# Optimal Risk Policies and Periodic Dividend Strategies for an Insurance
Company

Mark Kelbert
Harold A. Moreno-Franco
Nikolai P. Pogorelov

###### Abstract

We study the problem of optimal risk policies and dividend strategies for an
insurance company operating under the constraint that the timing of
shareholder payouts is governed by the arrival times of a Poisson process.
Concurrently, risk control is continuously managed through proportional
reinsurance. Our analysis confirms the optimality of a periodic-classical
barrier strategy for maximizing the expected net present value until the first
instance of bankruptcy across all admissible periodic-classical strategies.

Keywords: Proportional reinsurance, optimal periodic dividend, cheap and non-
cheap reinsurance.

2000 MR Subject Classification 60J60; 91B70; 91G50; 93E20.

labelinglabel

Optimal Risk Policies and Periodic Dividend Strategies for an Insurance
Company
Mark Kelbert111M. Kelbert and H. A. Moreno-Franco made equal contributions to
this paper, while N. P. Pogorelov worked only on Subsection 3.2 of this paper
as part of his master’s thesis under the supervision of H. A. Moreno-
Franco.,222M. Kelbert was supported by the Russian Science Foundation Grant
number 23-21-00052.

Harold A. Moreno-Franco11footnotemark: 1,333Corresponding author:
<EMAIL_ADDRESS>
Nikolai P. Pogorelov11footnotemark: 1
National Research University Higher School of Economics
Department of Statistics and Data Analysis
Laboratory of Stochastic Analysis and its Applications
Moscow, Russian Federation

##  1 Introduction

In the field of actuarial risk theory, significant research has focused on
optimal risk policies and/or dividend strategies, especially within the
context of a diffusion approximation of the classical risk model (e.g., see
[3, 6, 12]). This approach has been widely adopted when both the dividend
stream and the risk exposure are continuously modelled over time (e.g., see
[5, 4, 9, 13]).

This is exemplified in the works of Højgaard and Taksar [5], as well as Taksar
[13], where the authors delve into the control of risk through proportional
reinsurance and the management of cumulative dividend payments via singular
control. In both papers, it is demonstrated that a two-barrier strategy is
optimal for maximizing expected net discounted dividend payouts until the
first time of bankruptcy. One barrier signifies the threshold for maximum
risk-taking by the insurance company, while the other designates the level at
which the excess between the surplus process and the threshold is paid out to
the shareholders (after discounts) if the surplus process exceeds this
threshold. The authors explored this problem in the context of cheap and non-
cheap reinsurance, respectively.

In light of the recent wide-ranging exploration of periodic observation models
in insurance literature (e.g., [1, 2, 10]), this paper extends the
aforementioned problem with a periodic dividend constraint. This modification
dictates that the decision times for paying out to the shareholders are
governed by the arrivals times of a Poisson process with intensity $\gamma>0$,
while the control of risk is continuously managed by proportional reinsurance
as in [5, 13]. The primary objective of this paper is to establish that the
periodic-classical barrier strategy is optimal for the problem of maximizing
the expected net present value (NPV) until the first time of bankruptcy across
all the admissible periodic-classical strategies.

A notable contrast with the results obtained by Højgaard and Taksar [5], and
Taksar [13] is that the level of maximum risk-taking by the company, as
determined by the optimal barrier, does not consistently remain lower than the
level at which payouts should be made, as these levels are dependent on
$\gamma$. Moreover, for certain values of $\gamma$, the strategy of making the
payout equal to the entire amount of assets that the company holds at the
first arrival of the Poisson process is considered optimal.

The remainder of this document is organized as follows: in Section 2, we
consistently formulate the problem of optimal risk policies and periodic
dividend strategies. This section introduce the HJB equation associated with
the value function as defined in (2.4); see (2.5). Subsequently, in Subsection
2.1, the classical-barrier strategy is well-defined. Assuming that the
solution $v$ to (2.5) satisfies certain properties, we verify that $v$ is also
a solution to the non-linear partial differential system (NLPDS) given by
(2.2)–(2.10); see Proposition 3.9. Following this, a verification theorem is
presented; see Theorem 2.3. Moving to Section 3, by solving the NLPDS
(2.2)–(2.10), we obtain different explicit solutions to the HJB equation
(2.5). These explicit solutions are dependent on the various scenarios,
covering cases of cheap reinsurance and non-cheap reinsurance; see
Propositions 3.4, 3.8 and 3.13. Additionally, explicit forms of the levels
associated with the optimal periodic-classical barrier strategy proposed in
Subsection 2.1 are provided. Subsequently, in Section 4, we discuss the
behaviour of our solutions in relation to the parameter $\gamma$.
Specifically, as $\gamma$ tends to infinity, we recover the value functions
for the case of singular dividend strategies obtained by Hôgaard and Taksar
[5], and Taksar [13], in the cases of cheap and non-cheap reinsurance,
respectively. In conclusion, Section 5 presents numerical examples for further
illustration.

##  2 Formulation of the problem

In order to mitigate the risk, the cedent (or primary insurance) enters into a
contract with a reinsurance firm. According to this contract, at time
$\mathpzc{t}$, the insurer assumes the responsibility to pay a fraction
$u^{\pi}_{\mathpzc{t}}$ of each claim, while the reinsurance firm takes care
of the remaining payment $1-u^{\pi}_{\mathpzc{t}}$ for each claim. The
starting point is the classical Cramér-Lundberg ruin problem with reinsurance
and dividend payments, i.e. the capital of the primary insurer is governed by

$X^{\pi}_{\mathpzc{t}}=x_{0}+\lambda a[1+\eta]\mathpzc{t}-\lambda
a[1-u^{\pi}_{\mathpzc{t}}][1+\mu]\mathpzc{t}-u^{\pi}_{\mathpzc{t}}\sum_{j=1}^{N^{\lambda}_{\mathpzc{t}}}Y_{j}-L^{\pi}_{\mathpzc{t}},\quad\text{for}\
\mathpzc{t}\geq 0,$ (2.1)

where $X_{0}=x_{0}>0$ is the initial surplus of the insurance firm. Here
$\\{N^{\lambda}_{\mathpzc{t}},\mathpzc{t}\geq 0\\}$ stands for a Poisson
process of rate $\lambda$ , and $\\{Y_{j}\\}_{j\geq 1}$ represents i.i.d. non-
negative random variables corresponding to the sizes of individual claims,
which are also independent of $N^{\lambda}$. The premiums are paid
continuously at constant rates $\lambda a[1+\eta]$ and $\lambda
a[1-u^{\pi}_{\mathpzc{t}}][1+\mu]$ based on the expected value principle,
where $u^{\pi}_{\mathpzc{t}}\in[0,1]$, $a=\mathbbm{E}[Y_{1}]$,
$\bar{\sigma}^{2}=\operatorname{Var}[Y_{1}]$, and safety loadings
$\mu\geq\eta>0$ are involved. The payment rate $\eta$ represents the input
payment rate of the insurance firm, while $\mu$ is the insurer’s payment rate
to the reinsurance firm as per the contract mentioned earlier. Additionally,
$L^{\pi}_{\mathpzc{t}}$ is a non-decreasing random function that quantifies
the total amount of dividends paid by the primary insurer to the shareholders
and will be formally defined later. When $\mu>\eta$, we call it non-cheap
reinsurance, and in the case of $\mu=\eta$, it is usually referred as a cheap
reinsurance.

Notice that (2.1) can be rewritten in a different form as follows

$X^{\pi}_{\mathpzc{t}}=x_{0}+\lambda
a\\{\eta-[1-u^{\pi}_{\mathpzc{t}}]\mu\\}\mathpzc{t}+u^{\pi}_{\mathpzc{t}}Z_{\mathpzc{t}}-L^{\pi}_{\mathpzc{t}},\quad\text{for}\
\mathpzc{t}\geq 0,$

with $Z_{\mathpzc{t}}\raisebox{0.4pt}{$:$}=\lambda
a\mathpzc{t}-\sum_{j=1}^{N^{\lambda}_{\mathpzc{t}}}Y_{j}$. Then
$\mathbbm{E}[Z_{\mathpzc{t}}]=0$,
$\operatorname{Var}[Z_{\mathpzc{t}}]=\bar{\sigma}^{2}\lambda\mathpzc{t}$ and
for$\lambda\gg 1$, $a\ll 1$ and $\lambda a\approx 1$, the process
$Z_{\mathpzc{t}}$ is well approximated by the Brownian motion $\sigma
W_{\mathpzc{t}}$, with
$\sigma\raisebox{0.4pt}{$:$}=\bar{\sigma}\lambda^{1/2}$; for more details, see
for example [7]. This approximation is valid under the condition that the
random variables $\\{Y_{j}\\}_{j\geq 1}$ are stochastically small, and it
serves as the intuitive background for the following model

$\displaystyle\mathrm{d}X^{\pi}_{\mathpzc{t}}$
$\displaystyle=\\{\eta-[1-u^{\pi}_{\mathpzc{t}}]\mu\\}\mathrm{d}\mathpzc{t}+\sigma
u^{\pi}_{\mathpzc{t}}\mathrm{d}W_{\mathpzc{t}}-\mathrm{d}L^{\pi}_{\mathpzc{t}},\quad\text{for}\
\mathpzc{t}\geq 0,$ (2.2)

which is the main object of our study. Additionally, let us consider that the
insurance firm can pay out to its shareholders at the arrival times
$\mathcal{T}^{\gamma}\raisebox{0.4pt}{$:$}=\\{T_{i}:i\geq 1\\}$ of a Poisson
process
$N^{\gamma}\raisebox{0.4pt}{$:$}=\\{N^{\gamma}_{\mathpzc{t}}:\mathpzc{t}\geq
0\\}$ with intensity $\gamma>0$. Then, in order that (2.2) is well-defined,
consider that $W,\ N^{\gamma}$ are defined on a filtered probability space
$(\Omega,\mathcal{F},\mathbbm{H},\mathbbm{P})$, where
$\mathbbm{H}=\\{\mathcal{H}_{\mathpzc{t}}\\}_{\mathpzc{t}\geq 0}$ is the
right-continuous complete filtration generated by $(W,N^{\gamma})$, and the
processes $W$ and $N^{\gamma}$ are independent. Then, a dividend strategy
$L^{\pi}$ has the following form

$\displaystyle
L_{\mathpzc{t}}^{\pi}=\int_{[0,\mathpzc{t}]}\nu^{\pi}_{\mathpzc{s}}\mathrm{d}N^{\gamma}_{\mathpzc{s}},\quad\text{for
$\mathpzc{t}\geq 0$,}$ (2.3)

for some non-negative càglàd process
$\nu^{\pi}\raisebox{0.4pt}{$:$}={\\{\nu^{\pi}_{\mathpzc{t}}:\mathpzc{t}\geq
0\\}}$ adapted to the filtration $\mathbbm{H}$, where $\nu^{\pi}_{T_{i}}$
represents the payout made at time $T_{i}$, with
$T_{i}\in\mathcal{T}^{\gamma}$.

Therefore, under the strategy $\pi=(u^{\pi},L^{\pi})$, the surplus process
$X^{\pi}\raisebox{0.4pt}{$:$}=\\{X^{\pi}_{\mathpzc{t}}:\mathpzc{t}\geq 0\\}$
evolves as in (2.2), with $X^{\pi}_{0}=x_{0}$. Let $\mathcal{A}$ be the
collection of strategies $\pi=(u^{\pi},L^{\pi})$ such that $u^{\pi},\
\nu^{\pi}$ are $\mathbbm{H}$-adapted such that $u^{\pi}\in[0,1]$ and
$\nu^{\pi}$ satisfies (2.3) and $0\leq\nu^{\pi}_{\mathpzc{t}}\leq
X^{\pi}_{\mathpzc{t}-}$. We say that a strategy $\pi$ is admissible if
$\pi\in\mathcal{A}$.

For each $\pi\in\mathcal{A}$, the expected NPV is given by

$V^{\pi}(x)\raisebox{0.4pt}{$:$}=\mathbbm{E}_{x}\left[\int_{0}^{\tau^{\pi}}\operatorname{e}^{-\delta\mathpzc{t}}\mathrm{d}L^{\pi}_{\mathpzc{t}}\right],\quad
x\geq 0.$

Here $\mathbbm{E}_{x}$ represents the conditional expectation given an initial
surplus $x$, $\delta>0$ denotes the discount rate, and
$\tau^{\pi}\raisebox{0.4pt}{$:$}=\inf\\{\mathpzc{t}>0:X^{\pi}_{\mathpzc{t}}<0\\}$
is the first time of bankruptcy. Then, our goal is to find the value function
$V$ for the problem, which is

$V(x)\raisebox{0.4pt}{$:$}=\sup_{\pi\in\mathcal{A}}V^{\pi}(x),\quad x\geq 0.$
(2.4)

Furthermore, we aim to find an optimal strategy $\pi^{*}\in\mathcal{A}$ for
which the expected NPV $V^{\pi^{*}}$ coincides with $V$, if such a strategy
exists.

Using standard dynamic programming arguments, we can associate the value
function $V$ with the Hamilton Jacobi Bellman (HJB) equation given by

$\displaystyle\begin{split}\max_{0\leq u\leq 1,\,0\leq\xi\leq
x}\big{\\{}\mathcal{L}^{u}v+\gamma[\xi+v(x-\xi)-v(x)]\big{\\}}&=0\ \text{for}\
x>0,\\\ \text{s.t.}\ v(0)&=0.\end{split}$ (2.5)

with

$\mathcal{L}^{u}v(x)\raisebox{0.4pt}{$:$}=\frac{1}{2}[\sigma
u]^{2}v^{\prime\prime}(x)+a\lambda[\eta-[1-u]\mu]v^{\prime}(x)-\delta v(x).$
(2.6)

######  Remark 2.1.

It is noteworthy that optimal risk policies and periodic dividend strategies
with debt liabilities can be reformulated as a non-cheap reinsurance problem.
This is achieved by a simply reparametrization in (2.2) and (2.6).
Essentially, this implies that the results obtained in this paper are
applicable to this scenario as well, provided that we replace $\eta$ with
$\mu-\hat{\delta}$, where $\hat{\delta}$ represents the debt repayment.

###  2.1 Choosing an optimal strategy and verification theorem

To find an optimal solution to the problem (2.4), we focus our study on the
type of periodic-classical barrier strategies. We define a periodic-classical
barrier strategy $\pi^{b}=(u^{b},L^{b})\in\mathcal{A}$ as having dividend
payments given by

$\nu^{b}_{T_{i}}=\max\\{X^{b}_{T_{i}}-b,0\\},\quad\text{for}\
T_{i}\in\mathcal{T}^{\gamma},$

where
$L^{b}_{\mathpzc{t}}=\int_{[0,\mathpzc{t}]}\nu^{b}_{\mathpzc{s}}\mathrm{d}N^{\gamma}_{\mathpzc{s}}$
and $X^{b}\raisebox{0.4pt}{$:$}=X^{\pi^{b}}$ is the solution to the SDE

$\displaystyle\begin{split}\mathrm{d}X^{b}_{\mathpzc{t}}&=\Big{\\{}\eta-\Big{[}1-u^{b}_{\mathpzc{t}}\Big{]}\mu\Big{\\}}\mathrm{d}\mathpzc{t}+\sigma
u^{b}_{\mathpzc{t}}\mathrm{d}W_{\mathpzc{t}}-\mathrm{d}L^{b}_{\mathpzc{t}},\quad\text{for}\
\mathpzc{t}>0,\\\ X^{b}_{0}&=x.\end{split}$

To choose our candidate periodic-classical barrier strategy
$\pi^{b_{\gamma}}=(u^{b_{\gamma}},L^{b_{\gamma}})$ to be optimal, we need to
first look at solving equation (2.5). This is the same as finding a solution
to the non-linear partial differential system (NLPDS) (2.2)–(2.10), under
certain assumptions on $v$.

######  Proposition 2.2.

Let $v$ be a $\operatorname{C}^{2}$-continuous, concave and increasing
function on $(0,\infty)$ such that ${v^{\prime}}/{v^{\prime\prime}}$ is
decreasing on $(0,\infty)$ and

$\begin{split}&x_{\gamma}\raisebox{0.4pt}{$:$}=\inf\bigg{\\{}x>0:-\frac{\mu
v^{\prime}(x)}{\sigma^{2}v^{\prime\prime}(x)}\geq 1\bigg{\\}}<\infty,\\\
&b_{\gamma}\raisebox{0.4pt}{$:$}=\inf\\{x>0:v^{\prime}(x)<1\\}<\infty.\end{split}$
(2.7)

Then, $v$ is a solution to (2.5) if and only if $v$ is a solution to

$\displaystyle-\frac{\mu^{2}[v^{\prime}(x)]^{2}}{2\sigma^{2}v^{\prime\prime}(x)}+[\eta-\mu]v^{\prime}(x)-\delta
v(x)\hskip 156.49014pt$
$\displaystyle+\gamma\\{x-b_{\gamma}+v(b_{\gamma})-v(x)\\}\mathbbm{1}_{\\{x-b_{\gamma}>0\\}}$
$\displaystyle=0\quad\text{for}\ x\in(0,x_{\gamma}),$ (2.8)
$\displaystyle\frac{1}{2}\sigma^{2}v^{\prime\prime}(x)+\eta
v^{\prime}(x)-\delta
v(x)+\gamma\\{x-b_{\gamma}+v(b_{\gamma})-v(x)\\}\mathbbm{1}_{\\{x-b_{\gamma}>0\\}}$
$\displaystyle=0\quad\text{for}\ x\in(x_{\gamma},\infty),$ (2.9)
$\displaystyle\text{s.t.}\ v(0)$ $\displaystyle=0.$ (2.10)

###### Proof.

Since $v$ is $\operatorname{C}^{2}$-continuous, increasing and concave on
$(0,\infty)$, which satisfies $b_{\gamma}\in[0,\infty)$, it is easy to check
that

$\max_{0\leq\xi\leq x}\\{\xi+v(x-\xi)-v(x)\\}=\begin{cases}0&\text{if}\ x\leq
b_{\gamma},\\\ x-b_{\gamma}+v(b_{\gamma})-v(x)&\text{if}\
x>b_{\gamma}.\end{cases}$ (2.11)

Substituting (2.11) into (2.5), it yields

$\displaystyle\begin{split}\max_{0\leq u\leq
1}\big{\\{}\mathcal{L}^{u}v(x)\big{\\}}+\gamma\\{x-b_{\gamma}+v(b_{\gamma})-v(x)\\}\mathbbm{1}_{\\{x-b_{\gamma}>0\\}}&=0\
\text{for}\ x>0,\\\ \text{s.t.}\ v(0)&=0.\end{split}$ (2.12)

Maximizing w.r.t. $u$ in (2.12) we see that the maximizer operator $u^{*}$ is
given by

$u^{*}(x)\raisebox{0.4pt}{$:$}=\begin{cases}-\frac{\mu
v^{\prime}(x)}{\sigma^{2}v^{\prime\prime}(x)},&\text{if}\
x\in(0,x_{\gamma}),\\\ 1&\text{if}\ x\in(x_{\gamma},\infty),\end{cases}$
(2.13)

which is non-negative and increasing on $(0,x_{\gamma})$ since
$v^{\prime}/v^{\prime\prime}$ is decreasing. Therefore, applying (2.13) in
(2.12), we get that the double implication presented in the proposition above
holds. ∎

####  2.1.1  Verification Theorem

Consider $v$ as a solution to the HJB equation (2.5), satisfying the
hypotheses in Proposition 2.2. Then, the SDE

$\displaystyle\begin{split}\mathrm{d}X^{b_{\gamma}}_{\mathpzc{t}}&=\Big{\\{}\eta-\Big{[}1-u^{b_{\gamma}}_{\mathpzc{t}}\Big{]}\mu\Big{\\}}\mathrm{d}\mathpzc{t}+\sigma
u^{b_{\gamma}}_{\mathpzc{t}}\mathrm{d}W_{\mathpzc{t}}-\mathrm{d}L^{b_{\gamma}}_{\mathpzc{t}},\quad\text{for}\
\mathpzc{t}>0.\\\ X^{b_{\gamma}}_{0}&=x,\end{split}$ (2.14)

with

$\displaystyle\begin{split}&u^{b_{\gamma}}_{\mathpzc{t}}\raisebox{0.4pt}{$:$}=\
\begin{cases}-\frac{\mu
v^{\prime}(X^{b_{\gamma}}_{\mathpzc{t}})}{\sigma^{2}v^{\prime\prime}(X^{b_{\gamma}}_{\mathpzc{t}})}&\text{if}\
X^{b_{\gamma}}_{\mathpzc{t}}\in(0,x_{\gamma}),\\\ 1&\text{if}\
X^{b_{\gamma}}_{\mathpzc{t}}\in(x_{\gamma},\infty),\end{cases}\\\
&L^{b_{\gamma}}_{\mathpzc{t}}=\int_{[0,\mathpzc{t}]}\nu^{b_{\gamma}}_{\mathpzc{s}}\mathrm{d}N^{\gamma}_{\mathpzc{s}}\
\text{where}\
\nu^{b_{\gamma}}_{\mathpzc{t}}=\max\\{X^{b_{\gamma}}_{\mathpzc{t}}-b_{\gamma},0\\},\end{split}$
(2.15)

admits a unique solution
$X^{b_{\gamma}}=\\{X^{b_{\gamma}}_{\mathpzc{t}}:\mathpzc{t}\geq 0\\}$; see
[8]. Thus, the candidate strategy to be optimal is given by the barrier
strategy $\pi^{b_{\gamma}}=(u^{b_{\gamma}},L^{b_{\gamma}})$ where
$u^{b_{\gamma}}$ and $L^{b_{\gamma}}$ are as in (2.15). We will verify below
that its expected NPV

$V^{b_{\gamma}}(x)\raisebox{0.4pt}{$:$}=V^{\pi^{b_{\gamma}}}(x)=\mathbbm{E}_{x}\left[\int_{0}^{\tau^{b_{\gamma}}}\operatorname{e}^{-\delta\mathpzc{t}}\mathrm{d}L^{b_{\gamma}}_{\mathpzc{t}}\right],\quad\text{for}\
\mathpzc{t}\geq 0,$ (2.16)

with
$\tau^{b_{\gamma}}\raisebox{0.4pt}{$:$}=\inf\\{\mathpzc{t}>0:X^{b_{\gamma}}_{\mathpzc{t}}<0\\}$,
satisfies the HJB equation (2.5), proving that $V^{b_{\gamma}}=V$. From here,
it follows that the strategy $\pi^{b_{\gamma}}$ is indeed an optimal strategy.

######  Theorem 2.3 (Verification theorem).

Let $v$ be a solution to the HJB equation (2.5), satisfying the hypotheses in
Proposition 2.2. Then, $v$ agrees with the value function $V$ defined in
(2.4). Furthermore, the strategy
$\pi^{b_{\gamma}}=(u^{b_{\gamma}},L^{b_{\gamma}})$ given by (2.15) is an
optimal strategy and $V^{b_{\gamma}}=v=V$.

