Consumption, investment and healthcare with aging

Abstract

This paper solves the problem of optimal dynamic investment, consumption and healthcare spending with isoelastic utility, when natural mortality grows exponentially to reflect the Gompertz law and investment opportunities are constant. Healthcare slows the natural growth of mortality, indirectly increasing utility from consumption through longer lifetimes. Optimal consumption and healthcare imply an endogenous mortality law that is asymptotically exponential in the old-age limit, with lower growth rate than natural mortality. Healthcare spending steadily increases with age, both in absolute terms and relative to total spending. The optimal stochastic control problem reduces to a nonlinear ordinary differential equation with a unique solution, which has an explicit expression in the old-age limit. The main results are obtained through a novel version of Perron’s method.

Appendices

Appendix A: Proofs of main results

In this section, we prove the main results in Sect. 3 by a verification argument, relying on Theorem B.1 and Proposition B.3. Given $f:\mathbb{R}_{+}\to \mathbb{R}$, consider its Legendre transform $\widetilde{f}(y):=\sup _{x\ge 0}\{f(x)-xy\}$ for $y\in \mathbb{R}_{+}$. A heuristic derivation as in Sect. 4.1 shows that the Hamilton–Jacobi equation associated with $V(x,m)$ in (2.8) is

$$\begin{aligned} &\widetilde{U}\big(w_{x}(x,m)\big) +m w(\zeta x,m)-\left (\delta +m\right )w(x,m) + rxw_{x}(x,m) \\ &\quad{}+\beta m w_{m}(x,m)+\sup _{h\ge 0}\left \{ -m w_{m}(x,m)g(h)-h x w _{x}(x,m)\right \}=0. \end{aligned}$$

(A.1)

This is simply (4.9), without the second-order term that stems from the additional risky asset in Sect. 4.

1.1 A.1 Neither aging nor healthcare ($\beta =0$ and $g \equiv 0$)

Recall the setup in Sect. 3.1. Since $V(x,m)$ is nondecreasing in $x$ by definition, the supremum in the last term of (A.1) vanishes, leading to the equation

$$ \widetilde{U}\big(w_{x}(x,m)\big) +m w(\zeta x,m)-(\delta +m)w(x,m) + rxw_{x}(x,m) =0. $$

If $V$ is of the form $V(x,m) = \frac{x^{1-\gamma }}{1-\gamma } v(m)$, the above equation reduces to

$$ v(m)^{1-\frac{1}{\gamma }}- c_{0}(m) v(m)=0, $$

where $c_{0}(m)$ is defined as in (3.4). Setting $v(m)=u(m)^{- \gamma }$, we obtain from the above equation

$$ u^{2}(m) - c_{0}(m)u(m)=0, $$

whence $u(m) = c_{0}(m)$. We then prove Proposition 3.1 by verification.

Proof of Proposition 3.1

Set $w(x,m):=\frac{x^{1- \gamma }}{1-\gamma } c_{0}(m)^{-\gamma }$. Note that (3.3) implies that $c_{0}(m) >0$ for all $\gamma >0$, $\gamma \neq 1$.

Case I:$0<\gamma <1$. By Theorem B.1, it suffices to verify (B.2) and (B.3). For any $x\ge 0$, $c\in \mathcal{C}$ and $n\in \mathbb{N}$, since $0<\gamma <1$,

$$\begin{aligned} 0 &\le \mathbb{E}[e^{-(\delta +m)(t-\tau _{n})} w(\zeta ^{n} X^{0,x,c} _{t},m) \mid Z_{1},\dots ,Z_{n}] \\ &\le e^{-(\delta +m)(t-\tau _{n})} \frac{(\zeta ^{n} X^{0,x,c}_{\tau _{n}})^{1-\gamma }}{1-\gamma } e^{(1-\gamma )r(t-\tau _{n})} c_{0}(m)^{- \gamma }\longrightarrow \ 0 \ \ \hbox{a.s.} \qquad \hbox{as}\ t\to \infty , \end{aligned}$$

where the convergence follows from (3.3). This already verifies (B.2). On the other hand, since $\tau _{n}$ is the sum of $n$ independent, identically distributed exponential random variables with mean $1/m$,

$$\begin{aligned} 0 &\le \mathbb{E}[e^{-\delta \tau _{n}} w(\zeta ^{n} X^{0,x,c}_{\tau _{n}},m)] \le \zeta ^{(1-\gamma )n}\frac{x^{1-\gamma }}{1- \gamma } c_{0}(m)^{-\gamma } \mathbb{E}[e^{-\delta \tau _{n}}e^{(1- \gamma ) r\tau _{n}} ] \\ & = \zeta ^{(1-\gamma )n}\frac{x^{1-\gamma }}{1-\gamma } c_{0}(m)^{- \gamma } \int _{0}^{\infty }e^{-(\delta +(\gamma -1)r) t} m^{n} e^{-mt} \frac{t^{n-1}}{(n-1)!}\,dt \\ &= \frac{(m\zeta ^{(1-\gamma )})^{n}}{(n-1)!}\frac{x^{1-\gamma }}{1- \gamma } c_{0}(m)^{-\gamma } \int _{0}^{\infty }e^{-(\delta +m+(\gamma -1)r) t} t^{n-1} \,dt \\ &= \bigg(\frac{m\zeta ^{(1-\gamma )}}{\delta +m+(\gamma -1)r}\bigg)^{n} \frac{x^{1-\gamma }}{1-\gamma } c_{0}(m)^{-\gamma }, \end{aligned}$$

(A.2)

where the third equality requires $\delta +m+(\gamma -1)r>0$, which is true under (3.3). Since (3.3) also implies $\frac{m\zeta ^{(1-\gamma )}}{\delta +m+(\gamma -1)r}{\in} (0,1)$, we deduce $\mathbb{E}[e^{-\delta \tau ^{m}_{n}} w(\zeta ^{n} X^{0,x,c}_{ \tau ^{m}_{n}},m)]{\to} 0$ as $n\to \infty$, which verifies (B.3).

Case II:$\gamma >1$. By Proposition B.3, it suffices to establish (B.16)–(B.18) and show that $\hat{c}_{t} \equiv c_{0}(m)$ satisfies (B.2) and (B.3). For any $\varepsilon >0$, since $g\equiv 0$, the sequence $(\tau ^{\varepsilon }_{n})_{n\in \mathbb{N}_{0}}$ constructed in Appendix B coincides with $(\tau _{n})_{n\in \mathbb{N}_{0}}$. The counting process $N^{\varepsilon }$ in (B.14) is therefore the same as $N$ in (2.5). It follows that (B.18) trivially holds. Given $x\ge 0$, $c\in \mathcal{C}$ and $\varepsilon >0$, consider

$$ c^{\varepsilon }_{t} := \frac{c_{t} X^{0,x,c}_{t}}{X^{0,x,c}_{t} + \varepsilon e^{rt}}, \qquad \forall t\ge 0. $$

(A.3)

By construction, $X^{0,x+\varepsilon , c^{\varepsilon }}_{t} = X^{0,x, c} + \varepsilon e^{rt}$ for all $t\ge 0$. This together with $\gamma >1$ implies that for any $n\in \mathbb{N}$,

$$\begin{aligned} 0 &\ge \mathbb{E}[e^{-(\delta +m)(t-\tau _{n})} w(\zeta ^{n} X^{0,x+ \varepsilon , c^{\varepsilon }}_{t},m) \mid Z_{1},\dots ,Z_{n}] \\ &\ge e^{-(\delta +m)(t-\tau _{n})} \frac{(\zeta ^{n}\varepsilon )^{1- \gamma }e^{(1-\gamma )rt}}{1-\gamma } c_{0}(m)^{-\gamma }. \end{aligned}$$

Since the right-hand side converges to 0 a.s. as $t\to \infty$, the above inequality in particular implies that (B.16) holds. On the other hand, a calculation similar to (A.2) shows that

$$\begin{aligned} 0 &\ge \mathbb{E}[e^{-\delta \tau _{n}} w(\zeta ^{n} X^{0,x+\varepsilon , c^{\varepsilon }}_{\tau _{n}},m)] \ge \zeta ^{(1-\gamma )n}\frac{ \varepsilon ^{1-\gamma }}{1-\gamma } c_{0}(m)^{-\gamma } \mathbb{E}[e ^{-\delta \tau _{n}}e^{(1-\gamma ) r\tau _{n}} ] \\ &= \bigg(\frac{m\zeta ^{(1-\gamma )}}{\delta +m+(\gamma -1)r}\bigg)^{n} \frac{\varepsilon ^{1-\gamma }}{1-\gamma } c_{0}(m)^{-\gamma }. \end{aligned}$$

By (3.3), $\frac{m\zeta ^{(1-\gamma )}}{\delta +m+( \gamma -1)r}\in (0,1)$. We then have $\mathbb{E}[e^{-\delta \tau _{n}} w(\zeta ^{n} X^{0,x,c}_{\tau _{n}},m)]\to 0$ as $n\to \infty$, which verifies (B.17). Now, with $\hat{c}_{t} \equiv c_{0}(m)$, for any $n\in \mathbb{N}$,

$$\begin{aligned} 0 &\ge e^{-(\delta +m) (t-\tau _{n})} w(X^{0,x,\hat{c}}_{t},m) \\ &= \frac{(X^{0,x,\hat{c}}_{\tau _{n}})^{1-\gamma }}{1-\gamma } c_{0}(m)^{- \gamma } e^{-(\delta +m+(\gamma -1)(r- c_{0}(m)))(t-\tau _{n})}, \qquad \hbox{if $t>\tau _{n}$}. \end{aligned}$$

Observing that (3.3) implies

$$ \delta +m+(\gamma -1)\big(r- c_{0}(m)\big)= c_{0}(m)+ m\zeta ^{1- \gamma }>0, $$

(A.4)

we conclude that $e^{-(\delta +m) (t-\tau _{n})} w(X^{0,x,\hat{c}} _{t},m)\to 0$ a.s. as $t\to \infty$. This shows that $\hat{c}$ satisfies (B.2). Thanks to (A.4), a calculation similar to (A.2) yields

$$\begin{aligned} 0 &\ge \mathbb{E}[e^{-\delta \tau _{n}} w(\zeta ^{n} X^{0,x,\hat{c}} _{\tau _{n}},m)] = \zeta ^{(1-\gamma )n}\frac{x^{1-\gamma }}{1-\gamma } c_{0}(m)^{-\gamma } \mathbb{E}[e^{-\delta \tau _{n}}e^{(1-\gamma ) (r- c_{0}(m))\tau _{n}} ] \\ &= \bigg(\frac{m\zeta ^{1-\gamma }}{ c_{0}(m)+ m\zeta ^{1-\gamma }} \bigg)^{n} \frac{x^{1-\gamma }}{1-\gamma } c_{0}(m)^{-\gamma }\longrightarrow 0 \qquad \hbox{as}\ n\to \infty , \end{aligned}$$

which shows that $\hat{c}$ satisfies (B.3). □

1.2 A.2 Aging without healthcare ($g\equiv 0$)

Recall the setup in Sect. 3.2. The Hamilton–Jacobi equation (A.1) associated with the value function $V(x,m)$ now becomes

$$ \widetilde{U}\big(w_{x}(x,m)\big) +m w(\zeta x,m)-(\delta +m)w(x,m)+ rxw _{x}(x,m)+\beta m w_{m}(x,m) =0. $$

If $V$ is of the form $V(x,m) = \frac{x^{1-\gamma }}{1-\gamma } v(m)$, the above equation reduces to

$$ v(m)^{1-\frac{1}{\gamma }}- c_{0}(m) v(m) + \frac{\beta m}{\gamma } v'(m)=0, $$

where $c_{0}(m)$ is given by (3.4). Setting $v(m)=u(m)^{- \gamma }$, we obtain from the above equation

$$ u^{2}(m) - c_{0}(m)u(m)-\beta m u'(m)=0, $$

(A.5)

which admits the general solution

$$ u(m) = \beta e^{-\frac{(1-\zeta ^{1-\gamma })m}{\beta \gamma }} \bigg(C m^{\frac{\delta +(\gamma -1)r}{\beta \gamma }}+ \int _{1}^{\infty }e ^{-\frac{(1-\zeta ^{1-\gamma })mu}{\beta \gamma }} u^{-(1+ \frac{ \delta +(\gamma -1)r}{\beta \gamma })} \,du\bigg)^{-1}, $$

where $C\in \mathbb{R}$ is a constant to be determined. Proposition 3.2 states that taking $C=0$, which turns $u(m)$ into $u_{0}(m)$ in (3.6), leads to the value function $V(x,m)$. In the following, we separate the proof of Proposition 3.2 into two parts.

Lemma A.1

Under the assumptions of Proposition 3.2, $u_{0}$defined in (3.6) is a strictly increasing function on$(0,\infty )$satisfying (a) and (b) in Proposition 3.2.

