# A Unified Approach to Retirement and Consumption-Portfolio Choice111 Junkee
Jeon is supported by the National Research Foundation of Korea (NRF) grant
funded by the Korea government [Grant No. NRF-2020R1C1C1A01007313]. Hyeng Keun
Koo is supported by NRF grant [Grat No. NRF-2020R1A2C1A01006134].

Junkee Jeon 222E-mail<EMAIL_ADDRESS>Department of Applied Mathematics,
Kyung Hee University, Korea. Hyeng Keun Koo333E-mail<EMAIL_ADDRESS>Department of Financial Engineering, Ajou University, Korea.

###### Abstract

In this study we propose a unified model of optimal retirement, consumption
and portfolio choice of an individual agent, which encompasses a large class
of the models in the literature and provide a general methodology to solve the
model. Different from the traditional approach, we consider the problems
before and after retirement simultaneously and identify the difference in the
dual value functions as the utility value of lifetime labor. The utility value
has an option nature, namely, it is the maximized value of choosing the
retirement time optimally and we discover it by solving a variational
inequality. Then, we discover the dual value functions by using the utility
value. We discover the value function and optimal policies by establishing a
duality between the value function and the dual value function.

JEL Classification Codes: D14, D15, E21, G11

Keywords : Consumption, Human Wealth, Option Value, Portfolio, Retirement,
Utility Function

##  1 Introduction

In this study, we propose and investigate a general model of optimal
retirement, consumption and portfolio choice of an individual agent. The
optimal retirement decision is an important determinant of labor supply and
human capital. The option of voluntary retirement makes the beta of human
capital negative, since one can work longer after poor realizations of
investment outcomes and has an important effect on risk taking attitudes
(Dybvig and Liu, (2010)). Researchers have developed economic models of the
retirement decision in conjunction with the consumption and portfolio choice.
In particular, they have shown that the retirement option induces people to
increase savings and to make aggressive investments before retirement and to
increase risky positions in a stock market boom even after making large
profits (Choi and Shim, (2006), Fahri and Panageas, (2007), Choi et al.,
(2008), and Yang and Koo, (2018)). The models, thus, have provided a profound
insight into the interaction between the retiement decision and the life cycle
choice of consumption and investment.

The models in the literature, however, are highly specialized and often make
contradicting predictions. For example, the models of Fahri and Panageas,
(2007) and Dybvig and Liu, (2010) predict jumps of consumption at retirement,
whereas those of Choi and Shim, (2006) and Yang and Koo, (2018) do not. The
difference in the models originates from the special utility functions they
employ, and hence, the predictions are not robust to changes in the utility
functions.

Our objective is (i) to propose a unified model, which encompasses a large
class of the models in the literature and (ii) to provide a general
methodology to solve the model. For the first objective, we propose a model
with minimum assumptions among those employing time separable expected utility
functions with a constant subjective discount rate and stationary felicity
functions before and after retirement in a single-good economy. The minimum
assumptions consist of two; Assumption 1 is the condition for the problem to
be well-defined, which is much less restrictive than those in the literature,
and Assumption 2 is the necessary and sufficient condition for the retirement
option to have a positive value. We next solve the problem by the dual
martingale approach. However, we do not take the traditional approach of
taking sequential steps; solving the problem after retirement in the first
step and then solving the problem before retirement in the second step, taking
the solution in the second step as an essential ingredient.444Most of the
papers in the literature take this two-step approach. See Yang and Koo, (2018)
for a systematic treatment. Instead we consider the problems before and after
retirement simultaneously and identify the difference in the dual value
functions as the utility value of lifetime labor. The utility value has an
option nature, namely, it is the maximized value of choosing the retirement
time optimally, and can be obtained as a solution to a variational inequality.
Then, we find the dual value functions by using the utility value. We discover
the value function and optimal policies by establishing a duality between the
value function and the dual value function. In this way we show the total
wealth of the agent is composed of financial wealth and human wealth and human
wealth can be derived as a part of the derivative of utility value of lifetime
labor.

As an illustrative example we provide a model which subsumes major models in
the literature as special cases. We obtain optimal policies in closed form
when the felicity function has constant relative risk aversion.

Yang and Koo, (2018) have considered the utility value of lifetime labor and
derived its value by solving a variational inequality. Their method is,
however, still the two-step method. Jeon and Park, (2020) make an observation
that the dual value functions before and after retirement differ by the the
utility value of lifetime labor. Their utility functions before and after
retirement differ by an additive disutility, and we consider generally
different utility functions in this paper. In the absence of voluntary
retirement option Bodie et al., (1992) and Bodie et al., (2004) study flexible
labor supply with an exogenously fixed retirement date.

The paper is organized as follows. Section 2 explains the model. Section 3
sets up the dual optimization problem and provides a solution. Section 4
establishes the duality between the value function and the dual value function
and derives the optimal policies. Section 5 discusses examples and Section 6
concludes.

##  2 Model

We consider an agent who lives in a single good economy. The agent is
currently working and has an option to retire voluntarily. The agent cannot
come back to work after retirement, and hence, retirement is an irreversible
decision. The agent’s preference over consumption is represented by the
following expected utility function:

$U\equiv\mathbb{E}\left[\int_{0}^{\tau}e^{-\rho
t}u_{B}(c_{t})dt+\int_{\tau}^{\infty}e^{-\rho t}u_{A}(c_{t})dt\right].$ (1)

Here $\rho>0$ is the subjective discount rate, $\tau$ is the time of
retirement, $c_{t}$ is the rate of consumption, and
$u_{B}:{\mathcal{D}}\rightarrow\mathbb{R}$ and
$u_{A}:{\mathcal{D}}\rightarrow\mathbb{R}$ (${\mathcal{D}}=[0,\infty)$ or
$(0,\infty)$) are the agent’s felicity function of before and after
retirement, respectively.555 We consider an infinite horizon model for
simplicity of exposition, and extension to a finite horizon is technically
complex, but can be done following the methods in Yang and Koo, (2018).  We
define the class $\Sigma$ of all set of felicity function $u$ satisfying the
following conditions:

  * (i)

The felicity function $u:{\mathcal{D}}\to\mathbb{R}$ is strictly increasing,
strictly concave and continuously differentiable, and
$\lim_{c\to+\infty}u^{\prime}(c)=0$.

The strictly decreasing and continuous function
$u^{\prime}:(0,\infty)\overset{\textrm{onto}}{\longrightarrow}(0,u^{\prime}(0))$
has a strictly decreasing, continuous inverse
$I_{u}:(0,u^{\prime}(0))\overset{\textrm{onto}}{\longrightarrow}(0,\infty)$.
We extend $I_{u}$ by setting $I_{u}(y)=0$ for $y\geq u^{\prime}(0)$. Then, we
have

$\displaystyle
u^{\prime}(I_{u}(y))=\begin{cases}y,\;\;\;&0<y<u^{\prime}(0),\\\
u^{\prime}(0),\;\;\;&y\geq u^{\prime}(0),\end{cases}$ (2)

and $I_{u}(u^{\prime}(c))=c$ for $0<c<\infty$. Note that
$\lim_{y\to\infty}I_{u}(y)=0$.

  * (ii)

For any $y>0$,

$\displaystyle\int_{0}^{y}\eta^{-n_{2}}I_{u}(\eta)d\eta<\infty.$ (3)

where $n_{1}>0$ and $n_{2}<0$ are two roots of the quadratic equation:

$\dfrac{\theta^{2}}{2}n^{2}+\left(\rho-r-\frac{\theta^{2}}{2}\right)n-\rho=0.\;\;\;$
(4)

To guarantee that the problem is well-defined, we make the following
assumption:

######  Assumption 1.

$u_{B},u_{A}\in\Sigma$,

for all $y>0$, the following inequality holds:

$u_{B}(I_{u_{B}}(y))<u_{A}(I_{u_{A}}(y)).$

There are two assets in the financial market, a risk-free asset and a risky
asset. The return on the risk-free asset is a constant $r>0$. The cum-dividend
price $S_{t}$ of the risky asset follows the dynamics:

$\displaystyle\dfrac{dS_{t}}{S_{t}}=\mu dt+\sigma dB_{t},$ (5)

where $\mu>0$ and $\sigma$ are constants, describing the mean and the standard
deviation of the returns on the risky asset, and $B_{t}$ is a standard
Brownian motion defined on a filtered probability space
$(\Omega,{\mathcal{F}},P)$. We assume that filtration
${\mathcal{F}}=({\mathcal{F}}_{t})_{t\geq 0}$ is the augmented filtration
generated by $B_{t}$; it describes the information available to the agent at
each instance. The two asset model is a simplification of that of $n$-risky
assets and a single risk-free asset with constant covariance matrix of risky
asset returns, where the capital asset pricing model(CAPM) and the two-fund
separation theorem are valid (Grossman and Laroque, (1990).

We assume that the agent receives constant wage income equal to $\epsilon>0$
until retirement. The constant wage rate assumption is standard in
literature.666For example, Fahri and Panageas, (2007) states, “we maintain the
assumption that agents receive a constant wage. This is done not only for
simplicity, but more importantly because it makes the results more
surprising.” (p.91) Let $\pi_{t}$ denote the agent’s investment in the risky
asset at time $t$. Then, her wealth $X_{t}$ at time $t$ satisfies the
dynamics:

$dX_{t}=[rX_{t}+(\mu-r)\pi_{t}+\epsilon{\bf
1}_{\\{t<\tau\\}}-c_{t}]dt+\sigma\pi_{t}dB_{t}\;\;\mbox{with}\;\;X_{0}=x,$ (6)

where $x$ is her initial wealth and ${\bf 1}_{A}$ denotes the indicator
function of set $A$. The natural limit for the wealth is the following:

$\displaystyle X_{t}\geq-\frac{\epsilon}{r}{\bf 1}_{\\{t<\tau\\}},$ (7)

that is the agent cannot borrow more than the maximum possible present value
of labor income.

We give technical admissibility conditions for $(c,\pi,\tau)$. Throughout the
paper, $(c,\pi,\tau)$ belongs to the admissible class ${\cal A}(x)$ if they
are ${\cal F}_{t}$-progressively measurable processes satisfying the following
conditions:

  * (a)

$\tau$ belongs to ${\cal S}$, the set of all ${\cal F}$-stopping times taking
values in $(0,\infty)$,

  * (b)

$(c,\pi,\tau)$ satisfies the agent’s wealth dynamics in (6) together with
constraint (7),

  * (c)

$c_{t}$ and $\pi_{t}$ satisfy

$\int_{0}^{t}c_{s}ds<\infty,\;\;\mbox{a.s}\;\;\mbox{and}\;\;\int_{0}^{t}\pi_{s}^{2}ds<\infty\;\;\mbox{a.s},\;\;\;\forall\;t\geq
0.$ (8)

We now state the agent’s problem:

######  Problem 1.

Given $x>-\frac{\epsilon}{r}$, we consider the following optimization problem:

$V(x)=\sup_{(c,\pi,\tau)\in{\cal A}(x)}\mathbb{E}\left[\int_{0}^{\tau}e^{-\rho
t}u_{B}(t)dt+\int_{\tau}^{\infty}e^{-\rho t}u_{A}(c_{t})dt\right].$

##  3 Optimization

We derive a solution to the agent’s optimization problem by the standard dual
martingale approach developed by Karatzas et al., (1987) and Cox and Huang,
(1989). They show that the stochastic discount factor of the economy takes the
form:

$\displaystyle{\xi}_{t}=e^{-rt-\frac{1}{2}\theta^{2}t-\theta
B_{t}},\quad\theta\equiv\frac{\mu-r}{\sigma}.$ (9)

Note that the constant $\theta$ is equal to the Sharpe ratio of the risky
asset.

In the following proposition, we transform the wealth dynamics (6) into a
budget constraint in static form.

######  Proposition 3.1.

Let $x>-\frac{\epsilon}{r}.$ Suppose $(c,\pi,\tau)\in{\cal A}(x).$ Then
$c=(c_{t})_{t=0}^{\infty}$ satisfies the following static budget constraint

$\mathbb{E}\left[\int_{0}^{\infty}\xi\left(c_{t}-\epsilon{\bf
1}_{\\{t<\tau\\}}\right)dt\right]=x.$ (10)

Conversely, suppose that $c=(c_{t})_{t=0}^{\infty}$ satisfies the first
condition in (8) and $\tau\in{\cal S}$ such that (10) is satisfied. Then there
exists a portfolio process $\pi_{t}$ such that $(c,\pi,\tau)\in{\cal A}(x)$.
The corresponding wealth process $X^{x,c,\pi}$ is

$dX_{t}^{x,c,\pi}=[rX_{t}^{x,c,\pi}+(\mu-r)\pi_{t}-c_{t}+\epsilon{\bf
1}_{\\{t<\tau\\}}]dt+\sigma\pi_{t}dB_{t}\;\;\;t\geq 0$

and

$X_{t}^{x,c,\pi}=\mathbb{E}_{t}\left[\int_{t}^{\infty}\dfrac{\xi_{s}}{\xi_{t}}(c_{s}-\epsilon
1_{\\{s<\tau\\}})ds\right],$ (11)

where $\mathbb{E}_{t}[\cdot]=\mathbb{E}\left[\cdot\mid{\cal F}_{t}\right]$ is
the conditional expectation at time $t$ on the filtration ${\cal F}_{t}$.

