# Robust dividend policy: Equivalence of Epstein-Zin and Maenhout preferences

Kexin Chen  Department of Applied Mathematics, The Hong Kong Polytechnic
University, Hung Hom, Hong Kong<EMAIL_ADDRESS>,  Kyunghyun Park
Division of Mathematical Sciences Nanyang Technological University, Singapore
<EMAIL_ADDRESS>and  Hoi Ying Wong  (Corresponding author)
Department of Statistics, The Chinese University of Hong Kong, Shatin, N.T.,
Hong Kong<EMAIL_ADDRESS>
###### Abstract.

In a continuous-time economy, this study formulates the Epstein-Zin (EZ)
preference for the discounted dividend (or cash payouts) of stockholders as an
EZ singular control utility. We show that such a problem is well-defined and
equivalent to the robust dividend policy set by the firm’s executive in the
sense of Maenhout’s ambiguity-averse preference. While the firm’s executive
announces the expected future earnings in financial reports, they also signal
the firm’s confidence in the expected earnings through dividend or cash
payouts. The robust dividend policy can then be characterized by a Hamilton-
Jacobi-Bellman (HJB) variational inequality (VI). By constructing a novel
shooting method for the HJB-VI, we theoretically prove that the robust
dividend policy is a threshold strategy on the firm’s surplus process.
Therefore, dividend-caring investors can choose firms that match their
preferences by examining stock’s dividend policies and financial statements,
whereas executives can make use of dividend to signal their confidence, in the
form of ambiguity aversion, on realizing the earnings implied by their
financial statements.

JEL Classification. C61, G35, G33.

Key words. dividend policy; Epstein-Zin preference; Maenhout’s ambiguity-
aversion; robust singular control; backward stochastic differential equations;
variational inequality; shooting method.

Funding: The work described in this paper is substantially supported by grants
from the Research Grants Council of the Hong Kong Special Administrative
Region, China (Project No. GRF15305422, GRF15224923 RMG8601495 and
GRF14308422). K. Chen is also financially supported by a research grant from
the Hong Kong Polytechnic University (Project No. PolyU P0038553). K. Park
acknowledges support of the Presidential Postdoctoral Fellowship of Nanyang
Technological University.

##  1\. Introduction

Dividend (or cash payout) policy is an important topic in both corporate
finance and asset pricing. The Miller-Modigliani [49] dividend irrelevance
theory suggests that a firm’s dividend policy does not affect its value and
stock prices in a perfect market. However, the underlying incentives for
stockholders to receive dividends and for the firm’s executives to pay
dividends are largely unknown in reality. For instance, a stockholder can
liquidate some of his stocks for consumption instead of receiving cash
dividend, which is subject to a higher tax rate in the USA. Hence, Black [13]
proposes the dividend puzzle which has generated a great of attentions in the
literature.

A school of thought is the dividend signaling theory (DST) [12, 38, 50] that a
firm’s executive passes signals about the firm’s information to the public
through dividends. Black [14] wrote,

> “The idea that dividends convey information beyond that conveyed by the
> firm’s financial statements and public announcements stretches the
> imagination…. I think we must assume that investors care about dividends
> directly. We must put dividends into the utility functions.”

Although most empirical studies support DST, the survey in [17] shows that
executives viewed the mechanism behind DST as broadly misguided. However,
dividend changes do contain information about future earnings [31].
Additionally, payout announcements signal changes in cash flow volatility (in
the opposite direction) but not stock volatility, which the authors refer to
as ’signaling safety’ [48]. A firm’s cashflow is related to its earnings so
that the cashflow volatility indicates the earnings’ volatility. By modelling
the firm’s surplus process as diffusion models with a stochastic drift, when
there is no dividend, there is a mathematical toolkit [60] to compute the
optimal dividend policy under a random profitability.

In this paper, we consider the suggestions by Black [14]. We assume that the
firm’s financial reports contain information on the firm’s expected earnings
and risk. Thus, we model the firm’s surplus by Itô process with known
parameters if there is no dividend payment. We put dividends into the utility
functions for investors. Investors who care directly about dividends have a
value function that considers the short-term cash inflow from discounted cash
dividends and the certainty equivalence of long-term stock holdings subject to
bankruptcy risk. This consideration requires us to separate risk aversion from
elasticity of intertemporal substitutions (EIS) in the decision making
process. The Epstein-Zin (EZ) [25] preference is a natural candidate designed
for this purpose in the economic literature. In our formulation, we increase
the degree of risk aversion of intertemprol preference ordering via a positive
parameter $R$, without affecting “certainty preferences” of a dividend payment
policy. We also follow the continuous-time dividend models that the discounted
dividend stream is paid in a singular control manner [65, 36, 19, 60]. This
problem formulation turns out to be a singular control problem with recursive
utilities for which we have to develop the well-definedness that there exists
a unique value function for the EZ investor within the recursion in singular
control setting.

Alternatively, the firm’s executive aims to maximize the expected discounted
dividend stream subject to the bankruptcy cost. Although the financial reports
indicate the firm’s expected earnings and risk, the executive encounters
uncertainty in the predicted earnings. We view this uncertainty as model
ambiguity so that the executive’s value function is postulated as the Maenhout
[44] robust preference. Specifically, the executive specifies ambiguity
aversion on the expected earnings, such that the robust dividend decision is
made within a maximin singular control problem subject to bankruptcy risk and
Maenhout’s regularity.

Our study has four key contributions. The first one is the proof for the well-
definedness of the EZ singular control utility. We prove the existence and
uniqueness of the value function satisfying the recursion by using backward
stochastic differential equations (BSDEs) in Theorem 2.3. The proof is non-
trivial because our BSDE possesses a non-decreasing process (i.e., singular
control) in a non-Lipschitz aggregator together with the bankruptcy stopping
time. As the EZ preference model is based on a power function defined by a
positive parameter $R$, the analysis of the existence and uniqueness of
results diverges into two cases: when $R>1$ and when $R<1$. This paper
concentrates on the latter case, where, unlike the case of $R>1$, we cannot
leverage the advantage of priori bounds for the solution of classical BSDEs
(see e.g. [3, 58]). Despite this challenge, our research brings new insights
into the feasibility of the EZ singular control utility. In addition, we
characterize some important properties of the associated BSDEs. In other
words, our problem formulation is shown to be well-defined.

By those properties of such BSDEs, our second contribution establishes the
equivalence between EZ singular control utility for dividends and the
Maenhout’s counterpart once the parameter $R$ of the former is equal to the
ambiguity aversion parameter of the latter (Theorem 3.3). Although the
equivalence between EZ and Maenhout preferences has been addressed in [67]
under a regular control setting, we are the first one developing this
equivalence under a singular control setting, to the best of our knowledge,
even considering the bankruptcy stopping time. Unlike the regular control
situation in [44], our BSDE reveals that the singular control problem with the
Maenhout preference cannot be reduced to a classic utility maximization with a
singular control. Consequently, we cannot rely on existing methods to solve
our problem.

Fortunately, we find that the Maenhout’s singular control problem is more
tractable. Our third contribution develops a novel shooting method to
theoretically show that under certain regularities on the reference parameters
of the firm’s surplus process, the robust dividend policy is a threshold
strategy. When the firm’s surplus hits the upside threshold, the dividend is
paid. The threshold depends on the expected earnings, volatility and the
ambiguity aversion of the executive. However, the development of this result
is highly non-trivial. We first show that the Maenhout’s singular control
problem with bankruptcy can be transformed into a Hamilton-Jacobi-Bellman
(HJB) variational inequality (VI) subject to the bankruptcy boundary
condition. The nonlinear HJB differential operator appears in the VI
formulation because the robust control problem considers worst-case scenarios.
When there is no bankruptcy, the shooting method is shown useful in solving
the HJB-VI in [21] through searching for a suitable threshold to fit the
condition of the HJB-VI. The bankruptcy consideration in our problem
complicates the analysis because the bankruptcy time interacts with the
dividend decision. Specifically, we have to prove that there is a single
threshold on the firm’s surplus process that can shoot two targets on the
boundary condition and the HJB-VI, simultaneously.

Our fourth contribution is attributed to insightful economic interpretations
for DST. While investors are informed with the firm’s expected earnings from
the firm’s financial statements, executives set dividend threshold to showcase
their confidence on realizing the expected earnings. The level of confidence
is reflected by the ambiguity aversion parameter in our model. Therefore, our
model predicts that dividend policy conveys information about earnings
uncertainty or, equivalently, the executive’s confidence in realizing the
expected earnings. We call this notion ‘signaling confidence’.

Related literature. In the context of regular stochastic control, most of the
fundamental and often technically demanding mathematical questions in
recursive utility maximization problems are understood fairly well by now. For
general background, we refer to [24, 64, 41, 69, 46, 47, 32, 33, 34, 51].
While the dominant mathematical framework therein is the BSDE approach, it
does not involve a non-decreasing process in the aggregator. Although certain
proof techniques in the present article bear similarities to those in the
above references, our choice of singular control with random terminal time
makes certain BSDE’s arguments intricate.

On the other hand, considerable efforts have been made to characterize
classical (i.e., non-robust) singular optimal control from both dynamic
programming and probabilistic perspectives. Specifically, we refer to [68, 66,
29, 26, 23] for corresponding free boundary problems and to [7, 5, 4, 6] for
stochastic representation theorem approach. Recently, there has been an
intensive interest in robust analogue of the above references (e.g., [28, 27,
21, 20, 55]). Moreover, we also refer to [62, 8, 9, 10, 53, 57, 56] for
relations to optimal stopping time under ambiguity.

Finally, for completeness, let us mention that the Maenhout preference, which
can be viewed as a form of relative entropy deriving ambiguity-aversion (e.g.,
[22, 11, 2, 43]), has been utilized in robust optimization problems in finance
and economics [45, 37, 16, 70, 30].

The rest of the paper is organized as follows. Section 2 introduces the EZ
singular control utility process along with its BSDE representation and
presents corresponding well-definedness results. Section 3 formulates
Maenhout’s robust dividend-value utility process and establishes its
equivalence with the EZ singular control utility. Section 4 is dedicated to
the characterization of the robust dividend policy, with the detailed proof
based on the shooting method provided in Section 5. The verification theorem
for the robust dividend policy is provided in Section 6. Other preliminaries
and auxiliary proofs for Sections 2 and 4 are given in Appendices A and B.

##  2\. Stochastic differential utility on singular dividend flows

###  2.1. Formulation

Following the suggestion of Black [13], we put dividends into investors’
utility functions through recursive utilities in this section. We first lay
out the setup of a stochastic differential utility (SDU) on singular dividend
flows, and then show that it is not only mathematically well-defined, but also
leads to economically meaningful interpretations.

Consider a probability space $(\Omega,\mathcal{F},\mathbb{P})$ that supports a
one-dimensional Brownian motion $(W_{t})_{t\geq 0}$. Let
$\mathbb{F}:=(\mathcal{F}_{t})_{t\geq 0}$ be the augmented filtration
generated by $W$. We assume that ${\mathcal{}F}_{0}$ is trivial. We write
$\mathbb{E}[\cdot]$ for the expectation under $\mathbb{P}$, while for every
$t\geq 0$ the conditional expectations denoted by
$\mathbb{E}_{t}[\cdot]:=\mathbb{E}[\cdot|{\mathcal{}F}_{t}]$. $\mathcal{P}$ is
the set of all real-valued $\mathbb{F}$-progressively measurable processes,
and ${\mathcal{}P}_{+}$ the restrictions of $\mathcal{P}$ to processes that
take nonnegative values. $L^{p}(\mathcal{F}_{t})$ is the space of
$\mathcal{F}_{t}$-measurable random variable with norm
$\|X\|_{L^{p}}^{p}:=\mathbb{E}[|X|^{p}]<\infty$ for every $p\geq 1$ and $t\geq
0$. The following notations are useful in the rest of the paper.

$\displaystyle\mathcal{T}:=$ $\displaystyle\\{\tau:\tau\text{ is a
$\mathbb{P}$-a.s.~{}finite $\mathbb{F}$-stopping time}\\},$
$\displaystyle\mathcal{A}:=$ $\displaystyle\\{D:(D_{t})_{t\geq 0}\text{ is
$\mathbb{F}$-progressively measurable and $\mathbb{P}$-a.s.~{}continuous,}$
$\displaystyle\text{non-decreasing, with $D_{0-}=0$}\\}.$

In the definitions above, we use $D_{0-}=0$ to indicate that $D_{0}>0$ can
only be achieved by a jump of the process at time zero, followed by a
continuous path for all $t\geq 0$.

For the classical optimal dividend problem [1], the dividend-value objective
function $K=(K_{t})_{t\geq 0}$ is defined as

(2.1) $\displaystyle
K_{t}:=\mathbb{E}_{t}\bigg{[}\int_{t\wedge\tau}^{\tau}e^{-\rho
s}dD_{s}+\xi^{K}_{\tau}\bigg{]},\quad t\geq 0,$

where $\rho>0$ is the constant discount rate. Here, $D=(D_{t})_{t\geq
0}\in\mathcal{A}$ represents the accumulated dividend flows and
$\tau\in\mathcal{T}$ is the ruin time, which potentially depends on either $D$
or certain surplus process driven by $D$. One usually makes the convention
that for $t\geq\tau$, $D_{t}=D_{\tau}$ $\mathbb{P}$-a.s. The term
$\xi^{K}_{\tau}\in L^{1}({\mathcal{}F}_{\tau})$ represents a lump-sum payment
at the ruin time and is conventionally assumed the form of
$\xi^{K}_{\tau}=e^{-\rho\tau}\xi_{0}$, for some constant $\xi_{0}$, as the
time value of ruin; cf.[71].

In preparation, we introduce integrals of the form

$\displaystyle\int_{0}^{\tau}g_{t}dD_{t}=g_{0}D_{0}+\int_{(0,\tau]}g_{t}dD_{t},$

where $g:=(g_{t})_{t\geq 0}$ is $\mathbb{F}$-progressively measurable and
locally bounded, so the above stochastic integral exists in the Stieltjes
sense; cf. [59, Theorem IV.15]. In this context, the jump of $D$ at zero is
accounted and therefore $\int_{0}^{t}dD_{s}=D_{t}$ for $t\geq 0$.

We incorporate an intertemporal aggregator, denoted as $g$, to define a
dividend-value utility $V^{D}=(V^{D}_{t})_{t\geq 0}$ by

(2.2) $\displaystyle
V^{D}_{t}=\mathbb{E}_{t}\bigg{[}\int_{t\wedge\tau}^{\tau}g(s,V^{D}_{s})dD_{s}+\xi_{\tau}\bigg{]},\quad
t\geq 0,$

where the term $\xi_{\tau}\in L^{1}({\mathcal{}F}_{\tau})$ is different from
$\xi^{K}_{\tau}$ appears in (2.1) in general.

######  Definition 2.1.

Given $(\tau,D)\in\mathcal{T}\times\mathcal{A}$ and $\xi_{\tau}\in
L^{1}({\mathcal{}F}_{\tau})$, denote by $\mathbb{I}(g,D)$ the set of all
processes $V\in{\mathcal{}P}$ such that stochastic integral $\int
g(t,V_{t})dD_{t}$ exists and
$\mathbb{E}[\int_{0}^{\tau}|g(s,V_{s})|dD_{s}]<\infty$.

  * (i)

$V\in\mathbb{I}(g,D)$ is a utility process associated to $(g,D)$ if it
satisfies (2.2);

  * (ii)

Denote by $\mathbb{UI}(g,D)$ the subset of all utility processes $V$ that are
uniformly integrable.

This formulation is inspired by the concept of SDU from regular control
problems [25, 24, 64], which creates a feedback effect such that the dividend-
value utility at time $t$ may depend in a nonlinear way on its value at future
times. To proceed further, we introduce the Epstein–Zin (EZ) [25, 24]
aggregator:

(2.3) $\displaystyle g_{\text{EZ}}(s,v):=(1-R)e^{-\rho s}v^{\frac{-R}{1-R}}.$

Throughout the paper, we focus on the case where $\rho>0$ and $R\in(0,1)$. It
is straightforward that when $R=0$, $V$ defined in (2.4) below reduces to $K$
in (2.1). In this context, we define the utility process
$V=V^{D}=(V_{t})_{t\geq 0}$ as the EZ singular control utility that solves

(2.4) $\displaystyle
V_{t}=\mathbb{E}_{t}\bigg{[}\int_{t\wedge\tau}^{\tau}e^{-\rho
s}(1-R)(V_{s})^{\frac{-R}{1-R}}dD_{s}+\xi_{\tau}\bigg{]},\quad t\geq 0.$

We remark that the above definition takes a normalized or the ordinal
equivalent utility form [24], which implies that the economic meaningful one
should be $V^{\frac{1}{1-R}}$. As a result, in accordance to the terminal
condition $\xi^{K}_{\tau}$ in (2.1), we have
$\xi_{\tau}=(\xi_{\tau}^{K})^{1-R}$. In our formulation, we increase the
degree of risk aversion of intertemprol preference ordering, that is $R>0$,
without affecting “certainty preferences”.

Although the EZ singular control utility is similar to the EZ preference for
regular consumption, it distinctively focuses on the risk of ruin rather than
just cash flow utility. Our formulation emphasizes the timing of ruin, which
happens in a finite time, generating key difficulties in our analysis. This is
in a sharp contrast to the ergodic harvesting problem [21], which assumes that
the population does not go extinct in a finite time. The ruin risk results in
the set of dividend strategies that are constant after ruin is nonconvex. This
nonconvexity, in turn, contributes to the lack of concavity in our new EZ
singular control utility.

###  2.2. Well-definedness of EZ singular control utility

We aim to establish the existence and uniqueness of the EZ singular control
utility (2.4) in this subsection. In other words, we want to show that given
$(\tau,D)\in{\mathcal{}T}\times\mathcal{A}$ and $\xi_{\tau}\in
L^{1}({\mathcal{}F}_{\tau}$) satisfying certain reasonable conditions, there
is a unique solution $(Y,Z)$ of the following BSDE involving the non-
decreasing process $D$ in the EZ aggregator:

(2.5) $\displaystyle
Y_{t}=\xi_{\tau}+\int_{t\wedge\tau}^{\tau}g_{\text{EZ}}(s,Y_{s})dD_{s}-\int_{t\wedge\tau}^{\tau}Z_{s}dW_{s},\quad
t\geq 0.$

Referring to the definition of a solution of classical BSDE with random
terminal times (see, e.g., [18, 58]), we give the following definition.

######  Definition 2.2.

Let $(\tau,D)\in{\mathcal{}T}\times\mathcal{A}$ and $\xi_{\tau}\in
L^{1}({\mathcal{}F}_{\tau})$. A solution of the BSDE (2.5) is a pair
$(Y_{t},Z_{t})_{t\geq 0}\in{\mathcal{}P}_{+}\times{\mathcal{}P}$ satisfying
the following conditions $\mathbb{P}$-a.s.:

  * (i)

$Y_{t}=\xi_{\tau}$ and $Z_{t}=0$ on $\\{t\geq\tau\\}$;

  * (ii)

$t\mapsto\mathds{1}_{t\leq\tau}g_{\text{EZ}}(t,Y_{t})$ belongs to
$L^{1}_{\text{loc}}(0,\infty)$, and $t\mapsto Z_{t}$ belongs to
$L^{2}_{\text{loc}}(0,\infty)$;111For each $p\geq 1$,
$L_{\mathrm{loc}}^{p}(0,\infty)$ denotes the class of locally $p$-integrable
functions on $(0,\infty)$.

  * (iii)

For every $T\geq 0$, it holds that for $t\in[0,T]$,

$\displaystyle
Y_{t\wedge\tau}=Y_{T\wedge\tau}+\int_{t\wedge\tau}^{T\wedge\tau}g_{\text{EZ}}(s,Y_{s})dD_{s}-\int_{t\wedge\tau}^{T\wedge\tau}Z_{s}dW_{s}.$

We call the solution $(Y,Z)=(Y_{t},Z_{t})_{t\geq 0}$ of (2.5) an
$L^{2}$-solution if

$\displaystyle\mathbb{E}\bigg{[}\sup_{t\geq
0}|Y_{t\wedge\tau}|^{2}+\int_{0}^{\tau}|Z_{t}|^{2}dt\bigg{]}<\infty$

holds. It worth noting that $L^{2}$-solution particularly implies that $Y$ is
of class $\mathbb{UI}(g_{\text{EZ}},D)$ (see Definition 2.1).

######  Condition A.

Let $(\tau,D)\in{\mathcal{}T}\times\mathcal{A}$ and $\xi_{\tau}\in
L^{1}({\mathcal{}F}_{\tau})$. The triplet $(\tau,\xi_{\tau},D)$ satisfies that
$\mathbb{E}[(\int_{0}^{\tau}e^{-\rho s}dD_{s})^{2}]<\infty,$
$\mathbb{E}[(\xi_{\tau})^{\frac{2}{1-R}}]<\infty$, and $\xi_{\tau}>0$
$\mathbb{P}$-a.s..

######  Theorem 2.3.

Let $g_{\text{EZ}}(\cdot,\cdot)$ be given in (2.3) and $(\tau,\xi_{\tau},D)$
satisfy Condition A. Then the following statements hold:

  * (i)

There exists a unique utility process $V\in\mathbb{UI}(g_{\text{EZ}},D)$ which
has continuous paths and is strictly positive, with $\mathbb{E}[\sup_{t\geq
0}(V_{t})^{2}]<\infty$. In particular, there exists $Z\in\mathcal{P}$ such
that $\int_{0}^{\tau}Z_{t}^{2}dt<\infty$ $\mathbb{P}$-a.s. and $(V,Z)$ solves
the BSDE given in (2.5).

  * (ii)

Let us further assume that there is a real constant $C>0$ such that
$\xi_{\tau}>C$ $\mathbb{P}$-a.s.. Then there exists a unique $L^{2}$-solution
$(V,Z)\in{\mathcal{}P}_{+}\times{\mathcal{}P}$ of the BSDE given in (2.5). In
particular, $V$ has continuous paths and satisfies that $V_{t}\geq C$
$\mathbb{P}$-a.s. for every $t\geq 0$, and $\mathbb{E}[\sup_{t\geq
0}V_{t}^{\frac{2}{1-R}}]<\infty$. Therefore, $V$ is in
$\mathbb{UI}(g_{\text{EZ}},D)$ and is the unique utility process associated to
$(g_{\text{EZ}},D)$.

Theorem 2.3.(ii) is collected in Theorem A.10 with a stand-alone proof
presented in Appendix A. While the BSDE given in (2.5) is closely related to
EZ SDU for regular control problems, its existence and uniqueness result is
intricate. This is because the finite-variation part of (2.5) is integrated
with respect to a non-decreasing process, and we are dealing with a non-
Lipschitz aggregator $g_{\text{EZ}}$ and a random horizon setting.

