Now we consider our optimization problem. In this problem, the infinitely-lived agent wants to maximize her expected lifetime utility:

$$ V(x):=\sup_{(c,\pi)\in\mathcal{A}(x)} \mathbb{E} \biggl[ \int _{0}^{\infty}e^{-\rho t}\bigl(c_{t}-Rc^{2}_{t} \bigr)\,dt \biggr] $$

(3.1)

subject to the budget constraint (2.2) and the negative wealth constraint (2.3). Here, \(\rho>0\) is a subjective discount factor, \(R>0\) is the constant coefficient of the quadratic utility function, and \(\mathcal{A}(x)\) is the class of all admissible controls \((c,\pi)\) at *x*.

### Remark 3.1

Due to the feature of the quadratic utility function, there is a bliss consumption level \(\bar{c}:={1}/{(2R)}\), and the negative wealth constraint level should be lower than the bliss wealth level, that is,

$$ -\nu\frac{I}{r}< \frac{1}{2rR}-\frac{I}{r}. $$

(3.2)

This assumption is deduced from the bliss level which is one of the aspects of quadratic utility. The right-hand side of inequality (3.2) is bliss wealth level *x̄* which is obtained as follows:

$$ \bar{x}= \int_{0}^{\infty}e^{-rt}(\bar{c}-I)\,dt= \frac{1}{2rR}-\frac{I}{r}. $$

(3.3)

We denote by *ũ* the dual utility function of a quadratic utility function, which is defined by

$$\begin{aligned} \tilde{u}(y)&:=\sup_{c} \bigl\{ c-Rc^{2}-y(c-I) \bigr\} \\ &= \biggl\{ \frac{(1-y)^{2}}{4R} +yI \biggr\} \mathbf{1}_{\{0< y< 1\}}+yI \mathbf{1}_{\{y\geq1\}}, \end{aligned}$$

where \(\mathbf{1}_{A}\) is an indicator function. For a domain \(\{0< y<1\} \), the optimal consumption of the agent is \((1-y)/(2R)\) and, for a domain \(\{y\geq1\}\), the agent consumes nothing, that is, the optimal consumption is zero.

For a Lagrange multiplier \(\lambda>0\), we define a dual value function as follows:

$$\begin{aligned} \tilde{V}(\lambda)&=\sup_{c} \mathbb{E} \biggl[ \int_{0}^{\infty}e^{-\rho t}\bigl\{ c_{t}-Rc_{t}^{2}\bigr\} \,dt-\lambda \int_{0}^{\infty}H_{t}(c_{t}-I)\,dt \biggr] \\ &=\sup_{c} \mathbb{E} \biggl[ \int_{0}^{\infty}e^{-\rho t}\bigl\{ c_{t}-Rc_{t}^{2}-\lambda e^{\rho t} H_{t}(c_{t}-I)\bigr\} \,dt \biggr] \\ &=\mathbb{E} \biggl[ \int_{0}^{\infty}e^{-\rho t}\tilde{u}(y_{t}) \,dt \biggr], \end{aligned}$$

where \(y_{t}:=\lambda e^{\rho t} H_{t}\) and the process \(y_{t}\) follows the SDE

$$dy_{t}=(\rho-r)y_{t}\, dt-\theta y_{t}\, dB_{t}. $$

### Remark 3.2

For later use, we define the following quadratic equation:

$$ g(m):=\frac{1}{2}\theta^{2}m^{2}+ \biggl(\rho-r-\frac{1}{2}\theta ^{2} \biggr)m-\rho=0, $$

(3.4)

with two roots \(m_{+}>1\) and \(m_{-}<0\).

Now we consider the function

$$ \phi(t,y):=\mathbb{E} \biggl[ \int_{t} ^{\infty}e^{-\rho s} \biggl\{ \biggl( \frac{(1-y_{s})^{2}}{4R} +y_{s}I \biggr) \mathbf{1}_{\{0< y_{s}< 1\} }+y_{s}I \mathbf{1}_{\{y_{s}\geq1\}} \biggr\} \,ds\Big| y_{t}=y \biggr]. $$

By the Feymann-Kac formula, we derive the partial differential equations (PDEs) as follows:

$$\textstyle\begin{cases} \mathcal{L} \phi(t,y)+e^{-\rho t} [\frac{(1-y)^{2}}{4R} +yI ]=0 & \text{if }0< y< 1, \\ \mathcal{L} \phi(t,y)+e^{-\rho t}yI=0 & \text{if } y>1, \end{cases} $$

where the partial differential operator is given by

$$\mathcal{L}:=\frac{\partial}{\partial t}+(\rho-r)y\frac{\partial }{\partial y}+\frac{1}{2} \theta^{2} y^{2}\frac{\partial^{2}}{\partial y^{2}}. $$

