Consumption and portfolio decisions with uncertain lifetimes

Abstract

We study the consumption and portfolio decisions by incorporating mortality risk and altruistic factor in the classical model of Merton (Rev Econ Stat 51:247–257, 1969; J Econ Theory 3:373–413, 1971) and Yaari (Rev Econ Stud 32(2):137–150, 1965). We find that besides the present-biased preference, the process of updating mortality information may be another underlying cause of dynamically time-inconsistent consumption behavior. We use the game-theoretic approach to obtain the extended Hamilton–Jacobi–Bellman equation. Furthermore, we obtain the closed-form solution for the logarithmic utility and explore comparative statics and implications for dynamic behavior. We present numerical results for the power utility that shows the sophisticated individual enjoys higher expected discounted utility than the naive. Our analytical solution enables us to generate a set of testable predictions that are consistent with existing empirical evidence. In particular, we show that for a moderate range of expected investment return, individuals will exhibit a “hump-shaped” consumption pattern, as widely documented in the empirical literature.

Appendices

Appendix A

Define $MRS_t(t'+\tau _1,t'+\tau _2)$ as the marginal rate of substitution of utility at $t'+\tau _1$ for utility at $t'+\tau _2$ from the perspective of time t, $t'+\tau _1>t$, $\tau _2>\tau _1$, where $MRS_t(t'+\tau _1,t'+\tau _2)=\frac{D(t,t'+\tau _2)}{D(t,t'+\tau _1)}$. Following Halevy [35] and Frederick et al. [26], we term this marginal rate of substitution as time discounting-the value at $t'+\tau _1$ of receiving a unit utility at $t'+\tau _2$ from the perspective of time t. According to Read [56], Read and Roelofsma [57], Scholten and Read [60], Kinari et al. [43] and Halevy [35], in the case when the perspective of time is current, i.e. $t=0$, if an individual’s marginal rate of substitution $MRS_0(t'+\tau _1,t'+\tau _2)$ increases with $t'$, the individual is characterized by increasing patience, which is commonly called hyperbolic discounting.

For a wear-out individual characterized by the Weibull distribution with a constant pure time preference, i.e., the hazard rate of death is $\pi (\tau )=\lambda \gamma \tau ^{\gamma -1}$ and the pure time preference factor is $\rho$, then the discount function

$$\begin{aligned} D(0,t'+\tau _1)=(1-\beta )e^{-\lambda {(t'+\tau _1)}^{\gamma }-\rho (t'+\tau _1)}+\beta e^{-\rho (t'+\tau _1)} \end{aligned}$$

(A1)

where $\gamma >1$.

From the definition $MRS_0(t'+\tau _1,t'+\tau _2)$ and Eq. (A1), we have

$$\begin{aligned} MRS_0(t'+\tau _1,t'+\tau _2)=\frac{D(0,t'+\tau _2)}{D(0,t'+\tau _1)}=e^{-\rho (\tau _2-\tau _1)}\frac{(1-\beta )e^{-\lambda (t'+\tau _2)^\gamma }+\beta }{(1-\beta )e^{-\lambda (t'+\tau _1)^\gamma }+\beta } \end{aligned}$$

Therefore,

$$\begin{aligned} \frac{dMRS_0(t'+\tau _1,t'+\tau _2)}{dt'}=e^{-\rho (\tau _2-\tau _1)}\frac{g(t')}{[(1-\beta )e^{-\lambda (t'+\tau _1)^\gamma }+\beta ]^2} \end{aligned}$$

(A2)

where

$$\begin{aligned} g(t')= & {} (1-\beta )\lambda \gamma \{(1-\beta )e^{-\lambda (t'+\tau _2)^\gamma -\lambda (t'+\tau _1)^\gamma } [(t'+\tau _1)^{\gamma -1}-(t'+\tau _2)^{\gamma -1}]\nonumber \\&+\, \beta [e^{-\lambda (t'+\tau _1)^\gamma }(t'+\tau _1)^{\gamma -1}-e^{-\lambda (t'+\tau _2)^{\gamma }}(t'+\tau _2)^{\gamma -1}]\} \end{aligned}$$

(A3)

From Eq. (A3), we can see that $(1-\beta )e^{-\lambda (t'+\tau _2)^\gamma -\lambda (t'+\tau _1)^\gamma } [(t'+\tau _1)^{\gamma -1}-(t'+\tau _2)^{\gamma -1}]$ is less than zero; $\beta [e^{-\lambda (t'+\tau _1)^\gamma }(t'+\tau _1)^{\gamma -1}-e^{-\lambda (t'+\tau _2)^{\gamma }}(t'+\tau _2)^{\gamma -1}]$ is less than zero when $t'$ is sufficiently large. As the time $t'$ increases, $g(t')$ is first less than zero and then eventually greater than zero. Therefore, from Eq. (A2), as time $t'$ increases, the marginal rate of substitution $MRS_0(t'+\tau _1,t'+\tau _2)$ first decreases and then increases. Hence, by the definition of the hyperbolic discounting, Eq. (10) is not a hyperbolic discount function since it does not monotonically increases with $t'$.

Halevy [34] first formally define stationarity and time invariance. We use the concept of marginal rate of substitution to define them, an equivalent way to Halevy’s [34] (see Chen et al. [14]). A preference is stationary if and only if at any fixed decision time, the marginal rate of substitution over two future rewards depends only on the time distance and reward difference between these two alternatives, i.e., $MRS_{t} (t+\Delta _1, t+\Delta _2) = MRS_{t} (t'+\Delta _1, t'+\Delta _2)$. Since $MRS_{0}(t'+\tau _1,t'+\tau _2)$ first decreases and then increases as time $t'$ increases, the discount function of Eq. (10) is non-stationary.

A preference is time invariant if and only if the marginal rate of substitution of any two future two rewards with the same time distance and reward difference does not change long as the time distance between the decision time and rewards remains the same, i.e., $MRS_{t}(t+\Delta _1, t+\Delta _2) = MRS_{t'}(t'+\Delta _1, t'+\Delta _2)$. Since the hazard rate of death is non-constant, we can see that $MRS_0(\tau _1,\tau _2)\ne MRS_t(t+\tau _1,t+\tau _2)$ for $t>0$. Therefore, the discount function of Eq. (10) is time varying.