###### Proof of Theorem 2.3.(i).

For each $n\in\mathbb{N}$, set $\xi_{\tau}^{n}:=\frac{1}{n}\vee\xi_{\tau}$. By
Theorem A.10, there exists a unique $L^{2}$-solution
$(Y^{n},Z^{n})\in{\mathcal{}P}_{+}\times{\mathcal{}P}$ of the BSDE (2.5) (with
$\xi_{\tau}$ replaced by $\xi_{\tau}^{n}$) such that

$\displaystyle
Y^{n}_{t}=\mathbb{E}_{t}\bigg{[}\int_{t\wedge\tau}^{\tau}e^{-\rho
s}(1-R)(Y^{n}_{s})^{\frac{-R}{1-R}}dD_{s}+\xi^{n}_{\tau}\bigg{]},\quad t\geq
0.$

Note that $Y_{t}^{n}$ is non-increasing in $n\in\mathbb{N}$ for every $t\geq
0$ (see Proposition A.5) and $Y^{n}_{t}>0$ $\mathbb{P}$-a.s. for every $t\geq
0$ and $n\in\mathbb{N}$. Hence $V_{t}:=\lim_{n\rightarrow\infty}Y_{t}^{n}$ for
$t\geq 0$ is well-defined. Moreover, the monotone convergence theorem ensures
that for $t\geq 0$,

$\displaystyle V_{t}=$
$\displaystyle\lim_{n\rightarrow\infty}\mathbb{E}_{t}\bigg{[}\int_{t\wedge\tau}^{\tau}e^{-\rho
s}(1-R)(Y^{n}_{s})^{\frac{-R}{1-R}}dD_{s}+\xi^{n}_{\tau}\bigg{]}$
$\displaystyle=$
$\displaystyle\mathbb{E}_{t}\bigg{[}\int_{t\wedge\tau}^{\tau}e^{-\rho
s}(1-R)(V_{s})^{\frac{-R}{1-R}}dD_{s}+\xi_{\tau}\bigg{]}.$

Since $\xi_{\tau}>0$ $\mathbb{P}$-a.s.,
$V_{t}\geq\mathbb{E}_{t}[\xi_{\tau}]>0$ $\mathbb{P}$-a.s. for every $t\geq 0$.

We claim that $V\in\mathbb{UI}(g_{\text{EZ}},D)$. Indeed, since $Y_{0}^{n}$ is
non-increasing in $n\in\mathbb{N}$ and
$V_{0}=\lim_{n\rightarrow\infty}Y^{n}_{0}$, we have that
$\mathbb{E}[\int_{0}^{\tau}e^{-\rho
s}(1-R)(V_{s})^{\frac{-R}{1-R}}dD_{s}]<V_{0}\leq Y_{0}^{n}<\infty$ for every
$n\in\mathbb{N}$. Similarly, we have $\mathbb{E}[\sup_{t\geq
0}(Y_{t})^{\frac{2}{1-R}}]\leq\mathbb{E}[\sup_{t\geq
0}(Y^{n}_{t})^{\frac{2}{1-R}}]<\infty$. It particularly ensures that
$\mathbb{E}[\sup_{t\geq 0}(V_{t})^{2}]<\infty$ by Jensen’s inequality with
exponent $\frac{1}{1-R}>1$. Hence, this concludes that $V$ is uniformly
integrable and therefore is of class $\mathbb{UI}(g_{\text{EZ}},D)$.

It remains to show that $V$ is the unique utility process. By the martingale
representation theorem, there exists $Z\in\mathcal{P}$ such that
$\int_{0}^{\tau}Z_{t}^{2}dt<\infty$ $\mathbb{P}$-a.s. and $(V,Z)$ solves
(2.5). Let $V^{\prime}$ be another utility process of class
$\mathbb{UI}(g_{\text{EZ}},D)$ and $Z^{\prime}\in\mathcal{P}$ be such that
$\int_{0}^{\tau}(Z^{\prime}_{t})^{2}dt<\infty$ $\mathbb{P}$-a.s. and
$(V^{\prime},Z^{\prime})$ solves (2.5). Note that
$V_{t}=V^{\prime}_{t}=\xi_{\tau}$ on $\\{t\geq\tau\\}$.

Set $(\Delta V,\Delta Z):=(V-V^{\prime},Z-Z^{\prime})$. Then ($\Delta V,\Delta
Z)$ solves for every $t\geq 0$,

$\displaystyle\Delta
V_{t}=\int_{t\wedge\tau}^{\tau}\frac{g_{\text{EZ}}(s,V_{s})-g_{\text{EZ}}(s,V^{\prime}_{s})}{\Delta
V_{s}}\Delta V_{s}\mathds{1}_{\Delta V_{s}\neq
0}dD_{s}-\int_{t\wedge\tau}^{\tau}\Delta Z_{s}dW_{s}.$

Let $\alpha=(\alpha_{s})_{s\geq 0}$ be defined by
$\alpha_{s}:=\mathds{1}_{\Delta V_{s}\neq
0}(g_{\text{EZ}}(s,V_{s})-g_{\text{EZ}}(s,V^{\prime}_{s}))/\Delta V_{s}$.
Since $y\mapsto g_{\text{EZ}}(\cdot,y)$ is decreasing, we have $\alpha\leq 0$.
It then follows that for every $t\geq 0$

$\displaystyle e^{\int_{0}^{t}\alpha_{u}du}\Delta
V_{t}=-\int_{t\wedge\tau}^{\tau}e^{\int_{0}^{s}\alpha_{u}du}\Delta
Z_{s}dW_{s}.$

Since $\alpha\leq 0$ and $V,V^{\prime}\in\mathbb{UI}(g_{\text{EZ}},D)$, the
local martingale $\int_{0}^{t}e^{\int_{0}^{s}\alpha_{u}du}\Delta Z_{s}dW_{s}$
is a martingale. Hence, $\Delta V_{t}=0$ $\mathbb{P}$-a.s. for every $t\geq
0$. This completes the proof.∎

##  3\. Robust formulation of dividend flows

In this section, we stand at the perspective of the firm’s executives. We
consider that the executives (the agent) do not work on a particular utility
function but aim to maximize the expected discounted dividend payout subject
to the bankruptcy cost. However, executives have insider information on the
possibility to realize the earnings implied by the firm’s financial
statements. Actually, executives encounter uncertainty on earnings or on the
drift estimate of the diffusion model for the firm’s surplus. We aim to link
an investor’s EZ singular control utility to the robust dividend policy of the
firm’s executives, establishing an equilibrium condition that both parties are
satisfied with the dividend policy.

To introduce the robust singular control criterion analyzed in the main part
of the paper, let us recall the risk neutral utility $K$ defined in (2.1),
where the expectation is taken with respect to the underlying probability
measure $\mathbb{P}$. In the context of model ambiguity, the agent prefers a
family of unspecified alternative models that are in close proximity to the
reference model and do not deviate too far from it. With the aim of
introducing the family of alternative measures, we start with defining a
collection of processes that serves as the Girsanov kernels. Throughout this
section, we fix a constant $\mathcal{R}\in(0,1)$ as the ambiguity-averse
parameter in the robust singular control criterion.

######  Definition 3.1.

A Girsanov kernel process $(\theta_{t})_{t\geq 0}\in\mathcal{P}$ is said to be
admissible if the stochastic exponential
$(\eta^{\theta}_{t})_{t\in[0,\infty)}$ defined as

(3.1)
$\displaystyle\eta_{t}^{\theta}=\exp\bigg{(}\int_{0}^{t}\theta_{s}dW_{s}-\frac{1}{2}\int_{0}^{t}\theta_{s}^{2}ds\bigg{)},\quad
t\in[0,\infty)$

is a martingale. We denote by $\Theta$ the set that collects all admissible
Girsanov kernels $\theta$.

We introduce a new probability measure $\mathbb{Q}^{\theta}$ defined on
$(\Omega,\mathcal{F},\mathbb{F})$ by

(3.2)
$\displaystyle\frac{d\mathbb{Q}^{\theta}}{d\mathbb{P}}\bigg{|}_{\mathcal{F}_{T}}=\eta^{\theta}_{T}$

for each $T>0$. Under the new measure $\mathbb{Q}^{\theta}$, we have that

(3.3) $\displaystyle W_{t}^{\theta}:=W_{t}-\int_{0}^{t}\theta_{s}ds,\quad
t\in[0,T],$

is a Brownian motion. Let $\mathbb{E}^{\theta}_{t}[\cdot]$ denote the
conditional expectation under $\mathbb{Q}^{\theta}$ given $\mathcal{F}_{t}$,
it follows [67, Propositions 2 & 4] that the discounted relative entropy
process associated to $\mathbb{Q}^{\theta}$ is well-defined and
$\mathbb{P}$-a.s. valued in $[0,\infty]$, taking the form
$\frac{1}{2}\mathbb{E}^{\theta}_{t}[\int_{t\wedge\tau}^{\tau}e^{-\rho(s-t)}\theta_{s}^{2}ds]$
for $t>0$.

Consider $(\tau,\xi_{\tau},D)$ that satisfy Condition A. According to
Maenhout’s ambiguity-averse preference [44], we define the robust singular
control criterion $(V_{t}^{\text{rob}})_{t\geq 0}$ as

(3.4) $\displaystyle
V^{\text{rob}}_{t}:=\operatorname*{ess\,inf}\\{V_{t}^{\theta}:\theta\in\Theta^{D}\\},\quad
t\geq 0,$

where

(3.5) $\displaystyle
V_{t}^{\theta}:=\mathbb{E}^{\theta}_{t}\bigg{[}\int_{t\wedge\tau}^{\tau}e^{-\rho
s}dD_{s}+(\xi_{\tau})^{\frac{1}{1-\mathcal{R}}}\bigg{]}+\frac{1}{2\mathcal{R}}\mathbb{E}^{\theta}_{t}\bigg{[}\int_{t\wedge\tau}^{\tau}V^{\theta}_{s}\theta_{s}^{2}ds\bigg{]},$
(3.6)
$\displaystyle\Theta^{D}:=\bigg{\\{}\theta\in\Theta:\mathbb{E}\bigg{[}\bigg{(}\int_{0}^{\tau}\eta_{t}^{\theta}e^{\int_{0}^{t}\frac{\theta^{2}_{s}}{2\mathcal{R}}ds-\rho
t}dD_{t}\bigg{)}^{2}+(\eta_{\tau}^{\theta})^{2}e^{\int_{0}^{\tau}\frac{\theta^{2}_{u}}{\mathcal{R}}du}\big{(}(\xi_{\tau})^{\frac{2}{1-\mathcal{R}}}\vee
1\big{)}\bigg{]}<\infty\bigg{\\}},$

and $\eta^{\theta}$ is the stochastic exponential defined in (3.1). The
objective function $V^{\theta}$ consists of two expected terms: i. present
value of the future dividend payments under alternative probability measure
$\mathbb{Q}^{\theta}$ and ii. a penalty term reflecting the agent’s ambiguity
aversion, scaled with respect to the objective function. The agent considers
the worst-case scenario by choosing $\theta$ and thus the measure
$\mathbb{Q}^{\theta}$ by minimizing the objective function $V^{\theta}$,
resulting in the robust criterion $V^{\text{rob}}$. The ambiguity aversion of
the agent is thus increasing in $\mathcal{R}$.

######  Proposition 3.2.

Let $(\tau,\xi_{\tau},D)$ satisfy Condition A. For arbitrary
$\theta\in\Theta^{D}$, there exists a utility process $V^{\theta}$ that
satisfies (3.5) and is unique in the class
$\\{Y\in\mathcal{P}:\mathbb{E}[\sup_{t\geq 0}Y_{t\wedge\tau}^{2}]<\infty\\}$.
Hence, $V^{\operatorname{rob}}$ in (3.4) is well-defined. Furthermore,
$V^{\theta}$ admits the representation as

(3.7) $\displaystyle
V_{t}^{\theta}=\mathbb{E}^{\theta}_{t}\bigg{[}\int_{t\wedge\tau}^{\tau}e^{\int_{t}^{s}\frac{\theta^{2}_{u}}{2\mathcal{R}}du-\rho
s}dD_{s}+e^{\int_{t\wedge\tau}^{\tau}\frac{\theta^{2}_{u}}{2\mathcal{R}}du}(\xi_{\tau})^{\frac{1}{1-\mathcal{R}}}\bigg{]},\quad
t\geq 0.$

###### Proof.

We start with the construction of $V^{\theta}$ given in (3.7) and show that it
also satisfies (3.5). Since $\theta\in\Theta^{D}$, by Hölder’s inequality
(with exponent 2),

$\mathbb{E}\left[\eta_{\tau}^{\theta}e^{\int_{0}^{\tau}\frac{\theta_{u}^{2}}{2\mathcal{R}}du}(\xi_{\tau})^{\frac{1}{1-\mathcal{R}}}\right]\leq\mathbb{E}\left[(\eta_{\tau}^{\theta})^{2}e^{\int_{0}^{\tau}\frac{\theta^{2}_{u}}{\mathcal{R}}du}\right]^{\frac{1}{2}}\mathbb{E}\left[(\xi_{\tau})^{\frac{2}{1-\mathcal{R}}}\right]^{\frac{1}{2}}<\infty.$

Furthermore, Jensen’s inequality (with exponent 2) ensures that
$\mathbb{E}[\int_{0}^{\tau}\eta_{s}^{\theta}e^{\int_{0}^{s}\frac{\theta^{2}_{u}}{2\mathcal{R}}du-\rho
s}dD_{s}]<\infty$. Hence, we have that

$\displaystyle\mathbb{E}^{\theta}\bigg{[}\int_{0}^{\tau}e^{\int_{0}^{s}\frac{\theta^{2}_{u}}{2\mathcal{R}}du-\rho
s}dD_{s}+e^{\int_{0}^{\tau}\frac{\theta^{2}_{u}}{2\mathcal{R}}du}(\xi_{\tau})^{\frac{1}{1-\mathcal{R}}}\bigg{]}<\infty.$

As a result, $V_{t}^{\theta}$ constructed via (3.7) is a well-defined utility
process.

By Jensen’s inequality, we have $\mathbb{E}[\sup_{t\geq
0}(V^{\theta}_{t\wedge\tau})^{2}]<\infty$. It further implies that there
exists ${\iota}\in\mathcal{P}$ such that
$\int_{0}^{\infty}\iota^{2}_{t}dt<\infty$, $\mathbb{Q}^{\theta}$-a.s. and
$({V}^{\theta},{\iota})$ solves

(3.8)
$\displaystyle{V}^{\theta}_{t}=(\xi_{\tau})^{\frac{1}{1-\mathcal{R}}}+\int_{t\wedge\tau}^{\tau}e^{-\rho
s}dD_{s}+\int_{t\wedge\tau}^{\tau}\frac{1}{2\mathcal{R}}{V}^{\theta}_{s}\theta^{2}_{s}ds-\int_{t\wedge\tau}^{\tau}{\iota}_{s}dW_{s}^{\theta}.$

Since $(M_{t})_{t\geq 0}=(\int_{0}^{t}{\iota}_{s}dW_{s}^{\theta})_{t\geq 0}$
is a local martingale under $\mathbb{Q}^{\theta}$, we shall find an increasing
sequence of stopping times $\\{T_{n}\\}_{n\in\mathbb{N}}$ such that $T_{n}\geq
t$, $\mathbb{Q}^{\theta}(\lim_{n\rightarrow\infty}T_{n}=\infty)=1$ and the
stopped process $(M_{t\wedge T_{n}})_{t\geq 0}$ is a martingale under
$\mathbb{Q}^{\theta}$. Applying $\mathbb{E}_{t}^{\theta}$ to both sides of
(3.8), we have

$\displaystyle
V_{t}^{\theta}=\mathbb{E}^{\theta}_{t}\bigg{[}\int_{t\wedge\tau}^{T_{n}\wedge\tau}e^{-\rho
s}dD_{s}\bigg{]}+\frac{1}{2\mathcal{R}}\mathbb{E}^{\theta}_{t}\bigg{[}\int_{t\wedge\tau}^{T_{n}\wedge\tau}V^{\theta}_{s}\theta_{s}^{2}ds\bigg{]}+\mathbb{E}^{\theta}_{t}[V_{T_{n}\wedge\tau}^{\theta}].$

By sending $n\rightarrow\infty$, the first two terms on the right-hand side
converge by monotone convergence theorem, while the last term converges by the
dominated convergence theorem. We conclude that $V_{t}^{\theta}$ defined via
(3.7) satisfies (3.5). The uniqueness follows from standard localization
arguments, and we omit it here. ∎

The main result of this section is stated below, which extends [67, Theorem 5,
Sect. 6] to a random-horizon singular control problem.

######  Theorem 3.3.

Let $g_{\text{EZ}}$ be given in (2.3) with a fixed $R\in(0,1)$ and
$(\tau,\xi_{\tau},D)$ satisfy Condition A. Set $\mathcal{R}=R$ and
$\theta\in\Theta^{D}$, and let $V^{\operatorname{rob}}$ and $V^{\theta}$ be
defined in (3.4) and (3.5) respectively. Then the following statements hold:

  * (i)

Let $V^{D}$ be the unique utility process associated with $(g_{\text{EZ}},D)$
given by Theorem 2.3 (i), then
$(V^{D})^{\frac{1}{1-R}}=V^{\operatorname{rob}}$.

  * (ii)

Let us further assume that there is a real constant $C>0$ such that
$\xi_{\tau}>C$ $\mathbb{P}$-a.s. Let $(V^{D},Z)$ be the unique
$L^{2}$-solution to the BSDE with parameter
$(\tau,\xi_{\tau},g_{\text{EZ}},D)$ given by Theorem 2.3 (ii), and set
$\mathbb{Y}:=(V^{D})^{\frac{1}{1-R}}$ and
${\mathbb{Z}}:=(\mathbb{Y})^{R}{Z}/(1-R)$. Then it holds

(3.9) $\displaystyle
V^{\theta}_{t}=\mathbb{Y}_{t}+\mathbb{E}_{t}^{\theta}\bigg{[}\int_{t\wedge\tau}^{\tau}\frac{R}{2{\mathbb{Y}}_{s}}\Big{(}{\mathbb{Z}}_{s}+\frac{{\mathbb{Y}}_{s}}{R}\theta_{s}\Big{)}^{2}e^{\frac{1}{2R}\int_{t}^{s}\theta^{2}_{u}du}ds\bigg{]}.$

Hence $V^{\operatorname{rob}}\geq\mathbb{Y}$. In addition, we have
$\theta^{*}:=(-R\mathbb{Z}_{t}/\mathbb{Y}_{t})_{t\in[0,\infty)}\in\Theta^{D}$
hence
$V^{\operatorname{rob}}=V^{\theta^{*}}=\mathbb{Y}=(V^{D})^{\frac{1}{1-R}}$.

###### Proof of Theorem 3.3..

We first prove Theorem 3.3 (ii) with the restrictive condition on
$\xi_{\tau}$. By Theorem 2.3 (ii) and Itô’s lemma, $(\mathbb{Y},\mathbb{Z})$
solves

$\displaystyle\mathbb{Y}_{t}=$
$\displaystyle(\xi_{\tau})^{\frac{1}{1-R}}+\int_{t\wedge\tau}^{\tau}e^{-\rho
s}dD_{s}-\frac{R}{2}\int_{t\wedge\tau}^{\tau}\frac{(\mathbb{Z}_{s})^{2}}{\mathbb{Y}_{s}}ds-\int_{t\wedge\tau}^{\tau}\mathbb{Z}_{s}dW_{s}$
$\displaystyle=$
$\displaystyle(\xi_{\tau})^{\frac{1}{1-R}}+\int_{t\wedge\tau}^{\tau}e^{-\rho
s}dD_{s}-\frac{R}{2}\int_{t\wedge\tau}^{\tau}\frac{(\mathbb{Z}_{s})^{2}}{\mathbb{Y}_{s}}ds-\int_{t\wedge\tau}^{\tau}\mathbb{Z}_{s}\theta_{s}ds-\int_{t\wedge\tau}^{\tau}\mathbb{Z}_{s}dW_{s}^{\theta},\quad
t\geq 0.$

For $\theta\in\Theta^{D}$, denote by $({V}^{\theta},{\iota})$ a solution of
(3.8). We obtain

$\displaystyle{V}^{\theta}_{t}-\mathbb{Y}_{t}=\int_{t\wedge\tau}^{\tau}\bigg{(}\frac{R}{2\mathbb{Y}_{s}}\Big{(}\mathbb{Z}_{s}+\frac{\mathbb{Y}_{s}}{R}\theta_{s}\Big{)}^{2}+\frac{1}{2R}\theta_{t}^{2}({V}^{\theta}_{t}-\mathbb{Y}_{t})\bigg{)}ds-\int_{t\wedge\tau}^{\tau}({\iota}_{s}-\mathbb{Z}_{s})dW_{s}^{\theta}.$

Let $\beta_{t}:=\exp(\frac{1}{{2R}}\int_{0}^{t}{\theta^{2}_{s}}ds)$. Then
Itô’s lemma gives

(3.10)
$\displaystyle\beta_{t}({V}^{\theta}_{t}-\mathbb{Y}_{t})=\int_{t\wedge\tau}^{\tau}\beta_{s}\frac{R}{2\mathbb{Y}_{s}}\Big{(}\mathbb{Z}_{s}+\frac{\mathbb{Y}_{s}}{R}\theta_{s}\Big{)}^{2}ds-(M_{\tau}-M_{t\wedge\tau}),$

where
$M_{t}:=\int_{0}^{t}\beta_{s}({\iota}^{\theta}_{s}-\mathbb{Z}_{s})dW^{\theta}_{s}$.
Since $(M_{s})_{s\in[0,\infty)}$ is a local martingale under
$\mathbb{Q}^{\theta}$, we shall find an increasing sequence of stopping times
$\\{T_{n}\\}_{n\in\mathbb{N}}$ such that $T_{n}\geq t$,
$\mathbb{Q}^{\theta}(\lim_{n\rightarrow\infty}T_{n}=\infty)=1$ and the stopped
process $(M_{s\wedge T_{n}})_{s\in[t,\infty)}$ is a martingale under
$\mathbb{Q}^{\theta}$. Applying $\mathbb{E}_{t}^{\theta}$ to both sides of
(3.10), we have for $t\geq 0$

(3.11)
$\displaystyle{V}^{\theta}_{t}-\mathbb{Y}_{t}=\mathbb{E}_{t}^{\theta}\left[\int_{t\wedge\tau}^{T_{n}\wedge\tau}\frac{R}{2\mathbb{Y}_{s}}\Big{(}\mathbb{Z}_{s}+\frac{\mathbb{Y}_{s}}{R}\theta_{s}\Big{)}^{2}e^{\int_{t\wedge\tau}^{s}\frac{\theta^{2}_{u}}{2R}du}ds+e^{\int_{t\wedge\tau}^{T_{n}\wedge\tau}\frac{\theta^{2}_{u}}{2R}du}({V}^{\theta}_{T_{n}\wedge\tau}-\mathbb{Y}_{T_{n}\wedge\tau})\right].$

In order to pass the limit of $n\rightarrow\infty$ under the expectation, we
check that each terms of the right-hand side of above equation converges. The
first expectation converges by the monotone convergence since the integrand is
nonnegative. For the second term, we observe that