Proof

The definition of $u_{0}$ in (3.6) directly implies that $u_{0}$ is strictly increasing, $u_{0}(0)= \frac{\delta +(\gamma -1)r}{ \gamma }$ and $u_{0}'(0+)=\infty$. Since $u_{0}$ solves (A.5), for any $m\in (0,\infty )$,

$$ u_{0}(m) - c_{0}(m) = \frac{\beta m u_{0}'(m)}{u_{0}(m)} >0, $$

where the inequality follows from $u_{0}$ being positive and strictly increasing. On the other hand, using $y+1 < e^{y}$ for $y > 0$, (3.6) yields

$$\begin{aligned} u_{0}(m) &< \beta \bigg(\int _{0}^{\infty }\exp \Big(-\Big(\frac{(1- \zeta ^{1-\gamma })m}{\beta \gamma }+1+ \frac{\delta +(\gamma -1)r}{ \beta \gamma }\Big) y\Big) \,dy\bigg)^{-1} \\ & = \beta \bigg(\frac{\delta +(1-\zeta ^{1-\gamma })m+(\gamma -1)r}{ \beta \gamma }+1\bigg) = c_{0}(m) +\beta . \end{aligned}$$

Finally, the asymptotic expansion of $u_{0}(m)$ at infinity is

$$ u_{0}(m) = c_{0}(m) +\beta + O(1/m). $$

This implies $u_{0}(m) - ( c_{0}(m) +\beta ) \to 0$ as $m\to \infty$ and

$$ \lim _{m\to \infty } u_{0}'(m) = \lim _{m\to \infty } \frac{u_{0}(m)}{m} = \frac{1-\zeta ^{1-\gamma }}{\gamma }. $$

 □

Proof of Proposition 3.2

Set $w(x,m) := \frac{x^{1- \gamma }}{1-\gamma } u_{0}(m)^{-\gamma }$. By Lemma A.1, it remains to show that $V(x,m) = w(x,m)$ and $\hat{c}_{t} := u_{0}(m e^{\beta t})$, $t\ge 0$, is an optimal control for (3.2). First, we observe that $\hat{c}$ is an element of $\mathcal{C}$. Indeed, for any compact subset $K$ of $\mathbb{R}_{+}$, thanks to $u_{0} \le c_{0} +\beta$ in Lemma A.1,

$$ \int _{K} \hat{c}_{t} \,dt \le \int _{K} \bigg(\frac{\delta +(1- \zeta ^{1-\gamma }) me^{\beta t}+(\gamma -1)r}{\gamma } + \beta \bigg)\,dt < \infty . $$

Now we deal with two cases separately.

Case I: Condition (i) holds. By Theorem B.1, it suffices to verify (B.2) and (B.3). For any $(x,m)\in \mathbb{R}^{2}_{+}$, $c\in \mathcal{C}$ and $n\in \mathbb{N}$, as $X^{0,x,c}_{t} \le X^{0,x,c}_{\tau _{n}}\exp (r(t-\tau _{n}))$ on the set $\{t\ge \tau _{n}\}$ and as $\gamma \in (0,1)$, we get

$$\begin{aligned} 0 &\le \mathbb{E}\bigg[\exp \bigg(-\int _{\tau _{n}}^{t} (\delta + m e ^{\beta s}) \,ds\bigg) w(\zeta ^{n} X^{0,x,c}_{t}, me^{\beta t}) \Bigm| Z _{1},\dots ,Z_{n}\bigg] \\ &\le e^{-(\delta +m)(t-\tau _{n})} \frac{(\zeta ^{n} X^{0,x,c}_{\tau _{n}})^{1-\gamma }}{1-\gamma } e^{(1-\gamma )r(t-\tau _{n})} u_{0}(me ^{\beta t})^{-\gamma }\longrightarrow 0 \qquad \hbox{a.s.}\ \hbox{as}\ t\to \infty , \end{aligned}$$

where the convergence follows from $\delta + (\gamma -1) r >0$ and $u_{0}$ being nondecreasing by definition. This in particular implies (B.2). On the other hand,

$$\begin{aligned} 0 &\le \mathbb{E}[e^{-\delta \tau _{n}} w(\zeta ^{n} X^{0,x,c}_{\tau _{n}},me ^{\beta \tau _{n}})] \\ &\le \zeta ^{(1-\gamma )n}\frac{x^{1-\gamma }}{1-\gamma } \mathbb{E}[e ^{-\delta \tau _{n}}e^{(1-\gamma ) r\tau _{n}} u_{0}(me^{\beta \tau _{n}})^{- \gamma }] \\ &\le \zeta ^{(1-\gamma )n}\frac{x^{1-\gamma }}{1-\gamma } u_{0}(m)^{- \gamma } \mathbb{E}[e^{-(\delta +(\gamma -1)r) \tau _{n}}] \\ &\le \zeta ^{(1-\gamma )n}\frac{x^{1-\gamma }}{1-\gamma } u_{0}(m)^{- \gamma } \longrightarrow 0 \qquad \hbox{as}\ n\to \infty , \end{aligned}$$

where the fourth inequality follows from $\delta + (\gamma -1) r >0$ and the convergence from $\gamma ,\zeta \in (0,1)$. Thus (B.3) is satisfied.

Case II: Condition (ii) holds. Due to Proposition B.3, it suffices to establish (B.16)–(B.18) and show that $\hat{c} _{t} := u_{0}(me^{\beta t})$ satisfies (B.2) and (B.3). For any $\varepsilon >0$, since $g\equiv 0$, the sequence $(\tau ^{\varepsilon }_{n})_{n\in \mathbb{N} _{0}}$ constructed in Appendix B coincides with $(\tau _{n})_{n\in \mathbb{N}_{0}}$. The counting process $N^{\varepsilon }$ in (B.14) is therefore the same as $N$ in (2.5). Thus (B.18) trivially holds in the current context. Given $(x,m)\in \mathbb{R}^{2}_{+}$, $c\in \mathcal{C}$ and $\varepsilon >0$, consider the consumption policy $c^{\varepsilon }$ as in (A.3) and the property $X^{0,x+ \varepsilon , c^{\varepsilon }}_{t} = X^{0,x,c}_{t} + \varepsilon e ^{rt}$ for all $t\ge 0$. We then deduce from $\gamma >1$ and $u_{0}$ being a nondecreasing function that for any $n\in \mathbb{N}$,

$$\begin{aligned} 0 &\ge \mathbb{E}\bigg[\exp \left (-\int _{\tau _{n}}^{t} (\delta + m e ^{\beta s}) \,ds\right ) w(\zeta ^{n} X^{0,x+\varepsilon , c^{\varepsilon }}_{t},me^{\beta t}) \biggm| Z_{1},\dots ,Z_{n}\bigg] \\ &\ge e^{-(\delta +m)(t-\tau _{n})} \frac{(\zeta ^{n}\varepsilon )^{1- \gamma }e^{(1-\gamma )rt}}{1-\gamma } u_{0}(m)^{-\gamma }\longrightarrow 0 \ \ \hbox{a.s.} \qquad \hbox{as}\ t\to \infty , \end{aligned}$$

which in particular implies (B.16). Since $\gamma >1$ ensures $\delta +(\gamma -1)r>0$,

$$\begin{aligned} 0 &\ge \mathbb{E}[e^{-\delta \tau _{n}} w(\zeta ^{n} X^{0,x+\varepsilon , c^{\varepsilon }}_{\tau _{n}},me^{\beta \tau _{n}})]\ge \zeta ^{(1-\gamma )n}\frac{\varepsilon ^{1-\gamma }}{1-\gamma } u_{0}(m)^{- \gamma } \mathbb{E}[e^{-(\delta +(\gamma -1)r)\tau _{n}}] \\ & \ge \zeta ^{(1-\gamma )n}\frac{\varepsilon ^{1-\gamma }}{1-\gamma } u _{0}(m)^{-\gamma }\longrightarrow 0 \qquad \hbox{as}\ n\to \infty , \end{aligned}$$

where the convergence follows from $\gamma ,\zeta >1$. This verifies (B.17). Now for any $n\in \mathbb{N}$, applying $\hat{c}_{t} := u_{0}(me^{\beta t})$, $t\ge 0$, yields

$$\begin{aligned} 0 &\ge \mathbb{E}\bigg[\exp \left (-\int _{\tau _{n}}^{t} (\delta + m e ^{\beta s}) \,ds\right ) w(X^{0,x,\hat{c}}_{t},me^{\beta t}) \biggm| Z _{1},\dots ,Z_{n}\bigg] \\ &\ge e^{-(\delta +(\gamma -1)r) (t-\tau _{n})}\exp \bigg( - \int _{\tau _{n}}^{t} \big(m e^{\beta s}-(\gamma -1)u_{0}(me^{\beta s}) \big) \,ds\bigg) \\ & \phantom{=}{}\times \frac{(X^{0,x,\hat{c}}_{\tau _{n}})^{1-\gamma }}{1- \gamma } u_{0}(m)^{-\gamma } \end{aligned}$$

(A.6)

on the set $\{t \ge \tau _{n}\}$. By direct calculation, $u_{0}(m)$ defined in (3.6) equals

$$ u_{0}(m) = \beta \frac{e^{-\frac{m(1-\zeta ^{1-\gamma })}{\beta \gamma }}(\frac{m(1-\zeta ^{1-\gamma })}{\beta \gamma })^{-\frac{\delta +( \gamma -1)r}{\beta \gamma }}}{\overline{\varGamma }(-\frac{\delta +( \gamma -1)r}{\beta \gamma },\frac{m(1-\zeta ^{1-\gamma })}{\beta \gamma })}, $$

where $\overline{\varGamma }(s,z):=\int _{z}^{\infty }t^{s-1} e^{-t} \,dt$ is the upper incomplete gamma function. Recalling the property $\frac{\overline{\varGamma }(s,z)}{e^{-z}z^{s-1}}\to 1$ as $z\to \infty$, it follows that

$$ \lim _{m\to \infty } \frac{m}{\gamma u_{0}(m)}= \frac{1}{1- \zeta ^{1-\gamma }} >1, $$

(A.7)

whence $me^{\beta s} > \gamma u_{0}(m e^{\beta s})$ for $s$ large enough. This together with $u_{0}$ being a positive nonincreasing function shows that $\int _{\tau _{n}}^{t} (m e^{\beta s}-(\gamma -1)u _{0}(me^{\beta s})) \,ds\to \infty$ a.s. as $t\to \infty$. We then conclude from (A.6) that $\hat{c}$ satisfies (B.2). It remains to show that $\hat{c}$ satisfies (B.3). Observe that

$$\begin{aligned} 0 \ge & \mathbb{E}[e^{-\delta \tau _{n}} w(\zeta ^{n} X^{0,x,\hat{c}} _{\tau _{n}},me^{\beta \tau _{n}})] \\ \ge & \zeta ^{(1-\gamma )n}\frac{x^{1-\gamma }}{1-\gamma } u_{0}(m)^{- \gamma } \mathbb{E}\left [e^{-(\delta +(\gamma -1)r) \tau _{n}} \exp \left (\int _{0}^{\tau _{n}} (\gamma -1)u_{0}(me^{\beta s})\,ds\right ) \right ] \\ \ge & \zeta ^{(1-\gamma )n}\frac{x^{1-\gamma }}{1-\gamma } u_{0}(m)^{- \gamma } \int _{0}^{\infty }\mathbb{P}[\tau _{n}\in \,dt] \exp \left (\int _{0}^{t} (\gamma -1)u_{0}(me^{\beta s})\,ds\right ) \,dt, \end{aligned}$$

(A.8)

where the third line follows from $e^{-(\delta +(\gamma -1)r) \tau _{n}}\le 1$ as $\gamma >1$. For each $n\in \mathbb{N}$, note that $\sum _{i=1}^{n} Z_{i}$ has a gamma distribution with the law $\mathbb{P}[\sum _{i=1}^{n} Z_{i}\le z] = \frac{1}{\varGamma (n)} \underline{\varGamma}(n, z)$, where $\varGamma (s):=\int _{0}^{\infty }t^{s-1} e^{-t} \,dt$ is the gamma function and $\underline{\varGamma}(s,z):=\int _{0}^{z} t ^{s-1} e^{-t} \,dt$ the lower incomplete gamma function. We then observe from (3.5) that

$$ \mathbb{P}[\tau _{n}\le t] = \mathbb{P}\bigg[ \frac{m}{\beta } (e^{ \beta t}-1) \ge \sum _{i=1}^{n} Z_{i} \bigg] = \frac{1}{\varGamma (n)} \underline{ \varGamma }\left (n, \frac{m}{\beta }(e^{\beta t}-1)\right ), \qquad \ \forall t\ge 0. $$

Differentiating $\mathbb{P}[\tau _{n}\le t]$ with respect to $t$ gives

$$ \mathbb{P}[\tau _{n}\in dt] = \frac{1}{\varGamma (n)} \left (\frac{m}{ \beta }\right )^{n} (e^{\beta t}-1)^{n-1} e^{-\frac{m}{\beta }(e^{ \beta t}-1)} \beta e^{\beta t}, \qquad \forall t\ge 0. $$

(A.9)

Also, for any

$$ \alpha \in \big((1-1/\gamma )(1-\zeta ^{1-\gamma }),1\big), $$

(A.10)

(A.7) yields $\frac{m}{\gamma u_{0}(m)}> \frac{\alpha }{1- \zeta ^{1-\gamma }}$ for $m$ large enough. Thus there exists $t^{*}>0$ such that

$$ \frac{1-\zeta ^{1-\gamma }}{\alpha \gamma }m e^{\beta t}> u_{0}(me^{ \beta t}) \qquad \hbox{for}\ t\ge t^{*}. $$

(A.11)