Appendix B

If the individual does not care about her bequests, then $\beta =0$ and the optimization problem of Eq. (3) becomes

$$\begin{aligned} \max _{c_t(\tau ),\alpha _t(\tau )}E_t\left[ \int _s^{\infty }\Omega _t(\tau )e^{-\rho (\tau -t)}u(c_t(\tau )w(\tau ))d\tau \right] \end{aligned}$$

(B1)

s.t. Eq. (2) and the initial condition (s, w(s)).

Define $\{c_t^*(\tau ),\alpha _t^*(\tau )\}$, $0\le t\le s\le \tau$ as the optimal consumption and portfolio choices and the value function is Q(t, s, w(s)), then

$$\begin{aligned} Q(t,s,w(s))=E_t\left[ \int _s^{\infty }\Omega _t(\tau )e^{-\rho (\tau -t)}u(c_t^*(\tau )w(\tau ))d\tau \right] \end{aligned}$$

(B2)

Furthermore, the value functions Q(0, s, w(s)) and Q(t, s, w(s)) satisfy the following equation

$$\begin{aligned} Q(0,s,w(s))= & {} \max _{c_0(\tau ),\alpha _0(\tau )}E_0\left[ \int _s^{\infty }\Omega (\tau )e^{-\rho \tau } u(c_0(\tau )w(\tau ))d\tau \right] \nonumber \\= & {} \Omega (t)e^{-\rho t}\max _{c_t(\tau ),\alpha _t(\tau )}E_t\left[ \int _s^{\infty }\Omega _t(\tau )e^{-\rho (\tau -t)}u(c_t(\tau )w(\tau ))d\tau \right] \nonumber \\= & {} \Omega (t)e^{-\rho t}Q(t,s,w(s)) \end{aligned}$$

(B3)

Equation (B2) can be written as

$$\begin{aligned} Q(t,s,w(s))=\Omega _t(s)e^{-\rho (s-t)}u(c_t^*(s)w(s))ds+Q(t,s+ds,w(s+ds)) \end{aligned}$$

(B4)

By the Taylor expansion, we have

$$\begin{aligned}&E_t[Q(t,s+ds,w(s+ds))-Q(t,s,w(s))]\nonumber \\&\quad =[r+\alpha _t^*(s)a-c_t^*(s)]w(s)Q_wds +Q_sds+\frac{1}{2}(\alpha _t^*(s))^2{\bar{\sigma }}^2w^2(s)Q_{ww}ds \end{aligned}$$

(B5)

From Eqs. (B4)–(B5), we have

$$\begin{aligned}&Q_s+\left[ r+\alpha _t^*(s)a-c_t^*(s)\right] w(s)Q_w+\frac{1}{2}(\alpha _t^*(s))^2{\bar{\sigma }}^2w^2Q_{ww}\nonumber \\&\quad +\, \Omega _t(s)e^{-\rho (s-t)}u(c_t^*(s)w(s))=0 \end{aligned}$$

(B6)

From Eq. (B6), the optimal instantaneous consumption rate $c_0^*(s)$ and $c_t^*(s))$ must satisfy

$$\begin{aligned} u'(c_t^*(s)w(s))=\frac{Q_w(t,s,w(s))w(s)}{\Omega _t(s)e^{-\rho (s-t)}} \end{aligned}$$

(B7)

and

$$\begin{aligned} u'(c_0^*(s)w(s))=\frac{Q_w(0,s,w(s))w(s)}{\Omega (s)e^{-\rho s}} \end{aligned}$$

(B8)

The optimal instantaneous consumption rate is time consistent if and only if $c_0^*(s)=c_t^*(s)$ for $0<t\le s$. Therefore, from Eqs. (B7)–(B8), we have

$$\begin{aligned} \frac{Q_w(t,s,w(s))}{\Omega _t(s)e^{-\rho (s-t)}}=\frac{Q_w(0,s,w(s))}{\Omega (s)e^{-\rho s}} \end{aligned}$$

(B9)

From Eq. (B3), we obtain Eq. (B9).

By same analysis, the optimal instantaneous portfolio rule is also time consistent.

Appendix C

We first show that

$$\begin{aligned} \frac{dMRS_{t}(\tau _1,\tau _2)}{dt}<0,\quad \tau _2>\tau _1\ge s\ge t \end{aligned}$$

(C1)

and then use Eq. (C1) to show that the planned optimal instantaneous consumption rate is time inconsistent.

By the definition of marginal rate of substitution $MRS_{t}(\tau _1,\tau _2)$, we have

$$\begin{aligned} MRS_{t}(\tau _1,\tau _2)=\frac{D(t,\tau _2)}{D(t,\tau _1)}=\Omega _{\tau _1}(\tau _2)e^{-\rho (\tau _2-\tau _1)}\frac{1+\frac{\beta \Omega (t)}{\Omega (\tau _2)}-\beta }{1+\frac{\beta \Omega (t)}{\Omega (\tau _1)}-\beta } \end{aligned}$$

Therefore,

$$\begin{aligned} \frac{dMRS_{t}(\tau _1,\tau _2)}{dt}=\frac{\Omega _{\tau _1}(\tau _2)e^{-\rho (\tau _2-\tau _1)}}{({1+\frac{\beta \Omega (t)}{\Omega (\tau _1)}-\beta })^2} \frac{\beta (1-\beta )\Omega '(t)[\Omega (\tau _1)-\Omega (\tau _2)]}{\Omega (\tau _2)\Omega (\tau _1)}<0 \end{aligned}$$

since $\Omega '(t)<0$ and $\Omega (\tau _2)<\Omega (\tau _1)$.