$\displaystyle\mathbb{E}_{t}^{\theta}[e^{\int_{t\wedge\tau}^{T_{n}\wedge\tau}\frac{\theta^{2}_{u}}{2R}du}{V}^{\theta}_{T_{n}\wedge\tau}]=$
$\displaystyle\mathbb{E}_{t}^{\theta}\bigg{[}\int_{T_{n}\wedge\tau}^{\tau}e^{\int_{t\wedge\tau}^{s}\frac{\theta^{2}_{u}}{2R}du-\rho
s}dD_{s}+e^{\int_{t\wedge\tau}^{\tau}\frac{\theta^{2}_{u}}{2R}du}(\xi_{\tau})^{\frac{1}{1-R}}\bigg{]}.$

The r.h.s of the above equation converges by the monotone convergence theorem
since the integrand is nonnegative. It then follows that
$\mathbb{E}_{t}^{\theta}[e^{\int_{t\wedge\tau}^{T_{n}\wedge\tau}\frac{\theta^{2}_{u}}{2R}du}{V}^{\theta}_{T_{n}\wedge\tau}]$
converges to
$\mathbb{E}_{t}^{\theta}[e^{\int_{t\wedge\tau}^{\tau}\frac{\theta^{2}_{u}}{2R}du}\xi_{\tau}]$
a.s. Lastly, we show that the term
$\mathbb{E}_{t}^{\theta}[e^{\int_{t\wedge\tau}^{T_{n}\wedge\tau}\frac{\theta^{2}_{u}}{2R}du}\mathbb{Y}_{T_{n}\wedge\tau}]$
converges to
$\mathbb{E}_{t}^{\theta}[e^{\int_{t\wedge\tau}^{\tau}\frac{\theta^{2}_{u}}{2R}du}\xi_{\tau}]$.
Under the conditions $\theta\in\Theta^{D}$ and $\xi_{\tau}\geq C$ a.s., we
have

$\displaystyle\mathbb{E}^{\theta}\Big{[}e^{\int_{0}^{\tau}\frac{\theta^{2}_{u}}{2R}du}\sup_{t\geq
0}\mathbb{Y}_{t\wedge\tau}\Big{]}$
$\displaystyle\leq\mathbb{E}\Big{[}\eta^{\theta}_{\tau}e^{\int_{0}^{\tau}\frac{\theta^{2}_{u}}{2R}du}\xi_{\tau}^{\frac{1}{1-R}}\sup_{t\geq
0}\Big{(}\mathbb{Y}_{t\wedge\tau}\xi_{\tau}^{-\frac{1}{1-R}}\Big{)}\Big{]}$
$\displaystyle\leq
C^{-\frac{1}{1-R}}\mathbb{E}\Big{[}(\eta^{\theta}_{\tau})^{2}e^{\int_{0}^{\tau}\frac{\theta^{2}_{u}}{R}du}\xi_{\tau}^{\frac{2}{1-R}}\Big{]}^{\frac{1}{2}}\mathbb{E}\Big{[}\sup_{t\geq
0}\mathbb{Y}_{t\wedge\tau}^{2}\Big{]}^{\frac{1}{2}}<\infty.$

As a result,
$\mathbb{E}_{t}^{\theta}[e^{\int_{t\wedge\tau}^{T_{n}\wedge\tau}\frac{\theta^{2}_{u}}{2R}du}\mathbb{Y}_{T_{n}\wedge\tau}]$
is dominated by
$\mathbb{E}_{t}^{\theta}[e^{\int_{0}^{\tau}\frac{\theta^{2}_{u}}{2R}du}\sup_{t\geq
0}\mathbb{Y}_{t\wedge\tau}]$ which is finite from above analysis. The claimed
equation (3.9) then follows by taking limit of $n\rightarrow\infty$ of (3.11)
under the expectation. (3.9) implies that
$V^{\theta}_{t}\geq\mathbb{Y}_{t}=(V^{D}_{t})^{\frac{1}{1-R}}$ and equality
holds for $\theta^{*}=-R\mathbb{Z}/\mathbb{Y}$ if $\theta^{*}\in\Theta^{D}$.

It remains to prove that $\theta^{*}=-R\mathbb{Z}/\mathbb{Y}\in\Theta^{D}$. We
first show that $\eta^{\theta^{*}}_{t}$ is a martingale for $t\in[0,\infty)$.
Applying Itô’s lemma to $\ln(\mathbb{Y})$, we have

(3.12)
$\displaystyle\ln(\mathbb{Y}_{t\wedge\tau})-\ln(\mathbb{Y}_{0})=-\int_{0}^{t\wedge\tau}\frac{1}{\mathbb{Y}_{s}}e^{-\rho
s}dD_{s}+\int_{0}^{t\wedge\tau}\frac{R-1}{2}\Big{(}\frac{\mathbb{Z}_{s}}{\mathbb{Y}_{s}}\Big{)}^{2}ds+\int_{0}^{t\wedge\tau}\frac{\mathbb{Z}_{s}}{\mathbb{Y}_{s}}dW_{s}.$

(Note that since $(\mathbb{Y}_{t})_{t\geq 0}$ is continuous and never $0$,
$1/\mathbb{Y}$ is locally bounded.) Further note that $\theta^{*}=0$ on
$\\{t\geq\tau\\}$ by Definition 2.2 (i), we have for $t\geq 0$,

$\displaystyle-\frac{1}{2}\int_{0}^{t}\theta^{*}_{s}ds+\int_{0}^{t}\theta^{*}_{s}dW_{s}=R\bigg{(}\ln(\mathbb{Y}_{0})-\ln(\mathbb{Y}_{t\wedge\tau})-\int_{0}^{t\wedge\tau}\frac{1}{\mathbb{Y}_{s}}e^{-\rho
s}dD_{s}-\frac{1}{2}\int_{0}^{t\wedge\tau}\Big{(}\frac{\mathbb{Z}_{s}}{\mathbb{Y}_{s}}\Big{)}^{2}ds\bigg{)}.$

By definition of $\eta^{\theta^{*}}$ in (3.1), we have for $t\geq 0$,

(3.13)
$\displaystyle\eta^{\theta^{*}}_{t}=\Big{(}\frac{\mathbb{Y}_{0}}{\mathbb{Y}_{t\wedge\tau}}\Big{)}^{R}e^{-R\int_{0}^{t\wedge\tau}\frac{1}{\mathbb{Y}_{s}}e^{-\rho
s}dD_{s}-R\int_{0}^{t}(\frac{\mathbb{Z}_{s}}{\mathbb{Y}_{s}})^{2}ds}\leq\mathbb{Y}_{0}^{R}C^{-\frac{R}{1-R}},\quad\mathbb{P}\text{-a.s..}$

Here we have used the condition that $\xi_{\tau}>C$ $\mathbb{P}$-a.s. and
hence $V^{D}\geq C$ $\mathbb{P}$-a.s. (see Theorem 2.3 (ii)). For
$t\in[0,\infty)$, we know that $\eta_{t}$ is a local martingale, the uniform
bound (3.13) further implies that $\eta_{t}$ is a martingale.

Next we verify $\theta^{*}\in\Theta^{D}$ by condition (3.6). From (3.12), we
have

$\displaystyle\mathbb{E}\Big{[}(\eta^{\theta^{*}}_{\tau})^{2}e^{\int_{0}^{\tau}\frac{(\theta^{*}_{s})^{2}}{R}ds}(\xi_{\tau})^{\frac{2}{1-R}}\Big{]}=\mathbb{E}\Big{[}e^{-2R\int_{0}^{\tau}\frac{1}{\mathbb{Y}_{s}}e^{-\rho
s}dD_{s}}\mathbb{Y}_{\tau}^{2(1-R)}Y_{0}^{2R}\Big{]}\leq(V^{D}_{0})^{\frac{2R}{1-R}}\mathbb{E}[\xi_{\tau}^{2}]<\infty,$

where we have used the fact that ${\mathcal{}F}_{0}$ is trivial (see Section
2.1) in the inequality.

On the other hand,

$\displaystyle\mathbb{E}\left[\bigg{(}\int_{0}^{\tau}e^{\int_{0}^{t}\frac{(\theta^{*}_{s})^{2}}{2R}ds-\rho
t}\eta_{t}^{\theta^{*}}dD_{t}\bigg{)}^{2}\right]=$
$\displaystyle\mathbb{E}\left[\bigg{(}\int_{0}^{\tau}e^{R\Big{(}\ln(\mathbb{Y}_{0})-\ln(\mathbb{Y}_{t})-\int_{0}^{t}\frac{1}{\mathbb{Y}_{s}}e^{-\rho
s}dD_{s}\Big{)}-\rho t}dD_{t}\bigg{)}^{2}\right]$ $\displaystyle\leq$
$\displaystyle\mathbb{Y}_{0}^{2R}\mathbb{E}\left[\bigg{(}\sup_{t\geq
0}\mathbb{Y}_{t\wedge\tau}^{1-R}\int_{0}^{\tau}e^{-R\int_{0}^{t}\frac{1}{\mathbb{Y}_{s}}e^{-\rho
s}dD_{s}}\frac{1}{\mathbb{Y}_{t}}e^{-\rho t}dD_{t}\bigg{)}^{2}\right]$
$\displaystyle\leq$
$\displaystyle\mathbb{Y}_{0}^{2R}\mathbb{E}\left[\sup_{t\geq
0}(V^{D}_{t\wedge\tau})^{2}\frac{1}{R}(1-e^{-R\int_{0}^{\tau}\frac{1}{\mathbb{Y}_{s}}e^{-\rho
s}dD_{s}})^{2}\right]<\infty.$

This completes the proof for Theorem 3.3 (ii). Theorem 3.3 (i) follows by
considering a non-increasing sequence
$\xi_{\tau}^{n}=\frac{1}{n}\vee\xi_{\tau}$ for $n\in\mathbb{N}$, and by the
monotone convergence theorem. ∎

##  4\. A class of solvable dividend optimization problems

In the preceding sections, we show that EZ singular control utility is well-
defined and its equivalence to Meanhout’s robust dividend criterion. Now, we
turn our attention to determining a robust dividend policy amidst ambiguity in
earning ability. We propose a set of conditions that can characterize the
optimal dividend policy for a wide range of diffusion models for the
underlying surplus dynamic.

###  4.1. Dynamics, controls and the objective function

We start with the state space $\mathcal{X}:=(\ell,\infty)$,
$-\infty\leq\ell<0$, and model the uncontrolled surplus process as
$\mathcal{X}$-valued diffusion $X^{x}=(X^{x}_{t})_{t\geq 0}$ denote the
solution of the stochastic differential equation (SDE)

(4.1) $\displaystyle
X^{x}_{t}=x+\int_{0}^{t}\mu(X_{s})ds+\int_{0}^{t}\sigma(X_{s})dW_{t},\quad
t\geq 0,~{}X_{0}=x\in\mathcal{X},$

where functions $\mu:\mathcal{X}\rightarrow\mathbb{R}$,
$\sigma:\mathcal{X}\rightarrow\mathbb{R}/\\{0\\}$ are Borel functions. The
value $\mu(X_{t})/X_{t}$ is the expected per surplus growth rate and
$\sigma(X_{t})/X_{t}$ stands for infinitesimal volatility of fluctuations in
the per surplus growth rate.

######  Condition B.

The SDE (4.1) admits a unique in law weak solution, which is a regular
diffusion with unattainable boundaries $\\{\ell,\infty\\}$. Moreover, there is
a constant $\overline{\mu}$ such that

(4.2) $\displaystyle\sup_{x\geq 0}\mu(x)-\rho x\leq\overline{\mu}.$

Financially speaking, Condition B asserts that the net appreciation rate of
the surplus is uniformly bounded for all $x\geq 0$. This condition plays a
crucial role in deriving several fundamental properties for the value function
of the robust dividend problem. Specifically, it leads to a priori continuity
of the optimal value function on $[0,\infty)$, particularly at ${0}$ (see
Proposition 4.1 below). This continuity is essential when characterizing our
problem as a VI, as it provides a boundary condition at bankruptcy, i.e., when
$x=0$, that guarantees the uniqueness of our solution (see (4.8) below).
Condition B also implies that the uncontrolled surplus process will not
explode at $\infty$ and that the point $\\{0\\}$ is a regular point. This is
consistent with our emphasis on the ruin risk, that is, the bankruptcy happens
in our model.

For each $D\in\mathcal{A}$, the dynamics of the controlled surplus process
$X^{x,D}:=(X_{t}^{x,D})_{t\geq 0}$ is governed by

(4.3) $\displaystyle
X_{t}^{x,D}=x+\int_{0}^{t}\mu(X_{s}^{x,D})ds+\int_{0}^{t}\sigma(X_{s}^{x,D})dW_{t}-D_{t},\quad
X_{0-}^{x,D}=x.$

By the definition of $\mathcal{A}$ introduced in Section 2.1, the above
condition implies that a dividend distribution may take place at the initial
time, hence $X_{0-}^{D}$ denotes the initial surplus level before such a
distribution has ever occurred. We also define the stopping time
$\tau^{x,D}\in\mathcal{T}$ as the time of bankruptcy as

(4.4) $\displaystyle\tau^{x,D}:=\inf\\{t\geq 0:X_{t}^{x,D}\leq 0\\}.$

Moreover, we make the convention that $X_{t}^{x,D}=0$ on
$\\{t\geq\tau^{x,D}\\}$. At bankruptcy, we assume there is a lump-sum payment
given by a positive constant $\xi_{0}$. We further require that the
$D_{t}-D_{t-}\leq X_{t-}^{x,D}$, for any $t\geq 0$, so that the controlled
surplus cannot be a negative value even at the time of bankruptcy. We denote
by $\mathcal{A}^{x}\subset\mathcal{A}$, which may depend on the initial
surplus level, the admissible set of $D$ that fulfills the above requirements
and that $(\tau^{x,D},e^{-\rho\tau^{x,D}}\xi_{0},D)$ satisfies Condition A.

Following definitions (3.4) and (3.5) in Sect. 3, the value function $J^{*}$
of the robust dividend problem is defined as

(4.5) $\displaystyle
J^{*}(x):=\sup_{D\in\mathcal{A}^{x}}\inf_{\theta\in\Theta^{D}}J^{\theta}(x;D),$
(4.6) $\displaystyle
J^{\theta}(x;D):=\mathbb{E}^{\theta}\bigg{[}\int_{0}^{\tau^{x,D}}e^{-\rho
t}dD_{t}+\frac{1}{2\mathcal{R}}\int_{0}^{\tau^{x,D}}e^{-\rho
t}J^{\theta}(X_{t}^{x,D};D)\theta_{t}^{2}dt+e^{-\rho\tau^{x,D}}\xi_{0}\bigg{]}.$

By Proposition 3.2, $J^{\theta}$ admits the representation as

$\displaystyle
J^{\theta}(x;D):=\mathbb{E}^{\theta}\bigg{[}\int_{0}^{\tau^{x,D}}e^{\frac{1}{2\mathcal{}R}\int_{0}^{t}\theta_{s}^{2}ds-\rho
t}dD_{t}+e^{\frac{1}{2\mathcal{}R}\int_{0}^{\tau^{x,D}}\theta_{s}^{2}ds-\rho\tau^{x,D}}\xi_{0}\bigg{]}.$

Priori properties of the optimal value function $J^{*}$ are summarized in the
following proposition.

######  Proposition 4.1.

Under Condition B, it holds that $\xi_{0}\leq J^{*}(x)\leq
x+\xi_{0}+{\overline{\mu}}/{\rho}$, $x\in\mathbb{R}_{+}$, where
$\overline{\mu}$ is the constant that dominates function $\mu(x)-\rho x$ in
Condition B. In addition, $J^{*}$ is continuous and nondecreasing subject to

(4.7) $\displaystyle J^{*}(0+)=\xi_{0}.$

The proof is technical and we collect it in Appendix B.

In the rest of the paper, we focus on candidate controls which are threshold
dividend strategies. In the classical dividend-paying problems, the optimal
strategy has always been of the threshold type (see e.g. [1]). Our main result
shows that these types of controls are also optimal in the more general
setting including more general diffusion processes and Maenhout’s type of
ambiguity, hence also for EZ singular control utility.

Following [21, 20], we first provide a rigorous definition of threshold
dividend strategies by Skorokhod map on an interval. Denote by
$\mathcal{D}(\mathbb{R}_{+})$ the the space of right-continuous functions with
left limits mapping $\mathbb{R}_{+}$ into $\mathbb{R}$. Fix $\beta>0$. Given
$\psi\in\mathcal{D}(\mathbb{R}_{+})$ there exists a unique pair of functions
$(\phi,\eta)\in\mathcal{D}(\mathbb{R}_{+})^{2}$ that satisfy the following two
properties:

  1. (i)

For every $t\in\mathbb{R}_{+},\phi(t)=\psi(t)+\eta(t)$;

  2. (ii)

$\eta(0-)=0$, $\eta$ is nondecreasing, and

$\int_{0}^{\infty}\mathds{1}_{\\{\phi(s)<\beta\\}}d\eta(s)=0.$

We define the map
$\Gamma_{\beta}[\psi]=(\Gamma_{\beta}^{1},\Gamma_{\beta}^{2})[\psi]:=(\phi,\eta)$.
See [42] for the existence and uniqueness of solutions, as well as the
continuity propertiy of the map.

######  Definition 4.2 (threshold dividend strategies).

Fix $x,~{}b\in\mathbb{R}_{+}$. We call $D=D^{(b)}$ a $b$-threshold dividend
strategy if for every continuous $\psi\in\mathcal{D}(\mathbb{R}_{+})$, one has
$(X^{D},D)(\psi)=\Gamma_{b}[\psi]$.

One can easily verify that any $b$-threshold control is admissible in the
sense of $D^{(b)}\in\mathcal{A}^{x}$.

###  4.2. The free-boundary problem

We aim to show that for any fixed ${\mathcal{}R}\in(0,1)$, there is positive
constant $b$ such that, for any initial state $x>0$,

$\displaystyle
J^{*}(x)=\inf_{\theta\in\Theta^{D^{(b)}}}J^{\theta}(x;D^{(b)}).$

Since the value function has been proved to be continuous, this suggests that
the threshold level $b$ can be characterized by an HJB equation by dynamic
programming principle. Motivated by the robust structure of objective
function, together with the dynamic given in (4.3), let ${\mathcal{}L}$ be the
nonlinear operator with Hamiltonian over $\theta\in\mathbb{R}$ which acts on
$v\in C^{2}([0,\infty))$ as

where the last equality is valid provided that $v>0$ on $(0,\infty)$.

Next, we present the relevant HJB equation. $J^{*}$ defined in (4.5) is
expected to be a solution of the following Hamilton-Jacobi-Bellman Variational
Inequality (HJB-VI)

(4.8) $\displaystyle\left\\{\begin{aligned}
&\max\\{{\mathcal{}L}v(x),1-v^{\prime}(x)\\}=0\qquad\mbox{on}\;\;(0,\infty);\\\
&v(0)=\xi_{0}.\end{aligned}\right.$

Technically speaking, the boundary condition in (4.8) should be
$v(0+)=\xi_{0}$ as (4.7) shows, rather than $v(0)=\xi_{0}$. Here and in what
follows, we always adopt a convention that the solution of the HJB-VI equation
is extended to 0 by continuity and the value of $v$ at 0 is its limit from the
right.

Following our educated guess that the optimal control is a threshold control,
we choose to work with the following more explicit free-boundary ODE. Namely,
we are looking for a constant $b>0$ for which there exists $v\in
C^{2}([0,\infty))$ such that

(4.9) $\displaystyle\left\\{\begin{aligned}
&{\mathcal{}L}v(x)=0,\;\;v^{\prime}(x)\geq 1\qquad\mbox{on}\;\;(0,b];\\\
&{\mathcal{}L}v(x)\leq 0,\;\;v^{\prime}(x)=1\qquad\mbox{on}\;\;(b,\infty);\\\
&v(0)=\xi_{0},\end{aligned}\right.$

as well as $v^{\prime\prime}(b)=0$ and $v^{\prime}(b)=1$. The rationale behind
this is as follows: when the initial surplus is $X_{0}^{x,D}=x\in(b,\infty)$,
then in order to keep the process $X^{x,D}$ between 0 and $b$, there is an
instantaneous dividend payment of size $x-b$. When $x\in(0,b)$, no action is
being taken. When $X^{x,D}$ hits the boundary $b$, the threshold policy is
taking action, leading to the Neumann boundary condition at $b$. The surplus
level will be kept in $(0,b)$, with an initial payment
$0\wedge(X_{0}^{x,D}-b)$ and then pay out dividend only when the surplus
$X^{x,D}$ is at level $b$.

###  4.3. Further assumptions and the main result

######  Assumption 4.3.

$\mu:{\mathcal{}X}\rightarrow\mathbb{R}$ is continuously differentiable on
$[0,\infty)$ with $\mu(0)>0$. Furthermore,
$\sigma:{\mathcal{}X}\rightarrow(0,\infty)$ is non-decreasing and continuously
differentiable on $[0,\infty)$. There exist
$\underline{b},\overline{b}\in[0,\infty]$ with $\underline{b}<\overline{b}$
and the following conditions hold.

  * (i)

$\mu(x)^{2}\geq 3\mathcal{R}\rho\sigma(x)^{2}$ on $x\in[0,\underline{b}]$ and
there is
$\overline{b}=\inf\\{x\in(\underline{b},\infty):\mu(x)^{2}=2\mathcal{R}\rho\sigma(x)^{2}\\}$,
with the usual convention $\inf\\{\varnothing\\}:=\infty$. In addition,
$\mu(x)^{2}<2\mathcal{R}\rho\sigma(x)^{2}$ for $x>\overline{b}$.

  * (ii)

Define functions $\psi^{\pm}:[0,\overline{b}]\ni
x\rightarrow\psi^{\pm}(x)\in\mathbb{R}$ by

$\displaystyle\psi^{\pm}(x):=\frac{\mu(x)\pm\sqrt{\mu^{2}(x)-2{\mathcal{}R}\rho\sigma^{2}(x)}}{2\rho},$

in paricular, $\psi^{\pm}$ are roots of the quadratic equation
$Q(\psi;x):=\rho\psi^{2}-\mu(x)\psi+\frac{{\mathcal{}R}}{2}\sigma^{2}(x)$. The
function $\psi^{+}(x)-x$ increases on $[0,\underline{b})$ and decreases on
$[\underline{b},\overline{b})$.

  * (iii)

$\sup_{x\in[0,\overline{b}]}\\{\psi^{-}(x)-x\\}<\xi_{0}<\psi^{+}(\underline{b})-\underline{b}$
and there is
$\hat{b}:=\inf\\{x\in(\underline{b},\overline{b}):\psi^{+}(x)-x=\xi_{0}\\}$.

Before discussing the assumption,we validate it within a standard mean-
reverting surplus model, an extension of Brownian motion with constant drift,
which aligns with substantial empirical support (see [19]).