Setting $C(t^{*}):=\exp (\int _{0}^{t^{*}} (\gamma -1)u_{0}(me^{\beta s})\,ds)$, we obtain from (A.8) that

$$\begin{aligned} 0 &\ge \mathbb{E}[e^{-\delta \tau _{n}} w(\zeta ^{n} X^{0,x,\hat{c}}_{ \tau _{n}},me^{\beta \tau _{n}})] \\ &\ge \zeta ^{(1-\gamma )n}\frac{x^{1-\gamma }}{1-\gamma } \frac{u_{0}(m)^{- \gamma }}{\varGamma (n)} \left (\frac{m}{\beta }\right )^{n} \bigg(C(t ^{*})\int _{0}^{t^{*}} (e^{\beta t}-1)^{n-1} e^{-\frac{m}{\beta }(e ^{\beta t}-1)} \beta e^{\beta t} \,dt \\ & \hspace{0.8in}{} +\int _{t^{*}}^{\infty }(e^{\beta t}-1)^{n-1} e^{-\frac{m}{\beta }(e ^{\beta t}-1)} \beta e^{\beta t} e^{(1-\frac{1}{\gamma })(1- \zeta ^{1-\gamma })\frac{m}{\alpha \beta }(e^{\beta t}-1)} \,dt\bigg) \\ &\ge \zeta ^{(1-\gamma )n}\frac{x^{1-\gamma }}{1-\gamma } \frac{u_{0}(m)^{- \gamma }}{\varGamma (n)} \left (\frac{m}{\beta }\right )^{n} \bigg(C(t ^{*})\int _{0}^{\infty }y^{n-1} e^{-\frac{m}{\beta }y} \,dy \\ & \phantom{::::\zeta ^{(1-\gamma )n}\frac{x^{1-\gamma }}{1-\gamma } \frac{u_{0}(m)^{-\gamma }}{\varGamma (n)} \bigg(\frac{m}{\beta }\bigg)^{n} \bigg(}+ \int _{0}^{\infty }y^{n-1} e^{-(1-\frac{1}{\alpha }(1-\frac{1}{ \gamma })(1-\zeta ^{1-\gamma }))\frac{m}{\beta }y}\,dy \bigg) \\ &= \zeta ^{(1-\gamma )n}\frac{x^{1-\gamma }}{1-\gamma }u_{0}(m)^{- \gamma } \bigg(C(t^{*}) + \Big(1-\frac{1}{\alpha }\Big(1-\frac{1}{ \gamma }\Big)(1-\zeta ^{1-\gamma })\Big)^{-n}\bigg), \end{aligned}$$

where the second line follows from (A.9) and (A.11), the fourth is due to $y:=e^{\beta t}-1$, and the last equality holds as $1-\frac{1}{\alpha }(1-\frac{1}{\gamma })(1- \zeta ^{1-\gamma })>0$ thanks to (A.10). Noting that (A.10) gives $1-\frac{1}{\alpha }(1-\frac{1}{\gamma })(1- \zeta ^{1-\gamma })> 1-(1-\zeta ^{1-\gamma }) =\zeta ^{1-\gamma }$, we conclude from the above inequality that $\mathbb{E}[e^{-\delta \tau _{n}} w(\zeta ^{n} X^{0,x,\hat{c}}_{\tau _{n}},me^{\beta \tau _{n}})] \to 0$ as $n\to \infty$, i.e., $\hat{c}$ satisfies (B.3). □

1.3 A.3 Aging with healthcare

Recall the setup in Sect. 3.3, where mortality increases naturally according to the Gompertz law ($\beta >0$) and at the same time healthcare is available (i.e., $g:\mathbb{R}_{+}\to \mathbb{R}_{+}$ is not constantly 0) to slow down the mortality growth. For the rest of this section, let Assumption 3.3 hold and denote by $I:\mathbb{R}_{+}\to \mathbb{R} _{+}$ the inverse function of $g'$. Note that $I$ is strictly decreasing.

If the value function (2.8) has the form $V(x,m) = \frac{x ^{1-\gamma }}{1-\gamma } v(m)$, (A.1) yields

$$\begin{aligned} &v(m)^{1-\frac{1}{\gamma }}- c_{0}(m) v(m) + \frac{\beta m}{\gamma } v'(m) \\ &\quad{} + \frac{1-\gamma }{\gamma }\sup _{h\ge 0}\left \{-\frac{v'(m)}{1- \gamma }\left (m g(h)+h\frac{(1-\gamma ) v(m)}{v'(m)}\right )\right \} =0, \end{aligned}$$

where $c_{0}(m)$ is given by (3.4). By setting $v(m)=u(m)^{- \gamma }$ and assuming $u'(m)\ge 0$, the above equation becomes

$$\begin{aligned} &u^{2}(m) - c_{0}(m)u(m)-\beta m u'(m) \\ &\quad{}+ \textstyle\begin{cases} m u'(m) \sup \limits _{h\ge 0}\{g(h) - \frac{1-\gamma }{\gamma } \frac{u(m)}{mu'(m)} h\}=0, &\quad \hbox{if}\ 0< \gamma < 1, \\ m u'(m) \inf \limits _{h\ge 0}\{g(h) - \frac{1-\gamma }{\gamma } \frac{u(m)}{mu'(m)} h\}=0, &\quad \hbox{if}\ \gamma >1. \end{cases}\displaystyle \end{aligned}$$

(A.12)

Since $g$ is a nondecreasing function with $g(0)=0$, the infimum above equals 0. That is, when $\gamma >1$, the above equation reduces to (A.5), and the associated value function and optimal consumption strategy are as described in Proposition 3.2.

In consequence, we focus on the case $0 < \gamma < 1$in the rest of this section. Equation (A.12) is now $\mathcal{L}u(m) = 0$ as in (3.8). In the following, we employ Perron’s method to construct solutions to (3.8) under the assumption that

$$ \bar{c}:= \frac{\delta }{\gamma } + \left (1-\frac{1}{\gamma }\right ) r >0. $$

(A.13)

Definition A.2

Let $\varPi$ be the collection of $(p,q)$, where $p,q:\mathbb{R}_{+} \to \mathbb{R}$ are continuous and satisfy

• 1. $c_{0}\le p\le q\le c_{0} + \beta$ on $(0,\infty )$.

• 2. $p$ and $q$ are strictly increasing and concave.

• 3. $p$ (resp. $q$) is a viscosity subsolution (resp. supersolution) to (3.8) on $(0,\infty )$.

For any $(p,q)\in \varPi$, let $\mathcal{S}(p,q)$ denote the collection of continuous $f:\mathbb{R}_{+}\to \mathbb{R}$ such that

• 1. $p\le f\le q$ on $(0,\infty )$.

• 2. $f$ is strictly increasing and concave.

• 3. $f$ is a viscosity supersolution to (3.8) on $(0,\infty )$.

Remark A.3

Under (A.13) and (3.7), $\varPi \neq \emptyset$. Indeed, $c_{0}+\beta$ is a supersolution to (A.5) and thus a supersolution to (3.8). Specifically, for any $m>0$,

$$ \mathcal{L}(c_{0}+\beta )(m)\ge \big(c_{0}(m)+\beta \big)\beta - \beta m \left (\frac{1-\zeta ^{1-\gamma }}{\gamma }\right ) = \beta \bar{c}+\beta ^{2}>0. $$

On the other hand, $c_{0}$ is a subsolution to (3.8); indeed, for any $m>0$,

$$\begin{aligned} \mathcal{L}c_{0}(m) &= am\bigg(\sup _{h\ge 0}\bigg\{ g(h) - \frac{1- \gamma }{\gamma } \frac{\bar{c}}{a m} h\bigg\} -\beta \bigg) \\ &= am \bigg( g\Big(I\Big(\frac{1-\gamma }{\gamma } \big(1+\frac{ \bar{c}}{am}\big)\Big)\Big) \\ & \phantom{=:am \bigg(}{} - \frac{1-\gamma }{\gamma } \Big(1+\frac{\bar{c}}{am}\Big) I\Big(\frac{1- \gamma }{\gamma } \big(1+\frac{\bar{c}}{am}\big)\Big)-\beta \bigg) < 0, \end{aligned}$$

where $a:= \frac{1-\zeta ^{1-\gamma }}{\gamma }$ and the inequality follows from (3.7). Thus $\varPi$ contains at least $(c_{0},c _{0} +\beta )$. Also note that for each $(p,q)\in \varPi$, we have $\mathcal{S}(p,q)\neq \emptyset$ because by construction $q\in \mathcal{S}(p,q)$.

We first present a basic result for strictly increasing concave functions $f$ bounded by $c_{0}$ and $c_{0}+\beta$. Note that the concavity of $f$ implies $f'(\infty ) := \lim _{m\to \infty } f'(m-)$ is well defined.

Lemma A.4

Assume$0<\gamma <1$and (A.13). For any nonnegative, strictly increasing and concave$f:\mathbb{R}_{+}\to \mathbb{R}$, we have$\frac{f(m)}{m f'(m-)}\ge 1$for all$m\in (0,\infty )$. If$f$additionally satisfies$c_{0}\le f\le c_{0}+\beta$on$(0,\infty )$, then$f'(\infty ) =\frac{1-\zeta ^{1-\gamma }}{\gamma }$and$\frac{f(m)}{mf'(m-)}\to 1$as$m\to \infty$.

Proof

With $f$ being strictly increasing and $f(0)\ge 0$, the concavity of $f$ implies $\frac{f(m)}{m}\ge f'(m-) >0$ and thus $\frac{f(m)}{m f'(m-)}\ge 1$ for all $m\in (0,\infty )$. Suppose $f$ also lies between $c_{0}$ and $c_{0}+\beta$. Since $c_{0}$ is a linear function with slope $\frac{1-\zeta ^{1-\gamma }}{\gamma }$, if $f'( \infty ) \neq \frac{1-\zeta ^{1-\gamma }}{\gamma }$, then $f(m)\notin [c_{0}(m), c_{0}(m)+\beta ]$ for $m$ large enough, which is a contradiction. Moreover, we deduce from $c_{0}(m)\le f(m)\le c_{0}(m)+ \beta$ and $f'(\infty ) = \frac{1-\zeta ^{1-\gamma }}{\gamma }>0$ that

$$ 1+ \frac{\delta +(\gamma -1)r}{(1-\zeta ^{1-\gamma }) m}\le \frac{f(m)}{m f'(\infty )} \le 1+\frac{\delta +(\gamma -1)r+\beta \gamma }{(1- \zeta ^{1-\gamma }) m}, \qquad m>0. $$

This implies $\frac{f(m)}{mf'(\infty )}\to 1$ as $m\to \infty$. □

The next result shows that $c_{0}+\alpha$ is a supersolution to (3.8) on $(0,\infty )$ for $\alpha$ large enough.

Lemma A.5

Assume$0<\gamma <1$, (A.13) and (3.7). For any$\alpha \in [0,\beta ]$, $c_{0}+\alpha$is a supersolution to (3.8) on$(0,\infty )$if and only if$\alpha \in [\beta _{g}, \beta ]$, where$\beta _{g}$is defined in (3.10). Specifically,

$$\begin{aligned} \alpha \in [\beta _{g}, \beta ] &\quad \implies \quad \mathcal{L}({c_{0}+\alpha })(m)>0\textit{ for all }m>0, \\ \alpha \in [0,\beta _{g}) &\quad \implies \quad \mathcal{L}( {c_{0}+\alpha })(m)\to -\infty \textit{ as }m\to \infty . \end{aligned}$$

Proof

For any $a,b>0$, consider the function

$$\begin{aligned} \theta (m) &:= \mathcal{L}(am+b) \\ & \phantom{:}= (am+ b)\bigg(\Big(a-\frac{1-\zeta ^{1-\gamma }}{\gamma } \Big) m + (b-\bar{c})\bigg) + am \big(\ell (m)-\beta \big), \end{aligned}$$

(A.14)

where

$$\begin{aligned} \ell (m) &:= \sup _{h\ge 0}\bigg\{ g(h)-\frac{1-\gamma }{\gamma } \bigg(1+\frac{b}{am} \bigg) h\bigg\} \\ & \phantom{:}= g\bigg(I\Big(\frac{1-\gamma }{\gamma }\Big(1+ \frac{b}{am} \Big)\Big)\bigg)-\frac{1-\gamma }{\gamma }\bigg(1+ \frac{b}{am} \bigg) I\bigg(\frac{1-\gamma }{\gamma }\Big(1+ \frac{b}{am} \Big)\bigg). \end{aligned}$$

(A.15)

By direct calculation,

$$\begin{aligned} \ell '(m) &= \frac{1-\gamma }{\gamma }\frac{b}{a m^{2}} I\left (\frac{1- \gamma }{\gamma }\Big(1+\frac{b}{am} \Big)\right ), \end{aligned}$$

(A.16)

$$\begin{aligned} \theta '(m) &= 2a\bigg(a-\frac{1-\zeta ^{1-\gamma }}{\gamma }\bigg)m + a(b-\bar{c}) + b\left (a-\frac{1-\zeta ^{1-\gamma }}{\gamma }\right ) \\ & \phantom{=}{}+ a \bigg(\ell (m)-\beta + \frac{1-\gamma }{\gamma } \frac{b}{am} I\Big(\frac{1-\gamma }{\gamma }\Big(1+\frac{b}{a m} \Big) \Big)\bigg) , \end{aligned}$$

(A.17)

$$\begin{aligned} \theta ''(m) &= 2a\left (a-\frac{1-\zeta ^{1-\gamma }}{\gamma }\right ) - \left (\frac{1-\gamma }{\gamma }\right )^{2}\frac{b^{2}}{a m^{3}} I' \left (\frac{1-\gamma }{\gamma }\Big(1+\frac{b}{a m}\Big) \right ) \\ &>0, \end{aligned}$$

(A.18)

where the positivity follows from $I'(y) = \frac{d}{dy}(g')^{-1}(y) = \frac{1}{g''((g')^{-1}(y))}= \frac{1}{g''(I(y))}$ and $g''<0$.

For any $\alpha \in [0,\beta ]$, $c_{0}(m)+\alpha = am+b$ with $a=\frac{1-\zeta ^{1-\gamma }}{\gamma }$ and $b=\bar{c}+\alpha$. Then (A.14) reduces to

$$ \theta (m) = am \Big(\alpha -\big(\beta -\ell (m)\big)\Big) + \alpha b. $$

(A.19)

Observe from (A.15) that $\beta -\ell (m)\to \beta _{g}$ as $m\to \infty$. It follows that

$$ \lim _{m\to \infty } \mathcal{L}(c_{0}+\alpha )(m) = \lim _{m\to \infty }\theta (m) = \textstyle\begin{cases} -\infty ,\quad &\hbox{if}\ \alpha \in [0,\beta _{g}), \\ +\infty ,\quad &\hbox{if}\ \alpha \in (\beta _{g},\beta ]. \end{cases} $$

(A.20)

This already shows that if $\alpha \in [0,\beta _{g})$, $c_{0}+\alpha$ cannot be a supersolution to (3.8) on $(0,\infty )$.