###  3.1 Properties of the class $\Sigma$

For $y>0$, the conjugate function $\tilde{u}$ of $u\in\Sigma$ is defined as

$\tilde{u}(y)=\sup_{c\geq 0}\left(u(c)-yc\right)=u(I_{u}(y))-yI_{u}(y).$

######  Lemma 3.1.

For the felicity function $u\in\Sigma$ , the following integrability
conditions hold:

$\int_{0}^{y}\nu^{-n_{2}-1}|u(I_{u}(\nu))|d\nu+\int_{y}^{\infty}\nu^{-n_{1}-1}|u(I_{u}(\nu))|d\nu<\infty.$

and

$\int_{0}^{y}\nu^{-n_{2}-1}|\tilde{u}(\nu)|d\nu+\int_{y}^{\infty}\nu^{-n_{1}-1}|\tilde{u}(\nu)|d\nu<\infty.$

###### Proof.

It is easy to check that for any $u\in\Sigma$

$\int_{0}^{y}\nu^{-n_{2}}I_{u}(\nu)d\nu+\int_{y}^{\infty}\nu^{-n_{1}}I_{u}(\nu)d\nu<\infty.$

Note that for any $y>0$

$u(I_{u}(y))-yI_{u}(y)=\tilde{u}(y)\;\;\mbox{and}\;\;\tilde{u}^{\prime}(y)=-I_{u}(y),$

and thus

$u(I_{u}(\nu))=u(I_{u}(y))-yI_{u}(y)+\nu
I_{u}(\nu)+\int_{\nu}^{y}I_{u}(\eta)d\eta.$

It follows that

$\displaystyle\begin{split}&\int_{0}^{y}\nu^{-n_{2}-1}|u(I_{u}(\nu))|d\nu+\int_{y}^{\infty}\nu^{-n_{1}-1}|u(I_{u}(\nu))|d\nu\\\
\leq-&\dfrac{1}{n_{2}}y^{-n_{2}}|u(I_{u}(y))-yI_{u}(y)|+\int_{0}^{y}\nu^{-n_{2}}I_{u}(\nu)d\nu+\int_{0}^{y}\int_{\nu}^{y}\nu^{-n_{2}-1}I_{u}(\eta)d\eta
d\nu\\\
+&\dfrac{1}{n_{1}}y^{-n_{1}}|u(I_{u}(y))-yI_{u}(y)|+\int_{y}^{\infty}\nu^{-n_{1}}I_{u}(\nu)d\nu+\int_{y}^{\infty}\int_{y}^{\nu}\nu^{-n_{1}-1}I_{u}(\eta)d\eta
d\nu\\\
=-&\dfrac{1}{n_{2}}y^{-n_{2}}|u(I_{u}(y))-yI_{u}(y)|+\dfrac{1}{n_{1}}y^{-n_{1}}|u(I_{u}(y))-yI_{u}(y)|\\\
&+\left(1+\dfrac{1}{n_{1}}\right)\int_{y}^{\infty}\nu^{-n_{1}}I_{u}(\nu)d\nu+\left(1-\dfrac{1}{n_{2}}\right)\int_{0}^{y}\nu^{-n_{2}}I_{u}(\nu)d\nu<\infty.\end{split}$
(12)

where we have used Fubini’s theorem in last equality.

Since $u(I_{u}(\nu))=\tilde{u}(\nu)+\nu I_{u}(\nu)$, we have

$\int_{0}^{y}\nu^{-n_{2}-1}|\tilde{u}(\nu)|d\nu+\int_{y}^{\infty}\nu^{-n_{1}-1}|\tilde{u}(\nu)|d\nu<\infty.$
(13)

∎

For any measurable function $f:\mathbb{R}^{+}\to\mathbb{R}$, we define two
operators $\Xi$ and $\Gamma$ by

$\Xi_{f}(y)=\dfrac{2}{\theta^{2}(n_{1}-n_{2})}\left[y^{n_{2}}\int_{0}^{y}\nu^{-n_{2}-1}f(\nu)d\nu+y^{n_{1}}\int_{y}^{\infty}\nu^{-n_{1}-1}f(\nu)d\nu\right]$

and

$\Gamma_{f}(y)=\dfrac{2}{\theta^{2}(n_{1}-n_{2})}\left[y^{n_{2}-1}\int_{0}^{y}\nu^{-n_{2}}f(\nu)d\nu+y^{n_{1}-1}\int_{y}^{\infty}\nu^{-n_{1}}f(\nu)d\nu\right].$

For $u\in\Sigma$, $\Gamma_{I_{u}},\;\Xi_{u\circ I_{u}}$ and $\Xi_{\tilde{u}}$
are given by

$\displaystyle\Gamma_{I_{u}}(y)=$
$\displaystyle\dfrac{2}{\theta^{2}(n_{1}-n_{2})}\left[y^{n_{2}-1}\int_{0}^{y}\nu^{-n_{2}}I_{u}(\nu)d\nu+y^{n_{1}-1}\int_{y}^{\infty}\nu^{-n_{1}}I_{u}(\nu)d\nu\right],$
$\displaystyle\Xi_{u\circ I_{u}}(y)=$
$\displaystyle\dfrac{2}{\theta^{2}(n_{1}-n_{2})}\left[y^{n_{2}}\int_{0}^{y}\nu^{-n_{2}-1}u(I_{u}(\nu))d\nu+y^{n_{1}}\int_{y}^{\infty}\nu^{-n_{1}-1}u(I_{u}(\nu))d\nu\right]$

and

$\Xi_{\tilde{u}}(y)=\dfrac{2}{\theta^{2}(n_{1}-n_{2})}\left[y^{n_{2}}\int_{0}^{y}\nu^{-n_{2}-1}\tilde{u}(\nu)d\nu+y^{n_{1}}\int_{y}^{\infty}\nu^{-n_{1}-1}\tilde{u}(\nu)d\nu\right],$

respectively.

By Lemma 3.1 and Proposition B.1, we can directly obtain the following
proposition:

######  Proposition 3.2.

For any $u\in\Sigma$, $\Gamma_{I_{u}},\;\Xi_{u\circ I_{u}}$ and
$\Xi_{\tilde{u}}$ are well-defined and the following equalities hold:

$\Gamma_{I_{u}}(y)=\dfrac{1}{y}\mathbb{E}\left[\int_{0}^{\infty}e^{-\rho
t}{\cal Y}_{t}^{y}I_{u}({\cal Y}_{t}^{y})dt\right],\;\;\Xi_{u\circ
I_{u}}(y)=\mathbb{E}\left[\int_{0}^{\infty}e^{-\rho t}u(I_{u}({\cal
Y}_{t}^{y}))dt\right]$

and

$\Xi_{\tilde{u}}(y)=\mathbb{E}\left[\int_{0}^{\infty}e^{-\rho
t}\tilde{u}({\cal Y}_{t}^{y})dt\right],$

where ${\cal Y}_{t}^{y}\equiv ye^{\rho t}\xi_{t}$.

######  Lemma 3.2.

For any $u\in\Xi$, the following statements are true:

  * (a)

$\Xi_{\tilde{u}}^{\prime}(y)=-\Gamma_{I_{u}}(y)$.

  * (b)

$\Gamma_{I_{u}}(y)$ is strictly decreasing in $y>0$. Thus,
$\Xi_{\tilde{u}}(y)$ is strictly convex in $y>0$.

  * (c)

$\lim_{y\to\infty}\Gamma_{I_{u}}(y)=0$ and $\lim_{y\to
0+}\Gamma_{I_{u}}(y)=\infty$.

  * (d)

$\lim_{t\to\infty}e^{-\rho t}\mathbb{E}\left[{\cal
Y}_{t}^{y}\Gamma_{I_{u}}({\cal Y}_{t}^{y^{*}})\right]=0$.

  * (e)

There exist positive constants $\zeta_{1}$ and $\zeta_{2}$ such that

$|y\Gamma_{I_{u}}^{\prime}(y)|\leq\zeta_{1}(y^{\zeta_{2}}+y^{-\zeta_{2}}).$

###### Proof.

(a): By the definition of $\Xi_{\tilde{u}}(y)$, we have

$\displaystyle\Xi_{\tilde{u}}^{\prime}(y)=\dfrac{2}{\theta^{2}(n_{1}-n_{2})}\left[n_{2}y^{n_{2}-1}\int_{0}^{y}\nu^{-n_{2}-1}\tilde{u}(\nu)d\nu+n_{1}y^{n_{1}-1}\int_{y}^{\infty}\nu^{-n_{1}-1}\tilde{u}(\nu)d\nu\right].$

It follows from Lemma 3.1 and Proposition B.1 that

$\displaystyle\liminf_{y\downarrow
0}y^{-n_{2}}\tilde{u}(y)=\liminf_{y\uparrow\infty}y^{-n_{1}}\tilde{u}(y)=0.$
(14)

By applying the integration by parts for Riemann-Stieltjes integral, it
follows from the limiting behavior of $\tilde{u}$ in (14) that

$\displaystyle\Xi_{\tilde{u}}^{\prime}(y)=$
$\displaystyle\dfrac{2}{\theta^{2}(n_{1}-n_{2})}\left[n_{2}y^{n_{2}-1}\int_{0}^{y}\nu^{-n_{2}-1}\tilde{u}(\nu)d\nu+n_{1}y^{n_{1}-1}\int_{y}^{\infty}\nu^{-n_{1}-1}\tilde{u}(\nu)d\nu\right]$
$\displaystyle=$
$\displaystyle-\dfrac{2}{\theta^{2}(n_{1}-n_{2})}\left[y^{n_{2}-1}\int_{0}^{y}\nu^{-n_{2}}I_{u}(\nu)d\nu+y^{n_{1}-1}\int_{y}^{\infty}\nu^{-n_{1}}I_{u}(\nu)d\nu\right]$
$\displaystyle=$ $\displaystyle-\Gamma_{I_{u}}(y),$

where we have used the fact $(\tilde{u}(y))^{\prime}=-I_{u}(y)$.

(b): It follows from $u\in\Sigma$ and Proposition B.1 that

$\displaystyle\liminf_{y\downarrow
0}y^{-n_{2}}I_{u}(y)=\liminf_{y\uparrow\infty}y^{-n_{1}}I_{u}(y)=0.$ (15)

The integration by parts for Riemann-Stieltjes integral implies that

$\displaystyle\Gamma_{I_{u}}^{\prime}(y)=$
$\displaystyle\dfrac{2}{\theta^{2}(n_{1}-n_{2})}\left[(n_{2}-1)y^{n_{2}-2}\int_{0}^{y}\nu^{-n_{2}}I_{u}(\nu)d\nu+(n_{1}-1)y^{n_{1}-2}\int_{y}^{\infty}\nu^{-n_{1}}I_{u}(\nu)d\nu\right]$
$\displaystyle=$
$\displaystyle\dfrac{2}{\theta^{2}(n_{1}-n_{2})}\left[y^{n_{2}-2}\int_{0}^{y}\nu^{1-n_{2}}I^{\prime}_{u}(\nu)d\nu+y^{n_{1}-2}\int_{y}^{\infty}\nu^{1-n_{1}}I^{\prime}_{u}(\nu)d\nu\right]<0,$

where we have used the fact $I_{u}$ is strictly decreasing in $y>0$.

(c): Note that

$\Gamma_{I_{u}}(y)=\dfrac{1}{y}\mathbb{E}\left[\int_{0}^{\infty}e^{-\rho
t}{\cal Y}_{t}^{y}I_{u}({\cal
Y}_{t}^{y})dt\right]=\mathbb{E}\left[\int_{0}^{\infty}\xi_{t}I_{y}({\cal
Y}_{t})dt\right].$

Since $\mathbb{E}\left[\int_{0}^{\infty}\xi_{t}I_{u}({\cal
Y}_{t}^{y})dt\right]<\infty$, the dominated convergence theorem implies that

$\displaystyle\lim_{y\downarrow
0}\Gamma_{I_{u}}(y)=\mathbb{E}\left[\int_{0}^{\infty}\lim_{y\downarrow
0}\xi_{t}I_{u}(ye^{\rho t}{\xi}_{t})dt\right]=\infty$

and

$\displaystyle\lim_{y\uparrow\infty}\Gamma_{I_{u}}(y)=\mathbb{E}\left[\int_{0}^{\infty}\lim_{y\uparrow\infty}\xi_{t}I_{u}(ye^{\rho
t}{\xi}_{t})dt\right]=0.$

Proof of (d): Note that

$y\Gamma_{I_{u}}(y)=\mathbb{E}\left[\int_{0}^{\infty}e^{-\rho t}{\cal
Y}_{t}^{y}I_{u}({\cal Y}_{t}^{y})dt\right]<\infty.$

Thus, it follows from Proposition B.1 (e) that

$\lim_{t\to\infty}e^{-\rho t}\mathbb{E}\left[{\cal
Y}_{t}^{y}\Gamma_{I_{u}}({\cal Y}_{t}^{y^{*}})\right]=0.$

Proof of (e): Since $\Gamma_{I_{u}}(y)=-\Xi_{\tilde{u}}^{\prime}(y)$,
Proposition B.1 (d) implies that there exits a constant $C_{1}>0$ satisfying

$|\Gamma_{I_{u}}(y)|\leq C_{1}(y^{n_{1}-1}+y^{n_{2}-1}).$ (16)

Since $y\Gamma_{I_{u}}(y)=\mathbb{E}\left[\int_{0}^{\infty}e^{-\rho t}{\cal
Y}_{t}^{y}I_{u}({\cal Y}_{t}^{y})dt\right]$, there exists a constant $C_{2}>0$
such that

$|(y\Gamma_{I_{u}}(y))^{\prime}|\leq C_{2}(y^{n_{1}-1}+y^{n_{2}-1}).$ (17)

Thus, we have

$\displaystyle|y\Gamma_{I_{u}}^{\prime}(y)|\leq|(y\Gamma_{I_{u}}(y))^{\prime}|+|\Gamma_{I_{u}}(y)|\leq(C_{1}+C_{2})(y^{n_{1}-1}+y^{n_{2}-1}).$

This completes the proof.