######  Example 4.4.

In mean-reverting surplus mode, the dynamic (4.1) of the uncontrolled surplus
is given by

$\displaystyle
dX_{t}=\overline{\gamma}(\overline{\mu}-X_{t})dt+\overline{\sigma}dW_{t},$

where $\overline{\gamma}>0$, $\overline{\mu}>0$ and $\overline{\sigma}>0$ are
constants. The $\overline{\gamma}$ is the speed of mean-reversion,
$\overline{\mu}$ is the long-term mean of the process, and $\overline{\sigma}$
is the volatility. Suppose the following conditions hold:

$\displaystyle(\overline{\gamma}\overline{\mu})^{2}>3\mathcal{R}\rho\overline{\sigma}^{2},\quad\sqrt{\dfrac{\mathcal{R}}{2\rho}}<\xi_{0}<\frac{\overline{\gamma}\overline{\mu}+\sqrt{(\overline{\gamma}\overline{\mu})^{2}-2\mathcal{R}\rho\overline{\sigma}^{2}}}{2\rho}.$

Then Assumption 4.3 is satisfied with $\underline{b}=0$ and
$\overline{b}=\overline{\mu}-\overline{\sigma}\sqrt{2\mathcal{R}\rho}/\overline{\gamma}$.

Assumption 4.3 is the generalization of [1, Lemma 3.1.(ii)] to the setting
that includes ambiguity. We note that when $\mathcal{R}=0$ and $\xi_{0}=0$,
conditions in Assumption 4.3 are reduced to those in [1, Lemma 3.1.(ii)].
Indeed, when $\mathcal{R}=0$, functions $\psi^{+}_{0}(x)=\mu(x)/\rho$ and
$\psi^{-}_{0}=0$ by our assumption that $\mu(0)>0$. Assumption 4.3 (ii) imply
that there exists $\underline{b}\geq 0$ such that $\mu(x)-\rho x$ is
increasing on $[0,\underline{b}]$ and decreasing on $(\underline{b},\infty)$
with $\overline{b}=\infty$. When $\xi_{0}=0$, Assumption 4.3 (iii) is
equivalent to $\lim_{x\rightarrow\infty}\mu(x)-\rho x<0$ in [1, Lemma
3.1.(ii)], so that there exists $\hat{b}>\underline{b}$ such that
$\mu(\hat{b})-\rho\hat{b}=0$.

We end this section by concluding our main results in the following theorems.
We prove that an optimal dividend strategy exists and that it is a threshold
control. Moreover, we show that the threshold level and the value function
$J^{*}$ are characterized by the free-boundary problem (4.9). The proof of the
main results is presented in the subsequent sections.

######  Theorem 4.5 (Existence).

Suppose that Condition B and Assumption 4.3 hold. Then there exists a solution
${v}^{*}:[0,\infty)\rightarrow[0,\infty)$ of the free boundary problem (4.9)
such that it is in $C^{2}([0,\infty))$ and the corresponding free boundary
$b^{*}$ takes value in $(\underline{b},\hat{b})$ (with $\underline{b},\hat{b}$
appearing in Assumption 4.3).

######  Theorem 4.6 (Verification).

Suppose that Condition B and Assumption 4.3 hold. Let
$v^{*}:[0,\infty)\rightarrow[0,\infty)$ be a solution of the free boundary
problem (4.9) with the corresponding free boundary
${b}^{*}\in(\underline{b},\hat{b})$ (ensured by Theorem 4.5) and
$J^{*}:[0,\infty)\rightarrow[0,\infty)$ be defined in (4.5). Furthermore, let
$D^{*}:=D^{({{b}^{*}})}$ be the $b^{*}$-threshold dividend strategy (see
Definition 4.2). Then, $v^{*}$ coincides with $J^{*}$ and $D^{*}$ is the
robust dividend strategy of $J^{*}$, i.e., for every $x\in[0,\infty)$,

$v^{*}(x)=J^{*}(x)=\inf_{\theta\in\Theta^{D^{*}}}J^{\theta}(x;D^{*}).$

In addition, $J^{*}(x)=v^{*}(x)$ satisfies

$\displaystyle
v^{*}(x)=\sup_{D\in{\mathcal{}A}^{x}}\inf_{\theta\in\Theta_{D}}\mathbb{E}^{\theta}\bigg{[}\int_{0}^{\tau^{x,D}}e^{-\rho
t}dD_{t}+\frac{1}{2\mathcal{R}}\int_{0}^{\tau^{x,D}}e^{-\rho
t}\theta_{t}^{2}v^{*}(X_{t}^{x,D})dt+e^{-\rho\tau^{x,D}}\xi_{0}\bigg{]}.$

We note that while we have restricted the admissible set ${\mathcal{}A}^{x}$
to dividend policies with continuous paths, it is straightforward to confirm
that our derived $D^{*}$ remains optimal, even when considering a broader
admissible set that permits right continuous left limit processes.

##  5\. Proof of Theorem 4.5: Shooting Method

The proof of Theorem 4.5 adopts the shooting method. This is a method that
allows us to solve boundary value problems by reducing them to initial value
problems. We adapt it to our free-boundary problem. In our case, we set up a
parameter the boundary point $x=b$ such that (4.9) holds true. Our approach
requires some preliminary ODE results that stem from (4.9).

Recall the function $\psi^{+}$ (see Assumption 4.3), we claim that for every
$b>0$ and $\gamma\in\mathbb{R}$ that there is a function $v_{b,\gamma}\in
C^{2}((0,b)\cup(b,\infty))\cap C^{1}(\mathbb{R}_{+})$ (when $\gamma=0$,
$v_{b,0}\in C^{2}((0,\infty))$) solving the following nonlinear ODE

(5.1) $\displaystyle\left\\{\begin{aligned}
&\dfrac{\sigma^{2}(x)}{2}v^{\prime\prime}_{b,\gamma}(x)+\mu(x)v_{b,\gamma}^{\prime}(x)-{\mathcal{}R}\dfrac{\sigma^{2}(x)}{2}\dfrac{\big{(}v_{b,\gamma}^{\prime}(x)\big{)}^{2}}{v_{b,\gamma}(x)}=(\rho+\gamma)v_{b,\gamma}(x)\quad\mbox{on}\;\;{(0,\infty)},\\\
&v_{b,\gamma}^{\prime}(b)=1,\quad
v_{b,\gamma}(b)=\psi^{+}(b).\end{aligned}\right.$

Indeed, for any given $b>0$ and $\gamma\in\mathbb{R}$, one-to-one
correspondence

(5.2) $\displaystyle h_{b,\gamma}:=(v_{b,\gamma})^{1-{\mathcal{}R}}$

enables to consider the following linear ODE (which is equivalent to (5.1))
given by

(5.3) $\displaystyle\left\\{\begin{aligned}
&\dfrac{1}{2}\sigma^{2}(x)h^{\prime\prime}_{b,\gamma}(x)+\mu(x)h_{b,\gamma}^{\prime}(x)=(1-{\mathcal{}R})(\rho+\gamma)h_{b,\gamma}(x)\quad\mbox{on}\;\;[a_{1},a_{2}],\\\
&h_{b,\gamma}^{\prime}(b)=(1-{\mathcal{}R})\big{(}\psi^{+}(b)\big{)}^{-{\mathcal{}R}},\quad
h_{b,\gamma}(b)=\big{(}\psi^{+}(b)\big{)}^{1-{\mathcal{}R}},\end{aligned}\right.$

with an interval $[a_{1},a_{2}]\subset\mathbb{R}$ with $a_{1}<b<a_{2}$. Since
the linear ODE (5.3) admits a unique solution $h_{b,\gamma}\in
C^{2}([a_{1},b)\cup(b,a_{2}])\cap C^{1}([a_{1},a_{2}])$ ($h_{b,0}\in
C^{2}([a_{1},a_{2}])$ when $\gamma=0$) for any
$[a_{1},a_{2}]\subset\mathbb{R}$ with $a_{1}<b<a_{2}$ (The Picard-Lindelöf’s
Existence and Uniqueness Theorem, see e.g., [52, Chapter 13, Section 2,
Theorem 3]), this ensures our claim to hold.

To facilitate the analysis, define $g_{b,\gamma}:\mathbb{R}_{+}\ni
x\rightarrow g_{b,\gamma}(x)\in\mathbb{R}$ for every $b>0$ and
$\gamma\in\mathbb{R}$ by

(5.4) $\displaystyle
g_{b,\gamma}(x):=\frac{v^{\prime}_{b,\gamma}(x)}{v_{b,\gamma}(x)}.$

Since $v_{b,\gamma}\in C^{2}((0,b)\cup(b,\infty))\cap C^{1}({(0,\infty)})$
(when $\gamma=0$, $v_{b,0}\in C^{2}({(0,\infty)})$) is the solution of (5.1)
and satisfies

(5.5) $\displaystyle
v_{b,\gamma}(x)=\psi^{+}(b)\exp\left(\int_{b}^{x}g_{b,\gamma}(u)du\right),$

it follows that $g_{b,\gamma}$ is of $C^{1}((0,b)\cup(b,\infty))\cap
C(\mathbb{R}_{+})$ (when $\gamma=0$, $g_{b,0}\in C^{1}((0,\infty))$) solves
the following ODE

(5.6) $\displaystyle\left\\{\begin{aligned}
&\dfrac{1}{2}\sigma^{2}(x)g_{b,\gamma}^{\prime}(x)+\mu(x)g_{b,\gamma}(x)+\dfrac{(1-{\mathcal{}R})}{2}\sigma^{2}(x)g_{b,\gamma}^{2}(x)=\rho+\gamma\quad\mbox{on}\;\;{(0,\infty)},\\\
&g_{b,\gamma}(b)={1}/{\psi^{+}(b)}.\end{aligned}\right.$

For notational simplicity, we set $v_{b}:=v_{b,0}$, $h_{b}:=h_{b,0}$, and
$g_{b}:=g_{b,0}$ for every $b>0$. Targeting at proving Theorem 4.5, our main
task is to show the following proposition. The proof of this result is based
on a sequence of preliminary lemmas (Lemmas 5.2-5.6), which we will discuss in
the subsequent subsection.

######  Proposition 5.1.

Suppose that Assumption 4.3 holds. For every $b>0$, let $v_{b}\in
C^{2}((0,\infty))$ be the solution of the nonlinear ODE given in (5.1) (when
$\gamma=0$). Then there exists $b^{\ast}\in(\underline{b},\hat{b})$ such that

$v_{b^{*}}(0)=\xi_{0},\quad v^{\prime}_{b^{*}}(x)\geq
1\quad\mbox{on}\;\;(0,b^{*}],$

with $\underline{b},\hat{b}$ appearing in Assumption 4.3.

At this point, we see that $v^{*}(x)=v_{b^{*}}(x)$ to the left of $b^{*}$
satisfies the first line and the third line of (4.9). Setting up $v^{*}(x)=1$
on $(b^{*},\infty)$, it remains to show the second line of (4.9) holds. We
will present this final piece of the proof for Theorem 4.5 here.

###### Proof of Theorem 4.5.

From Proposition 5.1, define $v^{*}:[0,\infty)\rightarrow\mathbb{R}$ by

(5.7) $\displaystyle v^{*}(x):=\left\\{\begin{aligned}
&v_{b^{*}}(x)\quad&&\mbox{if}\;\;x\in[0,b^{*}];\\\
&x-b^{*}+\psi^{+}(b^{*})\quad&&\mbox{if}\;\;x\in(b^{*},\infty).\end{aligned}\right.$

Then, since $v_{b^{*}}$ is in ${\mathcal{}C}^{2}({(0,\infty)})$ and satisfies
(5.1) (when $\gamma=0$), by the definition given in (5.7) and $v^{*}$ is in
${\mathcal{}C}^{2}([0,\infty))$ and satisfies the free boundary problem (4.9)
to the left of $b^{*}$ . For $x\in(b^{*},\overline{b})$, $\psi^{+}(x)-x$ is
decreasing by Assumption 4.3 (ii), hence
$v^{*}(x)=x-b^{*}+\psi^{+}(b^{*})\geq\psi^{+}(x)$. Therefore
$\mathcal{L}v^{*}(x)=-\frac{Q(v^{*}(x);x)}{v^{*}(x)}\leq 0$ by recalling
$Q(\psi;x)$ defined in Assumption 4.3 (ii). If $\overline{b}=\infty$, the
proof ends. Otherwise, for $x\in[\overline{b},\infty)$, we have
$Q(v^{*}(x);x)<0$, hence $\mathcal{L}v^{*}(x)<0$. We conclude that $v^{*}(x)$
satisfies $\mathcal{L}v^{*}\leq 0$ on $(b^{*},\infty)$, so it solves the free
boundary problem (4.9) with free boundary $b^{*}\in(\underline{b},\hat{b})$.
This completes the proof. ∎

The above analysis indicates that most of our efforts should be focused on
selecting a point $b$ that satisfies the boundary condition $v_{b}(0)$.
Concurrently, it is also necessary to demonstrate that the challenging bound
$v_{b}^{\prime}\geq 1$ holds on the interval $[0,b]$.

###  5.1. Preliminary lemmas for shooting method

In what follows, we often make use of the following elementary properties that
are based on [21, Lemma 1].

######  Lemma 5.2.

Let $(a_{1},a_{2})\subseteq\mathbb{R}$ and $a_{3}\in(a_{1},a_{2})$, and let
$f:(a_{1},a_{2})\ni x\rightarrow f(x)\in\mathbb{R}$ be a function of class
$C^{1}((a_{1},a_{2}))$. Then the following hold:

  * (i)

If $f(a_{3})>c$ (resp. $<c$) with some $c\in\mathbb{R}$ and there exist
$y_{1}:=\sup\\{y\in(a_{1},a_{3}):f(y)=c\\}$ and
$y_{2}:=\inf\\{y\in(a_{3},a_{2}):f(y)=c\\}$, then

$f^{\prime}(y_{1})\geq 0\;\;\mbox{(resp.~{}$\leq 0$)},\quad
f^{\prime}(y_{2})\leq 0\;\;\mbox{(resp.~{}$\geq 0$)}.$

  * (ii)

If $f^{\prime}(a_{3})>0$ (resp., $<0$) and there exist
$y_{3}:=\sup\\{y\in(a_{1},a_{3}):f(y)=f(a_{3})\\}$ and
$y_{4}:=\inf\\{y\in(a_{3},a_{2}):f(y)=f(a_{3})\\}$, then

$f^{\prime}(y_{3})\leq 0\;\;\mbox{(resp.~{}$\geq 0$)},\quad
f^{\prime}(y_{4})\leq 0\;\;\mbox{(resp.~{}$\geq 0$)}.$

We start with a perturbation result that enables us to get estimations for
$g_{b}$ via the estimates of $g_{b,\gamma}$.

######  Lemma 5.3.

Suppose that Assumption 4.3 holds. For every $b>0$ and $\gamma\in\mathbb{R}$,
let $v_{b,\gamma}\in C^{2}((0,b)\cup(b,\infty))\cap C^{1}({(0,\infty)})$
($v_{b}=v_{b,0}\in C^{2}({(0,\infty)})$ when $\gamma=0$) be the solution of
(5.1) and $g_{b,\gamma}\in C^{1}((0,b)\cup(b,\infty))\cap C({(0,\infty)})$
($g_{b}=g_{b,0}\in C^{1}({(0,\infty)})$ when $\gamma=0$) be the solution of
(5.6). Then, the followings hold for every $b>0$:

  * (i)

Let $b\in(\underline{b},\overline{b})$ and $\gamma<0$ (with
$\underline{b},\overline{b}$ appearing in Assumption 4.3). Define
$y_{b,\gamma}$ by

$\displaystyle\qquad y_{b,\gamma}:=\left\\{\begin{aligned} &0\quad&&\mbox{if
$~{}g_{b,\gamma}(x)>{1}/{\psi^{+}(x)}$ for every $x\in(0,b)$};\\\
&\sup\\{x\in(0,b):g_{b,\gamma}(x)\leq
1/\psi^{+}(x)\\}\quad&&\mbox{else}.\end{aligned}\right.$

Then $y_{b,\gamma}\in[0,\underline{b}]$, and for every $x\in(y_{b,\gamma},b)$,
$g_{b,\gamma}(x)>1/\psi^{+}(x)$.

  * (ii)

Let $b_{1},b_{2}\in(\underline{b},\overline{b})$ with $b_{1}\leq b_{2}$ and
$\gamma_{1},\gamma_{2}\in(-\infty,0]$ with $\gamma_{1}>\gamma_{2}$. Then the
following holds: for every $x\in(0,b_{1})$,

$g_{b_{1},\gamma_{1}}(x)<g_{b_{2},\gamma_{2}}(x).$

  * (iii)

As $\gamma\rightarrow 0$

$\displaystyle\sup_{x\in[0,b]}|g_{b,\gamma}(x)-g_{b}(x)|=O(\gamma),$

where $O(\cdot)$ denotes the Landau symbol. Moreover, for every $x\in[0,b]$,

$\displaystyle\lim_{\gamma\rightarrow
0}|v_{b,\gamma}(x)-v_{b}(x)|=0,\qquad\lim_{\gamma\rightarrow
0}|v^{\prime}_{b,\gamma}(x)-v^{\prime}_{b}(x)|=0.$

###### Proof.

We start by proving (i). We first note that since $g_{b,\gamma}$ is the
solution of (5.6), the following inequality holds:

(5.8) $\displaystyle
g_{b,\gamma}^{\prime}(b)=\frac{2}{(\sigma(b)\psi^{+}(b))^{2}}Q(\psi^{+}(b))+\frac{2\gamma}{(\sigma(b))^{2}}-\frac{1}{(\psi^{+}(b))^{2}}<-\frac{1}{(\psi^{+}(b))^{2}},$

where we have used that $Q(\psi^{+}(b))=0$ (see Assumption 4.3 (ii)) and
$\gamma<0$.

Define $\phi_{b,\gamma}:\mathcal{X}\rightarrow\mathbb{R}$ by
$\phi_{b,\gamma}(x):=g_{b,\gamma}(x)-1/\psi^{+}(x)$, $x\in\mathcal{X}$. Since
$g_{b,\gamma}(b)={1}/{\psi^{+}(b)}$ and $(\psi^{+})^{\prime}(b)\leq 1$
(because $\psi^{+}(x)-x$ is decreasing on $[\underline{b},\overline{b})$ and
$b$ is in $(\underline{b},\overline{b})$; see Assumption 4.3 (ii)), the
inequality given in (5.8) ensures the following properties to hold:

(5.9)
$\displaystyle\phi_{b,\gamma}(b)=g_{b,\gamma}(b)-\frac{1}{\psi^{+}(b)}=0,\quad\phi_{b,\gamma}^{\prime}(b)=g^{\prime}_{b,\gamma}(b)+\frac{(\psi^{+})^{\prime}(b)}{(\psi^{+}(b))^{2}}<\frac{(\psi^{+})^{\prime}(b)-1}{(\psi^{+}(b))^{2}}\leq
0.$

We claim that $y_{b,\gamma}\leq\underline{b}$. Since the case with
$y_{b,\gamma}=0$ is trivial, we can and do consider only the case with
$y_{b,\gamma}=\sup\\{x\in(0,b):\phi_{b,\gamma}(x)=0\\}$. Arguing by
contradiction, assume that $y_{b,\gamma}>\underline{b}$. Then by the
continuity of $\phi_{b,\gamma}$ (due to that of $g_{b,\gamma}$ and
$\psi^{+}$), we have that
$\phi_{b,\gamma}(y_{b,\gamma})=g_{b,\gamma}(y_{b,\gamma})-{1}/{\psi^{+}(y_{b,\gamma})}=0.$
Furthermore, using the same arguments devoted for the second property given in
(5.9) (along with $y_{b,\gamma}>\underline{b}$), we have
$(\phi_{b,\gamma})^{\prime}(y_{2,\gamma})=g^{\prime}_{b,\gamma}(y_{2,\gamma})+\frac{(\psi^{+})^{\prime}(y_{2,\gamma})}{(\psi^{+}(y_{2,\gamma}))^{2}}<0$,
which contradicts Lemma 5.2 (ii). Hence, the claim holds true.

Furthermore, from the definition of $y_{b,\gamma}$ and the fact that
$(\phi_{b,\gamma})^{\prime}(b)=(g_{b,\gamma}-1/\psi^{+})^{\prime}(b)<0$ (see
(5.9)), this completes the proof.

Now let us prove (ii). To that end, let $y_{b_{2},\gamma_{2}}$ be defined as
Lemma 5.3 (i) (by assigning $b=b_{2}$ and $\gamma=\gamma_{2}$). Define $\Delta
g:[0,b]\ni x\rightarrow\Delta g(x)\in\mathbb{R}$ by $\Delta
g(x):=g_{b_{1},\gamma_{1}}(x)-g_{b_{2},\gamma_{2}}(x)$ for $x\in[0,b_{1}]$.
Since $g_{b_{i},\gamma_{i}}$, $i=1,2$, is the solution of (5.6) (when
$b=b_{i}$, $\gamma=\gamma_{i}$), the following holds: for every
$x\in[0,b_{1}]$,

(5.10) $\displaystyle\begin{aligned} \dfrac{1}{2}\sigma^{2}(x)(\Delta
g)^{\prime}(x)+\Delta
g(x)\left(\mu(x)+(1-{\mathcal{}R})\sigma^{2}(x)\left(\dfrac{\Delta
g(x)}{2}+g_{b_{2},\gamma_{2}}(x)\right)\right)=\gamma_{1}-\gamma_{2},\end{aligned}$

and $\Delta g(b_{1})=1/{\psi^{+}(b_{1})}-g_{b_{2},\gamma_{2}}(b_{1})<0$ which
follows from Lemma 5.3 (i) with fact that
$b_{1}\in(y_{b_{2},\gamma_{2}},b_{2})$ and
$g_{b_{1},\gamma_{1}}(b_{1})=1/{\psi^{+}(b_{1})}$.

Here, we claim that $\Delta g(x)<0$ on $(0,b_{1})$. Arguing by contradiction,
assume it does not hold; then there exists $y_{3}:=\sup\\{x\in(0,b_{1}):\Delta
g(x)=0\\}$. Since $\Delta g(y_{3})=0$ (by its continuity) and
$y_{3}\in[0,b_{1}]$, assigning $x=y_{3}$ into (5.10) ensures that $\Delta
g^{\prime}(y_{3})=2(\gamma_{1}-\gamma_{2})/\sigma^{2}(y_{3})>0$. This together
with the fact that $\Delta g(b_{1})<0$ contradicts Lemma 5.2 (i). This
completes the proof.

The proof for Lemma 5.3 (iii) is similar to the one given in [21, Lemma 2], so
we omit the details here. ∎

######  Lemma 5.4.

Suppose that Assumption 4.3 holds. For every $b>0$, let $v_{b}\in
C^{2}({(0,\infty)})$ be the solution of (5.1) (when $\gamma=0$) and $g_{b}\in
C^{1}({(0,\infty)})$ be the solution of (5.6) (when $\gamma=0$). Then the
following hold:

  * (i)

Let $b\in(\underline{b},\overline{b})$. Then there exists
$y_{b}^{*}\in[0,\underline{b}]$ satisfying that for every $x\in[y_{b}^{*},b]$,

$g_{b}(x)\geq 1/\psi^{+}(x),$

with $\underline{b}$ appearing in Assumption 4.3.