If we conjecture that \(\phi(t,y)=e^{-\rho t} v(y)\), then we derive the following ordinary differential equations (ODEs):

$$ \textstyle\begin{cases} \frac{1}{2}\theta^{2} y^{2} v''(y)+(\rho-r)y v'(y)-\rho v(y)+\frac {(1-y)^{2}}{4R} +yI=0 & \text{if } 0< y< 1, \\ \frac{1}{2}\theta^{2} y^{2} v''(y)+(\rho-r)y v'(y)-\rho v(y)+yI=0 & \text{if } y>1 . \end{cases} $$

(3.5)

We define \(\hat{y}>0\) as the level of a dual variable of the negative wealth constraint level \(-\nu{I}/{r}\).

### Remark 3.3

We define the positive constant *K* for indicating that *ŷ* is less or greater than 1.

$$ K:=\frac{1}{2R(m_{+}-1)}\frac{2-m_{+}}{\rho-2r+\theta^{2}}>0. $$

(3.6)

By comparison with the range from wealth constraint to the bliss level of wealth and *K*, we can check that \(\hat{y}>1\) or \(\hat{y}<1\). And we will solve ODEs (3.5) with two cases: one is \(\hat {y}>1\), and the other is \(\hat{y}<1\).

### 3.1 In the case of \(\hat{y}>1\)

If \(\hat{y}>1\), we can rewrite ODEs (3.5) as

$$ \textstyle\begin{cases} \frac{1}{2}\theta^{2} y^{2} v''(y)+(\rho-r)y v'(y)-\rho v(y)+\frac {(1-y)^{2}}{4R} +yI=0 & \text{if } 0< y< 1, \\ \frac{1}{2}\theta^{2} y^{2} v''(y)+(\rho-r)y v'(y)-\rho v(y)+yI=0 & \text{if } 1< y< \hat{y}. \end{cases} $$

(3.7)

### Proposition 3.1

*The solutions to ODEs* (3.7) *are given by*

$$ v(y)= \textstyle\begin{cases} D_{1}y^{m_{+}}-\frac{1}{4R(\rho-2r+\theta^{2})}y^{2}+ (\frac {I}{r}-\frac{1}{2rR} )y+\frac{1}{4\rho R} &\textit {if } 0< y< 1 , \\ C_{1} y^{m_{+}}+C_{2}y^{m_{-}}+\frac{I}{r}y &\textit{if } 1< y< \hat{y} , \end{cases} $$

(3.8)

*where*

$$\begin{aligned}& C_{2} =\frac{1}{m_{+}-m_{-}} \biggl(-\frac{m_{+}-2}{4R(\rho-2r+\theta ^{2})}- \frac{m_{+}-1}{2r R}+\frac{m_{+}}{4\rho R} \biggr)>0, \\& \hat{y} = \biggl(-\frac{(1-\nu)(m_{+}-1)I}{C_{2} m_{-}(m_{+}-m_{-})r} \biggr)^{\frac{1}{m_{-} -1}}, \\& C_{1} =-\frac{C_{2} m_{-}(m_{-}-1)}{m_{+}(m_{+}-1)}\hat{y}^{m_{-}-m_{+}}< 0, \\& D_{1} =C_{1}+C_{2}+\frac{1}{4R(\rho-2r+\theta^{2})}+ \frac{1}{2r R}-\frac {1}{4\rho R}. \end{aligned}$$

### Proof

From ODEs (3.7), we obtain the solution as follows:

$$ v(y)= \textstyle\begin{cases} D_{1}y^{m_{+}}+D_{2}y^{m_{-}}-\frac{y^{2}}{4R(\rho-2r+\theta^{2})}+ (\frac {I}{r}-\frac{1}{2rR} )y+\frac{1}{4\rho R} &\text{if } 0< y< 1 , \\ C_{1} y^{m_{+}}+C_{2}y^{m_{-}}+\frac{I}{r}y &\text{if } 1< y< \hat{y} , \end{cases} $$