∎

###  3.2 Dual formulation

We formulate the following Lagrangian:

$\displaystyle\begin{split}\mathfrak{L}=&\mathbb{E}\left[\int_{0}^{\tau}e^{-\rho
t}u_{B}(c_{t})dt+\int_{\tau}^{\infty}e^{-\rho
t}u_{A}(c_{t})dt\right]+y\left(x-\mathbb{E}\left[\int_{0}^{\infty}{\xi}_{t}(c_{t}-\epsilon{\bf
1}_{\\{t<\tau\\}})dt\right]\right)\\\
=&\mathbb{E}\left[\int_{0}^{\tau}e^{-\rho t}(u_{B}(c_{t})-{\cal
Y}_{t}^{y}c_{t}+{\cal
Y}_{t}^{y}\epsilon)dt+e^{-\rho\tau}{\mathbb{E}_{\tau}\left[\int_{\tau}^{\infty}e^{-\rho(t-\tau)}\left(u_{A}(c_{t})-{\cal
Y}_{t}^{y}c_{t}\right)dt\right]}\right]+yx\\\
\leq&\mathbb{E}\left[\int_{0}^{\tau}e^{-\rho t}(u_{B}(c_{t})-{\cal
Y}_{t}^{y}c_{t}+{\cal
Y}_{t}^{y}\epsilon)dt+e^{-\rho\tau}{\mathbb{E}_{\tau}\left[\int_{\tau}^{\infty}e^{-\rho(t-\tau)}\tilde{u}_{A}({\cal
Y}_{t}^{y})dt\right]}\right]+yx\\\
\leq&\mathbb{E}\left[\int_{0}^{\tau}e^{-\rho t}\left(\tilde{u}_{B}({\cal
Y}_{t}^{y})+{\cal Y}_{t}^{y}\epsilon\right)dt+\int_{\tau}^{\infty}e^{-\rho
t}\tilde{u}_{A}({\cal Y}_{t}^{y})dt\right]+yx\\\
=&\mathbb{E}\left[\int_{0}^{\tau}e^{-\rho t}\left(\tilde{u}_{B}({\cal
Y}_{t}^{y})-\tilde{u}_{A}({\cal Y}_{t}^{y})+{\cal
Y}_{t}^{y}\epsilon\right)dt+\int_{0}^{\infty}e^{-\rho t}\tilde{u}_{A}({\cal
Y}_{t}^{y})dt\right]+yx\\\ =&J_{A}(y)+\mathbb{E}\left[\int_{0}^{\tau}e^{-\rho
t}\left(\tilde{u}_{B}({\cal Y}_{t}^{y})-\tilde{u}_{A}({\cal Y}_{t}^{y})+{\cal
Y}_{t}^{y}\epsilon\right)dt\right]+yx,\end{split}$ (18)

where $y>0$ is a Lagrangian multiplier of the static budget constraint (10),
${\cal Y}_{t}^{y}=ye^{\rho t}{\xi}_{t}$,

$\displaystyle J_{A}(y)=\mathbb{E}\left[\int_{0}^{\infty}e^{-\rho
t}\tilde{u}_{A}({\cal Y}_{t}^{y})dt\right],$ (19)

and $\tilde{u}_{B}$, $\tilde{u}_{A}$ are the conjugate function of $u_{B}$,
$u_{A}$ given by

$\displaystyle\tilde{u}_{B}(y)\equiv$ $\displaystyle\sup_{c\geq
0}\left(u_{B}(c)-yc\right)=u_{B}(I_{u_{B}}(y))-yI_{u_{B}}(y),$
$\displaystyle\tilde{u}_{A}(y)\equiv$ $\displaystyle\sup_{c\geq
0}\left(u_{A}(c)-yc\right)=u_{A}(I_{u_{A}}(y))-yI_{u_{B}}(y),$

respectively. The dual variable ${\cal Y}_{t}$ is the marginal value of
wealth, as will be shown in Theorem 4.1.

Thus, for $y>0$ the candidate of optimal consumption $(\hat{c}({\cal
Y}_{t}^{y}))_{t=0}^{\infty}$ is given by

$\displaystyle\begin{split}\hat{c}({\cal
Y}_{t}^{y})=\begin{cases}I_{u_{B}}({\cal Y}_{t}^{y})\;\;\;&\mbox{for}\;\;0\leq
t<\tau,\\\ I_{u_{A}}({\cal
Y}_{t}^{y})\;\;\;&\mbox{for}\;\;t\geq\tau.\end{cases}\end{split}$ (20)

Considering the Lagrangian (18), we formulate the dual problem which chooses
the optimal time of retirement by observing the changes of the marginal value
of wealth ${\cal Y}_{t}$.

######  Problem 2.

$J(y)=J_{A}(y)+\sup_{\tau\in{\cal S}}\mathbb{E}\left[\int_{0}^{\tau}e^{-\rho
t}\left(\tilde{u}_{B}({\cal Y}_{t}^{y})-\tilde{u}_{A}({\cal Y}_{t}^{y})+{\cal
Y}_{t}^{y}\epsilon\right)dt\right],$ (21)

where ${\cal S}$ denotes the set of ${\mathcal{F}}-$stopping times. We call
$J(y)$ the dual value function.

Problem 2 states that the dual value function consists of $J_{A}$ and the
optimized value of the optimal stopping problem. If $\tau=0$, i.e., the agent
retires immediately, then the dual value function is equal to $J_{A}$, and
hence, $J_{A}$ can be regarded as the dual value function after retirement.
The quantity

$\displaystyle\tilde{u}_{B}({\cal Y}_{t}^{y})-\tilde{u}_{A}({\cal
Y}_{t}^{y})+{\cal Y}_{t}^{y}\epsilon$ (22)

inside the integral in the optimal stopping problem is composed of two
components: the first is the difference in the conjugate felicity functions,
$\tilde{u}_{B}({\cal Y}_{t}^{y})-\tilde{u}_{A}({\cal Y}_{t}^{y}),$ which can
be interpreted as the difference in the utility values before and after
retirement, and the second is ${\cal Y}_{t}^{y}\epsilon$, labor income
adjusted by the marginal utility of wealth and can be interpreted as the
utility value of labor income. Accordingly, quantity (22) is the marginal
benefit of work relative to retirement. The agent chooses the retirement time
$\tau$ that maximizes the present value of the marginal benefit. In this sense
we will call the optimized value of the optimal stopping problem as the
utility value of lifetime labor.

Let us denote the utility value of lifetime labor by ${\cal P}(y),$ i.e.,

${\cal P}(y)=\sup_{\tau\in{\cal S}}\mathbb{E}\left[\int_{0}^{\tau}e^{-\rho
t}\left(\tilde{u}_{B}({\cal Y}_{t}^{y})-\tilde{u}_{A}({\cal Y}_{t}^{y})+{\cal
Y}_{t}^{y}\epsilon\right)dt\right].$ (23)

Let us denote the current normalized marginal benefit of work by $\Psi(y),$
i.e.,

$\Psi(y)=\dfrac{1}{y}\left(\tilde{u}_{B}(y)-\tilde{u}_{A}(y)\right)+\epsilon.$

We will now investigate properties of the normalized marginal benefit.

######  Lemma 3.3.

$\Psi(y)$ is a strictly increasing function of $y>0$ and

$\lim_{y\to\infty}\Psi(y)=\epsilon.$

###### Proof.

It follows from Assumption 1 that

$\displaystyle\Psi^{\prime}(y)=$
$\displaystyle-\dfrac{1}{y^{2}}\left(\tilde{u}_{B}(y)-\tilde{u}_{A}(y)\right)+\dfrac{1}{y}(-I_{u_{B}}(y)+I_{u_{A}}(y))$
$\displaystyle=$
$\displaystyle-\dfrac{1}{y^{2}}\left((\tilde{u}_{B}(y)+yI_{u_{B}}(y))-(\tilde{u}_{A}(y)+yI_{u_{A}}(y))\right)$
$\displaystyle=$
$\displaystyle-\left(u(I_{u_{B}}(y))-u(I_{u_{A}}(y))\right)>0.$

Hence, $\Psi(y)$ is a strictly increasing function of $y>0$.

  * (i)

If $\lim_{y\to\infty}|\left(\tilde{u}_{B}(y)-\tilde{u}_{A}(y)\right)|=\infty$,

it follows form L’Hospital’s rule that

$\displaystyle\lim_{y\to\infty}\dfrac{\tilde{u}_{B}(y)-\tilde{u}_{A}(y)}{y}=\lim_{y\to\infty}\dfrac{-I_{u_{B}}(y)+I_{u_{A}}(y)}{1}=0.$

  * (ii)

If $\lim_{y\to\infty}|\left(\tilde{u}_{B}(y)-\tilde{u}_{A}(y)\right)|<\infty$,

it is clear that

$\displaystyle\lim_{y\to\infty}\dfrac{\tilde{u}_{B}(y)-\tilde{u}_{A}(y)}{y}=0.$

By (i) and (ii), we deduce that

$\lim_{y\to\infty}\Psi(y)=\epsilon.$

∎

If $\lim_{y\to 0+}\Psi(y)\geq 0$, Lemma 3.3 implies that $\Psi(y)>0$ for all
$y>0$. Thus, we deduce that for any $\tau\in{\cal S}$

$\mathbb{E}\left[\int_{0}^{\tau}e^{-\rho t}\left(\tilde{u}_{B}({\cal
Y}_{t}^{y})-\tilde{u}_{A}({\cal Y}_{t}^{y})+{\cal
Y}_{t}^{y}\epsilon\right)dt\right]\leq\mathbb{E}\left[\int_{0}^{\infty}e^{-\rho
t}\left(\tilde{u}_{B}({\cal Y}_{t}^{y})-\tilde{u}_{A}({\cal Y}_{t}^{y})+{\cal
Y}_{t}^{y}\epsilon\right)dt\right].$

That is, the agent does not choose the option to retire in the case when
$\lim_{y\to 0+}\Psi(y)\geq 0$, that is, the marginal benefit of work is always
positive. Consequently, the necessary condition for the agent to retire at a
finite time $\tau$ is the following assumption:

######  Assumption 2.

$\lim_{y\to 0+}\Psi(y)<0.$

Under Assumption 2, Lemma 3.3 implies that there exists a unique $\bar{z}>0$
such that

$\Psi(\bar{z})=0.$

By the standard theory of the optimal stopping problem (Peskir and Shiryaev,
(2006)), ${\cal P}(y)$ satisfies the following variational inequality:

$\displaystyle\begin{cases}\mathbb{L}{\cal P}+h(y)\leq
0\;\;\;&\mbox{if}\;\;{\cal P}(y)=0,\vspace{2mm}\\\ \mathbb{L}{\cal
P}+h(y)=0\;\;\;&\mbox{if}\;\;{\cal P}(y)>0,\vspace{2mm}\\\ \end{cases}$ (24)

where $h(y)\equiv y\Psi(y)=(\tilde{u}_{B}(y)-\tilde{u}_{A}(y)+\epsilon y)$ and
the differential operator $\mathbb{L}$ is given by

$\mathbb{L}\equiv\dfrac{\theta^{2}}{2}y^{2}\dfrac{d^{2}}{dy^{2}}+(\rho-r)y\dfrac{d}{dy}-\rho.$

######  Theorem 3.1 (Verification theorem).

Suppose that variational inequality (24) has a solution ${\cal Q}(y)$ which is
continuously differentiable in $y>0$ and twice continuously differentiable in
$(0,\infty)\backslash\\{b\\}$ for some point $b>0$, and there exist positive
constants $\zeta_{1}$ and $\zeta_{2}$ such that

$|{\cal Q}^{\prime}(y)|\leq\zeta_{1}\left(y^{-\zeta
2}+y^{\zeta_{2}}\right)\;\;\mbox{for}\;\;y>0.$

Then,

  * (a)

${\cal P}(y)\leq{\cal Q}(y)$.

  * (b)

If $\lim_{t\to\infty}e^{-\rho t}\mathbb{E}\left[{\cal Q}({\cal
Y}_{t})\right]=0$, then ${\cal Q}(y)={\cal P}(y)$ and the optimal solution to
the problem in (23) is given by

$\tau_{R}(y)=\inf\\{t\geq 0\mid{\cal Y}_{t}^{y}\in\\{y>0\mid{\cal
P}(y)=0\\}\\}.$

###### Proof.

Even though ${\cal Q}(y)$ is only continuously differentiable, we can still
apply Itô’s lemma (see Exercise 6.24 in Karatzas and Shreve, (1991)) to the
process $\\{e^{-\rho t}{\cal Q}({\cal Y}_{t}^{y})\\}$. The proof of this
theorem is almost similar to that of Theorem 3.2 and Lemma 3.4 in Knudsen et
al., (1998) and we omit the proof. ∎

From Theorem 3.1, we know that the transversality condition is necessary to
guarantee that the solution to the HJB equation is the option value of
voluntary retirement:

$\lim_{t\to\infty}e^{-\rho t}\mathbb{E}\left[{\cal P}({\cal
Y}_{t}^{y})\right]=0.$ (25)

We provide the utility value of labor in the following proposition:

######  Proposition 3.3.

The utility value of lifetime labor, ${\cal P}(y)$, is given by

$\displaystyle\begin{split}{\cal P}(y)=\sup_{\tau\in{\cal
S}}\mathbb{E}\left[\int_{0}^{\tau}e^{-\rho t}\left(\tilde{u}_{B}({\cal
Y}_{t}^{y})-\tilde{u}_{A}({\cal Y}_{t}^{y})+{\cal
Y}_{t}^{y}\epsilon\right)dt\right]=\begin{cases}Dy^{n_{2}}+\Xi_{h}(y)\;\;\;&\mbox{for}\;\;y\geq
z_{R},\vspace{2mm}\\\ 0\;\;\;&\mbox{for}\;\;\;0<y\leq
z_{R},\end{cases}\end{split}$ (26)

where $z_{R}\in(0,\bar{z})$ is a unique solution of the following equation

$\int_{z_{R}}^{\infty}\nu^{-n_{1}-1}{h}(\nu)d\nu=0$

and

$D=-\dfrac{2}{\theta^{2}(n_{1}-n_{2})}\int_{0}^{z_{R}}\nu^{-n_{2}-1}{h}(\nu)d\nu.$

Moreover, the optimal stopping time $\tau_{R}(y)$ is

$\tau_{R}(y)=\inf\\{t\geq 0\mid{\cal Y}_{t}^{y}\leq z_{R}\\}.$

That is,

${\cal P}(y)=\sup_{\tau\in{\cal S}}\mathbb{E}\left[\int_{0}^{\tau}e^{-\rho
t}\left(\tilde{u}_{B}({\cal Y}_{t}^{y})-\tilde{u}_{A}({\cal Y}_{t}^{y})+{\cal
Y}_{t}^{y}\epsilon\right)dt\right]=\mathbb{E}\left[\int_{0}^{\tau_{R}(y)}e^{-\rho
t}\left(\tilde{u}_{B}({\cal Y}_{t}^{y})-\tilde{u}_{A}({\cal Y}_{t}^{y})+{\cal
Y}_{t}^{y}\epsilon\right)dt\right].$

We first derive the utility value of lifetime labor, ${\cal P}(y),$ in a
heuristic and intuitive manner. Suppose that there exists a boundary $z_{R}$
such that if the agent’s marginal utility ${\cal Y}_{t}^{y}$ hits the
boundary, the agent retires. The option value becomes zero after the agent
retires. Hence, by the smooth pasting condition we have

$\displaystyle{\cal P}(z_{R})={\cal P}^{\prime}(z_{R})=0,\quad{\cal
P}(y)=0\;\;\;\mbox{for}\;\;0<y\leq z_{R}.$ (27)

When the agent is still working, ${\cal P}(y)$ satisfies the Hamilton-Jacobi-
Bellman(HJB) equation

$\displaystyle\dfrac{\theta^{2}}{2}y^{2}{\cal
P}^{\prime\prime}(y)+(\rho-r)y{\cal P}(y)-\rho{\cal P}+{h}(y)=0.$ (28)

That is, for $y>z_{R}$, a general solution to the HJB equation (28) can be
represented as the sum of a general solution to the homogeneous equation and a
particular solution as

${\cal P}(y)=D_{1}y^{n_{1}}+D_{2}y^{n_{2}}+\Xi_{h}(y),$

where the particular solution $\Xi_{h}(y)$ is given by

$\displaystyle\Xi_{h}(y)=\dfrac{2}{\theta^{2}(n_{1}-n_{2})}\left[y^{n_{2}}\int_{0}^{y}\nu^{-n_{2}-1}{h}(\nu)d\nu+y^{n_{1}}\int_{y}^{\infty}\nu^{-n_{1}-1}{h}(\nu)d\nu\right]$
(29)

To satisfy the transversality condition (25) for ${\cal P}(y)$, the
coefficient $D_{1}$ of $y^{n_{1}}$ should be zero, i.e,

$D_{1}=0.$

Thus, we can write down ${\cal P}(y)$ as follows:

${\cal P}(y)=Dy^{n_{2}}+\Xi_{h}(y).$

By using the smooth pasting condition (${\cal P}(z_{R})={\cal
P}^{\prime}(z_{R})=0$), we have

$\int_{z_{R}}^{\infty}\nu^{-n_{1}-1}{h}(\nu)d\nu=0\;\;\;\mbox{and}\;\;\;D=-\dfrac{2}{\theta^{2}(n_{1}-n_{2})}\int_{0}^{z_{R}}\nu^{-n_{2}-1}{h}(\nu)d\nu.$

We will now proceed to give a formal proof of the proposition.