  * (ii)

Let $b_{1},b_{2}\in(\underline{b},\overline{b})$ with $b_{1}<b_{2}$. Then the
following holds: for every $x\in(0,b_{1}]$,

$g_{b_{1}}(x)\leq g_{b_{2}}(x).$

  * (iii)

Let $b\in(\underline{b},\overline{b})$ and $\gamma\leq 0$. As
$\varepsilon\rightarrow 0$

$\displaystyle\sup_{x\in[0,b]}|g_{b+\varepsilon,\gamma}(x)-g_{b,\gamma}(x)|=O(\varepsilon).$

Furthermore, the following also hold: for every $x\in[0,b]$,

(5.11) $\displaystyle\lim_{\varepsilon\rightarrow
0}|v_{b+\varepsilon,\gamma}(x)-v_{b}(x)|=0,\qquad\lim_{\varepsilon\rightarrow
0}|v_{b+\varepsilon,\gamma}^{\prime}(x)-v_{b,\gamma}^{\prime}(x)|=0.$

###### Proof.

We start by proving (i). Let $y_{b,\gamma}\in[0,\underline{b}]$ for every
$\gamma<0$ be defined as in Lemma 5.3 (i). Then $y_{b,\gamma}$ increases in
$\gamma<0$. Indeed, Lemma 5.3 (ii) ensures that for any
$\gamma_{1},\gamma_{2}\in(-\infty,0)$ such that $\gamma_{1}>\gamma_{2}$, the
following holds that $g_{b,\gamma_{1}}(x)<g_{b,\gamma_{2}}(x)$ for every
$x\in(0,b)$. Combining this with the definition of $y_{b,\gamma_{i}}$,
$i=1,2$, (see Lemma 5.3 (i)), we have $y_{b,\gamma_{1}}>y_{b,\gamma_{2}}$.
This ensures that the (increasing) monotonicity of $y_{b,\gamma}$ with
resepect to $\gamma<0$.

Consider an increasing sequence
$(\gamma_{n})_{n\in\mathbb{N}}\subseteq(-\infty,0)$ such that
$\gamma_{n}\uparrow 0$ as $n\rightarrow\infty$. Then the monotonicity of
$(y_{b,\gamma_{n}})_{n\in\mathbb{N}}$ together with uniformly boundedness
within $[0,\underline{b}]$ ensures that there exists
$y_{b}^{*}:=\lim_{n\rightarrow\infty}y_{b,\gamma_{n}}$ satisfying
$y_{b}^{*}\in[0,\underline{b}]$ and $y_{b}^{*}\geq y_{b,\gamma_{n}}$ for every
$n\in\mathbb{N}$. Combined with Lemma 5.3 (i), this ensures that for every
$n\in\mathbb{N}$ and every $x\in[y_{b}^{*},b)$, $g_{b,\gamma_{n}}(x)\geq
1/\psi^{+}(x).$ Therefore combined with Lemma 5.3 (iii) and letting
$n\rightarrow\infty$, this completes the proof.

Now let us prove (ii). Lemma 5.3 (ii) ensures that for every $\gamma<0$ and
every $x\in(0,b_{1}]$, $g_{b_{1},0}(x)=g_{b_{1}}(x)<g_{b_{2},\gamma}(x).$
Combined with Lemma 5.3 (iii) and letting $\gamma\uparrow 0$, this completes
the proof.

Lastly, we prove (iii). We establish proof for the case where $\gamma=0$, as
the analysis can be readily extended to accommodate the case where $\gamma\neq
0$. We start with proving the first convergence therein. Let
$\varepsilon\in[0,1\wedge(\overline{b}-b))$. Since $g_{b+\varepsilon}\in
C^{2}({(0,\infty)})$ (resp. $g_{b}\in C^{2}({(0,\infty)})$) is the solution of
(5.6) (when $\gamma=0$ with $b+\varepsilon$ (resp. $b$)), the following holds:
for every $x\in{(0,\infty)}$,

(5.12) $\displaystyle\begin{aligned}
&g_{b+\varepsilon}(x)-g_{b}(x)-\big{(}g_{b+\varepsilon}(b)-g_{b}(b)\big{)}=-\int_{x}^{b}\big{(}g^{\prime}_{b+\varepsilon}(y)-g^{\prime}_{b}(y)\big{)}dy\\\
&\qquad=\int_{x}^{b}\bigg{(}(1-{\mathcal{}R})\sigma^{2}(x)\big{(}g_{b+\varepsilon}(y)+g_{b}(y)\big{)}+\frac{2\mu(x)}{\sigma^{2}(x)}\bigg{)}\big{(}g_{b+\varepsilon}(y)-g_{b}(y)\big{)}dy.\end{aligned}$

Lemma 5.4 (ii) ensures that $g_{b}(x)\leq g_{b+\varepsilon}(x)\leq
g_{b+1}(x)\leq C_{0}$ for every $x\in[0,b]$, where $C_{0}>0$ is a constant
(that depends on $1\wedge(\overline{b}-b)$ but not on $\varepsilon$).
Furthermore, $\sigma$ and $\mu$ are uniformly bounded on $[0,b]$ (see
Assumption 4.3). Hence, combined with (5.12) these properties ensure that
there is a constant $C_{1}>0$ (that depends on $C_{0}$ but not on
$\varepsilon$) such that for every $x\in[0,b)$, $0\leq
g_{b+\varepsilon}(x)-g_{b}(x)\leq
g_{b+\varepsilon}(b)-g_{b}(b)+C_{1}\int_{b}^{x}\big{(}g_{b+\varepsilon}(y)-g_{b}(y)\big{)}dy.$
An application of Grönwall’s inequality with this guarantees that there is a
constant $C_{2}>0$ (that depends on $C_{1},C_{0}$ but not on $\varepsilon$)
satisfying

(5.13) $\displaystyle
0\leq\sup_{x\in[0,b]}\Big{(}g_{b+\varepsilon}(x)-g_{b}(x)\Big{)}\leq
C_{2}\big{(}g_{b+\varepsilon}(b)-g_{b}(b)\big{)}.$

Furthermore, we claim that there is a constant $C_{3}$ (that depends on
$1\wedge(\overline{b}-b)$ but not on $\varepsilon$) such that $0\leq
g_{b+\varepsilon}(b)-g_{b}(b)\leq C_{3}\cdot\varepsilon.$ Indeed, we use the
similar arguments devoted for (5.12) with the fact that
$g_{b+\varepsilon}(b+\varepsilon)=1/\psi^{+}(b+\varepsilon)$ and
$g_{b}(b)=1/\psi^{+}(b)$ to have that

$\displaystyle\begin{aligned} &\left|g_{b+\varepsilon}(b)-g_{b}(b)\right|\\\
&\leq\bigg{|}\frac{1}{\psi^{+}(b+\varepsilon)}-\frac{1}{\psi^{+}(b)}\bigg{|}+\bigg{|}\int_{b}^{b+\varepsilon}\bigg{(}\frac{2}{\sigma^{2}(x)}\big{(}\mu(x)g_{b+\varepsilon}(x)-\rho\big{)}+(1-{\mathcal{}R})g^{2}_{b+\varepsilon}(x)\bigg{)}dx\bigg{|}\\\
&\leq\varepsilon\cdot\bigg{(}\sup_{x\in[b,b+\varepsilon]}\left|\frac{(\psi^{+})^{\prime}(x)}{(\psi^{+})^{2}(x)}\right|+\sup_{x\in[b,b+\varepsilon]}\bigg{|}\frac{2\mu(x)+\rho}{\sigma^{2}(x)}\bigg{|}\sup_{x\in[b,b+\varepsilon]}|g_{b+\varepsilon}(x)|\vee
1+\sup_{x\in[b,b+\varepsilon]}|g_{b+\varepsilon}(x)|^{2}\bigg{)}\\\ &\leq
C_{3}\cdot\varepsilon,\end{aligned}$

where we have used that $\mu,\sigma\in C^{1}({(0,\infty)})$ and hence
$\psi^{+}\in C^{1}([0,\overline{b}])$ (see Assumption 4.3 (ii), (iii)) and
that $1-{\mathcal{}R}\in(0,1)$. Combined with (5.13), this ensures that

(5.14) $\displaystyle\lim_{\varepsilon\downarrow
0}\sup_{x\in[0,b]}\Big{(}g_{b+\varepsilon}(x)-g_{b}(x)\Big{)}=0.$

(5.11) follows by the transformations
$v_{b}(x)=\psi^{+}(b)\exp(\int_{b}^{x}g_{b}(u)du)$ and
$v_{b}^{\prime}(x)=v_{b}(x)g_{b}(x)$.

The proof in the case $\varepsilon\uparrow 0$ is similar therefore we omit it.
∎

######  Lemma 5.5.

Suppose that Assumption 4.3 holds. Let $b\in(0,\underline{b}]$ and $v_{b}\in
C^{2}({(0,\infty)})$ be the solution of (5.1) (when $\gamma=0$), with
$\underline{b}$ appearing in Assumption 4.3. Then for every $x\in(0,b]$,

$v_{b}^{\prime}(x)\leq 1.$

###### Proof.

Let $\gamma>0$ and $v_{b,\gamma}$ be the solution of (5.1). Since
$v_{b,\gamma}^{\prime}(b)=1$ and $v_{b,\gamma}(b)=\psi^{+}(b)$, the nonlinear
ODE given in (5.1) ensures that
$v^{\prime\prime}_{b,\gamma}(b)=2\gamma\psi^{+}(b)/\sigma^{2}(b)>0$.

We claim that for every $x\in(0,b)$, $v^{\prime}_{b,\gamma}(x)<1$. Arguing by
contradiction, we assume that it does not hold. Then together with
$v_{b,\gamma}^{\prime}(b)=1$, the following supremum is attained:
$x_{1}:=\sup\\{x\in(0,b):v^{\prime}_{b,\gamma}(x)=1\\}$. By definition of
this, it is clear that for every $x\in[x_{1},b]$,
$v^{\prime}_{b,\gamma}(x)\leq 1,$ which ensures that
$v_{b,\gamma}(x)-v_{b,\gamma}(b)\geq x-b$ for every $x\in[x_{1},b]$.

Furthermore, from Assumption 4.3 (ii) (with $b<\underline{b}$) and
$v_{b,\gamma}=\psi^{+}(b)$, it follows that for every $x\in[x_{1},b]$,
$v_{b,\gamma}(x_{1})\geq\psi^{+}(b)-(b-x_{1})\geq\psi^{+}(x_{1})$. This
implies that $Q(v_{b,\gamma}(x_{1}))\geq Q(\psi^{+}(x_{1}))=0$ (see Assumption
4.3 (ii)). From the nonlinear ODE (5.1), it follows that

$\displaystyle\frac{1}{2}\sigma^{2}(x_{1})v^{\prime\prime}_{b,\gamma}(x_{1})=Q(v_{b,\gamma}(x_{1}))\frac{1}{v_{b,\gamma}(x_{1})}+\gamma
v_{b,\gamma}(x_{1})\geq\gamma\psi^{+}(x_{1})>0.$

Combining this with the fact that $v^{\prime\prime}_{b,\gamma}(b)>0$
contradicts Lemma 5.2 (ii). Hence we have shown that the claim holds.

The proof is completed by combining the claim with Lemma 5.3 (iii) and letting
$\gamma\downarrow 0$. ∎

######  Lemma 5.6.

Suppose that Assumption 4.3 holds. For every $b>0$ and $\gamma\in\mathbb{R}$,
let $v_{b,\gamma}\in C^{2}((0,b)\cup(b,\infty))\cap C^{1}({(0,\infty)})$
($v_{b}=v_{b,0}\in C^{2}({(0,\infty)})$ when $\gamma=0$) be the solution of
(5.1). Then the following hold:

  * (i)

Let $b\leq\hat{b}$ (with $\underline{b},\hat{b}$ appearing in Assumption 4.3)
and $\gamma\leq 0$. If $v_{b,\gamma}(0)\geq\xi_{0}$, then for every
$x\in[0,b]$,

$v_{b,\gamma}(x)\geq x+\xi_{0}.$

  * (ii)

Let $\gamma<0$, then there exists $b^{*}(\gamma)$ defined by

$\displaystyle
b^{*}(\gamma):=\inf\\{b\in(\underline{b},\hat{b}):v_{b,\gamma}(0)=\xi_{0}\\}.$

Furthermore, $b^{*}(\gamma)$ increases in $\gamma<0$, which hence ensures that
$b^{*}:=\lim_{\gamma\uparrow 0}b^{*}(\gamma)$ is well-defined, and the limit
$b^{*}$ satisfies that

$b^{*}\in(\underline{b},\hat{b}],\qquad v_{b^{*}}(0)=\xi_{0}.$

###### Proof.

We start by proving (i). Arguing by contradiction, we assume that it does not
hold. Recalling the one-to-one correspondence
$h_{b,\gamma}=(v_{b,\gamma})^{1-{\mathcal{}R}}$ (see (5.2)), we obtain that

$\max_{x\in[0,b]}\left\\{(x+\xi_{0})^{1-{\mathcal{}R}}-h_{b,\gamma}(x)\right\\}>0.$

We note that from Assumption 4.3 (iii) (with $b\leq\hat{b}$) and the condition
$v_{b,\gamma}(0)\geq\xi_{0}$, it follows that
$h_{b,\gamma}(x)\geq(x+\xi_{0})^{1-{\mathcal{}R}}$ when $x=0$ and $x=b$. Hence
this implies that $(x+\xi_{0})^{1-\epsilon}-h_{b,\gamma}(x)$ takes its maximum
at $x_{0}\in(0,b)$.

Furthermore, since $h_{b,\gamma}\in C^{2}((0,b)\cup(b,\infty))$ and the
maximality of $x_{0}$, the following hold:

(5.15)
$\displaystyle-{\mathcal{}R}(1-{\mathcal{}R})(x_{0}+\xi)^{-1-{\mathcal{}R}}-h_{b,\gamma}^{\prime\prime}(x_{0})\leq
0,\quad(1-{\mathcal{}R})(x_{0}+\xi)^{-{\mathcal{}R}}-h_{b,\gamma}^{\prime}(x_{0})=0.$

From the linear ODE given in (5.3), we get to a contradiction

$\displaystyle 0=$
$\displaystyle\dfrac{1}{2}\sigma^{2}(x_{0})h_{b,\gamma}^{\prime\prime}(x_{0})+\mu(x_{0})h_{b,\gamma}^{\prime}(x_{0})-(1-{\mathcal{}R})(\rho+\gamma)h_{b,\gamma}(x_{0})$
$\displaystyle>$
$\displaystyle(1-{\mathcal{}R})(x_{0}+\xi_{0})^{-1-{\mathcal{}R}}\bigg{(}-\frac{{\mathcal{}R}}{2}\sigma^{2}(x_{0})+\mu(x_{0})(x_{0}+\xi)-(\rho+\gamma)(x_{0}+\xi)^{2}\bigg{)}\geq
0,$

where the first inequality follows from (5.15) and the fact that
$h_{b,\gamma}(x_{0})<(x_{0}+\xi)^{1-{\mathcal{}R}}$, and the second inequality
follows from Assumption 4.3 (ii) and $\gamma\leq 0$. This completes the proof.

We now prove (ii). That will be achieved in three steps.

Step 1. We claim that $v_{\underline{b},\gamma}(0)>\xi_{0}$ and
$v_{\hat{b},\gamma}(0)<\xi_{0}$.

Lemma 5.5 (when $b=\underline{b}$) ensures that
$v_{\underline{b},\gamma}(0)\geq
v_{\underline{b},\gamma}(\underline{b})-(\underline{b}-x)$ for every
$x\in[0,\underline{b}]$. In particular,
$v_{\underline{b},\gamma}(0)>v_{\underline{b},\gamma}(\underline{b})-\underline{b}$
because it cannot be the case that $v_{\underline{b},\gamma}^{\prime}(x)\equiv
0$ for $x\in[0,\underline{b}]$. Furthermore, from the fact that
$v_{\underline{b},\gamma}(\underline{b})=\psi^{+}(\underline{b})$ and
Assumption 4.3 (ii) (iii) (with $\underline{b}<\hat{b}$), it follows that
$v_{\underline{b},\gamma}(0)>\psi^{+}(\underline{b})-\underline{b}>\psi^{+}(\hat{b})-\hat{b}=\xi_{0}$.

It remains to show $v_{\hat{b},\gamma}(0)<\xi_{0}$. Arguing by contradiction,
we assume that it does not hold. From Lemma 5.6 (i), it follows that
$v_{\hat{b},\gamma}(x)\geq x+\xi_{0}$ for every $x\in[0,\hat{b}]$. Since
$v_{\hat{b},\gamma}$ is the solution of (5.1), assigning $b=\hat{b}$ into
(5.1) ensures that
$v^{\prime\prime}_{\hat{b},\gamma}(\hat{b})=2\gamma\sigma^{2}(\hat{b})\psi^{+}(\hat{b})<0$.

Here we claim that $v_{\hat{b},\gamma}^{\prime}(x)>1$ for every
$x\in(\underline{b},\hat{b})$. Arguing by contradiction, we assume that it
does not hold. Then from the fact that
$v_{\hat{b},\gamma}^{\prime}(\hat{b})=1$ and
$v_{\hat{b},\gamma}^{\prime\prime}(\hat{b})<0$, the following supremum is
attained:
$x_{1}:=\sup\\{x\in(\underline{b},\hat{b}):v^{\prime}_{\hat{b},\gamma}(x)=1\\}$.
Furthermore, since $v_{\hat{b},\gamma}^{\prime}(x_{1})=1$ (noting that
$v_{\hat{b},\gamma}\in C({(0,\infty)})$), we have
$v_{\hat{b},\gamma}^{\prime}(x)>1$ on $(x_{1},\hat{b})$. Combined with the
fact that $\psi^{+}(x)-x$ is decreasing on $(x_{1},\hat{b})$ (see Assumption
4.3 (ii) and $\hat{b}\in[\underline{b},\overline{b}]$) and
$v_{\hat{b},\gamma}(\hat{b})=\psi^{+}(\hat{b})$, this guarantees that

(5.16) $\displaystyle
v_{\hat{b},\gamma}(x_{1})\leq\psi^{+}(\hat{b})-(\hat{b}-x_{1})\leq\psi^{+}(x_{1}).$

On the other hand, Lemma 5.6 (i) (together with our assumption that
$v_{\hat{b},\gamma}(0)\geq\xi_{0}$) ensures that
$v_{\hat{b},\gamma}(x_{1})\geq x_{1}+\xi_{0}$. Hence, combined with (5.16),
this ensures that

$\displaystyle\frac{1}{2}\sigma^{2}(x_{1})v^{\prime\prime}_{\hat{b},\gamma}(x_{1})=(\rho+\gamma)v_{\hat{b},\gamma}(x_{1})-\mu(x_{1})+\dfrac{{\mathcal{}R}}{2}\dfrac{\sigma^{2}(x_{1})}{v_{\hat{b},\gamma}(x_{1})}\leq\gamma(x_{1}+\xi_{0})<0.$

This together with the fact that
$v^{\prime\prime}_{\hat{b},\gamma}(\hat{b})<0$ contradicts Lemma 5.2 (ii). We
hence have shown the claim that $v_{\hat{b},\gamma}^{\prime}>1$ on
$(\underline{b},\hat{b})$.

From the claim, the fact that $v_{\hat{b},\gamma}(\hat{b})=\psi^{+}(\hat{b})$,
and Assumption 4.3 (ii) (with $\hat{b}\in[\underline{b},\overline{b}]$), it
follows that
$v_{\hat{b},\gamma}(\underline{b})<\psi^{+}(\hat{b})-(\hat{b}-\underline{b})=\xi_{0}+\underline{b}$.
This contradicts with Lemma 5.6 (i). Therefore, we conclude that
$v_{\hat{b},\gamma}(0)<\xi_{0}$.

Step 2. We claim that there exists $b^{*}(\gamma)$ defined by
$b^{*}(\gamma):=\inf\\{b\in(\underline{b},\hat{b}):v_{b,\gamma}(0)=\xi_{0}\\}$
and that the following holds: $v_{b,\gamma}(0)<\xi_{0}$ for every
$b\in(b^{*}(\gamma),\hat{b})$.

By Lemma 5.4 (iii), we have the continuity of $b\mapsto v_{b,\gamma}(0)$ with
fixed $\gamma\leq 0$. There
$b^{*}(\gamma):=\inf\\{b\in(\underline{b},\hat{b}):v_{b,\gamma}(0)=\xi_{0}\\}$
is well-defined.

Now we claim that $v_{b,\gamma}(0)<\xi_{0}$ for every
$b\in(b^{*}(\gamma),\hat{b})$. From Lemma 5.4 (i) together with
$b^{*}(\gamma)>\underline{b}$, it follows that $g_{b,\gamma}(x)\geq
1/\psi^{+}(x)$ for every $x\in[b^{*}(\gamma),b]$. Combining this with the fact
that $(\psi^{+})^{\prime}(x)<1$ for every $x\in[b^{*}(\gamma),b]$ (see
Assumption 4.3 (ii)) ensures that
$g_{b,\gamma}(x)>(\psi^{+})^{\prime}(x)\psi^{+}(x)$ for every
$x\in[b^{*}(\gamma),b]$.

Furthermore, since $b>b^{*}(\gamma)$, Lemma 5.4 (ii) ensures that
$g_{b,\gamma}(x)>g_{b^{*}(\gamma),\gamma}(x)$ for $x\in(0,b^{*}(\gamma)]$.
Hence we conclude that

$\displaystyle
v_{b,\gamma}(0)=\psi^{+}(b)\exp\left({\int_{b}^{b^{*}(\gamma)}g_{b,\gamma}(y)dy+\int_{b^{*}(\gamma)}^{0}g_{b,\gamma}(y)dy}\right)$
$\displaystyle<\psi^{+}(b^{*}(\gamma))\exp\left(\int_{b^{*}(\gamma)}^{0}g_{b^{*}(\gamma),\gamma}(y)dy\right)=v_{b^{*}(\gamma),\gamma}(0)=\xi_{0}.$

Step 3. We claim that $b^{*}(\gamma)$ increases in $\gamma<0$, which ensures
that $b^{*}:=\lim_{\gamma\uparrow 0}b^{*}(\gamma)$ is well-defined by monotone
convergence. Furthermore, we also claim that $b^{*}\in(\underline{b},\hat{b})$
and $v_{b^{*}}(0)=\xi_{0}.$

Let $b\in(\underline{b},\overline{b})$ and
$\gamma_{1},\gamma_{2}\in(-\infty,0]$ with $\gamma_{1}\geq\gamma_{2}$. Then
Lemma 5.3 (ii) ensures that $g_{b,\gamma_{1}}(x)<g_{b,\gamma_{2}}(x)$ for
every $[0,b]$. From this and the relationship given in (5.5), it follows that
$v_{b,\gamma_{1}}(x)\geq v_{b,\gamma_{2}}(x)$ on $[0,b]$. In particular, since
it cannot be the case that $g_{b,\gamma_{1}}\equiv v_{b,\gamma_{2}}$ for
$x\in[0,b]$, it follows that $v_{b,\gamma_{1}}(0)>v_{b,\gamma_{2}}(0)$.