###### Proof.

Let $\pi=(u^{\pi},L^{\pi})$ be in $\mathcal{A}$, where $X^{\pi}$ evolves as in
(2.2), with $X^{\pi}_{0}=x$. For each $n\in\mathbb{N}$ such that
$v^{\prime}(n)<1$, take $\tau_{n}$ in the following manner
$\tau_{n}\raisebox{0.4pt}{$:$}=\inf\\{t>0:X^{\pi}_{\mathpzc{t}}>n\ \text{or}\
X^{\pi}_{\mathpzc{t}}<1/n\\}$. Notice that there is an $\bar{n}\in\mathbb{N}$
such that $v^{\prime}(n)<1$ for each $n>\bar{n}$, because of (2.7). Using
integration by parts and Itô’s formula in
$\operatorname{e}^{-\delta[\mathpzc{t}\wedge\tau_{n}]}v(X^{\pi}_{\mathpzc{t}\wedge\tau_{n}})$
(since $X^{\pi}$ is a semi-martingale and $v$ is a
$\operatorname{C}^{2}$-continuous function); see [11, Ch. II, Theorem 33], and
taking into account (2.2), it gives

$\displaystyle\operatorname{e}^{-\delta[\mathpzc{t}\wedge\tau_{n}]}v(X^{\pi}_{\mathpzc{t}\wedge\tau_{n}})-v(x)$
$\displaystyle=\int_{0}^{\mathpzc{t}\wedge\tau_{n}}\operatorname{e}^{-\delta\mathpzc{s}}\mathcal{L}^{u^{\pi}_{\mathpzc{s}}}v(X^{\pi}_{\mathpzc{s}})\mathrm{d}\mathpzc{s}+M_{\mathpzc{t}\wedge\tau_{n}}$
$\displaystyle\quad+\int_{[0,\mathpzc{t}\wedge\tau_{n}]}\operatorname{e}^{-\delta\mathpzc{s}}[v(X^{\pi}_{\mathpzc{s}-}-\nu^{\pi}_{\mathpzc{s}})-v(X^{\pi}_{\mathpzc{s}-})]\mathrm{d}N^{\gamma}_{\mathpzc{s}},$
(2.17)

where
$M_{\mathpzc{t}\wedge\tau_{n}}\raisebox{0.4pt}{$:$}=\sigma\int_{0}^{\mathpzc{t}\wedge\tau_{n}}\operatorname{e}^{-\delta\mathpzc{s}}u^{\pi}_{\mathpzc{s}}v^{\prime}(X^{\pi}_{\mathpzc{s}})\mathrm{d}W_{\mathpzc{s}}$.
Notice that

$\displaystyle\int_{[0,\mathpzc{t}\wedge\tau_{n}]}\operatorname{e}^{-\delta\mathpzc{s}}[v(X^{\pi}_{\mathpzc{s}-}-\nu^{\pi}_{\mathpzc{s}})-v(X^{\pi}_{\mathpzc{s}-})]\mathrm{d}N^{\gamma}_{\mathpzc{s}}$
$\displaystyle\quad=-\int_{[0,\mathpzc{t}\wedge\tau_{n}]}\operatorname{e}^{-\delta\mathpzc{s}}\nu^{\pi}_{\mathpzc{s}}\mathrm{d}N^{\gamma}_{\mathpzc{s}}+\int_{0}^{\mathpzc{t}\wedge\tau_{n}}\operatorname{e}^{-\delta\mathpzc{s}}\gamma[\nu^{\pi}_{\mathpzc{s}}+v(X^{\pi}_{\mathpzc{s}-}-\nu^{\pi}_{\mathpzc{s}})-v(X^{\pi}_{\mathpzc{s}-})]\mathrm{d}\mathpzc{s}+H_{\mathpzc{t}\wedge\tau_{n}},$

with
$H_{\mathpzc{t}\wedge\tau_{n}}\raisebox{0.4pt}{$:$}=\int_{[0,\mathpzc{t}\wedge\tau_{n}]}\operatorname{e}^{-\delta\mathpzc{s}}[\nu^{\pi}_{\mathpzc{s}}+v(X^{\pi}_{\mathpzc{s}-}-\nu^{\pi}_{\mathpzc{s}})-v(X^{\pi}_{\mathpzc{s}-})]\mathrm{d}\widetilde{N}^{\gamma}_{\mathpzc{s}}$
and
$\widetilde{N}\raisebox{0.4pt}{$:$}=N^{\gamma}_{\mathpzc{s}}-\gamma\mathpzc{s}$
is the compensated Poisson process. Applying (2.1.1) in (2.1.1), it gives

$\displaystyle\operatorname{e}^{-\delta[\mathpzc{t}\wedge\tau_{n}]}v(X^{\pi}_{\mathpzc{t}\wedge\tau_{n}})-v(x)$
$\displaystyle=\int_{0}^{\mathpzc{t}\wedge\tau_{n}}\operatorname{e}^{-\delta\mathpzc{s}}\Big{\\{}\mathcal{L}^{u^{\pi}_{\mathpzc{s}}}v(X^{\pi}_{\mathpzc{s}})+\gamma[\nu^{\pi}_{\mathpzc{s}}+v(X^{\pi}_{\mathpzc{s}-}-\nu^{\pi}_{\mathpzc{s}})-v(X^{\pi}_{\mathpzc{s}-})]\Big{\\}}\mathrm{d}\mathpzc{s}$
$\displaystyle\quad-\int_{[0,\mathpzc{t}\wedge\tau_{n}]}\operatorname{e}^{-\delta\mathpzc{s}}\nu^{\pi}_{\mathpzc{s}}\mathrm{d}N^{\gamma}_{\mathpzc{s}}+M_{\mathpzc{t}\wedge\tau_{n}}+H_{\mathpzc{t}\wedge\tau_{n}}.$
(2.19)

Considering that for each $x>0$, $\tilde{u}\in[0,1]$ and
$\tilde{\xi}\in[0,x]$,

$\displaystyle\mathcal{L}^{\tilde{u}}v(x)+\gamma[\tilde{\xi}+v(x-\tilde{\xi})-v(x)]\leq\max_{0\leq
u\leq 1,\,0\leq\xi\leq
x}\big{\\{}\mathcal{L}^{u}v(x)+\gamma[\xi+v(x-\xi)-v(x)]\big{\\}}=0,$

it implies that

$\displaystyle\operatorname{e}^{-\delta[\mathpzc{t}\wedge\tau_{n}]}v(X^{\pi}_{\mathpzc{t}\wedge\tau_{n}})-v(x)$
$\displaystyle\leq-\int_{[0,\mathpzc{t}\wedge\tau_{n}]}\operatorname{e}^{-\delta\mathpzc{s}}\nu^{\pi}_{\mathpzc{s}}\mathrm{d}N^{\gamma}_{\mathpzc{s}}+M_{\mathpzc{t}\wedge\tau_{n}}+H_{\mathpzc{t}\wedge\tau_{n}}.$
(2.20)

Since $v^{\prime}\in(0,v^{\prime}(0+))$ is decreasing on $(0,\infty)$, observe
that
$\operatorname{e}^{-\delta\mathpzc{s}}u^{\pi}_{\mathpzc{s}}v^{\prime}(X^{\pi}_{\mathpzc{s}})\leq
v^{\prime}(1/n)<\infty$ for $\mathpzc{s}\in(0,\mathpzc{t}\wedge\tau_{n})$.
Meanwhile, since $v$ is concave, it follows that
$\operatorname{e}^{-\delta\mathpzc{s}}[\nu^{\pi}_{\mathpzc{s}}+v(X^{\pi}_{\mathpzc{s}-}-\nu^{\pi}_{\mathpzc{s}})-v(X^{\pi}_{\mathpzc{s}-})]\leq
X^{\pi}_{\mathpzc{s}-}[1-v^{\prime}(n)]<n[1-v^{\prime}(n)]<\infty$ for
$\mathpzc{s}\in[0,\mathpzc{t}\wedge\tau_{n}]$. Thus, it follows that the
processes $\\{M_{t\wedge\tau_{n}}:t\geq 0\\}$ and
$\\{H_{t\wedge\tau_{n}}:t\geq 0\\}$ are zero-mean
$\mathbbm{P}_{x}$-martingales. Taking expectation in (2.20), it gives

$\displaystyle v(x)$
$\displaystyle\geq\mathbbm{E}_{x}\Big{[}\operatorname{e}^{-\delta[\mathpzc{t}\wedge\tau_{n}]}v(X^{\pi}_{\mathpzc{t}\wedge\tau_{n}})\Big{]}+\mathbbm{E}_{x}\bigg{[}\int_{[0,\mathpzc{t}\wedge\tau_{n}]}\operatorname{e}^{-\delta\mathpzc{s}}\nu^{\pi}_{\mathpzc{s}}\mathrm{d}N^{\gamma}_{\mathpzc{s}}\bigg{]}.$
(2.21)

Taking into account again the concavity property of $v$, it is known that
$v(X^{\pi}_{\mathpzc{t}\wedge\tau_{n}})\leq
K[1+X^{\pi}_{\mathpzc{t}\wedge\tau_{n}}]$, for some positive constant $K$
independent of $n$. Moreover, notice that

$\displaystyle\mathbbm{E}_{x}[X^{\pi}_{\mathpzc{t}\wedge\tau_{n}}]$
$\displaystyle=\mathbbm{E}_{x}\bigg{[}x+\int_{0}^{\mathpzc{t}\wedge\tau_{n}}\\{\eta-[1-u^{\pi}_{\mathpzc{s}}]\mu\\}\mathrm{d}\mathpzc{s}+\sigma\int_{0}^{\mathpzc{t}\wedge\tau_{n}}u^{\pi}_{\mathpzc{t}}\mathrm{d}W_{\mathpzc{t}}-\int_{[0,\mathpzc{t}]}\nu^{\pi}_{\mathpzc{s}}\mathrm{d}N^{\gamma}_{\mathpzc{s}}\bigg{]}$
$\displaystyle=\mathbbm{E}_{x}\bigg{[}x+\int_{0}^{\mathpzc{t}\wedge\tau_{n}}\\{\eta-[1-u^{\pi}_{\mathpzc{s}}]\mu\\}\mathrm{d}\mathpzc{s}-\int_{[0,\mathpzc{t}]}\nu^{\pi}_{\mathpzc{s}}\mathrm{d}N^{\gamma}_{\mathpzc{s}}\bigg{]}\leq
x+\eta\mathbbm{E}_{x}[\mathpzc{t}\wedge\tau_{n}].$

Then, it follows that
$\lim_{t,n\rightarrow\infty}\mathbbm{E}_{x}\Big{[}\operatorname{e}^{-\delta[\mathpzc{t}\wedge\tau_{n}]}v(X^{\pi}_{\mathpzc{t}\wedge\tau_{n}})\Big{]}\leq\lim_{t,n\rightarrow\infty}K\mathbbm{E}_{x}[\operatorname{e}^{-\delta[\mathpzc{t}\wedge\tau_{n}]}\\{1+x+\eta[\mathpzc{t}\wedge\tau_{n}]\\}]=0$.
Letting $\mathpzc{t},n\rightarrow\infty$ in (2.21), considering the previous
identity and using dominated convergence theorem, we get that

$v(x)\geq\mathbbm{E}_{x}\bigg{[}\int_{0}^{\tau^{\pi}}\operatorname{e}^{-\delta\mathpzc{s}}\nu^{\pi}_{\mathpzc{s}}\mathrm{d}N^{\gamma}_{\mathpzc{s}}\bigg{]}=V^{\pi}(x),$
(2.22)

due to $\tau_{n}\xlongrightarrow[n\rightarrow\infty]{}\tau^{\pi}$
$\mathbbm{P}_{x}$-a.s.. Therefore, $v(x)\geq V(x)$ for $x\geq 0$. Now,
consider the strategy $\pi^{b_{\gamma}}=(u^{b_{\gamma}},L^{b_{\gamma}})$ given
by (2.15) where the controlled process $X^{b_{\gamma}}$ associated to this
strategy evolves as (2.14). Defining
$\bar{\tau}_{n}=\inf\\{t>0:X^{b_{\gamma}}_{\mathpzc{t}}>n\ \text{or}\
X^{b_{\gamma}}_{\mathpzc{t}}<1/n\\}$, it can be observed that (2.1.1) holds,
but with the substitution of $X^{\pi}$, $u^{\pi}$ and $\tau^{\pi}_{n}$ with
$X^{b_{\gamma}}$, $u^{b_{\gamma}}$ and $\tau^{b_{\gamma}}_{n}$, because of
$\pi^{b_{\gamma}}\in\mathcal{A}$. Considering that $v$ satisfies (2.5) and
(2.11)–(2.13), we have that

$\displaystyle\operatorname{e}^{-\delta[\mathpzc{t}\wedge\bar{\tau}_{n}]}v(X^{b_{\gamma}}_{\mathpzc{t}\wedge\tau_{n}})-v(x)$
$\displaystyle=-\int_{[0,\mathpzc{t}\wedge\bar{\tau}_{n}]}\operatorname{e}^{-\delta\mathpzc{s}}\nu^{\pi}_{\mathpzc{s}}\mathrm{d}N^{\gamma}_{\mathpzc{s}}+M_{\mathpzc{t}\wedge\bar{\tau}_{n}}+H_{\mathpzc{t}\wedge\bar{\tau}_{n}},$

with
$M_{\mathpzc{t}\wedge\bar{\tau}_{n}}=\sigma\int_{0}^{\mathpzc{t}\wedge\bar{\tau}_{n}}\operatorname{e}^{-\delta\mathpzc{s}}u^{b_{\gamma}}_{\mathpzc{s}}v^{\prime}(X^{b_{\gamma}}_{\mathpzc{s}})\mathrm{d}W_{\mathpzc{s}}$
and
$H_{\mathpzc{t}\wedge\bar{\tau}_{n}}=\int_{[0,\mathpzc{t}\wedge\bar{\tau}_{n}]}\operatorname{e}^{-\delta\mathpzc{s}}[\nu^{b_{\gamma}}_{\mathpzc{s}}+v(X^{b_{\gamma}}_{\mathpzc{s}-}-\nu^{b_{\gamma}}_{\mathpzc{s}})-v(X^{b_{\gamma}}_{\mathpzc{s}-})]\mathrm{d}\widetilde{N}^{\gamma}_{\mathpzc{s}}$.
From here and by similar arguments seen in (2.21)–(2.22), it follows that

$v(x)=\mathbbm{E}_{x}\bigg{[}\int_{0}^{\tau^{b_{\gamma}}}\operatorname{e}^{-\delta\mathpzc{s}}\nu^{b_{\gamma}}_{\mathpzc{s}}\mathrm{d}N^{\gamma}_{\mathpzc{s}}\bigg{]}=V^{b_{\gamma}}(x)\leq
V(x)\qquad\text{for}\ \mathpzc{t}\geq 0.$

Therefore, we conclude that $V^{b_{\gamma}}=v=V$, and that $\pi^{b_{\gamma}}$
is an optimal strategy. ∎

##  3 Constructing a solution to the HJB equation and describing the optimal
barrier strategy

In this section, we will solve the NLPDS (2.2)–(2.10), providing an explicit
solution to the HJB equation (2.5). We will consider the cases of non-cheap
($\mu>\eta$) and cheap ($\mu=\eta$) reinsurance separately. The proofs of the
results presented in this section can be found in the Appendix.

###  3.1 Non-cheap reinsurance

####  3.1.1 The case $\boldsymbol{\mu\geq 2\eta}$

Let us consider the scenario where the insurance company takes the maximum
risk at any time first, which implies that $u^{*}\equiv 1$ on the interval
$(0,\infty)$, and

$\max_{0\leq u\leq
1}\big{\\{}\mathcal{L}^{u}v(x)\big{\\}}=\frac{1}{2}\sigma^{2}v^{\prime\prime}(x)+\eta
v^{\prime}(x)-\delta v(x),\quad\text{for}\ x>0.$

To ensure that this scenario holds, we must have $\mu\geq 2\eta$ as we will
show in Proposition 3.4. Recall that $\mu>\eta$.

Now, let us proceed by finding a solution $v_{\gamma,b}$ to (3.1), which
depends on the parameter $b>0$.

######  Proposition 3.1.

Let $b>0$ be fixed. Then, a solution of the NLPDS

$\displaystyle\begin{split}\frac{1}{2}\sigma^{2}v^{\prime\prime}(x)+\eta
v^{\prime}(x)-\delta
v(x)+\gamma\\{(x-b)+v(b)-v(x)\\}\mathbbm{1}_{\\{x-b\\}>0\\}}&=0,\quad\text{for}\
x>0,\\\ \text{s.t.}\ v(0)&=0,\end{split}$ (3.1)

is a $\operatorname{C}^{2}$-continuous function, whose form is given by

$v_{\gamma,b}(x)=\begin{cases}c_{1,1}(b)h_{1}(x)&\text{for}\ x\in(0,b),\\\
c_{1,2}(b)\operatorname{e}^{\lambda_{\gamma}[x-b]}+\frac{\gamma}{\gamma+\delta}\bigg{[}x-b+v_{\gamma,b}(b)+\frac{\eta}{\gamma+\delta}\bigg{]}&\text{for}\
x\in[b,\infty),\end{cases}$ (3.2)

where $\lambda_{\gamma}$ is the negative root of

$\frac{\sigma^{2}}{2}r^{2}+\eta r-(\delta+\gamma)=0,$ (3.3)

$h_{1}$ is defined as

$h_{1}(x)=\operatorname{e}^{\theta_{+}x}-\operatorname{e}^{\theta_{-}x}\quad\text{for}\
x\in(0,\infty),$ (3.4)

with $\theta_{+},\ \theta_{-}$ as the positive and negative root,
respectively, of

$\frac{\sigma^{2}}{2}r^{2}+\eta r-\delta=0,$ (3.5)

and

$\displaystyle
c_{1,1}(b)=\frac{\frac{\gamma}{\gamma+\delta}\big{[}1-\frac{\eta\lambda_{\gamma}}{\gamma+\delta}\big{]}}{h^{\prime}_{1}(b)-\frac{\delta\lambda_{\gamma}}{\gamma+\delta}h_{1}(b)}\quad\text{and}\quad
c_{1,2}(b)=\frac{1}{\gamma+\delta}\bigg{[}\delta
c_{1,1}(b)h_{1}(b)-\frac{\gamma\eta}{\gamma+\delta}\bigg{]}.$ (3.6)

Supposing that $b_{\gamma}$ as in (2.7) belongs in $(0,\infty)$, and
$v_{\gamma,b_{\gamma}}$ given by (3.2) (when $b=b_{\gamma}$) is both concave
and increasing on $(0,\infty)$, it can be immediately observed that
$v^{\prime}_{\gamma,b_{\gamma}}$ is decreasing on $(0,\infty)$. Then, in order
to determine when

$\displaystyle v^{\prime}_{\gamma,b_{\gamma}}(b_{\gamma}-)=1$ (3.7)

holds, calculating the first derivative of (3.2) on $(0,b_{\gamma})$, we
obtain that (3.7) is equivalent to verifying whether

$\displaystyle
g_{1}(b_{\gamma})=\frac{1}{\delta\lambda_{\gamma}}\bigg{[}\delta+\frac{\gamma\eta\lambda_{\gamma}}{\gamma+\delta}\bigg{]}$
(3.8)

is true, where

$g_{1}(b)\raisebox{0.4pt}{$:$}=\frac{h_{1}(b)}{h^{\prime}_{1}(b)}\quad\text{for}\
b>0.$ (3.9)

To further analyse (3.8), let us first establish some properties of
$g_{1}(b)$, with $b>0$.

######  Lemma 3.2.

Let $g_{1}$ be as in (3.9). Then, $g_{1}$ is strictly increasing on
$(0,\infty)$ satisfying

$\lim_{b\downarrow
0}g_{1}(b)=0\quad\text{and}\quad\lim_{b\uparrow\infty}g_{1}(b)=\frac{1}{\theta_{+}}.$

So, by (3.8) and Lemma 3.2, we deduce that there exists a $b_{\gamma}>0$ such
that (3.7) holds if and only if

$0<\frac{1}{\delta\lambda_{\gamma}}\bigg{[}\delta+\frac{\gamma\eta\lambda_{\gamma}}{\gamma+\delta}\bigg{]}<\frac{1}{\theta_{+}}$
(3.10)

is true. Taking $f_{1}$ as follows

$\displaystyle f_{1}(\gamma)$
$\displaystyle\raisebox{0.4pt}{$:$}=\frac{1}{\lambda_{\gamma}\delta}\bigg{[}\delta+\frac{\lambda_{\gamma}\eta\gamma}{\delta+\gamma}\bigg{]}=\frac{\eta\gamma}{\delta[\delta+\gamma]}-\frac{\sigma^{2}}{\eta+\sqrt{\eta^{2}+2\sigma^{2}(\delta+\gamma)}}\quad\text{for}\
\gamma>0,$ (3.11)

the next lemma, provides the necessary conditions on $\gamma$ so that (3.10)
is satisfied.

######  Lemma 3.3.

Let $f_{1}$ be as in (3.11). Then $f_{1}$ is strictly increasing on
$(0,\infty)$ satisfying

$\lim_{\gamma\downarrow
0}f_{1}(\gamma)=\frac{1}{\theta_{-}}\quad\text{and}\quad\lim_{\gamma\uparrow\infty}f_{1}(\gamma)=\frac{\eta}{\delta}.$
(3.12)

Since $\frac{1}{\theta_{-}}<\frac{\eta}{\delta}<\frac{1}{\theta_{+}}$ and by
Lemma 3.3, we conclude that there exists a unique $\gamma_{0}>0$ such that
$f_{1}(\gamma_{0})=0$. Thus, for any $\gamma\in(\gamma_{0},\infty)$ fixed,
(3.10) is satisfied, and there exists a unique $b_{\gamma}\in(0,\infty)$ where
(3.7) holds. On the other hand, calculating the first and second derivatives
of $v_{\gamma,b_{\gamma}}$ on the interval $(0,b_{\gamma})$, notice that

$\displaystyle\lim_{x\downarrow 0}\frac{\mu
v^{\prime}_{\gamma,b_{\gamma}}(x)}{\sigma^{2}v^{\prime\prime}_{\gamma,b_{\gamma}}(x)}=\lim_{x\downarrow
0}\frac{\mu
h^{\prime}_{1}(x)}{\sigma^{2}h^{\prime\prime}_{1}(x)}=-\frac{\mu}{2\eta}.$
(3.13)

Then, assuming
$v^{\prime}_{\gamma,b_{\gamma}}/v^{\prime\prime}_{\gamma,b_{\gamma}}$ is
decreasing on $(0,\infty)$, by (3.13), it follows immediately that if $\mu\geq
2\eta$, $u^{*}\equiv 1$, with $u^{*}$ as in (2.13).

In the case that $\gamma\in(0,\gamma_{0}]$, we take $b_{\gamma}=0$, and the
solution to (2.5), when $b=0$, is given by

$v_{\gamma,0}(x)=-\frac{\gamma\eta}{[\gamma+\delta]^{2}}\operatorname{e}^{\lambda_{\gamma}x}+\frac{\gamma}{\gamma+\delta}\bigg{\\{}x+\frac{\eta}{\gamma+\delta}\bigg{\\}}\quad\text{for}\
x>0.$ (3.14)

Then, calculating the first and second derivatives of (3.14), it gives that
$\lim_{x\downarrow 0}\frac{\mu
v^{\prime}_{\gamma,0}(x)}{\sigma^{2}v^{\prime\prime}_{\gamma,0}(x)}=\frac{\mu}{\sigma^{2}}\Big{\\{}\frac{1}{\lambda_{\gamma_{0}}}-\frac{[\gamma_{0}+\delta]}{\eta[\lambda_{\gamma_{0}}]^{2}}\Big{\\}}=-\frac{\mu}{2\eta}$.
We have again that $u^{*}\equiv 1$ if $\mu\geq 2\eta$ and
$v^{\prime}_{\gamma,0}/v^{\prime\prime}_{\gamma,0}$ is decreasing on
$(0,\infty)$.

######  Proposition 3.4.

Let $\mu\geq 2\eta$ and $\gamma_{0}=f^{-1}_{1}(0)$.

  1. (i)

If $\gamma>\gamma_{0}$, then (3.2) is a $\operatorname{C}^{2}$-continuous,
increasing and concave solution to (2.5), when

$b=b_{\gamma}=g_{1}^{-1}(f_{1}(\gamma)).$ (3.15)

Furthermore,
$v^{\prime}_{\gamma,b_{\gamma}}/v^{\prime\prime}_{\gamma,b_{\gamma}}$ is
decreasing on $(0,\infty)$ and $x_{\gamma}=0$.

  2. (ii)

If $\gamma<\gamma_{0}$, then (3.14) is a $\operatorname{C}^{2}$-continuous,
increasing and concave solution to (2.5), with linear growth, when
$b=b_{\gamma}=0$. Furthermore,
$v^{\prime}_{\gamma,0}/v^{\prime\prime}_{\gamma,0}$ is decreasing on
$(0,\infty)$ and $x_{\gamma}=0$.

Optimal strategy. By Theorem 2.3 and Proposition 3.4, we conclude that if
$\mu\geq 2\eta$ and $\gamma>\gamma_{0}$, the optimal strategy is given by
(2.15), where $x_{\gamma}=0$ and $b_{\gamma}$ is determined by (3.15). The
controlled process $X^{b_{\gamma}}$ is given as the solution to the SDE

$\displaystyle\mathrm{d}X^{b_{\gamma}}_{\mathpzc{t}}$
$\displaystyle=\eta\mathrm{d}\mathpzc{t}+\sigma\mathrm{d}W_{\mathpzc{t}}-\mathrm{d}L^{b_{\gamma}}_{\mathpzc{t}},\quad\text{for}\
\mathpzc{t}>0.$ $\displaystyle X^{b_{\gamma}}_{0}$ $\displaystyle=x.$

Moreover, the value function $V$ and the expected NPV $V^{b_{\gamma}}$
(defined in (2.4) and (2.16) respectively) are given by (3.2). In this
context, the optimal strategy for the insurance company is to take the maximum
risk while the payout $\max\\{X^{b_{\gamma}}_{T_{i}}-b_{\gamma},0\\}$ is made
at time $T_{i}<\tau^{b_{\gamma}}$.