where \(m_{+}\) and \(m_{-}\) are two roots of quadratic equation (3.4). Since, for \(0< y<1\), the first equation has to satisfy the well-definedness, \(D_{2}\) should be zero. By using the free boundary conditions of the negative wealth constraint, \(v'(\hat{y})=\nu\frac {I}{r}\) and \(v''(\hat{y})=0\), and \(C^{1}\)-condition of \(v(y)\) at \(y=1\), we derive the coefficients and *ŷ* as follows:

$$\begin{aligned}& C_{2} =\frac{1}{m_{+}-m_{-}} \biggl(-\frac{m_{+}-2}{4R(\rho-2r+\theta ^{2})}- \frac{m_{+}-1}{2r R}+\frac{m_{+}}{4\rho R} \biggr)>0, \\& \hat{y} = \biggl(-\frac{(1-\nu)(m_{+}-1)I}{C_{2} m_{-}(m_{+}-m_{-})r} \biggr)^{\frac{1}{m_{-} -1}}, \\& C_{1} =-\frac{C_{2} m_{-}(m_{-}-1)}{m_{+}(m_{+}-1)}\hat{y}^{m_{-}-m_{+}}< 0, \\& D_{1} =C_{1}+C_{2}+\frac{1}{4R(\rho-2r+\theta^{2})}+ \frac{1}{2r R}-\frac {1}{4\rho R} \end{aligned}$$

(we will show that \(C_{2}>0\) in Proposition 3.3). □

By the Legendre inverse transform formula, the value function \(V(\cdot )\) can be obtained as follows:

$$ V(x)=\min_{\lambda>0} \bigl\{ v(\lambda)+\lambda x \bigr\} . $$

(3.9)

### Theorem 3.1

*The value function*
\(V(x)\)
*of optimization problem* (3.1) *is given by*

$$ V(x)= \textstyle\begin{cases} (1-m_{+})C_{1} \zeta^{m_{+}}+(1-m_{-})C_{2}\zeta^{m_{-}} &\textit{if } -\nu\frac {I}{r} \leq x< \tilde{x}, \\ (1-m_{+})D_{1}\xi^{m_{+}}+\frac{\xi^{2}}{4R(\rho-2r+\theta^{2})}+\frac {1}{4\rho R} &\textit{if } \tilde{x} \leq x< \bar{x} , \\ \frac{1}{4\rho R} & \textit{if } x\geq\bar{x} , \end{cases} $$

*where*

$$\begin{aligned}& \tilde{x}=-C_{1}m_{+}-C_{2}m_{-} -\frac{I}{r}, \\& \bar{x}=\frac{1}{2rR}-\frac{I}{r}, \end{aligned}$$

*ξ*
*and*
*ζ*
*are the solutions to the algebraic equations*

$$\begin{aligned}& x=-C_{1} m_{+}\zeta^{m_{+} -1}-C_{2} m_{-} \zeta^{m_{-} -1} -\frac{I}{r}, \\& x=-D_{1} m_{+} \xi^{m_{+}-1}+\frac{1}{2R(\rho-2r+\theta^{2})}\xi+ \frac {1}{2rR}-\frac{I}{r}, \end{aligned}$$

*respectively*.

### Proof

The first order condition for equation (3.9) takes the following form:

where the function \(v(\cdot)\) is described in Proposition 3.1. Substituting the first order condition into equation (3.9), we can complete the proof. □

When \(\hat{y}>1\), we can provide the optimal strategy \((c,\pi)\).

### Theorem 3.2

*The optimal strategies are given by*

$$ c^{*}= \textstyle\begin{cases} 0 &\textit{if } -\nu\frac{I}{r} \leq x< \tilde{x} , \\ \frac{1-\xi}{2R} &\textit{if } \tilde{x} \leq x< \bar{x} , \\ \frac{1}{2R}& \textit{if } x\geq\bar{x} , \end{cases} $$

*and*

$$ \pi^{*}= \textstyle\begin{cases} \frac{\theta}{\sigma} (C_{1}m_{+}(m_{+} -1)\zeta^{m_{+} -1}+C_{2}m_{-}(m_{-} -1)\zeta^{m_{-} -1} ) &\textit{if } -\nu\frac{I}{r} \leq x< \tilde {x} , \\ \frac{\theta}{\sigma} (D_{1}m_{+}(m_{+} -1)\xi^{m_{+} -1}-\frac {1}{2R(\rho-2r+\theta^{2})} \xi ) &\textit{if } \tilde{x} \leq x< \bar{x} , \\ 0& \textit{if } x\geq\bar{x} . \end{cases} $$

