Maximizing expected exponential utility of consumption with a constraint on expected time in poverty

Abstract

We compute the optimal investment and consumption strategies for an individual who wishes to maximize her expected discounted exponential utility of lifetime consumption, while imposing a constraint on the expected time her wealth spends below a poverty threshold b. First, we compute the optimal strategies for the corresponding (unconstrained) problem with a running penalty for time that wealth spends below b. This penalty acts as a Lagrange multiplier for our original constrained problem, so we recover the optimal strategies for our original problem from the recast problem. We show that (1) if the current wealth is greater than b, then the optimal investment strategy becomes more conservative as the poverty constraint becomes sharper; and (2) if the current wealth is less than b, then the optimal investment strategy is either independent of the poverty constraint or becomes more aggressive as the poverty constraint becomes sharper, depending on the value b. We also show that the optimal rate of consumption (weakly) decreases as the poverty constraint becomes sharper.

Appendix: Proof of Proposition 1

Note that depending on the value of $y_b$, (3.1) can be divided into two cases, specifically, $y_b> 1$ and $y_b\le 1$.

First, we consider the case for which $y_b> 1$. In this case, the FBP (3.1) becomes

$$\begin{aligned} {\left\{ \begin{array}{ll} \lambda {{\hat{\psi }}}= {\left\{ \begin{array}{ll} \delta y^2 {{\hat{\psi }}}_{yy} + (\lambda - r) y {{\hat{\psi }}}_y - \big (1 - \ln y \big ) \dfrac{y}{\eta }, &{}y \in [0, 1), \\ \delta y^2 {{\hat{\psi }}}_{yy} + (\lambda - r) y {{\hat{\psi }}}_y - \dfrac{1}{\eta }, &{}y \in [1, y_b), \\ \delta y^2{{\hat{\psi }}}_{yy} + (\lambda - r) y {{\hat{\psi }}}_y - \dfrac{1}{\eta } + P, &{}y \in [y_b, \infty ), \end{array}\right. } \\ \lim _{y \rightarrow \infty }{{\hat{\psi }}}(y; P) = \dfrac{P - 1/\eta }{\lambda }, \quad \lim _{y \rightarrow \infty } {{\hat{\psi }}}_y(y; P) = 0, \quad {{\hat{\psi }}}_y(y_b; P) = - \, b. \end{array}\right. } \end{aligned}$$

(A.1)

Define the function ${\mathcal {H}}_1$ by

$$\begin{aligned} {\mathcal {H}}_1(y; P) = b y + \dfrac{P}{\delta (\rho _{+} - \rho _{-})} - \dfrac{1}{\delta \eta \rho _{-}^2 (\rho _{+} - \rho _{-})} \, y^{\kappa _{-}}, \end{aligned}$$

for $y > 0$. It follows that

$$\begin{aligned} \left\{ \begin{array} [c]{l} ({\mathcal {H}}_1)_y(y; P) = b - \dfrac{\kappa _{-}}{\delta \eta \rho _{-}^2 (\rho _{+} - \rho _{-})} \, y^{\rho _{-}} > 0, \\ {\mathcal {H}}_1(1; P) = b + \dfrac{P}{\delta (\rho _{+} - \rho _{-})} - \dfrac{1}{\delta \eta \rho _{-}^2 (\rho _{+} - \rho _{-})} , \quad \lim _{y \rightarrow \infty }{\mathcal {H}}_1(y; P) = \infty . \end{array} \right. \end{aligned}$$

Note that if $P \le \min \big (\, {{\overline{P}}}, 0 \big )$, in which ${{\overline{P}}}$ is given in (3.7), then ${\mathcal {H}}_1(1; P) \le 0$, which implies that ${\mathcal {H}}_1$ has a unique zero $y_b= y_b(P) \in [1, \infty )$.

