_{t}^{a}\,(\zeta-2\gamma\,\delta_{t}^{a})+b_{t}\,\beta_{t}^{b}\,(\zeta-2\gamma\,\delta_{t}^{b})\big{)}\,dt-2\,\int_{0}^{T}\phi_{t}\,V_{t}\,Q_{t}\,dt-2\,A\,V_{T}\,Q_{T}\Big{]},$

where
$V_{t}=\gamma\int_{0}^{t}(a_{s}\,\beta_{s}^{a}-b_{s}\,\beta_{s}^{b})\,ds$. A
control $(\delta^{*,a},\delta^{*,b})\in\mathbb{H}^{8}\times\mathbb{H}^{8}$
maximises the functional $J$ if it can be represented by

$\delta_{t}^{b,*}=\frac{\zeta}{2\gamma}-\frac{1}{2}Y_{t},\text{\; and
\;}\delta_{t}^{a,*}=\frac{\zeta}{2\gamma}+\frac{1}{2}Y_{t},$ (41)

where $Y$ solves the following FBSDE:

$\left\\{\begin{aligned}
\;&dQ_{t}=\zeta\,(b_{t}-a_{t})\,dt\,/\,2+\gamma\,(a_{t}+b_{t})\,Y_{t}\,dt\,/\,2,\\\
&dY_{t}=2\,\phi_{t}\,Q_{t}\,dt+Z_{t}\,dW_{t},\\\ &Q_{0}=q_{0},\quad
Y_{T}=-2A\,Q_{T}.\end{aligned}\right.$

The representation (41) is also necessary when $a,b>0$.

###### PROOF.

Fix an admissible control
$\boldsymbol{\delta}=(\delta^{a},\delta^{b})\in\mathbb{H}^{8}\times\mathbb{H}^{8}$
and denote by $Q=Q^{\boldsymbol{\delta}}\in\mathbb{S}^{2}$ the corresponding
controlled inventory. Next, consider
$\boldsymbol{\beta}=(\beta^{a},\beta^{b})\in\mathbb{H}^{8}\times\mathbb{H}^{8}$—that
is also uniformly bounded—as the direction in which we are computing the
Gâteaux derivative of $J$. For each $\epsilon>0$, we consider the admissible
control $\boldsymbol{\delta}^{\epsilon}\in\mathbb{H}^{8}\times\mathbb{H}^{8}$
defined by
$\boldsymbol{\delta}^{\epsilon}_{t}=\boldsymbol{\delta}_{t}+\epsilon\,\boldsymbol{\beta}_{t}$,
and the corresponding controlled state is denoted by
$Q^{\epsilon}:=Q^{\boldsymbol{\delta}^{\epsilon}}\in\mathbb{S}^{2}$. Note that
$V\in\mathbb{S}^{8\mathscr{D}}$. Direct calculations yield

$\displaystyle\frac{1}{\epsilon}$
$\displaystyle\big{[}J(\boldsymbol{\delta}+\epsilon\boldsymbol{\beta})-J(\boldsymbol{\delta})\big{]}$
(42)
$\displaystyle=\mathbb{E}\Big{[}\int_{0}^{T}a_{t}\,\beta_{t}^{a}\,(\zeta-2\,\gamma\,\delta_{t}^{a}-\epsilon\,\gamma\,\beta_{t}^{a})\,dt\Big{]}+\mathbb{E}\Big{[}\int_{0}^{T}b_{t}\,\beta_{t}^{b}\,(\zeta-2\,\gamma\,\delta_{t}^{b}-\epsilon\,\gamma\,\beta_{t}^{b})\,dt\Big{]}$
$\displaystyle\hskip
28.45274pt-\mathbb{E}\Big{[}\int_{0}^{T}\phi_{t}\,(Q_{t}^{\epsilon}+Q_{t})\,V_{t}\,dt\Big{]}-\mathbb{E}\Big{[}A\,(Q_{T}^{\epsilon}+Q_{T})\,V_{T}\Big{]}.$

We look at the first term on the right hand side of (42). To perform the
limiting procedure, note that

$\displaystyle
a_{t}\,\beta_{t}^{a}\,(\zeta-2\,\gamma\,\delta_{t}^{a}-\epsilon\,\gamma\,\beta_{t}^{a})\,dt$
$\displaystyle\leq
a_{t}\,|\beta_{t}^{a}|\,\big{(}\zeta+2\,\gamma\,|\delta_{t}^{a}|+\gamma\,|\beta_{t}^{a}|\big{)},$

the right hand side of which is $d\mathbb{P}\times dt$-integrable. Therefore,
it follows from the dominated convergence theorem that

$\lim_{\epsilon\to
0}\,\mathbb{E}\Big{[}\int_{0}^{T}a_{t}\,\beta_{t}^{a}\,(\zeta-2\,\gamma\,\delta_{t}^{a}-\epsilon\,\gamma\,\beta_{t}^{a})\,dt\Big{]}=\mathbb{E}\Big{[}\int_{0}^{T}a_{t}\,\beta_{t}^{a}\,(\zeta-2\,\gamma\,\delta_{t}^{a})\,dt\Big{]}.$

The second term of (42) can be verified in the same way. With respect to third
term, we are aware of

$\displaystyle\int_{0}^{T}\phi_{t}\,(Q_{t}^{\epsilon}+Q_{t})\,V_{t}\,dt$
$\displaystyle\leq
C\int_{0}^{T}\phi_{t}^{2}\,\big{(}|Q_{t}^{\epsilon}|+|Q_{t}|\big{)}\,dt+C\int_{0}^{T}V_{t}^{2}\,\big{(}|Q_{t}^{\epsilon}|+|Q_{t}|\big{)}\,dt,$
$\displaystyle\leq
C\sup_{t\in[0,T]}\big{(}|Q_{t}^{\epsilon}|^{2}+|Q_{t}|^{2}\big{)}+C\,\Big{(}\int_{0}^{T}\phi_{t}^{2}\,dt\Big{)}^{2}+C\,\Big{(}\int_{0}^{T}V_{t}^{2}\,dt\Big{)}^{2}$
$\displaystyle\leq
C\sup_{t\in[0,T]}|Q_{t}|^{2}+C\,\Big{(}\int_{0}^{T}(a_{t}+b_{t})\,dt\Big{)}^{2}+C\,\Big{(}\int_{0}^{T}\phi_{t}^{2}\,dt\Big{)}^{2}$
$\displaystyle\hskip 79.6678pt+C\,\Big{(}\int_{0}^{T}V_{t}^{2}\,dt\Big{)}^{2}$
$\displaystyle\phi_{t}\,(Q_{t}^{\epsilon}+Q_{t})\,V_{t}$ $\displaystyle\leq
C\,|Q_{t}|^{2}+C\,\Big{(}\int_{0}^{T}(a_{t}+b_{t})\,dt\Big{)}^{2}+C\,\phi_{t}^{2}+C\,V_{t}^{2}$

and the integrability on the right hand sides, and hence can conclude

$\lim_{\epsilon\to
0}\,\mathbb{E}\Big{[}\int_{0}^{T}\phi_{t}\,(Q_{t}^{\epsilon}+Q_{t})\,V_{t}\,dt\Big{]}=2\,\mathbb{E}\Big{[}\int_{0}^{T}\phi_{t}\,Q_{t}\,V_{t}\,dt\Big{]}.$