In the case that $\mu\geq 2\eta$ and $\gamma\in(0,\gamma_{0}]$, by Theorem 2.3
and Proposition 3.4, we get that $x_{\gamma}=b_{\gamma}=0$. Furthermore, the
optimal strategy for the company in this scenario is to take the maximum risk
while making the payout equal to the entire amount of the company’s assets at
the first arrival of the Poisson process $N^{\gamma}$. Additionally,
$V^{b_{\gamma}}$ and $V$ are given by (3.14).

####  3.1.2 The case $\boldsymbol{\mu<2\eta}$

Now, assuming $\mu<2\eta$, we will examine the scenarios where $x_{\gamma}\leq
b_{\gamma}$ and $x_{\gamma}>b_{\gamma}$. Recall that $x_{\gamma}$ indicates
the level at which maximum retention is reached, whereas $b_{\gamma}$
represents the comparison level at which the insurer determines if payouts to
shareholders are appropriate at the arrival times $\mathcal{T}^{\gamma}$.

Given a $b>0$ fixed, we will determine an explicit solution $v_{\gamma,b}$ to
the NLPDS (3.20)–(3.22) and (3.39)–(3.41) first. To derive this solution, we
will consider that $v_{\gamma,b}$ must have linear growth, and there exists an
$x_{b}\in(0,\infty)$ such that

$u^{*}_{b}(x)=\begin{cases}-\frac{\mu
v^{\prime}_{\gamma,b}(x)}{\sigma^{2}v^{\prime\prime}_{\gamma,b}(x)},&\text{if}\
x\in(0,x_{b}),\\\ 1&\text{if}\ x\in[x_{b},\infty).\end{cases}$ (3.16)

Subsequently, we will derive sufficient conditions regarding the parameter
$\gamma$ to ensure the existence of $b_{\gamma}$, which is defined in (2.7).
After, it will be shown that
$v^{\prime}_{\gamma,b_{\gamma}}/v^{\prime\prime}_{\gamma,b_{\gamma}}$ is
decreasing on $(0,\infty)$ satisfying

$\lim_{x\downarrow 0}\frac{\mu
v^{\prime}_{\gamma,b_{\gamma}}(x)}{\sigma^{2}v^{\prime\prime}_{\gamma,b_{\gamma}}(x)}>-1,$
(3.17)

which is enough to obtain the existence of $x_{\gamma}$ given in (2.7). The
conditions will be presented in Propositions 3.8 and 3.13.

For the case where $x_{\gamma}\leq b_{\gamma}$, we take

$\bar{x}\raisebox{0.4pt}{$:$}=\frac{c_{2,1}}{1+\delta\bar{\eta}}\ln[c_{2,2}]+\frac{\bar{\eta}[2\eta-\mu]}{2[1+\delta\bar{\eta}]},$
(3.18)

where

$\displaystyle\bar{\eta}\raisebox{0.4pt}{$:$}=\frac{2\sigma^{2}}{\mu^{2}},\
c_{2,1}\raisebox{0.4pt}{$:$}=\frac{\bar{\eta}[\mu-\eta]}{1+\delta\bar{\eta}},\
\text{and}\
c_{2,2}\raisebox{0.4pt}{$:$}=\frac{\delta\bar{\eta}\mu+2\eta-\mu}{2\delta\bar{\eta}[\mu-\eta]}.$
(3.19)

######  Proposition 3.5.

Let $\mu<2\eta$. If $b>\bar{x}$ is fixed, a solution to the NLPDS

$\displaystyle-\frac{\mu^{2}[v^{\prime}(x)]^{2}}{2\sigma^{2}v^{\prime\prime}(x)}+[\eta-\mu]v^{\prime}(x)-\delta
v(x)$ $\displaystyle=0\quad\text{for}\ x\in(0,x_{b}),$ (3.20)
$\displaystyle\frac{1}{2}\sigma^{2}v^{\prime\prime}(x)+\eta
v^{\prime}(x)-\delta v(x)+\gamma\\{x-b+v(b)-v(x)\\}\mathbbm{1}_{\\{x-b>0\\}}$
$\displaystyle=0\quad\text{for}\ x\in(x_{b},\infty),$ (3.21)
$\displaystyle\text{s.t.}\ v(0)$ $\displaystyle=0,$ (3.22)

is a $\operatorname{C}^{2}$-continuous function on $(0,\infty)$, whose form is
given by

$\displaystyle
v_{\gamma,b}(x)=\begin{cases}c_{2,1}\operatorname{e}^{-\mathpzc{x}_{\,1}^{-1}(x)}\big{[}\operatorname{e}^{[1+\delta\bar{\eta}][M(b)+\mathpzc{x}_{\,1}^{-1}(x)]}-1\big{]}&\text{if}\
x\in(0,x_{b}),\\\ \vspace{-0.4cm}&\\\
\operatorname{e}^{M(b)}c_{2,2}^{-1/[1+\delta\bar{\eta}]}h_{2}(x-x_{b})&\text{if}\
x\in(x_{b},b),\\\ \vspace{-0.4cm}&\\\
c_{2,3}(b)\operatorname{e}^{\lambda_{\gamma}[x-b]}+\frac{\gamma}{\gamma+\delta}\Big{[}x-b+v_{\gamma,b}(b)+\frac{\eta}{\delta+\gamma}\Big{]}&\text{if}\
x\in(b,\infty),\end{cases}$ (3.23)

where $\mathpzc{x}_{\,1}^{-1}$ is the inverse function of

$\mathpzc{x}_{\,1}(z)=k_{2,1}\operatorname{e}^{[1+\delta\bar{\eta}][z+M(b)]}+c_{2,1}z+k_{2,2}(b),\
\text{for}\ z\in[-M(b),\bar{z}(b)],$ (3.24)

with

$\displaystyle\begin{split}k_{2,1}&\raisebox{0.4pt}{$:$}=\frac{\delta\bar{\eta}c_{2,1}}{1+\delta\bar{\eta}},\quad
k_{2,2}(b)\raisebox{0.4pt}{$:$}=c_{2,1}\bigg{[}M(b)-\frac{\delta\bar{\eta}}{1+\delta\bar{\eta}}\bigg{]},\\\
\bar{z}(b)&\raisebox{0.4pt}{$:$}=\frac{1}{1+\delta\bar{\eta}}\ln[c_{2,2}]-M(b),\end{split}$
(3.25)

the function $h_{2}$ is taken as

$\displaystyle
h_{2}(x)\raisebox{0.4pt}{$:$}=\frac{1}{\theta_{+}-\theta_{-}}\Big{[}a_{1}\operatorname{e}^{\theta_{+}x}-a_{2}\operatorname{e}^{\theta_{-}x}\Big{]},$
(3.26)

with $\theta_{+}$, $\theta_{-}$, $\lambda_{\gamma}$ as in Proposition 3.1, and

$\displaystyle\begin{split}x_{b}&=\bar{x},\quad
a_{1}\raisebox{0.4pt}{$:$}=1-\theta_{-}\frac{[2\eta-\mu]}{2\delta},\quad
a_{2}\raisebox{0.4pt}{$:$}=1-\theta_{+}\frac{[2\eta-\mu]}{2\delta},\\\
M(b)&=\ln\left[\frac{\frac{\gamma}{\gamma+\delta}-\frac{\lambda_{\gamma}\eta\gamma}{[\delta+\gamma]^{2}}}{c_{2,2}^{-1/[1+\delta\bar{\eta}]}\Big{[}h^{\prime}_{2}(b-\bar{x})-\frac{\lambda_{\gamma}\delta}{\delta+\gamma}h_{2}(b-\bar{x})\Big{]}}\right],\\\
c_{2,3}(b)&\raisebox{0.4pt}{$:$}=\frac{\operatorname{e}^{M(b)}\delta
c_{2,2}^{-1/[1+\delta\bar{\eta}]}}{\delta+\gamma}h_{2}(b-\bar{x})-\frac{\eta\gamma}{[\delta+\gamma]^{2}}.\end{split}$
(3.27)

######  Remark 3.6.

Observe that $\mathpzc{x}_{\,1}$, given in (3.24), is a positive and
increasing function on $(0,\bar{x})$. Then, by the construction of
$v_{\gamma,b}$, which is given by (3.23), it obtains immediately that
$v_{\gamma,b}$ is concave and increasing on $(0,\bar{x})$, since
$v_{\gamma,b}$ satisfies

$v^{\prime}_{\gamma,b}(x)=\operatorname{e}^{-\mathpzc{x}_{\,1}^{-1}(x)}\quad\text{and}\quad
v^{\prime\prime}_{\gamma,b}(x)=-\frac{\operatorname{e}^{-\mathpzc{x}_{\,1}^{-1}(x)}}{\mathpzc{x}_{\,1}^{\prime}(\mathpzc{x}_{\,1}^{-1}(x))},\quad\text{for}\
x\in(0,\bar{x}).$ (3.28)

It implies that $u^{*}_{b}(x)<1$ for $x\in(0,\bar{x})$, and $u^{*}_{b}(x)=1$
at $x=\bar{x}$. For a more detail exposition, see Subsection A.2.

Taking the first derivative in (3.23) on $(\bar{x},b)$, we see that (3.7) is
achieved if and only if

$\displaystyle
g_{2}(b_{\gamma}-\bar{x})=\frac{1}{\lambda_{\gamma}\delta}\bigg{[}\delta+\frac{\lambda_{\gamma}\eta\gamma}{\delta+\gamma}\bigg{]}$
(3.29)

is true, where

$g_{2}(b)=\frac{h_{2}(b)}{h^{\prime}_{2}(b)}\quad\text{for}\ b\in(0,\infty).$
(3.30)

In order to guarantee the veracity of (3.29), let us see some properties of
the function $g_{2}$ first.

######  Lemma 3.7.

Let $g_{2}$ be as in (3.30). Then, $g_{2}$ is increasing on $(0,\infty)$,
satisfying

$\lim_{b\downarrow
0}g_{2}(b)=\frac{2\eta-\mu}{2\delta}\quad\text{and}\quad\lim_{b\uparrow\infty}g_{2}(b)=\frac{1}{\theta_{+}}.$

By Lemma 3.7, we get that $b\mapsto g_{2}(b-\bar{x})$ is increasing on
$(\bar{x},\infty)$ satisfying
$\lim_{b\downarrow\bar{x}}g_{2}(b-\bar{x})=\frac{2\eta-\mu}{2\delta}$ and
$\lim_{b\uparrow\infty}g_{2}(b-\bar{x})=\frac{1}{\theta_{+}}$. Then, we have
that (3.29) is equivalent to verify that
$\frac{2\eta-\mu}{2\delta}\leq\frac{1}{\lambda_{\gamma}\delta}\big{[}\delta+\frac{\lambda_{\gamma}\eta\gamma}{\delta+\gamma}\big{]}<\frac{1}{\theta_{+}}$,
which is true only for
$\gamma>\gamma_{1}\raisebox{0.4pt}{$:$}=f^{-1}_{1}\big{(}\frac{2\eta-\mu}{2\delta}\big{)}$,
where $f^{-1}_{1}$ is the inverse of $f_{1}$ given in (3.11), because of
$\frac{1}{\theta_{-}}<\frac{2\eta-\mu}{2\delta}<\frac{\eta}{\delta}<\frac{1}{\theta_{+}}$
and Lemma 3.3. Thus, for $\gamma>\gamma_{1}$ fixed, there exists a unique
$b_{\gamma}\in(0,\infty)$ that satisfies (3.29). By the seen above, we get the
following proposition.

######  Proposition 3.8.

If $\mu<2\eta$ and $\gamma>\gamma_{1}$, then (3.23) is a
$\operatorname{C}^{2}$-continuous, increasing and concave solution to (2.5),
with

$b=b_{\gamma}=\bar{x}+g^{-1}_{2}(f_{1}(\gamma)).$ (3.31)

Furthermore,
$v^{\prime}_{\gamma,b_{\gamma}}/v^{\prime\prime}_{\gamma,b_{\gamma}}$ is
decreasing on $(0,\infty)$ satisfying (3.17). Then, $x_{\gamma}=\bar{x}$.

For the case that $b_{\gamma}<x_{\gamma}$, we will analyse the behaviour of
the solution $v_{\gamma,b_{\gamma}}$ to the NLPDS (3.39)–(3.41) for
$\gamma\in(0,\gamma_{1})$, as shown in Proposition 3.13. For
$\beta\in\mathbbm{R}$ fixed, take $H_{\beta}$ and $\bar{f}_{\beta}$ as

$\displaystyle H_{\beta}(z)$
$\displaystyle\raisebox{0.4pt}{$:$}=\int_{\operatorname{e}^{-\beta}}^{z}\frac{1}{y^{2}g(y)}\mathrm{d}y,\quad\text{for}\
z>\operatorname{e}^{-\beta},$ (3.32) $\displaystyle\bar{f}_{\beta}(z)$
$\displaystyle\raisebox{0.4pt}{$:$}=c_{3,1}(\beta)[G(z)-G(\operatorname{e}^{-\beta})]-\bar{\eta}[\mu-\eta]\bigg{[}G(z)H_{\beta}(z)-\int_{\operatorname{e}^{-\beta}}^{z}\frac{G(y)}{y^{2}g(y)}\mathrm{d}y\bigg{]},\quad\text{for}\
z>\operatorname{e}^{-\beta},$ (3.33)

with

$c_{3,1}(\beta)\raisebox{0.4pt}{$:$}=\frac{\sigma^{2}}{\mu\alpha_{\gamma}g(\alpha_{\gamma})}+\bar{\eta}[\mu-\eta]H_{\beta}(\alpha_{\gamma})\quad\text{and}\quad\alpha_{\gamma}\raisebox{0.4pt}{$:$}=\frac{\gamma+\delta}{\gamma}\bigg{[}1+\frac{\mu}{\sigma^{2}\lambda_{\gamma}}\bigg{]}.$
(3.34)

Here, $G$ is a gamma accumulative distribution function with parameters
$(\bar{\eta}[\delta+\gamma]+1,[\gamma\bar{\eta}]^{-1}])$, and its density
function is represented by $g$ which has the following form

$\displaystyle
g(x)=\frac{[\bar{\eta}\gamma]^{\bar{\eta}[\delta+\gamma]+1}}{\Gamma(\bar{\eta}[\delta+\gamma]+1)}x^{\bar{\eta}[\delta+\gamma]}\operatorname{e}^{-\gamma\bar{\eta}x}\quad\
\text{for}\ x>0,$ (3.35)

where $\Gamma(\cdot)$ is the gamma function. Observe that $\alpha_{\gamma}$ is
a positive constant due to $2\eta-\mu>0$, and $\bar{f}_{\beta}$ is an
invertible function for $\beta\in(0,\infty)$. For more details, refer to
Subsection A.3. Additionally, for each $\gamma\in(0,\gamma_{1})$ fixed, it is
worth noticing that $\beta\mapsto H_{\beta}(\alpha_{\gamma})$ is well-defined
on $(0,\infty)$, because of $\alpha_{\gamma}>1$; as stated in Lemma 3.11.(i).

Proposition 3.13 presents a solution $v_{\gamma,b}$ to the NPDS (3.39)–(3.41),
which depends on $b>0$ satisfying

$\displaystyle[1+\delta\bar{\eta}]\bigg{[}1+\frac{b}{c_{2,1}}\bigg{]}\in([1+\delta\bar{\eta}],\bar{g}(0+)],$
(3.36)

with

$\displaystyle\bar{g}(\beta)$
$\displaystyle\raisebox{0.4pt}{$:$}=\frac{\operatorname{e}^{-\beta}g(\operatorname{e}^{-\beta})}{c_{2,1}}c_{3,1}(\beta)+\ln\bigg{[}\frac{1}{\delta\bar{\eta}}\bigg{\\{}\frac{\operatorname{e}^{-\beta}g(\operatorname{e}^{-\beta})}{c_{2,1}}c_{3,1}(\beta)-1\bigg{\\}}\bigg{]},\quad\text{for}\
\beta\in(0,\underline{$b$}),$ (3.37)

and
$\underline{$b$}\raisebox{0.4pt}{$:$}=\inf\\{\beta>0:\bar{g}(\beta)<0\\}<\infty$.
The function $\bar{g}$ is a decreasing function on $(0,\underline{$b$})$, as
is verified in Subsection A.3. Also $\bar{g}(0+)>1+\delta\bar{\eta}$ holds if
and only if
$\frac{1}{\delta\bar{\eta}}\big{\\{}\frac{g(1)}{c_{2,1}}c_{3,1}(0)-1\big{\\}}>1$,
which, using (3.19) and (3.32)–(3.34), is rewritten as follows

$\displaystyle\frac{\sigma^{2}\operatorname{e}^{\bar{\eta}\gamma\alpha_{\gamma}}}{\mu[\alpha_{\gamma}]^{\bar{\eta}[\delta+\gamma]+1}}+\bar{\eta}[\mu-\eta]\int_{1}^{\alpha_{\gamma}}\frac{\operatorname{e}^{\bar{\eta}\gamma
y}}{y^{\bar{\eta}[\delta+\gamma]+2}}\mathrm{d}y>\bar{\eta}[\mu-\eta]\operatorname{e}^{\bar{\eta}\gamma}.$
(3.38)

######  Proposition 3.9.

Let $\mu<2\eta$ and $\gamma\in(0,\gamma_{1})$ satisfying $\eqref{e12}$. If
$b>0$ is such that (3.36) holds, a solution to the NLPDS

$\displaystyle-\frac{\mu^{2}[v^{\prime}(x)]^{2}}{2\sigma^{2}v^{\prime\prime}(x)}+[\eta-\mu]v^{\prime}(x)-\delta
v(x)+\gamma\\{x-b+v(b)-v(x)\\}\mathbbm{1}_{\\{x-b>0\\}}$
$\displaystyle=0\quad\text{for}\ x\in(0,x_{b}),$ (3.39)
$\displaystyle\frac{1}{2}\sigma^{2}v^{\prime\prime}(x)+\eta
v^{\prime}(x)-[\delta+\gamma]v(x)+\gamma\\{x-b+v(b)\\}$
$\displaystyle=0\quad\text{for}\ x\in(x_{b},\infty),$ (3.40)
$\displaystyle\text{s.t.}\ v(0)$ $\displaystyle=0,$ (3.41)

is a $\operatorname{C}^{2}$-continuous, whose form is given by

$\displaystyle
v_{\gamma,b}(x)=\begin{cases}c_{2,1}\operatorname{e}^{-\mathpzc{x}_{\,1}^{-1}(x)}\Big{[}\operatorname{e}^{[1+\delta\bar{\eta}][M_{\gamma}+\mathpzc{x}_{\,1}^{-1}(x)]}-1\Big{]}&\text{if}\
x\in(0,b),\\\ \vspace{-0.4cm}&\\\
\frac{\operatorname{e}^{-\mathpzc{x}_{\,2}^{-1}(x)}}{\delta+\gamma}\Big{\\{}\frac{g(\operatorname{e}^{\mathpzc{x}_{\,2}^{-1}(x)})\operatorname{e}^{\mathpzc{x}_{\,2}^{-1}(x)}}{\bar{\eta}}&\\\
\vspace{-0.4cm}&\\\
\times[c_{3,1}(M_{2})-\bar{\eta}[\mu-\eta]H_{M_{2}}(\operatorname{e}^{\mathpzc{x}_{\,2}^{-1}(x)})]-[\mu-\eta]\Big{\\}}+\frac{\gamma[v_{\gamma,b}(b)+x-b]}{\delta+\gamma}&\text{if}\
x\in(b,x_{b}),\\\ \vspace{-0.4cm}&\\\
c_{3,2}(\alpha_{\gamma})\operatorname{e}^{\lambda_{\gamma}[x-x_{b}]}+\frac{\gamma}{\gamma+\delta}\bigg{[}x-b+v_{\gamma,b}(b)+\frac{\eta}{\delta+\gamma}\bigg{]}&\text{if}\
x\in(x_{b},\infty),\end{cases}$ (3.42)

where $\bar{\eta}$, $c_{2,1}$, $c_{2,2}$ are as in (3.19),
$\mathpzc{x}_{\,1}^{-1}$ is the inverse function of
$\mathpzc{x}_{\,1}(z)=k_{2,1}\operatorname{e}^{[1+\delta\bar{\eta}][z+M_{\gamma}]}+c_{2,1}z+k_{2,2}^{(\gamma)}$,
for$z\in[-M_{\gamma},-M_{2}]$, with $k_{2,1}$ is as in (3.25),
$\mathpzc{x}_{\,2}^{-1}$ is the inverse function of
$\mathpzc{x}_{\,2}(z)=\bar{f}_{M_{2}}(\operatorname{e}^{z})+b$ for
$z\in[-M_{2},\ln[\alpha_{\gamma}]]$, with $\bar{f}_{M_{2}}$ defined as in
(3.33) when $\beta=M_{2}$, and

$\displaystyle\begin{split}x_{b}&=\bar{f}_{M_{2}}(\alpha_{\gamma})+b,\quad
c_{3,2}(\alpha_{\gamma})\raisebox{0.4pt}{$:$}=-\frac{\mu}{\sigma^{2}\alpha_{\gamma}\lambda_{\gamma}^{2}},\quad
k_{2,2}^{(\gamma)}\raisebox{0.4pt}{$:$}=c_{2,1}\bigg{[}M_{\gamma}-\frac{\delta\bar{\eta}}{1+\delta\bar{\eta}}\bigg{]},\\\
M_{\gamma}&=\frac{1}{1+\delta\bar{\eta}}\ln\bigg{[}\frac{1}{\delta\bar{\eta}}\bigg{\\{}\frac{\operatorname{e}^{-M_{2}}g(\operatorname{e}^{-M_{2}})}{c_{2,1}}c_{3,1}(M_{2})-1\bigg{\\}}\bigg{]}+M_{2},\end{split}$
(3.43)

where $M_{2}\in(0,\underline{$b$})$ is the solution to

$\bar{g}(M_{2})=[1+\delta\bar{\eta}]\bigg{[}1+\frac{b}{c_{2,1}}\bigg{]}.$
(3.44)

######  Remark 3.10.

Taking $\gamma\in(0,\gamma_{1})$ such that (3.38) holds, we get that $\bar{g}$
is a positive and decreasing function on $(0,\underline{$b$})$. It implies
that $M_{2}$ is the unique solution to (3.44), for any $b>0$ satisfying
(3.36). Additionally, by the construction of $v_{\gamma,b}$, which is given by
(3.42), it is known that $v_{\gamma,b}$ is concave and increasing on
$(0,\infty)$, since $c_{3,2}(\alpha_{\gamma})<0$ and $v_{\gamma,b}$ satisfies
(3.28) on $(0,b)$, and

$v^{\prime}_{\gamma,b}(x)=\operatorname{e}^{-\mathpzc{x}_{\,2}^{-1}(x)}\quad\text{and}\quad
v^{\prime\prime}_{\gamma,b}(x)=-\frac{\operatorname{e}^{-\mathpzc{x}_{\,2}^{-1}(x)}}{\mathpzc{x}_{\,2}^{\prime}(\mathpzc{x}_{\,2}^{-1}(x))},\quad\text{for}\
x\in(b,x_{b}).$ (3.45)

It implies also that $u^{*}_{b}(x)<1$ for $x\in(0,x_{b})$, and
$u^{*}_{b}(x)=1$ at $x=x_{b}$. For a more detail exposition, see Subsection
A.3.