It is tedious, but straightforward, to verify by substitution that ${\mathcal {V}}_1$ given by

$$\begin{aligned} {\mathcal {V}}_1(y; P) = {\left\{ \begin{array}{ll} C_1 y^{\kappa _{+}} + \left( \dfrac{\lambda + \delta - 2r}{r} + \ln y \right) \dfrac{y}{r \eta }, &{}y \in [0, 1), \\ C_2 y^{\kappa _{+}} + C_3 y^{\kappa _{-}} - \dfrac{1}{\lambda \eta }, &{}y \in [1, y_b), \\ C_4y^{\kappa _{-}} + \dfrac{P - 1/\eta }{\lambda }, &{}y \in [y_b, \infty ), \end{array}\right. } \end{aligned}$$

classically solves FBP (A.1) and is ${\mathcal {C}}^{1}$ at $y = y_b$. Here, the coefficients are given in (3.4).

Next, we show that ${\mathcal {V}}_1(y; P)$ is decreasing and convex with respect to y on $\mathbb {R}_{+}$, for $P \in \big [ {\underline{P}}, \, \min \big (\, {{\overline{P}}}, 0 \big ) \big ]$, in which ${\underline{P}}$ is given in (3.3). Rearranging ${\underline{P}}$ yields the useful identity

$$\begin{aligned} {\underline{P}} = \frac{1}{\eta \rho _{-}^2}\left( \frac{- r \eta \, {\underline{P}}}{\delta } \right) - b \left( \frac{- r \eta \, {\underline{P}}}{\delta }\right) ^{1/\kappa _{-}}\delta (\rho _{+} - \rho _{-}). \end{aligned}$$

(A.2)

Define ${\mathcal {P}}_1:\mathbb {R}_{+}/\{0\} \rightarrow \mathbb {R}$ by

$$\begin{aligned} {\mathcal {P}}_1(y) = \frac{1}{\eta \rho _{-}^2} \, y^{\kappa _{-}} - b y \delta (\rho _{+} - \rho _{-}). \end{aligned}$$

(A.3)

Note that ${\mathcal {P}}_1(y_b(P)) = P$ when $P \le \min \big (\, {{\overline{P}}}, 0 \big )$. Because ${\mathcal {P}}_1$ is decreasing, it has an inverse ${\mathcal {Y}}_1 = {\mathcal {P}}_1^{-1}: \mathbb {R} \rightarrow \mathbb {R}_{+}/\{0\}$, which is also decreasing, with ${\mathcal {Y}}_1(P) = y_b(P)$ when $P \le \min \big (\, {{\overline{P}}}, 0 \big )$. Next, define ${\mathcal {G}}_1:\mathbb {R} \rightarrow \mathbb {R}$ by

$$\begin{aligned} {\mathcal {G}}_1(P) = P + \dfrac{\delta }{r \eta } \, {\mathcal {Y}}_1(P) ^{\kappa _{-}}, \end{aligned}$$

which increases with P because $({\mathcal {G}}_1)_P = 1 + \frac{\delta \kappa _{-}}{r \eta } \, {\mathcal {Y}}_1^{\rho _{-}} \cdot ({\mathcal {Y}}_1)_P > 0$. From (A.2) and (A.3), we deduce that ${\mathcal {Y}}_1^{\kappa _{-}}({\underline{P}}) = -r \eta \, {\underline{P}}/\delta$; thus, if $P \ge {\underline{P}}$, then

$$\begin{aligned} {\mathcal {G}}_1(P) \ge {\mathcal {G}}_1({\underline{P}}) = {\underline{P}} + \dfrac{\delta }{r \eta } \, {\mathcal {Y}}_1({\underline{P}})^{\kappa _{-}} = 0. \end{aligned}$$

(A.4)

The second derivative of ${\mathcal {V}}_1$ with respect to y equals

$$\begin{aligned} ({\mathcal {V}}_1)_{yy}(y; P) = {\left\{ \begin{array}{ll} C_1 \kappa _{+} \rho _{+} y^{\rho _{+} - 1} + \dfrac{1}{r \eta y}, &{}y \in (0,1), \\ C_2 \kappa _{+} \rho _{+} y^{\rho _{+} - 1} + C_3 \kappa _{-} \rho _{-} y^{\rho _{-} - 1}, &{}y \in [1, y_b), \\ C_4 \kappa _{-} \rho _{-} y^{\rho _{-} - 1}, &{}y \in (y_b, \infty ). \end{array}\right. } \end{aligned}$$