###### Proof.

Let us denote ${\cal G}(y)$ by

$\displaystyle{\cal
G}(y)=\int_{y}^{\infty}\nu^{-n_{1}-1}{h}(\nu)d\nu=\int_{y}^{\infty}\nu^{-n_{1}}{\Psi}(\nu)d\nu.$

By Lemma 3.3, we deduce that ${\cal G}(y)$ is strictly increasing on
$(0,\bar{z})$ and strictly decreasing on $(\bar{z},\infty)$. Moreover,

${\cal G}(z)>0\;\;\mbox{for all}\;\;z\geq\bar{z}.$

By Assumption 2, there exist $\delta>0$ and $y_{\delta}>0$ such that

$\Psi(y)<-\delta\;\;\;\mbox{for all}\;\;y\in(0,y_{\delta}).$

For sufficiently small $y<y_{\delta}$, we have

$\displaystyle{\cal G}(y)=\int_{y}^{\infty}\nu^{-n_{1}}\Psi(\nu)d\nu=$
$\displaystyle\int_{y}^{y_{\delta}}\nu^{-n_{1}}\Psi(\nu)d\nu+\int_{y_{\delta}}^{\infty}\nu^{-n_{1}}\Psi(\nu)d\nu$
$\displaystyle<$
$\displaystyle-\delta\int_{y}^{y_{\delta}}\nu^{-n_{1}}d\nu+\int_{y_{\delta}}^{\infty}\nu^{-n_{1}}\Psi(\nu)d\nu$
$\displaystyle=$
$\displaystyle-\delta\dfrac{1}{1-n_{1}}({y_{\delta}}^{1-n_{1}}-y^{1-n_{1}})+\int_{y_{\delta}}^{\infty}\nu^{-n_{1}}\Psi(\nu)d\nu.$

Since
$\displaystyle\int_{y_{\delta}}^{\infty}\nu^{-n_{1}}|\Psi(\nu)|d\nu<\infty$,

$\lim_{y\to 0+}{\cal
G}(y)<\delta\dfrac{1}{n_{1}-1}({y_{\delta}}^{1-n_{1}}-\lim_{y\to
0+}y^{1-n_{1}})+\int_{y_{\delta}}^{\infty}\nu^{-n_{1}}\Psi(\nu)d\nu=-\infty.$

Thus, there exists a unique $z_{R}\in(0,\bar{z})$ such that ${\cal
G}(z_{R})=0$.

Claim 1: ${\cal P}(y)$, defined by

$\displaystyle{\cal
P}(y)=\begin{cases}Dy^{n_{2}}+\Xi_{h}(y)\;\;\;&\mbox{for}\;\;y>z_{R}\vspace{2mm}\\\
0\;\;\;&\mbox{for}\;\;0<y\leq z_{R},\end{cases}$

is continuously differentiable in $(0,\infty)$ and twice continuously
differentiable in $(0,\infty)\backslash\\{z_{R}\\}$ and satisfies the
variational inequality (24).

Proof of Claim 1: By the construction of ${\cal P}(y)$, it is clear that
${\cal P}(y)$ is continuously differentiable in $(0,\infty)$ and twice
continuously differentiable in $(0,\infty)\backslash\\{z_{R}\\}$.

Note that for $y\in(0,z_{R}]$

$\displaystyle\mathbb{L}{\cal P}(y)+h(y)=h(y)=y\Psi(y)<0,$

where we have used that fact that $\Psi(y)$ is strictly increasing $y>0$ and
$\Psi(\bar{z})=0$.

Since $h(y)=\tilde{u}_{B}(y)-\tilde{u}_{A}(y)+\epsilon y$, we have

$\Xi_{h}(y)=\Xi_{\tilde{u}_{B}}(y)-\Xi_{\tilde{u}_{A}}(y)+\dfrac{\epsilon}{r}y.$

It follows from Lemma 3.2 that

$\Xi_{h}^{\prime}(y)=\Gamma_{I_{u_{A}}}(y)-\Gamma_{I_{u_{B}}}(y)+\dfrac{\epsilon}{r}$

and

$\lim_{y\uparrow\infty}\Xi_{h}^{\prime}(y)=\lim_{y\uparrow\infty}\Gamma_{I_{u_{A}}}(y)-\lim_{y\uparrow\infty}\Gamma_{I_{u_{B}}}(y)+\dfrac{\epsilon}{r}=\dfrac{\epsilon}{r}.$

Since ${\cal P}(y)=Dy^{n_{2}}+\Xi_{h}(y)$ for $y>z_{R}$, we have

$\displaystyle\lim_{y\uparrow\infty}{\cal
P}^{\prime}(y)=\lim_{y\uparrow\infty}(n_{2}Dy^{n_{2}-1})+\lim_{y\uparrow\infty}\Xi_{h}^{\prime}(y)=\dfrac{\epsilon}{r}>0.$

Note that for $y>z_{R}$,

$\displaystyle{\cal P}(y)$ $\displaystyle=Dy^{n_{2}}+\Xi_{h}(y)$
$\displaystyle=-\dfrac{2}{\theta^{2}(n_{1}-n_{2})}y^{n_{2}}\int_{0}^{z_{R}}\nu^{-n_{2}-1}{h}(\nu)d\nu+\dfrac{2}{\theta^{2}(n_{1}-n_{2})}\left[y^{n_{2}}\int_{0}^{y}\nu^{-n_{2}-1}{h}(\nu)d\nu+y^{n_{1}}\int_{y}^{\infty}\nu^{-n_{1}-1}{h}(\nu)d\nu\right]$
$\displaystyle=\dfrac{2}{\theta^{2}(n_{1}-n_{2})}\left[y^{n_{2}}\int_{z_{R}}^{y}\nu^{-n_{2}-1}{h}(\nu)d\nu+y^{n_{1}}\int_{y}^{\infty}\nu^{-n_{1}-1}{h}(\nu)d\nu\right].$

Thus,

${\cal
P}^{\prime}(y)=\dfrac{2}{\theta^{2}(n_{1}-n_{2})}\left[n_{2}y^{n_{2}-1}\int_{z_{R}}^{y}\nu^{-n_{2}-1}{h}(\nu)d\nu+n_{1}y^{n_{1}-1}\int_{y}^{\infty}\nu^{-n_{1}-1}{h}(\nu)d\nu\right]\;\;\mbox{for}\;\;y>z_{R}.$

Since $h(y)\geq 0$ for $y\geq\bar{z}$ and $h(y)<0$ for $y<\bar{z}$, we deduce
that ${\cal P}^{\prime}(y)$ is strictly increasing in $y\in(z_{R},\bar{z})$
and strictly decreasing in $y>\bar{z}$.

It follows from ${\cal P}^{\prime}(z_{R})=0$ and $\lim_{y\uparrow\infty}{\cal
P}^{\prime}(y)=\frac{\epsilon}{r}$ that

${\cal P}^{\prime}(y)>0\;\;\mbox{for}\;\;y>z_{R}.$

That is, ${\cal P}(y)$ is strictly increasing in $y\in(z_{R},\infty)$. Since
${\cal P}(z_{R})=0$, we have

${\cal P}(y)>0\;\;\mbox{for}\;\;y>z_{R}.$

Therefore, ${\cal P}(y)$ satisfies the variational inequality (24). This
completes the proof of Claim 1.

Claim 2: There exists positive constant $\zeta$ such that

$|{\cal P}^{\prime}(y)|\leq\zeta\left(y^{n_{1}-1}+y^{n_{2}-1}\right)$

and $\lim_{y\to\infty}e^{-\rho t}\mathbb{E}\left[{\cal P}({\cal
Y}_{t}^{y})\right]=0$.

Proof of Claim 2: Since

${\cal P}(y)=\left(Dy^{n_{2}}+\Xi_{h}(y)\right){\bf
1}_{\\{y>z_{R}\\}}=\left(Dy^{n_{2}}+\Xi_{\tilde{u}_{B}}(y)-\Xi_{\tilde{u}_{A}}(y)+\dfrac{\epsilon}{r}y\right){\bf
1}_{\\{y>z_{R}\\}},$

we have

$\displaystyle|{\cal P}^{\prime}(y)|\leq$
$\displaystyle\left|n_{2}Dy^{n_{2}-1}+\Xi_{\tilde{u}_{B}}^{\prime}(y)-\Xi_{\tilde{u}_{A}}^{\prime}(y)+\dfrac{\epsilon}{r}\right|$
$\displaystyle\leq$ $\displaystyle
n_{2}|D|y^{n_{2}-1}+|\Xi_{\tilde{u}_{B}}^{\prime}(y)|+|\Xi_{\tilde{u}_{A}}^{\prime}(y)|+\dfrac{\epsilon}{r}.$

Since $u_{A},u_{B}\in\Sigma$, it follows from Lemma 3.1 and Proposition B.1
(d) that

$|{\cal P}^{\prime}(y)|\leq\zeta\left(y^{n_{1}-1}+y^{n_{2}-1}\right)$

for some constant $\zeta>0$.

Moreover,

$\displaystyle|{\cal P}(y)|=$
$\displaystyle|\left(Dy^{n_{2}}+\Xi_{h}(y)\right){\bf 1}_{\\{y>z_{R}\\}}|$
$\displaystyle\leq$ $\displaystyle|D|y^{n_{2}}{\bf
1}_{\\{y>z_{R}\\}}+|\Xi_{\tilde{u}_{A}}(y)|+|\Xi_{\tilde{u}_{B}}(y)|+\dfrac{\epsilon}{r}y.$
$\displaystyle\leq$
$\displaystyle|D|z_{R}^{n_{2}}+|\Xi_{\tilde{u}_{A}}(y)|+|\Xi_{\tilde{u}_{B}}(y)|+\dfrac{\epsilon}{r}y.$

By Proposition B.1 (e), we have

$\lim_{t\to\infty}e^{-\rho t}\mathbb{E}\left[|\Xi_{\tilde{u}_{A}}({\cal
Y}_{t}^{y})|\right]=\lim_{t\to\infty}e^{-\rho
t}\mathbb{E}\left[|\Xi_{\tilde{u}_{B}}({\cal Y}_{t}^{y})|\right]=0.$

Since $\lim_{t\to\infty}e^{-\rho t}|D|z_{R}^{n_{2}}=0$ and
$\lim_{t\to\infty}e^{-\rho t}\mathbb{E}\left[{\cal
Y}_{t}^{y}\right]=\lim_{t\to\infty}e^{-rt}y=0$, it follows that

$\lim_{t\to\infty}e^{-\rho t}\mathbb{E}\left[|{\cal P}({\cal
Y}_{t}^{y})|\right]=0.$

This completes the proof of Claim 2.

By Claim 1, Claim 2 and Theorem 3.1, we have proved the desired results. ∎

Note that

$\Xi_{h}(y)=\Xi_{\tilde{u}_{B}}(y)-\Xi_{\tilde{u}_{A}}(y)+\dfrac{\epsilon}{r}y\;\;\mbox{and}\;\;J_{A}(y)=\Xi_{\tilde{u}_{A}}(y).$

From (21), we derive the following corollary.

######  Corollary 3.1.

The dual value function $J(y)$ is given by

$\displaystyle\begin{split}J(y)=\begin{cases}Dy^{n_{2}}+\Xi_{\tilde{u}_{B}}(y)+\dfrac{\epsilon}{r}y\;\;\;&\mbox{for}\;\;y\geq
z_{R},\vspace{2mm}\\\ \Xi_{\tilde{u}_{A}}(y)\;\;\;&\mbox{for}\;\;\;0<y\leq
z_{R}.\end{cases}\end{split}$ (30)

##  4  Optimal Policies and Human Wealth

In this section we derive the optimal policies of the agent and discuss their
properties.

######  Lemma 4.1.

Let us denote ${\cal X}(y)$ and ${\Pi}(y)$ by

${\cal
X}(y)=-J^{\prime}(y)\;\;\;\mbox{and}\;\;\;\Pi(y)=\dfrac{\theta}{\sigma}yJ^{\prime\prime}(y),$

respectively.