By the seen in the remark above and considering that (3.7) must be true, it
follows that
$\operatorname{e}^{-\mathpzc{x}_{\,1}^{-1}(b_{\gamma})}=\operatorname{e}^{-\mathpzc{x}_{\,2}^{-1}(b_{\gamma})}=\operatorname{e}^{M_{2}}=1$,
which implies that $M_{2}=0$ and

$\displaystyle\mathpzc{x}_{\,1}^{-1}(b_{\gamma})=0\quad\Longleftrightarrow\quad
b_{\gamma}=c_{2,1}\bigg{[}\frac{\bar{\eta}\delta}{1+\delta\bar{\eta}}[\operatorname{e}^{M_{\gamma}[1+\delta\bar{\eta}]}-1]+M_{\gamma}\bigg{]}.$
(3.46)

We have also that

$M_{\gamma}=\frac{1}{1+\delta\bar{\eta}}\ln\bigg{[}\frac{1}{\delta\bar{\eta}}\bigg{\\{}\frac{g(1)}{c_{2,1}}c_{3,1}(0)-1\bigg{\\}}\bigg{]}.$
(3.47)

To establish the values of $\gamma\in(0,\gamma_{1})$ for which the condition
$M_{\gamma}>0$ holds, due to $v^{\prime}_{\gamma,b_{\gamma}}(0+)>1$, we must
study when (3.38) is true. To achieve this, let us highlight some properties
of the following functions

$\displaystyle\begin{split}f_{2}(\gamma)&\raisebox{0.4pt}{$:$}=\alpha_{\gamma}=\frac{\gamma+\delta}{\gamma}\bigg{[}1-\frac{\mu}{\eta+\sqrt{\eta^{2}+2\sigma^{2}[\delta+\gamma]}}\bigg{]},\\\
f_{3}(\gamma)&\raisebox{0.4pt}{$:$}=\frac{\sigma^{2}\operatorname{e}^{\bar{\eta}\gamma
f_{2}(\gamma)}}{\mu[f_{2}(\gamma)]^{\bar{\eta}[\delta+\gamma]+1}}+\bar{\eta}[\mu-\eta]\int_{1}^{f_{2}(\gamma)}\frac{\operatorname{e}^{\bar{\eta}\gamma
y}}{y^{\bar{\eta}[\delta+\gamma]+2}}\mathrm{d}y,\\\
f_{4}(\gamma)&\raisebox{0.4pt}{$:$}=\bar{\eta}[\mu-\eta]\operatorname{e}^{\bar{\eta}\gamma},\end{split}\quad\
\text{for}\ \gamma\in(0,\gamma_{1}).$ (3.48)

######  Lemma 3.11.

Let $f_{2},f_{3},f_{4}$ be as in (3.48). Then,

  1. (i)

$f_{2}$ is decreasing on $(0,\gamma_{1})$ satisfying

$\lim_{\gamma\downarrow
0}f_{2}(\gamma)=\infty\quad\text{and}\quad\lim_{\gamma\uparrow\gamma_{1}}f_{2}(\gamma)=1.$
(3.49)

  2. (ii)

$f_{3}$ and $f_{4}$ are increasing on $(0,\gamma_{1})$ satisfying

$\displaystyle\begin{split}&\lim_{\gamma\downarrow
0}f_{3}(\gamma)=\frac{\bar{\eta}[\mu-\eta]}{\bar{\eta}\delta+1}\quad\text{and}\quad\lim_{\gamma\uparrow\gamma_{1}}f_{3}(\gamma)=\frac{\sigma^{2}\operatorname{e}^{\bar{\eta}\gamma_{1}}}{\mu},\\\
&\lim_{\gamma\downarrow
0}f_{4}(\gamma)=\bar{\eta}[\mu-\eta]\quad\text{and}\quad\lim_{\gamma\uparrow\gamma_{1}}f_{4}(\gamma)=\bar{\eta}[\mu-\eta]\operatorname{e}^{\bar{\eta}\gamma_{1}}.\end{split}$
(3.50)

######  Remark 3.12.

By (3.19), (3.49)–(3.50) and since $\mu>\eta$ and $2\eta>\mu$, it is easy to
check that $f_{3}(0+)<f_{4}(0+)$ and $f_{3}(\gamma_{1}-)>f_{4}(\gamma_{1}-)$.
It implies that there exists a unique $\gamma_{2}\in(0,\gamma_{1})$ such that

$f_{3}(\gamma_{2})=f_{4}(\gamma_{2}),$ (3.51)

due to the increasing property of $f_{3}$ and $f_{4}$. Therefore, for any
$\gamma\in(\gamma_{2},\gamma_{1})$, (3.38) holds, demonstrating that $M>0$. it
is also observed that $\bar{g}(0+)>1+\delta\bar{\eta}$ for
$\gamma\in(\gamma_{2},\gamma_{1})$.

In case that $\gamma\in(0,\gamma_{2})$, we take $b=0$ and
$x_{0}=\bar{f}_{-M_{\gamma}}(\alpha_{\gamma})$, where

$\displaystyle\bar{f}_{\beta}(z)$
$\displaystyle=c_{3,3}(\beta)[G(z)-G(\operatorname{e}^{-\beta})]-\bar{\eta}[\mu-\eta]\bigg{[}G(z)H_{\beta}(z)-\int_{\operatorname{e}^{-\beta}}^{z}\frac{G(y)}{y^{2}g(y)}\mathrm{d}y\bigg{]},$
$\displaystyle c_{3,3}(\beta)$
$\displaystyle\raisebox{0.4pt}{$:$}=\frac{\bar{\eta}[\mu-\eta]}{\operatorname{e}^{-\beta}g(\operatorname{e}^{-\beta})},$

and $M_{\gamma}$ is the unique positive solution to

$\displaystyle\frac{\mu-\eta}{\operatorname{e}^{M_{\gamma}}g(\operatorname{e}^{M_{\gamma}})}=\frac{\mu}{2\alpha_{\gamma}g(\alpha_{\gamma})}+[\mu-\eta]H_{-M_{\gamma}}(\alpha_{\gamma}).$
(3.52)

Then, a solution to (3.39)–(3.41) is given by

$v_{\gamma,0}(x)=\begin{cases}\frac{\operatorname{e}^{-\mathpzc{x}_{\,2}^{-1}(x)}}{\delta+\gamma}\Big{\\{}\frac{g(\operatorname{e}^{\mathpzc{x}_{\,2}^{-1}(x)})\operatorname{e}^{\mathpzc{x}_{\,2}^{-1}(x)}}{\bar{\eta}}&\\\
\times\big{[}c_{3,3}(M_{\gamma})-\bar{\eta}[\mu-\eta]H_{-M_{\gamma}}(\operatorname{e}^{\mathpzc{x}_{\,2}^{-1}(x)})\big{]}-[\mu-\eta]\Big{\\}}+\frac{\gamma
x}{\delta+\gamma}&\text{if}\ x\in(0,x_{0}),\\\
c_{3,2}(\alpha_{\gamma})\operatorname{e}^{\lambda_{\gamma}[x-x_{0}]}+\frac{\gamma}{\gamma+\delta}\Big{[}x+\frac{\eta}{\delta+\gamma}\Big{]}&\text{if}\
x\in(x_{0},\infty),\end{cases}$ (3.53)

where $c_{3,2}(\alpha_{\gamma})$ is as in (3.43) and $\mathpzc{x}_{\,2}^{-1}$
is the inverse function of
$\mathpzc{x}_{\,2}:[M_{\gamma},\bar{z}]\longrightarrow[0,x_{\gamma}]$, with
$\bar{z}=\ln[\alpha_{\gamma}]$, which has the following form
$\mathpzc{x}_{\,2}(z)=\bar{f}_{-M_{\gamma}}(\operatorname{e}^{z})$. For more
details about the construction of (3.53), see the proof of Proposition 3.13 in
Section A.3.

######  Proposition 3.13.

Let $\mu<2\eta$ and $\gamma_{2}\in(0,\gamma_{1})$ satisfying (3.51).

  1. (i)

If $\gamma\in(\gamma_{2},\gamma_{1})$, then (3.42) is a
$\operatorname{C}^{2}$-continuous, increasing and concave solution to (2.5),
with $M_{2}=0$, $b=b_{\gamma}$ and $M_{\gamma}$ as in (3.46) and (3.47).
Furthermore,
$v^{\prime}_{\gamma,b_{\gamma}}/v^{\prime\prime}_{\gamma,b_{\gamma}}$ is
decreasing on $(0,\infty)$. Then,

$x_{\gamma}=\bar{f}_{0}(\alpha_{\gamma})+b_{\gamma}$ (3.54)

where $\bar{f}_{0}$ is as in (3.33) when $\beta=0$.

  2. (ii)

If $\gamma\in(0,\gamma_{2})$, then $v_{\gamma,0}$ as in (3.53) is a solution
to the equation (2.5), with $b_{\gamma}=0$. Furthermore,
$v^{\prime}_{\gamma,0}/v^{\prime\prime}_{\gamma,0}$ is decreasing on
$(0,\infty)$. Then,

$x_{\gamma}=\bar{f}_{-M}(\alpha_{\gamma}).$ (3.55)

Optimal strategy. In light of Theorem 2.3 and Proposition 3.8, we deduce under
the conditions $\mu<2\eta$ and $\gamma>\gamma_{1}$, the optimal strategy is
determined by (2.15), where $x_{\gamma}$ and $b_{\gamma}$ are given by (3.18)
and (3.31), respectively. The controlled process $X^{b_{\gamma}}$ is given as
the solution to the SDE described in (2.14). Moreover, the value function $V$
and the expected NPV $V^{b_{\gamma}}$ are given by (3.23) with $b=b_{\gamma}$
and $x_{b}=\bar{x}$,. In this context, the optimal strategy for the insurance
company is to assume the maximum risk exclusively when
$X^{b_{\gamma}}_{\mathpzc{t}}\geq x_{\gamma}$, while the payout
$\max\\{X^{b_{\gamma}}_{T_{i}}-b_{\gamma},0\\}$ is made at time
$T_{i}<\tau^{b_{\gamma}}$. It is noteworthy that in the strategy
$\pi^{b_{\gamma}}$, shareholder payments are made precisely when the insurance
company is exposed to the maximum risk, as a consequence of
$b_{\gamma}>x_{\gamma}$.

On the other hand, if $\gamma\in(\gamma_{2},\gamma_{1})$ , by Theorem 2.3 and
Proposition 3.13.(i), we have that the optimal strategy is determined by
(3.46) and (3.54), respectively. The value function $V$ and the expected NPV
$V^{b_{\gamma}}$ are given by (3.42), with $b=b_{\gamma}$ and
$x_{\gamma}=\bar{f}_{0}(\alpha_{\gamma})+b_{\gamma}$. It is remarkable that in
the strategy $\pi^{b_{\gamma}}$, shareholder payments are made precisely when
the insurance company does not wish to be exposed to the maximum risk, as a
consequence of $b_{\gamma}<x_{\gamma}$.

In the scenario where $\mu<2\eta$ and $\gamma\in(0,\gamma_{2}]$, according to
Theorem 2.3 and Proposition 3.13.(ii), we observe that $b_{\gamma}=0$ and
$x_{\gamma}=\bar{f}_{0}(\alpha_{\gamma})$. Furthermore, the optimal strategy
for the company in this scenario is to accept the maximum risk when
$X^{0}_{\mathpzc{t}}>\bar{f}_{0}(\alpha_{\gamma})$, while making the payout
equal to the entire amount that the company has at the first arrival of the
Poisson process $N^{\gamma}$. Additionally, $V^{0}$ and $V$ are given by
(3.53).

###  3.2 Cheap reinsurance

In this subsection, we analyse the case where $\eta=\mu$. Similar to our
previous analysis, we examine the scenarios $x_{\gamma}\leq b_{\gamma}$ and
$x_{\gamma}>b_{\gamma}$. However, since $\mu<2\mu$, we can leverage the
arguments presented in Sub-subsection 3.1.2 to verify the validity of this
subsection. Consequently, we will omit the proofs of the main results of this
part.

For the first case, let us consider $b>\hat{x}$, with

$\hat{x}\raisebox{0.4pt}{$:$}=\frac{\sigma^{2}}{\mu[1+\delta\bar{\eta}]}.$
(3.56)

######  Proposition 3.14.

Let $\eta=\mu$. If $b>\hat{x}$ is fixed, a solution to the NLPDS

$\displaystyle-\frac{\mu^{2}[v^{\prime}(x)]^{2}}{2\sigma^{2}v^{\prime\prime}(x)}-\delta
v(x)$ $\displaystyle=0\quad\text{for}\ x\in(0,x_{b}),$ (3.57)
$\displaystyle\frac{1}{2}\sigma^{2}v^{\prime\prime}(x)+\mu
v^{\prime}(x)-\delta v(x)+\gamma\\{x-b+v(b)-v(x)\\}\mathbbm{1}_{\\{x-b>0\\}}$
$\displaystyle=0\quad\text{for}\ x\in(x_{b},\infty),$ (3.58)
$\displaystyle\text{s.t.}\ v(0)$ $\displaystyle=0,$ (3.59)

is a $\operatorname{C}^{2}$-continuous function on $(0,\infty)$, whose form is
given by

$\displaystyle
v_{\gamma,b}(x)=\begin{cases}c_{4,1}(b)x^{\delta\bar{\eta}/[1+\delta\bar{\eta}]}&\text{if}\
x\in(0,x_{b}),\\\ \vspace{-0.4cm}&\\\ c_{4,1}(b)h_{3}(x-x_{b})&\text{if}\
x\in(x_{b},b),\\\ \vspace{-0.4cm}&\\\
c_{4,2}(b)\operatorname{e}^{\lambda_{\gamma}[x-b]}+\frac{\gamma}{\gamma+\delta}\bigg{[}x-b+v_{\gamma,b}(b)+\frac{\mu}{\delta+\gamma}\bigg{]}&\text{if}\
x\in(b,\infty),\end{cases}$ (3.60)

where the function $h_{3}$ is taken as
$h_{3}(x)\raisebox{0.4pt}{$:$}=\frac{1}{\theta_{+}-\theta_{-}}\Big{[}\bar{a}_{1}\operatorname{e}^{\theta_{+}x}-\bar{a}_{2}\operatorname{e}^{\theta_{-}x}\Big{]}$,
with $\theta_{+}$, $\theta_{-}$, $\lambda_{\gamma}$ as in Proposition 3.1, and

$\displaystyle\begin{split}x_{b}&=\hat{x},\quad\bar{a}_{1}\raisebox{0.4pt}{$:$}=\bigg{[}\frac{\delta\bar{\eta}}{1+\delta\bar{\eta}}\bigg{]}x_{b}^{-1/[1+\delta\bar{\eta}]}-\theta_{-}x_{b}^{\delta\bar{\eta}/[1+\delta\bar{\eta}]},\\\
\bar{a}_{2}&\raisebox{0.4pt}{$:$}=\bigg{[}\frac{\delta\bar{\eta}}{1+\delta\bar{\eta}}\bigg{]}x_{b}^{-1/[1+\delta\bar{\eta}]}-\theta_{+}x_{b}^{\delta\bar{\eta}/[1+\delta\bar{\eta}]},\\\
c_{4,1}(b)&=\frac{\frac{\gamma}{\gamma+\delta}-\frac{\mu\gamma\lambda_{\gamma}}{[\delta+\gamma]^{2}}}{h^{\prime}_{3}(b-x_{b})-\frac{\lambda_{\gamma}\delta}{\gamma+\delta}h_{3}(b-x_{b})},\\\
c_{4,2}(b)&\raisebox{0.4pt}{$:$}=\frac{\delta}{\delta+\gamma}c_{4,1}(b)h_{3}(b-{x}_{b})-\frac{\mu\gamma}{[\delta+\gamma]^{2}}.\end{split}$

######  Remark 3.15.

By the construction of $v_{\gamma,b}$, which is given by (3.60), for any
$b>\hat{x}$, it is easy to observe that $v_{\gamma,b}$ is concave and
increasing on $(0,\hat{x})$, due to $c_{4,1}(b)>0$. Moreover,
$u^{*}_{b}(x)=\frac{\mu}{\sigma^{2}}[1+\delta\bar{\eta}]x$, which increases on
$(0,\hat{x})$ and satisfies $u^{*}_{b}(\hat{x})=1$. For a more detail
exposition, see Subsection A.3.

Taking the first derivative in (3.60) on $(\hat{x},b_{\gamma})$, we see that
(3.7) is achieved if and only if

$\displaystyle
g_{4}(b_{\gamma}-\hat{x})=\frac{1}{\lambda_{\gamma}\delta}\bigg{\\{}\delta+\frac{\mu\gamma\lambda_{\gamma}}{\delta+\gamma}\bigg{\\}}$
(3.61)

is true, where

$\displaystyle
g_{4}(b)\raisebox{0.4pt}{$:$}=\frac{h_{3}(b)}{h^{\prime}_{3}(b)}\quad\text{for}\
b\in(0,\infty).$ (3.62)

In order to guarantee the veracity of (3.61), let us mention some properties
of the function $g_{4}$ first.

######  Lemma 3.16.

Let $g_{4}$ be as in (3.62). Then, $g_{4}$ is increasing on $(0,\infty)$,
satisfying

$\lim_{b\downarrow
0}g_{4}(b)=\frac{\mu}{2\delta}\quad\text{and}\quad\lim_{b\uparrow\infty}g_{4}(b)=\frac{1}{\theta_{+}}.$

By Lemma 3.16, we have that (3.61) is equivalent to verify that
$\frac{\mu}{2\delta}\leq\frac{1}{\lambda_{\gamma}\delta}\big{[}\delta+\frac{\lambda_{\gamma}\mu\gamma}{\delta+\gamma}\big{]}<\frac{1}{\theta_{+}}$,
which is true only for
$\gamma>\bar{\gamma}_{1}\raisebox{0.4pt}{$:$}=f^{-1}_{1}\big{(}\frac{\mu}{2\delta}\big{)}$,
where $f^{-1}_{1}$ is the inverse of $f_{1}$ given in (3.11) when $\eta=\mu$,
because of
$\frac{1}{\theta_{-}}<\frac{\mu}{2\delta}<\frac{\mu}{\delta}<\frac{1}{\theta_{+}}$
and Lemma 3.3. Thus, for $\gamma>\bar{\gamma}_{1}$ fixed, there exists a
unique $b_{\gamma}\in(0,\infty)$ that satisfies (3.61). By the seen above, we
get the following proposition.

######  Proposition 3.17.

If $\eta=\mu$ and $\gamma>\bar{\gamma}_{1}$, then (3.60) is a
$\operatorname{C}^{2}$-continuous, increasing and concave solution to
(3.57)–(3.59), with

$b=b_{\gamma}=\hat{x}+g^{-1}_{4}\bigg{(}\frac{1}{\lambda_{\gamma}\delta}\bigg{[}\delta+\frac{\lambda_{\gamma}\mu\gamma}{\delta+\gamma}\bigg{]}\bigg{)}.$
(3.63)

Furthermore,
$v^{\prime}_{\gamma,b_{\gamma}}/v^{\prime\prime}_{\gamma,b_{\gamma}}$ is
decreasing on $(0,\infty)$ satisfying (3.17). Then, $x_{\gamma}=\hat{x}$.

The following proposition presents a solution $v_{\gamma,b}$ to the NPDS
(3.65)–(3.67), which depends on $b\in(0,\bar{b})$ where

$\displaystyle\bar{b}_{1}\raisebox{0.4pt}{$:$}=\frac{c_{5,1}(\alpha_{\gamma})g(1)}{1+\delta\bar{\eta}},\quad\text{and}\quad
c_{5,1}(\alpha_{\gamma})\raisebox{0.4pt}{$:$}=\frac{\sigma^{2}}{\mu\alpha_{\gamma}g(\alpha_{\gamma})},$
(3.64)

with $\alpha_{\gamma}$ as in (3.34) when $\eta=\mu$. Recall that $G$ is the
gamma accumulative distribution whose density function $g$ is given by (3.35),
and its inverse function is represented by $G^{-1}$.

######  Proposition 3.18.

Let $\eta=\mu$. If $b\in(0,\bar{b}_{1})$ is fixed, with $\bar{b}_{1}$ as in
(3.64), a solution to the NLPDS

$\displaystyle-\frac{\mu^{2}[v^{\prime}(x)]^{2}}{2\sigma^{2}v^{\prime\prime}(x)}-\delta
v(x)+\gamma\\{x-b+v(b)-v(x)\\}\mathbbm{1}_{\\{x-b>0\\}}$
$\displaystyle=0\quad\text{for}\ x\in(0,x_{b}),$ (3.65)
$\displaystyle\frac{1}{2}\sigma^{2}v^{\prime\prime}(x)+\mu
v^{\prime}(x)-[\delta+\gamma]v(x)+\gamma\\{x-b+v(b)\\}$
$\displaystyle=0\quad\text{for}\ x\in(x_{b},\infty),$ (3.66)
$\displaystyle\text{s.t.}\ v(0)$ $\displaystyle=0,$ (3.67)

is a $\operatorname{C}^{2}$-continuous, whose form is given by

$\displaystyle
v_{\gamma,b}(x)=\begin{cases}\frac{[1+\delta\bar{\eta}]\operatorname{e}^{M_{2}}b^{1/[1+\delta\bar{\eta}]}}{\delta\bar{\eta}}x^{\delta\bar{\eta}/[1+\delta\bar{\eta}]}&\text{if}\
x\in(0,b),\\\ \vspace{-0.4cm}&\\\
\frac{1}{\delta+\gamma}\big{\\{}\frac{c_{5,1}(\alpha_{\gamma})}{\bar{\eta}}g\big{(}G^{-1}\big{(}\frac{x-b}{c_{5,1}(\alpha_{\gamma})}+G(\operatorname{e}^{-M_{2}})\big{)}\big{)}+\gamma[v_{\gamma,b}(b)+x-b]\big{\\}}&\text{if}\
x\in(b,x_{b}),\\\ \vspace{-0.4cm}&\\\
c_{3,2}(\alpha_{\gamma})\operatorname{e}^{\lambda_{\gamma}[x-x_{b}]}+\frac{\gamma}{\gamma+\delta}\big{[}x-b+v_{\gamma,b}(b)+\frac{\mu}{\delta+\gamma}\big{]}&\text{if}\
x\in(x_{b},\infty),\end{cases}$ (3.68)

where $c_{3,2}(\alpha_{\gamma})$, $c_{5,1}(\alpha_{\gamma})$ are given in
(3.43) and (3.64), respectively, and
$x_{b}=c_{5,1}(\alpha_{\gamma})[G(\alpha_{\gamma})-G(\operatorname{e}^{-M_{2}})]+b$.
The parameter $M_{2}>0$ is the unique solution to
$\operatorname{e}^{-M_{2}}g(\operatorname{e}^{-M_{2}})=\frac{b[1+\delta\bar{\eta}]}{c_{5,1}(\alpha_{\gamma})}$.

######  Remark 3.19.

Since $\beta\mapsto\operatorname{e}^{-\beta}g(\operatorname{e}^{-\beta})$ is a
positive and decreasing function on $(0,\infty)$, we have a unique solution
$M_{2}$ to (3.44), for any $b\in(0,\bar{b}_{1})$. Additionally,
$v_{\gamma,b}$, given in (3.68), is concave and increasing on $(0,x_{b})$, due
to its construction. Additionally, we have also that $u^{*}_{b}(x)<1$ for
$x\in(0,\bar{x})$, and $u^{*}_{b}(x)=1$ at $x=x_{b}$. For a more detail
exposition, see Subsection A.3.

Since (3.7) should hold, it follows that $\operatorname{e}^{M_{2}}=1$, which
implies that $M_{2}=0$ and

$\displaystyle
b_{\gamma}=\frac{c_{5,1}(\alpha_{\gamma})g(1)}{1+\delta\bar{\eta}}.$ (3.69)

Notice that $b_{\gamma}$ as before is a positive real value because
$\gamma\mapsto\alpha_{\gamma}$ is a positive and decreasing function on
$(0,\bar{\gamma}_{1})$, due to Lemma 3.11.(i) when $\eta\downarrow\mu$.

######  Proposition 3.20.

If $\eta=\mu$ and $\gamma\in(0,\bar{\gamma}_{1})$, then (3.68) is a
$\operatorname{C}^{2}$-continuous, increasing and concave solution to
(3.65)–(3.67), with
$b=b_{\gamma}=\frac{c_{5,1}(\alpha_{\gamma})g(1)}{1+\delta\bar{\eta}}$.
Furthermore,
$v^{\prime}_{\gamma,b_{\gamma}}/v^{\prime\prime}_{\gamma,b_{\gamma}}$ is
decreasing on $(0,\infty)$ satisfying (3.17). Then,

$\displaystyle
x_{\gamma}=c_{5,1}(\alpha_{\gamma})[G(\alpha_{\gamma})-G(1)]+b_{\gamma}.$
(3.70)

Optimal strategy. When $\gamma\in(\bar{\gamma}_{1},\infty)$, according to
Theorem 2.3 and Proposition 3.17, the optimal strategy $\pi^{b_{\gamma}}$ is
given by equations (3.56) and (3.63). In this scenario, the expressions for
$V$ and $V^{b_{\gamma}}$ are represented by (3.60) when $b=b_{\gamma}$ as in
(3.63). It is worth noting that in the strategy $\pi^{b_{\gamma}}$,
shareholder payments occur precisely when the insurance company is exposed to
the maximum risk, as a consequence of $b_{\gamma}>x_{\gamma}$.