Given the inequality in (A.4), for $y \in (0, 1)$ and $P \in \big [ {\underline{P}}, \, \min \big (\, {{\overline{P}}}, 0 \big ) \big ]$,

$$\begin{aligned} ({\mathcal {V}}_1)_{yy}(y; P)&= \dfrac{1}{r \eta y} \, \dfrac{\rho _{+}}{\rho _{+} - \rho _{-}} \left[ \dfrac{r \eta P}{\delta y_b^{\kappa _{+}}} \, y^{\rho _{+}} +\dfrac{\rho _{-}}{\rho _{+}} \, y^{\rho _{+}} + \dfrac{\rho _{+} - \rho _{-}}{\rho _{+}} \right] , \\&\ge \dfrac{1}{r \eta y} \, \dfrac{\rho _{+}}{\rho _{+}-\rho _{-}} \left[ \dfrac{r \eta P}{\delta y_b^{\kappa _{+}}} + 1 \right] \ge \frac{1}{\delta y \, y_b^{\kappa _{-}}}\, \dfrac{\rho _{+}}{\rho _{+} - \rho _{-}} \left[ P + \dfrac{\delta }{r \eta } \, y_b^{\kappa _{-}} \right] \ge 0, \end{aligned}$$

in which the second inequality follows from $y_b= y_b(P) > 1$. For $y \in [1, y_b)$ and $P \in \big [ {\underline{P}}, \, \min \big (\, {{\overline{P}}}, 0 \big ) \big ]$,

$$\begin{aligned} ({\mathcal {V}}_1)_{yy}(y; P)&= \dfrac{1}{\delta y^2} \dfrac{\rho _{+}}{\rho _{+}- \rho _{-}}\left[ P \left( \dfrac{y}{y_b} \right) ^{\kappa _{+}} + \dfrac{\delta }{r \eta } \, y^{\kappa _{-}}\right] \\&\ge \dfrac{1}{\delta y^2} \dfrac{\rho _{+}}{\rho _{+}- \rho _{-}}\left[ P + \dfrac{\delta }{r \eta } \, y_b^{\kappa _{-}}\right] \ge 0, \end{aligned}$$

in which the first inequality follows from the fact that the expression in square brackets decreases with respect to y; recall that $P \le 0$. Clearly, for $y \in (y_b, \infty )$, $({\mathcal {V}}_1)_{yy}(y; P) > 0$ because $C_4 > 0$. Because $\lim _{y \rightarrow \infty } ({\mathcal {V}}_1)_y(y; P) = \lim _{y \rightarrow \infty } C_4 \kappa _{-} y^{\rho _{-}} = 0$, ${\mathcal {V}}_1$ is decreasing and convex with respect to y on $\mathbb {R}_{+}$, for $P \in \big [ {\underline{P}}, \, \min \big (\, {{\overline{P}}}, 0 \big ) \big ]$.

To summarize, if $b \le \frac{1}{\delta \eta \rho _{-}^2 (\rho _{+} - \rho _{-})} \,$, then ${\overline{P}} \ge 0$; thus, ${\mathcal {V}}_1$ is decreasing and convex with y for all $P \in \left[ {\underline{P}}, 0 \right]$. If $\frac{1}{\delta \eta \rho _{-}^2 (\rho _{+} - \rho _{-})} < b \le - \, \frac{1}{r \eta \rho _{-}} \,$, then ${\underline{P}} \le {{\overline{P}}} < 0$; thus ${\mathcal {V}}_1$ is decreasing and convex with y for all $P \in \left[ {\underline{P}},{\overline{P}} \right]$. Finally, if $b > - \, \frac{1}{r \eta \rho _{-}} \,$, then there does not exist $P \le 0$ such that ${\mathcal {V}}_1$ is decreasing and convex because ${\underline{P}} > {\overline{P}}$ in that case.