It remains to show that if $\alpha \in [\beta _{g},\beta ]$, $c_{0}+\alpha$ satisfies $\mathcal{L}({c_{0}+\alpha })(m)>0$ for all $m>0$. This is true for $\alpha = \beta$, as explained in Remark A.3. For any $\alpha \in (\beta _{g},\beta )$, using $a=\frac{1-\zeta ^{1-\gamma }}{\gamma }$ and $b=\bar{c}+\alpha$ in the current setting and (A.15), (A.17) becomes

$$ \theta '(m) = a\bigg(\alpha -\beta +g\Big(I\Big(\frac{1-\gamma }{ \gamma }\big(1+\frac{b}{a m} \big)\Big)\Big)-\frac{1-\gamma }{\gamma } I\Big(\frac{1-\gamma }{\gamma }\big(1+\frac{b}{a m} \big)\Big) \bigg), $$

(A.21)

implying $\theta '(m)\to \alpha -\beta <0$ as $m\downarrow 0$. This together with $\lim _{m\to \infty }\theta (m)=\infty$ in (A.20) and $\theta ''(\cdot )>0$ in (A.18) shows that $\theta$ must attain a global minimum at some $m^{*}\in (0,\infty )$. Using $\theta '(m^{*})=0$, we obtain from (A.21) that

$$ \alpha - \frac{1-\gamma }{\gamma } I\left (\frac{1-\gamma }{\gamma } \Big(1+\frac{b}{a m^{*}} \Big)\right ) = \beta -g\bigg(I\Big(\frac{1- \gamma }{\gamma }\Big(1+\frac{b}{am^{*}} \Big)\Big)\bigg). $$

(A.22)

The global minimum is then computed as

$$\begin{aligned} \theta (m^{*}) &= am^{*}\big(\alpha -\beta + \ell (m)\big)+ \alpha b\\ &= am^{*}\bigg(\alpha - \beta +g\Big(I\Big(\frac{1-\gamma }{\gamma } \big(1+\frac{b}{am^{*}} \big)\Big)\Big) \\ & \phantom{=:am^{*}\bigg(}{}-\frac{1-\gamma }{\gamma } \Big(1+\frac{b}{am ^{*}} \Big)I\Big(\frac{1-\gamma }{\gamma }\big(1+\frac{b}{am^{*}} \big)\Big)\bigg) +\alpha b \\ &= b\bigg(\alpha -\frac{1-\gamma }{\gamma } I\Big(\frac{1-\gamma }{ \gamma }\big(1+\frac{b}{am^{*}} \big)\Big) \bigg) \\ &= b\bigg(\beta -g\Big(I\Big(\frac{1-\gamma }{\gamma }\big(1+\frac{b}{am ^{*}} \big)\Big)\Big)\bigg) >0, \end{aligned}$$

where the second equality comes from (A.15), the third and fourth follow from (A.22), and the final inequality is due to (3.7). We thus conclude that for any $\alpha \in (\beta _{g}, \beta )$, $\mathcal{L}(c_{0}+\alpha )(m) = \theta (m) \ge \theta (m ^{*}) >0$ for all $m\in (0,\infty )$. Finally, for the case $\alpha =\beta _{g}$, since $c_{0}+\beta _{g}$ is the pointwise infimum of the supersolutions $c_{0}+\alpha$, $\alpha \in (\beta _{g},\beta ]$, it must also be a supersolution thanks to Crandall et al. [7, Lemma 4.2]. Observe from (A.21) that $\lim _{m\uparrow \infty }\theta '(m) = a( \beta _{g}-\beta _{g}) =0$. This together with $\theta ''>0$ in (A.18) and $c_{0}+\beta _{g}$ being a supersolution shows that $\theta (m)$ must be strictly decreasing on $(0,\infty )$ with $\lim _{m\uparrow \infty }\theta (m)\ge 0$. As a result, $\mathcal{L}(c _{0}+\beta _{g})(m)=\theta (m)>0$ for all $m>0$. □

Lemma A.6

Assume$0<\gamma <1$, (A.13) and (3.7). For any$(p,q)\in \varPi$, we then have$p < c_{0}+\beta _{g}$on$\mathbb{R} _{+}$, with$\beta _{g}$defined in (3.10).

Proof

By contradiction, suppose that ‘$p < c_{0}+\beta _{g}$ on $\mathbb{R} _{+}$’ does not hold. If there exists $m_{0}>0$ such that $p(m)= c _{0}(m)+\beta _{g}$ for all $m\ge m_{0}$, then the subsolution property of $p$ is violated, thanks to Lemma A.5. Thus it remains to deal with the second case: there exists $m_{0}>0$ such that $p(m)> c_{0}(m)+\beta _{g}$ for $m>m_{0}$.

Since $p'(\infty )=\frac{1-\zeta ^{1-\gamma }}{\gamma }$ by Lemma A.4 (i), there must exist $\alpha _{0} \in (\beta _{g},\beta ]$ such that $p\le c_{0}+\alpha _{0}$ and $(c_{0}(m)+\alpha )-p_{0}(m)\downarrow 0$ as $m\to \infty$. Consider the collection of functions $\{c_{0}+\alpha :\alpha \in (\frac{\alpha _{0}+\beta _{g}}{2},\alpha _{0})\}$. For each $\alpha \in (\frac{\alpha _{0}+\beta _{g}}{2},\alpha _{0})$, we introduce $\theta ^{\alpha }(m):= \mathcal{L}(c_{0}+\alpha )(m)$ and recall the formula for $\theta ^{\alpha }$ in (A.19). Then we get

$$ \frac{\theta ^{\alpha }(m)}{m} = a\Big(\alpha -\big(\beta -\ell (m) \big)\Big)+\frac{\alpha (\bar{c}+\alpha )}{m} \longrightarrow a( \alpha -\beta _{g}) \qquad \hbox{as}\ m\to \infty , $$

where $a:=\frac{1-\zeta ^{1-\gamma }}{\gamma }$. Moreover, in view of (A.15) with $b$ replaced by $\bar{c}+\alpha$, the above convergence is uniform in $\alpha \in (\frac{\alpha _{0}+\beta _{g}}{2}, \alpha _{0})$. That is, for any $\delta >0$, there is $M(\delta )>0$ such that for all $m\ge M(\delta )$ and $\alpha \in (\frac{\alpha _{0}+\beta _{g}}{2},\alpha _{0})$, $|\frac{\theta ^{\alpha }(m)}{m}- a(\alpha - \beta _{g})|<\delta$. Taking $\delta := \frac{a(\alpha _{0}-\beta _{g})}{4}$, we get for any $m>M(\delta )$ and $\alpha \in (\frac{\alpha _{0}+\beta _{g}}{2},\alpha _{0})$ that

$$ \frac{\theta ^{\alpha }(m)}{m} > a(\alpha -\beta _{g}) - \delta > a \left (\frac{\alpha _{0}+\beta _{g}}{2}-\beta _{g}\right ) -\delta = a \left (\frac{\alpha _{0}-\beta _{g}}{4}\right ). $$

(A.23)

Fix $\varepsilon \in (0,\frac{a(\alpha _{0}-\beta _{g})}{4\beta })$. We can take $\alpha \in (\frac{\alpha _{0}+\beta _{g}}{2},\alpha _{0})$ large enough such that $c_{0}+\alpha$ intersects $p$ at $m^{*}> M(\delta )$ and $a< p'(m^{*})< a+\varepsilon$. It follows that

$$\begin{aligned} \mathcal{L}(p)(m^{*}) &= p(m^{*})\big(p(m^{*})-c_{0}(m^{*})\big) \\ & \phantom{=}{}+ m^{*} p'(m^{*}) \bigg(\sup _{h\ge 0}\bigg\{ g(h)-\frac{1- \gamma }{\gamma }\frac{p(m^{*})}{m^{*}p'(m^{*})}h\bigg\} -\beta \bigg) \\ &= \alpha \big(c_{0}(m^{*})+\alpha \big) + \sup _{h\ge 0}\left \{g(h)m ^{*}p'(m^{*})-\frac{1-\gamma }{\gamma }p(m^{*})h\right \} \\ & \phantom{=}{}-\beta m^{*}p'(m^{*}) \\ &\ge \alpha \big(c_{0}(m^{*})+\alpha \big) + \sup _{h\ge 0}\left \{g(h)m ^{*} a -\frac{1-\gamma }{\gamma }\big(c_{0}(m^{*})+\alpha \big)h\right \} \\ & \phantom{=}{}-\beta m^{*}(a+\varepsilon ) \\ &= \theta ^{\alpha }(m^{*}) - \beta m^{*}\varepsilon = m^{*}\bigg(\frac{ \theta ^{\alpha }(m^{*})}{m^{*}}-\beta \varepsilon \bigg) >0, \end{aligned}$$

where the third equality follows from $\theta ^{\alpha }(m^{*})= \mathcal{L}(c_{0}+\alpha )(m^{*})$ and the last inequality is due to (A.23) and the choice of $\varepsilon$. This implies that $p$ cannot be a subsolution to (3.8) on $(0,\infty )$, which is a contradiction. □

Following Perron’s method, we introduce for each $(p,q)\in \varPi$ the function

$$ u^{*}_{p,q}(m) := \inf _{f\in \mathcal{S}(p,q)}f(m), \qquad m\ge 0. $$

Proposition A.7

(Supersolution property)

Assume$0<\gamma <1$and (A.13). For any$(p,q)\in \varPi$, we have$u^{*}_{p,q}\in \mathcal{S}(p,q)$.

Proof

As a pointwise infimum of concave nondecreasing functions between $c_{0}$ and $c_{0}+\beta$, $u^{*}_{p,q}$ is by definition concave, nondecreasing and between $c_{0}$ and $c_{0}+\beta$. The concavity of $u^{*}_{p,q}$ yields the desired continuity. Then by Crandall et al. [7, Lemma 4.2], $u^{*}_{p,q}$, being continuous and a pointwise infimum of viscosity supersolutions, is again a viscosity supersolution. It remains to show that $u^{*}_{p,q}$ is strictly increasing. Suppose to the contrary that $u^{*}_{p,q}\equiv \kappa >0$ in a neighbourhood of some $m^{*}\in (0,\infty )$. The concavity of $u^{*}_{p,q}$ then implies that $u^{*}_{p,q}\equiv \kappa$ on $[m^{*},\infty )$. It follows that $\mathcal{L}u ^{*}_{p,q}(m) = \kappa (\kappa -c_{0}(m))<0$ as $m$ large enough. This contradicts the supersolution property of $u^{*}_{p,q}$. □

Proposition A.8

(Subsolution property)

Let$0<\gamma <1$and assume (A.13). Fix$(p,q)\in \varPi$. Suppose$u^{*}_{p,q}$is strictly concave at$m_{0}\in (0,\infty )$in the following sense:

$$\begin{aligned} &\textit{for any }m_{1},m_{2}\in (0,\infty )\textit{ and }\lambda \in (0,1)\textit{ such that } m_{0} = \lambda m_{1} + (1-\lambda ) m_{2}, \\ &\textit{we have }u^{*}_{p,q}(m_{0}) > \lambda u^{*}_{p,q}(m_{1}) + (1- \lambda ) u^{*}_{p,q}(m_{2}). \end{aligned}$$

(A.24)

Then$u^{*}_{p,q}$is a viscosity subsolution to (3.8) at$m_{0}$.

Proof

If $u^{*}_{p,q}$ is strictly concave at $m_{0}\in (0,\infty )$ as in (A.24), there are three cases:

(i) $(u^{*}_{p,q})'(m_{0} -)\neq (u^{*}_{p,q})'(m_{0}+)$;

(ii) $(u^{*}_{p,q})'(m_{0} -) = (u^{*}_{p,q})'(m_{0}+)$, and $u^{*}$ is strictly concave on $[m_{0}-\kappa ,m_{0}+\kappa ]$ for some $\kappa >0$;

(iii) $(u^{*}_{p,q})'(m_{0} -) = (u^{*}_{p,q})'(m_{0}+)$, and there exists $\kappa _{1},\kappa _{2}>0$ such that $u^{*}_{p,q}$ is linear on $[m_{0}-\kappa _{1},m_{0}]$ and strictly concave on $[m_{0},m_{0}+ \kappa _{2}]$, or strictly concave on $[m_{0}-\kappa _{1},m_{0}]$ and linear on $[m_{0},m_{0}+\kappa _{2}]$.

Assume by contradiction that there exists a test function $\psi \in C^{1}((0,\infty ))$ such that $0 = (u^{*}_{p,q}-\psi )(m_{0}) > (u ^{*}_{p,q}-\psi )(m)$ for all $m\in (0,\infty )\setminus \{m_{0}\}$ and $\mathcal{L}\psi (m_{0})>0$. For the cases (i) and (ii), we can assume without loss of generality that $\psi$ is strictly increasing and concave on $(0,\infty )$. Take $\delta >0$ small enough such that $\mathcal{L}\psi (m)>0$ for all $m\in (m_{0}-\delta ,m_{0}+\delta )$. Then for small enough $\varepsilon >0$, one can take $0<\delta _{1} \le \delta$ such that for each $0<\eta \le \varepsilon$, $\mathcal{L}(\psi -\eta )(m)>0$ for all $m\in (m_{0}-\delta _{1},m_{0}+ \delta _{1})$. Consider the function

$$ u^{\eta }(m) := \textstyle\begin{cases} \min \{u^{*}_{p,q}(m),\psi (m)-\eta \} &\quad \hbox{for}\ m\in [m _{0}-\delta _{1}, m_{0}+\delta _{1}], \\ u^{*}_{p,q}(m) &\quad \hbox{for}\ m\notin [m_{0}-\delta _{1}, m_{0}+ \delta _{1}]. \end{cases} $$

When $\eta$ is small enough, $u^{\eta }$ is by construction a concave, strictly increasing viscosity supersolution to (3.8) on $(0,\infty )$, and $u^{*}_{p,q}-\eta \le u^{\eta }\le u^{*}_{p,q}$. That is, $u^{\eta }\in \mathcal{S}(p,q)$ when $\eta$ is small enough. However, by definition, $u^{\eta }< u^{*}_{p,q}$ in some small neighbourhood of $m_{0}$, which contradicts the definition of $u^{*}_{p,q}$.