Then, the dynamics of ${\cal X}({\cal Y}_{t}^{y})$ follows

$d{\cal X}({\cal Y}_{t}^{y})=\left(r{\cal X}({\cal
Y}_{t}^{y})+(\mu-r)\Pi({\cal Y}_{t}^{y})-\hat{c}({\cal
Y}_{t}^{y})+\epsilon{\bf 1}_{\\{t<\tau_{R}(y)\\}}\right)dt+\sigma\Pi({\cal
Y}_{t}^{y})dB_{t}$

and

${\cal X}(y)=\mathbb{E}\left[\int_{0}^{\infty}\xi_{t}(\hat{c}({\cal
Y}_{t}^{y})-\epsilon{\bf 1}_{\\{t<\tau_{R}(y)\\}})dt\right].$

###### Proof.

Since $J(y)$ is smooth in $y>z_{R}$, it follows that

$\dfrac{\theta^{2}}{2}y^{2}{\cal
X}^{\prime\prime}(y)+(\rho-r+\theta^{2})y{\cal X}^{\prime}(y)-r{\cal
X}(y)+I_{u_{B}}(y)-\epsilon=0\;\;\mbox{for}\;\;y>z_{R}.$

By Proposition B.1, the explicit-form of $J_{A}(y)$ is given by

$J_{A}(y)=\dfrac{2}{\theta^{2}(n_{1}-n_{2})}\left[y^{n_{2}}\int_{0}^{y}\nu^{-n_{2}-1}\tilde{u}_{A}(\nu)d\nu+y^{n_{1}}\int_{y}^{\infty}\nu^{-n_{1}-1}\tilde{u}_{A}(\nu)d\nu\right].$

Since $J_{A}(y)=\Xi_{\tilde{u}_{A}}(y)$ , Lemma 3.2 implies that $J_{A}(y)$ is
strictly convex in $y>0$ and

$\lim_{y\uparrow\infty}J_{A}^{\prime}(y)=0\;\;\mbox{and}\;\;\lim_{y\downarrow
0+}J_{A}^{\prime}(y)=-\infty.$

Moreover, $J_{A}(y)$ satisfies the following ordinary differential
equation(ODE):

$\dfrac{\theta^{2}}{2}y^{2}J_{A}^{\prime\prime}(y)+(\rho-r)yJ_{A}^{\prime}(y)-\rho
J_{A}+\tilde{u}_{A}(y)=0.$ (31)

For $0<y\leq z_{R}$, it follows from the ODE (31) that

$\dfrac{\theta^{2}}{2}y^{2}{\cal
X}^{\prime\prime}(y)+(\rho-r+\theta^{2})y{\cal X}^{\prime}(y)-r{\cal
X}(y)+I_{u_{A}}(y)=0\;\;\mbox{for}\;\;0<y\leq z_{R}.$

Since

$\displaystyle\begin{split}{\cal
X}(y)=\begin{cases}-Dn_{2}y^{n_{2}-1}-\Xi_{\tilde{u}_{B}}^{\prime}(y)-\dfrac{\epsilon}{r}\;\;\;&\mbox{for}\;\;y\geq
z_{R},\vspace{2mm}\\\
-\Xi_{\tilde{u}_{A}}^{\prime}(y)\;\;\;&\mbox{for}\;\;\;0<y\leq
z_{R},\end{cases}\end{split}$ (32)

it is easy to show that

${\cal X}(\cdot)\in{\cal W}_{loc}^{2,p}(0,\infty)\;\;\mbox{for any}\;\;p\geq
1,$

where $\mathcal{W}^{2,p}(0,\infty),\,p\geq 1,$ is the completion of
$C^{\infty}(0,\infty)$ under the norm

$\|\,V\,\|\,_{\mathcal{W}^{2,p}(0,\infty)}\triangleq\left[\;\int_{0}^{\infty}\,\left(\,|\,V\,|\,^{p}+|\,\partial_{y}V\,|\,^{p}+|\,\partial_{yy}V\,|\,^{p}\;\right)\,dy\,\right]^{1\over
p}.$

and $\mathcal{W}^{2,p}_{{loc}}(0,\infty),\,p\geq 1,$ is the set of all
functions whose restrictions to the domain $\mathcal{K}$ belong to
$\mathcal{W}^{2,p}(\mathcal{K})$ for any compact subset $\mathcal{K}$ of
$(0,\infty)$.

For a given $T>0$, the generalized Itô’s lemma for $W_{loc}^{2,p}$ (see
Krylov, (1980)) to ${\cal X}({\cal Y}_{t}^{y})$ yields that

$\displaystyle\begin{split}d{\cal X}({\cal Y}_{t}^{y})=&{\cal
X}^{\prime}({\cal Y}_{t}^{y})d{\cal Y}_{t}^{y}+\dfrac{1}{2}{\cal
X}^{\prime\prime}({\cal Y}_{t}^{y})(d{\cal Y}_{t}^{y})^{2}\\\
=&\left(\dfrac{\theta^{2}}{2}({\cal Y}_{t}^{y})^{2}{\cal
X}^{\prime\prime}({\cal Y}_{t}^{y})+(\rho-r){\cal Y}_{t}^{y}{\cal
X}^{\prime}({\cal Y}_{t}^{y})\right)dt-\theta{\cal Y}_{t}^{y}{\cal
X}^{\prime}({\cal Y}_{t}^{y})dB_{t}\\\ =&\left(r{\cal X}({\cal
Y}_{t}^{y})-\hat{c}({\cal Y}_{t}^{y})+\epsilon{\bf
1}_{\\{t<\tau_{R}(y)\\}}+(\mu-r)\left(-\frac{\theta}{\sigma}{\cal
X}^{\prime}({\cal
Y}_{t}^{y})\right)\right)dt+\sigma\left(-\frac{\theta}{\sigma}{\cal
Y}_{t}^{y}{\cal X}^{\prime}({\cal Y}_{t}^{y})\right)dB_{t}\\\ =&\left(r{\cal
X}({\cal Y}_{t}^{y})-\hat{c}({\cal Y}_{t}^{y})+\epsilon{\bf
1}_{\\{t<\tau_{R}(y)\\}}+(\mu-r)\left(-\frac{\theta}{\sigma}{\cal
X}^{\prime}({\cal
Y}_{t}^{y})\right)\right)dt+\sigma\left(-\frac{\theta}{\sigma}{\cal
Y}_{t}^{y}{\cal X}^{\prime}({\cal Y}_{t}^{y})\right)dB_{t}\\\ =&\left(r{\cal
X}({\cal Y}_{t}^{y})-\hat{c}({\cal Y}_{t}^{y})+\epsilon{\bf
1}_{\\{t<\tau_{R}(y)\\}}\right)dt+\sigma\left(-\frac{\theta}{\sigma}{\cal
Y}_{t}^{y}{\cal X}^{\prime}({\cal
Y}_{t}^{y})\right)dB_{t}^{\mathbb{Q}},\end{split}$ (33)

where the measure $\mathbb{Q}$ is defined in (53).

Hence, we deduce that

$\displaystyle\begin{split}d\left(e^{-rt}{\cal X}({\cal
Y}_{t}^{y})\right)=e^{-rt}\left(\epsilon{\bf
1}_{\\{t<\tau_{R}(y)\\}}-\hat{c}({\cal
Y}_{t}^{y})\right)dt+e^{-rt}\sigma\left(-\frac{\theta}{\sigma}{\cal
Y}_{t}^{y}{\cal X}^{\prime}({\cal
Y}_{t}^{y})\right)dB_{t}^{\mathbb{Q}}.\end{split}$ (34)

From (32), we deduce that for any $y>0$

$\displaystyle{\cal X}(y)\leq$ $\displaystyle
D|n_{2}|z_{R}^{n_{2}-1}+|\Gamma_{I_{u_{A}}}(y)|+|\Gamma_{I_{u_{B}}}(y)|+\dfrac{\epsilon}{r},$

and

$\displaystyle|y{\cal X}^{\prime}(y)|\leq$ $\displaystyle
D|n_{2}(n_{2}-1)|z_{R}^{n_{2}-1}+|y\Gamma_{I_{u_{A}}}^{\prime}(y)|+|y\Gamma_{I_{u_{B}}}^{\prime}(y)|.$

Thus, it follows from Lemma 3.2 (d) and (e) that

$\displaystyle\lim_{T\to\infty}e^{-rT}\mathbb{E}^{\mathbb{Q}}\left[{\cal
X}({\cal Y}_{T}^{y})\right]=\dfrac{1}{y}\lim_{T\to\infty}e^{-\rho
T}\mathbb{E}\left[{\cal Y}_{T}^{y}{\cal X}({\cal Y}_{T}^{y})\right]=0$

and the stopped process for $T$

${\cal M}_{t\wedge T}=\int_{0}^{t\wedge T}e^{-rs}\theta{\cal Y}_{s}^{y}{\cal
X}^{\prime}({\cal Y}_{s}^{y})dB^{\mathbb{Q}}_{s}$

is a martingale (see Lemma 3.4 in Knudsen et al., (1998)).

By integrating the both sides of (34) with respect to $t$, we have

$\displaystyle\begin{split}{\cal
X}(y)=\mathbb{E}\left[\int_{0}^{T}{\xi_{t}}\left(\hat{c}({\cal
Y}_{t}^{y})-\epsilon){\bf
1}_{\\{t<\tau_{R}(y)\\}}\right)dt\right]+e^{-rT}\mathbb{E}^{\mathbb{Q}}\left[{\cal
X}({\cal Y}_{T}^{y})\right].\end{split}$

Since $u_{A},u_{B}\in\Sigma$, it follows from Proposition B.1 that

$\mathbb{E}\left[\int_{0}^{\infty}\xi_{t}I_{\tilde{u}_{A}}({\cal
Y}_{t}^{y})dt\right]<\infty\;\;\mbox{and}\;\;\mathbb{E}\left[\int_{0}^{\infty}\xi_{t}I_{\tilde{u}_{B}}({\cal
Y}_{t}^{y})dt\right]<\infty.$

Thus, letting $T\to\infty$, the dominated convergence theorem implies that

$\displaystyle\begin{split}{\cal
X}(y)=\mathbb{E}\left[\int_{0}^{\infty}{\xi_{t}}\left(\hat{c}({\cal
Y}_{t}^{y})-\epsilon{\bf 1}_{\\{t<\tau_{R}(y)\\}}\right)dt\right].\end{split}$

It follows from (33) that

$d{\cal X}({\cal Y}_{t}^{y})=\left(r{\cal X}({\cal
Y}_{t}^{y})+(\mu-r)\Pi({\cal Y}_{t}^{y})-\hat{c}({\cal
Y}_{t}^{y})+\epsilon{\bf 1}_{\\{t<\tau^{*}\\}}\right)dt+\sigma\Pi({\cal
Y}_{t}^{y})dB_{t}\;\;\;\mbox{for}\;\;t\geq 0.$

∎

We derive the agent’s optimal policies by establishing the duality between the
value function and the dual value function.

######  Theorem 4.1.

Given $x>-\frac{\epsilon}{r}$.

  * (a)

$V(x)$ and $J(y)$ satisfy the duality relationship:

$V(x)=\inf_{y>0}\left(J(y)+yx\right),\quad
J(y)=\sup_{x>-\frac{\epsilon}{r}}\left(V(x)-yx\right).$ (35)

There exists a unique $y^{*}>0$ such that

$x=-J^{\prime}(y^{*}).$

  * (b)

The optimal policies $(c^{*},\pi^{*},\tau^{*})$ are

$\displaystyle\begin{split}c_{t}^{*}=\hat{c}({\cal
Y}_{t}^{y^{*}})=\begin{cases}I_{B}({\cal
Y}_{t}^{y^{*}})\;\;\;&\mbox{for}\;\;0\leq t<\tau^{*},\vspace{1mm}\\\
I_{A}({\cal
Y}_{t}^{y^{*}})\;\;\;&\mbox{for}\;\;t\geq\tau^{*},\end{cases},\;\;\;\tau^{*}=\tau_{R}(y^{*})=\inf\\{t\geq
0\;\mid{\cal Y}_{t}^{y^{*}}\leq z_{R}\\},\end{split}$

and

$\displaystyle\pi_{t}^{*}=\dfrac{\theta}{\sigma}{\cal
Y}_{t}^{*}J^{\prime\prime}({\cal Y}_{t}^{y^{*}})$

where $X_{t}=-J^{\prime}({\cal Y}_{t}^{y^{*}})$ and ${\cal
Y}_{t}^{y^{*}}=y^{*}e^{\rho t}{\xi}_{t}$.

###### Proof.

Since $z_{R}<\bar{z}$ and $h(y)<0$ for $y\in(0,\bar{z})$, we have

$D=-\dfrac{2}{\theta^{2}(n_{1}-n_{2})}\int_{0}^{z_{R}}\nu^{-n_{2}-1}{h}(\nu)d\nu>0.$

By (30), it follows from Lemma 3.2 and $D>0$ that

$J^{\prime\prime}(y)=n_{2}(n_{2}-1)Dy^{n_{2}-2}+\Xi_{\tilde{u}_{B}}^{\prime\prime}(y)>0\;\;\mbox{for}\;\;y>z_{R}.$

Since $J(y)=J_{A}(y)=\Xi_{\tilde{u}_{A}}(y)$ for $0<y\leq z_{R}$, it is clear
that

$J^{\prime\prime}(y)=\Xi_{\tilde{u}_{A}}^{\prime\prime}(y)>0\;\;\mbox{for}\;\;0<y\leq
z_{R}.$

Hence, $J(y)$ is strictly convex in $y>0$.