On the other hand, when $\gamma\in(0,\bar{\gamma}_{1})$, based on Theorem 2.3
and Proposition 3.20, the optimal strategy $\pi^{b_{\gamma}}$ is given by
equations (3.69)–(3.70). At the same time, the representations for $V$ and
$V^{b_{\gamma}}$ are given by (3.68), when $b=b_{\gamma}$ as in (3.69). It is
remarkable that in the strategy $\pi^{b_{\gamma}}$, dividend payments to
shareholders occur precisely when the insurance company does not wish to be
exposed to the maximum risk, as a consequence of $b_{\gamma}<x_{\gamma}$.

##  4 Analysing the limit behaviour for $\boldsymbol{\gamma}$

###  4.1 Non-cheap reinsurance

####  4.1.1 The case $\boldsymbol{\mu\geq 2\eta}$

Notice that the trajectory $\gamma\mapsto
b_{\gamma}=g^{-1}_{1}(f_{1}(\gamma))$ is continuous and strictly increasing on
$(\gamma_{0},\infty)$, due to Lemmas 3.2–3.3, which satisfies
$b_{\gamma_{0}}=0$ and
$b_{\infty}\raisebox{0.4pt}{$:$}=\lim_{\gamma\uparrow\infty}b_{\gamma}=g^{-1}_{1}\big{(}\frac{\eta}{\delta}\big{)}=\frac{1}{\theta_{+}-\theta_{-}}\ln\big{[}\frac{\delta-\eta\theta_{-}}{\delta-\eta\theta_{+}}\big{]}$.
Furthermore, By (3.12) and taking into account that

$\displaystyle\frac{\delta\lambda_{\gamma}}{\gamma+\delta}\xrightarrow[\gamma\uparrow\infty]{}0\quad\text{and}\quad\frac{\gamma}{\gamma+\delta}\Big{[}1-\frac{\eta\lambda_{\gamma}}{\gamma+\delta}\Big{]}\xrightarrow[\gamma\uparrow\infty]{}1,$
(4.1)

we have that

$\displaystyle\lim_{\gamma\uparrow\infty}c_{1,1}(b_{\gamma})=\frac{1}{h^{\prime}_{1}(b_{\infty})}\
\text{and}\ \lim_{\gamma\uparrow\infty}c_{1,2}(b_{\gamma})=0.$ (4.2)

Then, taking $v_{\gamma,b_{\gamma}}$ as in (3.2), with $b=b_{\gamma}$, and
using (4.2), it follows that

$v_{b_{\infty}}(x)\raisebox{0.4pt}{$:$}=\lim_{\gamma\uparrow\infty}v_{\gamma,b_{\gamma}}(x)=\begin{cases}\frac{h_{1}(x)}{h^{\prime}_{1}(b_{\infty})}&\text{for}\
x\in(0,b_{\infty}),\\\ x-b_{\infty}+v_{b_{\infty}}(b_{\infty})&\text{for}\
x\in[b_{\infty},\infty).\end{cases}$ (4.3)

Here $x_{\infty}=0$. These asymptotic limits allow us to recover the value
function for the case of singular dividend strategies for the optimal problem
obtained by Taksar [13] in Theorem 5.2.

####  4.1.2 The case $\boldsymbol{\mu<2\eta}$

Notice that the trajectory $\gamma\mapsto
b_{\gamma}=\bar{x}+g^{-1}_{2}(f_{1}(\gamma))$ is continuous and strictly
increasing on $(\gamma_{1},\infty)$, due to Lemmas 3.2 and 3.7, which
satisfies
$b_{\gamma_{1}}=\bar{x}+g^{-1}_{2}\big{(}\frac{2\eta-\mu}{2\delta}\big{)}=\bar{x}$
and
$b_{\infty}\raisebox{0.4pt}{$:$}=\lim_{\gamma\uparrow\infty}b_{\gamma}=\bar{x}+g^{-1}_{2}\big{(}\frac{\eta}{\delta}\big{)}=\bar{x}+\frac{1}{\theta_{+}-\theta_{-}}\ln\Big{[}\frac{a_{2}[\delta-\eta\theta_{-}]}{a_{1}[\delta-\eta\theta_{+}]}\Big{]}$.
On the other hand, considering (4.1), we get that
$\lim_{\gamma\uparrow\infty}c_{2,3}(b_{\gamma})=0$ and
$M(b_{\infty})\raisebox{0.4pt}{$:$}=\lim_{\gamma\uparrow\infty}M(b_{\gamma})=\ln\Big{[}\frac{c_{2,2}^{1/[1+\delta\bar{\eta}]}}{h^{\prime}_{2}(b_{\infty}-\bar{x})}\Big{]}$.
Then, taking $v_{\gamma,b_{\gamma}}$ as in (3.42), with $b=b_{\gamma}$, it
follows that

$\displaystyle
v_{b_{\infty}}(x)=\lim_{\gamma\uparrow\infty}v_{\gamma,b}(x)=\begin{cases}c_{2,1}\operatorname{e}^{-\mathpzc{x}_{\,1}^{-1}(x)}\Big{[}\operatorname{e}^{[1+\delta\bar{\eta}][M(b_{\infty})+\mathpzc{x}_{\,1}^{-1}(x)]}-1\Big{]}&\text{if}\
x\in(0,\bar{x}),\\\ \vspace{-0.4cm}&\\\
\frac{h_{2}(x-\bar{x})}{h^{\prime}_{2}(b_{\infty}-\bar{x})}&\text{if}\
x\in(\bar{x},b_{\infty}),\\\ \vspace{-0.4cm}&\\\
x-b_{\infty}+v_{b_{\infty}}(b_{\infty})&\text{if}\
x\in(b_{\infty},\infty),\end{cases}$ (4.4)

where $\mathpzc{x}_{\,1}^{-1}$ is the inverse function of

$\mathpzc{x}_{\,1}(z)=k_{2,1}\operatorname{e}^{[1+\delta\bar{\eta}][z+M(b_{\infty})]}+c_{2,1}z+k_{2,2}(b_{\infty})\
\quad\text{for}\ z\in[-M(b_{\infty}),\bar{z}(b_{\infty})],$

with
$k_{2,2}(b_{\infty})=c_{2,1}\big{[}M(b_{\infty})-\frac{\delta\bar{\eta}}{1+\delta\bar{\eta}}\big{]}$
and
$\bar{z}(b_{\infty})=\frac{1}{1+\delta\bar{\eta}}\ln[c_{2,2}]-M(b_{\infty})$.
Here $x_{\infty}=\bar{x}$. These asymptotic limits allow us to recover the
value function for the case of singular dividend strategies for the optimal
problem obtained by Taksar [13] in Theorem 5.3.

Additionally, a solution to the HJB equation (2.5), when
$\gamma\downarrow\gamma_{1}$, is given by

$\displaystyle
v_{\gamma,\bar{x}}(x)=\begin{cases}c_{2,1}\operatorname{e}^{-\mathpzc{x}_{\,1}^{-1}(x)}\Big{[}\operatorname{e}^{[1+\delta\bar{\eta}][M(\bar{x})+\mathpzc{x}_{\,1}^{-1}(x)]}-1\Big{]}&\text{if}\
x\in(0,\bar{x}),\\\ \vspace{-0.4cm}&\\\
c_{2,3}(\bar{x})\operatorname{e}^{\lambda_{\gamma}[x-\bar{x}]}+\frac{\gamma}{\gamma+\delta}\Big{[}x-\bar{x}+v_{\gamma,\bar{x}}(\bar{x})+\frac{\eta}{\delta+\gamma}\Big{]}&\text{if}\
x\in(\bar{x},\infty),\end{cases}$ (4.5)

with

$\displaystyle M(\bar{x})$
$\displaystyle=\ln\left[\frac{\frac{\gamma_{1}}{\gamma_{1}+\delta}-\frac{\lambda_{\gamma_{1}}\eta\gamma_{1}}{[\delta+\gamma_{1}]^{2}}}{c_{2,2}^{-1/[1+\delta\bar{\eta}]}\bigg{[}1-\frac{\lambda_{\gamma_{1}}[2\eta-\mu]}{2[\delta+\gamma_{1}]}\bigg{]}}\right]=\frac{1}{1+\delta\bar{\eta}}\ln[c_{2,2}],$

because of

$\displaystyle
f_{1}(\gamma_{1})=\frac{2\eta-\mu}{2\delta}\quad\Longleftrightarrow\quad\frac{1}{\lambda_{\gamma_{1}}}=\frac{2\eta-\mu}{2\delta}-\frac{\eta\gamma_{1}}{\delta[\delta+\gamma_{1}]}.$
(4.6)

It implies that
$c_{2,3}(\bar{x})=\frac{2\eta-\mu}{2[\delta+\gamma_{1}]}-\frac{\eta\gamma_{1}}{[\delta+\gamma_{1}]^{2}}=-\frac{\mu}{\lambda^{2}_{\gamma_{1}}\sigma^{2}}=c_{3,2}(1)$,
due to

$\displaystyle
f_{2}(\gamma_{1})=1\quad\Longleftrightarrow\quad\frac{1}{\lambda_{\gamma_{1}}}=-\frac{\delta\sigma^{2}}{[\gamma_{1}+\delta]\mu}.$
(4.7)

Meanwhile, using (3.48), we see that $b_{\gamma}$ and $M$ as in (3.46) and
(3.47), respectively, can be written in the following way

$\displaystyle M_{\gamma}$
$\displaystyle=\frac{1}{1+\delta\bar{\eta}}\ln\bigg{[}\frac{\operatorname{e}^{-\gamma\bar{\eta}}[f_{3}(\gamma)-f_{4}(\gamma)]}{\delta\bar{\eta}c_{2,1}}+1\bigg{]},$
$\displaystyle b_{\gamma}$
$\displaystyle=\frac{\operatorname{e}^{-\gamma\bar{\eta}}[f_{3}(\gamma)-f_{4}(\gamma)]}{c_{2,1}^{-1}\bar{\eta}[\mu-\eta]}+\frac{c_{2,1}}{1+\delta\bar{\eta}}\ln\bigg{[}\frac{\operatorname{e}^{-\gamma\bar{\eta}}[f_{3}(\gamma)-f_{4}(\gamma)]}{\delta\bar{\eta}c_{2,1}}+1\bigg{]}.$
(4.8)

Then, the trajectory $\gamma\mapsto b_{\gamma}$ is continuous on
$(\gamma_{2},\gamma_{1})$. Moreover, by Remark 3.13, it follows that
$b_{\gamma_{2}}=0$. Additionally, using (3.50), it gives that
$f_{3}(\gamma_{1})-f_{4}(\gamma_{1})=\big{[}\frac{\sigma^{2}}{\mu}-\bar{\eta}[\mu-\eta]\big{]}\operatorname{e}^{\bar{\eta}\gamma_{1}}$.
From here and (4.8), it follows that $b_{\gamma_{1}}=\bar{x}$, with $\bar{x}$
as in (3.18), and $M_{\gamma}=M(\bar{x})$. On the other hand, notice that
$f_{0}(\alpha_{\gamma_{1}})=0$, which implies that
$x_{\gamma_{1}}=f_{0}(\alpha_{\gamma_{1}})+\bar{x}=\bar{x}$. Therefore, for
each $x\in(0,\infty)$ fixed,
$v_{\gamma,b_{\gamma}}(x)\xrightarrow[\gamma\uparrow\gamma_{1}]{}v_{\gamma,\bar{x}}(x)$
where $v_{\gamma,\bar{x}}$ is as in (4.5).

###  4.2 Cheap reinsurance

Observe that the trajectory $\gamma\mapsto
b_{\gamma}=\hat{x}+g^{-1}_{4}(f_{1}(\gamma))$ is continuous and strictly
increasing on $(\gamma_{1},\infty)$, due to Lemmas 3.2 and 3.16, which
satisfies
$b_{\gamma_{1}}=\hat{x}+g^{-1}_{4}\big{(}\frac{\mu}{2\delta}\big{)}=\hat{x}$
and
$b_{\infty}\raisebox{0.4pt}{$:$}=\lim_{\gamma\uparrow\infty}b_{\gamma}=\hat{x}+g^{-1}_{4}\big{(}\frac{\eta}{\delta}\big{)}=\hat{x}+\frac{1}{\theta_{+}-\theta_{-}}\ln\Big{[}\frac{\bar{a}_{2}[\delta-\eta\theta_{-}]}{\bar{a}_{1}[\delta-\eta\theta_{+}]}\Big{]}$.
Additionally, since

$\lim_{\gamma\uparrow\infty}c_{4,1}(b_{\gamma})=\frac{1}{h^{\prime}_{3}(b_{\infty}-\hat{x})}\quad\text{
and}\quad\ \lim_{\gamma\uparrow\infty}c_{4,2}(b_{\gamma})=0,$

it gives that

$\displaystyle
v_{b_{\infty}}(x)=\lim_{\gamma\uparrow\infty}v_{\gamma,b_{\gamma}}(x)=\begin{cases}\frac{x^{\delta\bar{\eta}/[1+\delta\bar{\eta}]}}{h^{\prime}_{3}(b_{\infty}-\hat{x})}&\text{if}\
x\in(0,\hat{x}),\\\ \vspace{-0.4cm}&\\\
\frac{h_{3}(x-\hat{x})}{{h^{\prime}_{3}(b_{\infty}-\hat{x})}}&\text{if}\
x\in(\hat{x},b_{\infty}),\\\ \vspace{-0.4cm}&\\\
x-b_{\infty}+v_{b_{\infty}}(b_{\infty})&\text{if}\
x\in(b_{\infty},\infty).\end{cases}$ (4.9)

These asymptotic limits allow us to recover the value function for the case of
singular dividend strategies for the optimal problem obtained by Højgaard and
Taksar [5] in Theorem 3.1. Furthermore, a solution to the HJB equation (2.5),
when $\gamma\downarrow\bar{\gamma}_{1}$, is given by

$\displaystyle
v_{\gamma,\hat{x}}(x)=\begin{cases}c_{4,1}(\hat{x})x^{\delta\bar{\eta}/[1+\delta\bar{\eta}]}&\text{if}\
x\in(0,\hat{x}),\\\ \vspace{-0.4cm}&\\\
c_{4,2}(\hat{x})\operatorname{e}^{\lambda_{\bar{\gamma}_{1}}[x-\hat{x}]}+\frac{\bar{\gamma}_{1}}{\bar{\gamma}_{1}+\delta}\Big{[}x-b+v_{\gamma,\hat{x}}(\hat{x})+\frac{\mu}{\delta+\bar{\gamma}_{1}}\Big{]}&\text{if}\
x\in(\hat{x},\infty),\end{cases}$

with
$c_{4,1}(\hat{x})=\frac{\hat{x}^{1/[1+\delta\bar{\eta}]}\big{[}\frac{\bar{\gamma}_{1}}{\bar{\gamma}_{1}+\delta}-\frac{\mu\bar{\gamma}_{1}\lambda_{\bar{\gamma}_{1}}}{[\delta+\bar{\gamma}_{1}]^{2}}\big{]}}{\frac{\delta\bar{\eta}}{1+\delta\bar{\eta}}\big{[}1-\frac{\mu\lambda_{\bar{\gamma}_{1}}}{2[\bar{\gamma}_{1}+\delta]}\big{]}}=\frac{[1+\delta\bar{\eta}]}{\delta\bar{\eta}}\hat{x}^{1/[1+\delta\bar{\eta}]}$
because of
$\frac{1}{\lambda_{\bar{\gamma}_{1}}}=\frac{\mu}{2\delta}-\frac{\mu\bar{\gamma}_{1}}{\delta[\delta+\bar{\gamma}_{1}]}$.
It implies that
$c_{4,2}(\hat{x})=\frac{\sigma^{2}}{\bar{\eta}\mu[\delta+\bar{\gamma}_{1}]}-\frac{\mu\bar{\gamma}_{1}}{[\delta+\bar{\gamma}_{1}]^{2}}=c_{3,2}(1)$,
due to
$\frac{1}{\lambda_{\gamma_{1}}}=-\frac{\delta\sigma^{2}}{\mu[\delta+\bar{\gamma}_{1}]}$.
Taking $b_{\gamma}$ and $\hat{x}$ as in (3.69) and (3.70), respectively, it is
easy to check that
$\lim_{\gamma\uparrow\gamma_{1}}b_{\gamma}=\lim_{\gamma\uparrow\gamma_{1}}x_{\gamma}=\hat{x}$.
Moreover, considering $v_{\gamma,b_{\gamma}}$ as in (3.68), we have that
$v_{\gamma,b_{\gamma}}(x)\xrightarrow[\gamma\uparrow\gamma_{1}]{}v_{\gamma,\hat{x}}(x)$.

##  5 Numerical results

This section presents the numerical results for the various scenarios
discussed in Section 3, taking into account the parameters outlined in Table
1. Each row of Table 1 represents a distinct scenario, as mentioned earlier.
In Figures 1–3, we illustrate the behaviours of different $v_{\gamma,b}$,
considering (3.2), (3.14), (3.23), (3.42), and (3.53).

$\delta$ | $\sigma$ | $\mu$ | $\eta$
---|---|---|---
0.5 | 0.3 | 1.2 | 0.2
1.5 | 0.3 | 0.8 | 0.5
1.5 | 0.5 | 0.7 | 0.7
Table 1: Parameter values of the HJB equation (2.5) where each row of the
table above represents the different scenarios presented in Section 3.

Figure 1: Plots of $v_{\gamma,b}$ considering (3.2) and (3.14) with respect
the parameters outlined in the first row of Table 1. Upper- left corner: plots
of the optimal solution $v_{\gamma,b_{\gamma}}$ and $v_{\infty}$ (given by
(4.3)) with the points $(b_{\gamma},v_{\gamma,b_{\gamma}}(b_{\gamma}))$
indicated by the triangles for $\gamma=2^{N}$ as $N\in[-4,50]$ grows at a rate
of 0.2. Upper-right corner and center below: plots of $v_{\gamma,b_{\gamma}}$
when $\gamma$ is equal to $2^{-0.2}$ and $2^{-2.4}$, respectively, are being
compared with $v_{\gamma,b}$ as $b\in[0,0.5]$ grows at a rate of 0.02.

Taking into account Propositions 3.1 and 3.4, in the upper-left corner of
Figure 1 depicts the behaviours of the optimal solution
$v_{\gamma,b_{\gamma}}$ concerning the parameter $\gamma=2^{N}$ as
$N\in[-4,50]$ with a growth rate of 0.2. It is evident that as $N$ increases,
$v_{\gamma,b_{\gamma}}$ (blue line) closely approaches $v_{\infty}$ (red
line). Here, $v_{\infty}$, given by (4.3), represents the value function for
the singular dividend strategies for the optimal problem derived by Taksar
[13] in Theorem 5.2.

In addition, recalling that $b_{\gamma}$ is given by (3.15) if
$\gamma>\gamma_{0}\approx 0.2812$, otherwise $b_{\gamma}=0$, we can appreciate
the behaviour of the trajectory
$\gamma\mapsto(b_{\gamma},v_{\gamma,b_{\gamma}}(b_{\gamma}))$ (triangular
points) in this graphic. It exhibits an increasing trend and is converging to
the point $(b_{\infty},v_{\infty}(b_{\infty}))\approx(0.3121,0.4)$ as
$N\uparrow\infty$, where $b_{\infty}$ is as in Sub-subsection 4.1.1. Moreover,
according to Proposition 3.4, in the case where $\mu\geq 2\eta$ (with
$\mu>\eta$), the optimal strategy is to take the maximum risk, resulting in
$x_{\gamma}=0$ for any value of $\gamma$. Conversely, in the upper-right
corner and in the center below of Figure 1, the plots of
$v_{\gamma,b_{\gamma}}$ (red line) are compared when $\gamma$ takes values of
$2^{-0.2}$ and $2^{-2.4}$, respectively, with the plots of $v_{\gamma,b}$
(blue line) as $b\in[0,0.5]$ grows at a rate of 0.02. For the former,
$b_{2^{-0.2}}\approx 0.1082$, and for the latter, $b_{2^{-2.4}}=0$.

Figure 2:  Plots of $v_{\gamma,b}$ considering (3.23), (3.42) and (3.53) with
respect the parameters outlined in the second row of Table 1. Upper- left
corner: plots of the optimal solution $v_{\gamma,b_{\gamma}}$ and $v_{\infty}$
(given by (4.4)) with the points
$(x_{\gamma},v_{\gamma,b_{\gamma}}(x_{\gamma}))$ and
$(b_{\gamma},v_{\gamma,b_{\gamma}}(b_{\gamma}))$ indicated by the circles and
triangles, respectively, for $\gamma=2^{N}$ as $N\in[-4,50]$ grows at a rate
of 0.2. Upper-right, left-below and right-below corners: plots of
$v_{\gamma,b_{\gamma}}$ when $\gamma$ is equal to $2$, $2^{-0.4}$ and
$2^{-1.4}$, respectively, are compared with $v_{\gamma,b}$ as $b\in[0,0.5]$
grows at a rate of 0.02.

Figure 3: Plots of $v_{\gamma,b}$ considering (3.60) and (3.68) with respect
the parameters outlined in the first row of Table 1. Upper-left corner: plots
of the optimal solution $v_{\gamma,b_{\gamma}}$ and $v_{\infty}$ (given by
(4.9)) with the points $(x_{\gamma},v_{\gamma,b_{\gamma}}(x_{\gamma}))$ and
$(b_{\gamma},v_{\gamma,b_{\gamma}}(b_{\gamma}))$ indicated by the circles and
triangles, respectively, for $\gamma=2^{N}$ as $N\in[-4,50]$ grows with at a
rate of 0.2. Upper-right corner and center below: plots of
$v_{\gamma,b_{\gamma}}$ when $\gamma$ is equal to $2^{2.2}$ and $2^{0.8}$,
respectively, are compared with $v_{\gamma,b}$ as $b\in[0,0.5]$ grows at a
rate of 0.02.

Utilizing Propositions 3.5, 3.8, 3.14 and 3.17 and considering the second and
the third row of Table 1, in the upper-left corner of Figures 2 and 3 depict
the behaviours of the optimal solution $v_{\gamma,b_{\gamma}}$ with respect to
$\gamma=2^{N}$ with $N$ as previously defined. We observe that
$v_{\gamma,b_{\gamma}}$ (blue line) goes to $v_{\infty}$ (red line) when
$N\uparrow\infty$. Here, $v_{\infty}$, as described in (4.4) and (4.9)
respectively, stands the value function for the singular dividend strategies
for the optimal problem obtained by Taksar [13] in Theorem 5.3 and Højgaard
and Taksar [5] in Theorem 3.1, respectively.

In the case that $\mu<2\eta$ (with $\mu>\eta$), as indicated in Propositions
3.8 and 3.13, the values of $b_{\gamma}$ is determined by (3.31) if
$\gamma>\gamma_{1}\approx 1.0078$, and by (3.46) if
$\gamma\in(\gamma_{2},\gamma_{1})$, with $\gamma_{2}\approx 0.3979$;
otherwise, it is zero. In the upper-left corner of Figure 2, we observe the
trajectory $\gamma\mapsto(b_{\gamma},v_{\gamma,b_{\gamma}}(b_{\gamma}))$
(triangular points), which exhibits an increasing trend and converges to the
point $(b_{\infty},v_{\infty}(b_{\infty}))\approx(0.2131,0.3333)$ as
$N\uparrow\infty$, where $b_{\infty}$ is as in Sub-subsection 4.1.2.
Additionally, we can also observe the behaviour of the trajectory
$\gamma\mapsto(x_{\gamma},v_{\gamma,b_{\gamma}}(x_{\gamma}))$ (circular point)
plotted with $x_{\gamma}$ determined by (3.18) if $\gamma>\gamma_{1}$, by
(3.54) if $\gamma\in(\gamma_{2},\gamma_{1})$ and by (3.55) if
$\gamma\in(0,\gamma_{2})$. It is noteworthy that
$(x_{\gamma},v_{\gamma,b_{\gamma}}(x_{\gamma}))$ converges to the point
$(x_{\infty},v_{b_{\gamma}}(x_{\infty}))\approx(0.0512,0.1299)$ as
$N\uparrow\infty$. On the other hand, in the upper-right, left-below and
right-below corners of Figure 2, the plots of $v_{\gamma,b_{\gamma}}$ (red
line) when $\gamma$ is equal to $2$, $2^{-0.4}$, and $2^{-1.4}$, respectively,
are compared with $v_{\gamma,b}$ (blue line) as $b\in[0,0.5]$ increases at a
rate of 0.02. In the first case $x_{2}\approx 0.0512$ and $b_{2}\approx
0.0862$; in the second case $x_{2^{-0.4}}\approx 0.0516$ and
$b_{2^{-0.4}}=0.0357$; and in the last case $x_{2^{-1.4}}\approx 0.0553$ and
$b_{2^{-1.4}}=0$.