Since the final term can be proved similarly, the equation (42) is verified
and an integration by parts further yields

$\displaystyle\frac{d}{d\epsilon}J(\boldsymbol{\delta}+\epsilon$
$\displaystyle\boldsymbol{\beta})\big{|}_{\epsilon=0}$ (43)
$\displaystyle=\mathbb{E}\bigg{\\{}\int_{0}^{T}\Big{[}a_{t}\,\beta_{t}^{a}\,\big{(}\zeta-2\gamma\,\delta_{t}^{a}-2\gamma\int_{t}^{T}\phi_{s}\,Q_{s}\,ds-2\gamma\,A\,Q_{T}\big{)}\Big{]}\,dt\bigg{\\}}$
$\displaystyle\hskip
56.9055pt+\mathbb{E}\bigg{\\{}\int_{0}^{T}\Big{[}b_{t}\,\beta_{t}^{b}\,\big{(}\zeta-2\gamma\,\delta_{t}^{b}-2\gamma\int_{t}^{T}\phi_{s}\,Q_{s}\,ds+2\gamma\,A\,Q_{T}\big{)}\Big{]}\,dt\bigg{\\}}.$

Since the functional $J$ is concave, a control
$\boldsymbol{\delta}^{*}=(\delta^{*,a},\delta^{*,b})\in\mathbb{H}^{8}\times\mathbb{H}^{8}$
is optimal if and only if (43) is equal to $0$ for any bounded
$\boldsymbol{\beta}\in\mathbb{H}^{8}\times\mathbb{H}^{8}$. With the law of
total expectation, this infers that

$\displaystyle\gamma\,Y_{t}=\zeta-2\gamma\,\delta_{t}^{*,a}$
$\displaystyle=\mathbb{E}_{t}\Big{[}2\gamma\int_{t}^{T}\phi_{s}\,Q_{s}\,ds+2\gamma\,A\,Q_{T}\Big{]},$
$\displaystyle\gamma\,Y_{t}=2\gamma\,\delta_{t}^{*,b}-\zeta$
$\displaystyle=\mathbb{E}_{t}\Big{[}2\gamma\int_{t}^{T}\phi_{s}\,Q_{s}\,ds+2\gamma\,A\,Q_{T}\Big{]}$

$d\mathbb{P}\times dt$-almost everywhere. The FBSDE can be obtained by
‘differentiating’ $Y$. ∎

The remaining of this section is devoted the well-posedness of the FBSDE.
Before looking at the FBSDE associated with the stochastic maximum principle,
we first investigate the cornerstone of its well-posedness theory—a quadratic
BSDE:

$dY_{t}=\big{(}-\gamma\,(a_{t}+b_{t})\,(Y_{t})^{2}+2\,\phi_{t}\big{)}\,dt+Z_{t}\,dW_{t},\text{\quad
with \quad}Y_{T}=-2\,A,$ (44)

the coefficients $a,b,\phi,A$ of which are not bounded. Note that the $(Y,Z)$
here is different from the one in the FBSDE, and here we only focus on the
existence of solutions. We localize this equation via the truncation:

$dY_{t}=\big{(}-\gamma\,\big{(}(a_{t}+b_{t})\wedge
m\big{)}\,(Y_{t})^{2}+2\,(\phi_{t}\wedge
n)\big{)}\,dt+Z_{t}\,dW_{t},\text{\quad with \quad}Y_{T}=-2\,(A\wedge n),$
(45)

where $m,n\in\mathbb{N}$. First, we try to remove the constant $n$.

######  Lemma 4.3.

The BSDE

$dY_{t}=\big{(}-\gamma\,\big{(}(a_{t}+b_{t})\wedge
m\big{)}\,(Y_{t})^{2}+2\,\phi_{t}\big{)}\,dt+Z_{t}\,dW_{t},\text{\quad with
\quad}Y_{T}=-2\,A,$ (46)

accepts a solution $(Y,Z)\in\mathbb{S}^{8\mathscr{D}}\times\mathbb{H}^{2}$.

###### PROOF.

Recall that (45) is well-posed for any pair $(m,n)$ by Lemma 3.6. Fixing any
$m\in\mathbb{N}$, equation (45) accepts a unique solution
$(Y^{(n)},Z^{(n)})\in\mathbb{S}^{2}\times\mathbb{H}^{2}$ for all
$n\in\mathbb{N}$. Because $|Y_{t}^{(n)}|\leq 2n+2nT$, the standard comparison
theorem for Lipschitz BSDEs implies

$0\geq Y_{t}^{(n)}\geq Y_{t}^{(n+1)},\qquad 0\leq t\leq
T,\qquad\mathbb{P}\text{-a.s.}$

Consequently, for almost every $\omega\in\Omega$, we can define the limit as
follows:

$U_{t}(\omega):=\lim_{n\to\infty}Y_{t}^{(n)}(\omega),\qquad 0\leq t\leq T.$

Note that $-U_{t}(\omega)$ may be infinity for some $t$ and $\omega$. Thanks
to the monotonicity of $(Y^{(n)})_{n\in\mathbb{N}}$, we observe that

$\lim_{n\to\infty}|Y_{t}^{(n)}|^{8\mathscr{D}}\leq\lim_{n\to\infty}\,\sup_{0\leq
u\leq T}|Y_{u}^{(n)}|^{8\mathscr{D}},\qquad 0\leq t\leq T,$

the right hand side of which is well-defined (possibly infinite). It further
reveals the integrability of $U$ as follows