Note that

$\lim_{y\uparrow\infty}{\cal
P}^{\prime}(y)=\dfrac{\epsilon}{r},\;\;\lim_{y\downarrow 0}{\cal
P}^{\prime}(y)=0$

and

$\lim_{y\uparrow\infty}J_{A}^{\prime}(y)=0,\;\;\lim_{y\downarrow
0}J_{A}^{\prime}(y)=-\infty.$

Thus, it follows that

$\displaystyle\lim_{y\uparrow\infty}{J}^{\prime}(y)=\dfrac{\epsilon}{r},\vspace{2mm}$
$\displaystyle\lim_{y\downarrow 0}{J}^{\prime}(y)=-\infty.$

Therefore, there exists a unique $y^{*}>0$ such that for given
$x>-\frac{\epsilon}{r}$

$x=-J^{\prime}(y^{*}).$

It follows from Lemma 4.1 that

$x=-J^{\prime}(y^{*})={\cal
X}(y^{*})=\mathbb{E}\left[\int_{0}^{\infty}{\xi_{t}}\left(\hat{c}({\cal
Y}_{t}^{y^{*}})-\epsilon{\bf 1}_{\\{t<\tau^{*}\\}}\right)dt\right].$

Hence, we have

$\displaystyle\begin{split}y^{*}x=&y^{*}\mathbb{E}\left[\int_{0}^{\infty}{\xi_{t}}\left(\hat{c}({\cal
Y}_{t}^{y^{*}})-\epsilon{\bf 1}_{\\{t<\tau^{*}\\}}\right)dt\right]\\\
=&\mathbb{E}\left[\int_{0}^{\infty}{\cal Y}_{t}^{y^{*}}\left(\hat{c}({\cal
Y}_{t}^{y^{*}})-\epsilon{\bf 1}_{\\{t<\tau^{*}\\}}\right)dt\right]\\\
=&\mathbb{E}\left[\int_{0}^{\tau^{*}}e^{-\rho t}u_{A}(\hat{c}({\cal
Y}_{t}^{y^{*}}))dt+\int_{\tau^{*}}^{\infty}e^{-\rho t}u_{B}(\hat{c}({\cal
Y}_{t}^{y^{*}}))dt\right]\\\ -&\mathbb{E}\left[\int_{0}^{\tau^{*}}e^{-\rho
t}\left(\tilde{u}_{B}({\cal Y}_{t}^{y^{*}})+{\cal
Y}_{t}^{y^{*}}\epsilon\right)dt+\int_{0}^{\tau^{*}}e^{-\rho
t}\tilde{u}_{A}({\cal Y}_{t}^{y^{*}})dt\right]\\\
=&\mathbb{E}\left[\int_{0}^{\tau^{*}}e^{-\rho t}u_{A}(\hat{c}({\cal
Y}_{t}^{y^{*}}))dt+\int_{\tau^{*}}^{\infty}e^{-\rho t}u_{B}(\hat{c}({\cal
Y}_{t}^{y^{*}}))dt\right]-J(y^{*}).\end{split}$

Hence,

$\displaystyle J(y^{*})+y^{*}x=$
$\displaystyle\mathbb{E}\left[\int_{0}^{\tau^{*}}e^{-\rho
t}u_{A}(\hat{c}({\cal Y}_{t}^{y^{*}}))dt+\int_{\tau^{*}}^{\infty}e^{-\rho
t}u_{B}(\hat{c}({\cal Y}_{t}^{y^{*}}))dt\right]$ $\displaystyle\leq$
$\displaystyle\sup_{(c,\pi,\tau\in{\cal
A}(x)}\mathbb{E}\left[\int_{0}^{\tau}e^{-\rho
t}u_{A}(c_{t})dt+\int_{\tau}^{\infty}e^{-\rho t}u_{B}(c_{t})dt\right].$

For any $y>0$ and $(c,\pi,\tau)\in{\cal A}(x)$,

$\displaystyle\mathbb{E}\left[\int_{0}^{\tau}e^{-\rho
t}u_{A}(c_{t})dt+\int_{\tau}^{\infty}e^{-\rho t}u_{B}(c_{t})dt\right]$
$\displaystyle\leq$ $\displaystyle\mathbb{E}\left[\int_{0}^{\tau}e^{-\rho
t}u_{A}(c_{t})dt+e^{-\rho\tau}V_{A}(X_{\tau})\right]+y\left(x-\mathbb{E}\left[\int_{0}^{\tau}{\xi_{t}}(c_{t}-\epsilon)dt+\xi_{\tau}X_{\tau}\right]\right)$
$\displaystyle\leq$ $\displaystyle\mathbb{E}\left[\int_{0}^{\tau}e^{-\rho
t}\left(\tilde{u}_{A}({\cal Y}_{t}^{y})+{\cal
Y}_{t}^{y}\epsilon\right)dt+e^{-\rho\tau}J_{A}({\cal
Y}_{\tau}^{y})\right]+yx.$

This implies that

$\displaystyle\sup_{(c,\pi,\tau)\in{\cal
A}(x)}\mathbb{E}\left[\int_{0}^{\tau}e^{-\rho
t}u_{A}(c_{t})dt+\int_{\tau}^{\infty}e^{-\rho t}u_{B}(c_{t})dt\right]$
$\displaystyle\leq$ $\displaystyle\sup_{\tau\in{\cal
S}}\inf_{y>0}\left(\mathbb{E}\left[\int_{0}^{\tau}e^{-\rho
t}\left(\tilde{u}_{A}({\cal Y}_{t}^{y})+{\cal
Y}_{t}^{y}\epsilon\right)dt+e^{-\rho\tau}J_{A}({\cal
Y}_{\tau}^{y})\right]+yx\right)$ $\displaystyle\leq$
$\displaystyle\inf_{y>0}\sup_{\tau\in{\cal
S}}\left(\mathbb{E}\left[\int_{0}^{\tau}e^{-\rho t}\left(\tilde{u}_{A}({\cal
Y}_{t}^{y})+{\cal Y}_{t}^{y}\epsilon\right)dt+e^{-\rho\tau}J_{A}({\cal
Y}_{\tau}^{y})\right]+yx\right)=\inf_{y>0}\left(J(y)+yx\right).$

Overall, we have

$\displaystyle J(y^{*})+y^{*}x\leq\inf_{y>0}\left(J(y)+yx\right)\leq
J(y^{*})+y^{*}x$

and

$\displaystyle\mathbb{E}\left[\int_{0}^{\tau^{*}}e^{-\rho
t}u_{A}(\hat{c}({\cal Y}_{t}^{y^{*}}))dt+\int_{\tau^{*}}^{\infty}e^{-\rho
t}u_{B}(\hat{c}({\cal Y}_{t}^{y^{*}}))dt\right]=\sup_{(c,\pi,\tau)\in{\cal
A}(x)}\mathbb{E}\left[\int_{0}^{\tau}e^{-\rho
t}u_{A}(c_{t})dt+\int_{\tau}^{\infty}e^{-\rho t}u_{B}(c_{t})dt\right].$

That is, $c_{t}^{*}=\hat{c}({\cal Y}_{t}^{y^{*}})$ and
$\tau^{*}=\tau^{*}(y^{*})$ are optimal.

By Proposition 3.1, there exist a portfolio $\pi_{t}^{*}$ such that
$(c^{*},\pi^{*},\tau^{*})\in{\cal A}(x)$ and the corresponding wealth process
$X^{x,c^{*},\pi^{*}}$ is

$dX_{t}^{x,c^{*},\pi^{*}}=[rX_{t}^{x,c^{*},\pi^{*}}+(\mu-r)\pi_{t}^{*}-c_{t}^{*}+\epsilon{\bf
1}_{\\{t<\tau^{*}\\}}]dt+\sigma\pi_{t}^{*}dB_{t}\;\;\;t\geq 0$ (36)

and

$X_{t}^{x,c^{*},\pi^{*}}=\mathbb{E}_{t}\left[\int_{t}^{\infty}\dfrac{\xi_{s}}{\xi_{t}}(c_{s}^{*}-\epsilon
1_{\\{s<\tau^{*}\\}})ds\right].$

The strong Markov property implies that

$\xi_{t}{\cal X}({\cal
Y}_{t}^{y^{*}})=\mathbb{E}\left[\int_{t}^{\infty}\xi_{s}\left(\hat{c}({\cal
Y}_{s}^{y^{*}})-\epsilon{\bf 1}_{\\{s<\tau_{R}({\cal
Y}_{t}^{y^{*}})\\}}\right)dt\right]=\xi_{t}X_{t}^{x,c^{*},\pi^{*}}.$

That is,

${\cal X}({\cal
Y}_{t}^{y^{*}})=X_{t}^{x,c^{*},\pi^{*}}\;\;\;\mbox{for}\;\;t\geq 0.$ (37)

By Proposition 3.1,

$d{\cal X}({\cal Y}_{t}^{y^{*}})=\left(r{\cal X}({\cal
Y}_{t}^{y^{*}})+(\mu-r)\Pi({\cal Y}_{t}^{y^{*}})-\hat{c}({\cal
Y}_{t}^{y^{*}})+\epsilon{\bf
1}_{\\{t<\tau_{R}(y^{*})\\}}\right)dt+\sigma\Pi({\cal Y}_{t}^{y^{*}})dB_{t}$
(38)

It follows from (36), (37), and (38) that

$\pi_{t}^{*}=\Pi({\cal Y}_{t}^{y^{*}})=\frac{\theta}{\sigma}{\cal
Y}_{t}^{y^{*}}J^{\prime\prime}({\cal Y}_{t}^{y^{*}}),$

∎

Lemma 4.1 and Theorem 4.1 imply

$X_{t}=-J^{\prime}({\cal
Y}_{t}^{y^{*}})=\mathbb{E}\left[\int_{t}^{\infty}e^{-\rho(s-t)}\frac{\xi_{s}}{\xi_{t}}\left(\hat{c}({\cal
Y}_{s}^{y^{*}})-\epsilon{\bf 1}_{\\{s<\tau_{R}(y^{*})\\}}\right)ds\right].$

Thus, we establish the following relationship:

$\displaystyle\mathbb{E}\left[\int_{t}^{\infty}e^{-\rho(s-t)}\frac{\xi_{s}}{\xi_{t}}c_{s}^{*}ds\right]=X_{t}+\mathbb{E}\left[\int_{t}^{\tau^{*}}e^{-\rho(s-t)}\frac{\xi_{s}}{\xi_{t}}\epsilon
ds\right].$ (39)

The left-hand side is the present value of lifetime consumption. The right-
hand side consists of two components, financial wealth $X_{t}$ and the present
value of labor income, which can be regarded as human wealth. Thus, the right-
hand side is equal to the agent’s total wealth, comprised of financial wealth
and human wealth. Equation (39) says that life time consumption is financed by
financial wealth and human wealth.

Theorem 4.1 also implies that the agent retires when the marginal utility of
wealth reaches threshold level $z_{R}$, or equivalently, the agent’s wealth
level reaches threshold $x_{R}$, satisfying

$x_{R}=-J^{\prime}(z_{R})=-J_{A}^{\prime}(z_{R}),$ (40)

where we have used ${\cal P}^{\prime}(z_{R})=0$, i.e., human wealth is equal
to 0 at retirement.

######  Proposition 4.1.

If $\epsilon$ increases, then the optimal retirement threshold $x_{R}$
increases.

###### Proof.

Let $\Psi_{i}(y)$ be the current normalized marginal benefit of work for given
$\epsilon_{i}>0$ $(i=1,2)$.

Suppose that $\epsilon_{1}>\epsilon_{2}$. Then,

$\Psi_{1}(y)>\Psi_{2}(y)\;\;\;\mbox{for any}\;\;y>0.$

Note that there exists a unique $\bar{z}_{i}>0$ such that
$\Psi_{i}(\bar{z}_{i})=0$ for $i=1,2$.

Since $0=\Psi_{1}(\bar{z}_{1})=\Psi_{2}(\bar{z}_{2})>\Psi_{2}(\bar{z}_{1})$,
it follows from Lemma 3.3 that

$\bar{z}_{2}>\bar{z}_{1}.$

For $i=1,2,$ there exists a unique $z_{R,i}\in(0,\bar{z}_{i})$ such that

${\cal G}_{i}(z_{R,i})=0,$

where

${\cal G}_{i}(y)\equiv\int_{y}^{\infty}\nu^{-n_{1}}\Psi_{i}(\nu)d\nu.$

It follows from $\Psi_{1}(y)>\Psi_{2}(y)$ for all $y>0$ that

$0={\cal G}_{1}(z_{R,1})={\cal G}_{2}(z_{R,2})>{\cal G}_{2}(z_{R,1}).$

Since ${\cal G}_{2}(y)$ is strictly increasing in $y\in(0,\bar{z}_{2})$, it
follows from $\bar{z}_{2}>\bar{z}_{1}>z_{R,1}$ that

$z_{R,2}>z_{R,1}.$

That is, the free boundary $z_{R}$ decreases as $\epsilon$ increases.

Let $x_{R,i}$ be the optimal retirement threshold corresponding to $z_{R,i}$.

Since $J_{A}(y)$ is strictly convex in $y>0$, it follows from
$x_{R,i}=-J_{A}^{\prime}(z_{R,i})$ that

$x_{R,1}>x_{R,2}.$

Therefore, if $\epsilon>0$ increases, the optimal retirement threshold $x_{R}$
increases.

∎

######  Proposition 4.2.

If $\mu-r>0\;(\mu-r<0)$, the optimal portfolio $\pi^{*}$ jumps downward
(upward) just after retirement. In other words,

$\pi_{\tau^{*}-}-\pi_{\tau^{*}+}=\lim_{y\to z_{R}+}\Pi(y)-\lim_{y\to
z_{R}-}\Pi(y)=-\dfrac{2}{\mu-r}\Psi(z_{R}).$

###### Proof.

Since $J(y)=J_{A}(y)+{\cal P}(y)$, it is easy to obtain that

$\displaystyle\lim_{y\to z_{R}+}\Pi(y)-\lim_{y\to z_{R}-}\Pi(y)$
$\displaystyle=\dfrac{\theta}{\sigma}z_{R}{\cal P}^{\prime\prime}(z_{R}).$
(41)

Note that for $y>z_{R}$, ${\cal P}(y)$ satisfies

$\dfrac{\theta^{2}}{2}y^{2}{\cal P}^{\prime\prime}(y)+(\rho-r)y{\cal
P}(y)-\rho{\cal P}+{h}(y)=0.$

Since ${\cal P}(z_{R})={\cal P}^{\prime}(z_{R})=0$, we have

$\displaystyle\lim_{y\to z_{R}+}{\cal P}^{\prime\prime}(y)=$
$\displaystyle-\lim_{y\to
z_{R}+}\dfrac{2}{\theta^{2}}\dfrac{1}{y^{2}}\left((\rho-r)y{\cal
P}(y)-\rho{\cal
P}+{h}(y)\right)=-\dfrac{2}{\theta^{2}}\dfrac{1}{z_{R}}h(z_{R})=-\dfrac{2}{\theta^{2}}\dfrac{1}{z_{R}}\Psi(z_{R}).$
(42)

By (41) and (42), we deduce that

$\lim_{y\to z_{R}+}\Pi(y)-\lim_{y\to
z_{R}-}\Pi(y)=-\dfrac{2}{\mu-r}\Psi(z_{R}).$

∎

##  5 Example

In this section we provide an example.