In the case where $\mu=\eta$, similar results to those described in the
paragraph above are obtained. However, it should be noted that the strategy
involving $b=0$ is not considered optimal for any value of $\gamma$, as
depicted in Figure 3. This scenario is characterized by
$\bar{\gamma}_{1}\approx 3.7959$, along with the values
$(x_{\infty},v_{\infty}(x_{\infty}))\approx(0.1411,0.2888)$ and
$(b_{\infty},v_{\infty}(b_{\infty}))\approx(0.3071,0.4667)$. Furthermore, in
the upper-right corner and in the center below of Figure 3, the plots of
$v_{\gamma,b_{\gamma}}$ (red line) when $\gamma$ is equal to $2^{2.2}$ and
$2^{0.8}$, respectively, are compared with $v_{\gamma,b}$ (blue line) when $b$
is as before. In the former case $x_{2^{2.2}}\approx 0.1411$ and
$b_{2^{2.2}}\approx 0.1542$, and in the latter $x_{2^{0.8}}\approx 0.1472$ and
$b_{2^{0.8}}=0.0833$.

##  Appendix A Proofs of some technical results

In the upcoming subsections, we will present the proofs for the results
discussed in Section 3. To establish $v_{\gamma,b}$ as in (3.2), (3.23) and
(3.42), we will utilize the following results. Given the assumption that
$v_{\gamma,b}$ is concave on some open set $\mathcal{O}$, there exists a
function $\mathpzc{x}$ satisfying
$-\ln[v^{\prime}_{\gamma,b}(\mathpzc{x}(z))]=z$. This transformation has been
applied in numerous optimal control problems; see e.g. [13, 5] and references
therein. Additionally, it is worth noting that considering
$\lambda_{\gamma},\lambda_{+}$ as the negative and positive roots of (3.3),
respectively, the function

$v(x)=c_{3}\operatorname{e}^{\lambda_{\gamma}[x-\max\\{x_{b},b\\}]}+c_{4}\operatorname{e}^{\lambda_{+}[x-\max\\{x_{b},b\\}]}+\frac{\gamma}{\gamma+\delta}\bigg{[}x-b+v(b)+\frac{\eta}{\gamma+\delta}\bigg{]}$
(A.1)

satisfies the equation $\frac{1}{2}\sigma^{2}v^{\prime\prime}(x)+\eta
v^{\prime}(x)-\delta v(x)+\gamma\\{x-b_{\gamma}+v(b_{\gamma})-v(x)\\}=0$, for
$x>\max\\{x_{b},b\\}$. Here, $c_{3}$ and $c_{4}$ are parameters to be found.
However, since our solution must have a linear growth, we conclude that
$c_{4}=0$.

###  A.1 Proofs of Propositions 3.1, 3.4, and Lemmas 3.2, 3.3

###### Proof of Proposition 3.1.

Since $v_{\gamma,b}(0)=0$, a solution of (3.1) has the form
$v_{\gamma,b}(x)=k_{1}h_{1}(x)$ for $x\in(0,b)$, otherwise, i.e. for $x>b$,
$v_{\gamma,b}(x)$ is given by (A.1), with $\max\\{x_{b},b\\}=b$ and $c_{4}=0$.
Here $k_{1},c_{3}$ are parameters to be found, and $h_{1}$ is as in (3.4). To
determine the values of the parameters, applying smooth fit at $b>0$, we have
that

$k_{1}h_{1}(b)=c_{3}+\frac{\gamma}{\gamma+\delta}\bigg{\\{}k_{1}h_{1}(b)+\frac{\eta}{\gamma+\delta}\bigg{\\}}\quad\text{and}\quad
k_{1}h^{\prime}_{1}(b)=c_{3}\lambda_{\gamma}+\frac{\gamma}{\gamma+\delta}.$
(A.2)

Solving the system of equations above w.r.t. $k_{1},c_{3}$, it can be easily
checked that $c_{1,1}(b)$ and $c_{1,2}(b)$ as in (3.6) satisfy (A.2). From
here, we have immediately that $v_{\gamma,b}$ given as in (3.2) is
$\operatorname{C}^{2}$-continuous on $(0,\infty)\setminus\\{b\\}$. By (3.1),
we have for $x>b$,
$v^{\prime\prime}_{\gamma,b}(x)=-\frac{2}{\sigma^{2}}\\{\eta
v^{\prime}_{\gamma,b}(x)-\delta
v_{\gamma,b}(x)+\gamma[x-b+v_{\gamma,b}(b)-v_{\gamma,b}(x)]\\}$. Then, letting
$x\downarrow b$ in the equation above, it follows that
$v_{\gamma,b}^{\prime\prime}(b+)=-\frac{2}{\sigma^{2}}\\{\eta
v_{\gamma,b}^{\prime}(b)-[\delta+\gamma]v_{\gamma,b}(b)=v_{\gamma,b}^{\prime\prime}(b-)$.
Therefore, $v$ is $\operatorname{C}^{2}$-continuous at $b$. ∎

###### Proof of Lemma 3.2.

Let $g_{1}$ be as in (3.9). Calculating the first derivative of $g_{1}$ and
considering $h_{1}$ as in (3.4) and its first and second derivative, we get
that
$g^{\prime}_{1}(b)=\frac{[h^{\prime}_{1}(b)]^{2}-h_{1}(b)h^{\prime\prime}_{1}(b)}{[h^{\prime}_{1}(b)]^{2}}=\frac{[\theta_{+}-\theta_{-}]^{2}\operatorname{e}^{[\theta_{+}-\theta_{-}]b}}{[h^{\prime}_{1}(b)]^{2}}$
which is positive. Then, by the seen before, we have that $g^{\prime}_{1}>0$
on $(0,\infty)$, which is enough to conclude that $g$ is strictly increasing
on $(0,\infty)$. By definition of $h_{1}$, it is easy to verify that
$\lim_{b\downarrow 0}g_{1}=0$ and
$\lim_{b\uparrow\infty}g_{1}(x)=\lim_{b\rightarrow\infty}\frac{1-\operatorname{e}^{[\theta_{-}-\theta_{+}]b}}{\theta_{+}-\theta_{-}\operatorname{e}^{[\theta_{-}-\theta_{+}]b}}=\frac{1}{\theta_{+}}$.∎

###### Proof of Lemma 3.3.

Let $f_{1}$ be as in (3.11). Since the functions
$\gamma\mapsto\frac{\eta\gamma}{\delta[\delta+\gamma]}$ and
$\gamma\mapsto\frac{\sigma^{2}}{\eta+\sqrt{\eta^{2}+2\sigma^{2}(\delta+\gamma)}}$
are strictly increasing and strictly decreasing on $(0,\infty)$, respectively,
it immediately follows that $f_{1}$ is strictly increasing on $(0,\infty)$.
The reader can quickly verify that the limits given in Lemma 3.3 hold. ∎

###### Proof of Proposition 3.4.

Assuming that $\mu\geq 2\eta$ and $\gamma>\gamma_{0}$, it is known that there
exists $b_{\gamma}>0$ satisfying (3.8). Take $v_{\gamma,b_{\gamma}}$ as in
(3.2) when $b=b_{\gamma}$. Observe that $c_{1,1}(b_{\gamma})$, defined in
(3.6), is positive because of $\lambda_{\gamma}<0$ and
$h^{\prime}_{1}(b_{\gamma})-\frac{\delta\lambda_{\gamma}}{\gamma+\delta}h_{1}(b_{\gamma})=\frac{h^{\prime}_{1}(b_{\gamma})}{\gamma+\delta}\big{[}\gamma-\frac{\gamma\eta\lambda_{\gamma}}{\gamma+\delta}\big{]}>0$.
It is worth noting that $v_{\gamma,b_{\gamma}}$ inherits the increasing
property of $h_{1}$ on $(0,b_{\gamma})$, due to $h^{\prime}_{1}>0$ on
$(0,b_{\gamma})$. Since $h^{\prime}_{1}$ attains its unique minimum value at
$x_{1}=\frac{2}{\theta_{+}-\theta_{-}}\ln\big{[}-\frac{\theta_{-}}{\theta_{+}}\big{]}$,
it is sufficient to verify that
$h^{\prime\prime}_{1}(b_{\gamma})<h^{\prime\prime}_{1}(x_{1})=0$ in order to
establish that $b_{\gamma}<x_{1}$, and consequently conclude that
$v_{\gamma,b_{\gamma}}$ is concave on $(0,b_{\gamma})$. It is known

$\frac{\sigma^{2}}{2}h^{\prime\prime}_{1}+\eta h^{\prime}_{1}-\delta
h_{1}=0\quad\text{on}\ (0,\infty).$ (A.3)

From here and using (3.8), it gives that
$h^{\prime\prime}_{1}=\frac{2h^{\prime}_{1}}{\sigma^{2}}[\delta g_{1}-\eta]$
on $(0,\infty)$. Then
$h^{\prime\prime}_{1}(b_{\gamma})=\frac{2h^{\prime}_{1}(b_{\gamma})}{\sigma^{2}}\big{[}\frac{1}{\lambda_{\gamma}}\big{[}\delta+\frac{\gamma\eta\lambda_{\gamma}}{\gamma+\delta}\big{]}-\eta\big{]}=\frac{2\delta
h^{\prime}_{1}(b_{\gamma})}{\sigma^{2}}\big{[}\frac{1}{\lambda_{\gamma}}-\frac{\eta}{\gamma+\delta}\big{]}<0$,
because of $h^{\prime}_{1}(b_{\gamma})>0$ and $\lambda_{\gamma}<0$. Now, let
us see that $v_{\gamma,b_{\gamma}}$ is increasing and concave on
$(b_{\gamma},\infty)$. For this purpose, it is sufficient to check that

$c_{1,2}(b_{\gamma})<0,$ (A.4)

because if (A.4) holds, it follows immediately that
$v^{\prime}_{\gamma,b_{\gamma}}>0$ and
$v^{\prime\prime}_{\gamma,b_{\gamma}}<0$ on $(b_{\gamma},\infty)$. Using
(3.6), (3.9) and (3.15), we see that (A.4) is equivalent to
$0<\delta[\delta+\gamma-\delta\eta\lambda_{\gamma}]$, which is true due to
$\lambda_{\gamma}<0<\delta$. By the seen before, it is known that
$v^{\prime}_{\gamma,b_{\gamma}}$ is strictly decreasing on $(0,\infty)$.
Moreover, considering that $b_{\gamma}$ is the unique solution to (3.8), it
follows that
$v^{\prime}_{\gamma,b_{\gamma}}(b_{\gamma}-)=\frac{\frac{\gamma}{\gamma+\delta}\big{[}1-\frac{\eta\lambda_{\gamma}}{\gamma+\delta}\big{]}}{1-\frac{\delta\lambda_{\gamma}}{\gamma+\delta}g_{1}(b_{\gamma})}=1$.
It implies that
$b_{\gamma}=\inf\\{x>0:v^{\prime}_{\gamma,b_{\gamma}}(x)<1\\}<\infty$. Let us
prove now that
$\hat{f}_{1}\raisebox{0.4pt}{$:$}=v^{\prime}_{\gamma,b_{\gamma}}/v^{\prime\prime}_{\gamma,b_{\gamma}}$
is decreasing on $(0,\infty)$. Calculating the first and second derivatives of
$v_{\gamma,b_{\gamma}}$, and taking into account (A.3), it gives

$\displaystyle\hat{f}_{1}(x)=\begin{cases}\frac{\sigma^{2}}{2[\delta
g_{1}(x)-\eta]}&\text{if}\ x\in(0,b_{\gamma}),\\\ \vspace{-0.4cm}&\\\
\frac{\gamma\operatorname{e}^{-\lambda_{\gamma}[x-b_{\gamma}]}}{c_{1,2}(b_{\gamma})\lambda_{\gamma}^{2}[\gamma+\delta]}+\frac{1}{\lambda_{\gamma}}&\text{if}\
x\in(b_{\gamma},\infty).\\\ \end{cases}$

From here we see that $\hat{f}_{1}$ is decreasing on $(0,\infty)$ since
$\delta g_{1}-\eta$ is increasing in $(0,b_{\gamma})$, and
$\hat{f}^{\prime}_{1}(x)=-\frac{\gamma\operatorname{e}^{-\lambda_{\gamma}[x-b_{\gamma}]}}{c_{1,2}(b_{\gamma})\lambda_{\gamma}[\gamma+\delta]}<0$
for $x\in(b_{\gamma},\infty).$ By the seen before, we get that
$-\frac{\mu}{\sigma^{2}}\hat{f}_{1}$ is increasing on $(0,\infty)$ and
$-\frac{\mu}{\sigma^{2}}\hat{f}_{1}(0+)=\frac{\mu}{2\eta}\geq 1$. Therefore
$x_{\gamma}=0$. Since $v_{\gamma,b_{\gamma}}$ satisfies the hypotheses of
Proposition 2.2, we conclude that $v_{\gamma,b_{\gamma}}$ is a solution to the
HJB equation (2.5).

Now, let us check the veracity of Item (ii). For each
$\gamma\in(0,\gamma_{0})$ fixed, let us take $v_{\gamma,0}$ as in (3.14). By
similar arguments seen before, it is easy to check that $v_{\gamma,0}$ is an
increasing and concave solution to (3.1). Notice that
$v^{\prime}_{\gamma,0}(0+)=-\frac{\gamma\eta\lambda_{\gamma}}{[\gamma+\delta]^{2}}+\frac{\gamma}{\gamma+\delta}$,
which is less or equal than one if and only if

$-\lambda_{\gamma}<\frac{\delta[\gamma+\delta]}{\eta\gamma}.$ (A.5)

Since

$\lambda_{\gamma}=\frac{-\eta-\sqrt{\eta^{2}+2\sigma^{2}[\delta+\gamma]}}{\sigma^{2}},$
(A.6)

we see that (A.5) is equivalent to verify that
$\gamma<\frac{1}{2}\big{[}\frac{\sigma\delta}{\eta}\big{]}^{2}$. Meanwhile,
observe that
$f_{1}\big{(}\frac{1}{2}\big{[}\frac{\sigma\delta}{\eta}\big{]}^{2}\big{)}=0$,
with $f_{1}$ as in (3.11). It implies that
$\gamma_{0}=\frac{1}{2}\big{[}\frac{\sigma\delta}{\eta}\big{]}^{2}$, since
$f_{1}$ is strictly decreasing on $(0,\infty)$ and $\gamma_{0}$ is the unique
point where $f_{1}(\gamma_{0})=0$. From here (A.5) is true. Concluding that
$v^{\prime}_{\gamma,0}(0+)\leq 1$ and
$b_{\gamma}=\inf\\{x>0:v^{\prime}_{\gamma,0}(x)\leq 1\\}=0$. On the other
hand,
$\frac{v^{\prime}_{\gamma,0}(x)}{v^{\prime\prime}_{\gamma,0}(x)}=\frac{1}{\lambda_{\gamma}}-\frac{[\gamma+\delta]\operatorname{e}^{-\lambda_{\gamma}x}}{\eta\lambda_{\gamma}^{2}}$
is decreasing on $(0,\infty)$. Thus, $-\frac{\mu
v^{\prime}_{\gamma,0}}{\sigma^{2}v^{\prime\prime}_{\gamma,0}}$ is increasing
on the same domain mentioned before, and

$\displaystyle-\frac{\mu
v^{\prime}_{\gamma,0}(0+)}{\sigma^{2}v^{\prime\prime}_{\gamma,0}(0+)}=-\frac{\mu}{\sigma^{2}\lambda_{\gamma}}\bigg{[}1-\frac{\gamma+\delta}{\lambda_{\gamma}\eta}\bigg{]}=\frac{\mu}{2\eta}.$
(A.7)

The last equality in (A.7) is obtained because of (A.6). By the seen before
and invoking Proposition 2.2, we conclude that $v_{\gamma,0}$, given in (3.14)
is a solution to the HJB equation (2.5).

To finish, let us see that $v_{\gamma,0}$ as in (3.14), when
$\gamma=\gamma_{0}$, satisfies also (2.5). Consider $v_{\gamma,b_{\gamma}}$ as
in (3.2) when $b=b_{\gamma}$. Then, by the seen in Subsection 4.1, we get that
for each $x\in(0,\infty)$,
$v_{\gamma,b_{\gamma}}(x)\xrightarrow[\gamma\downarrow\gamma_{0}]{}v_{\gamma_{0},0}(x)$,
with $v_{\gamma_{0},0}$ as in (3.14). Furthermore,
$v_{\gamma,b_{\gamma}}(b_{\gamma})\xrightarrow[\gamma\downarrow\gamma_{0}]{}v_{\gamma_{0},0}(0)=0$
and
$v^{\prime}_{\gamma,b_{\gamma}}(b_{\gamma})\xrightarrow[\gamma\downarrow\gamma_{0}]{}v^{\prime}_{\gamma_{0},0}(0)=1$.
By the same arguments seen in the case $\gamma<\gamma_{0}$, it follows that
$v_{\gamma,0}$ as in (3.14), when $\gamma=\gamma_{0}$, satisfies also (2.5). ∎

###  A.2 Proofs of Propositions 3.5, 3.8, and Lemma 3.7

###### Proof of Proposition 3.5.

Consider $b>\bar{x}$, with $\bar{x}$ as in (3.18). Let us suppose that the
solution $v_{\gamma,b}$ to (3.20)–(3.22) is concave on $(0,x_{b})$ which will
be proven later on. Here, $x_{b}\leq b$ is a parameter to be determined. Then,
by this assumption, it is known that there exists
$\mathpzc{x}_{\,1}:(-\infty,\bar{z})\longrightarrow[0,\infty)$ satisfying
$-\ln(v^{\prime}_{\gamma,b}(\mathpzc{x}_{\,1}(z)))=z$, for some
$\bar{z}\in\mathbbm{R}$. It implies (3.28). Applying (3.28) in (3.20), it
follows that

$\displaystyle\frac{\mu^{2}\mathpzc{x}_{\,1}^{\prime}(z)\operatorname{e}^{-z}}{2\sigma^{2}}-[\mu-\eta]\operatorname{e}^{-z}-\delta
v_{\gamma,b}(\mathpzc{x}_{\,1}(z))=0.$ (A.8)

Taking first derivatives w.r.t. $z$ in (A.8),

$\displaystyle\mathpzc{x}_{\,1}^{\prime\prime}(z)-[1+\delta\bar{\eta}]\mathpzc{x}_{\,1}^{\prime}(z)+\bar{\eta}[\mu-\eta]=0,$
(A.9)

where $\bar{\eta}$ is defined in (3.19). A solution to (A.9) is given by

$\mathpzc{x}_{\,1}(z)=k_{1}\operatorname{e}^{[1+\delta\bar{\eta}]z}+\frac{\bar{\eta}[\mu-\eta]}{1+\delta\bar{\eta}}z+k_{2}.$
(A.10)

where $k_{1}$ and $k_{2}$ are positive constants that shall be given after.
Taking $M>0$ such that $\mathpzc{x}_{\,1}(-M+)=0$ and
$v^{\prime}_{\gamma,b}(\mathpzc{x}_{\,1}(-M+))=\operatorname{e}^{M}$, the
function $\mathpzc{x}_{\,1}$ is from $[-M,\bar{z})$ to $[0,x_{b})$, where
$\bar{z}>-M$ such that $\mathpzc{x}_{\,1}(\bar{z})=x_{b}$. It implies that
$k_{1}=-\operatorname{e}^{M[1+\delta\bar{\eta}]}\big{[}k_{2}-\frac{\bar{\eta}M[\mu-\eta]}{1+\delta\bar{\eta}}\big{]}$.
By the assumption of concavity of $v_{\gamma,b}$ on $(0,x_{b})$, it implies
that $\mathpzc{x}_{\,1}^{-1}$ is an increasing function on $[0,x_{b})$. Notice
that

$\mathpzc{x}_{\,1}^{\prime}(z)=k_{1}[1+\delta\bar{\eta}]\operatorname{e}^{[1+\delta\bar{\eta}]z}+\frac{\bar{\eta}[\mu-\eta]}{1+\delta\bar{\eta}}.$
(A.11)

Then, from here and (A.8), we have that a solution to (3.20) is represented by

$\displaystyle v_{\gamma,b}(x)$
$\displaystyle=\frac{\operatorname{e}^{-\mathpzc{x}_{\,1}^{-1}(x)}}{\bar{\eta}\delta}[\mathpzc{x}_{\,1}^{\prime}(\mathpzc{x}_{\,1}^{-1}(x))+\bar{\eta}[\eta-\mu]]$
$\displaystyle=\frac{\operatorname{e}^{-\mathpzc{x}_{\,1}^{-1}(x)}}{\bar{\eta}\delta}\bigg{[}k_{1}[1+\delta\bar{\eta}]\operatorname{e}^{[1+\delta\bar{\eta}]\mathpzc{x}_{\,1}^{-1}(x)}-\frac{\bar{\eta}[\eta-\mu]}{1+\delta\bar{\eta}}+\bar{\eta}[\eta-\mu]\bigg{]}$
$\displaystyle=\frac{\operatorname{e}^{-\mathpzc{x}_{\,1}^{-1}(x)}[\eta-\mu]}{\delta}\bigg{[}-\bigg{[}k_{2}+\frac{\bar{\eta}M[\eta-\mu]}{1+\delta\bar{\eta}}\bigg{]}\bigg{[}\frac{1+\delta\bar{\eta}}{\bar{\eta}[\eta-\mu]}\bigg{]}\operatorname{e}^{[1+\delta\bar{\eta}][M+\mathpzc{x}_{\,1}^{-1}(x)]}+\frac{\delta\bar{\eta}}{1+\delta\bar{\eta}}\bigg{]}.$

Since $v_{\gamma,b}(0)=0$, we have that

$\displaystyle
k_{2}=c_{2,1}\bigg{[}M-\frac{\delta\bar{\eta}}{1+\delta\bar{\eta}}\bigg{]}.$
(A.12)

From here notice that

$k_{1}=\frac{\delta\bar{\eta}c_{2,1}}{1+\delta\bar{\eta}}\operatorname{e}^{M[1+\delta\bar{\eta}]}.$
(A.13)

Then,

$\displaystyle v_{\gamma,b}(x)$
$\displaystyle=\frac{\operatorname{e}^{-\mathpzc{x}_{\,1}^{-1}(x)}[\eta-\mu]}{\delta}\bigg{[}-\frac{\delta\bar{\eta}}{1+\delta\bar{\eta}}\operatorname{e}^{[1+\delta\bar{\eta}][M+\mathpzc{x}_{\,1}^{-1}(x)]}+\frac{\delta\bar{\eta}}{1+\delta\bar{\eta}}\bigg{]}$
$\displaystyle=c_{2,1}\operatorname{e}^{-\mathpzc{x}_{\,1}^{-1}(x)}\Big{[}\operatorname{e}^{[1+\delta\bar{\eta}][M+\mathpzc{x}_{\,1}^{-1}(x)]}-1\Big{]}\quad\text{for}\
x\in(0,x_{b}),$ (A.14)

with $c_{2,1}$ as in (3.19). Observe that $v_{\gamma,b}$ is concave on
$(0,x_{b})$, since $v^{\prime\prime}_{\gamma,b}$ is non-positive on
$(0,x_{b})$, because of (3.28) and $\mathpzc{x}_{\,1}^{\prime}\geq 0$ on
$(-M,\bar{z})$. From (3.16), (A.11) and (A.2), it yields that $u^{*}_{b}$ has
the following form for each $x\in(0,x_{b})$,
$u^{*}_{b}(x)=\frac{\mu}{\sigma^{2}}\mathpzc{x}_{\,1}^{\prime}(\mathpzc{x}_{\,1}^{-1}(x))=\frac{2[\mu-\eta]}{\mu[1+\delta\bar{\eta}]}\\{\delta\bar{\eta}\operatorname{e}^{[1+\delta\bar{\eta}][M+\mathpzc{x}_{\,1}^{-1}(x)]}+1\\}$.
We see that $u^{*}_{b}$ is increasing on $(0,x_{b})$ and

$\displaystyle\lim_{x\downarrow 0}u^{*}_{b}(x)=\frac{2[\mu-\eta]}{\mu}<1,$
(A.15)

because of $\mu<2\eta$. Then, by (3.16), we have that $x_{b}$ is determined by
the condition $u^{*}_{b}(x_{b}-)=1$, which is equivalent to
$\frac{2[\mu-\eta]}{\mu[1+\delta\bar{\eta}]}\\{\delta\bar{\eta}\operatorname{e}^{[1+\delta\bar{\eta}][M+\mathpzc{x}_{\,1}^{-1}(x_{b})]}+1\\}=1$.
From here, using (A.10) and considering that $\mathpzc{x}_{\,1}^{-1}$ is the
inverse function of $\mathpzc{x}_{\,1}$, it gives that $\bar{z}$ is as in
(3.25) and