Now we deal with the case (iii). Set $a:= (u^{*}_{p,q})'(m_{0} -) = (u ^{*}_{p,q})'(m_{0}+)$. In view of (3.8), to get the desired subsolution property, it suffices to prove

$$\begin{aligned} &(u^{*}_{p,q})^{2}(m_{0}) - c_{0}(m_{0})u^{*}_{p,q}(m_{0}) + a m_{0} \bigg(\sup \limits _{h\ge 0}\bigg\{ g(h) - \frac{1-\gamma }{ \gamma } \frac{u^{*}_{p,q}(m_{0})}{a m_{0}} h\bigg\} -\beta \bigg) \\ &\quad\le 0. \end{aligned}$$

(A.25)

We assume without loss of generality that $u^{*}_{p,q}$ is linear on $[m_{0}-\kappa _{1},m_{0}]$ and strictly concave on $[m_{0},m_{0}+ \kappa _{2}]$. Take $(\ell _{n})_{n\in \mathbb{N}}$ in $(m_{0},m_{0}+ \kappa _{2}]$ such that $\ell _{n}\downarrow m_{0}$ and $u^{*}_{p,q}$ is differentiable at $\ell _{n}$. Then the subsolution property established above for case (ii) implies that $\mathcal{L} u^{*}_{p,q}(\ell _{n}) \le 0$ for all $n\in \mathbb{N}$. Observe that the map

$$ m\mapsto \sup \limits _{h\ge 0}\bigg\{ g(h) - \frac{1-\gamma }{\gamma } \frac{u^{*}_{p,q}(m)}{m(u^{*}_{p,q})'(m)} h\bigg\} \qquad \hbox{is continuous around $m_{0}$}, $$

(A.26)

as $g$ is strictly concave, nondecreasing and $g'(\infty )=0$. By the continuity of $u^{*}_{p,q}$ and (A.26), $\mathcal{L} u ^{*}_{p,q}(\ell _{n}) \le 0$ implies (A.25) as $n\to \infty$. □

We next establish the strict concavity of $u^{*}_{p,q}$. Recall that $I$ denotes the inverse function of $g'$.

Proposition A.9

(Strict concavity)

Assume$0<\gamma <1$, (A.13) and (3.7). For any$(p,q)\in \varPi$, $u^{*}_{p,q}$is strictly concave on$(0,\infty )$.

Proof

Assume by contradiction that $u^{*}_{p,q}$ is linear, i.e., $u^{*}_{p,q}(m) =am+b$, on some interval of $\mathbb{R}_{+}$. Since $u^{*}_{p,q}\in \mathcal{S}(p,q)$, we deduce from Lemma A.4 that $a\ge \frac{1-\zeta ^{1-\gamma }}{ \gamma }$ and $b\in [\bar{c}, \bar{c}+\beta ]$. Recall $\theta (m):= \mathcal{L}(am+b)$ from (A.14).

Case I:$a=\frac{1-\zeta ^{1-\gamma }}{\gamma }$ and $b\in [\bar{c},\bar{c}+\beta _{g})$: Then we have $u^{*}_{p,q}(m) = c _{0}(m)+\alpha$ for $m$ large enough, where $\alpha := b-\bar{c} \in [0,\beta _{g})$. Thanks to Lemma A.5, we have $\lim _{m\uparrow \infty }\mathcal{L}(u^{*}_{p,q})(m) = \lim _{m\uparrow \infty }\mathcal{L}(c_{0}+\alpha )(m)=-\infty$. This contradicts the supermartingale property of $u^{*}_{p,q}$.

Case II:$a=\frac{1-\zeta ^{1-\gamma }}{\gamma }$ and $b\in [\bar{c}+\beta _{g}, \bar{c}+\beta ]$: There are three sub-cases.

Case II-1:$u^{*}_{p,q}(m) = am+b$ for all $m\ge 0$, with $b\in (\bar{c}+\beta _{g}, \bar{c}+\beta ]$: Let us write $u^{*}_{p,q}(m) = c_{0}(m)+\alpha$ with $\alpha := b-\bar{c}\in (\beta _{g},\beta ]$. For any $\bar{\alpha }\in (\beta _{g},\alpha )$, Lemmas A.5 and A.6 imply that $c_{0}+\bar{\alpha }$ belongs to $\mathcal{S}({p,q})$ and is strictly less than $u^{*}_{p,q}$, which contradicts the definition of $u^{*}_{p,q}$.

Case II-2:$u^{*}_{p,q}(m) = am+(\bar{c} +\beta _{g})$ for all $m\ge 0$: By (A.14) and (A.21), $\lim _{m\downarrow 0}\theta (m)= b(b-\bar{c})>0$ and $\lim _{m\downarrow 0}\theta '(m)= b- \bar{c}-\beta <0$. Thus we can take $m^{*}>0$ small enough such that $\theta (m^{*})>0$ and $\theta '(m^{*})<0$. In view of the continuous dependence of $\theta (m^{*})$ and $\theta '(m^{*})$ on $a,b$ in (A.14) and (A.17), there exists $\delta > 0$ small enough such that when $a,b$ are replaced by $\bar{a} \in (a,a+\delta )$ and $\bar{b} \in (b-\delta ,b)$, $\theta (m^{*})>0$ and $\theta '(m^{*})<0$ still hold. Take suitable $\bar{a} \in (a,a+\delta )$ and $\bar{b} \in (b-\delta ,b)$ such that $\bar{a} m^{*}+\bar{b} = u^{*}_{p,q}(m ^{*})$ and $\bar{a} m +\bar{b} > p(m)$ for $m\in (0,m^{*}]$ (such $\bar{a}$ and $\bar{b}$ exist by Lemma A.6). For clarity, let $\bar{\theta }$ and $\bar{\theta }'$ denote $\theta$ and $\theta '$ with $a,b$ replaced by $\bar{a},\bar{b}$. Now we deduce from $\lim _{m\downarrow 0}\bar{\theta }(m)=\bar{b}(\bar{b}-\bar{c})>0$ (obtained from (A.14) as above), $\bar{\theta }(m^{*})>0$, $\bar{\theta }'(m^{*})<0$ and $\bar{\theta }''(m)>0$ for all $m> 0$ (by (A.18)) that $\bar{\theta }(m)>0$ for all $m\in (0, m^{*})$. Consider the function $\psi (m):= \bar{a} m+ \bar{b}$. By definition, $\mathcal{L}\psi (m)=\bar{\theta }(m)>0$ for $m\in (0,m^{*})$. Thus $\psi \wedge u^{*}_{p,q}$ belongs to $\mathcal{S}(p,q)$ and is strictly less than $u^{*}_{p,q}$ for $m\in (0, m^{*})$. This contradicts the definition of $u^{*}_{p,q}$.

Case II-3: There exists $m_{0}>0$ such that $u^{*}_{p,q}(m) = am+b$ for all $m\ge m_{0}$ and $u^{*}_{p,q}$ is strictly concave at $m_{0}$ in the sense of (A.24): By Proposition A.8, $u^{*}_{p,q}$ is a viscosity subsolution to (3.8) at $m_{0}$. Consider the test function $\psi (m) := am+ b$, $m\in (0,\infty )$, of $u^{*}_{p,q}$ at $m_{0}$. The subsolution property of $u^{*}_{p,q}$ yields $\mathcal{L}\psi (m_{1})\le 0$. Note that $\psi (m) = c_{0}(m)+\alpha$ with $\alpha :=b-\bar{c}\in [\beta _{g},\beta ]$. Thus by Lemma A.5, $\mathcal{L}{\psi }(m)>0$ for all $m>0$, which is a contradiction.

Case III:$a>\frac{1-\zeta ^{1-\gamma }}{\gamma }$ and $b=\bar{c}$: There exists $m_{0}>0$ with the property that $u ^{*}_{p,q}(m) = am+b$ for $m\in [0,m_{0}]$ and $u^{*}_{p,q}$ is strictly concave at $m_{0}$ in the sense of (A.24). By Proposition A.8, $u^{*}_{p,q}$ is a viscosity subsolution to (3.8) at $m_{0}$. For all $m\in (0,\infty )$, define

$$\begin{aligned} \eta (m) &:= \left (a+\frac{b}{m}\right )\left (\Big(a-\frac{1- \zeta ^{1-\gamma }}{\gamma }\Big) m + (b-\bar{c})\right ) + a \big( \ell (m)-\beta \big), \end{aligned}$$

with $\ell$ as in (A.15). Note that $\theta (m) = m\eta (m)$. By direct calculation and (A.16),

$$ \eta '(m) = a\left (a-\frac{1-\zeta ^{1-\gamma }}{\gamma }\right ) - \frac{b}{m ^{2}}\Bigg((b-\bar{c}) - \frac{1-\gamma }{\gamma } I\Big(\frac{1- \gamma }{\gamma }\Big(1+\frac{b}{a m} \Big)\Big)\Bigg). $$

Since we currently have $b = \bar{c}$, $\eta '(m) >0$ for all $m\in (0,\infty )$. Take $\psi (m) := am+b$, $m\in (0,\infty )$, as a test function of $u^{*}_{p,q}$ at $m_{0}$. The subsolution property of $u^{*}_{p,q}$ implies $0\ge \mathcal{L}\psi (m_{0})=\theta (m_{0})=m _{0}\eta (m_{0})$. We therefore have $\eta (m)<0$ for all $m\in (0,m _{0})$. The supersolution property of $u^{*}_{p,q}$, however, entails $0\le \mathcal{L}u^{*}_{p,q}(m)=\theta (m)=m\eta (m)$ for all $m\in (0,m_{0})$, which is a contradiction.

Case IV:$a> \frac{1-\zeta ^{1-\gamma }}{\gamma }$ and $b\in (\bar{c}, \bar{c}+\beta )$: There are two sub-cases.

Case IV-1: There exists $m_{0}>0$ such that $u^{*}_{p,q}(m) = am+b$ for all $m\in [0,m_{0}]$ and $u^{*}_{p,q}$ is strictly concave at $m_{0}$ in the sense of (A.24): We first show that $p(0)$ has to be strictly less than $u^{*}_{p,q}(0)$. If $p(0)= u^{*}_{p,q}(0)$, then $\lim _{m\downarrow 0}p'(m) \le a$; otherwise, $p(m)>u^{*}_{p,q}(m)$ for $m> 0$ small enough, which contradicts $u^{*}_{p,q}\in \mathcal{S}(p,q)$. By the concavity of $p$, we can take a real sequence $(\ell _{n})$ such that $\ell _{n} \downarrow 0$ and $p$ is differentiable at $\ell _{n}$. The subsolution property of $p$ then implies $\mathcal{L}p(\ell _{n})\le 0$ for all $n\in \mathbb{N}$. As $n\to \infty$, we get $p(0)(p(0)-\bar{c}) \le 0$, thanks to the finiteness of $\lim _{m\downarrow 0}p'(m)$. This shows that $p(0)<\bar{c}$, contradicting $p\ge c_{0}$.

By Proposition A.8, $u^{*}_{p,q}$ is a viscosity subsolution to (3.8) at $m_{0}$. Consider the test function $\psi (m) := am+b$, $m\in (0,\infty )$, of $u^{*}_{p,q}$ at $m_{0}$. The subsolution property of $u^{*}_{p,q}$ implies $0\ge \mathcal{L}\psi (m_{0})=\theta (m_{0})$. By (A.14), $\lim _{m\downarrow 0}\theta (m)= b(b-\bar{c})>0$. If $\lim _{m\downarrow 0}\theta '(m)\ge 0$, then $\theta ''>0$ on $(0,\infty )$ (by (A.18)) implies that $\theta (m)>\theta (0)>0$ for all $m>0$, which contradicts $\theta (m_{0})\le 0$. If $\lim _{m\downarrow 0} \theta '(m)< 0$, we can follow the argument in Case II-2. Take $0< m^{*}< m_{0}$ small enough such that $\theta (m^{*})>0$ and $\theta '(m^{*})<0$. By the continuous dependence of $\theta (m^{*})$ and $\theta '(m^{*})$ on $a,b$, there exists $\delta > 0$ such that when $a,b$ are replaced by $\bar{a} \in (a,a+\delta )$ and $\bar{b} \in (b- \delta ,b)$, $\theta (m^{*})>0$ and $\theta '(m^{*})<0$ still hold. Choose suitable $\bar{a} \in (a,a+\delta )$ and $\bar{b} \in (b- \delta ,b)$ such that $\bar{a} m^{*}+\bar{b} = u^{*}_{p,q}(m^{*})$ and $\bar{a} m +\bar{b} > p(m)$ for $m\in (0,m^{*}]$ (such $\bar{a}$ and $\bar{b}$ exist because $p(0)< u^{*}_{p,q}(0)$). For clarity, let $\bar{\theta }$ and $\bar{\theta }'$ denote $\theta$ and $\theta '$ with $a,b$ replaced by $\bar{a},\bar{b}$. Now we deduce from $\lim _{m\downarrow 0}\bar{\theta }(m)>0$, $\bar{\theta }(m^{*})>0$, $\bar{\theta }'(m^{*})<0$ and $\bar{\theta }''>0$ on $(0,\infty )$ that $\bar{\theta }(m)>0$ for all $m\in (0, m^{*})$. Consider the function $\phi (m):= \bar{a} m+\bar{b}$. By definition, $\mathcal{L}\phi (m)=\bar{ \theta }(m)>0$ for $m\in (0,m^{*})$. Thus $\phi \wedge u^{*}_{p,q}$ belongs to $\mathcal{S}(p,q)$ and is strictly less than $u^{*}_{p,q}$ for $m\in (0, m^{*})$. This contradicts the definition of $u^{*}_{p,q}$.