For $u\in\Sigma$, let us consider the following utility functions:

$\displaystyle u_{B}(c)=u(c)-l\;\;\;\mbox{and}\;\;u_{A}(c)=u(kc+b),\;\;l\geq
0,k\geq 1,b\geq 0,\;\;\mbox{with}\;\;{(k-1)^{2}+l^{2}\neq 0}.$ (43)

The case $l>0,k=1,b=0$ is the model of Choi and Shim, (2006), the case
$l=0,k>1,b=0$ is that of Fahri and Panageas, (2007) and Dybvig and Liu,
(2010), and the case $l>0,b>0$ is that of Bae et al., (2020). Thus, the
example encompasses major models in the literature777Fahri and Panageas,
(2007), Dybvig and Liu, (2010) and Bae et al., (2020) only consider the
constant relative risk aversion(CRRA) felicity function..

######  Proposition 5.1.

$u_{A}$ and $u_{B}$ in (43) satisfy Assumptions 1-2.

###### Proof.

We have

$\displaystyle\tilde{u}_{B}(y)$ $\displaystyle=\sup_{c\geq
0}\left(u(c)-l-yc\right)=u(I_{u}(y))-yI_{u}(y)-l=\tilde{u}(y)-l,$
$\displaystyle\tilde{u}_{A}(y)=\sup_{c\geq 0}\left(u(kc+b)-yc\right)=$
$\displaystyle\left(u(I_{u}(\dfrac{y}{k}))-\dfrac{y}{k}\left(I_{u}(\dfrac{y}{k})-b\right)\right){\bf
1}_{\\{y/k\leq u^{\prime}(b)\\}}+u(b){\bf 1}_{\\{y/k>u^{\prime}(b)\\}}$
$\displaystyle=$
$\displaystyle\left(\tilde{u}(\dfrac{y}{k})+b\dfrac{y}{k}\right){\bf
1}_{\\{y/k\leq u^{\prime}(b)\\}}+u(b){\bf 1}_{\\{y/k>u^{\prime}(b)\\}}$

and

$\displaystyle
I_{u_{B}}(y)=I_{u}(y)\;\;\;\mbox{and}\;\;\;I_{u_{A}}(y)=\dfrac{1}{k}\left(I_{u}(\dfrac{y}{k})-b\right){\bf
1}_{\\{y/k\leq u^{\prime}(b)\\}}.$ (44)

Since $u\in\Sigma$, it is easy to check that

$u_{A},\;u_{B}\in\Sigma.$ (45)

Note that

$\displaystyle u_{A}(I_{A}(y))-u_{B}(I_{A}(y))=$
$\displaystyle\;u\left(I_{u}(\dfrac{y}{k}){\bf 1}_{\\{y/k\leq
u^{\prime}(b)\\}}+b{\bf
1}_{\\{y/k>u^{\prime}(b)\\}}\right)-\left(u(I_{u}(y))-l\right)$
$\displaystyle\geq$ $\displaystyle\;u(I_{u}(\dfrac{y}{k}))-u(I_{u}(y))+l$ (46)
$\displaystyle>$ $\displaystyle\;0,$

where we have used the fact $(k-1)^{2}+l^{2}\neq 0$ in the second inequality.
Since $\tilde{u}_{A}(y)=\sup_{c\geq 0}\left(u(kc+b)-yc\right)\geq\sup_{c\geq
0}\left(u(kc)-yc\right)=\tilde{u}(\dfrac{y}{k})$, we deduce that

$\displaystyle\Psi(y)=$
$\displaystyle\dfrac{1}{y}\left(\tilde{u}_{B}(y)-\tilde{u}_{A}(y)\right)+\epsilon$
(47) $\displaystyle\leq$
$\displaystyle\dfrac{1}{y}\left(\tilde{u}(y)-l-\tilde{u}(\frac{y}{k})\right)+\epsilon.$

If $k=1$ and $l>0$, then

$\lim_{y\to 0+}\Psi(y)\leq\lim_{y\to
0+}\dfrac{1}{y}\left(\tilde{u}(y)-l-\tilde{u}(y)\right)+\epsilon=-\infty.$
(48)

If $k\neq 1$, then it follows from the mean-value theorem that for any $y>0$
there exists $y_{\delta}\in(y,y/k)$ such that

$\Psi(y)\leq\frac{1}{y}(\tilde{u}(y)-l-\tilde{u}(\frac{y}{k}))+\epsilon=-\left(1-\frac{1}{k}\right){I}_{u}(y_{\delta})-\dfrac{l}{y}+\epsilon.$

Since $\lim_{y\to 0+}I_{u}(y)=+\infty$, we deduce that

$\lim_{y\to 0+}\Psi(y)\leq\lim_{y\to
0+}\left[\frac{1}{y}(\tilde{u}(y)-l-\tilde{u}(\frac{y}{k}))+\epsilon\right]=\lim_{y\to
0+}\left[-\left(1-\frac{1}{k}\right){I}_{u}(y_{\delta})-\dfrac{l}{y}+\epsilon\right]=-\infty.$
(49)

From (45), (5), (48), and (49), we conclude that $u_{A}$ and $u_{B}$ in (43)
satisfy Assumptions 1-2. ∎

By Proposition 5.1, all results in Sections 3-4 can be applied when $u_{A}$,
$u_{B}$ are given by (43).

Recall that $z_{R}\in(0,\bar{z})$ is a unique solution satisfying

$\displaystyle{\cal G}(y)=0,$

where

${\cal
G}(y)=\int_{y}^{\infty}\nu^{-n_{1}-1}{h}(\nu)d\nu=\int_{y}^{\infty}\nu^{-n_{1}}{\Psi}(\nu)d\nu.$

Moreover, ${\cal G}(y)$ is strictly increasing and decreasing in
$y\in(0,\bar{z})$ and $y\in(\bar{z},\infty)$, respectively. Since $\lim_{y\to
0+}{\cal G}(y)=-\infty$ and ${\cal G}(y)>0\;\;\mbox{for}\;\;y\geq\bar{z}$, we
can easily derive the following lemma:

######  Lemma 5.1.

$z_{R}<ku^{\prime}(b)$ if and only if ${\cal G}(ku^{\prime}(b))>0$.

###  5.1 CRRA felicity function

We assume that the agent has constant relative risk aversion(CRRA), i.e.,

$u(c)=\dfrac{c^{1-\gamma}}{1-\gamma},\;\;\;\gamma\neq 1,\;\gamma>0,$ (50)

where $\gamma$ denotes the coefficient of relative risk aversion.

Then,

$u_{B}(c)=\dfrac{c^{1-\gamma}}{1-\gamma}-l\;\;\;\mbox{and}\;\;u_{A}(c)=\dfrac{(kc+b)^{1-\gamma}}{1-\gamma},$

where $l\geq 0,k\geq 1,b\geq 0,\;\;\mbox{with}\;\;{(k-1)^{2}+l^{2}\neq 0}$.

Note that, under the CRRA utility (50), the integrabiity condition (3) is
equivalent to the following assumption:

######  Assumption 3.

The Merton constant $M$ defined by

$M\equiv
r+\dfrac{\rho-r}{\gamma}+\dfrac{\gamma-1}{\gamma^{2}}\dfrac{\theta^{2}}{2}>0$
(51)

is positive.

We can easily obtain that

$\displaystyle I_{u_{B}}(y)$
$\displaystyle=y^{-\frac{1}{\gamma}},\;\;\;I_{u_{A}}(y)=\dfrac{1}{k}\left(\left(\dfrac{y}{k}\right)^{-\frac{1}{\gamma}}-b\right){\bf
1}_{\\{y\leq kb^{-\gamma}\\}},\vspace{2mm}$ $\displaystyle\tilde{u}_{B}(y)$
$\displaystyle=\dfrac{\gamma}{1-\gamma}y^{-\frac{1-\gamma}{\gamma}}-l,\;\;\tilde{u}_{A}(y)=\left(\dfrac{\gamma}{1-\gamma}\left(\dfrac{y}{k}\right)^{-\frac{1-\gamma}{\gamma}}+b\dfrac{y}{k}\right){\bf
1}_{\\{y\leq kb^{-\gamma}\\}}+\dfrac{b^{1-\gamma}}{1-\gamma}{\bf
1}_{\\{y>kb^{-\gamma}\\}}.$

By Corollary 3.1, the dual value function $J(y)$ is given by

$\displaystyle
J(y)=\begin{cases}Dy^{n_{2}}+\Xi_{\tilde{u}_{B}}(y)+\dfrac{\epsilon}{r}y\;\;\;&\mbox{for}\;\;y\geq
z_{R},\vspace{2mm}\\\ \Xi_{\tilde{u}_{A}}(y)\;\;\;&\mbox{for}\;\;\;0<y\leq
z_{R}.\end{cases}$

Since

$\displaystyle\Xi_{\tilde{u}_{B}}(y)$
$\displaystyle=\dfrac{2}{\theta^{2}(n_{1}-n_{2})}\left[y^{n_{2}}\int_{0}^{y}\nu^{-n_{2}-1}\tilde{u}_{B}(\nu)d\nu+y^{n_{1}}\int_{y}^{\infty}\nu^{-n_{1}-1}\tilde{u}_{B}(\nu)d\nu\right],$
$\displaystyle\Xi_{\tilde{u}_{A}}(y)$
$\displaystyle=\dfrac{2}{\theta^{2}(n_{1}-n_{2})}\left[y^{n_{2}}\int_{0}^{y}\nu^{-n_{2}-1}\tilde{u}_{A}(\nu)d\nu+y^{n_{1}}\int_{y}^{\infty}\nu^{-n_{1}-1}\tilde{u}_{A}(\nu)d\nu\right],$

we have

$\displaystyle\Xi_{\tilde{u}_{B}}(y)=\dfrac{2}{\theta^{2}(n_{1}-n_{2})}\left[y^{n_{2}}\int_{0}^{y}\nu^{-n_{2}-1}\tilde{u}_{B}(\nu)d\nu+y^{n_{1}}\int_{y}^{\infty}\nu^{-n_{1}-1}\tilde{u}_{B}(\nu)d\nu\right]=\dfrac{1}{M}\dfrac{\gamma}{1-\gamma}y^{-\frac{1-\gamma}{\gamma}}-\dfrac{l}{\rho},$

and

$\displaystyle\begin{split}\Xi_{\tilde{u}_{A}}(y)=\begin{cases}\phi_{1}(y)\;\;&\mbox{for}\;\;\;0<y\leq
kb^{-\gamma},\vspace{1mm}\\\
\phi_{2}(y)\;\;&\mbox{for}\;\;y>kb^{-\gamma},\end{cases}\end{split}$ (52)

where $\phi_{1}(y)$ and $\phi_{2}(y)$ are given by

$\displaystyle\phi_{1}(y)=$
$\displaystyle\left(\dfrac{\frac{1}{M}\left(\frac{\gamma
n_{2}}{1-\gamma}+1\right)+\frac{n_{2}-1}{r}-\frac{n_{2}}{\rho(1-\gamma)}}{n_{1}-n_{2}}\right)b^{1-\gamma+\gamma
n_{1}}\left(\dfrac{y}{k}\right)^{n_{1}}+\dfrac{1}{K}\dfrac{\gamma}{1-\gamma}(\dfrac{y}{k})^{-\frac{1-\gamma}{\gamma}}+\dfrac{b}{r}\dfrac{y}{k}$

and

$\displaystyle\phi_{2}(y)=$
$\displaystyle\left(\dfrac{\frac{1}{M}\left(\frac{\gamma
n_{1}}{1-\gamma}+1\right)+\frac{n_{1}-1}{r}-\frac{n_{1}}{\rho(1-\gamma)}}{n_{1}-n_{2}}\right)b^{1-\gamma+\gamma
n_{2}}\left(\dfrac{y}{k}\right)^{n_{2}}+\dfrac{b^{1-\gamma}}{\rho(1-\gamma)},$

respectively.

Since the current normalized marginal benefit of work $\Psi(y)$ is given by

$\Psi(y)=\dfrac{1}{y}\left[\dfrac{\gamma}{1-\gamma}y^{-\frac{1-\gamma}{\gamma}}-l-\left(\dfrac{\gamma}{1-\gamma}\left(\dfrac{y}{k}\right)^{-\frac{1-\gamma}{\gamma}}+b\dfrac{y}{k}\right){\bf
1}_{\\{y\leq kb^{-\gamma}\\}}-\dfrac{b^{1-\gamma}}{1-\gamma}{\bf
1}_{\\{y>kb^{-\gamma}\\}}\right]+\epsilon,$

we have

$\displaystyle{\cal G}(kb^{-\gamma})=$
$\displaystyle\int_{kb^{-\gamma}}^{\infty}\nu^{-n_{1}}\Psi(\nu)d\nu$
$\displaystyle=$
$\displaystyle\int_{kb^{-\gamma}}^{\infty}\nu^{-n_{1}-1}\left(\dfrac{\gamma}{1-\gamma}\nu^{-\frac{1-\gamma}{\gamma}}-l-\dfrac{b^{1-\gamma}}{1-\gamma}+\epsilon\nu\right)d\nu$
$\displaystyle=$
$\displaystyle\left[\dfrac{\gamma}{1-\gamma}\dfrac{1}{1-n_{1}-\frac{1}{\gamma}}\nu^{1-n_{1}-\frac{1}{\gamma}}+l\dfrac{\nu^{-n_{1}}}{n_{1}}+\dfrac{b^{1-\gamma}}{1-\gamma}\dfrac{\nu^{-n_{1}}}{n_{1}}-\epsilon\dfrac{\nu^{1-n_{1}}}{n_{1}-1}\right]_{\nu=kb^{-\gamma}}^{\infty}$
$\displaystyle=$
$\displaystyle-\left[\dfrac{\gamma}{1-\gamma}\dfrac{1}{1-n_{1}-\frac{1}{\gamma}}(kb^{-\gamma})^{1-n_{1}-\frac{1}{\gamma}}+l\dfrac{(kb^{-\gamma})^{-n_{1}}}{n_{1}}+\dfrac{b^{1-\gamma}}{1-\gamma}\dfrac{(kb^{-\gamma})^{-n_{1}}}{n_{1}}-\epsilon\dfrac{(kb^{-\gamma})^{1-n_{1}}}{n_{1}-1}\right].$

Lemma 5.1 implies that

${\cal G}(kb^{-\gamma})>0$ (or equivalently, $z_{R}<kb^{-\gamma}$)