Assume that $c_0^*(\tau )$ is the optimal instantaneous consumption rate and $w_0(\tau )$ is the wealth at time $\tau$ from the perspective of time 0, then

$$\begin{aligned} D(0,\tau _1)E_0[u'(c_0^*(\tau _1)w_0(\tau _1))]=D(0,\tau _2)E_0[u'(c_0^*(\tau _2)w_0(\tau _2))] \end{aligned}$$

(C2)

Equation (C1) implies that

$$\begin{aligned} MRS_{t}(\tau _1,\tau _2)<MRS_{0}(\tau _1,\tau _2)\quad \mathrm{for}\quad t>0 \end{aligned}$$

(C3)

Equations (C2)–(C3) imply that

$$\begin{aligned} MRS_{t}(\tau _1,\tau _2)=\frac{D(t,\tau _2)}{D(t,\tau _1)}<MRS_{0}(\tau _1,\tau _2)=\frac{D(0,\tau _2)}{D(0,\tau _1 )}=\frac{E_0[u'(c_0^*(\tau _1)w_0(\tau _1))]}{E_0[u'(c_0^*(\tau _2)w_0(\tau _2))]} \end{aligned}$$

Therefore,

$$\begin{aligned} D(t,\tau _1)E_0[u'(c_0^*(\tau _1)w_0(\tau _1))]>D(t,\tau _2)E_0[u'(c_0^*(\tau _2)w_0(\tau _2) )] \end{aligned}$$

(C4)

Equation (C4) implies that the planned optimal instantaneous consumption rate $c_0^*(\tau )$ from the perspective of time 0 is not optimal from the perspective of time t and the individual will increase her welfare by deviating from it.

Since $u'(\cdot )>0$ and $u''(\cdot )<0$, the individual increases her welfare by raising the instantaneous consumption rate at $\tau _1$ and lowering the instantaneous consumption rate at $\tau _2$ until

$$\begin{aligned} D(t,\tau _1)E_t[u'(c_t^*(\tau _1)w_t(\tau _1))]=D(t,\tau _2)E_t[u'(c_t^*(\tau _2)w_t(\tau _2) )] \end{aligned}$$

(C5)

Therefore, when an individual can re-optimize, she has an incentive to deviate from her ex-ante plan and increase the instantaneous consumption rate, i.e., $c_\tau ^*(\tau )>c_t^*(\tau )>c_0^*(\tau )$, where $\tau \ge s\ge t>0$.

Appendix D

Since

$$\begin{aligned} D(t,\tau )=e^{-\rho \epsilon }D(t+\epsilon ,\tau )+(1-\beta )e^{-\rho (\tau -t-\epsilon )}e^{-\int _{t+\epsilon }^\tau \pi (y)dy}[e^{-\int _t^{t+\epsilon }\pi (y)dy}-1]e^{-\rho \epsilon }\nonumber \\ \end{aligned}$$

(D1)

we have

$$\begin{aligned} \begin{aligned} V(t,w(t))&=\max _{c(\tau ),\alpha (\tau )}E_t\left[ \int _t^{\infty }D(t,\tau )u(c(\tau )w(\tau ))d\tau \right] \\&=\max _{c(t),\alpha (t)}E_t\left[ u(c(t)w(t))\epsilon +e^{-\rho \epsilon }V(t+\epsilon ,w(t+\epsilon )) -K(t+\epsilon ,w(t+\epsilon ))\epsilon +o(\epsilon )\right] \\ \end{aligned} \end{aligned}$$

(D2)

where

$$\begin{aligned} K(t,w(t))=(1-\beta )\pi (t)\int _{t}^{\infty }e^{-\rho (\tau -t)}e^{-\int _{t}^\tau \pi (y)dy}u({\hat{c}}(\tau ){\hat{w}}(\tau ))d\tau \end{aligned}$$

(D3)

and ${\hat{c}}(\tau )$ is the solution of Eq. (D2) and ${\hat{w}}(\tau )$ is the corresponding Markovian state.

From Eq. (D2), we have

$$\begin{aligned} \begin{aligned} V(t,w(t))&=\max _{c(t),\alpha (t)}E_t\left[ u(c(t)w(t))\epsilon +(1-\rho \epsilon )V(t+\epsilon ,w(t+\epsilon ))\right. \\&\left. \quad -K(t+\epsilon ,w(t+\epsilon ))\epsilon +o(\epsilon )\right] \\ \end{aligned} \end{aligned}$$

(D4)

From Eq. (D4), we have

$$\begin{aligned} \begin{aligned} 0&=\max _{c(t),\alpha (t)}E_t[u(c(t)w(t))+\frac{V(t+\epsilon ,w(t+\epsilon ))-V(t,w(t))}{\epsilon }\\&\quad -\rho V(t+\epsilon ,w(t+\epsilon )) -K(t+\epsilon ,w(t+\epsilon ))+o(\epsilon )]\\ \end{aligned} \end{aligned}$$

(D5)

From Eq. (2), $w(t+\epsilon )$ satisfies the following equation

$$\begin{aligned} w(t+\epsilon )=w(t)+\mu (t,c(t),\alpha (t))\epsilon +{\bar{\sigma }}\alpha (t)w(t)(z(t+\epsilon )-z(t)) \end{aligned}$$

(D6)

where $\mu (t,c(t),\alpha (t))=\alpha (t)aw(t)+rw(t)-c(t)w(t)$. From Eq. (D6), we have

$$\begin{aligned}&\lim _{\epsilon \rightarrow 0}\frac{E_t[V((t+\epsilon ),w(t+\epsilon ))-V(t,w(t))]}{\epsilon }\nonumber \\&\quad =V_t+V_w \mu (t,{\hat{c}}(t),{\hat{\alpha }}(t))+\frac{1}{2}V_{ww}{\hat{\alpha }}^2(t){\bar{\sigma }}^2 w^2(t) \end{aligned}$$

(D7)

Combining Eqs. (D5) and (D7) and letting $\epsilon \rightarrow 0$, we obtain

$$\begin{aligned}&\rho V(t,w(t))-V_t+K(t,w(t))\nonumber \\&\quad =\max _{\alpha (t),c(t)}\left[ u(c(t)w(t))+V_w \mu (t,c(t),\alpha (t))+\frac{1}{2}V_{ww}\alpha ^2(t){\bar{\sigma }}^2 w^2(t)\right] \end{aligned}$$

(D8)

where

$$\begin{aligned} K(t,w(t))=(1-\beta )\pi (t)\int _t^{\infty } e^{-\int _t^y\pi (m)dm}e^{-\rho (y-t)}E_t\left[ u({\hat{c}}(y){\hat{w}}(y))\right] dy \end{aligned}$$

(D9)

and ${\hat{c}}(t)$ satisfies Eq. (D8) and ${\hat{w}}(t)$ is the corresponding Markovian state.