From this and the fact that
$b^{*}(\gamma_{1})=\inf\\{b\in(\underline{b},\hat{b}):v_{b,\gamma_{1}}(0)=\xi_{0}\\}$,
it follows that
$\xi_{0}=v_{b^{*}(\gamma_{1}),\gamma_{1}}(0)>v_{b^{*}(\gamma_{1}),\gamma_{2}}(0)$.
Furthermore, Step 2. ensures that $v_{b,\gamma_{2}}(0)<\xi_{0}$ for every
$b\in(b^{*}(\gamma_{2}),\hat{b})$. We hence have
$b^{*}(\gamma_{1})>b^{*}(\gamma_{2})$.

From the monotonicity of $b^{*}(\gamma)$ and its uniform boundedness, we can
apply the monotone convergence theorem to have the existence of
$\lim_{\gamma\uparrow 0}b^{*}(\gamma)=:b^{*}$.

It remains to show that $v_{b^{*}}(0)=\xi_{0}$ and
$b^{*}\in(\underline{b},\hat{b})$. To that end, consider an increasing
sequence $(\gamma_{n})_{n\in\mathbb{N}}$ such that $\gamma_{n}\in(-\infty,0)$
for every $n\in\mathbb{N}$ and $\lim_{n\rightarrow\infty}\gamma_{n}=0$.

Since $\lim_{n\rightarrow\infty}b^{*}(\gamma_{n})=b^{*}$, Lemma 5.4 (iii)
ensures that for any $\varepsilon>0$, there exists
$N_{\varepsilon}\in\mathbb{N}$ satisfying that for every $n\geq
N_{\varepsilon}$, $|v_{b^{*}}(0)-v_{b^{*}(\gamma_{n})}(0)|<\varepsilon.$
Furthermore Lemma 5.3 (iii) ensures that there exists
$\widetilde{N}_{\varepsilon}\in\mathbb{N}$ satisfying that for every
$n\geq\widetilde{N}_{\varepsilon}$,
$|v_{b^{*}(\gamma_{n}),\gamma_{n}}(0)-v_{b^{*}(\gamma_{n})}(0)|<\varepsilon.$
Since $v_{b^{*}(\gamma_{n}),\gamma_{n}}(0)=\xi_{0}$ for every $n\in\mathbb{N}$
(see Step 2.), we hence obtain that for every
$n\geq\widetilde{N}_{\varepsilon}\vee N_{\varepsilon}$,

$\displaystyle|v_{b^{*}}(0)-\xi_{0}|\leq|v_{b^{*}}(0)-v_{b^{*}(\gamma_{n})}(0)|+|v_{b^{*}(\gamma_{n}),\gamma_{n}}(0)-v_{b^{*}(\gamma_{n})}(0)|<2\varepsilon.$

By letting $\varepsilon\downarrow 0$, we show $v_{b^{*}}(0)=\xi_{0}$. Finally,
we know that $\underline{b}<b^{*}(\gamma)<\hat{b}$ for every $\gamma<0$ by
Step 1, therefore $\underline{b}\leq\lim_{\gamma\uparrow
0}b^{*}(\gamma)=b^{*}\leq\hat{b}$. The strict inequality $\underline{b}<b^{*}$
follows by the increasing monotonicity of $b^{*}(\gamma)$.

This completes the proof. ∎

###  5.2. Proof of Proposition 5.1

###### Proof of Proposition 5.1.

Given the existence of $b^{*}\in(\underline{b},\hat{b}]$ satisfying
$v_{b^{*}}(0)=\xi_{0}$ by Lemma 5.6 (ii), we first prove that
$v_{b^{*}}^{\prime}(x)\geq 1$ for every $x\in[0,b^{*}]$.

From Lemma 5.6 (ii), define $b^{\gamma}$ for every $\gamma<0$ by

(5.17) $\displaystyle
b^{\gamma}:=b^{*}(\gamma)=\inf\\{b\in(\underline{b},\hat{b}):v_{b,\gamma}(0)=\xi_{0}\\}.$

Furthermore, denote by $v_{b^{\gamma},\gamma}\in
C^{2}((l,b^{\gamma})\cup(b^{\gamma},\infty))\cap C^{1}({(0,\infty)})$ the
solution of (5.1) (when $b=b^{\gamma}$).

The proof will be achieved in two steps. Step 1. We star with perturbation
analysis with a fixed $\gamma\in(-\rho,0)$. We aim to show that

(5.18) $\displaystyle v_{b^{\gamma},\gamma}^{\prime}(b^{\gamma})(x)\geq
1,\forall x\in(0,b^{\gamma}].$

From the boundary conditions $v_{b^{\gamma},\gamma}^{\prime}(b^{\gamma})=1$
and $v_{b^{\gamma},\gamma}(b^{\gamma})=\psi^{+}(b^{\gamma})$, the $C^{2}$
property of $v_{b^{\gamma},\gamma}$ and nonlinear ODE (5.1) gives
$v^{\prime\prime}_{b^{\gamma},\gamma}(b^{\gamma})=2\gamma\sigma^{2}(b^{\gamma})\psi^{+}(b^{\gamma})<0.$
Argue by contradiction, assume (5.18) doesn’t hold. Then combined with the
fact that $v_{b^{\gamma},\gamma}^{\prime}(b^{\gamma})=1$, we have the
existence of

$x_{1}:=\sup\\{x\in(0,b^{\gamma}):v^{\prime}_{b^{\gamma},\gamma}(x)=1\\}.$

In the following, we prove that the existence of $x_{1}$ leads to a
contradiction, which proves that (5.18) holds.

First, we consider the case $x_{1}\in(\underline{b},{b}^{\gamma})$. Then since
$v_{b^{\gamma},\gamma}^{\prime}(x_{1})=1$ (noting that
$v_{b^{\gamma},\gamma}\in C^{1}({(0,\infty)})$), it follows that
$v_{b^{\gamma},\gamma}^{\prime}(x)>1$ for every $x\in(x_{1},b^{\gamma})$.
Furthermore, since $\psi^{+}(x)-x$ decreases on
$[x_{1},b^{\gamma}]\subset[\underline{b},\hat{b}]$ (see Assumption 4.3 (ii)
and $x_{1}\in(\underline{b},{b}^{\gamma})$) and
$v_{b^{\gamma},\gamma}(b^{\gamma})=\psi^{+}(b^{\gamma})$, we hence have that

(5.19) $\displaystyle
v_{b^{\gamma},\gamma}(x_{1})\leq\psi^{+}(b^{\gamma})-(b^{\gamma}-x_{1})\leq\psi^{+}(x_{1}).$

Furthermore, since $x_{1}\in(\underline{b},\hat{b})$ and
$v_{b^{\gamma},\gamma}(0)=\xi_{0}$ (by definition of $b^{\gamma}$), Lemma 5.6
(i) ensures that $v_{b^{\gamma},\gamma}(x_{1})\geq
x_{1}+\xi_{0}\geq\psi^{-}(x_{1})$, so that
$Q(v_{b^{\gamma},\gamma}(x_{1});x_{1})\leq 0$ by recalling that
$Q(\cdot;x_{1})$ defined in Assumption 4.3 (ii). This implies that

$\displaystyle
v^{\prime\prime}_{b^{\gamma},\gamma}(x_{1})=\frac{2}{\sigma^{2}(x_{1})}\left(\gamma
v_{b^{\gamma},\gamma}(x_{1})+\frac{1}{v_{b^{\gamma},\gamma}(x_{1})}Q(v_{b^{\gamma},\gamma}(x_{1});x_{1})\right)\leq\frac{2}{\sigma^{2}(x_{1})}\gamma(x_{1}+\xi_{0})<0.$

This together with the fact that
$v^{\prime\prime}_{b^{\gamma},\gamma}(b^{\gamma})<0$ contradicts Lemma 5.2
(ii).

Now it remains to check the other case when $x_{1}\in(0,\underline{b}]$. Here
we will also verify a contradiction. Recalling
$x_{1}=\sup\\{x\in(0,b^{\gamma}):v^{\prime}_{b^{\gamma},\gamma}(x)=1\\}$ and
the fact that $v_{b^{\gamma},\gamma}^{\prime}(b^{\gamma})=1$ and
$v_{b^{\gamma},\gamma}^{\prime\prime}(b^{\gamma})<0$, we employ Lemma 5.2 (ii)
to obtain that $v^{\prime\prime}_{b^{\gamma},\gamma}(x_{1})\geq 0$, which can
be rewritten by

(5.20) $\displaystyle
v^{\prime\prime}_{b^{\gamma},\gamma}(x_{1})=\frac{2}{\sigma^{2}(x_{1})}\frac{1}{v_{b^{\gamma},\gamma}(x_{1})}\widetilde{Q}_{\gamma}(v_{b^{\gamma},\gamma}(x_{1});x_{1})\geq
0,$

where
$\widetilde{Q}_{\gamma}(\psi;x_{1}):=(\rho+\gamma)\psi^{2}-\mu(x_{1})\psi+\frac{\mathcal{}R}{2}\sigma^{2}(x_{1})$
(which is well-defined by Assumption 4.3 (ii) with the fact that
$\rho>\rho+\gamma>0$) and we have used the nonlinear ODE of
$v_{b^{\gamma},\gamma}$ given in (5.1).

Lemma 5.6 (i) together with Assumption 4.3 (iii) (and
$x_{1}\in(0,\underline{b}]\subset[0,\hat{b}]$) ensures that
$v_{b^{\gamma},\gamma}(x_{1})\geq x_{1}+\xi_{0}\geq\psi^{-}(x_{1})$.
Furthermore, recalling that $\gamma$ satisfies $\rho>\rho+\gamma>0$, we have

(5.21) $\displaystyle
v_{b^{\gamma},\gamma}(x_{1})\geq\frac{\mu(x_{1})-\sqrt{\mu^{2}(x_{1})-2{\mathcal{}R}\rho\sigma^{2}(x_{1})}}{2\rho}>\frac{\mu(x_{1})-\sqrt{\mu^{2}(x_{1})-2{\mathcal{}R}(\rho+\gamma)\sigma^{2}(x_{1})}}{2(\rho+\gamma)}:=\widetilde{\psi^{-}}(x),$

where we note that the second term equals $\psi^{-}(x_{1})$ and that the last
term $\widetilde{\psi^{-}}(x_{1})$ is the smaller one among two roots of the
quadratic function $\widetilde{Q}_{\gamma}(\cdot;x_{1})$.

Since $\widetilde{Q}_{\gamma}(v_{b^{\gamma},\gamma}(x_{1});x_{1})\geq 0$ (see
(5.20)), from (5.21) it follows that

(5.22) $\displaystyle
v_{b^{\gamma},\gamma}(x_{1})\geq\frac{\mu(x_{1})+\sqrt{\mu^{2}(x_{1})-2{\mathcal{}R}(\rho+\gamma)\sigma^{2}(x_{1})}}{2(\rho+\gamma)}:=\widetilde{\psi^{+}}(x_{1}),$

where the right-hand side $\widetilde{\psi^{+}}(x_{1})$ is the other (larger)
root of $\widetilde{Q}_{\gamma}(\cdot;x_{1})$.

Now we claim that for $x$ fixed, if $(\psi^{+})^{\prime}(x)\geq 1$ then
$(\widetilde{\psi^{+}})^{\prime}(x)\geq 1$. By taking derivative of
$\psi^{+}(x)$ w.r.t. $x$, we have

$\displaystyle(\psi^{+})^{\prime}(x)=\frac{1}{2\rho}\mu^{\prime}(x)+\frac{\mu(x)\mu^{\prime}(x)}{2\rho\sqrt{\mu(x)^{2}-2\mathcal{R}\rho\sigma^{2}(x)}}-\frac{\mathcal{R}\sigma(x)\sigma^{\prime}(x)}{\sqrt{\mu(x)^{2}-2\mathcal{R}\rho\sigma^{2}(x)}}.$

When $(\psi^{+})^{\prime}(x)\geq 1$, it is clear that $\mu^{\prime}(x)>0$
since $\sigma^{\prime}(x)\geq 0$ by Assumption 4.3. Therefore both the terms
$\frac{1}{2\rho}\mu^{\prime}(x)$ and
$-\frac{\mathcal{R}\sigma(x)\sigma^{\prime}(x)}{\sqrt{\mu(x)^{2}-2\mathcal{R}\rho\sigma^{2}(x)}}$
are decreasing w.r.t $\rho$. It remains to show the same monotonicity for the
second term in $(\psi^{+})^{\prime}(x)$. Take derivative of the function
$h(\rho;x):=\rho\sqrt{\mu(x)^{2}-2\mathcal{R}\rho\sigma^{2}(x)}$ w.r.t $\rho$,
we have

$\displaystyle\partial_{\rho}h(\rho;x)=\sqrt{\mu(x)^{2}-2\mathcal{R}\rho\sigma^{2}(x)}-\frac{\mathcal{R}\rho\sigma^{2}(x)}{\sqrt{\mu(x)^{2}-2\mathcal{R}\rho\sigma^{2}(x)}}\geq
0,$

where the inequality is followed by Assumption 4.3 (i) with
$(x_{2},x_{1})\subset[0,\underline{b}]$. As a result, we show that
$1\leq(\psi^{+})^{\prime}(x)\leq(\widetilde{\psi^{+}})^{\prime}(x)$ since
$\rho+\gamma<\rho$ for $\gamma<0$. Combined with Assumption 4.3 (ii), which
says that $\psi^{+}(x)-x$ increases on $[0,\underline{b}]$, we have
$\widetilde{\psi^{+}}(x)-x$ increases on $[0,\underline{b}]$.

To proceed, recalling that $v^{\prime}_{b^{\gamma},\gamma}(x_{1})=1$ and
$v^{\prime\prime}_{b^{\gamma},\gamma}(x_{1})\geq 0$ (see (5.20)), we claim
that there exists

$x_{2}:=\sup\\{x\in(0,x_{1}):v^{\prime}_{b^{\gamma},\gamma}(x)=1\\}.$

Arguing by contradiction, we assume that this does not hold. This implies that
$v_{b^{\gamma},\gamma}^{\prime}(x)<1$ for every $x\in(0,x_{1})$. Furthermore,
since $v_{b^{\gamma},\gamma}\in C({(0,\infty)})$ and
$v_{b^{\gamma},\gamma}(0)=\xi_{0}$ (by definition of $b^{\gamma}$),
$v_{b^{\gamma},\gamma}(x_{1})<x_{1}+v_{b^{\gamma},\gamma}(0)=x_{1}+\xi_{0},$
which contradicts Lemma 5.6 (i). Hence, the existence is guaranteed.

Since $v^{\prime}_{b^{\gamma},\gamma}(x)<1$ for every $x\in(x_{2},x_{1})$
(from the definition of $x_{2}$ and the fact that
$v^{\prime}_{b^{\gamma},\gamma}(x_{1})=1$ and
$v^{\prime\prime}_{b^{\gamma},\gamma}(x_{1})\geq 0$), the following hold:
$v^{\prime}_{b^{\gamma},\gamma}(x_{2})=1$ and

(5.23) $\displaystyle\begin{aligned}
v_{b^{\gamma},\gamma}(x_{2})>v_{b^{\gamma},\gamma}(x_{1})-(x_{1}-x_{2})&\geq\frac{\mu(x_{1})+\sqrt{\mu^{2}(x_{1})-2{\mathcal{}R}(\rho+\gamma)\sigma^{2}(x_{1})}}{2(\rho+\gamma)}-(x_{1}-x_{2})\\\
&\geq\frac{\mu(x_{2})+\sqrt{\mu^{2}(x_{2})-2{\mathcal{}R}(\rho+\gamma)\sigma^{2}(x_{2})}}{2(\rho+\gamma)},\end{aligned}$

where the second inequality follows from (5.22) and the last inequality
follows from the fact that $\widetilde{\psi^{+}}(x)-x$ increases on
$(x_{2},x_{1})$ since $(x_{2},x_{1})\subset[0,\underline{b}]$

Recalling the nonlinear ODE (5.1), the inequality given in (5.23) ensures that

$v^{\prime\prime}_{b^{\gamma},\gamma}(x_{2})=\frac{2}{\sigma^{2}(x_{2})}\frac{1}{v_{b^{\gamma},\gamma}(x_{2})}\widetilde{Q}_{\gamma}(v_{b^{\gamma},\gamma}(x_{2});x_{2})>0.$

This together with (5.20) contradicts Lemma 5.2. (Lemma 2.2 (ii) requires
$v^{\prime\prime}_{b^{\gamma},\gamma}(x_{1})>0$, but we can see it cannot be
the case that $v^{\prime\prime}_{b^{\gamma},\gamma}(x_{2})>0$,
$v^{\prime\prime}_{b^{\gamma},\gamma}(x_{1})=0$ and
$v^{\prime\prime}_{b^{\gamma},\gamma}(b^{\gamma})<0$)

Therefore, $x_{1}$ is not well-defined, we conclude that (5.18) holds.

Step 2. Recall $b^{\gamma}$ for every $\gamma<0$ given in (5.17) and let
$b^{*}$ be defined by $b^{*}=\lim_{\gamma\uparrow 0}b^{\gamma}$ (see Lemma 5.6
(ii)). We claim that $v^{\prime}_{b^{*}}(x)\geq 1$ for every $x\in(0,b^{*}]$.

To that end, let us consider an increasing sequence
$(\gamma_{n})_{n\in\mathbb{N}}$ such that $-\rho<\gamma_{n}<0$ for every
$n\in\mathbb{N}$ and $\lim_{n\rightarrow\infty}\gamma_{n}=0$. Since
$\lim_{j\rightarrow\infty}b^{\gamma_{n}}=b^{*}$ and
$(b^{\gamma_{n}})_{n\in\mathbb{N}}$ is a increasing sequence (see Lemma 5.6
(ii)), the following hold: $b^{*}\geq b^{\gamma_{n}}\geq b^{\gamma_{0}}$ for
every $n\in\mathbb{N}$.

Combining this with Step 1. (along with $v_{b^{\gamma_{n}},\gamma_{n}}\in
C^{1}({(0,\infty)})$ for every $n\in\mathbb{N}$) ensures that for every
$n\in\mathbb{N}$ and every $x\in(0,b^{*}]$,

(5.24) $\displaystyle v_{b^{\gamma_{n}},\gamma_{n}}^{\prime}(x)\geq 1.$

Fix $y\in(0,b^{\gamma_{0}}]\subset(0,b^{*}]$. For arbitrary $\varepsilon>0$,
then Lemma 5.3 (iii) ensures that there exists
${N}_{\varepsilon}\in\mathbb{N}$ such that
$|v^{\prime}_{b^{\gamma_{n}},\gamma_{n}}(x)-v^{\prime}_{b^{\gamma_{n}}}(x)|<\varepsilon$
for every $n\geq{N}_{\varepsilon}$. Furthermore, Lemma 5.4 (iii) ensures that
there exists $\widetilde{N}_{\varepsilon}\in\mathbb{N}$ such that
$|v^{\prime}_{b^{*}}(x)-v^{\prime}_{b^{\gamma_{n}}}(x)|<\varepsilon$ for every
$n\geq\widetilde{N}_{\varepsilon}$. Therefore, from (5.24), it follows that
for every $n\geq\widetilde{N}_{\varepsilon}\vee N_{\varepsilon}$,

$\displaystyle
v^{\prime}_{b^{*}}(x)>v^{\prime}_{b^{\gamma_{n}}}(x)-\varepsilon>v^{\prime}_{b^{\gamma_{n}},\gamma_{n}}(x)-2\varepsilon\geq
1-2\varepsilon.$

By letting $\varepsilon\rightarrow 0$, we conclude that
$v_{b^{*}}^{\prime}(x)\geq 1$ for $y\in(0,b^{\gamma_{0}}]$.

It remains to show $v_{b^{*}}^{\prime}(y)\geq 1$ on $(b^{\gamma_{0}},b^{*}]$.
Assume the contrary, we have
$\min_{y\in[b^{\gamma_{0}},b^{*}]}v_{b^{*}}^{\prime}(y)-1<0$. Then
$v_{b^{*}}^{\prime}(y)-1$ takes its minimum at
$x_{0}\in(b^{\gamma_{0}},b^{*})$ because $v^{\prime}_{b^{*}}(b^{*})=1$ and
$v^{\prime}_{b^{*}}(b^{\gamma_{0}})\geq 1$ by above analysis. Then we have
that $v_{b^{*}}^{\prime\prime}(x_{0})=0$ and $v^{\prime}_{b^{*}}(x_{0})<1$. By
the continuity of $v^{\prime}_{b^{*}}$, let
$\epsilon=(1-v^{\prime}_{b^{*}}(x_{0}))/2$ there exists $x_{0}^{\prime}$ in a
local neighborhood of $x_{0}$ such that
$v_{b^{*}}^{\prime\prime}(x_{0}^{\prime})>0$ and
$v_{b^{*}}^{\prime}(x_{0}^{\prime})<v^{\prime}_{b^{*}}(x_{0})+\epsilon<1$.

Hence

$g_{b^{*}}(x_{0}^{\prime})=\dfrac{v^{\prime}_{b^{*}}(x_{0}^{\prime})}{v_{b^{*}}(x_{0}^{\prime})}<\frac{1}{x_{0}^{\prime}+\xi}<\frac{1}{\psi^{-}(x_{0}^{\prime})},$

where the first inequality follows Lemma 5.6 (i) that
$v_{b^{*}}(x_{0}^{\prime})\geq x_{0}^{\prime}+\xi$ since $v_{b^{*}}(0)=\xi$,
while the last inequality follows Assumption 4.3 (iii). Recall the ODE (5.1)
of $v_{b^{*}}$, we have

$\displaystyle\frac{1}{2}\sigma^{2}(x_{0})v^{\prime\prime}_{b^{*}}(x_{0}^{\prime})=\frac{1}{v_{b^{*}}(x_{0}^{\prime})}Q(v_{b^{*}}(x_{0}^{\prime});x_{0}^{\prime})>0.$

Combined with the above arguments, we conclude that
$g_{b^{*}}(x_{0}^{\prime})<1/\psi^{+}(x_{0}^{\prime})$. This contradicts with
Lemma 5.4 (i) which gives
$g_{b^{*}}(x_{0}^{\prime})\geq{1}/{\psi^{+}(x_{0}^{\prime})}$ as
$x_{0}>b^{\gamma_{0}}>\underline{b}$. As a result, we see
$v_{b^{*}}^{\prime}(y)\geq 1$ on $(b^{\gamma_{0}},b^{*}]$.