$\displaystyle\mathbb{E}\Big{[}\sup_{0\leq t\leq
T}|U_{t}|^{8\mathscr{D}}\Big{]}$
$\displaystyle\leq\lim_{n\to\infty}\,\mathbb{E}\Big{[}\sup_{0\leq t\leq
T}|Y_{t}^{(n)}|^{8\mathscr{D}}\Big{]}$ (47)
$\displaystyle=\lim_{n\to\infty}\,\mathbb{E}\Big{[}\sup_{0\leq u\leq
T}\Big{|}\,\mathbb{E}_{u}\big{[}-2\,(A\wedge n)-2\int_{t}^{T}(\phi_{s}\wedge
n)\,ds$ $\displaystyle\hskip
99.58464pt+\int_{t}^{T}\gamma\,\big{(}(a_{u}+b_{u})\wedge
m\big{)}\,(Y_{u})^{2}\,du\big{]}\,\Big{|}^{8\mathscr{D}}\Big{]}$
$\displaystyle\leq\lim_{n\to\infty}\,\mathbb{E}\Big{[}\sup_{0\leq u\leq
T}\mathbb{E}_{u}\big{[}2\,(A\wedge n)+2\int_{t}^{T}(\phi_{s}\wedge
n)\,ds\big{]}^{8\mathscr{D}}\Big{]}$
$\displaystyle\leq\lim_{n\to\infty}\,\mathbb{E}\Big{[}\sup_{0\leq u\leq
T}\mathbb{E}_{u}\big{[}2\,A+2\int_{0}^{T}\phi_{s}\,ds\big{]}^{8\mathscr{D}}\Big{]}$
$\displaystyle\leq
C\,\mathbb{E}\Big{[}\big{(}2\,A+2\int_{0}^{T}\phi_{s}\,ds\big{)}^{8\mathscr{D}}\Big{]}$
$\displaystyle<\infty,$

where we have applied the monotone convergence theorem, the fact
$Y_{t}^{(n)}\leq 0$, and the Doob’s inequality. For any $u\in[0,T]$, since two
random variables

$\mathbb{E}_{u}\Big{[}2\,A+2\int_{u}^{T}\phi_{s}\,ds\Big{]}\text{\quad and
\quad}\mathbb{E}_{u}\Big{[}\int_{u}^{T}\big{(}(a_{s}+b_{s})\wedge
m\big{)}\,(U_{s})^{2}\,ds\Big{]}$

are both integrable, then they are almost surely finite and one can use the
monotone convergence theorem again to see

$\displaystyle U_{t}$ $\displaystyle=\lim_{n\to\infty}Y_{t}^{(n)}$
$\displaystyle=\lim_{n\to\infty}\mathbb{E}_{t}\Big{[}-2\,(A\wedge
n)-2\int_{t}^{T}(\phi_{s}\wedge
n)\,ds+\gamma\int_{t}^{T}\big{(}(a_{s}+b_{s})\wedge
m\big{)}\,\big{(}Y_{s}^{(n)}\big{)}^{2}\,ds\Big{]}$
$\displaystyle=\mathbb{E}_{t}\Big{[}-2\,A-2\int_{t}^{T}\phi_{s}\,ds+\gamma\int_{t}^{T}\big{(}(a_{s}+b_{s})\wedge
m\big{)}\,(U_{s})^{2}\,ds\Big{]}.$

While the random variable

$-2\,A-2\int_{0}^{T}\phi_{s}\,ds+\gamma\int_{0}^{T}\big{(}(a_{s}+b_{s})\wedge
m\big{)}\,(Y_{s})^{2}\,ds$

is square integrable, we finally conclude that $U$ satisfies the BSDE

$dU_{t}=\big{(}-\gamma\,\big{(}(a_{t}+b_{t})\wedge
m\big{)}\,(U_{t})^{2}+2\,\phi_{t}\big{)}\,dt+Z_{t}^{U}\,dW_{t},\text{\quad
with \quad}U_{T}=-2\,A,$

where $Z^{U}\in\mathbb{H}^{2}$ is obtained via the martingale representation
theorem. ∎

Note that the solution of (46) may not be bounded, and hence the standard
comparison theorem for Lipschitz BSDEs can not be applied. We summarize the
comparison result in the following lemma.

######  Lemma 4.4.

For each $m\in\mathbb{N}$, denote by $(Y^{(m)},Z^{(m)})$ the solution of
equation (46). Then, $(Y^{(m)})_{m\in\mathbb{N}}$ is non-decreasing with
respect to $m$.

###### PROOF.

Define $\Delta Y:=Y^{(m+1)}-Y^{(m)}$ and $\Delta Z:=Z^{(m+1)}-Z^{(m)}$. The
proof starts with taking the difference of two BSDEs:

$\displaystyle d\Delta Y_{t}$
$\displaystyle=\Big{(}-\gamma\,\big{(}(a_{t}+b_{t})\wedge(m+1)\big{)}\,(Y_{t}^{(m+1)})^{2}+\gamma\,\big{(}(a_{t}+b_{t})\wedge(m+1)\big{)}\,(Y_{t}^{(m)})^{2}$
$\displaystyle\hskip
28.45274pt-\gamma\,\big{(}(a_{t}+b_{t})\wedge(m+1)\big{)}\,(Y_{t}^{(m)})^{2}+\gamma\,\big{(}(a_{t}+b_{t})\wedge
m\big{)}\,(Y_{t}^{(m)})^{2}\Big{)}\,dt+\Delta Z_{t}\,dW_{t}$
$\displaystyle=\Big{(}-\gamma\,\big{(}(a_{t}+b_{t})\wedge(m+1)\big{)}\,(Y_{t}^{(m+1)}+Y_{t}^{(m)})\,\Delta
Y_{t}$ $\displaystyle\hskip
28.45274pt-\gamma\,(Y_{t}^{(m)})^{2}\,\big{[}(a_{t}+b_{t})\wedge(m+1)-(a_{t}+b_{t})\wedge
m\big{]}\Big{)}\,dt+\Delta Z_{t}\,dW_{t}.$

Set
$\iota_{t}=\gamma\,\big{(}(a_{t}+b_{t})\wedge(m+1)\big{)}\,(Y_{t}^{(m+1)}+Y_{t}^{(m)})$,
and observe that $\iota_{t}\leq 0$ since any $Y^{(m)}$ is non-positive. The
integration factor gives

$\Delta
Y_{t}e^{\int_{0}^{t}\iota_{u}\,du}=\gamma\,\mathbb{E}_{t}\Big{[}\int_{t}^{T}e^{\int_{0}^{s}\iota_{u}\,du}\,(Y_{s}^{(m)})^{2}\,\big{[}(a_{s}+b_{s})\wedge(m+1)-(a_{s}+b_{s})\wedge
m\big{]}\,ds\Big{]}\geq 0,$

where we have used the fact that $\Delta Y_{T}=0$. ∎

Given the comparison result, we can finally remove the constant $m$ as well.

######  Theorem 4.5.

The BSDE (44) accepts a solution
$(Y,Z)\in\mathbb{S}^{8\mathscr{D}}\times\mathbb{H}^{2}$.