Case IV-2: There are $m_{1}, m_{2}\in (0,\infty )$ with the property that $u^{*}_{p,q}(m) = am+b$ for $m\in [m_{1},m_{2}]$ and $u^{*}_{p,q}$ is strictly concave at $m_{1}$ and $m_{2}$ in the sense of (A.24): By Proposition A.8, $u^{*}_{p,q}$ is a viscosity subsolution to (3.8) at both $m_{1}$ and $m_{2}$. Take $\psi (m) := am+b$, $m\in (0,\infty )$, as a test function of $u^{*}_{p,q}$ at $m_{1}$ and $m_{2}$. Then the subsolution property of $u^{*}_{p,q}$ implies $0\ge \mathcal{L}\psi (m_{1})=\theta (m_{1})$ and $0\ge \mathcal{L}\psi (m_{2})=\theta (m_{2})$. As $\theta ''>0$ on $(0,\infty )$ by (A.18), we must have $\theta (m_{3})<0$ for some $m_{3}\in (m_{1},m_{2})$. The supersolution property of $u^{*}_{p,q}$, however, entails $0\le \mathcal{L}\psi (m)=\theta (m)$ for all $m\in (m_{1},m_{2})$, which is a contradiction. □

Proposition A.10

(Regularity)

Let$0<\gamma <1$and assume (A.13) and (3.7). For any$(p,q)\in \varPi$, $u^{*}_{p,q}$is a strictly concave classical solution to (3.8) on$(0,\infty )$.

Proof

For any $(p,q)\in \varPi$, Propositions A.7–A.9 imply that $u^{*}_{p,q}$ is a strictly concave viscosity solution to (3.8) on $(0,\infty )$. It remains to show that $u^{*}_{p,q}$ is differentiable everywhere on $(0,\infty )$. Assume by contradiction that there exists $m_{0}\in (0,\infty )$ such that $a:=(u^{*})'(m_{0}+) < (u^{*})'(m_{0}-)=:b$. Take $(k_{n})_{n\in \mathbb{N}}$ and $(\ell _{n})_{n\in \mathbb{N}}$ in $(0,\infty )$ such that $k_{n} \uparrow m_{0}$, $\ell _{n}\downarrow m_{0}$ and $u^{*}_{p,q}$ is differentiable at $k_{n}$ and $\ell _{n}$ for all $n\in \mathbb{N}$. By the viscosity solution property of $u^{*}_{p,q}$, $\mathcal{L}u^{*}(k _{n}) = \mathcal{L}u^{*}(\ell _{n})=0$ for all $n\in \mathbb{N}$. As $n\to \infty$, it follows that

$$ \sup _{h\ge 0}\bigg\{ \big(g(h)-\beta \big)a - \frac{1-\gamma }{\gamma } \frac{u^{*}_{p,q}(m_{0})}{m_{0} } h\bigg\} =\sup _{h\ge 0}\bigg\{ \big(g(h)-\beta \big)b - \frac{1-\gamma }{\gamma } \frac{u^{*}_{p,q}(m _{0})}{m_{0} } h\bigg\} , $$

which implies that $a=b$, a contradiction. □

Proposition A.11

(Verification)

Let$0 < \gamma < 1$and suppose (A.13) and (3.7) hold. If$u:\mathbb{R}_{+}\to \mathbb{R}_{+}$is a nonnegative, strictly increasing and concave classical solution to (3.8) on$(0,\infty )$, then

$$ V(x,m)=\frac{x^{1-\gamma }}{1-\gamma } u(m)^{-\gamma } \qquad \textit{for all}\ (x,m)\in \mathbb{R}^{2}_{+}. $$

Furthermore, $(\hat{c},\hat{h})$defined by

$$ \hat{c}_{t} := u(M_{t})\quad \textit{and} \quad \hat{h}_{t} := I\left (\frac{1- \gamma }{\gamma }\frac{u(M_{t})}{M_{t} \,u'(M_{t})}\right ) \qquad \textit{for all}\ t\ge 0 $$

is an optimal control for (2.8).

Proof

Set $w(x,m) := \frac{x^{1-\gamma }}{1-\gamma } u(m)^{-\gamma }$. In view of Theorem B.1, it suffices to show that (B.2) and (B.3) hold and $(\hat{c},\hat{h})$ belongs to $\mathcal{A}$. Since $u$ is nonnegative, strictly increasing and concave, Lemma A.4 implies that $\hat{h} _{t} \le I(\frac{1-\gamma }{\gamma })$ for all $t\ge 0$. Moreover, there exist $a,b>0$ such that $u(m)< am+b$ for all $m\ge 0$. It follows that for any compact subset $K\subset \mathbb{R}_{+}$,

$$ \int _{K} \hat{c}_{t} \,dt \le \int _{K} (aM_{t} +b) \, dt \le \int _{K} (a m e^{\beta t} +b)\, dt < \infty . $$

This already shows that $(\hat{c},\hat{h})\in \mathcal{A}$.

As $u$ is a classical solution to (3.8), (3.7) implies $u^{2}(m) - u(m) c_{0}(m) \ge 0$ and thus $u(m)\ge c_{0}(m)$ for all $m\in (0,\infty )$. Now, given $(x,m)\in \mathbb{R}^{2}_{+}$, $(c,h)\in \mathcal{A}$ and $n\in \mathbb{N}$, using $0<\gamma <1$ and $X^{0,x,c,h}_{t} \le X^{0,x,c,h}_{\tau _{n}}\exp \left (r(t-\tau _{n})\right )$ yields

$$\begin{aligned} 0 &\le \mathbb{E}\bigg[\exp \left (-\int _{\tau _{n}}^{t} (\delta + M ^{0,m,h}_{s}) \,ds\right ) w(\zeta ^{n} X^{0,x,c,h}_{t}, M^{0,m,h} _{t}) \biggm| Z_{1},\dots ,Z_{n}\bigg] \\ &\le e^{-\delta (t-\tau _{n})} \frac{(X^{0,x,c,h}_{\tau _{n}})^{1- \gamma }}{1-\gamma } e^{(1-\gamma )r(t-\tau _{n})} \mathbb{E}[u(M^{0,m,h} _{t})^{-\gamma } \mid Z_{1},\dots ,Z_{n}] \\ & \le e^{-(\delta +(\gamma -1)r)(t-\tau _{n})} \frac{(X^{0,x,c,h}_{ \tau _{n}})^{1-\gamma }}{1-\gamma } (\bar{c})^{-\gamma }\longrightarrow 0 \ \ \hbox{a.s.} \qquad \hbox{as}\ t\to \infty , \end{aligned}$$

where the second line is due to $M^{0,m,h}_{t} \ge 0$ by definition, the third follows from $u$ being strictly increasing with $u(m)> u(0) \ge c_{0}(0)=\bar{c}>0$, and the convergence is a consequence of $\delta + (\gamma -1) r = \gamma \bar{c}>0$. This in particular implies (B.2). On the other hand, for each $n\in \mathbb{N}$,

$$ \begin{aligned} 0 &\le \mathbb{E}[e^{-\delta \tau _{n}} w(\zeta ^{n} X^{0,x,c,h}_{\tau _{n}},M^{0,m,h}_{\tau _{n}})] \\ &\le \zeta ^{(1-\gamma )n}\frac{x^{1-\gamma }}{1-\gamma } \mathbb{E}[e ^{-\delta \tau _{n}}e^{(1-\gamma ) r\tau _{n}} u(M^{0,m,h}_{\tau _{n}})^{- \gamma }] \\ &\le \zeta ^{(1-\gamma )n}\frac{x^{1-\gamma }}{1-\gamma } (\bar{c})^{- \gamma } \mathbb{E}[e^{-(\delta +(\gamma -1)r) \tau _{n}}] \\ &\le \zeta ^{(1-\gamma )n}\frac{x^{1-\gamma }}{1-\gamma } (\bar{c})^{- \gamma }\longrightarrow 0 \qquad \hbox{as}\ t\to \infty , \end{aligned} $$

where the last inequality is due to $\delta +(\gamma -1)r =\gamma \bar{c}>0$. This shows that (B.3) also holds. □

Proposition A.11 together with Propositions A.10 and A.7 leads to

Corollary A.12

Assume$0 < \gamma < 1$, (A.13) and (3.7). Then$u^{*}_{p,q}$is independent of the choice of$(p,q)\in \varPi$, and it is the unique nonnegative, strictly increasing and concave classical solution to (3.8) on$(0,\infty )$.

Remark A.13

Proposition A.11 and Corollary A.12 yield Theorem 3.4.

In the following, we simply denote by $u^{*}$ the function $u^{*}_{p,q}$ for any $(p,q) \in \varPi$.

Corollary A.14

(Strict concavity of $u_{0}$)

Assume$0 < \gamma < 1$and (A.13). Then$u_{0}$defined in (3.6) is strictly concave on$(0,\infty )$.

Proof

With $g\equiv 0$, Eq. (3.8) reduces to (A.5). We can repeat the same arguments as in this section (with much simpler proofs) to show that the strictly concave $u^{*}$ constructed via Perron’s method coincides with $u_{0}$. □

We are now ready to prove Theorem 3.5.

Proof of Theorem 3.5

Observe that $\beta _{g} = \beta - g(I(\frac{1- \gamma }{\gamma })) + \frac{1-\gamma }{\gamma } I(\frac{1-\gamma }{ \gamma })$. Then (3.7) implies $\beta _{g}>0$. Since $u_{0}$ is a solution to (A.5) and $u_{0}'(m)\ge 0$, it is a supersolution to (3.8). This together with Lemma A.1, Corollary A.14 and Remark A.3 shows that $(c_{0},u_{0})\in \varPi$. As a result, $u^{*} = u^{*}_{c_{0},u_{0}}\le u_{0}$. Similarly, $(c_{0},c _{0}+\beta _{g})\in \varPi$ by Lemma A.5, which implies $u^{*}=u^{*}_{c_{0},c_{0}+\beta _{g}}\le c_{0}+\beta$. This already yields the relation $u^{*}\le \min \{u_{0}, c_{0}+\beta _{g}\}$. On the other hand, thanks again to Lemma A.1 and Corollary A.14 with $\beta$ replaced by $\beta _{g}$, $u_{0}^{g}$ is nonnegative, strictly increasing, concave and bounded from below and above by $c_{0}$ and $c_{0}+ \beta _{g}$, respectively. Then Lemma A.4 implies $\frac{u^{g}_{0}(m)}{m (u ^{g}_{0})'(m)}\ge 1$ for all $m>0$. It follows that

$$ \beta -\sup _{h\ge 0}\bigg\{ g(h)-\frac{1-\gamma }{\gamma }\frac{u^{g} _{0}(m)}{m \,(u^{g}_{0})'(m)} h\bigg\} \ge \beta _{g}, \qquad \forall m>0. $$

Since $u_{0}^{g}$ is by construction a solution to (A.5) with $\beta$ replaced by $\beta _{g}$, the above inequality gives

$$\begin{aligned} 0 &= \big(u^{g}_{0}(m)\big)^{2} - u^{g}_{0}(m) c_{0}(m) - \beta _{g} m (u^{g}_{0})'(m) \\ &\ge \big(u^{g}_{0}(m)\big)^{2} - u^{g}_{0}(m) c_{0}(m) \\ & \phantom{=}{} + m (u^{g}_{0})'(m) \bigg(\sup _{h\ge 0}\bigg\{ g(h)-\frac{1-\gamma }{ \gamma }\frac{u^{g}_{0}(m)}{m \,(u^{g}_{0})'(m)} h\bigg\} - \beta \bigg) = \mathcal{L}u^{g}_{0}(m) \end{aligned}$$

for all $m>0$. This shows $(u^{g}_{0},c_{0}+\beta _{g})\in \varPi$, whence $u^{*}=u^{*}_{u^{g}_{0},c_{0}+\beta _{g}}\ge u^{g}_{0}$. □

Appendix B: Verification

In this section, we provide a general verification theorem for the value function $V(x,m)$ in (2.8). Given $(c,h)\in \mathcal{A}$, we introduce for each $n\in \mathbb{N}$ the truncated policies $(c^{(n)},h^{(n)})\in \mathcal{A}$ as

$$ \begin{aligned} c^{(n)}_{t} &:= \bigg(\sum _{k=0}^{n-1} c_{k}(t) 1_{\{\tau _{k}\le t< \tau _{{k+1}}\}}\bigg) + c_{n}(t) 1_{\{t\ge \tau _{{n}}\}}, \\ h^{(n)}_{t} &:= \bigg(\sum _{k=0}^{n-1} h_{k}(t) 1_{\{\tau _{k}\le t< \tau _{{k+1}}\}}\bigg) + h_{n}(t) 1_{\{t\ge \tau _{{n}}\}}. \end{aligned} $$

(B.1)

Theorem B.1

Let$w\in C^{1,1}(\mathbb{R}_{+}\times \mathbb{R}_{+})$satisfy (A.1). Suppose for any$(x,m)\in \mathbb{R}^{2}_{+}$and$(c,h)\in \mathcal{A}$that

$$\begin{aligned} &\lim _{t\to \infty } \mathbb{E}\bigg[\exp \left (-\int _{\tau _{n}}^{t} (\delta + M^{0,m,h^{(n)}}_{s}) \,ds\right ) \\ & \phantom{\lim _{t\to \infty } \mathbb{E}\bigg[}{}\times w\big(\zeta ^{n} X^{0,x,c^{(n)},h^{(n)}}_{t},M^{0,m,h^{(n)}}_{t}\big) \biggm| Z _{1},\dots ,Z_{n} \bigg] = 0, \qquad \forall n\ge 0, \\ \end{aligned}$$

(B.2)

$$\begin{aligned} &\lim _{n\to \infty } \mathbb{E}\big[e^{-\delta \tau _{n}} w\big(\zeta ^{n} X^{0,x,c^{(n)},h^{(n)}}_{\tau _{n}},M^{0,m,h^{(n)}}_{\tau _{n}} \big)\big] = 0. \end{aligned}$$

(B.3)

Then:

(i) $w(x,m) \ge V(x,m)$on$\mathbb{R}_{+}\times \mathbb{R} _{+}$.