$\displaystyle x_{b}$
$\displaystyle=\mathpzc{x}_{\,1}\bigg{(}\frac{1}{1+\delta\bar{\eta}}\ln[c_{2,2}]-M\bigg{)}$
$\displaystyle=k_{1}c_{2,2}\operatorname{e}^{-[1+\delta\bar{\eta}]M}+\frac{\bar{\eta}[\mu-\eta]}{[1+\delta\bar{\eta}]^{2}}\ln[c_{2,2}]-\frac{\bar{\eta}[\mu-\eta]}{1+\delta\bar{\eta}}\frac{\delta\bar{\eta}}{1+\delta\bar{\eta}}$
$\displaystyle=\frac{c_{2,1}}{1+\delta\bar{\eta}}\ln[c_{2,2}]+\frac{\bar{\eta}[2\eta-\mu]}{2[1+\delta\bar{\eta}]}=\bar{x}.$

Recall that $c_{2,2}$ and $k_{1}$ are as in (3.19) and (A.13). On the other
hand, The NLPD equation (3.21) admits the following solution

$v_{\gamma,b}(x)=\begin{cases}c_{1}\operatorname{e}^{\theta_{-}[x-x_{b}]}+c_{2}\operatorname{e}^{\theta_{+}[x-x_{b}]}&\text{for}\
x\in(x_{b},b),\\\ \vspace{-0.4cm}&\\\
c_{3}\operatorname{e}^{\lambda_{\gamma}[x-b]}+\frac{\gamma}{\gamma+\delta}\big{[}x-b+v_{\gamma,b}(b)+\frac{\eta}{\delta+\gamma}\big{]}&\text{for}\
x\in(b,\infty),\end{cases}$

where $c_{1},c_{2},c_{3}$ are free constants. Notice that

$\displaystyle v_{\gamma,b}(x_{b}-)$
$\displaystyle=\frac{\operatorname{e}^{-\mathpzc{x}_{\,1}^{-1}(x_{b})}\bar{\eta}[\mu-\eta]}{1+\delta\bar{\eta}}\Big{[}\operatorname{e}^{[1+\delta\bar{\eta}][M+\mathpzc{x}_{\,1}^{-1}(x_{b})]}-1\Big{]}=\operatorname{e}^{M}\frac{c_{2,2}^{-1/[1+\delta\bar{\eta}]}[2\eta-\mu]}{2\delta},$
$\displaystyle v^{\prime}_{\gamma,b}(x_{b}-)$
$\displaystyle=\operatorname{e}^{-\mathpzc{x}_{\,1}^{-1}(x_{b})}=\operatorname{e}^{M}c_{2,2}^{-1/[1+\delta\bar{\eta}]},$

because of

$\displaystyle\mathpzc{x}_{\,1}^{-1}(x_{b})$
$\displaystyle=\mathpzc{x}_{\,1}^{-1}\bigg{(}\mathpzc{x}_{\,1}\bigg{(}\frac{1}{1+\delta\bar{\eta}}\ln[c_{2,2}]-M\bigg{)}\bigg{)}=\frac{1}{1+\delta\bar{\eta}}\ln[c_{2,2}]-M,$
$\displaystyle\operatorname{e}^{-\mathpzc{x}_{\,1}^{-1}(x_{b})}$
$\displaystyle=\exp\bigg{[}-\frac{1}{1+\delta\bar{\eta}}\ln[c_{2,2}]+M\bigg{]}=c_{2,2}^{-1/[1+\delta\bar{\eta}]}\operatorname{e}^{M},$
$\displaystyle\operatorname{e}^{[1+\delta\bar{\eta}]\mathpzc{x}_{\,1}^{-1}(x_{b})}$
$\displaystyle=\exp\bigg{[}\ln[c_{2,2}]-[1+\delta\bar{\eta}]M\bigg{]}=c_{2,2}\operatorname{e}^{-[1+\delta\bar{\eta}]M}.$

Meanwhile, $v_{\gamma,b}(x_{b}+)=c_{1}+c_{2}$ and
$v^{\prime}_{\gamma,b}(x_{b}+)=c_{1}\theta_{-}+c_{2}\theta_{+}$. In order that
$v_{\gamma,b}$ is $\operatorname{C}^{1}$-continuous at $x_{b}$, by smooth fit,
we have the following system of equations

$\displaystyle
c_{1}+c_{2}=\operatorname{e}^{M}\frac{c_{2,2}^{-1/[1+\delta\bar{\eta}]}[2\eta-\mu]}{2\delta}\quad\text{and}\quad
c_{1}\theta_{-}+c_{2}\theta_{+}=\operatorname{e}^{M}c_{2,2}^{-1/[1+\delta\bar{\eta}]},$

whose solution is given by
$c_{1}=-\frac{\operatorname{e}^{M}c_{2,2}^{-1/[1+\delta\bar{\eta}]}}{\theta_{+}-\theta_{-}}\Big{[}1-\theta_{+}\frac{[2\eta-\mu]}{2\delta}\Big{]}$
and
$c_{2}=\frac{\operatorname{e}^{M}c_{2,2}^{-1/[1+\delta\bar{\eta}]}}{\theta_{+}-\theta_{-}}\Big{[}1-\theta_{-}\frac{[2\eta-\mu]}{2\delta}\Big{]}$.
From here, we have that for $x\in(x_{b},b)$,
$v_{\gamma,b}(x)=\operatorname{e}^{M}c_{2,2}^{-1/[1+\delta\bar{\eta}]}h_{2}(x-x_{b})$,
where $h_{2}$ is defined in (3.26). Then,

$\displaystyle
v_{\gamma,b}(b-)=\operatorname{e}^{M}c_{2,2}^{-1/[1+\delta\bar{\eta}]}h(b-x_{b}),\quad
v^{\prime}_{\gamma,b}(b-)=\operatorname{e}^{M}c_{2,2}^{-1/[1+\delta\bar{\eta}]}h^{\prime}(b-x_{b}),$

and

$\displaystyle v_{\gamma,b}(b+)$
$\displaystyle=c_{3}+\frac{\gamma}{\gamma+\delta}\bigg{[}\operatorname{e}^{M}c_{2,2}^{-1/[1+\delta\bar{\eta}]}h(b-x_{b})+\frac{\eta}{\delta+\gamma}\bigg{]},$
$\displaystyle v^{\prime}_{\gamma,b}(b+)$
$\displaystyle=c_{3}\lambda_{\gamma}+\frac{\gamma}{\gamma+\delta}.$ (A.16)

Since $v_{\gamma,b}(b-)=v_{\gamma,b}(b+)$, we have that

$\displaystyle c_{3}=\frac{\operatorname{e}^{M}\delta
c_{2,2}^{-1/[1+\delta\bar{\eta}]}}{\delta+\gamma}h(b-x_{b})-\frac{\eta\gamma}{[\delta+\gamma]^{2}}.$
(A.17)

Substituting (A.17) into (A.16), and taking into account that
$v^{\prime}_{\gamma,b}(b-)=v^{\prime}(b+)$ must be occurred, it gives that

$\displaystyle\operatorname{e}^{M}c_{2,2}^{-1/[1+\delta\bar{\eta}]}h^{\prime}(b-x_{b})=\lambda_{\gamma}\bigg{\\{}\frac{\operatorname{e}^{M}\delta
c_{2,2}^{-1/[1+\delta\bar{\eta}]}}{[\delta+\gamma]}h(b-x_{b})-\frac{\eta\gamma}{[\delta+\gamma]^{2}}\bigg{\\}}+\frac{\gamma}{\gamma+\delta}.$

Solving the previous equation w.r.t. $M$, it yields that $M=M(b)$, where
$M(b)$ is given by (3.27), which depends on $b$. Therefore, (3.23) is a
solution to the NLPDS (3.20)–(3.22). By the construction seen above, we know
that $v_{\gamma,b}$ is is
$\operatorname{C}^{1}((0,\infty))\cap\operatorname{C}^{2}((0,\infty)\setminus\\{x_{b},b\\})$.
However, since $v_{\gamma,b}$ is $\operatorname{C}^{1}$-continuous on
$(0,\infty)$ and satisfies (3.20)–(3.21) in the neighbourhoods of $x_{b}$ and
$b$, it follows $v^{\prime\prime}_{\gamma,b}$ is continuous at $x_{b}$ and at
$b$. ∎

###### Proof of Lemma 3.7.

Let $g_{2}$ be as in (3.30). In order to demonstrate that $g_{2}$ is an
increasing function on $(0,\infty)$, it suffices to show that
$g_{2}^{\prime}>0$ on $(0,\infty)$. Notice that
$g^{\prime}_{2}(b)=\frac{[h^{\prime}_{2}(b)]^{2}-h_{2}(b)h^{\prime\prime}_{2}(b)}{[h^{\prime}_{2}(b)]^{2}}=\frac{a_{1}a_{2}\operatorname{e}^{[\theta_{+}+\theta_{-}]b}}{[h^{\prime}_{2}(b)]^{2}}$
for $b\in(0,\infty)$, where $\theta_{-}$ and $\theta_{+}$ are the negative and
positive root of (3.5), and $a_{1}$ and $a_{2}$ are defined in (3.27).
Considering that $-\mu[2\eta-\mu]<2\delta\sigma^{2}$, it can be verified
easily that $\theta_{+}<\frac{2\delta}{2\eta-\mu}$. From here it gives that
$a_{1}a_{2}>0$. Thus, $g^{\prime}_{2}>0$ on $(0,\infty)$. By definition of
$h_{2}$, $a_{1}$ and $a_{2}$, we get that

$\displaystyle\lim_{b\downarrow
0}\frac{h_{2}(b)}{h^{\prime}_{2}(b)}=\frac{a_{1}-a_{2}}{a_{1}\theta_{+}-a_{2}\theta_{-}}=\frac{2\eta-\mu}{2\delta},$
$\displaystyle\lim_{b\uparrow\infty}\frac{h_{2}(b)}{h^{\prime}_{2}(b)}=\lim_{b\uparrow\infty}\frac{a_{1}-a_{2}\operatorname{e}^{[\theta_{-}-\theta_{+}]x}}{a_{1}\theta_{+}-a_{2}\theta_{-}\operatorname{e}^{[\theta_{-}-\theta_{+}]x}}=\frac{1}{\theta_{+}}.\qed$

###### Proof of Proposition 3.8.

We know that $b_{\gamma}$ is given by (3.31), due to $\mu<2\eta$ and
$\gamma>\gamma_{1}$. Consider $v_{\gamma,b_{\gamma}}$ as in (3.23) (when
$b=b_{\gamma}$). By (3.28) we see that (3.28) holds. Therefore
$v_{\gamma,b_{\gamma}}$ is concave and increasing on $(0,\bar{x})$, with
$\bar{x}$ as in (3.18). notice that $h^{\prime}_{2}>0$ on $(0,\infty)$. It
implies that $v_{\gamma,b_{\gamma}}$ is increasing on $(\bar{x},b_{\gamma})$.
To verify that $v_{\gamma,b_{\gamma}}$ is concave on $(b_{\gamma},\bar{x})$,
by similar arguments seen in the proof of Proposition 3.4, we only need to
verify that
$h^{\prime\prime}_{2}(b_{\gamma}-\bar{x})<h^{\prime\prime}_{2}(x_{2}-\bar{x})=0$,
where
$x_{2}\raisebox{0.4pt}{$:$}=\bar{x}+\frac{1}{\theta_{+}-\theta_{-}}\ln\Big{[}\frac{a_{2}\theta_{-}^{2}}{a_{1}\theta_{+}^{2}}\Big{]}$
is the point where $h_{2}^{\prime}(x-\bar{x})$ attains its global minimum.
Since $h_{2}$ satisfies (A.3), it follows that
$h^{\prime\prime}_{2}(b_{\gamma}-\bar{x})=\frac{2h^{\prime}_{2}(b_{\gamma}-\bar{x})}{\sigma^{2}}\big{[}\frac{1}{\lambda_{\gamma}}\big{[}\delta+\frac{\gamma\eta\lambda_{\gamma}}{\gamma+\delta}\big{]}-\eta\big{]}=\frac{2\delta
h^{\prime}_{2}(b_{\gamma}-\bar{x})}{\sigma^{2}}\big{[}\frac{1}{\lambda_{\gamma}}-\frac{\eta}{\gamma+\delta}\big{]}<0$.
To verify that $v_{\gamma,b_{\gamma}}$ is increasing and concave on
$(b,\infty)$, we only need to check that $c_{2,3}(b_{\gamma})<0$, which is
true because of $0<\delta[\delta+\gamma-\delta\eta\lambda_{\gamma}]$. By the
seen before, $v^{\prime}_{\gamma,b_{\gamma}}$ is decreasing on $(0,\infty)$.
Furthermore, since $b_{\gamma}$ as in (3.31) satisfies (3.29), it immediately
follows that
$v^{\prime}_{\gamma,b_{\gamma}}(b_{\gamma}-)=\frac{\frac{\gamma}{\gamma+\delta}\big{[}1-\frac{\eta\lambda_{\gamma}}{\gamma+\delta}\big{]}}{1-\frac{\delta\lambda_{\gamma}}{\gamma+\delta}g_{2}(b_{\gamma}-\bar{x})}=1$.
Thus, $b_{\gamma}=\inf\\{x>0:v^{\prime}_{\gamma,b_{\gamma}}(x)<1\\}<\infty$.
Let us prove now that
$\hat{f}_{1}=v^{\prime}_{\gamma,b_{\gamma}}/v^{\prime\prime}_{\gamma,b_{\gamma}}$
is decreasing on $(0,\infty)$. Calculating the first and second derivatives of
$v_{\gamma,b_{\gamma}}$, and taking into account (3.28) and (A.3), it gives

$\displaystyle\hat{f}_{1}(x)$
$\displaystyle=\begin{cases}-\mathpzc{x}_{\,1}^{\prime}(\mathpzc{x}_{\,1}^{-1}(x))&\text{if}\
x\in(0,\bar{x}),\\\ \vspace{-0.4cm}&\\\ \frac{\sigma^{2}}{2[\delta
g_{2}(x)-\eta]}&\text{if}\ x\in(\bar{x},b_{\gamma}),\\\ \vspace{-0.4cm}&\\\
\frac{\gamma\operatorname{e}^{-\lambda_{\gamma}[x-b_{\gamma}]}}{c_{2,3}(b_{\gamma})\lambda_{\gamma}^{2}[\gamma+\delta]}+\frac{1}{\lambda_{\gamma}}&\text{if}\
x\in(b_{\gamma},\infty).\\\ \end{cases}$

Using similar arguments seen in the Proof of Proposition 3.4, it can be seen
that $\hat{f}_{1}$ is decreasing on $(0,\infty)$. It implies that
$-\frac{\mu}{\sigma^{2}}\hat{f}_{1}$ is increasing on $(0,\infty)$ and
$-\frac{\mu}{\sigma^{2}}\hat{f}_{1}(0+)=\frac{2[\mu-\eta]}{\mu}<1$. Therefore
$x_{\gamma}=\bar{x}$. Since $v_{\gamma,b_{\gamma}}$ satisfies the hypotheses
of Proposition 2.2, we conclude that $v_{\gamma,b_{\gamma}}$ is a solution to
the HJB equation (2.5). ∎

###  A.3 Proofs of Proposition 3.9, 3.13 and Lemma 3.11

###### Proof of Proposition 3.9.

Let $b\in(0,\underline{$b$})$, with $\underline{$b$}$ as in (3.36). Let us
assume that $b<x_{b}$, which will be proven later on. Here, $x_{b}$ is a
parameter to be determined. By the seen in the proof of Proposition 3.5, we
know that
$v_{\gamma,b}(x)=c_{2,1}\operatorname{e}^{-\mathpzc{x}_{\,1}^{-1}(x)}\big{[}\operatorname{e}^{[1+\delta\bar{\eta}][M+\mathpzc{x}_{\,1}^{-1}(x)]}-1\big{]}$
for $x\in(0,b)$, is a solution to (3.39) on $(0,b)$ where
$\mathpzc{x}_{\,1}^{-1}$ is the inverse function of
$\mathpzc{x}_{\,1}:[-M,-M_{2}]\longrightarrow[0,b]$ which is given by (A.10)
with $k_{2}$ and $k_{1}$ as in (A.12) and (A.13), respectively. Here, $M$ and
$M_{2}$ (with $M_{2}\leq M$) are unknown parameters where
$\mathpzc{x}_{\,1}(-M)=0$ and $\mathpzc{x}_{\,1}(-M_{2})=b$. So, we shall
construct the solution to (3.39) on $(b,x_{b})$. Denoting $\bar{v}$ as a
solution to

$\displaystyle-\frac{\mu^{2}[v^{\prime}(x)]^{2}}{2\sigma^{2}v^{\prime\prime}(x)}-[\mu-\eta]\bar{v}^{\prime}(x)-[\delta+\gamma]v(x)+\gamma
x=0\quad\text{for}\ x\in(b,x_{b}),$ (A.18)

we see easily that

$v_{\gamma,b}(x)=\bar{v}(x)+\frac{\gamma[v_{\gamma,b}(b)-b]}{\delta+\gamma}$
(A.19)

is a solution to (3.39) for $x\in(b,x_{b})$. Assuming concavity of $\bar{v}$,
which will be proven later on, it is known that there exists
$\mathpzc{x}_{\,2}:\mathbbm{R}\longrightarrow[0,\infty)$ satisfying
$-\ln(\bar{v}^{\prime}(\mathpzc{x}_{\,2}(z)))=z$. Then, (3.45) holds.
Substituting (3.45) in (A.18), it follows that

$\displaystyle\frac{\mu^{2}\mathpzc{x}_{\,2}^{\prime}(z)\operatorname{e}^{-z}}{2\sigma^{2}}-[\mu-\eta]\operatorname{e}^{-z}-[\delta+\gamma]\bar{v}(\mathpzc{x}_{\,2}(z))+\gamma\mathpzc{x}_{\,2}(z)=0.$
(A.20)

Taking first derivatives w.r.t. $z$ in (A.20), it yields

$\displaystyle\mathpzc{x}_{\,2}^{\prime\prime}(z)-\\{1+\bar{\eta}[\delta+\gamma]-\gamma\bar{\eta}\operatorname{e}^{z}\\}\mathpzc{x}_{\,2}^{\prime}(z)+\bar{\eta}[\mu-\eta]=0$
(A.21)

where $\bar{\eta}$ is as in (3.19). Taking $M_{2}\geq 0$ such that
$\mathpzc{x}_{\,2}(-M_{2}+)=b$ and
$v^{\prime}(\mathpzc{x}_{\,2}(-M_{2}+))=\operatorname{e}^{M_{2}}$, the
function $\mathpzc{x}_{\,2}$ is from $[-M_{2},\bar{z}_{1})$ to $[b,x_{b})$,
with $\bar{z}_{1}>-M_{2}$ such that $\mathpzc{x}_{\,2}(\bar{z}_{1})=x_{b}$. It
implies that $\mathpzc{x}_{\,2}^{\prime}$ has the following solution

$\displaystyle\mathpzc{x}_{\,2}^{\prime}(z)$
$\displaystyle=k_{3}\exp[\\{1+\bar{\eta}[\delta+\gamma]\\}z-\gamma\bar{\eta}\operatorname{e}^{z}]$
$\displaystyle\quad-\bar{\eta}[\mu-\eta]\exp[\\{1+\bar{\eta}[\delta+\gamma]\\}z-\gamma\bar{\eta}\operatorname{e}^{z}]\int_{-M_{2}}^{z}\exp[-\\{1+\bar{\eta}[\delta+\gamma]\\}y+\gamma\bar{\eta}\operatorname{e}^{y}]\mathrm{d}y$
$\displaystyle=\operatorname{e}^{z}g(\operatorname{e}^{z})\bigg{[}k_{3}-\bar{\eta}[\mu-\eta]\int_{-M_{2}}^{z}\frac{1}{\operatorname{e}^{y}g(\operatorname{e}^{y})}\mathrm{d}y\bigg{]},$
(A.22)

where $k_{3}$ is a parameter to be determined, and $g$ is the density function
of a gamma distribution given by (3.35). From here, it follows that

$\mathpzc{x}_{\,2}(z)=\bar{f}_{M_{2}}(\operatorname{e}^{z})+b$ (A.23)

is a solution to (A.21), with $H_{M_{2}}$ as in (3.32) when $\beta=M_{2}$, and

$\displaystyle\bar{f}_{M_{2}}(z)$
$\displaystyle\raisebox{0.4pt}{$:$}=\int_{\operatorname{e}^{-M_{2}}}^{z}\big{\\{}-\bar{\eta}[\mu-\eta]g(y)H_{M_{2}}(y)+k_{3}g(y)\big{\\}}\mathrm{d}y$
$\displaystyle=k_{3}[G(z)-G(\operatorname{e}^{-M_{2}})]-\bar{\eta}[\mu-\eta]\bigg{[}G(z)H_{M_{2}}(z)-\int_{\operatorname{e}^{-M_{2}}}^{z}\frac{G(y)}{y^{2}g(y)}\mathrm{d}y\bigg{]}.$
(A.24)

Recall that $G$ represents the gamma accumulative distribution function of
$g$. By the assumption of concavity of $\bar{v}$, it implies that
$\mathpzc{x}_{\,2}^{-1}$ is an increasing function on $[0,\infty)$. From here
and since $\operatorname{e}^{z}$ is also increasing, wee have that
$\bar{f}^{-1}_{M_{2}}$ is an increasing function. Using (A.20) and (A.3), we
have that for $x\in(b,x_{b})$

$\displaystyle\bar{v}(x)$
$\displaystyle=\frac{\operatorname{e}^{-\mathpzc{x}_{\,2}^{-1}(x)}}{\delta+\gamma}\bigg{\\{}\frac{\mu^{2}\mathpzc{x}_{\,2}^{\prime}(\mathpzc{x}_{\,2}^{-1}(x))}{2\sigma^{2}}-[\mu-\eta]\bigg{\\}}+\frac{\gamma
x}{\delta+\gamma}$
$\displaystyle=\frac{\operatorname{e}^{-\mathpzc{x}_{\,2}^{-1}(x)}}{\delta+\gamma}\bigg{\\{}\frac{\mu^{2}g(\operatorname{e}^{\mathpzc{x}_{\,2}^{-1}(x)})\operatorname{e}^{\mathpzc{x}_{\,2}^{-1}(x)}}{2\sigma^{2}}\big{[}k_{3}-\bar{\eta}[\mu-\eta]H_{M_{2}}(\operatorname{e}^{\mathpzc{x}_{\,2}^{-1}(x)})\big{]}-[\mu-\eta]\bigg{\\}}+\frac{\gamma
x}{\delta+\gamma}$ (A.25)

is a solution to (A.18). Applying (A.3) in (A.19), it follows that for
$x\in(b,x_{b})$,

$\displaystyle v_{\gamma,b}(x)$
$\displaystyle=\frac{\operatorname{e}^{-\mathpzc{x}_{\,2}^{-1}(x)}}{\delta+\gamma}\bigg{\\{}\frac{\mu^{2}g(\operatorname{e}^{\mathpzc{x}_{\,2}^{-1}(x)})\operatorname{e}^{\mathpzc{x}_{\,2}^{-1}(x)}}{2\sigma^{2}}[k_{3}-\bar{\eta}[\mu-\eta]H_{M_{2}}(\operatorname{e}^{\mathpzc{x}_{\,2}^{-1}(x)})]-[\mu-\eta]\bigg{\\}}+\frac{\gamma[v_{\gamma,b}(b)+x-b]}{\delta+\gamma}$

is a solution to (3.39). Since $v_{\gamma,b}^{\prime}$ must be continuous at
$b$, by smooth fit, (3.45), (3.28), and (A.23)–(A.24), we get that
$\operatorname{e}^{-\mathpzc{x}_{\,1}^{-1}(b)}=\operatorname{e}^{-\mathpzc{x}_{\,2}^{-1}(b)}=\operatorname{e}^{M_{2}}$,
where $\mathpzc{x}_{\,2}^{-1}$ is the inverse function of $\eqref{eqo11.0}$.
Thus,