###### PROOF.

Denote by $(Y^{(m)},Z^{(m)})\in\mathbb{S}^{8\mathscr{D}}\times\mathbb{H}^{2}$
the solution of equation (46) for each $m\in\mathbb{N}$. Given the
monotonicity of $(Y^{(m)})_{m\in\mathbb{N}}$, for almost every
$\omega\in\Omega$, let us define the limit similarly as

$Y_{t}(\omega):=\lim_{m\to\infty}Y_{t}^{(m)}(\omega)\leq 0,\qquad 0\leq t\leq
T.$

Its integrability can be directly seen from

$\mathbb{E}\Big{[}\sup_{0\leq t\leq
T}|Y_{t}|^{8\mathscr{D}}\Big{]}\leq\mathbb{E}\Big{[}\sup_{0\leq t\leq
T}|Y_{t}^{(1)}|^{8\mathscr{D}}\Big{]}<\infty,$

since $Y^{(m)}\leq Y^{(m+1)}\leq 0$ for any $m\in\mathbb{N}$. Thanks to the
integrability of the random variable

$\mathbb{E}_{u}\Big{[}\int_{u}^{T}(a_{s}+b_{s})\,\big{(}Y_{s}^{(1)}\big{)}^{2}\,ds\Big{]},$

the dominated convergence theorem implies

$\displaystyle Y_{t}=\lim_{m\to\infty}Y_{t}^{(m)}$
$\displaystyle=\lim_{m\to\infty}\mathbb{E}_{t}\Big{[}-2\,A-2\int_{t}^{T}\phi_{s}\,ds+\gamma\int_{t}^{T}\big{(}(a_{s}+b_{s})\wedge
m\big{)}\,\big{(}Y_{s}^{(m)}\big{)}^{2}\,ds\Big{]}$
$\displaystyle=\mathbb{E}_{t}\Big{[}-2\,A-2\int_{t}^{T}\phi_{s}\,ds+\gamma\int_{t}^{T}(a_{s}+b_{s})\,(Y_{s})^{2}\,ds\Big{]}.$

Finally, the square integrability of the random variable

$-2\,A-2\int_{0}^{T}\phi_{s}\,ds+\gamma\int_{0}^{T}(a_{s}+b_{s})\,(Y_{s})^{2}\,ds$

helps us conclude that $Y$ solves the BSDE

$dY_{t}=\big{(}-\gamma\,(a_{t}+b_{t})\,(Y_{t})^{2}+2\,\phi_{t}\big{)}\,dt+Z_{t}\,dW_{t},\text{\quad
with \quad}Y_{T}=-2\,A.$

Note that $Z$ is again obtained via the martingale representation theorem. ∎

Subsequently, we look at the FBSDE associated with the stochastic maximum
principle

$\left\\{\begin{aligned}
\;&dQ_{t}=\zeta\,(b_{t}-a_{t})\,dt\,/\,2+\gamma\,(a_{t}+b_{t})\,Y_{t}\,dt\,/\,2,\\\
&dY_{t}=2\,\phi_{t}\,Q_{t}\,dt+Z_{t}\,dW_{t},\\\ &Q_{0}=q_{0},\quad
Y_{T}=-2A\,Q_{T}.\end{aligned}\right.$ (48)

The next theorem presents the well-posedness result of the main FBSDE, and
then specifies the value of $\mathscr{D}$ in line with the admissibility of
the control.

######  Theorem 4.6.

The FBSDE (48) accepts a unique solution
$(Q,Y,Z)\in\mathbb{S}^{2\mathscr{D}}\times\mathbb{S}^{\mathscr{D}}\times\mathbb{H}^{2}$.
It suffices to let $\mathscr{D}=8$ to ensure the resulting control is
admissible.

###### PROOF.

The linear structure of (48) suggests the affine ansatz

$Y_{t}=\mathscr{P}_{t}\,Q_{t}+\mathscr{H}_{t},$ (49)

where processes $\mathscr{P}$ and $\mathscr{H}$ solve the BSDEs

$d\mathscr{P}_{t}=\big{(}-\gamma\,(a_{t}+b_{t})\,(\mathscr{P}_{t})^{2}\,/\,2+2\,\phi_{t}\big{)}\,dt+Z_{t}^{1}\,dW_{t},\text{\quad
with \quad}\mathscr{P}_{T}=-2\,A,$
$d\mathscr{H}_{t}=\big{(}-\gamma\,(a_{t}+b_{t})\,\mathscr{P}_{t}\,\mathscr{H}_{t}-\zeta\,(b_{t}-a_{t})\,\mathscr{P}_{t}\big{)}\,dt\,/\,2+Z_{t}^{2}\,dW_{t},\text{\quad
with \quad}\mathscr{H}_{T}=0.$

The well-posedness of $\mathscr{P}\in\mathbb{S}^{8\mathscr{D}}$ is guaranteed
by Theorem 4.5, and the linear BSDE of $\mathscr{H}$ can be solved explicitly
by

$\displaystyle\mathscr{H}_{t}=\mathbb{E}_{t}\Big{[}\int_{t}^{T}\zeta\,(b_{s}-a_{s})\,\mathscr{P}_{s}\,e^{\int_{s}^{T}\gamma\,(a_{u}+b_{u})\,\mathscr{P}_{u}\,du}\,ds\Big{]}.$

Direct estimations and Doob’s inequality then yield

$\displaystyle|\mathscr{H}_{t}|$ $\displaystyle\leq C\sup_{0\leq u\leq
T}\mathbb{E}_{u}\Big{[}\int_{0}^{T}(a_{s}-b_{s})^{2}\,ds\Big{]}+C\sup_{0\leq
u\leq T}\mathbb{E}_{u}\Big{[}\int_{0}^{T}(\mathscr{P}_{s})^{2}\,ds\Big{]},$
$\displaystyle\mathbb{E}\big{[}\sup_{0\leq t\leq
T}|\mathscr{H}_{t}|^{4\mathscr{D}}]$ $\displaystyle\leq
C\,\mathbb{E}\Big{[}\big{(}\int_{0}^{T}(a_{s}-b_{s})^{2}\,ds\big{)}^{4\mathscr{D}}\Big{]}+C\,\mathbb{E}\Big{[}\big{(}\int_{0}^{T}(\mathscr{P}_{s})^{2}\,ds\big{)}^{4\mathscr{D}}\Big{]}<\infty,$

from which one can see $\mathscr{H}\in\mathbb{S}^{4\mathscr{D}}$. On the other
hand, we can calculate the corresponding inventory

$Q_{t}=q_{0}\,e^{\gamma\,\int_{0}^{t}(a_{s}+b_{s})\,\mathscr{P}_{s}\,ds\,/\,2}+\int_{0}^{t}e^{\gamma\,\int_{s}^{t}(a_{u}+b_{u})\,\mathscr{P}_{u}\,du\,/\,2}\big{[}\zeta\,(b_{s}-a_{s})\,/\,2+\gamma\,(a_{s}+b_{s})\,\mathscr{H}_{s}\,/\,2\big{]}\,ds,$

and deduce that $Q\in\mathbb{S}^{2\mathscr{D}}$ from the following estimation:

$\displaystyle\mathbb{E}\big{[}\sup_{0\leq t\leq
T}|Q_{t}|^{2\mathscr{D}}\big{]}$ $\displaystyle\leq
C\,q_{0}^{2\mathscr{D}}+C\,\mathbb{E}\Big{[}\big{(}\int_{0}^{T}(b_{t}-a_{t})^{2}\,dt\big{)}^{\mathscr{D}}\Big{]}+C\,\mathbb{E}\Big{[}\big{(}\int_{0}^{T}(a_{t}+b_{t})\,|\mathscr{H}_{t}|\,dt\big{)}^{2\mathscr{D}}\Big{]}$
$\displaystyle\leq
C\,q_{0}^{2\mathscr{D}}+C\,\mathbb{E}\Big{[}\big{(}\int_{0}^{T}(b_{t}-a_{t})^{2}\,dt\big{)}^{\mathscr{D}}\Big{]}+C\,\mathbb{E}\Big{[}\big{(}\int_{0}^{T}(a_{t}+b_{t})^{2}\,dt\big{)}^{2\mathscr{D}}\Big{]}$
$\displaystyle\hskip
85.35826pt+C\,\mathbb{E}\Big{[}\big{(}\int_{0}^{T}|\mathscr{H}_{t}|^{2}\,dt\big{)}^{{2\mathscr{D}}}\Big{]}$
$\displaystyle<\infty.$

Since $Y_{t}=\mathscr{P}_{t}\,Q_{t}+\mathscr{H}_{t}$ and

$\mathbb{E}\big{[}\sup_{0\leq t\leq
T}|\mathscr{P}_{t}\,Q_{t}|^{\mathscr{D}}\big{]}\leq
C\,\mathbb{E}\big{[}\sup_{0\leq t\leq
T}|\mathscr{P}_{t}|^{2\mathscr{D}}\big{]}+C\,\mathbb{E}\big{[}\sup_{0\leq
t\leq T}|Q_{t}|^{2\mathscr{D}}\big{]}<\infty,$

it yields $Y\in\mathbb{S}^{\mathscr{D}}$. The uniqueness of solution again
follows from the continuation argument in Lemma 3.6.

Recalling that the optimal control is given by (41) and the admissibility
requires that the control is in $\mathbb{H}^{8}$, it suffices to take
$\mathscr{D}=8$. ∎

##  5\. Appendix: Stochastic Maximum Principle

This section is devoted to the stochastic maximum principle for the general
intensity function. We follow the argument in [7], and start with the
necessary condition. Fix an admissible control
$\boldsymbol{\delta}\in\mathbb{A}\times\mathbb{A}$ and denote by
$Q=Q^{\boldsymbol{\delta}}\in\mathbb{S}^{2}$ the corresponding controlled
inventory, which is also uniformly bounded. Next, consider
$\boldsymbol{\beta}\in\mathbb{A}\times\mathbb{A}$—that is uniformly bounded
and satisfies that
$\boldsymbol{\delta}+\boldsymbol{\beta}\in\mathbb{A}\times\mathbb{A}$—as the
direction in which we are computing the Gâteaux derivative of $J$. For each
$\epsilon>0$ small enough, we consider the admissible control
$\boldsymbol{\delta}^{\epsilon}\in\mathbb{A}\times\mathbb{A}$ defined by
$\boldsymbol{\delta}^{\epsilon}_{t}=\boldsymbol{\delta}_{t}+\epsilon\,\boldsymbol{\beta}_{t}$,
and the corresponding controlled state
$Q^{\epsilon}:=Q^{\boldsymbol{\delta}^{\epsilon}}\in\mathbb{S}^{2}$. Define
$V$ as the solution of the equation

$dV_{t}=\partial_{\boldsymbol{\delta}}\mu(t,\boldsymbol{\delta}_{t})\cdot\boldsymbol{\beta}_{t}\,dt,$

with the initial condition $V_{0}=0$, where

$\mu(t,\boldsymbol{\delta}):=-a_{t}\,\Lambda^{a}(\delta_{t}^{a})+b_{t}\,\Lambda^{b}(\delta_{t}^{b}).$

It is clear that $V$ is uniformly bounded in the finite time horizon. We then
look at the Gâteaux differentiability of the functional $J$.

######  Lemma 5.1.

The functional $\boldsymbol{\delta}\hookrightarrow J(\boldsymbol{\delta})$ is
Gâteaux differentiable and the derivative reads

$\displaystyle\frac{d}{d\epsilon}J(\boldsymbol{\delta}+\epsilon\boldsymbol{\beta})\big{|}_{\epsilon=0}:$
$\displaystyle=\lim_{\epsilon\searrow
0}\frac{1}{\epsilon}\,\big{[}J(\boldsymbol{\delta}+\epsilon\boldsymbol{\beta})-J(\boldsymbol{\delta})\big{]}$
(50)
$\displaystyle=\mathbb{E}\Big{[}\int_{0}^{T}\big{(}\partial_{q}f(t,Q_{t},\boldsymbol{\delta}_{t})\,V_{t}+\partial_{\boldsymbol{\delta}}f(t,Q_{t},\boldsymbol{\delta}_{t})\cdot\beta_{t}\big{)}\,dt-2\,A\,V_{T}\,Q_{T}\Big{]},$

where $f$ is the running payoff defined by

$f(t,Q_{t},\boldsymbol{\delta}_{t})=a_{t}\,\delta_{t}^{a}\,\Lambda^{a}(\delta_{t}^{a})+b_{t}\,\delta_{t}^{b}\,\Lambda^{b}(\delta_{t}^{b})-\phi_{t}\,Q_{t}^{2}.$

######  PROOF.

Regarding the ask side, we compute that

$\displaystyle\frac{1}{\epsilon}$
$\displaystyle\mathbb{E}\Big{[}\int_{0}^{T}a_{t}\,(\delta_{t}^{a}+\epsilon\,\beta_{t}^{a})\,\Lambda^{a}(\delta_{t}^{a}+\epsilon\,\beta_{t}^{a})\,dt-\int_{0}^{T}a_{t}\,\delta_{t}^{a}\,\,\Lambda^{a}(\delta_{t}^{a})\,dt\Big{]}$
$\displaystyle=\mathbb{E}\Big{[}\int_{0}^{T}a_{t}\,\delta_{t}^{a}\,\frac{1}{\epsilon}\,\big{(}\Lambda^{a}(\delta_{t}^{a}+\epsilon\,\beta_{t}^{a})-\Lambda^{a}(\delta_{t}^{a})\big{)}\,dt\Big{]}+\mathbb{E}\Big{[}\int_{0}^{T}a_{t}\,\beta_{t}^{a}\,\Lambda^{a}(\delta_{t}^{a}+\epsilon\,\beta_{t}^{a})\,dt\Big{]}.$