This completes the proof for the statement $v_{b^{*}}^{\prime}(x)\geq 1$ on
$[0,b^{*}]$.

Finally, we show that $b^{*}<\hat{b}$, where $\hat{b}$ defined in Assumption
4.3 (iii). We only need to show $b^{*}\neq\hat{b}$ since $b^{*}\leq\hat{b}$ by
Lemma 5.6 (ii). When $b^{*}=\hat{b}$, by definition of $\hat{b}$, we have
$v_{b^{*}}(0)=\xi_{0}$ while $v_{b^{*}}(b^{*})=\psi^{+}(b^{*})=b^{*}+\xi_{0}$.
Together with our previous result that $v_{b^{*}}^{\prime}(x)\geq 1$ on
$(0,b^{*}]$, we conclude $v_{b^{*}}^{\prime}(x)\equiv 1$ on $(0,b^{*}]$, which
cannot be the case due to the structure of ODE (5.1). Therefore, we have
$b^{*}<\hat{b}$. ∎

##  6\. Proof of Theorem 4.6

###### Proof of Theorem 4.6.

Step 1. We claim that for every $x\in[0,\infty)$,

(6.1) $\displaystyle
v^{*}(x)\geq\sup_{D\in{\mathcal{}A}^{x}}\inf_{\theta\in\Theta_{D}}\mathbb{E}^{\theta}\bigg{[}\int_{0}^{\tau^{x,D}}e^{-\rho
t}dD_{t}+\frac{1}{2\mathcal{R}}\int_{0}^{\tau^{x,D}}e^{-\rho
t}\theta_{t}^{2}v^{*}(X_{t}^{x,D})dt+e^{-\rho\tau^{x,D}}\xi_{0}\bigg{]},$

and

(6.2) $\displaystyle v^{*}(x)\geq J^{*}(x),$

where $J^{*}$ is defined in (4.5). As the proof for the case $x=0$ is obvious,
we assume $x\in(0,\infty)$. For simplicity, we will omit the superscript $x$
in $X^{x,D}$ and $\tau^{x,D}$, as defined in (4.3) and (4.4), respectively.

Set an arbitrary $D\in{\mathcal{}A}^{x}$ and define the stopping times
$(\hat{\tau}_{n})_{n\in\mathbb{N}}$ by

(6.3) $\displaystyle\hat{\tau}_{n}:=\inf\\{t\geq
0:X^{D}_{t}\notin[1/n,n]\\}\wedge n.$

Recall the structure of the operator $\mathcal{L}$ from (4.2), and define the
measure $\mathbb{Q}^{*}$ via (3.2) with the Girsanov kernel
$\theta^{*}:=(\theta^{*}(D)_{s})_{s\geq 0}$ given by

(6.4)
$\displaystyle\theta^{*}_{s}:=-{\mathcal{}R}\frac{\sigma(X_{s}^{D})(v^{*})^{\prime}(X_{s}^{D})}{v^{*}(X_{s}^{D})}.$

Note that by (5.7) and Proposition 5.1, $(v^{*})^{\prime}$ is bounded on
$[0,b^{*}]$ and is identical to $1$ on $(b^{*},\infty)$ while $v^{*}$ is
bounded below by $\xi_{0}$ on $[0,\infty)$. Following similar arguments of in
Theorem 3.3, it follows that $\theta^{*}\in\Theta^{D}$. From this, we can
denote by $\mathbb{Q}^{*}$ the corresponding probability measure obtained from
$\theta^{*}$.

An application of Dynkin formula into the process $e^{-\rho
t}v^{*}(X_{t}^{D})$ on the random time interval
$[0,\tau^{D}\wedge\hat{\tau}_{n}]$ under $\mathbb{Q}^{*}$ (corresponding to
$\theta^{*}$) ensures that

(6.5) $\displaystyle\begin{aligned}
&e^{-\rho(\tau^{D}\wedge\hat{\tau}_{n})}v^{*}(X_{\tau^{D}\wedge\hat{\tau}_{n}}^{D})-v^{*}(x)=\int_{0}^{\tau^{D}\wedge\hat{\tau}_{n}}e^{-\rho
s}\;\hat{\mathcal{}L}^{\theta^{*}}v^{*}(X_{s}^{D})ds\\\
&+\int_{0}^{\tau^{D}\wedge\hat{\tau}_{n}}e^{-\rho
s}\sigma(X_{s}^{D})(v^{*})^{\prime}(X_{s}^{D})dW^{\mathbb{Q}^{*}}_{s}-\int_{0}^{\tau^{D}\wedge\hat{\tau}_{n}}e^{-\rho
s}(v^{*})^{\prime}(X_{s}^{D})dD_{s},\end{aligned}$

where $\hat{\mathcal{}L}^{\theta^{*}}$ is the infinitesimal operator under
$\mathbb{Q}^{*}$, i.e.,

$\displaystyle\hat{\mathcal{}L}^{\theta^{*}}v^{*}(X_{s}^{D}):=\frac{\sigma^{2}(X_{s}^{D})}{2}(v^{*})^{\prime\prime}(X_{s}^{D})+(\mu(X_{s}^{D})+\sigma(X_{s}^{D})\theta_{s}^{*})(v^{*})^{\prime}(X_{s}^{D})-\rho
v^{*}(X_{s}^{D}),$

and $D^{C}$ denotes the continuous part of $D$ and $W^{\mathbb{Q}^{*}}$ is the
$\mathbb{Q}^{*}$-Brownian motion. Here, we note that the summation term in the
last line is non-zero only for at mostly countably many times $s$ such that
$D_{s}-D_{s-}>0$;

By similary arguements, an application of Dynkin formula to the process
$e^{-\rho
t\int_{0}^{t}({\theta}^{*}_{s})^{2}/2{\mathcal{}R}ds}v^{*}(X_{t}^{D})$ on the
random time interval $[0,\tau^{D}\wedge\hat{\tau}_{n}]$ gives

(6.6) $\displaystyle\begin{aligned}
&e^{-\rho(\tau^{D}\wedge\hat{\tau}_{n})+\frac{1}{2{\mathcal{}R}}\int_{0}^{\tau^{D}\wedge\hat{\tau}_{n}}(\theta_{s}^{*})^{2}ds}v^{*}(X_{\tau^{D}\wedge\hat{\tau}_{n}}^{D})-v^{*}(x)=\int_{0}^{\tau^{D}\wedge\hat{\tau}_{n}}e^{-\rho
s+\frac{1}{2{\mathcal{}R}}\int_{0}^{s}(\theta_{u}^{*})^{2}du}\;{\mathcal{}L}v^{*}(X_{s}^{D})ds\\\
&+\int_{0}^{\tau^{D}\wedge\hat{\tau}_{n}}e^{-\rho
s+\frac{1}{2{\mathcal{}R}}\int_{0}^{s}(\theta_{u}^{*})^{2}du}\sigma(X_{s}^{D})(v^{*})^{\prime}(X_{s}^{D})dW^{\mathbb{Q}^{*}}_{s}-\int_{0}^{\tau^{D}\wedge\hat{\tau}_{n}}e^{-\rho
s+\frac{1}{2{\mathcal{}R}}\int_{0}^{s}(\theta_{u}^{*})^{2}du}(v^{*})^{\prime}(X_{s}^{D})dD_{s},\end{aligned}$

where recall that ${\mathcal{}L}$ is defined in (4.2).

Based on the definitions of the nonlinear operator ${\mathcal{}L}$ in (4.2),
the process $\theta^{*}$ in (6.4), and given that $v^{*}(X_{s}^{D})>0$ for
$s\in[0,\tau^{D})$ and $v^{*}$ solves (4.9), it follows that for every
$s\in[0,\tau^{D})$:

(6.7)
$\displaystyle\hat{\mathcal{}L}^{\theta^{*}}v^{*}(X_{s}^{D})={\mathcal{}L}v^{*}(X_{s}^{D})-\frac{(\theta_{s}^{*})^{2}}{2{\mathcal{}R}}v^{*}(X_{s}^{D})\leq-\frac{(\theta_{s}^{*})^{2}}{2{\mathcal{}R}}v^{*}(X_{s}^{D}).$

Furthermore, Since $v^{*}\in C^{2}([0,\infty))$ (see Theorem 4.5), $\sigma\in
C^{1}({(0,\infty)})$ (see Assumption 4.3 (i)) and $X^{D}$ is bounded in
$[0,\tau^{D}\wedge\hat{\tau}_{n}]$ (see (6.3)), the Brownian local martingale
terms given in (6.5) and (LABEL:eq:thm_veri_pf_1.5) are martingales. Lastly,
since $(v^{*})^{\prime}(x)\geq 1$ for $x\in[0,\infty)$ and $v^{*}(0)=\xi_{0}$,
it follows that $v^{*}(X_{\tau_{D}\wedge\hat{\tau}_{n}}^{D})\geq\xi_{0}$.
Taking the expectation over (6.5) under $\mathbb{Q}^{*}$ gives

$\displaystyle v^{*}(x)\geq$
$\displaystyle\mathbb{E}^{\theta^{*}}\left[\int_{0}^{(\tau_{D}\wedge\hat{\tau}_{n})}e^{-\rho
s}dD_{s}+\int_{0}^{(\tau_{D}\wedge\hat{\tau}_{n})}\frac{(\theta^{*}_{s})^{2}}{2{\mathcal{}R}}v^{*}(X^{D}_{s})ds+e^{-\rho(\tau_{D}\wedge\hat{\tau}_{n})}\xi_{0}\right].$

Since $\tau_{D}\wedge\hat{\tau}_{n}\rightarrow\tau_{D}$ as
$n\rightarrow\infty$ (see (6.3)), an application of Fatou’s lemma together
with $\theta^{*}\in\Theta^{D}$ (see (6.4)) ensures that

$\displaystyle v^{*}(x)\geq$
$\displaystyle\mathbb{E}^{\theta^{*}}\left[\int_{0}^{\tau_{D}}e^{-\rho
s}dD_{s}+\int_{0}^{\tau_{D}}e^{-\rho
s}\frac{(\theta_{s}^{*})^{2}}{2{\mathcal{}R}}v^{*}(X^{D}_{s})ds+e^{-\rho\tau_{D}}\xi_{0}\right]$
$\displaystyle\geq$
$\displaystyle\inf_{\theta\in\Theta_{D}}\mathbb{E}^{\theta}\left[\int_{0}^{\tau_{D}}e^{-\rho
s}dD_{s}+\int_{0}^{\tau_{D}}e^{-\rho
s}\frac{\theta_{s}^{2}}{2{\mathcal{}R}}v^{*}(X^{D}_{s})ds+e^{-\rho\tau_{D}}\xi_{0}\right].$

As the inequality holds for every $D\in{\mathcal{}A}^{x}$, the claim (6.1)
holds. With similar arguments applied to (LABEL:eq:thm_veri_pf_1.5), we obtain
that

$\displaystyle
v^{*}(x)\geq\mathbb{E}^{\theta^{*}}\left[\int_{0}^{\tau_{D}}e^{-\rho
s+\frac{1}{2{\mathcal{}R}}\int_{0}^{s}(\theta_{u}^{*})^{2}du}dD_{s}+e^{-\rho\tau_{D}+\frac{1}{2{\mathcal{}R}}\int_{0}^{\tau_{D}}(\theta_{u}^{*})^{2}du}\xi_{0}\right]\geq
J^{\theta^{*}}(x;D)\geq\inf_{\theta\in\Theta^{D}}J^{\theta}(x;D),$

where $J^{\theta}$ is defined in (4.6). We obtain (6.2) by taking supreme over
${D\in\mathcal{A}^{x}}$ on both sides of the above equation.

Step 2. Let $D^{*}=D(b^{*})$ be the $b^{*}$-threshold strategy. We show that
for every $x\in[0,\infty)$,

(6.8) $\displaystyle
v^{*}(x)\leq\inf_{\theta\in\Theta_{D^{*}}}\mathbb{E}^{\theta}\bigg{[}\int_{0}^{\tau^{D^{*}}}e^{-\rho
t}dD^{*}_{t}+\int_{0}^{\tau^{D^{*}}}e^{-\rho
t}\frac{\theta_{t}^{2}}{2\mathcal{R}}v^{*}(X_{t}^{D^{*}})dt+e^{-\rho\tau^{D^{*}}}\xi_{0}\bigg{]},$

and

(6.9) $\displaystyle v^{*}(x)\leq J^{*}(x).$

Without loss of generality, we assume that $x\in(0,\infty)$ which will be
divided into two cases (i.e., $x\in(0,b^{*}]$ or $x\in(b^{*},\infty)$). Before
starting the proof, we first note that from the definition of
$b^{*}$-threshold strategy (see Definition 4.2), the following properties hold
for every $s\in(0,\tau^{D^{*}}]$:

(6.10) $\displaystyle X^{D^{*}}_{s}\in[0,b^{*}],\;\;\quad
dD^{*}_{s}>0\quad\mbox{only when $X_{s}^{D^{*}}=b^{*}$}.$

Furthermore, if $x\in(0,b^{*}]$, the above properties hold for every
$s\in[0,\tau^{D^{*}}]$.

We first consider the case $x\in(0,b^{*}]$. Let $\theta\in\Theta({D^{*}})$ and
$\mathbb{Q}$ be the corresponding probability measure induced by the kernel
$\theta$. Furthermore, as in (6.3), we define for every $n\in\mathbb{N}$ by
$\hat{\tau}_{n}:=\inf\\{t\geq 0:X^{D^{*}}_{t}\notin[1/n,n]\\}.$ An application
of Dynkin formula to the process $e^{-\rho t}v^{*}(X_{t}^{D^{*}})$ on the
random time interval $[0,\tau^{D^{*}}\wedge\hat{\tau}_{n}]$ under $\mathbb{Q}$
gives that

(6.11) $\displaystyle\begin{aligned}
v^{*}(x)&=\mathbb{E}^{\theta}\left[e^{-\rho(\tau^{D^{*}}\wedge\hat{\tau}_{n})}v^{*}(X_{\tau^{D^{*}}\wedge\hat{\tau}_{n}}^{D^{*}})-\int_{0}^{\tau^{D^{*}}\wedge\hat{\tau}_{n}}e^{-\rho
s}\left(\hat{\mathcal{}L}^{\theta}v^{*}(X_{s}^{D^{*}})ds-
dD^{*}_{s}\right)\right],\end{aligned}$

where $\hat{\mathcal{}L}^{\theta}$ is the infinitesimal operator under
$\mathbb{Q}^{*}$, i.e.,

$\displaystyle\hat{\mathcal{}L}^{\theta}v^{*}(X_{s}^{D^{*}}):=\frac{1}{2}\sigma^{2}(X_{s}^{D^{*}})(v^{*})^{\prime\prime}(X_{s}^{D^{*}})+(\mu(X_{s}^{D^{*}})+\sigma(X_{s}^{D^{*}})\theta_{s})(v^{*})^{\prime}(X_{s}^{D^{*}})-\rho
v^{*}(X_{s}^{D^{*}}),$

and we use the fact that
$\mathbb{E}^{\theta}[\int_{0}^{\tau^{D^{*}}\wedge\hat{\tau}_{n}}e^{-\rho
s}\sigma(X_{s}^{D^{*}})(v^{*})^{\prime}(X_{s}^{D^{*}})dW^{\mathbb{Q}}_{s}]=0$
and that $dD^{*}_{s}>0$ only when $(v^{*})^{\prime}(X_{s}^{D^{*}})=1$ for
$s\in[0,\tau^{D^{*}}]$. From (6.10) and the fact that
${\mathcal{}L}v^{*}(x)=0$ for every $x\in(0,b^{*}]$ (see (4.9) with the free
boundary $b^{*}$), it follows that for every $s\in[0,\tau^{D^{*}}]$,
${\mathcal{}L}v^{*}(X_{s}^{D^{*}})=0$. Recall the definition of
${\mathcal{}L}$ in (4.2), we obtain that for every $s\in[0,\tau^{D^{*}}]$

(6.12)
$\displaystyle\hat{\mathcal{}L}^{\theta}v^{*}(X_{s}^{D^{*}})\geq{\mathcal{}L}v^{*}(X_{s}^{D^{*}})-\frac{(\theta_{s})^{2}}{2{\mathcal{}R}}v^{*}(X_{s}^{D^{*}})=-\frac{(\theta_{s})^{2}}{2{\mathcal{}R}}v^{*}(X_{s}^{D^{*}}).$

By (6.12) and (6.11), we hence obtain that

(6.13) $\displaystyle\begin{aligned}
v^{*}(x)&\leq\mathbb{E}^{\theta}\left[\int_{0}^{\tau^{D^{*}}\wedge\hat{\tau}_{n}}e^{-\rho
s}\left(\frac{(\theta_{s})^{2}}{2{\mathcal{}R}}v^{*}(X_{s}^{D^{*}})ds+dD^{*}_{s}\right)+e^{-\rho(\tau^{D^{*}}\wedge\hat{\tau}_{n})}v^{*}(X_{\tau^{D^{*}}\wedge\hat{\tau}_{n}}^{D^{*}})\right].\end{aligned}$

From $v^{*}\in C^{2}([0,\infty))$ and the fact that
$X^{D^{*}}_{s}\in[0,b^{*}]$ for every $s\in[0,\tau^{D^{*}}]$ (see (6.10)), it
follows that $\sup_{s\in[0,\tau^{D^{*}}]}|v^{*}(X_{s}^{D^{*}})|<\infty.$
Hence, an application of the dominated convergence theorem (applicable to the
last term of right-hand side of (6.13)) and monotone convergence theorem
(applicable to the first term of right-hand side of (6.13)), together with
$\tau^{D^{*}}\wedge\hat{\tau}_{n}\rightarrow\tau^{D^{*}}$ as
$n\rightarrow\infty$ into (6.13) ensures that

(6.14) $\displaystyle v^{*}(x)$
$\displaystyle\leq\mathbb{E}^{\theta}\left[\int_{0}^{\tau^{D^{*}}}e^{-\rho
s}\left(\frac{(\theta_{s})^{2}}{2{\mathcal{}R}}v^{*}(X_{s}^{D^{*}})ds+dD^{*}_{s}\right)+e^{-\rho\tau^{D^{*}}}\xi_{0}\right],$

where we have used that $v^{*}(X_{\tau^{D^{*}}}^{D^{*}})=v^{*}(0)=\xi_{0}$
(see (4.9)). As the inequality holds for every $\theta\in\Theta_{D^{*}}$, the
claim (6.8) holds for every $x\in(0,b^{*}]$.

It remains to prove the case $x\in(b^{*},\infty)$. Note that for every
$x\in(b^{*},\infty)$, $v^{*}(x)=x-b^{*}+\psi^{+}(b^{*})$ (see (5.7)).
Furthermore, since the strategy $D^{*}=D(b^{*})$ starts with an instantaneous
increase with amount $x-b^{*}$, it follows that for every
$\theta\in\Theta_{D^{*}}$,

$\displaystyle\mathbb{E}^{\theta}\left[\int_{0}^{\tau^{D^{*}}}e^{-\rho
s}\left(\frac{(\theta_{s})^{2}}{2{\mathcal{}R}}v^{*}(X_{s}^{x,D^{*}})ds+dD^{*}_{s}\right)+e^{-\rho\tau^{D^{*}}}\xi_{0}\right]$
$\displaystyle=x-b^{*}+\mathbb{E}^{\theta}\left[\int_{0}^{\tau^{D^{*}}}e^{-\rho
s}\left(\frac{(\theta_{s})^{2}}{2{\mathcal{}R}}v^{*}(X_{s}^{b^{*},D^{*}})ds+dD^{*}_{s}\right)+e^{-\rho\tau^{D^{*}}}\xi_{0}\right]\geq
x-b^{*}+v^{*}(b^{*})$

where we have used the inequality (6.14) (since it holds when $x=b^{*}$),
which ensures

$\mathbb{E}^{\theta}\left[\int_{0}^{\tau^{D^{*}}}e^{-\rho
s}\left(\frac{(\theta_{s})^{2}}{2{\mathcal{}R}}v^{*}(X_{s}^{x,D^{*}})ds+dD^{*}_{s}\right)+e^{-\rho\tau^{D^{*}}}\xi_{0}\right]\geq
x-b^{*}+v^{*}(b^{*})=v^{*}(x),$

where we have used that $v^{*}(b^{*})=\psi^{+}(b^{*})$ with $v^{*}\in
C^{2}([0,\infty))$. As the inequality holds for every
$\theta\in\Theta_{D^{*}}$, the claim (6.8) holds for every
$x\in(b^{*},\infty)$.

To show (6.9), similar to the analysis in Step 1, an application of Dynkin
formula to the process $e^{-\rho
t\int_{0}^{t}({\theta}_{s})^{2}/2{\mathcal{}R}ds}v^{*}(X_{t}^{D^{*}})$ gives

$\displaystyle
v^{*}(x)\leq\inf_{\theta\in{\Theta_{D}^{*}}}J^{\theta}(x;D^{*})\leq\sup_{D\in\mathcal{A}^{x}}\inf_{\theta\in{\Theta_{D}}}J^{\theta}(x;D)\leq
J^{*}(x),\quad x\in[0,\infty).$

This completes the proof. ∎

##  Appendix A Proof for Section 2

Let us introduce the following sets that are used throughout the appendices:
for every $p\geq 1$ and $t\geq 0$,

  * $\cdot$

$\mathcal{S}_{p}$ is the set of the set of all real-valued,
$\mathbb{F}$-progressively measurable processes $Y$ such that
$\|Y\|_{{\mathcal{}S}_{p}}^{p}:=\mathbb{E}[\sup_{t\geq
0}|Y_{t\wedge\tau}|^{p}]<\infty$;

  * $\cdot$

${\mathcal{}S}_{\infty}$ is the set of all real-valued,
$\mathbb{F}$-progressively measurable processes $Y$ such that
$\|Y\|_{{\mathcal{}S}_{\infty}}:=\inf\\{C\geq 0:|Y_{t}|\leq C\;\mbox{for
all}\;t\geq 0\;\mbox{$\mathbb{P}$-a.s.}\\}$;

  * $\cdot$

$\mathcal{M}_{p}$ is the set of all real-valued, $\mathbb{F}$-progressively
measurable processes $Z$ such that
$\|Z\|_{\mathcal{M}_{p}}^{p}:=\mathbb{E}[(\int_{0}^{\tau}|Z_{t}|^{2}dt)^{p/2}]<\infty$.

  * $\cdot$

Denote by $\mathcal{B}_{p}:=\mathcal{S}_{p}\times\mathcal{M}_{p}$ the product
space equipped with the norm
$\|(Y,Z)\|_{\mathcal{B}_{p}}^{p}:=\|Y\|_{{\mathcal{}S}_{p}}^{p}+\|Z\|_{\mathcal{M}_{p}}^{p}$
for every $(Y,Z)\in\mathcal{B}_{p}$.