Thus, the HJB equation of problem (15) subject to Eq. (2) is Eqs. (D8)–(D9) with the following boundary condition:

$$\begin{aligned} \lim _{\tau \rightarrow \infty }E_t[e^{-\rho (\tau -t)}V(\tau ,{\hat{w}}(\tau ))]=0 \end{aligned}$$

(D10)

Appendix E

Since $V(t,w(t))=A(t)\ln w(t)+B(t)$, we have

$$\begin{aligned} V_t=A'(t)\ln w(t)+B'(t),\quad V_w=\frac{A(t)}{w(t)},\quad V_{ww}=-\frac{A(t)}{(w(t))^2} \end{aligned}$$

Thus, from Eq. (21), we obtain

$$\begin{aligned} {\hat{c}}(t)=\frac{1}{A(t)},\quad {\hat{\alpha }}(t)=\frac{-V_wa}{V_{ww}w{\bar{\sigma }}^2}=\frac{a}{{\bar{\sigma }}^2} \end{aligned}$$

(E1)

and the corresponding Markovian state trajectory becomes

$$\begin{aligned} d{\hat{w}}(y)= \left[ \frac{a^2}{{\bar{\sigma }}^2}+r-\frac{1}{A(y)}\right] {\hat{w}}(y)dy+\frac{a}{{\bar{\sigma }}}{\hat{w}}(y)dz(y) \end{aligned}$$

(E2)

Substituting Eq. (E1) into Eq. (22), we obtain

$$\begin{aligned} K(t,w(t))= & {} (1-\beta )\pi (t)\int _t^{\infty } e^{-\int _t^x\pi (m)dm}e^{-\rho (x-t)}E_t[u({\hat{c}}(x){\hat{w}}(x))]dx\nonumber \\= & {} (1-\beta )\pi (t)\int _t^{\infty } e^{-\int _t^x\pi (m)dm}e^{-\rho (x-t)}E_t[\ln {\hat{w}}(x)-\ln A(x)]dx \end{aligned}$$

(E3)

From Eq. (E2), we have

$$\begin{aligned} d\ln {\hat{w}}(x)=\left[ \frac{a^2}{2{\bar{\sigma }}^2}+r-\frac{1}{A(x)}\right] dx+\frac{a}{{\bar{\sigma }}}dz(x) \end{aligned}$$

(E4)

Hence

$$\begin{aligned} E_t[\ln {\hat{w}}(x)]=\ln w(t)+\int _t^x\left[ \frac{a^2}{2{\bar{\sigma }}^2}+r-\frac{1}{A(y)}\right] dy \end{aligned}$$

(E5)

Substituting Eq. (E5) into Eq. (E3), K(t, w(t)) can be expressed as

$$\begin{aligned} K(t,w(t))= & {} (1-\beta )\pi (t)\left[ \ln w(t)\int _t^{\infty } e^{-\int _t^x\pi (m)dm}e^{-\rho (x-t)}dx\right. \nonumber \\&\left. +\int _t^{\infty } e^{-\int _t^x\pi (m)dm}e^{-\rho (x-t)}\int _t^x\left( \frac{a^2}{2{\bar{\sigma }}^2}+r-\frac{1}{A(y)}\right) dydx\right. \nonumber \\&\left. -\int _t^{\infty } e^{-\int _t^x\pi (m)dm-\rho (x-t)}\ln A(x)dx\right] \end{aligned}$$

(E6)

Substituting Eqs. (E1) and (E6) into Eq. (21), we obtain

$$\begin{aligned}&\rho (A(t)\ln w(t)+B(t))+(1-\beta )\pi (t)\left[ \ln w(t)\int _t^{\infty } e^{-\int _t^x\pi (m)dm}e^{-\rho (x-t)}dx\right. \nonumber \\&\left. \qquad +\int _t^{\infty } e^{-\int _t^x\pi (m)dm}e^{-\rho (x-t)}\int _t^x\left( \frac{a^2}{2{\bar{\sigma }}^2}+r-\frac{1}{A(y)}\right) dydx\right. \nonumber \\&\left. \qquad -\int _t^{\infty } e^{-\int _t^x\pi (m)dm-\rho (x-t)}\ln A(x)dx\right] -A'(t)\ln w(t)-B'(t)\nonumber \\&=\quad \ln w(t)-\ln A(t)+\left( \frac{a^2}{2{\bar{\sigma }}^2}+r-\frac{1}{A(t)}\right) A(t) \end{aligned}$$

(E7)

Because Eq. (E7) must hold for every t and w(t), A(t) satisfy the following ordinary differential equation:

$$\begin{aligned} A'(t)=\rho A(t)+(1-\beta )\pi (t)\int _t^{\infty } e^{-\int _t^x\pi (m)dm}e^{-\rho (x-t)}dx-1 \end{aligned}$$

(E8)

We solve Eq. (E8) and obtain

$$\begin{aligned} A(t)=(1-\beta )\int _t^{\infty }e^{-\lambda (x^\gamma -t^\gamma )}e^{-\rho (x-t)} dx+\frac{\beta }{\rho } \end{aligned}$$

(E9)

From Eqs. (E1) and (E9), we obtain the consumption rate determined by Eqs. (21)–(24)

$$\begin{aligned} {\hat{c}}(t)=\frac{1}{(1-\beta )\int _t^{\infty }e^{-\lambda (x^\gamma -t^\gamma )}e^{-\rho (x-t)} dx+\frac{\beta }{\rho }} \end{aligned}$$

(E10)

Appendix F

Since $V(t,w(t))=h(t)\frac{(w(t))^{1-b }}{1-b}$, we have

$$\begin{aligned} V_t=h'(t)\frac{(w(t))^{1-b}}{1-b},\quad V_w=h(t)(w(t))^{-b},\quad V_{ww}=-bh(t)(w(t))^{-1-b} \end{aligned}$$