Second, we consider the case for which $y_b\le 1$. In this case, the FBP (3.1) becomes

$$\begin{aligned} {\left\{ \begin{array}{ll} \lambda {{\hat{\psi }}}= {\left\{ \begin{array}{ll} \delta y^2 {{\hat{\psi }}}_{yy} + (\lambda - r) y {{\hat{\psi }}}_y - \big (1 - \ln y \big ) \dfrac{y}{\eta }, &{}y \in [0, y_b), \\ \delta y^2 {{\hat{\psi }}}_{yy} + (\lambda - r) y {{\hat{\psi }}}_y - \big (1 - \ln y \big ) \dfrac{y}{\eta } + P, &{}y \in [y_b, 1), \\ \delta y^2{{\hat{\psi }}}_{yy} + (\lambda - r) y {{\hat{\psi }}}_y - \dfrac{1}{\eta } + P, &{}y \in [1, \infty ), \end{array}\right. } \\ \lim _{y \rightarrow \infty }{{\hat{\psi }}}(y; P) = \dfrac{P - 1/\eta }{\lambda }, \quad \lim _{y \rightarrow \infty } {{\hat{\psi }}}_y(y; P) = 0, \quad {{\hat{\psi }}}_y(y_b; P) = - \, b. \end{array}\right. } \end{aligned}$$

(A.5)

We introduce an identity to use in this proof.

$$\begin{aligned} \dfrac{1}{\delta \eta \rho _{-}^2(\rho _{+} - \rho _{-})} + \frac{\lambda + \delta - r}{r^2 \eta } = \frac{1}{\delta \eta \rho _{+}^2 (\rho _{+} - \rho _{-})} . \end{aligned}$$

(A.6)

Define the function ${\mathcal {H}}_2$ by

$$\begin{aligned} {\mathcal {H}}_2(y; P) = by + \left( \frac{\lambda + \delta - r}{r} + \ln y \right) \frac{y}{r \eta } + \frac{P}{\delta (\rho _{+} - \rho _{-})} - \frac{1}{\delta \eta \rho _{+}^2 (\rho _{+} - \rho _{-})} \, y^{\kappa _{+}}, \end{aligned}$$

for $y > 0$. By using (A.6), we compute the following quantities.

$$\begin{aligned} \left\{ \begin{array} [c]{l} \lim _{y \rightarrow 0+} {\mathcal {H}}_2(y; P) = \dfrac{P}{\delta (\rho _{+} - \rho _{-})} < 0 , \quad {\mathcal {H}}_2(1; P) = b - \dfrac{1}{\delta \eta \rho _{-}^2 (\rho _{+} - \rho _{-})} \\ \quad + \dfrac{P}{\delta (\rho _{+} - \rho _{-})} ,\\ ({\mathcal {H}}_2)_y(y; P) = b + \left( \dfrac{\lambda + \delta }{r} + \ln y \right) \dfrac{1}{r \eta } -\dfrac{\kappa _{+}}{\delta \eta \rho _{+}^2(\rho _{+} - \rho _{-})}\, y^{\rho _{+}}, \quad \\ \quad \lim _{y \rightarrow 0+} ({\mathcal {H}}_2)_y(y) = - \infty , \\ ({\mathcal {H}}_2)_{yy}(y; P) = \dfrac{1}{\eta y}\left( \dfrac{1}{r} - \dfrac{\kappa _{+}}{\delta \rho _{+}(\rho _{+} - \rho _{-})} \, y^{\rho _{+}} \right) . \end{array} \right. \end{aligned}$$