In this case, the coefficient $D$ is given by

$\displaystyle D=$
$\displaystyle-\dfrac{2}{\theta^{2}(n_{1}-n_{2})}\int_{0}^{z_{R}}\nu^{-n_{2}}\Psi(\nu)d\nu$
$\displaystyle=$
$\displaystyle-\dfrac{2}{\theta^{2}(n_{1}-n_{2})}\int_{0}^{z_{R}}\nu^{-n_{2}-1}\left[\dfrac{\gamma}{1-\gamma}\nu^{-\frac{1-\gamma}{\gamma}}-l-\left(\dfrac{\gamma}{1-\gamma}\left(\dfrac{\nu}{k}\right)^{-\frac{1-\gamma}{\gamma}}+b\dfrac{\nu}{k}\right)+\epsilon\nu\right]d\nu$
$\displaystyle=$
$\displaystyle-\dfrac{2}{\theta^{2}(n_{1}-n_{2})}\left[\dfrac{\gamma}{1-\gamma}\dfrac{1}{1-n_{2}-\frac{1}{\gamma}}\left(1-k^{\frac{1-\gamma}{\gamma}}\right)(z_{R})^{1-n_{2}-\frac{1}{\gamma}}+l\dfrac{(z_{R})^{-n_{2}}}{n_{2}}-\dfrac{(z_{R})^{1-n_{2}}}{n_{2}-1}\left(\epsilon-\dfrac{b}{k}\right)\right]$

and $z_{R}$ is a unique solution of the following algebraic equation:

$\displaystyle 0={\cal G}(z_{R})=$
$\displaystyle\int_{z_{R}}^{\infty}\nu^{-n_{1}}\Psi(\nu)d\nu$ $\displaystyle=$
$\displaystyle\int_{z_{R}}^{kb^{-\gamma}}\nu^{-n_{1}-1}\left[\dfrac{\gamma}{1-\gamma}\nu^{-\frac{1-\gamma}{\gamma}}-l-\left(\dfrac{\gamma}{1-\gamma}\left(\dfrac{\nu}{k}\right)^{-\frac{1-\gamma}{\gamma}}+b\dfrac{\nu}{k}\right)+\epsilon\nu\right]d\nu$
$\displaystyle+$
$\displaystyle\int_{kb^{-\gamma}}^{\infty}\nu^{-n_{1}-1}\left[\dfrac{\gamma}{1-\gamma}\nu^{-\frac{1-\gamma}{\gamma}}-l-\dfrac{b^{1-\gamma}}{1-\gamma}+\epsilon\nu\right]d\nu$
$\displaystyle=$
$\displaystyle\left[\dfrac{\gamma}{1-\gamma}\dfrac{1}{1-n_{1}-\frac{1}{\gamma}}\left(1-k^{\frac{1-\gamma}{\gamma}}\right)(kb^{-\gamma})^{1-n_{1}-\frac{1}{\gamma}}+l\dfrac{(kb^{-\gamma})^{-n_{1}}}{n_{1}}-\dfrac{(kb^{-\gamma})^{1-n_{1}}}{n_{1}-1}\left(\epsilon-\dfrac{b}{k}\right)\right]$
$\displaystyle-$
$\displaystyle\left[\dfrac{\gamma}{1-\gamma}\dfrac{1}{1-n_{1}-\frac{1}{\gamma}}\left(1-k^{\frac{1-\gamma}{\gamma}}\right)(z_{R})^{1-n_{1}-\frac{1}{\gamma}}+l\dfrac{(z_{R})^{-n_{1}}}{n_{1}}-\dfrac{(z_{R})^{1-n_{1}}}{n_{1}-1}\left(\epsilon-\dfrac{b}{k}\right)\right]$
$\displaystyle-$
$\displaystyle\left[\dfrac{\gamma}{1-\gamma}\dfrac{1}{1-n_{1}-\frac{1}{\gamma}}(kb^{-\gamma})^{1-n_{1}-\frac{1}{\gamma}}+l\dfrac{(kb^{-\gamma})^{-n_{1}}}{n_{1}}+\dfrac{b^{1-\gamma}}{1-\gamma}\dfrac{(kb^{-\gamma})^{-n_{1}}}{n_{1}}-\epsilon\dfrac{(kb^{-\gamma})^{1-n_{1}}}{n_{1}-1}\right].$

${\cal G}(kb^{-\gamma})\leq 0$ (or equivalently, $z_{R}\geq kb^{-\gamma}$)

In this case, the coefficient $D$ is given by

$\displaystyle D=$
$\displaystyle-\dfrac{2}{\theta^{2}(n_{1}-n_{2})}\int_{0}^{z_{R}}\nu^{-n_{2}}\Psi(\nu)d\nu$
$\displaystyle=$
$\displaystyle-\dfrac{2}{\theta^{2}(n_{1}-n_{2})}\int_{0}^{kb^{-\gamma}}\nu^{-n_{2}-1}\left[\dfrac{\gamma}{1-\gamma}\nu^{-\frac{1-\gamma}{\gamma}}-l-\left(\dfrac{\gamma}{1-\gamma}\left(\dfrac{\nu}{k}\right)^{-\frac{1-\gamma}{\gamma}}+b\dfrac{\nu}{k}\right)+\epsilon\nu\right]d\nu$
$\displaystyle-\dfrac{2}{\theta^{2}(n_{1}-n_{2})}\int_{kb^{-\gamma}}^{z_{R}}\nu^{-n_{2}-1}\left[\dfrac{\gamma}{1-\gamma}\nu^{-\frac{1-\gamma}{\gamma}}-l-\dfrac{b^{1-\gamma}}{1-\gamma}+\epsilon\nu\right]d\nu$
$\displaystyle=$
$\displaystyle-\dfrac{2}{\theta^{2}(n_{1}-n_{2})}\left[\dfrac{\gamma}{1-\gamma}\dfrac{1}{1-n_{2}-\frac{1}{\gamma}}\left(1-k^{\frac{1-\gamma}{\gamma}}\right)(kb^{-\gamma})^{1-n_{2}-\frac{1}{\gamma}}+l\dfrac{(kb^{-\gamma})^{-n_{2}}}{n_{2}}-\dfrac{(kb^{-\gamma})^{1-n_{2}}}{n_{2}-1}\left(\epsilon-\dfrac{b}{k}\right)\right]$
$\displaystyle-\dfrac{2}{\theta^{2}(n_{1}-n_{2})}\left[\dfrac{\gamma}{1-\gamma}\dfrac{1}{1-n_{2}-\frac{1}{\gamma}}(z_{R})^{1-n_{2}-\frac{1}{\gamma}}+l\dfrac{(z_{R})^{-n_{2}}}{n_{2}}+\dfrac{b^{1-\gamma}}{1-\gamma}\dfrac{(z_{R})^{-n_{2}}}{n_{2}}-\epsilon\dfrac{(z_{R})^{1-n_{2}}}{n_{2}-1}\right]$
$\displaystyle+-\dfrac{2}{\theta^{2}(n_{1}-n_{2})}\left[\dfrac{\gamma}{1-\gamma}\dfrac{1}{1-n_{2}-\frac{1}{\gamma}}(kb^{-\gamma})^{1-n_{2}-\frac{1}{\gamma}}+l\dfrac{(kb^{-\gamma})^{-n_{2}}}{n_{2}}+\dfrac{b^{1-\gamma}}{1-\gamma}\dfrac{(kb^{-\gamma})^{-n_{2}}}{n_{2}}-\epsilon\dfrac{(kb^{-\gamma})^{1-n_{2}}}{n_{2}-1}\right]$

and $z_{R}$ is a unique solution of the following algebraic equation:

$\displaystyle 0={\cal G}(z_{R})=$
$\displaystyle\int_{z_{R}}^{\infty}\nu^{-n_{1}}\Psi(\nu)d\nu$ $\displaystyle=$
$\displaystyle\int_{z_{R}}^{\infty}\nu^{-n_{1}-1}\left[\dfrac{\gamma}{1-\gamma}\nu^{-\frac{1-\gamma}{\gamma}}-l-\dfrac{b^{1-\gamma}}{1-\gamma}+\epsilon\nu\right]d\nu$
$\displaystyle=$
$\displaystyle-\left[\dfrac{\gamma}{1-\gamma}\dfrac{1}{1-n_{1}-\frac{1}{\gamma}}(z_{R})^{1-n_{1}-\frac{1}{\gamma}}+l\dfrac{(z_{R})^{-n_{1}}}{n_{1}}+\dfrac{b^{1-\gamma}}{1-\gamma}\dfrac{(z_{R})^{-n_{1}}}{n_{1}}-\epsilon\dfrac{(z_{R})^{1-n_{1}}}{n_{1}-1}\right].$

##  6 Concluding Remarks

We have studied a retirement and consumption/portfolio choice of an individual
economic agent. We have identified minimum assumptions for the problem well-
defined and the retirement option valuable. We have solved the problem by
formulating a Lagrangian which allows us to consider the difference between
the dual value functions before retirement and after retirement as the option
value of lifetime labor. We have shown that the option value is the present
value of the marginal benefit of work compared to retirement. We have shown
human wealth is a component in the derivative of the option value with respect
to the marginal value of wealth.

In this paper we have assumed a constant investment opportunity facing the
agent. Consideration of the model to allow general market environment as in
Yang et al., (2020) would be a useful topic for future research. We have not
considered borrowing constraints, which can play an important role for
individuals’ choice. The extension incorporating borrowing constraints would
also be an interesting topic for future research.

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##  Appendix A Proof of Proposition 3.1.

We derive budget constraint (10). For a given $T>0$, we define the equivalent
martingale measure $\mathbb{Q}$ by

$\dfrac{d\mathbb{Q}}{d\mathbb{P}}=e^{-\frac{1}{2}\theta^{2}T-\theta B_{T}}.$
(53)

Then, $B_{t}^{\mathbb{Q}}=B_{t}+\theta t,t\in[0,T]$ is a standard Brownian
motion under measure $\mathbb{Q}$. Let us consider the process
$N_{t}:=\int_{0}^{t}e^{-rs}(c_{s}-\epsilon{\bf
1}_{\\{s<\tau\\}})ds+e^{-rt}X_{t}$.

Then, Itô’s lemma implies $N_{t}$ is a continuous $\mathbb{Q}$-local
martingale, i.e.,

$dN_{t}=e^{-rt}\sigma\pi_{t}dB_{t}^{\mathbb{Q}}.$ (54)

By constraint (7) we know

$N_{t}\geq-\epsilon\int_{0}^{t}e^{-rs}ds-e^{-rt}\frac{\epsilon}{r}=-\dfrac{\epsilon}{r}.$

Since $N_{t}$ is bounded below, $N_{t}$ is a $\mathbb{Q}$-supermartingale. It
follows that

$x=\mathbb{E}^{\mathbb{Q}}[N_{0}]\geq\mathbb{E}^{Q}[N_{T}]$ (55)

Thus, we have

$\displaystyle x=X_{0}$
$\displaystyle\geq\mathbb{E}^{\mathbb{Q}}\left[\int_{0}^{T}e^{-rt}(c_{t}-\epsilon{\bf
1}_{\\{t<\tau\\}})ds\right]+\mathbb{E}^{\mathbb{Q}}\left[e^{-rT}X_{T}\right]$
$\displaystyle=\mathbb{E}\left[\int_{0}^{T}\xi_{t}(c_{t}-\epsilon{\bf
1}_{\\{t<\tau\\}})ds\right]+\mathbb{E}\left[\xi_{T}X_{T}\right]$
$\displaystyle\geq\mathbb{E}\left[\int_{0}^{T}\xi_{t}(c_{t}-\epsilon{\bf
1}_{\\{t<\tau\\}})ds\right]-\mathbb{E}\left[\xi_{T}\frac{\epsilon}{r}\right]$
$\displaystyle=\mathbb{E}\left[\int_{0}^{T}\xi_{t}(c_{t}-\epsilon{\bf
1}_{\\{t<\tau\\}})ds\right]-e^{-rT}\frac{\epsilon}{r},$ (56)

where the second inequality follows from constraint (7). By the dominated
convergence theorem the first expectation in the right-hand side of (56)
converges to $\mathbb{E}\left[\int_{0}^{\infty}\xi_{t}(c_{t}-\epsilon{\bf
1}_{\\{t<\tau\\}})dt\right]$. Now budget constraint (10) follows from (56) by
taking the limit as $T\to\infty$.

The converse can be obtained by slightly modifying the proof of Theorem 9.4 in
Karatzas and Shreve, (1998).

##  Appendix B Review of Proposition 4.1 in Knudsen et al., (1998)

######  Proposition B.1 (Proposition 4.1 in Knudsen et al., (1998)).

Let $f(y)$ be an arbitrary measurable function defined on $(0,\infty)$. Then
the following conditions are equivalent:

  * (i)

for every $y>0$

$\mathbb{E}\left[\int_{0}^{\infty}e^{-\rho t}|f({\cal
Y}_{t})|dt\right]<\infty,$

  * (ii)

for every $y>0$

$\int_{0}^{y}\nu^{-n_{2}-1}|f(\nu)|d\nu+\int_{y}^{\infty}\nu^{-n_{1}-1}|f(\nu)|d\nu<\infty.$

Let us denote $\Xi_{f}(y)$ by

$\Xi_{f}(y)=\mathbb{E}\left[\int_{0}^{\infty}e^{-\rho t}f({\cal
Y}_{t})dt\right].$

Under the condition (i) or (ii), the following statements are true:

  * (a)

$\liminf_{y\downarrow
0}y^{-n_{2}}|f(y)|=\liminf_{y\uparrow\infty}y^{-n_{1}}|f(y)|=0$,

  * (b)

$\Xi_{f}$ has a following form:

$\displaystyle\Xi_{f}(y)=\dfrac{2}{\theta^{2}(n_{1}-n_{2})}\left[y^{n_{2}}\int_{0}^{y}\nu^{-n_{2}-1}f(\nu)d\nu+y^{n_{1}}\int_{y}^{\infty}\nu^{-n_{1}-1}f(\nu)d\nu\right],$

  * (c)

$\Xi_{f}$ is twice differentiable and

$\dfrac{\theta^{2}}{2}y^{2}\Xi_{f}^{\prime\prime}(y)+(\rho-r)y\Xi_{f}^{\prime}(y)-\rho\Xi_{f}(y)+f(y)=0,$

  * (d)

there exists a positive constant $C$ such that

$|\Xi_{f}^{\prime}(y)|\leq C(y^{n_{1}-1}+y^{n_{2}-1})\;\;\;\mbox{for
all}\;\;y>0,$

  * (e)

$\lim_{t\to\infty}e^{-\rho t}\mathbb{E}\left[|\Xi_{f}({\cal
Y}_{t})|\right]=0$.