Therefore, from Eq. (21), we have

$$\begin{aligned} {\hat{c}}(t)=(h(t))^{-\frac{1}{b}},\quad {\hat{\alpha }}(t)=\frac{-V_wa}{V_{ww}w{\bar{\sigma }}^2}=\frac{a}{b{\bar{\sigma }}^2} \end{aligned}$$

(F1)

and the corresponding state trajectory becomes

$$\begin{aligned} d{\hat{w}}(x)=\left[ \frac{a^2}{b{\bar{\sigma }}^2}+r-(h(x))^{-\frac{ 1}{b}}\right] {\hat{w}}(x)dx+\frac{a}{b{\bar{\sigma }}}{\hat{w}}(x)dz(x) \end{aligned}$$

(F2)

Substituting Eq. (F1) into Eq. (22), we obtain

$$\begin{aligned} K(t,w(t))=(1-\beta )\pi (t)\int _t^{\infty } e^{-\int _t^x\pi (m)dm}e^{-\rho (x-t)}E_t\left[ \frac{(h(x))^{-\frac{1-b}{b}} ({\hat{w}}(x))^{1-b}}{1-b}\right] dx\nonumber \\ \end{aligned}$$

(F3)

From Eq. (F2), we have

$$\begin{aligned} E_t[(w(x))^{1-b}]=(w(t))^{1-b} \exp \left[ (1-b)\int _t^x\left( \frac{a^2}{2b{\bar{\sigma }}^2}+r-(h(y))^{-\frac{1}{b}}\right) dy\right] \end{aligned}$$

(F4)

Substituting Eq. (F4) into Eq. (F3), K(t, w(t)) can be expressed as

$$\begin{aligned} K(t,w(t))= & {} (1-\beta )\pi (t)\frac{(w(t))^{1-b}}{1-b}\left\{ \int _t^{\infty } e^{-\int _t^x\pi (m)dm}e^{-\rho (x-t)}(h(x))^{-\frac{1-b}{b}}\right. \nonumber \\&\left. \times \exp \left[ (1-b)\int _t^x\left( \frac{a^2}{2b{\bar{\sigma }}^2}+r-(h(y))^{-\frac{1}{b}}\right) dy\right] dx\right\} \end{aligned}$$

(F5)

Substituting Eqs. (F1) and (F5) into Eq. (21), we obtain

$$\begin{aligned} h'(t)= & {} \left[ \rho -(1-b)\left( \frac{a^2}{2b{\bar{\sigma }}^2}+r\right) \right] h(t)-b(h(t))^{1-\frac{1}{b}}\nonumber \\&+(1-\beta )\pi (t)\left\{ \int _t^{\infty } e^{-\int _t^x\pi (m)dm}e^{-\rho (x-t)} h(x))^{1-\frac{1}{b}}\right. \nonumber \\&\left. \times \exp \left[ (1-b)\int _t^x\left( \frac{a^2}{2b{\bar{\sigma }}^2}+r-(h(y))^{-\frac{1}{b}}\right) dy\right] dx\right\} \end{aligned}$$

(F6)

Because of the individual’s finite lifetime, her wealth will be inherited by her heirs after her death, the death time denoted by S. Following in Merton [49], we assume that the bequest function takes the form of $H(S,w(S))=\xi ^b\frac{(w(S))^{1-b}}{1-b}$, which means $V(S,w(S))=\beta H(S,w(S))$ and $H(S)=\beta \xi ^b$, where $\xi >0$. Therefore, $\lim _{t\rightarrow \infty }h(t)=\beta \xi ^b$.

Appendix G

For any consumption and portfolio rules $\varphi (\tau ,w(\tau ))=(\alpha (\tau ),c(\tau ))$, we denote

$$\begin{aligned} \mu ^{\varphi }(\tau ,w(\tau ))=\mu (\tau ,w(\tau ),\varphi )=[a\alpha (\tau )+r-c(\tau )]w(\tau ) \end{aligned}$$

(G1)

and

$$\begin{aligned} \sigma ^{\varphi }(\tau ,w(\tau ))=\sigma (\tau ,w(\tau ),\varphi )={\bar{\sigma }}\alpha (\tau )w(\tau ) \end{aligned}$$

(G2)

Denote $U^{t,w(t),\varphi }(\tau ,w(\tau ))$ as the discounted instantaneous utility when the initial condition is (t, w(t)) and the individual follows the strategy $\varphi$, where

$$\begin{aligned}&U^{t,w(t),\varphi _{t,\epsilon ,\varsigma }}(\tau ,w(\tau ))=D(t,\tau )u(c^{\epsilon }(\tau )w^{\epsilon }(\tau )),\quad U^{t,w(t),{\hat{\varphi }}}(\tau ,w(\tau ))\nonumber \\&\quad =D(t,\tau )u({\hat{c}}(\tau ){\hat{w}}(\tau )) \end{aligned}$$

(G3)

Denote ${\bar{u}}^{t,w(t),x}(\tau ,w(\tau ))$ as the expected discounted instantaneous utility when the individual follows the strategy ${\hat{\varphi }}$ and the wealth at time $\tau$ is $w(\tau )$, where

$$\begin{aligned}&{\bar{u}}^{t,w(t),x}(\tau ,w(\tau ))\nonumber \\&\quad =E_{\tau ,w(\tau )}[U^{t,w(t),{\hat{\varphi }}}(x,w(x))]=D(t,x)E_{\tau ,w(\tau )}[u({\hat{c}}(x){\hat{w}}(x))],\quad \tau \in [t,x]\nonumber \\ \end{aligned}$$

(G4)

Consider a function $\xi ^x(\tau ,w(\tau ))$ for $\tau \in [t,x]$ and $w(\tau )\in (0,\infty )$, where $\xi ^x(\tau ,w(\tau ))$ has continuous first-order partial derivatives with $\tau \in [t,x]$ and $w(\tau )\in (0,\infty )$ and a continuous second-order partial derivative with $x\in (0,\infty )$, and denote these derivatives as $\frac{\partial \xi ^x(\tau ,w(\tau ))}{\partial \tau }$, $\frac{\partial \xi ^x(\tau ,w(\tau ))}{\partial w}$, and $\frac{\partial ^2\xi ^x(\tau ,w(\tau ))}{\partial w^2}$, respectively. For any $\xi ^x(\tau ,w(\tau ))$, denote

$$\begin{aligned}&\Lambda ^{\varphi }\xi ^x(\tau ,w(\tau ))=\frac{\partial \xi ^x(\tau ,w(\tau ))}{\partial \tau }+\mu ^{\varphi }(\tau ,w(\tau ))\frac{\partial \xi ^x(\tau ,w(\tau ))}{\partial w}\nonumber \\&\quad +\frac{1}{2}(\sigma ^{\varphi }(\tau ,w(\tau )))^2\frac{\partial ^2\xi ^x(\tau ,w(\tau ))}{\partial w^2} \end{aligned}$$

(G5)

where $\tau \in [t,x]$, $w(\tau )\in (0,\infty )$.