To perform the limiting procedure, note that

$\displaystyle
a_{t}\,\delta_{t}^{a}\,\frac{1}{\epsilon}\,\big{(}\Lambda^{a}(\delta_{t}^{a}+\epsilon\,\beta_{t}^{a})-\Lambda^{a}(\delta_{t}^{a})\big{)}$
$\displaystyle\leq C\,a_{t}\,|\delta_{t}^{a}|,$ $\displaystyle
a_{t}\,\beta_{t}^{a}\,\Lambda^{a}(\delta_{t}^{a}+\epsilon\,\beta_{t}^{a})$
$\displaystyle\leq C\,a_{t},$

the right hand sides of which are $d\mathbb{P}\times dt$-integrable on the
time horizon. Therefore, it follows by the dominated convergence theorem that

$\displaystyle\lim_{\epsilon\to
0}\frac{1}{\epsilon}\,\mathbb{E}\Big{[}\int_{0}^{T}a_{t}\,(\delta_{t}^{a}+\epsilon\,\beta_{t}^{a})$
$\displaystyle\Lambda^{a}(\delta_{t}^{a}+\epsilon\,\beta_{t}^{a})\,dt-\int_{0}^{T}a_{t}\,\delta_{t}^{a}\,\,\Lambda^{a}(\delta_{t}^{a})\,dt\Big{]}$
$\displaystyle=\mathbb{E}\Big{[}\int_{0}^{T}a_{t}\,\delta_{t}^{a}\,\big{(}\Lambda^{a}(\delta_{t}^{a})\big{)}^{\prime}\,\beta_{t}^{a}\,dt\Big{]}+\mathbb{E}\Big{[}\int_{0}^{T}a_{t}\,\beta_{t}^{a}\,\Lambda^{a}(\delta_{t}^{a})\,dt\Big{]}$
$\displaystyle=\mathbb{E}\Big{[}\int_{0}^{T}\partial_{\delta^{a}}f(t,Q_{t},\boldsymbol{\delta}_{t})\,\beta_{t}^{a}\,dt\Big{]}.$

The bid side can be computed in the same way:

$\displaystyle\lim_{\epsilon\to
0}\frac{1}{\epsilon}\,\mathbb{E}\Big{[}\int_{0}^{T}b_{t}\,(\delta_{t}^{b}+\epsilon\,\beta_{t}^{b})$
$\displaystyle\Lambda^{b}(\delta_{t}^{b}+\epsilon\,\beta_{t}^{b})\,dt-\int_{0}^{T}b_{t}\,\delta_{t}^{b}\,\,\Lambda^{b}(\delta_{t}^{b})\,dt\Big{]}$
$\displaystyle=\mathbb{E}\Big{[}\int_{0}^{T}\partial_{\delta^{b}}f(t,Q_{t},\boldsymbol{\delta}_{t})\,\beta_{t}^{b}\,dt\Big{]}.$

With respect to the running penalty, we further calculate that

$\displaystyle\frac{1}{\epsilon}\,\mathbb{E}\Big{[}\int_{0}^{T}$
$\displaystyle\phi_{t}\,(Q_{t}^{\epsilon})^{2}\,dt-\int_{0}^{T}\phi_{t}\,(Q_{t})^{2}\,dt\Big{]}$
$\displaystyle=\mathbb{E}\Big{[}\int_{0}^{T}\phi_{t}\,(Q_{t}^{\epsilon}+Q_{t})\,\Big{(}-\int_{0}^{t}a_{u}\,\frac{1}{\epsilon}\,\big{[}\Lambda^{a}(\delta_{u}^{a}+\epsilon\,\beta_{u}^{a})-\Lambda^{a}(\delta_{u}^{a})\big{]}\,du$
$\displaystyle\hskip
113.81102pt+\int_{0}^{t}b_{u}\,\frac{1}{\epsilon}\,\big{[}\Lambda^{b}(\delta_{u}^{b}+\epsilon\,\beta_{u}^{b})-\Lambda^{b}(\delta_{u}^{b})\big{]}\,du\Big{)}\,dt\Big{]}.$

Being aware of the boundedness of $\phi$, $Q$, $Q^{\epsilon}$, $a$, $b$ as
well as the one of

$\frac{1}{\epsilon}\,\big{[}\Lambda^{a}(\delta_{u}^{a}+\epsilon\,\beta_{u}^{a})-\Lambda^{a}(\delta_{u}^{a})\big{]},\text{\quad
and
\quad}\frac{1}{\epsilon}\,\big{[}\Lambda^{b}(\delta_{u}^{b}+\epsilon\,\beta_{u}^{b})-\Lambda^{b}(\delta_{u}^{b})\big{]},$

we can conclude

$\lim_{\epsilon\to
0}\,\frac{1}{\epsilon}\,\mathbb{E}\Big{[}\int_{0}^{T}\phi_{t}\,(Q_{t}^{\epsilon})^{2}\,dt-\int_{0}^{T}\phi_{t}\,(Q_{t})^{2}\,dt\Big{]}=2\,\mathbb{E}\Big{[}\int_{0}^{T}\phi_{t}\,Q_{t}\,V_{t}\,dt\Big{]}.$

The terminal penalty part can be justified in the same way and the proof is
complete. ∎

Let $(Y,Z)$ be the adjoint processes associated with $\boldsymbol{\delta}$,
i.e., processes $(Y,Z)$ solve the BSDE

$dY_{t}=2\,\phi_{t}\,Q_{t}\,dt+Z_{t}\,dW_{t},\text{\quad and
\quad}Y_{T}=-2A\,Q_{T}.$

The well-posedness of such equation is guaranteed due to the boundedness of
$A$, $\phi$ and $Q$. The duality relation provides an expression of the
Gâteaux derivative of the cost functional in terms of the Hamiltonian of the
system.