$\displaystyle\mathpzc{x}_{\,1}(-M_{2})=b$
$\displaystyle\quad\Longleftrightarrow\quad\bigg{[}\frac{\delta\bar{\eta}}{1+\delta\bar{\eta}}\bigg{]}\operatorname{e}^{[1+\delta\bar{\eta}][M-M_{2}]}+M-M_{2}=\frac{b}{c_{2,1}}+\frac{\delta\bar{\eta}}{1+\delta\bar{\eta}},$
(A.26)

where $c_{2,1}$ is as in (3.19). On the other hand, in order that
$v_{\gamma,b}$ is $\operatorname{C}^{2}$-continuous at $b$, by smooth fit
again and using (3.28) and (A.13), we have that
$v_{\gamma,b}^{\prime\prime}(b-)=-\frac{\operatorname{e}^{M_{2}}}{\mathpzc{x}_{\,1}^{\prime}(-M_{2})}=-\frac{\operatorname{e}^{M_{2}}}{\mathpzc{x}_{\,2}^{\prime}(-M_{2})}=v_{\gamma,b}^{\prime\prime}(b+)$.
From here, and by (A.11) and (A.3), it follows that

$\displaystyle k_{3}$
$\displaystyle=\frac{c_{2,1}}{\operatorname{e}^{-M_{2}}g(\operatorname{e}^{-M_{2}})}\big{[}\delta\bar{\eta}\operatorname{e}^{[M-M_{2}][1+\delta\bar{\eta}]}+1\big{]}.$
(A.27)

A solution to the equation (3.40), is given by (A.1), with
$\max\\{x_{b},b\\}=x_{b}$ and $c_{4}=0$, where $c_{3}$ is a free constant. By
smooth fit, it gives the following identities

$\displaystyle
v^{\prime}_{\gamma,b}(x_{b}-)=c_{3}\lambda_{\gamma}+\frac{\gamma}{\gamma+\delta}\quad\text{and}\quad
v^{\prime\prime}_{\gamma,b}(x_{b}-)=c_{3}\lambda_{\gamma}^{2}.$

Since
$v^{\prime}_{\gamma,b}(x_{b}-)=\operatorname{e}^{-\mathpzc{x}_{\,2}^{-1}(x_{b})}=\frac{1}{\bar{f}^{-1}_{M_{2}}(x_{b}-b)}$
(because of $\mathpzc{x}_{\,2}^{-1}(x)=\ln[\bar{f}^{-1}_{M_{2}}(x-b)]$), and
considering that $x_{b}$ is a point where $-\frac{\mu
v^{\prime}_{\gamma,b}(x_{b})}{\sigma^{2}v^{\prime\prime}_{\gamma,b}(x_{b})}=1$,
we have the following system of equations

$\displaystyle\frac{1}{\hat{\alpha}}=c_{3}\lambda_{\gamma}+\frac{\gamma}{\gamma+\delta}\quad\text{and}\quad-\frac{\mu}{\sigma^{2}}\frac{1}{\hat{\alpha}}=c_{3}\lambda_{\gamma}^{2}.$
(A.28)

with $\hat{\alpha}=\bar{f}^{-1}_{M_{2}}(x_{b}-b)$. Then, solving the system
(A.28), we see that $\hat{\alpha}=\alpha_{\gamma}$ and
$c_{3}=c_{3,2}(\alpha_{\gamma})$ are as in (3.34) and (3.43), respectively. It
implies that $x_{b}=\bar{f}_{M_{2}}(\alpha_{\gamma})+b$. On the other hand,
since
$u^{*}_{b}(x_{b})=\frac{\mu}{\sigma^{2}}\mathpzc{x}_{\,2}^{\prime}(\mathpzc{x}_{\,2}^{-1}(x_{b}))=1$
must be occurred, it gives that

$\displaystyle\frac{\mu}{\sigma^{2}}\operatorname{e}^{\mathpzc{x}_{\,2}^{-1}(x_{b})}g(\operatorname{e}^{\mathpzc{x}_{\,2}^{-1}(x_{b})})\\{k_{3}-\bar{\eta}[\mu-\eta]H_{M_{2}}(\operatorname{e}^{\mathpzc{x}_{\,2}^{-1}(x_{b})})\\}=1\quad\Longleftrightarrow\quad
k_{3}=\frac{\sigma^{2}}{\mu\alpha_{\gamma}g(\alpha_{\gamma})}+\bar{\eta}[\mu-\eta]H_{M_{2}}(\alpha_{\gamma}).$
(A.29)

Notice that $\bar{z}_{1}=\ln[\alpha_{\gamma}]$. Substituting (A.27) in (A.29),
we obtain that

$\displaystyle\begin{split}\operatorname{e}^{[M-M_{2}][1+\delta\bar{\eta}]}&=\frac{1}{\delta\bar{\eta}}\bigg{\\{}\frac{\operatorname{e}^{-M_{2}}g(\operatorname{e}^{-M_{2}})}{c_{2,1}}\bigg{[}\frac{\sigma^{2}}{\mu\alpha_{\gamma}g(\alpha_{\gamma})}+\bar{\eta}[\mu-\eta]H_{M_{2}}(\alpha_{\gamma})\bigg{]}-1\bigg{\\}},\\\
[M-M_{2}]&=\frac{1}{1+\delta\bar{\eta}}\ln\bigg{[}\frac{1}{\delta\bar{\eta}}\bigg{\\{}\frac{\operatorname{e}^{-M_{2}}g(\operatorname{e}^{-M_{2}})}{c_{2,1}}\bigg{[}\frac{\sigma^{2}}{\mu\alpha_{\gamma}g(\alpha_{\gamma})}+\bar{\eta}[\mu-\eta]H_{M_{2}}(\alpha_{\gamma})\bigg{]}-1\bigg{\\}}\bigg{]}.\end{split}$
(A.30)

Applying (A.30) in (A.26), (3.44) holds. If there is a unique $M_{2}$ solving
(3.44), by the seen before, we conclude the results announced in Proposition
3.9.

Concavity property of $v_{\gamma,b}$. Let us now verify that $v_{\gamma,b}$ is
concave on $(0,\infty)$. For this, it is sufficient to check that
$v^{\prime\prime}_{\gamma,b}$ is non-positive on $(0,\infty)$, which is true
on $(0,b)$ because of (3.28) and $\mathpzc{x}_{\,1}^{\prime}\geq 0$ on
$(-M,-M_{2})$. To verify that $v^{\prime\prime}_{\gamma,b}\leq 0$ on
$(b,x_{b})$, by (3.45), it is sufficient to show that

$\displaystyle\mathpzc{x}_{\,2}^{\prime}(\mathpzc{x}_{\,2}^{-1}(x))\geq
0\quad\text{on}\ (b,x_{b}).$ (A.31)

Taking $k_{3}$ as in (A.29) and substituting this in (A.3), it gives that

$\displaystyle\mathpzc{x}_{\,2}^{\prime}(\mathpzc{x}_{\,2}^{-1}(x))=\operatorname{e}^{\mathpzc{x}_{\,2}^{-1}(x)}g(\operatorname{e}^{\mathpzc{x}_{\,2}^{-1}(x)})\bigg{[}\frac{\sigma^{2}}{\mu\alpha_{\gamma}g(\alpha_{\gamma})}+\bar{\eta}[\mu-\eta]\\{H_{M_{2}}(\alpha_{\gamma})-H_{M_{2}}(\operatorname{e}^{\mathpzc{x}_{\,2}^{-1}(x)})\\}\bigg{]}.$

From here it is clear that (A.31) is true, due to
$H_{M_{2}}(\operatorname{e}^{\mathpzc{x}_{\,2}^{-1}(x)})<H_{M_{2}}(\alpha_{\gamma})$
for $x\in(b,x_{b})$. Since $c_{3,2}(\alpha_{\gamma})$ given in (3.43) is
negative, we conclude that $v^{\prime\prime}<0$ on $(x_{b},\infty)$.

Increasing property of $u^{*}_{b}$ on $(0,x_{b})$. By the seen in the proof of
Proposition 3.8, it is clear that $u^{*}_{b}=-\frac{\mu
v^{\prime}_{\gamma,b}}{\sigma^{2}v^{\prime\prime}_{\gamma,b}}$ is increasing
on $(0,b)$ satisfying (A.15). Notice that

$u^{*}_{b}(x)=\frac{\mu}{\sigma^{2}}\mathpzc{x}_{\,2}^{\prime}(\mathpzc{x}_{\,2}^{-1}(x))=\frac{\mu}{\sigma^{2}}\operatorname{e}^{\mathpzc{x}_{\,2}^{-1}(x)}\bar{f}^{\prime}_{M_{2}}(\operatorname{e}^{\mathpzc{x}_{\,2}^{-1}(x)})=\frac{\mu}{\sigma}\bar{f}^{-1}_{M_{2}}(x-b)\bar{f}^{\prime}_{M_{2}}(\bar{f}^{-1}_{M_{2}}(x-b)).$
(A.32)

Then, to verify that $u^{*}_{b}$ is increasing on $(b,x_{b})$, it is enough to
check that
$f_{5}(x)\raisebox{0.4pt}{$:$}=\frac{\mu}{\sigma^{2}}x\bar{f}^{\prime}_{M_{2}}(x)$
is increasing on $(\operatorname{e}^{-M_{2}},\alpha_{\gamma})$, since
$u^{*}_{b}(x)=f_{5}(\bar{f}^{-1}_{M_{2}}(x-b))$, and
$\bar{f}^{-1}_{M_{2}}(x-b)$ is increasing on $(b,x_{b})$. Observe that
$f_{5}\geq 0$ on $(\operatorname{e}^{-M_{2}},\alpha_{\gamma})$ and
$f_{5}(\alpha_{\gamma}-)=1$. Furthermore,

$\displaystyle
f^{\prime}_{5}(x)=\frac{1}{x}\bigg{\\{}[\bar{\eta}[\delta+\gamma]-\bar{\eta}\gamma
x+1]f_{5}(x)-\frac{\mu\bar{\eta}[\mu-\eta]}{\sigma^{2}}\bigg{\\}},$ (A.33)

because of

$\displaystyle
xg^{\prime}(x)+g(x)=g(x)\\{\bar{\eta}[\delta+\gamma]-\bar{\eta}\gamma x+1\\}.$
(A.34)

From here
$f^{\prime\prime}_{5}(x)=\frac{1}{x}\big{\\{}[\bar{\eta}[\delta+\gamma]-\bar{\eta}\gamma
x]f^{\prime}_{5}(x)-\gamma f_{5}(x)\big{\\}}$. Assuming there exists a point
$x^{*}\in(\operatorname{e}^{-M_{2}},\alpha_{\gamma})$ such that
$f^{\prime}_{5}(x^{*})=0$, it gives $f^{\prime\prime}_{5}(x^{*})=-\frac{\gamma
f_{5}(x^{*})}{x^{*}}<0$, which implies that there are no local minimums on
$(\operatorname{e}^{-M_{2}},\alpha_{\gamma})$. Hence, $f_{5}$ is concave on
$(\operatorname{e}^{-M_{2}},\alpha_{\gamma})$. Moreover, letting
$x\uparrow\alpha_{\gamma}$ in (A.33), we see that

$\displaystyle
f^{\prime}_{5}(\alpha_{\gamma}-)=\frac{1}{\alpha_{\gamma}}\bigg{\\{}\bar{\eta}[[\delta+\gamma]-\gamma\alpha_{\gamma}]+1-\frac{\mu\bar{\eta}[\mu-\eta]}{\sigma^{2}}\bigg{\\}}>0,$
(A.35)

due to $1-\frac{\mu\bar{\eta}[\mu-\eta]}{\sigma^{2}}>0$ and
$\bar{\eta}[[\delta+\gamma]-\gamma\alpha_{\gamma}]=-\frac{2}{\mu\lambda_{\gamma}}>0$.
Therefore, $f_{5}$ is increasing on
$(\operatorname{e}^{-M_{2}},\alpha_{\gamma})$, concluding that $u^{*}_{b}$ is
increasing on $(0,x_{b})$, and $u^{*}_{b}<1$ on $(0,x_{b})$.

Decreasing property of $\bar{g}$ on $(0,\underline{$b$})$. To finish, let us
determine the well solvability of (3.44). For this, it is sufficient to check
that $\bar{g}$ defined in (3.37) is positive and decreasing on
$(0,\underline{$b$})$, concluding that for each $b>0$ satisfying (3.36), there
exists a unique $M_{2}>0$ such that (3.44) is true. For this purpose, let us
first check the following function
$f_{6}(\beta)\raisebox{0.4pt}{$:$}=\frac{\operatorname{e}^{-\beta}g(\operatorname{e}^{-\beta})}{c_{2,1}}c_{3,1}(\beta)$,
for $\beta>0$, is positive and decreasing. Calculating the first and second
derivatives and taking into account (A.34), it gives that

$\displaystyle\frac{\mathrm{d}}{\mathrm{d}\beta}f_{6}(\beta)=f_{7}(\beta)\bigg{[}\frac{\bar{\eta}[\mu-\eta]}{c_{2,1}f_{7}(\beta)}-f_{6}(\beta)\bigg{]}\quad\text{and}\quad\frac{\mathrm{d}^{2}}{\mathrm{d}\beta^{2}}f_{6}(\beta)=-\bar{\eta}\gamma\operatorname{e}^{-\beta}f_{6}(\beta)-f_{7}(\beta)\frac{\mathrm{d}}{\mathrm{d}\beta}f_{6}(\beta),$

with
$f_{7}(\beta)\raisebox{0.4pt}{$:$}=\bar{\eta}[\delta+\gamma]+1-\bar{\eta}\gamma\operatorname{e}^{-\beta}$.
Notice that $\beta\mapsto\frac{\bar{\eta}[\mu-\eta]}{c_{2,1}f_{7}(\beta)}$ is
decreasing and positive on $(0,\infty)$, satisfying $\lim_{\beta\downarrow
0}\frac{\bar{\eta}[\mu-\eta]}{c_{2,1}f_{7}(\beta)}=1$ and
$\lim_{\beta\uparrow\infty}\frac{\bar{\eta}[\mu-\eta]}{c_{2,1}f_{7}(\beta)}=\frac{\bar{\eta}[\mu-\eta]}{c_{2,1}[\bar{\eta}[\delta+\gamma]+1]}$.
Meanwhile, observe that $f_{6}(0+)>1+\delta\bar{\eta}>\lim_{\beta\downarrow
0}\frac{\bar{\eta}[\mu-\eta]}{c_{2,1}f_{7}(\beta)}$ due to (3.38), and by
L’Hôpital’s rule $\lim_{\beta\downarrow
0}f_{6}(\beta)=\frac{\bar{\eta}[\mu-\eta]}{c_{2,1}[\bar{\eta}[\delta+\gamma]+1]}$.
It implies that $f_{6}$ is decreasing on $(0,\bar{\beta})$, for some
$\bar{\beta}>0$. On the other hand, assuming there exists a point
$\beta^{*}\in(\bar{\beta},\infty)$ such that $f^{\prime}_{6}(\beta^{*})=0$, it
gives
$f^{\prime\prime}_{6}(\beta^{*})=-\bar{\eta}\gamma\operatorname{e}^{-\beta^{*}}f_{6}(\beta^{*})<0$,
which implies that there are no local minimums on $(\beta^{*},\infty)$. Thus,
$f_{6}$ is decreasing on $(0,\infty)$. Since there exists a unique point
$\bar{\beta}$ such that $f_{6}(\beta)>1$, for $\beta\in(0,\bar{\beta})$, and
$f_{6}(\bar{\beta})=0$, it gives that
$\lim_{\beta\uparrow\bar{\beta}}\bar{g}(\beta)=-\infty$, and thus
$\bar{b}<\infty$. Calculating the first derivatives and considering that
$\frac{\mathrm{d}}{\mathrm{d}\beta}f_{6}(\beta)<0$ for $\beta\in(0,\infty)$,
we have
$\frac{\mathrm{d}}{\mathrm{d}\beta}\bar{g}(\beta)=\frac{\mathrm{d}}{\mathrm{d}\beta}f_{6}(\beta)\big{(}1+\frac{1}{f_{6}(\beta)-1}\big{)}<0,\quad\text{for}\
\beta\in(0,\bar{\beta})$. Therefore $\bar{g}$ is decreasing and positive on
$(0,\bar{\beta})$. ∎

###### Proof of Lemma 3.11.

(i) Taking $f_{2}$ as in (3.48), it is easy to corroborate that
$f_{2}(\gamma)\xrightarrow[\gamma\downarrow 0]{}\infty$. In order that
$f_{2}(\gamma_{1})=1$ holds, we see that $\gamma_{1}$ must satisfy (4.7).
Moreover, since $f_{1}(\gamma_{1})=\frac{2\eta-\mu}{2\delta}$, with $f_{1}$ as
in (3.11), it follows that (4.6) holds. Applying (4.7) in (4.6), it follows
that
$\gamma_{1}=\frac{\delta}{\mu}\big{[}\frac{2\delta\sigma^{2}}{\mu}+2\eta-\mu\big{]}$.
With this representation of $\gamma_{1}$ we obtain easily that
$f_{2}(\gamma)\xrightarrow[\gamma\uparrow\gamma_{1}]{}1$.

(ii) Let $f_{3}$ be as in (3.48). Notice that $\gamma
f_{2}(\gamma)\xrightarrow[\gamma\downarrow
0]{}\delta\Big{[}1-\frac{\mu}{\eta+\sqrt{\eta^{2}+2\sigma^{2}\delta}}\Big{]}$.
It implies that $\lim_{\gamma\downarrow
0}f_{3}(\gamma)=\bar{\eta}[\mu-\eta]\int_{1}^{\infty}\frac{1}{y^{\bar{\eta}\delta+2}}\mathrm{d}y=\frac{\bar{\eta}[\mu-\eta]}{\bar{\eta}\delta+1}$.
On the other hand, considering that $f_{2}(\gamma_{1})=1$, it is easy to
verify that
$f_{3}(\gamma)\xrightarrow[\gamma\uparrow\gamma_{1}]{}\frac{\sigma^{2}\operatorname{e}^{\bar{\eta}\gamma_{1}}}{\mu}$.
Take
$\bar{g}_{1}(\gamma)\raisebox{0.4pt}{$:$}=\sqrt{\eta^{2}+2\sigma^{2}[\delta+\gamma]}$,
$\bar{g}_{2}(\gamma)\raisebox{0.4pt}{$:$}=\frac{\mu\sigma^{2}}{\bar{g}(\gamma)[\bar{g}_{1}(\gamma)+\eta][\bar{g}_{1}(\gamma)+\eta-\mu]}$
and
$\bar{g}_{3}(\gamma)\raisebox{0.4pt}{$:$}=\frac{\delta}{\gamma[\gamma+\delta]}$
. Then

$\displaystyle f^{\prime}_{2}(\gamma)$
$\displaystyle=\frac{f_{2}(\gamma)}{\gamma\bar{g}_{1}(\gamma)}[\gamma\bar{g}_{1}(\gamma)\bar{g}_{2}(\gamma)-\gamma\bar{g}_{1}(\gamma)\bar{g}_{3}(\gamma)].$
(A.36)

Observe that $\gamma\bar{g}_{1}(\gamma)\bar{g}_{2}(\gamma)$ and
$\gamma\bar{g}_{1}(\gamma)\bar{g}_{3}(\gamma)$ are increasing and decreasing
on $(0,\infty)$, respectively, due to

$\displaystyle\frac{\mathrm{d}}{\mathrm{d}\gamma}[\gamma\bar{g}_{1}(\gamma)\bar{g}_{2}(\gamma)]$
$\displaystyle=\frac{\mu\sigma^{2}[2\eta-\mu]\\{\eta[\eta+\bar{g}_{1}(\gamma)]+\sigma^{2}[\gamma+2\delta]\\}+2\mu\sigma^{4}\delta\bar{g}_{1}(\gamma)}{\bar{g}_{1}(\gamma)\\{[\bar{g}_{1}(\gamma)+\eta][\bar{g}_{1}(\gamma)+\eta-\mu]\\}^{2}}>0,$
$\displaystyle\frac{\mathrm{d}}{\mathrm{d}\gamma}[\gamma\bar{g}_{1}(\gamma)\bar{g}_{3}(\gamma)]$
$\displaystyle=-\frac{\eta^{2}+\delta^{2}[x+\delta]}{[x+\delta]^{2}\bar{g}_{1}(\gamma)}<0,$

for $\gamma>0$. From here, it implies that there exists $\bar{\gamma}>0$ such
that $f^{\prime}_{2}(\gamma)<0$ for $\gamma\in(0,\bar{\gamma})$. To finish the
proof, let us verify that $\gamma_{1}\leq\bar{\gamma}$, i.e.

$\displaystyle\gamma\bar{g}_{1}(\gamma_{1})\bar{g}_{2}(\gamma_{1})=\frac{\mu\sigma^{2}\gamma_{1}}{[\bar{g}_{1}(\gamma_{1})+\eta][\bar{g}_{1}(\gamma_{1})+\eta-\mu]}<\frac{\delta\bar{g}_{1}(\gamma_{1})}{\gamma_{1}+\delta}=\gamma_{1}\bar{g}_{1}(\gamma)\bar{g}_{3}(\gamma_{1}).$
(A.37)

Considering that
$\bar{g}_{1}(\gamma_{1})=\frac{\mu[\gamma_{1}+\delta]}{\delta}-\eta$, because
of $f_{2}(\gamma_{1})=1$, we see that (A.37) is equivalent to show that

$\frac{[\sigma\delta]^{2}}{\mu}<\frac{\mu\gamma_{1}+\delta[\mu-\eta]}{\delta}.$
(A.38)

Since
$\gamma_{1}=\frac{\delta}{\mu}\big{[}\frac{2\delta\sigma^{2}}{\mu}+2\eta-\mu\big{]}$,
it follows that (A.38) is true, due to $\eta<2\eta$. Therefore, by the seen
before, we conclude that $f_{2}$ is decreasing on $(0,\gamma_{1})$.

(ii) Taking first derivatives in $f_{3}$ and using (A.36), it gives that

$\displaystyle f^{\prime}_{3}(\gamma)$
$\displaystyle=\bar{\eta}^{2}[\mu-\eta]\int_{1}^{f_{2}(\gamma)}\frac{\operatorname{e}^{\bar{\eta}\gamma
y}}{y^{\bar{\eta}[\delta+\gamma]+2}}[y-\ln[y]]\mathrm{d}y+\frac{\operatorname{e}^{\bar{\eta}\gamma
f_{2}(\gamma)}}{[f_{2}(\gamma)]^{\bar{\eta}[\delta+\gamma]+1}}$
$\displaystyle\quad\times\Bigg{\\{}\frac{\sigma^{2}}{\mu}\bar{\eta}[f_{2}(\gamma)+\gamma
f^{\prime}_{2}(\gamma)]+\frac{\sigma^{2}}{\mu}\Bigg{[}-\frac{f^{\prime}_{2}(\gamma)}{f_{2}(\gamma)}\bar{g}_{4}(\gamma)-\bar{\eta}\ln[f_{2}(\gamma)]\Bigg{]}\Bigg{\\}}$
$\displaystyle=\bar{\eta}^{2}[\mu-\eta]\int_{1}^{f_{2}(\gamma)}\frac{\operatorname{e}^{\bar{\eta}\gamma
y}}{y^{\bar{\eta}[\delta+\gamma]+2}}[y-\ln[y]]\mathrm{d}y+\frac{\operatorname{e}^{\bar{\eta}\gamma
f_{2}(\gamma)}}{[f_{2}(\gamma)]^{\bar{\eta}[\delta+\gamma]+1}}$
$\displaystyle\quad\times\Bigg{\\{}\frac{\sigma^{2}}{\mu}\bar{\eta}[f_{2}(\gamma)+\gamma
f^{\prime}_{2}(\gamma)]+\frac{\sigma^{2}}{\mu}\Bigg{[}\bar{\eta}\bigg{[}\frac{\delta}{\gamma}-\ln[f_{2}(\gamma)]\bigg{]}+\frac{\delta[2\eta-\mu]}{\mu\gamma[\gamma+\delta]}-\bar{g}_{2}(\gamma)\bar{g}_{4}(\gamma)\Bigg{]}\Bigg{\\}}$
(A.39)

with
$\bar{g}_{4}(\gamma)\raisebox{0.4pt}{$:$}=\bar{\eta}[\delta+\gamma]+\frac{1}{\mu}[2\eta-\mu]$.
Observe that the terms
$\int_{1}^{f_{2}(\gamma)}\frac{\operatorname{e}^{\bar{\eta}\gamma
y}}{y^{\bar{\eta}[\delta+\gamma]+2}}\\{y-\ln[y]\\}\mathrm{d}y$,
$\frac{\operatorname{e}^{\bar{\eta}\gamma
f_{2}(\gamma)}}{[f_{2}(\gamma)]^{\bar{\eta}[\delta+\gamma]+1}}$ and