Following the same argument as in Lemma 3 in He and Jiang [37], for the logarithmic utility function and the power utility function, we have

$$\begin{aligned} J(t,w(t),\varphi _{t,\epsilon ,\varsigma })-J(t,w(t),{\hat{\varphi }})=\epsilon \Gamma ^{{\hat{\varphi }}}(t,w(t),\varsigma )+\epsilon o(1) \end{aligned}$$

(G6)

where

$$\begin{aligned}&\Gamma ^{{\hat{\varphi }}}(t,w(t),\varsigma )=U^{t,w(t),\varsigma }(t,w(t))-U^{t,w(t),{\hat{\varphi }}}(t,w(t))\nonumber \\&\quad +\int _t^\infty \Lambda ^{\varsigma }{\bar{u}}^{t,w(t),x}(t,w(t))dx \end{aligned}$$

(G7)

and o(1) is infinitesimal and $\lim _{\epsilon \rightarrow 0}\frac{o(1)}{\epsilon }=0$.

(1) The logarithmic utility function

For the logarithmic utility function, ${\hat{\varphi }}=({\hat{c}}(t),{\hat{\alpha }}(t))$ is the solution of Eqs. (21)–(24), where

$$\begin{aligned} {\hat{\alpha }}(t)=\frac{a}{{\bar{\sigma }}^2},\quad {\hat{c}}(t)=\frac{1}{(1-\beta )\int _t^{\infty }e^{-\lambda (y^\gamma -t^\gamma )}e^{-\rho (y-t)} dy+\frac{\beta }{\rho }} \end{aligned}$$

(G8)

Denote ${\hat{w}}(\tau )$ as the above Markovian state of consumption and portfolio rules. From Eqs. (2) and (G8), we have

$$\begin{aligned} d{\hat{w}}(\tau )=[a{\hat{\alpha }}(\tau )+r-{\hat{c}}(\tau )]{\hat{w}}(\tau )d\tau +{\bar{\sigma }}{\hat{\alpha }}(\tau ){\hat{w}}(\tau )dz(\tau ) \end{aligned}$$

(G9)

From Eq. (G9), we have

$$\begin{aligned} \ln {\hat{w}}(\tau )=\left( \frac{a^2}{{\bar{\sigma }}^2}+r\right) (\tau -t)-\int _t^\tau {\hat{c}}(y)dy+\frac{a}{{\bar{\sigma }}}(z(\tau )-z(t))+\ln w(t) \end{aligned}$$

(G10)

Denote $w^{\epsilon }(\tau )$ as the Markovian state of consumption and portfolio rules $\varphi _{t,\epsilon ,\varsigma }(\tau ,w(\tau ))$, which is defined in Eq. (17), $\tau \ge t$. If the individual follows the strategy $\varphi _{t,\epsilon ,\varsigma }$, the state trajectory $w^{\epsilon }(\tau )$ becomes

$$\begin{aligned} dw^\epsilon (\tau )=[a\alpha ^\epsilon (\tau )+r-c^\epsilon (\tau )]w^\epsilon (\tau )d\tau +{\bar{\sigma }}\alpha ^\epsilon (\tau )w^\epsilon (\tau )dz(\tau ) \end{aligned}$$

(G11)

From Eqs. (G4) and (G10), for the logarithmic utility function, we have

$$\begin{aligned}&{\bar{u}}^{t,w(t),x}(\tau ,w(\tau ))\nonumber \\&\quad =D(t,x)\left[ \ln {\hat{c}}(x)+\left( \frac{a^2}{{\bar{\sigma }}^2}+r\right) (x-\tau )-\int _{\tau }^x {\hat{c}}(y)dy+\ln {\hat{w}}(\tau )\right] \end{aligned}$$

(G12)

From Eq. (G3), we have

$$\begin{aligned} U^{t,w(t),\varsigma }(t,w(t))-U^{t,w(t),{\hat{\varphi }}}(t,w(t))=\ln \frac{c^\epsilon (t)}{{\hat{c}}(t)} \end{aligned}$$

(G13)

From Eqs. (G1), (G2), (G5) and (G12), it is easy to see that

$$\begin{aligned} \begin{aligned} \Lambda ^{\varsigma }{\bar{u}}^{t,w(t),x}(\tau ,w(\tau ))|_{t=\tau }&=\left\{ \frac{\partial {\bar{u}}}{\partial \tau }+ \mu ^{\varsigma }(\tau ,w(\tau ))\frac{\partial {\bar{u}}}{\partial w}+\frac{1}{2}(\sigma ^{\varsigma }(\tau ,w(\tau )))^2\frac{\partial ^2{\bar{u}}}{\partial w^2}\right\} |_{t=\tau }\\&=D(t,x)\left\{ a[\alpha ^\epsilon (t)-{\hat{\alpha }}(t)]-[c^\epsilon (t)-{\hat{c}}(t)]-\frac{1}{2}{\bar{\sigma }}^2(\alpha ^\epsilon (t))^2\right\} \end{aligned}\nonumber \\ \end{aligned}$$

(G14)