(ii) Suppose there exist two measurable functions$\bar{c}$, $\bar{h}:\mathbb{R}^{2}_{+}\to \mathbb{R}_{+}$such that$\bar{c}(x,m)$and$\bar{h}(x,m)$are maximisers of

$$ \sup _{c\ge 0}\left \{U(cx)-cxw_{x}(x,m)\right \} \qquad \textit{and} \qquad \sup _{h\ge 0}\left \{ -w_{m}(x,m)g(h)-h x w_{x}(x,m)\right \}, $$

respectively, for all$(x,m)\in \mathbb{R}^{2}_{+}$. Let$\bar{X}$, $\bar{M}$, $\bar{N}$denote the solutions to

$$\begin{aligned} dX_{s} &= X_{s}\Big(r - \big(\bar{c}(X_{s},M_{s})+\bar{h}(X_{s},M_{s}) \big)\Big) \,ds, \qquad X_{0} =x, \\ dM_{s} &= M_{s}\Big(\beta - g\big(\bar{h}(\zeta ^{N_{s}} X_{s},M_{s}) \big)\Big)\,ds, \qquad M_{0}=m, \\ N_{s} &= \sum _{k=0}^{\infty }k 1_{\{T_{k}\le t < T_{k+1}\}}, \end{aligned}$$

where$T_{0}:=0$and$T_{n+1}:= \inf \{t\ge T_{n} : \int _{T_{n}}^{t} M _{s} \,ds\ge Z_{n+1}\}$for$n\ge 1$. Define the processes$(\hat{c}, \hat{h})$by

$$ \hat{c}_{t} := \bar{c}(\zeta ^{\bar{N}_{t}} \bar{X}_{t}, \bar{M}_{t}) \quad \textit{and}\quad \hat{h}_{t} := \hat{h}(\zeta ^{\bar{N}_{t}} \bar{X}_{t},\bar{M}_{t}) \qquad \textit{for}\ t\ge 0. $$

(B.4)

If$(\hat{c},\hat{h})\in \mathcal{A}$, then$(\hat{c},\hat{h})$is an optimal control for (2.8) and$w(x,m)=V(x,m)$on$\mathbb{R}_{+}\times \mathbb{R}_{+}$.

Proof

(i) Given $(c,h)\in \mathcal{A}$, recall that

$$ c_{t} = \sum _{n=0}^{\infty }c_{n}(t) 1_{\{\tau _{n}\le t< \tau _{{n+1}} \}} \qquad \hbox{and} \qquad h_{t} = \sum _{n=0}^{\infty }h_{n}(t) 1_{\{\tau _{n}\le t< \tau _{n+1}\}} $$

for some $(c_{n})$, $(h_{n})\in \mathfrak{L}$. We claim that for all $n\in \mathbb{N}$, we have

$$ w(x,m)\ge \mathbb{E}\bigg[\int _{0}^{\tau _{{n}}} e^{-\delta t} U(c_{t} \zeta ^{N_{t}}{X}^{0,x,c,h}_{t})\,dt\bigg]+\mathbb{E}[e^{-\delta \tau _{{n}}}w(\zeta ^{n} X^{0,x,c,h}_{\tau _{{n}}},M^{0,m,h}_{\tau _{{n}}})]. $$

(B.5)

First, we prove this for $n=1$. Since $c_{0}$ and $h_{0}$ are deterministic functions and $w$ is a classical solution to (A.1),

$$\begin{aligned} &e^{-\int _{0}^{t} (\delta +M^{0,m,h_{0}}_{\nu }) \,d\nu } w(X^{0,x,c _{0},h_{0}}_{t},M^{0,m,h_{0}}_{t}) \\ &\quad\le w(x,m)-\int _{0}^{t} e^{-\int _{0}^{s}(\delta +M^{0,m,h_{0}}_{ \nu })\,d\nu } \\ & \phantom{=:w(x,m)-\int _{0}^{t} }{}\times \Big(U\big(c_{0}(s) X^{0,x,c _{0},h_{0}}_{s}\big) + M^{0,m,h_{0}}_{s} w(\zeta X^{0,x,c_{0},h_{0}} _{s},M^{0,m,h_{0}}_{s})\Big)\,ds \end{aligned}$$

(B.6)

for all $t\ge 0$. Letting $t\to \infty$ and in view of (B.2),

$$\begin{aligned} w(x,m) &\ge \int _{0}^{\infty }e^{-\int _{0}^{s}(\delta +M^{0,m,h_{0}} _{\nu })\,d\nu } U\big(c_{0}(s) X^{0,x,c_{0},h_{0}}_{s}\big) \,ds \\ & \phantom{=}{}+ \int _{0}^{\infty }e^{-\int _{0}^{s}(\delta +M^{0,m,h_{0}} _{\nu })\,d\nu } M^{0,m,h_{0}}_{s} w(\zeta X^{0,x,c_{0},h_{0}}_{s},M ^{0,m,h_{0}}_{s}) \,ds. \end{aligned}$$

(B.7)

Thanks to Fubini’s theorem and (2.6), observe that

$$\begin{aligned} &\mathbb{E}\bigg[\int _{0}^{\tau _{1}} e^{-\delta t}U(c_{t} \zeta ^{N _{t}}{X}^{0,x,c,h}_{t})\, dt\bigg] \\ &\quad= \mathbb{E}\left [\int _{0}^{\infty }1_{\{\tau _{1}>t\}} e^{-\delta t}U\big(c_{0}(t) X^{0,x,c_{0},h_{0}}_{t}\big) \,dt\right ] \\ &\quad=\int _{0}^{\infty }e^{-\int _{0}^{t}(\delta +M^{0,m,h_{0}}_{\nu })d \nu } U\big(c_{0}(t) X^{0,x,c_{0},h_{0}}_{t}\big) \,dt, \end{aligned}$$

(B.8)

$$\begin{aligned} &\mathbb{E}[e^{-\delta \tau _{1}} w(\zeta X^{0,x,c,h}_{\tau _{1}},M ^{0,m,h}_{\tau _{1}})] \\ &\quad= \mathbb{E}[e^{-\delta \tau _{1}} w(\zeta X^{0,x,c_{0},h_{0}}_{\tau _{1}},M^{0,m_{0},h_{0}}_{\tau _{1}})] \\ &\quad=\int _{0}^{\infty }e^{-\int _{0}^{t}(\delta +M^{0,m,h_{0}}_{\nu })\,d \nu } M^{0,m,h_{0}}_{t} w(\zeta X^{0,x,c_{0},h_{0}}_{t},M^{0,m,h_{0}} _{t}) \,dt , \end{aligned}$$

(B.9)

whence (B.5) holds true for $n=1$ in view of (B.7)–(B.9). Now suppose (B.5) holds true for $n=k>1$. That is,

$$ w(x,m)\ge \mathbb{E}\bigg[\int _{0}^{\tau _{{k}}} e^{-\delta t} U(c_{t} \zeta ^{N_{t}}{X}^{0,x,c,h}_{t})\,dt\bigg]+\mathbb{E}[e^{-\delta \tau _{{k}}}w(\zeta ^{k} X^{0,x,c,h}_{\tau _{{k}}},M^{0,m,h}_{\tau _{{k}}})]. $$

(B.10)

By writing $x_{k} = \zeta ^{k}X^{0,x,c,h}_{\tau _{k}}$ and $m_{k} = M ^{0,m,h}_{\tau _{k}}$, we get (B.6) with $(0,x,m,c_{0},h_{0})$ replaced by $(\tau _{k},x_{k},m_{k},c_{k},h_{k})$. This together with (B.2) gives

$$\begin{aligned} &w(\zeta ^{k}X^{0,x,c,h}_{\tau _{{k}}},M^{0,m,h}_{\tau _{{k}}}) \\ &\quad= w(x_{k},m_{k}) \\ &\quad\ge \mathbb{E}\bigg[\int _{\tau _{k}}^{\infty }e^{-\int _{\tau _{k}} ^{t}(\delta +M^{\tau _{k},m_{k},h_{k}}_{\nu })\,d\nu } \Big( U\big(c_{k}(t) X^{\tau _{k},x_{k},c_{k},h_{k}}_{t}\big) \\ & \qquad \qquad \qquad \qquad{} + M^{\tau _{k},m_{k},h_{k}}_{t} w(\zeta X^{\tau _{k},x_{k},c_{k},h_{k}} _{t},M^{\tau _{k},m_{k},h_{k}}_{t})\Big) \,dt \biggm| Z_{1},\dots ,Z_{k} \bigg]. \end{aligned}$$

(B.11)

Using Fubini’s theorem and (2.6) as in (B.8), (B.9), the above inequality implies that

$$\begin{aligned} \mathbb{E}[e^{-\delta \tau _{k}}w(\zeta ^{k}X^{0,x,c,h}_{\tau _{{k}}},M ^{0,m,h}_{\tau _{{k}}})] &\ge \mathbb{E}\bigg[\int _{\tau _{k}}^{\tau _{ {k+1}}} e^{-\delta t} U(c_{t} \zeta ^{N_{t}}{X}^{0,x,c,h}_{t}) \,dt \bigg] \\ & \phantom{=}{}+\mathbb{E}[ e^{-\delta \tau _{{k+1}}} w(\zeta ^{k+1} X^{0,x,c,h} _{\tau _{{k+1}}},M^{0,m,h}_{\tau _{{k+1}}})]. \end{aligned}$$

(B.12)

This together with (B.10) shows that

$$\begin{aligned} w(x,m) &\ge \mathbb{E}\bigg[\int _{0}^{\tau _{{k+1}}} e^{-\delta t} U(c _{t} \zeta ^{N_{t}}{X}^{0,x,c,h}_{t}) \,dt\bigg] \\ & \phantom{=}{}+\mathbb{E}[e^{-\delta \tau _{{k+1}}}w(\zeta ^{k+1} X^{0,x,c,h} _{\tau _{{k+1}}},M^{0,m,h}_{\tau _{{k+1}}})]. \end{aligned}$$

The claim (B.5) then holds by induction. As $n\to \infty$ in (B.5), monotone convergence and (B.3) yield $w(x,m)\ge \mathbb{E}[\int _{0}^{\infty } e^{-\delta t} U(c_{t} \zeta ^{N_{t}}{X}^{0,x,c,h}_{t}) \,dt]$ for all $(c,h)\in \mathcal{A}$. Taking the supremum over $(c,h)\in \mathcal{A}$ gives $w(x,m)\ge V(x,m)$.

(ii) With $(c,h)=(\hat{c},\hat{h})$, the inequality (B.6) turns into an equality, whence (B.5) holds with equality. As $n\to \infty$, monotone convergence and (B.3) imply

$$ w(x,m)= \mathbb{E}\bigg[\int _{0}^{\infty } e^{-\delta t} U(\hat{c} _{t} \zeta ^{N_{t}} {X}^{0,x,\hat{c},\hat{h}}_{t})\,dt\bigg] \le V(x,m). $$

This together with part (i) shows that $w(x,m)=V(x,m)$ and $(\hat{c}, \hat{h})$ is an optimal control. □

Theorem B.1 is now extended to include the risky asset $S$ in (4.1). Recall the setup in Sect. 4, especially $\mathcal{A}'$ in (4.4) and the value function $V$ in (4.5). For any $(c,h,\pi )\in \mathcal{A}'$, we can also consider the truncated version $(c^{(n)}, h^{(n)}, \pi ^{(n)}) \in \mathcal{A}'$ defined as in (B.1).

Theorem B.2

Let$w\in C^{1,2}(\mathbb{R}_{+}\times \mathbb{R}_{+})$satisfy (4.7). Suppose for any$(x,m)\in \mathbb{R}^{2}_{+}$and$(c,h,\pi )\in \mathcal{A}'$that (B.2) and (B.3) hold, where$X^{0,x,c^{(n)},h^{(n)}}$is replaced by$X^{0,x,c^{(n)},h^{(n)},\pi ^{(n)}}$and$\mathbb{E}[\ \cdot \mid Z _{1},\dots ,Z_{n}]$by$\mathbb{E}[\ \cdot \mid \mathcal{F}_{\tau _{n}}]$. Then:

(i) $w(x,m) \ge V(x,m)$on$\mathbb{R}_{+}\times \mathbb{R} _{+}$.

(ii) Suppose there exist measurable functions$\bar{c}$, $\bar{h}$, $\bar{\pi }:\mathbb{R}^{2}_{+}\to \mathbb{R}_{+}$with$\bar{c}$and$\bar{h}$as described in Theorem B.1 (ii) and$\bar{\pi }(x,m)$being the maximiser of

$$ \sup _{\pi \in \mathbb{R}}\left \{\pi \mu x w_{x}(x,m)+ \frac{1}{2} \sigma ^{2}\pi ^{2} x^{2} w_{xx}(x,m)\right \}, \qquad \forall \ (x,m)\in \mathbb{R}^{2}_{+}. $$

Let$\bar{X}$, $\bar{M}$, $\bar{N}$denote the solutions to

$$\begin{aligned} dX_{s} &= X_{s}\Big(r + \mu \bar{\pi }(X_{s},M_{s})- \big(\bar{c}(X _{s},M_{s})+\bar{h}(X_{s},M_{s})\big)\Big) \,ds \\ & \phantom{=}{}+ \sigma X_{s}\bar{\pi }(X_{s},M_{s}) \,dW_{s}, \qquad X_{0} =x, \\ dM_{s} &= M_{s} \Big(\beta - g\big(\bar{h}(\zeta ^{N_{s}} X_{s},M_{s}) \big)\Big)\,ds, \qquad M_{0}=m, \\ N_{s} &= \sum _{k=0}^{\infty }k 1_{\{T_{k}\le t < T_{k+1}\}}, \end{aligned}$$

where$T_{0}:=0$and$T_{n+1}:= \inf \{t\ge T_{n}: \int _{T_{n}}^{t} M _{s} \,ds\ge Z_{n+1}\}$for$n\ge 1$. Consider$(\hat{c},\hat{h}, \hat{\pi })$with$\hat{c}$and$\hat{h}$defined as in (B.4) and$\hat{\pi }_{t}:=\bar{\pi }(\zeta ^{\bar{N}_{t}} \bar{X}_{t}, \bar{M}_{t})$for$t\ge 0$. If$(\hat{c},\hat{h}, \hat{\pi })$is in$\mathcal{A}'$, defined in (4.4), then$(\hat{c},\hat{h}, \hat{\pi })$is an optimal control for (4.5) and$w(x,m)=V(x,m)$on$\mathbb{R}_{+} \times \mathbb{R}_{+}$.