######  Lemma 5.2.

The duality relation is given by

$\mathbb{E}[Y_{T}\,V_{T}]=\mathbb{E}\Big{[}\int_{0}^{T}\big{(}\partial_{\boldsymbol{\delta}}\mu(t,\boldsymbol{\delta}_{t})\cdot\boldsymbol{\beta}_{t}\,Y_{t}-\partial_{q}f(t,Q_{t},\boldsymbol{\delta}_{t})\,V_{t}\big{)}\,dt\Big{]}.$

The duality further implies

$\frac{d}{d\epsilon}\,J(\boldsymbol{\delta}+\epsilon\boldsymbol{\beta})\big{|}_{\epsilon=0}=\mathbb{E}\Big{[}\int_{0}^{T}\partial_{\boldsymbol{\delta}}\mathscr{H}(t,Q_{t},Y_{t},\boldsymbol{\delta}_{t})\cdot\boldsymbol{\beta}_{t}\,dt\Big{]},$

where $\mathscr{H}$ denotes the Hamiltonian as follows

$\mathscr{H}(t,q,y,\boldsymbol{\delta}):=\big{(}-a_{t}\,\Lambda^{a}(\delta^{a})+b_{t}\,\Lambda^{b}(\delta^{b})\big{)}\,y+a_{t}\,\delta^{a}\,\Lambda^{a}(\delta^{a})+b_{t}\,\delta^{b}\,\Lambda^{b}(\delta^{b})-\phi_{t}\,(q)^{2}.$

######  PROOF.

The integration by parts gives

$\displaystyle Y_{T}\,V_{T}$
$\displaystyle=Y_{0}\,V_{0}+\int_{0}^{T}Y_{t}\,dV_{t}+\int_{0}^{T}V_{t}\,dY_{t}$
$\displaystyle=\int_{0}^{T}Y_{t}\,\partial_{\boldsymbol{\delta}}b(t,\boldsymbol{\delta}_{t})\cdot\boldsymbol{\beta}_{t}\,dt+\int_{0}^{T}2\,\phi_{t}\,V_{t}\,Q_{t}\,dt+\int_{0}^{T}V_{t}\,Z_{t}\,dW_{t}.$

The duality is verified after taking the expectation, since the last term is a
square integrable martingale. Using this relation, we can further compute

$\displaystyle\frac{d}{d\epsilon}$ $\displaystyle
J(\boldsymbol{\delta}+\epsilon\boldsymbol{\beta})\big{|}_{\epsilon=0}$
$\displaystyle=\mathbb{E}\Big{[}\int_{0}^{T}\big{(}\partial_{q}f(t,Q_{t},\boldsymbol{\delta}_{t})\,V_{t}+\partial_{\boldsymbol{\delta}}f(t,Q_{t},\boldsymbol{\delta}_{t})\cdot\boldsymbol{\beta}_{t}\big{)}\,dt+V_{T}\,Y_{T}\Big{]}$
$\displaystyle=\mathbb{E}\Big{[}\int_{0}^{T}Y_{t}\,\partial_{\boldsymbol{\delta}}b(t,\boldsymbol{\delta}_{t})\cdot\boldsymbol{\beta}_{t}\,dt+\int_{0}^{T}\partial_{\boldsymbol{\delta}}f(t,Q_{t},\boldsymbol{\delta}_{t})\cdot\boldsymbol{\beta}_{t}\,dt\Big{]}$
$\displaystyle=\mathbb{E}\Big{[}\int_{0}^{T}\partial_{\boldsymbol{\delta}}\mathscr{H}(t,Q_{t},Y_{t},\boldsymbol{\delta}_{t})\cdot\boldsymbol{\beta}_{t}\,dt\Big{]}.$

∎

With these preliminaries, we finally reach the necessary condition for
optimality.

######  Theorem 5.3.

If the admissible control $\boldsymbol{\delta}$ is optimal, $Q$ is the
associated controlled inventory, and $(Y,Z)$ are the associated adjoint
processes, then

$\mathscr{H}(t,Q_{t},Y_{t},\boldsymbol{\delta}_{t})\geq\mathscr{H}(t,Q_{t},Y_{t},\beta),\qquad\text{a.e.
in }t\in[0,T],\;\mathbb{P}\text{-}a.s.,$

for any $\beta\in[-\xi,\xi]\times[-\xi,\xi]$.

###### PROOF.

Due to the convexity of the admissible control, given any admissible and
bounded $\boldsymbol{\beta}\in\mathbb{A}\times\mathbb{A}$, we can choose the
perturbation
$\boldsymbol{\delta}^{\epsilon}=\boldsymbol{\delta}+\epsilon\,(\boldsymbol{\beta}-\boldsymbol{\delta})$
which is still admissible. Since $\boldsymbol{\delta}$ is optimal, we have the
inequality

$\frac{d}{d\epsilon}\,J(\boldsymbol{\delta}+\epsilon\,(\boldsymbol{\beta}-\boldsymbol{\delta}))\big{|}_{\epsilon=0}=\mathbb{E}\Big{[}\int_{0}^{T}\partial_{\boldsymbol{\delta}}\mathscr{H}(t,Q_{t},Y_{t},\boldsymbol{\delta}_{t})\cdot(\boldsymbol{\beta}_{t}-\boldsymbol{\delta}_{t})\,dt\Big{]}\leq
0.$

From this we can see

$\partial_{\boldsymbol{\delta}}\mathscr{H}(t,Q_{t},Y_{t},\boldsymbol{\delta}_{t})\cdot(\boldsymbol{\beta}-\boldsymbol{\delta}_{t})\leq
0,\qquad\text{a.e. in }t\in[0,T],\;\mathbb{P}-a.s.,$ (51)

for all $\beta\in[-\xi,\xi]\times[-\xi,\xi]$. Regarding the condition (51), it
can only happen at time $t$ if one of the following three cases holds:

$\displaystyle\delta_{t}^{j}\in(-\xi,+\xi)$ $\displaystyle\text{\quad with
\quad}\partial_{\boldsymbol{\delta}^{j}}\mathscr{H}(t,Q_{t},Y_{t},\boldsymbol{\delta}_{t})=0;$
$\displaystyle\delta_{t}^{j}=-\xi\leq\beta^{j}$ $\displaystyle\text{\quad with
\quad}\partial_{\boldsymbol{\delta}^{j}}\mathscr{H}(t,Q_{t},Y_{t},\boldsymbol{\delta}_{t})\leq
0;$ $\displaystyle\delta_{t}^{j}=+\xi\geq\beta^{j}$ $\displaystyle\text{\quad
with
\quad}\partial_{\boldsymbol{\delta}^{j}}\mathscr{H}(t,Q_{t},Y_{t},\boldsymbol{\delta}_{t})\geq
0,$

where $j\in\\{a,b\\}$ representing the bid-ask side. Take the ask side for
example. If

$\partial_{\boldsymbol{\delta}^{a}}\mathscr{H}(t,Q_{t},Y_{t},\boldsymbol{\delta}_{t})=-a_{t}\,(\Lambda^{a}(\delta_{t}^{a}))^{\prime}\,\Big{[}Y_{t}-\delta_{t}^{a}-\frac{\Lambda^{a}(\delta_{t}^{a})}{(\Lambda^{a}(\delta_{t}^{a}))^{\prime}}\Big{]}=0,$

the monotonicity of
$\delta+\frac{\Lambda^{a}(\delta)}{(\Lambda^{a}(\delta))^{\prime}}$ (see
Assumption 3.2) implies that $\delta_{t}^{a}$ maximizes the corresponding part
of the Hamiltonian. On the other hand, we look at the case
$\delta_{t}^{a}=-\xi$ and
$\partial_{\boldsymbol{\delta}^{a}}\mathscr{H}(t,Q_{t},Y_{t},\boldsymbol{\delta}_{t})\leq
0$. Since $\mathscr{H}$ is first increasing and then decreasing, such
condition indicates that $\delta_{t}^{a}=-\xi$ is the maximizer of the
Hamiltonian on the interval $[-\xi,\xi]$. While the final case can be
discussed in the same way, all three cases imply that the optimal control
$\boldsymbol{\delta}$ maximizes the Hamiltonian along the optimal paths, as
stated by the theorem. ∎

The sufficient condition is more straightforward. We summarize the result in
the following theorem.