If $P > {\overline{P}}$, in which ${{\overline{P}}}$ is given in (3.7), then ${\mathcal {H}}_2(1; P) > 0$; thus, ${\mathcal {H}}_2$ has an odd number, $2k + 1$, of zeros in (0, 1), for some $k = 0, 1, \ldots$. It follows that $({\mathcal {H}}_2)_y$ has $2k + 1$ zeros in (0, 1), and $({\mathcal {H}}_2)_{yy}$ has 2k zeros in (0, 1). From the above expression for $({\mathcal {H}}_2)_{yy}$, it is clear that the second derivative changes sign at most once in (0, 1); thus, $k = 0$. Thus, for $P \in ( {\overline{P}}, 0)$, ${\mathcal {H}}_2$ has a unique zero $y_b= y_b(P)$ in (0, 1).³

As in the case for which $y_b> 1$, it is tedious, but straightforward, to verify that

$$\begin{aligned} {\mathcal {V}}_2(y; P) = {\left\{ \begin{array}{ll} D_1 y^{\kappa _{+}} + \left( \dfrac{\lambda + \delta - 2r}{r} + \ln y \right) \dfrac{y}{r \eta } , &{}y \in [0, y_b), \\ D_2 y^{\kappa _{+}} + D_3 y^{\kappa _{-}} + \left( \dfrac{\lambda + \delta - 2r}{r} + \ln y \right) \dfrac{y}{r \eta } + \dfrac{P}{\lambda } , &{}y \in [y_b,1), \\ D_4y^{\kappa _{-}} + \dfrac{P - 1/\eta }{\lambda } , &{}y \in [1,\infty ), \end{array}\right. } \end{aligned}$$

classically solves FBP (A.5) and is ${\mathcal {C}}^{1}$ at $y = y_b$. Here, the coefficients are given in (3.8).

Next, we show that ${\mathcal {V}}_2(y; P)$ is decreasing and convex with respect to y on $\mathbb {R}_{+}$ for appropriate ranges of P, specified below. The second derivative of ${\mathcal {V}}_2$ with respect to y is given by

$$\begin{aligned} ({\mathcal {V}}_2)_{yy}(y; P) = {\left\{ \begin{array}{ll} D_1 \kappa _{+} \rho _{+} y^{\rho _{+} - 1} + \dfrac{1}{r \eta y}, &{}y \in [0, y_b), \\ D_2 \kappa _{+} \rho _{+} y^{\rho _{+} - 1} + D_3 \kappa _{-} \rho _{-} y^{\rho _{-} - 1} + \dfrac{1}{r \eta y}, &{}y \in (y_b, 1),\\ D_4 \kappa _{-} \rho _{-} y^{\rho _{-} - 1}, &{}y \in [1, \infty ), \end{array}\right. } \end{aligned}$$

If $y \in [1, \infty )$, it is clear that $({\mathcal {V}}_2)_{yy}(y; P) > 0$ because $D_4 > 0$. If $y \in (y_b, 1)$, then, for all $P \in \big ( {\overline{P}}, 0 \big ]$,

$$\begin{aligned} ({\mathcal {V}}_2)_{yy}(y; P)&= \frac{1}{y} \left[ - \, \frac{1}{\delta \eta \rho _{+}(\rho _{+} - \rho _{-})} \, y^{\rho _{+}} + \frac{P \rho _{-}}{\delta (\rho _{+} - \rho _{-}) y_b^{\kappa _{-}}} \, y^{\rho _{-}} + \frac{1}{r \eta } \right] \\&> \frac{1}{y} \left[ - \, \frac{1}{\delta \eta \rho _{+}(\rho _{+} - \rho _{-})} + \frac{P \rho _{-}}{\delta (\rho _{+} - \rho _{-}) y_b^{\kappa _{-}}} + \frac{1}{r \eta } \right] \\&= \frac{1}{y} \left[ \frac{\rho _{+}}{r \eta (\rho _{+} - \rho _{-})} + \frac{P \rho _{-}}{\delta (\rho _{+} - \rho _{-}) y_b^{\kappa _{-}}} \right] > 0, \end{aligned}$$

in which the first inequality follows because the expression in square brackets decreases with respect to y.