Proof

(i) We follow the arguments in Theorem B.1, replacing $X^{0,x,c,h}$ by $X^{0,x,c,h,\pi }$ in (4.2). For any $(\hat{c}, \hat{h},\hat{\pi })\in \mathcal{A}'$, we now prove (B.5) for all $n\in \mathbb{N}$. For $n=1$, as $w$ is a solution to (4.7), Itô’s formula yields (B.6), with the left-hand side and the second line under the expectation $\mathbb{E} _{2}$. By (B.2), letting $t\to \infty$ gives (B.7), with the right-hand side under the expectation $\mathbb{E}_{2}$. Fubini’s theorem and (4.3) imply (B.8) and (B.9), with their second lines again under the expectation $\mathbb{E}_{2}$. Thus (B.5) holds for $n=1$. Now suppose (B.5) holds for $n=k>1$, i.e., (B.10) is true. As $w$ is a solution to (4.7) and in view of (B.2), by Itô’s formula, (B.11) holds with $\mathbb{E}[\ \cdot \mid Z_{1},\dots ,Z_{n}]$ replaced by $\mathbb{E}[\ \cdot \mid \mathcal{F}_{\tau _{n}}]$. By Fubini’s theorem and (4.3) as above, (B.12) follows. This together with (B.10) implies that (B.5) holds for $n=k+1$. Thus (B.5) follows by induction. Letting $n\to \infty$ in (B.5) and recalling (B.3), the same argument as at the end of the proof of Theorem B.1 (i) yields that $w(x,m)\ge V(x,m)$.

(ii) This follows from the same argument as in Theorem B.1 (ii). □

In the sequel, we relax the conditions in Theorem B.1. For any $(x,m)\in \mathbb{R}^{2}_{+}$ and $(c,h)\in \mathcal{A}$, suppose that the household is given additional wealth $\varepsilon >0$ at time 0 and decides not to spend it at all over time. Imagine that at time 0, the household deposits $x\ge 0$ in a standard account and $\varepsilon >0$ in a separate additional account. Then the household behaves as if there was no additional wealth: at each time $t\ge 0$, the amount it spends in consumption (resp. healthcare) is $c_{t}$ (resp. $h_{t}$) multiplied by the standard account balance. The rates of spending in consumption and healthcare therefore become

$$ c^{\varepsilon }_{t} = \frac{c_{t} X^{0,x,c,h}_{t}}{X^{0,x,c,h}_{t}+ \varepsilon e^{rt}}, \qquad h^{\varepsilon }_{t} = \frac{h_{t} X^{0,x,c,h}_{t}}{X^{0,x,c,h}_{t}+ \varepsilon e^{rt}}, \qquad \forall t\ge 0. $$

(B.13)

This new process $h^{\varepsilon }$ of spending rate in healthcare, different from $h$, changes the moments of deaths. More precisely, starting from time 0, the household takes

$$ h_{0}^{\varepsilon }(t) := \frac{h_{0}(t) X^{0,x,c_{0},h_{0}}_{t}}{X^{0,x,c_{0},h_{0}}_{t}+\varepsilon e^{rt}} $$

as instantaneous spending rates in healthcare. As in (2.4), the time of the first death is defined as

$$ \tau ^{\varepsilon }_{1} := \inf \bigg\{ t\ge 0 : \int _{0}^{t} M^{0,m,h ^{\varepsilon }_{0}}_{s} \,ds \ge Z_{1}\bigg\} \le \tau _{1}. $$

Starting from time $\tau ^{\varepsilon }_{1}$, the household takes

$$ h_{1}^{\varepsilon }(t) := \frac{h_{0}(t) X^{0,x,c_{0},h_{0}}_{t}}{X ^{0,x,c_{0},h_{0}}_{t} +\varepsilon e^{rt}}1_{\{t< \tau _{1}\}} + \frac{h _{1}(t) X^{0,x,c^{(1)},h^{(1)}}_{t}}{X^{0,x,c^{(1)},h^{(1)}}_{t} + \varepsilon e^{rt}}1_{\{t\ge \tau _{1}\}} $$

as instantaneous spending rates in healthcare. Set $m^{\varepsilon } _{1}:= M^{0,m,h^{\varepsilon }_{0}}_{\tau ^{\varepsilon }_{1}}$; then the time of the second death is defined as in (2.4) by

$$ \tau ^{\varepsilon }_{2} := \inf \bigg\{ t\ge \tau ^{\varepsilon }_{1} : \int _{\tau ^{\varepsilon }_{1}}^{t} M^{\tau ^{\varepsilon }_{1},m^{ \varepsilon }_{1},h^{\varepsilon }_{1}}_{s} \,ds \ge Z_{2}\bigg\} \le \tau _{2}. $$

In general, for each $n\in \mathbb{N}$, the household, starting from time $\tau ^{\varepsilon }_{n}$, takes

$$ h_{n}^{\varepsilon }(t) := \sum _{k=0}^{n-1} \frac{h_{k}(t) X^{0,x,c ^{(k)},h^{(k)}}_{t}}{X^{0,x,c^{(k)},h^{(k)}}_{t}+\varepsilon e^{rt}} 1_{ \{\tau _{{k}}\le t< \tau _{{k+1}}\}} + \frac{h_{n}(t) X^{0,x,c^{(n)},h ^{(n)}}_{t}}{X^{0,x,c^{(n)},h^{(n)}}_{t}+\varepsilon e^{rt}} 1_{\{t \ge \tau _{{n}}\}} $$

as instantaneous spending rates in healthcare. Set $m^{\varepsilon } _{n}:= M^{\tau ^{\varepsilon }_{n-1},m^{\varepsilon }_{n-1},h^{\varepsilon }_{n-1}}_{\tau ^{\varepsilon }_{n}}$; then the $(n+1)$th death moment is defined as in (2.4) by

$$ \tau ^{\varepsilon }_{n+1} := \inf \bigg\{ t\ge \tau ^{\varepsilon } _{n} : \int _{\tau ^{\varepsilon }_{n}}^{t} M^{\tau ^{\varepsilon }_{n},m ^{\varepsilon }_{n},h^{\varepsilon }_{n}}_{s} \,ds \ge Z_{n+1}\bigg\} \le \tau _{{n+1}}. $$

As in (2.5), we can introduce the counting process

$$ N^{\varepsilon }_{t} := n \qquad \hbox{for}\ t\in [\tau ^{\varepsilon }_{{n}},\tau ^{\varepsilon }_{ {n+1}}). $$

(B.14)

Similarly, define for each $n\in \mathbb{N}$

$$ c_{n}^{\varepsilon }(t) := \sum _{k=0}^{n-1} \frac{c_{k}(t) X^{0,x,c ^{(k)},h^{(k)}}_{t}}{X^{0,x,c^{(k)},h^{(k)}}_{t}+\varepsilon e^{rt}} 1_{ \{\tau _{{k}}\le t< \tau _{{k+1}}\}} + \frac{c_{n}(t) X^{0,x,c^{(n)},h ^{(n)}}_{t}}{X^{0,x,c^{(n)},h^{(n)}}_{t}+\varepsilon e^{rt}} 1_{\{t \ge \tau _{{n}}\}}. $$

Observe that $(c_{n}^{\varepsilon })$, $(h_{n}^{\varepsilon })\in \mathfrak{L}$, and it can be checked that

$$ c^{\varepsilon }_{t} = \sum _{k=0}^{\infty } c_{n}^{\varepsilon }(t) 1_{ \{\tau ^{\varepsilon }_{{k}}\le t< \tau ^{\varepsilon }_{{k+1}}\}}, \qquad h^{\varepsilon }_{t} = \sum _{k=0}^{\infty } h_{n}^{\varepsilon }(t) 1_{ \{\tau ^{\varepsilon }_{{k}}\le t< \tau ^{\varepsilon }_{{k+1}}\}}. $$

This in particular shows that $( c^{\varepsilon },h^{\varepsilon }) \in \mathcal{A}$. In view of (B.13), we have the identity

$$ {X}^{0,x+\varepsilon , c^{\varepsilon },h^{\varepsilon }}_{t} = {X} ^{0,x,c,h}_{t} + \varepsilon e^{rt}. $$

(B.15)

Proposition B.3

Let$w\in C^{1,1}(\mathbb{R}_{+}\times \mathbb{R}_{+})$satisfy (A.1). Suppose for any$(x,m)\in \mathbb{R}^{2}_{+}$, $(c,h)\in \mathcal{A}$and$\varepsilon >0$that

$$\begin{aligned} &\lim _{t\to \infty } \mathbb{E}\bigg[\exp \bigg(- \int _{\tau ^{\varepsilon }_{n}}^{t} (\delta + M^{0,m,h^{\varepsilon } _{n}}_{s}) \,ds\bigg) \\ & \phantom{\lim _{t\to \infty } \mathbb{E}\bigg[}\times w\big(\zeta ^{n} X^{0,x+\varepsilon ,c^{\varepsilon }_{n},h^{\varepsilon }_{n}} _{t},M^{0,m,h^{\varepsilon }_{n}}_{t}\big) \biggm| Z_{1},\dots ,Z_{n} \bigg]= 0, \qquad \forall n\ge 0, \end{aligned}$$

(B.16)

$$\begin{aligned} &\lim _{n\to \infty } \mathbb{E}\big[e^{-\delta \tau ^{\varepsilon } _{n}} w\big(\zeta ^{n} X^{0,x+\varepsilon ,c^{\varepsilon },h^{\varepsilon }}_{\tau ^{\varepsilon }_{n}},M^{0,m,h^{\varepsilon }}_{\tau ^{\varepsilon }_{n}}\big)\big] = 0, \end{aligned}$$

(B.17)

$$\begin{aligned} &\lim _{\varepsilon \to 0} \mathbb{E}\bigg[\int _{0}^{\infty } e^{- \delta t} U(c_{t} \zeta ^{N^{\varepsilon }_{t}} {X}^{0,x,c,h}_{t})\,dt \bigg] = \mathbb{E}\bigg[\int _{0}^{\infty } e^{-\delta t} U(c_{t} \zeta ^{N_{t}} {X}^{0,x,c,h}_{t})\,dt\bigg]. \end{aligned}$$

(B.18)

Then:

(i) $w(x,m)\ge V(x,m)$on$\mathbb{R}_{+}\times \mathbb{R} _{+}$.

(ii) Suppose the measurable functions$\bar{h}$and$\bar{c}$specified in Theorem B.1 (ii) exist so that we can define$(\hat{c},\hat{h})$as in (B.4). If$(\hat{c},\hat{h})$is in$\mathcal{A}$and satisfies (B.2) and (B.3), then$(\hat{c},\hat{h})$is an optimal control for (2.8) and$w(x,m)= V(x,m)$on$\mathbb{R}_{+}\times \mathbb{R}_{+}$.

Proof

We carry out the same arguments as in Theorem B.1. With the aid of (B.16), we obtain (B.5), with $(x,c,h,\tau _{n},N_{t})$ replaced by $(x+\varepsilon ,c^{\varepsilon },h^{\varepsilon },\tau ^{\varepsilon }_{n},N^{\varepsilon }_{t})$. Letting $n\to \infty$ and using (B.17), we get

$$\begin{aligned} w(x+\varepsilon ,m) &\ge \mathbb{E}\left [\int _{0}^{\infty } e^{- \delta t} U( c^{\varepsilon }_{t} \zeta ^{N^{\varepsilon }_{t}}{X}^{0,x+ \varepsilon ,c^{\varepsilon },h^{\varepsilon }}_{t})\,dt\right ] \\ & =\mathbb{E}\left [\int _{0}^{\infty } e^{-\delta t} U(c_{t} \zeta ^{N^{\varepsilon }_{t}} {X}^{0,x,c,h}_{t})\,dt\right ], \end{aligned}$$

where the equality follows from (B.15) and the definition of $c^{\varepsilon }$ in (B.13). Letting $\varepsilon \to 0$, we obtain from (B.18) that $w(x,m)\ge \mathbb{E}[ \int _{0}^{\infty } e^{-\delta t} U(c_{t} \zeta ^{N_{t}} {X}^{0,x,c,h} _{t})\,dt]$. Taking the supremum over $(c,h)\in \mathcal{A}$ leads to $w(x,m)\ge V(x,m)$. The proof of (ii) is the same as Theorem B.1 (ii). □

Theorem B.2 can also be relaxed in a similar fashion.

Proposition B.4

Let$w\in C^{1,2}(\mathbb{R}_{+}\times \mathbb{R}_{+})$satisfy (4.7). Suppose for any$(x,m)\in \mathbb{R}^{2}_{+}$, $(c,h,\pi )\in \mathcal{A}'$and$\varepsilon >0$that (B.16)–(B.18) hold with$X^{0,x,c_{n}^{\varepsilon },h_{n}^{\varepsilon }}$replaced by$X^{0,x,c_{n}^{\varepsilon },h _{n}^{\varepsilon },\pi _{n}^{\varepsilon }}$and$\mathbb{E}[\ \cdot \mid Z_{1},\dots ,Z_{n}]$by$\mathbb{E}[\ \cdot \mid \mathcal{F}_{ \tau _{n}}]$. Then:

(i) $w(x,m)\ge V(x,m)$on$\mathbb{R}_{+}\times \mathbb{R} _{+}$.

(ii) Suppose the measurable functions$(\bar{c}, \bar{h}, \bar{ \pi })$specified in Theorem B.2 (ii) exist so that we can define$(\hat{c},\hat{h}, \hat{\pi })$therein. If$(\hat{c},\hat{h},\hat{\pi })$is in$\mathcal{A}'$and satisfies (B.2) and (B.3) as specified in Theorem B.2, then$(\hat{c},\hat{h}, \hat{\pi })$is an optimal control for the problem (4.5) and$w(x,m)= V(x,m)$on$\mathbb{R}_{+} \times \mathbb{R}_{+}$.