######  Theorem 5.4.

Let $\boldsymbol{\delta}$ be an admissible control,
$Q=Q^{\boldsymbol{\delta}}$ be the corresponding controlled inventory, and
$(Y,Z)$ be the adjoint processes. If it holds $\mathbb{P}$-a.s. that

$\mathscr{H}(t,Q_{t},Y_{t},\boldsymbol{\delta}_{t})=\sup_{\boldsymbol{\alpha}\in[-\xi,\xi]^{2}}\mathscr{H}(t,Q_{t},Y_{t},\boldsymbol{\alpha}),\qquad\text{a.e.
in }t\in[0,T],$

then $\boldsymbol{\delta}$ is the optimal control, that is to say,
$J(\boldsymbol{\delta})=\sup_{\boldsymbol{\alpha}\in\mathbb{A}\times\mathbb{A}}J(\boldsymbol{\alpha})$.

###### PROOF.

Since the inventory process—that is associated with any admissible strategy—is
bounded, the adjoint BSDE

$dY_{t}=2\phi_{t}\,Q_{t}\,dt+Z_{t}\,dW_{t},\quad\text{with}\quad
Y_{T}=-2A\,Q_{T},$

admits a unique solution $(Y,Z)\in\mathbb{S}^{2}\times\mathbb{H}^{2}$. Let
$\boldsymbol{\alpha}\in\mathbb{A}\times\mathbb{A}$ be a generic admissible
strategy and $Q^{\prime}$ be the associated state process. Due to the
concaveness of the terminal penalty, we have

$\displaystyle\mathbb{E}[-A$
$\displaystyle(Q_{T})^{2}+A\,(Q^{\prime}_{T})^{2}]$
$\displaystyle\geq\mathbb{E}\big{[}-2A\,Q_{T}\,(Q_{T}-Q^{\prime}_{T})\big{]}$
$\displaystyle=\mathbb{E}\big{[}Y_{T}\,(Q_{T}-Q^{\prime}_{T})\big{]}$
$\displaystyle=\mathbb{E}\Big{[}\int_{0}^{T}2\phi_{t}\,Q_{t}\cdot(Q_{t}-Q^{\prime}_{t})\,dt+\int_{0}^{T}(Q_{t}-Q^{\prime}_{t})\cdot
Z_{t}\,dW_{t}+\int_{0}^{T}Y_{t}\,d(Q_{t}-Q^{\prime}_{t})\Big{]}$
$\displaystyle=\mathbb{E}\Big{[}\int_{0}^{T}2\phi_{t}\,Q_{t}\cdot(Q_{t}-Q^{\prime}_{t})\,dt+\int_{0}^{T}Y_{t}\,d(Q_{t}-Q^{\prime}_{t})\Big{]},$

where we have used the integration by parts and the fact that the stochastic
integral is a bona fide martingale. By the definition of the Hamiltonian, one
can also obtain

$\displaystyle\mathbb{E}\int_{0}^{T}\Big{[}a_{t}$
$\displaystyle\delta_{t}^{a}\,\Lambda^{a}(\delta_{t}^{a})+b_{t}\,\delta_{t}^{b}\,\Lambda^{b}(\delta_{t}^{b})$
$\displaystyle\hskip
56.9055pt-\phi_{t}\,(Q_{t})^{2}-\big{(}a_{t}\,\alpha_{t}^{a}\,\Lambda^{a}(\alpha_{t}^{a})+b_{t}\,\alpha_{t}^{b}\,\Lambda^{a}(\alpha_{t}^{a})-\phi_{t}\,(Q^{\prime}_{t})^{2}\big{)}\Big{]}\,dt$
$\displaystyle=\mathbb{E}\int_{0}^{T}\big{[}\mathscr{H}(t,Q_{t},Y_{t},\boldsymbol{\delta}_{t})-\mathscr{H}(t,Q^{\prime}_{t},Y_{t},\boldsymbol{\alpha}_{t})\big{]}\,dt-\mathbb{E}\int_{0}^{T}Y_{t}\,d(Q_{t}-Q^{\prime}_{t}).$

Combining two results above, finally we can get

$\displaystyle J$ $\displaystyle(\boldsymbol{\delta})-J(\boldsymbol{\alpha})$
$\displaystyle=\mathbb{E}\int_{0}^{T}\Big{[}a_{t}\,\delta_{t}^{a}\,\Lambda^{a}(\delta_{t}^{a})+b_{t}\,\delta_{t}^{b}\,\Lambda^{b}(\delta_{t}^{b})-\phi_{t}\,(Q_{t})^{2}$
$\displaystyle\hskip
28.45274pt-\big{(}a_{t}\,\alpha_{t}^{a}\,\Lambda^{a}(\alpha_{t}^{a})+b_{t}\,\alpha_{t}^{b}\,\Lambda^{b}(\alpha_{t}^{b})-\phi_{t}\,(Q^{\prime}_{t})^{2}\big{)}\Big{]}\,dt+\mathbb{E}\big{[}-2A\,(Q_{T})^{2}+2A\,(Q^{\prime}_{T})^{2}\big{]}$
$\displaystyle\geq\mathbb{E}\int_{0}^{T}\big{[}\mathscr{H}(t,Q_{t},Y_{t},\boldsymbol{\delta}_{t})-\mathscr{H}(t,Q^{\prime}_{t},Y_{t},\boldsymbol{\alpha}_{t})+2\phi_{t}\,Q_{t}\cdot(Q_{t}-Q^{\prime}_{t})\big{]}\,dt$
$\displaystyle\geq\mathbb{E}\int_{0}^{T}\big{[}\mathscr{H}(t,Q_{t},Y_{t},\boldsymbol{\alpha}_{t})-\mathscr{H}(t,Q^{\prime}_{t},Y_{t},\boldsymbol{\alpha}_{t})+2\phi_{t}\,Q_{t}\cdot(Q_{t}-Q^{\prime}_{t})\big{]}\,dt$
$\displaystyle=\mathbb{E}\int_{0}^{T}\phi_{t}\,(Q_{t}-Q^{\prime}_{t})^{2}\,dt$
$\displaystyle\geq 0.$

∎

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