It remains to show that $({\mathcal {V}}_2)_{yy}(y; P) > 0$ for $y \in (0, y_b)$. To that end, define ${\mathcal {P}}_2:\mathbb {R}_{+}/\{0\} \rightarrow \mathbb {R}$ by

$$\begin{aligned} {\mathcal {P}}_2(y) = \frac{1}{\eta \rho _{+}^2} \, y^{\kappa _{+}} - b y \delta (\rho _{+} - \rho _{-}) - \delta (\rho _{+} - \rho _{-})\left( \frac{\lambda + \delta - r}{r} + \ln y \right) \dfrac{y}{r \eta } . \nonumber \\ \end{aligned}$$

(A.7)

Note that ${\mathcal {P}}_2(y_b(P)) = P$ when $P \in \big (\, {{\overline{P}}}, 0 \big ]$. From ${\mathcal {P}}_2$’s expression, we obtain

$$\begin{aligned} \left\{ \begin{array} [c]{l} \lim _{y \rightarrow 0+} {\mathcal {P}}_2(y) = 0, \quad {\mathcal {P}}_2(1) = \dfrac{1}{\eta \rho _{-}^2} - b \delta (\rho _{+}-\rho _{-}) = {\overline{P}}< 0, \\ ({\mathcal {P}}_2)_y(y) = \dfrac{\kappa _{+}}{\eta \rho _{+}^2} \, y^{\rho _{+}} - b \delta (\rho _{+} - \rho _{-}) - \delta (\rho _{+} - \rho _{-})\left( \dfrac{\lambda + \delta }{r} + \ln y \right) \dfrac{1}{r \eta },\\ \lim _{y \rightarrow 0+} ({\mathcal {P}}_2)_y(y) = \infty , \quad ({\mathcal {P}}_2)_y(1) = \delta (\rho _{+} - \rho _{-})\left( \dfrac{\kappa _{-}}{\delta \eta \rho _{-}^2(\rho _{+} - \rho _{-})} - b \right) < 0, \\ ({\mathcal {P}}_2)_{yy}(y) = \dfrac{1}{y}\left[ \dfrac{\kappa _{+}}{\eta \rho _{+}} \, y^{\rho _{+}} - \dfrac{\delta (\rho _{+} - \rho _{-})}{r \eta } \right] , \end{array} \right. \end{aligned}$$

in which we use the identity in (A.6) to simplify $({\mathcal {P}}_2)_y(1)$. From the above information, we deduce that ${\mathcal {P}}_2$ has an odd number, $2k + 1$, of zeros in (0, 1), for some $k = 0, 1, \ldots$. It follows that $({\mathcal {P}}_2)_y$ has $2k + 1$ zeros in (0, 1), and $({\mathcal {P}}_2)_{yy}$ has 2k zeros in (0, 1). From the above expression for $({\mathcal {P}}_2)_{yy}$, it is clear that the second derivative changes sign at most once in (0, 1); thus, $k = 0$. Thus, for $P \in \big (\, {{\overline{P}}}, 0 \big ]$, ${\mathcal {P}}_2$ has a unique zero ${{\hat{y}}} = {{\hat{y}}}(P)$ in (0, 1), and ${\mathcal {P}}_2$ is decreasing on $\big [{{\hat{y}}}, 1 \big ]$.

We, then, define

$$\begin{aligned} {\mathcal {Y}}_2 = {\mathcal {P}}_2^{-1}: \big [{\overline{P}}, 0 \big ] \rightarrow \big [{\hat{y}}, 1 \big ], \end{aligned}$$