From Eq. (G14), we have

$$\begin{aligned} \begin{aligned} \int _t^\infty \Lambda ^{\varsigma }{\bar{u}}dx&=\left\{ a[\alpha ^\epsilon (t)-{\hat{\alpha }}(t)]-[c^\epsilon (t)-{\hat{c}}(t)]-\frac{1}{2}{\bar{\sigma }}^2(\alpha ^\epsilon (t))^2\right\} \int _t^\infty D(t,x)dx\\&=\left\{ a[\alpha ^\epsilon (t)-{\hat{\alpha }}(t)]-[c^\epsilon (t)-{\hat{c}}(t)]-\frac{1}{2}{\bar{\sigma }}^2(\alpha ^\epsilon (t))^2\right\} \frac{1}{{\hat{c}}(t)} \end{aligned} \end{aligned}$$

(G15)

From Eqs. (G7), (G13) and (G15), we have

$$\begin{aligned}&\Gamma ^{{\hat{\varphi }}}(t,w(t),\varsigma )=[\ln c^\epsilon (t)-\ln {\hat{c}}(t)]\nonumber \\&\quad +\left\{ a[\alpha ^\epsilon (t)-{\hat{\alpha }}(t)]+[{\hat{c}}(t)-c^\epsilon (t)]-\frac{1}{2}{\bar{\sigma }}^2(\alpha ^\epsilon (t))^2\right\} \frac{1}{{\hat{c}}(t)} \end{aligned}$$

(G16)

Since both ${\hat{\alpha }}(t)$ and ${\hat{c}}(t)$ are constant, it is straightforward to see the maximum $\Gamma ^{{\hat{\varphi }}}(t,w(t),\varsigma )\le 0$. Furthermore, $\Gamma ^{{\hat{\varphi }}}(t,w(t),\varsigma )=0$ if and only if ${\hat{\varphi }}(t,w(t))\equiv \varsigma (t,w(t))$.

From the above analysis, for the logarithmic utility function, the strategy ${\hat{\varphi }}(t,w(t))$ determined by Eqs. (21)–(24) satisfies Eq. (19) and thus is an equilibrium strategy, i.e.,

$$\begin{aligned} J(t,w(t),{\hat{\varphi }})\ge J(t,w(t),\varphi _{t,\epsilon ,\varsigma }) \end{aligned}$$

for any sufficiently small $\epsilon$.

Furthermore, following the same argument as in Theorem 1 in He and Jiang [37], the strategy is a weak and regular equilibrium strategy.

(2) The Power utility function

For the power utility function, ${\hat{\varphi }}=({\hat{c}}(t),{\hat{\alpha }}(t))$ is the solution of Eqs. (21)–(24),

$$\begin{aligned} {\hat{c}}(t)=(h(t))^{-\frac{1}{b}},\quad {\hat{\alpha }}(t)=\frac{a}{b{\bar{\sigma }}^2} \end{aligned}$$

(G17)

where h(t) satisfies the following non-linear integro-differential equation

$$\begin{aligned} \begin{aligned} h'(t)&=\left[ \rho -(1-b)\left( \frac{a^2}{2b{\bar{\sigma }}^2}+r\right) \right] h(t)-b((h(t))^{1-\frac{1}{b}}\\&\quad +(1-\beta )\pi (t)\left\{ \int _t^{\infty } e^{-\int _t^x\pi (m)dm}e^{-\rho (x-t)} (h(x))^{1-\frac{1}{b}}\right. \\&\left. \quad \times \exp \left[ (1-b)\int _t^x\left( \frac{a^2}{2b{\bar{\sigma }}^2}+r-(h(y))^{-\frac{1}{b}}\right) dy\right] dx\right\} \end{aligned} \end{aligned}$$

(G18)

and $\lim _{t\rightarrow \infty }h(t)=\beta \xi ^b$

Denote ${\hat{w}}(\tau )$ as the above Markovian state of consumption and portfolio rules. From Eqs. (2) and (G17), we have

$$\begin{aligned} d{\hat{w}}(\tau )=[a{\hat{\alpha }}(\tau )+r-{\hat{c}}(\tau )]{\hat{w}}(\tau )d\tau +{\bar{\sigma }}{\hat{\alpha }}(\tau ){\hat{w}}(\tau )dz(\tau ) \end{aligned}$$

(G19)

and therefore

$$\begin{aligned} E_t[({\hat{w}}(x))^{1-b}]=(w(t))^{1-b}\exp \left[ (1-b)\int _t^x\left( \frac{a^2}{2b{\bar{\sigma }}^2}+r-(h(y))^{-\frac{1}{b}}\right) dy\right] \end{aligned}$$

(G20)

Denote $w^{\epsilon }(\tau )$ as the Markovian state of consumption and portfolio rules $\varphi _{t,\epsilon ,\varsigma }(\tau ,w(\tau ))$, which is defined in Eq. (17), $\tau \ge t$. If the individual follows the strategy $\varphi _{t,\epsilon ,\varsigma }$, the state trajectory $w^{\epsilon }(\tau )$ becomes

$$\begin{aligned} dw^\epsilon (\tau )=[a\alpha ^\epsilon (\tau )+r-c^\epsilon (\tau )]w^\epsilon (\tau )d\tau +{\bar{\sigma }}\alpha ^\epsilon (\tau )w^\epsilon (\tau )dz(\tau ) \end{aligned}$$

(G21)

From Eqs. (G4) and (G20), for the power utility function, we have

$$\begin{aligned}&{\bar{u}}^{t,w(t),x}(\tau ,w(\tau ))\nonumber \\&\quad =D(t,x)\frac{({\hat{c}}(x))^{1-b}}{1-b}({\hat{w}}(\tau ))^{1-b}\exp \left[ (1-b)\int _\tau ^x\left( \frac{a^2}{2b{\bar{\sigma }}^2}+r-(h(y))^{-\frac{1}{b}}\right) dy\right] \nonumber \\ \end{aligned}$$

(G22)

From Eq. (G3), we have

$$\begin{aligned} U^{t,w(t),\varsigma }(t,w(t))-U^{t,w(t),{\hat{\varphi }}}(t,w(t))=\frac{(w(t))^{1-b}}{1-b}\left[ (c^\epsilon (t))^{1-b}-({\hat{c}}(t))^{1-b}\right] \nonumber \\ \end{aligned}$$

(G23)