### Proof

By the duality of value function and the Itô formula, we can obtain the following equation:

$$ dX_{t}= \biggl(-v''(y) ( \rho-r)y-\frac{1}{2}v'''(y) \theta^{2} y^{2} \biggr)\,dt+\theta y v''(y) \,dB_{t}. $$

(3.10)

By comparing (2.1) and (3.10), we derive the optimal consumption \(c_{t}^{*}\) and portfolio \(\pi_{t}^{*}\) as follows:

$$ \begin{aligned} &c_{t}^{*}=-rv'(y)+ \bigl(\rho-r+\theta^{2}\bigr)yv''(y)+ \frac{1}{2}\theta ^{2}y^{2}v'''(y)+I, \\ &\pi_{t}^{*}=\frac{\theta}{\sigma}yv''(y). \end{aligned} $$

(3.11)

By substituting equation (3.8) into (3.11), we can derive the optimal consumption and portfolio. □

### Proposition 3.2

*If the following inequality* (3.12) *holds*,

$$ \frac{1}{2rR}-\frac{I}{r}- \biggl(-\nu \frac{I}{r} \biggr)>\frac {1}{2R(m_{+}-1)}\frac{2-m_{+}}{\rho-2r+\theta^{2}}=:K, $$

(3.12)

*where*
*K*
*is given in* (3.6), *then*

$$\hat{y}= \biggl(-\frac{(1-\nu)(m_{+}-1)I}{C_{2} m_{-}(m_{+}-m_{-})r} \biggr)^{\frac{1}{m_{-} -1}}>1. $$

### Proof

Since \(m_{-}<0\), it is enough to show that

$${-\frac{C_{2} m_{-}(m_{+}-m_{-})r}{(1-\nu)(m_{+}-1)I}>1}. $$

This inequality implies

$$ \frac{1}{2rR}-\frac{I}{r}- \biggl(-\nu \frac{I}{r} \biggr)>\frac {1}{2rR} \biggl(\frac{r}{2} \frac{m_{-}(2-m_{+})}{(m_{+}-1)(\rho-2r+\theta ^{2})}+(1-m_{-})+\frac{rm_{+} m_{-}}{2\rho(m_{+}-1)} \biggr). $$

(3.13)

From the difference of the right-hand side of inequalities (3.12) and (3.13), we have

$$ \frac{1}{2rR} \biggl(\frac{r(2-m_{+})(m_{-}-2)}{2(m_{+}-1)(\rho-2r+\theta ^{2})}+(1-m_{-})+ \frac{rm_{+} m_{-}}{2\rho(m_{+}-1)} \biggr). $$

(3.14)

By the relation between roots and coefficients of quadratic equation (3.4), we can check that equation (3.14) should be zero, that is, \(\hat{y}>1\). □

### Proposition 3.3

*In Proposition *
3.1, \(C_{2}\geq0\).

### Proof

Refer to Koo et al. [4]. □

### 3.2 In the case of \(\hat{y}<1\)

If \(\hat{y}<1\), *y* does not exceed *ŷ*. So we obtain the following ODE from ODEs (3.5).

$$ \frac{1}{2}\theta^{2} y^{2} v''(y)+(\rho-r)y v'(y)-\rho v(y)+ \frac {(1-y)^{2}}{4R} +yI=0 \quad \text{if } 0< y< \hat{y}. $$

(3.15)

### Proposition 3.4

*The solution to ODE* (3.15) *is given by*

$$ v(y)=D_{1}y^{m_{+}}-\frac{y^{2}}{4R(\rho-2r+\theta^{2})}+ \biggl( \frac {I}{r}-\frac{1}{2rR} \biggr)y+\frac{1}{4\rho R} \quad \textit{if } 0< y< \hat{y}, $$

(3.16)

*where*

$$\begin{aligned}& \hat{y} =2R\bigl(\rho-2r+\theta^{2}\bigr) \biggl(\frac{m_{+}-1}{m_{+}-2} \biggr) \biggl((1-\nu)\frac{I}{r}-\frac{1}{2rR} \biggr), \\& D_{1} =\frac{1}{2R(\rho-2r+\theta^{2})(m_{+}-1)m_{+}}\hat{y}^{2-m_{+}}. \end{aligned}$$

### Proof

For \(0< y<\hat{y}\), similar to the proof of Proposition 3.1, we derive the solution \(v(\cdot)\) in (3.16). By the free boundary conditions at \(y=\hat{y}\), we obtain two equations \(v'(\hat {y})=\nu{I}/{r}\) and \(v''(\hat{y})=0\). Using these equations, we have

$$\begin{aligned}& \hat{y} =2R\bigl(\rho-2r+\theta^{2}\bigr) \biggl(\frac{m_{+}-1}{m_{+}-2} \biggr) \biggl((1-\nu)\frac{I}{r}-\frac{1}{2rR} \biggr), \\& D_{1} =\frac{1}{2R(\rho-2r+\theta^{2})(m_{+}-1)m_{+}}\hat{y}^{2-m_{+}}. \end{aligned}$$

□

By the duality of value function \(V(x)\), we derive the value function \(V(\cdot)\).