(A.8)

which is also decreasing, with ${\mathcal {Y}}_2\big ({\overline{P}} \big ) = 1$, ${\mathcal {Y}}_2(0) = {\hat{y}}$, and ${\mathcal {Y}}_2(P) = y_b(P)$ more generally. Then, for $y \in (0, y_b)$,

$$\begin{aligned} ({\mathcal {V}}_2)_{yy}(y; P)&= \dfrac{1}{y} \left[ \dfrac{1}{r \eta } + \frac{\rho _{+}}{\delta (\rho _{+} - \rho _{-})} \left( \frac{P}{y_b^{\kappa _{+}}} - \frac{1}{\eta \rho _{+}^2} \right) y^{\rho _{+}} \right] \\&= \dfrac{1}{y} \left[ \dfrac{1}{r \eta } - \rho _{+} \left( b + \dfrac{\lambda + \delta - r}{r^2 \eta } + \frac{1}{r \eta } \ln y_b\right) \left( \dfrac{y}{y_b} \right) ^{\rho _{+}} \right] \\&\ge \dfrac{1}{y} \left[ \dfrac{1}{r \eta } - \rho _{+} \left( b + \dfrac{\lambda + \delta - r}{r^2 \eta } + \frac{1}{r \eta } \, \ln y_b\right) \right] , \end{aligned}$$

in which the inequality follows because the expression in square brackets decreases with respect to y. Next, define ${\mathcal {G}}_2: \big [ {\overline{P}}, 0 \big ] \rightarrow \mathbb {R}$ by

$$\begin{aligned} {\mathcal {G}}_2(P) = \dfrac{1}{r \eta } - \rho _{+} \left( b + \dfrac{\lambda + \delta - r}{r^2 \eta } + \frac{1}{r \eta } \, \ln {\mathcal {Y}}_2(P) \right) . \end{aligned}$$

(A.9)

It follows that

$$\begin{aligned} \left\{ \begin{array} [c]{l} {\mathcal {G}}_2 \big ({\overline{P}} \big ) = \dfrac{1}{r \eta } - \rho _{+} \left( b + \dfrac{\lambda + \delta - r}{r^2 \eta } \right) , \\ {\mathcal {G}}_2(0) = \dfrac{1}{r \eta } - \rho _{+} \left( b + \dfrac{\lambda + \delta - r}{r^2 \eta } + \dfrac{1}{r \eta } \, \ln {{\hat{y}}} \right) ,\\ ({\mathcal {G}}_2)_P(P) = - \, \dfrac{\rho _{+}}{r \eta } \, \dfrac{1}{{\mathcal {Y}}_2(P)} \, ({\mathcal {Y}}_2)_P(P) > 0. \end{array} \right. \end{aligned}$$

If $b \le - \, \frac{1}{r \eta \rho _{-}}$,

$$\begin{aligned} {\mathcal {G}}_2 \big ({\overline{P}}\big ) \ge \dfrac{1}{r \eta } - \rho _{+} \left( - \, \frac{1}{r \eta \rho _{-}} + \dfrac{\lambda + \delta - r}{r^2 \eta } \right) = 0. \end{aligned}$$

Thus, because ${\mathcal {G}}_2$ increases with respect to P, if $b \le - \, \frac{1}{r \eta \rho _{-}}$ and $P \in \big [ {\overline{P}}, 0 \big ]$, then ${\mathcal {G}}_2(P) \ge 0$, or equivalently, $({\mathcal {V}}_2)_{yy}(y; P) \ge 0$ for $y \in (0, y_b)$. If $b > - \, \frac{1}{r \eta \rho _{-}}$, then ${\mathcal {G}}_2 \big ( {\overline{P}} \big ) < 0$, and there exists ${\hat{P}} \in ({\overline{P}}, 0]$ such that ${\mathcal {G}}_2\big ( {\hat{P}} \big ) = 0$, and for $P > {\hat{P}}$, ${\mathcal {G}}_2(P) > 0$.

To summarize, if $\frac{1}{\delta \eta \rho _{-}^2 (\rho _{+} - \rho _{-})} < b \le - \, \frac{1}{r \eta \rho _{-}} \,$, then there exists a decreasing, convex solution ${\mathcal {V}}_2$ of (A.5) for all $P \in \big ( {\overline{P}}, 0 \big ]$. If $b > - \, \frac{1}{r \eta \rho _{-}} \,$, then there exists such a solution for all $P \in \big [ {{\hat{P}}}, 0 \big ]$. $\square$