###  A.1. An overview of arguments

We first demonstrate the existence of a solution for BSDEs with any bounded
$D\in\mathcal{A}$ and a Lipschitz aggregator. We then present a comparison
theorem for these BSDEs, establishing that any subsolution always lies below
any supersolution, provided the aggregator is nonincreasing in its second
argument. Uniqueness of BSDE solutions follows by a standard argument, i.e.,
any solution is both a subsolution and a supersolution. Additionally, we
establish a priori lower and upper bounds for a candidate BSDE solution with
an (non-Lipschitz) EZ aggregator. Progressing further, we construct a sequence
of Lipschitz aggregators that converges to the EZ aggregator. The a priori
bounds allow us to prove that the sequence of the BSDE solutions with
Lipschitz aggregators converges in $\mathcal{B}_{2}$ space. Finally, for
general unbounded $D\in{\mathcal{}A}$, we adopt a technique motivated by [3].

###  A.2. Step 1. Lipschitz aggregator: fixed horizon and bounded controls

For any $0<T<\infty$, we define the spaces ${\mathcal{}S}_{2}([0,T])$,
$\mathcal{M}_{2}([0,T])$, and $\mathcal{B}_{2}([0,T])$ similarly, with $\tau$
replaced by $T$ in the definition of ${\mathcal{}S}_{2}$, ${\mathcal{}M}_{2}$,
and ${\mathcal{}B}_{2}$.

######  Lemma A.1.

Let $T<\infty$ and $(Y_{t},Z_{t})\in\mathcal{B}_{2}([0,T])$. Then the
continuous local martingale $(\int_{0}^{t}Y_{s}Z_{s}dW_{s})_{t\in[0,T]}$ is a
uniformly integrable martingale.

###### Proof.

This is exactly [3, Lemma A.1], we omit the proof here. ∎

######  Lemma A.2.

Let $T<\infty$. Let $A\in{\mathcal{}S}_{2}([0,T])$ and $D\in\mathcal{A}$ be
such that $D_{T}\leq K$ $\mathbb{P}$-a.s. for some constant $K$, and both $A$
and $D$ have continuous paths. If the following holds for every
$\zeta\in{\mathcal{}T}$ satisfying that $\zeta\leq T$ $\mathbb{P}$-a.s.,

(A.1)
$\displaystyle\mathbb{E}[A_{\zeta}]\leq\alpha+\mathbb{E}\bigg{[}\int_{\zeta}^{T}A_{s}dD_{s}\bigg{]},$

with some constant $\alpha$, then it holds that $\mathbb{E}[A_{t}]\leq\alpha
e^{K}$ for every $t\in[0,T]$.

###### Proof.

For fixed $t\in[0,T)$ and $0\leq k\leq K$, set $\zeta_{k}=\inf\\{s\geq
t:D_{s}\geq k\\}\wedge T$. Then each $\zeta_{k}$ is a stopping time bounded by
$T$. Moreover, since $\\{\zeta_{k}\leq s\\}=\\{D_{s}\geq k\\}$ for $s\geq t$
and $\zeta_{K}=T$ holds $\mathbb{P}$-a.s., it follows that

$\displaystyle\mathbb{E}[A_{\zeta_{k}}]\leq$
$\displaystyle\alpha+\mathbb{E}\bigg{[}\int_{\zeta_{k}}^{T}A_{s}dD_{s}\bigg{]}=\alpha+\mathbb{E}\bigg{[}\int_{t}^{T}A_{s}\mathds{1}_{s\geq\zeta_{k}}dD_{s}\bigg{]}$
$\displaystyle\leq$
$\displaystyle\alpha+\mathbb{E}\bigg{[}\int_{t}^{T}A_{s}\mathds{1}_{\zeta_{k}\leq
s\leq\zeta_{K}}dD_{s}\bigg{]}=\alpha+\mathbb{E}\bigg{[}\int_{k}^{K}A_{\zeta_{s}}ds\bigg{]}=\alpha+\int_{k}^{K}\mathbb{E}[A_{\zeta_{s}}]ds.$

Now set $u_{s}:=\mathbb{E}[A_{\zeta_{s}}]$ for $k\in[0,K]$. Then since
$u_{k}\leq\alpha+\int_{k}^{K}u_{s}ds$, for $k\in[0,K]$, the standard
(backward) Grönwall’s inequality ensures that $u_{k}\leq\alpha e^{K-k}$ for
$k\in[0,K]$. We complete the proof by letting $k=0$. ∎

######  Theorem A.3 (Lipschitz aggregator).

For fixed horizon $0<T<\infty$ and $D\in{\mathcal{}A}$ with
$\int_{0}^{T}e^{-\rho t}dD_{t}\leq K$ $\mathbb{P}$-a.s. for some real constant
$K$, and $\xi_{T}\in L^{2}(\mathcal{F}_{T})$ assume that $g$ satisfies the
following conditions:

  * •

There exists $C_{g}>0$ such that $|g(t,y_{1})-g(t,y_{2})|\leq C_{g}e^{-\rho
t}|y_{1}-y_{2}|$ for all $y_{1},y_{2}$.

  * •

$\mathbb{E}[(\int_{0}^{T}|g(t,0)|dD_{t})^{2}]<\infty$.

Then the BSDE with parameter $(T,\xi_{T},g,D)$ has a unique solution in
$\mathcal{B}_{2}$.

###### Proof.

The solution is constructed by modifying the proof of [54, Proposition 2.2].
The idea is constructing an approximating sequence by a kind of Picard
iteration, and we show that it is a Cauchy sequence in $\mathcal{B}_{2}$. Let
$y^{0}=0$, and $\\{(y^{n}_{t},z^{n}_{t})_{t\in[0,T]}\\}_{n\geq 1}$ be a
sequence in $\mathcal{B}_{2}$ define recursively by

$\displaystyle
y_{t}^{n+1}=\xi_{T}+\int_{t}^{T}g(s,y_{s}^{n})dD_{s}-\int_{t}^{T}z_{s}^{n}dW_{s},\quad
0\leq t\leq T.$

For any $y^{n}\in\mathcal{S}_{2}$, we construct $y^{n+1}$ and $z^{n}$ as
follows. From • ‣ A.3 and • ‣ A.3, we derive

$\displaystyle\mathbb{E}\bigg{[}\bigg{(}\xi_{T}+\int_{0}^{T}g(s,y^{n}_{s})dD_{s}\bigg{)}^{2}\bigg{]}\leq
2\mathbb{E}\bigg{[}\xi_{T}^{2}+\bigg{(}\int_{0}^{T}\big{(}|g(s,0)|+C_{g}e^{-\rho
s}|y^{n}_{s}|\big{)}dD_{s}\bigg{)}^{2}\bigg{]}$ $\displaystyle\leq
2\mathbb{E}[\xi_{T}^{2}]+4\mathbb{E}\bigg{(}\int_{0}^{T}|g(s,0)|dD_{s}\bigg{)}^{2}+C_{g}\mathbb{E}(\sup_{0\leq
t\leq T}|y^{n}_{t}|^{2})K^{2}<\infty.$

This implies that
$\\{\mathbb{E}_{t}[\xi_{T}+\int_{0}^{T}g(s,y^{n}_{s})dD_{s}]\\}_{t\geq 0}$ is
a square integrable martingale. By the martingale representation theorem, we
constrcuct a unique $z^{n}\in\mathcal{M}_{2}$ and $y^{n+1}\in\mathcal{S}_{2}$
such that

(A.2)
$\displaystyle\int_{0}^{t}z^{n}_{s}dW_{s}=\mathbb{E}_{t}\bigg{[}\xi_{T}+\int_{0}^{T}g(s,y^{n}_{s})dD_{s}\bigg{]}-\mathbb{E}\bigg{[}\xi_{T}+\int_{0}^{T}g(s,y^{n}_{s})dD_{s}\bigg{]},\quad
0\leq t\leq T;$ (A.3) $\displaystyle\text{ and
}y^{n+1}_{t}:=\mathbb{E}_{t}\bigg{[}\xi_{T}+\int_{t}^{T}g(s,y^{n}_{s})dD_{s}\bigg{]}.$

For stopping time $\tau\in\mathcal{T}$, it follows from Itô’s lemma applied to
$(u_{s\wedge\tau}^{n+1})^{2}:=|y_{s\wedge\tau}^{n+1}-y_{s\wedge\tau}^{n}|^{2}$
that

$\displaystyle(u_{t\wedge\tau}^{n+1})^{2}+\int_{{s\wedge\tau}}^{T\wedge\tau}|z_{s}^{n}-z_{s}^{n-1}|^{2}ds$
$\displaystyle=2\int_{{s\wedge\tau}}^{T\wedge\tau}\big{(}g(s,y_{s}^{n})-g(s,y_{s}^{n-1})\big{)}u_{s}^{n+1}dD_{s}-2\int_{{s\wedge\tau}}^{T\wedge\tau}u_{s}^{n+1}(z_{s}^{n}-z_{s}^{n-1})dW_{s}.$

It is clear by Lemma A.1 that the above stochastic integral is a uniformly
integrable martingale and has zero expectation. By taking expectations on both
sides, together with condition • ‣ A.3 it follows that

$\displaystyle\mathbb{E}\big{[}(u_{t\wedge\tau}^{n+1})^{2}\big{]}+\mathbb{E}\bigg{[}\int_{t\wedge\tau}^{T\wedge\tau}|z_{s}^{n}-z_{s}^{n-1}|^{2}ds\bigg{]}$
$\displaystyle\leq
2\mathbb{E}\bigg{[}\int_{t\wedge\tau}^{T\wedge\tau}C_{g}u_{s}^{n}u_{s}^{n+1}e^{-\rho
s}d{D}_{s}\bigg{]}\leq
C_{g}\mathbb{E}\bigg{[}\int_{t\wedge\tau}^{T\wedge\tau}(u_{s}^{n})^{2}e^{-\rho
s}d{D}_{s}\bigg{]}+C_{g}\mathbb{E}\bigg{[}\int_{t\wedge\tau}^{T\wedge\tau}(u_{s}^{n+1})^{2}e^{-\rho
s}d{D}_{s}\bigg{]}.$

Lemma A.2 implies

(A.4) $\displaystyle\mathbb{E}\big{[}(u_{t}^{n+1})^{2}\big{]}\leq
C_{g}e^{C_{g}K}\mathbb{E}\bigg{[}\int_{t}^{T}(u_{s}^{n})^{2}e^{-\rho
s}d{D}_{s}\bigg{]},\quad t\in[0,T].$

Now we introduce sequence of bounded stopping time in $\mathcal{T}$ as
$\zeta_{s}:=\inf\\{t\geq 0:\int_{0}^{t}e^{-\rho\tau}D_{\tau}\geq s\\}\wedge T$
for $s\in[0,K]$. From the proof of Lemma A.2, it is obvious that the
inequality

(A.5) $\displaystyle\mathbb{E}\big{[}(u_{\zeta_{t}}^{n+1})^{2}\big{]}\leq
C_{g}e^{C_{g}K}\mathbb{E}\bigg{[}\int_{\zeta_{t}}^{T}(u_{s}^{n})^{2}e^{-\rho
s}d{D}_{s}\bigg{]}$

holds for $t\in[0,K]$. Iterating (A.4) and (A.5) gives for $t\in[0,T]$,

$\displaystyle\mathbb{E}\big{[}(u_{t}^{n+1})^{2}\big{]}\leq
C_{g}e^{C_{g}K}\mathbb{E}\bigg{[}\int_{t}^{T}(u_{s}^{n})^{2}e^{-\rho
s}d{D}_{s}\bigg{]}\leq
C_{g}e^{C_{g}K}\mathbb{E}\bigg{[}\int_{0}^{T}(u_{s}^{n})^{2}e^{-\rho
s}\mathds{1}_{\zeta_{0}\leq s\leq\zeta_{K}}d{D}_{s}\bigg{]}$
$\displaystyle=C_{g}e^{C_{g}K}\mathbb{E}\bigg{[}\int_{0}^{K}(u_{\zeta_{s}}^{n})^{2}ds\bigg{]}=C_{g}e^{C_{g}K}\int_{0}^{K}\mathbb{E}\big{[}(u_{\zeta_{s}}^{n})^{2}\big{]}ds$
$\displaystyle\leq(C_{g}e^{C_{g}K})^{2}\int_{0}^{K}\mathbb{E}\bigg{[}\int_{\zeta_{s}}^{T}(u_{t}^{n-1})^{2}e^{-\rho
t}d{D}_{t}\bigg{]}ds=(C_{g}e^{C_{g}K})^{2}\int_{0}^{K}\int_{t_{1}}^{K}\mathbb{E}[(u_{\zeta_{t_{2}}}^{n-1})^{2}]dt_{2}dt_{1}$
$\displaystyle\cdots$
$\displaystyle\leq(C_{g}e^{C_{g}K})^{n}\int_{0}^{K}\int_{t_{1}}^{K}\cdots\int_{t_{n-1}}^{K}\mathbb{E}[(u_{\zeta_{t_{n}}}^{1})^{2}]dt_{n}dt_{n-1}\cdots
dt_{1}\leq(C_{g}e^{C_{g}K})^{n}\mathbb{E}\big{[}\sup_{t\in[0,T]}(u_{t}^{1})^{2}\big{]}\frac{K^{n}}{n!}.$

The above estimate implies that $\\{(y^{n},z^{n})\\}$ is a Cauchy sequence in
$\mathcal{S}_{2}\times\mathcal{M}_{2}$, due to the fact that
$y^{1}\in\mathcal{S}_{2}$ and so $\mathbb{E}[\sup_{t\in[0,T]}(u_{t}^{1})^{2}]$
is finite. It follows that
$\lim_{n\rightarrow\infty}(y^{n},z^{n})=(y,z)\in\mathcal{B}_{2}$. By standard
arguments, we show that $(y,z)$ solves BSDE (2.5) with parameter
$(T,\xi_{T},g,D)$. The uniqueness of $(y,z)$ follows standard arguments with
comparison theorem (see Proposition A.5 below). ∎

###  A.3. Step 2. Comparison theorem under general condition and priori bounds

We introduce the notion of subsolutions and supersolutions under the general
condition.

######  Definition A.4.

For $(\tau,\xi_{\tau},D)$ that satisfies Condition A, an adapted process
$(Y_{t},Z_{t})_{t\geq 0}$ is called a supersolution (resp. a subsolution) for
BSDE with parameter $(\tau,\xi_{\tau},g,D)$ if

$\displaystyle
Y_{t}=\xi_{\tau}+\int_{t}^{\tau}g(s,Y_{s})dD_{s}-\int_{t}^{\tau}Z_{s}dW_{s}+\int_{t}^{\tau}dA_{s}\bigg{(}\text{resp.}-\int_{t}^{\tau}dA_{s}\bigg{)},$

where $(A_{t})_{t\geq 0}$ is a right continuous left limit process in
${\mathcal{}P}$; equivalently,
$(Y_{t}+\int_{0}^{t\wedge\tau}g(s,Y_{s})dD_{s})_{t\geq 0}$ is a local
supermartingale (resp. a local submartingale). A solution of BSDE with
parameter $(\tau,\xi_{\tau},g,D)$ is both a supersolution and a subsolution.

The following result shows a comparison result for BSDE (2.5) without
restricting ourselves to the conditions given by Theorem A.3.

######  Proposition A.5.

Let $(Y,Z)$ (resp. $(\widetilde{Y},\widetilde{Z})$) be a supersolution (resp.
subsolution) to BSED (2.5) with parameter $(\tau,\xi_{\tau},g,D)$ (resp.
$(\tau,\tilde{\xi}_{\tau},\tilde{g},D)$). Assuming that both $(Y,Z)$ and
$(\widetilde{Y},\widetilde{Z})$ are of class $\mathcal{B}_{2}$ and one of the
following conditions hold:

  * (i)

$\tilde{g}(t,Y_{t})\leq g(t,Y_{t})$ and $\tilde{g}(t,y)$ is nonincreasing in
$y$, $d\mathbb{P}\times dt$-a.e., or

  * (ii)

$\tilde{g}(t,\widetilde{Y}_{t})\leq g(t,\widetilde{Y}_{t})$ and $g(t,y)$ is
nonincreasing in $y$, $d\mathbb{P}\times dt$-a.e.;

if $\xi_{\tau}\geq\tilde{\xi}_{\tau}$, then $\widetilde{Y}_{t}\leq Y_{t}$ for
any $t\geq 0$ a.s.

###### Proof.

The proof can be straightforwardly generalized from the approach presented in
[40, Theorem 2.6]; hence, we chose to omit the details here. ∎

We introduce two auxiliary utility processes that act as lower and upper
bounds for the solutions of our target BSDE. The well-defined nature of these
processe is established in the following lemmas.

######  Lemma A.6.

For $(\tau,\xi_{\tau},D)$ that satisfies Condition A, there exists a process
$Y\in{\mathcal{}P}$ such that $\mathbb{E}[\sup_{t\geq
0}\overline{Y}_{t}^{\frac{2}{1-R}}]<\infty$ (in particular
$(\overline{Y}_{t})_{t\geq 0}\in\mathcal{S}_{2}$) given by

(A.6)
$\displaystyle\overline{Y}_{t}:=\bigg{(}\mathbb{E}_{t}\bigg{[}\int_{t}^{\tau}e^{-\rho
s}dD_{s}+(\xi_{\tau})^{\frac{1}{1-R}}\bigg{]}\bigg{)}^{1-R},\quad t\geq 0,$

where recall that $R\in(0,1)$ is the risk-aversion coefficient in
$g_{\text{EZ}}$ defined in (2.3).

###### Proof.

Observe that $(\overline{Y})^{\frac{1}{1-R}}$ is the classical dividend-paying
utility process, it follows directly that $(\overline{Y})^{\frac{1}{1-R}}$ is
the unique solution to BSDE with parameter
$(\tau,(\xi_{\tau})^{\frac{1}{1-R}},\overline{g},D)$, with
$\overline{g}(t,v)=e^{-\rho t}$. Since $(\tau,\xi_{\tau},D)$ satisfies
Condition A, $\mathbb{E}[\sup_{t\geq
0}\overline{Y}_{t}^{\frac{2}{1-R}}]<\infty$ by Doob’s maximal inequality.
$\overline{Y}\in\mathcal{S}_{2}$ follows as a consequence of Jensen’s
inequality. ∎

The triplet $(T,\xi_{T},D)$ satisfying conditions in Theorem A.3 is a special
case of the above Lemma.

######  Lemma A.7.

For the triplet $(T,\xi_{T},D)$ satisfying conditions in Theorem A.3, there
exists $(\underline{Y}_{t})_{t\in[0,T]}\in\mathcal{S}_{2}$ satisfying

(A.7)
$\displaystyle\underline{Y}_{t}:=\mathbb{E}_{t}\bigg{[}\int_{t}^{T}\underline{g}(s,\underline{Y}_{s})dD_{s}+\xi_{T}\bigg{]},\quad\underline{g}(t,v)=e^{-\rho
t}(1-Rv),\quad t\in[0,T].$

In addition, $\underline{Y}_{t}\geq e^{-R\|\int_{0}^{T}e^{-\rho
s}dD_{s}\|_{\infty}}\mathbb{E}_{t}[\xi_{T}]$ for all $t\in[0,T]$ a.s.

###### Proof.

Since aggregator $\underline{g}(\cdot,y)$ satisfies condition in Theorem A.3,
we obtain the existence and uniqueness of the solution to BSDE (2.5) with
parameters $(\xi,\underline{g},T,D)$. It follows from the comparison theorem
(Proposition A.5) that $\underline{Y}_{t}\geq\tilde{Y}_{t}$, where
$(\tilde{Y},\tilde{Z})$ is the unique solution to BSDE

(A.8) $\displaystyle\tilde{Y}_{t}=\xi_{T}-\int_{t}^{T}e^{-\rho
s}R\tilde{Y}_{s}dD_{s}-\int_{t}^{\tau}\tilde{Z}_{t}dW_{s}.$

It follows that
$\underline{Y}_{t}\geq\tilde{Y}_{t}=\mathbb{E}_{t}[e^{-R\int_{t}^{T}e^{-\rho
s}dD_{s}}\xi_{T}]\geq e^{-R\|\int_{0}^{T}e^{-\rho
s}dD_{s}\|_{\infty}}\mathbb{E}_{t}[\xi_{T}]$, $t\in[0,T]$ a.s. ∎

###  A.4. Step 3. EZ aggregator: fixed horizon and bounded controls

We start by approximating the EZ aggregator $g_{\text{EZ}}$ in (2.3) using
Lipschitz aggregators that satisfy conditions • ‣ A.3 and • ‣ A.3. The proof
of the construction of approximation is straightforward with direct
calculation, so we omit it in the following lemma.

######  Lemma A.8.

The sequence of aggregators

(A.9) $\displaystyle
g_{n}(\omega,t,y)=\inf_{x>0}\\{g_{\text{EZ}}(\omega,t,x)+nRe^{-\rho
t}(x-y)\\}$

is well-defined for each $n\geq 1$, satisfying • ‣ A.3 and $d\mathbb{P}\times
dt$-a.s.:

  * •

Monotonicity in $n$: $\forall~{}y\geq 0$, $g_{n}(\omega,t,y)$ increase in $n$.

  * •

Lipshcitz condition • ‣ A.3: $\forall~{}y_{1},y_{2}\geq 0$,
$|g_{n}(\omega,t,y_{1})-g_{n}(\omega,t,y_{2})|\leq nRe^{-\rho
t}|y_{1}-y_{2}|$.

  * •

Monotonicity in $y$: $g_{n}$ is nonincreasing in $y$.

######  Theorem A.9.

For $(T,D)$ satisfying conditions in Theorem A.3 and
$\mathbb{E}[(\xi_{T})^{\frac{2}{1-R}}]<\infty$ and there is a real constant
$C>0$ such that $\xi_{T}>C$ $\mathbb{P}$-a.s. Then the BSDE (2.5) with
parameter $(T,\xi_{T},g_{\text{EZ}},D)$ has a unique solution
$(y_{t},z_{t})_{t\in[0,T]}\in\mathcal{B}_{2}$.

###### Proof.

For fixed $(\omega,t)$, consider the sequence $g_{n}(\omega,t,y)$ associated
to $g_{\text{EZ}}$ in (2.3) constructed by Lemma A.8. It follows by Theorem
A.3 that, for each $n\geq 1$, the BSDE with parameter $(T,\xi_{T},g_{n},D)$
has a unique solution
$(y_{t}^{n},z_{t}^{n})_{t\in[0,\tau]}\in\mathcal{B}_{2}$. Lemma A.8 • ‣ A.8
and • ‣ A.8, Proposition A.5 yield that for $n\geq 1$, $y_{t}^{1}(\omega)\leq
y_{t}^{n}(\omega)\leq y_{t}^{n+1}$, $d\mathbb{P}\times dt$-a.s. Note that
$y^{1}$ is identical to $\underline{Y}$ defined in Lemma A.7 (with current
triplet $(T,\xi_{T},D)$), since $g_{1}$ and $\underline{g}$ are identical.
Under the condition that $\xi_{T}\in L^{2}(\mathcal{F}_{T})_{+}$ and