### Theorem 3.3

*When*
\(\hat{y}<1\), *the value function*
\(V(x)\)
*of optimization problem* (3.1) *is given by*

$$ V(x)= \textstyle\begin{cases} (1-m_{+})D_{1}\xi^{m_{+}}+\frac{\xi^{2}}{4R(\rho-2r+\theta^{2})}+\frac {1}{4\rho R} &\textit{if } -\nu\frac{I}{r} \leq x< \bar{x} , \\ \frac{1}{4\rho R} & \textit{if } x\geq\bar{x} , \end{cases} $$

*where*

$$\bar{x}=\frac{1}{2rR}-\frac{I}{r}, $$

*and*
*ξ*
*is the solution to the algebraic equation*

$$ x=-D_{1} m_{+} \xi^{m_{+}-1}+\frac{1}{2R(\rho-2r+\theta^{2})}\xi+ \frac {1}{2rR}-\frac{I}{r}. $$

### Theorem 3.4

*The optimal strategies are given by*

$$ c^{*}= \textstyle\begin{cases} \frac{1-\xi}{2R} &\textit{if } -\nu\frac{I}{r} \leq x< \bar{x} , \\ \frac{1}{2R}& \textit{if } x\geq\bar{x}, \end{cases} $$

*and*

$$ \pi^{*}= \textstyle\begin{cases} \frac{\theta}{\sigma} (D_{1}m_{+}(m_{+} -1)\xi^{m_{+} -1}-\frac {1}{2R(\rho-2r+\theta^{2})} \xi ) &\textit{if } -\nu\frac{I}{r} \leq x< \bar{x} , \\ 0& \textit{if } x\geq\bar{x} . \end{cases} $$

### Proposition 3.5

*If the following inequality holds*, *then*
\(\hat{y}<1\).

$$ \frac{1}{2rR}-\frac{I}{r}- \biggl(-\nu \frac{I}{r} \biggr)< \frac {1}{2R(m_{+}-1)}\frac{2-m_{+}}{\rho-2r+\theta^{2}}=:K, $$

(3.17)

*where*
*K*
*is given in* (3.6).

### Proof

Since the signs of \(\rho-2r+\theta^{2}\) and \(2-m_{+}\) are the same, equation (3.17) is rewritten as follows:

$$\hat{y}=2R\bigl(\rho-2r+\theta^{2}\bigr)\frac{m_{+}-1}{2-m_{+}} \biggl( \frac {1}{2rR}-\frac{I}{r}- \biggl(-\nu\frac{I}{r} \biggr) \biggr)< 1. $$

□

### Remark 3.4

In Figure 1, we plot the optimal consumption and portfolio with respect to the wealth (especially, small figures in Figure 1(b) represent the specific region of wealth level from −12.5 to −12.2). From the figures, we can see the effects of the bliss wealth level and the threshold *ŷ* (\(\nu=1, 0.5\): \(\hat{y}>1\) and \(\nu=0\): \(\hat{y}<1\)). The left dotted line of each figure represents the wealth level at which optimal consumption is zero, and the right dotted line represents the bliss wealth level.

Basically, we obtain similar results to those of Koo et al. [4], that is, if \(\hat{y}>1\), then the optimal consumption is zero until wealth reaches the threshold wealth level corresponding to the dual variable \(y=1\). After wealth reaches this level, the optimal consumption increases as wealth increases. But after the bliss level *x̄*, the optimal consumption stays at \(\bar{c}=1/(2R)\). For the optimal portfolio, it increases from zero to the certain maximum until wealth reaches a certain wealth level. The optimal portfolio decreases above this level and becomes zero above the bliss level *x̄*.

If \(0<\hat{y}<1\), however, there is no zero consumption region. So the optimal consumption increases as wealth increases. But after the bliss level *x̄*, the optimal consumption stays at \(\bar{c}=1/(2R)\). For the optimal portfolio, we see behavior similar to the case of \(\hat{y}>1\).