From Eqs. (G1), (G2), (G5), (G17) and (G22), it is easy to see that

$$\begin{aligned} \begin{aligned}&\Lambda ^{\varsigma }{\bar{u}}^{t,w(t),x}(\tau ,w(\tau ))|_{t=\tau }=\left\{ \frac{\partial {\bar{u}}}{\partial \tau }+ \mu ^{\varsigma }(\tau ,w(\tau ))\frac{\partial {\bar{u}}}{\partial w}+\frac{1}{2}(\sigma ^{\varsigma }(\tau ,w(\tau )))^2\frac{\partial ^2{\bar{u}}}{\partial w^2}\right\} |_{t=\tau }\\&\quad =(w(t))^{1-b}\left[ -\frac{a^2}{2b{\bar{\sigma }}^2}+({\hat{c}}(t)-c^\epsilon (t))+\left( a\alpha ^\epsilon (t)-\frac{1}{2}b{\bar{\sigma }}^2(\alpha ^\epsilon (t))^2\right) \right] \\&\qquad \times D(t,x)({\hat{c}}(x))^{1-b}\exp \left[ (1-b)\int _t^x\left( \frac{a^2}{2b{\bar{\sigma }}^2}+r-(h(y))^{-\frac{1}{b}}\right) dy\right] \\&\quad =(w(t))^{1-b}\left[ -\frac{a^2}{2b{\bar{\sigma }}^2}+({\hat{c}}(t)-c^\epsilon (t))+\left( a\alpha ^\epsilon (t)-\frac{1}{2}b{\bar{\sigma }}^2(\alpha ^\epsilon (t))^2\right) \right] \\&\qquad \times D(t,x)(h(x))^{1-\frac{1}{b}}\exp \left[ (1-b)\int _t^x(\frac{a^2}{2b{\bar{\sigma }}^2}+r-(h(y))^{-\frac{1}{b}})dy\right] \end{aligned} \end{aligned}$$

(G24)

From Eq. (G24), we have

$$\begin{aligned} \begin{aligned}&\int _t^\infty \Lambda ^{\varsigma }{\bar{u}}^{t,w(t),x}(t,w(t))dx =(w(t))^{1-b}\left[ \frac{-a^2}{2b{\bar{\sigma }}^2}+({\hat{c}}(t)-c^\epsilon (t))+\left( a\alpha ^\epsilon (t)-\frac{1}{2}b{\bar{\sigma }}^2(\alpha ^\epsilon (t))^2\right) \right] \\&\quad \times \int _t^\infty D(t,x)(h(x))^{1-\frac{1}{b}}\exp \left[ (1-b)\int _t^x\left( \frac{a^2}{2b{\bar{\sigma }}^2}+r-(h(y))^{-\frac{1}{b}}\right) dy\right] dx \end{aligned}\nonumber \\ \end{aligned}$$

(G25)

Denote $f(t)=\int _t^\infty D(t,x)(h(x))^{1-\frac{1}{b}}\exp \left[ (1-b)\int _t^x(\frac{a^2}{2b{\bar{\sigma }}^2}+r-(h(y))^{-\frac{1}{b}})dy\right] dx$. It is easy to see that f(t) satisfies Eq. (28) and $\lim _{t\rightarrow \infty }f(t)=\beta \xi ^b$. Therefore, $f(t)\equiv h(t)$ and Eq. (G25) becomes

$$\begin{aligned} \begin{aligned}&\int _t^\infty \Lambda ^{\varsigma }{\bar{u}}^{t,w(t),x}(t,w(t))dx\\&\quad =(w(t))^{1-b}\left[ \frac{-a^2}{2b{\bar{\sigma }}^2}+({\hat{c}}(t)-c^\epsilon (t))+\left( a\alpha ^\epsilon (t)-\frac{1}{2}b{\bar{\sigma }}^2(\alpha ^\epsilon (t))^2\right) \right] h(t)\\&\quad =(w(t))^{1-b}\left[ \frac{-a^2}{2b{\bar{\sigma }}^2}+({\hat{c}}(t)-c^\epsilon (t))+\left( a\alpha ^\epsilon (t)-\frac{1}{2}b{\bar{\sigma }}^2(\alpha ^\epsilon (t))^2\right) \right] ({\hat{c}}(t))^{-b} \end{aligned} \end{aligned}$$

(G26)

From Eqs. (G7), (G23) and (G26), we have

$$\begin{aligned} \begin{aligned} \Gamma ^{{\hat{\varphi }}}(t,w(t),\varsigma )&=\frac{(w(t))^{1-b}}{1-b}\left[ (c^\epsilon (t))^{1-b}-({\hat{c}}(t))^{1-b}\right] \\&\quad +(w(t))^{1-b}\left[ \frac{-a^2}{2b{\bar{\sigma }}^2}+({\hat{c}}(t)-c^\epsilon (t))+\left( a\alpha ^\epsilon (t)-\frac{1}{2}b{\bar{\sigma }}^2(\alpha ^\epsilon (t))^2\right) \right] ({\hat{c}}(t))^{-b}\\ \end{aligned} \end{aligned}$$

(G27)

Since both ${\hat{\alpha }}(t)$ and ${\hat{c}}(t)$ are constant, it is straightforward to see the maximum $\Gamma ^{{\hat{\varphi }}}(t,w(t),\varsigma )\le 0$. Furthermore, $\Gamma ^{{\hat{\varphi }}}(t,w(t),\varsigma )=0$ if and only if ${\hat{\varphi }}(t,w(t))\equiv \varsigma (t,w(t))$.

From the above analysis, for the power utility function, the strategy ${\hat{\varphi }}(t,w(t))$ determined by Eqs. (21)–(24) satisfies Eq. (19) and thus is an equilibrium strategy, i.e.,

$$\begin{aligned} J(t,w(t),{\hat{\varphi }})\ge J(t,w(t),\varphi _{t,\epsilon ,\varsigma }) \end{aligned}$$

for any sufficiently small $\epsilon$.

Furthermore, following the same argument as in Theorem 1 in He and Jiang [37], the strategy is a weak and regular equilibrium strategy.

Appendix H

We prove $\frac{\partial {\hat{c}}(t)}{\partial t}>0$ and ${\hat{c}}(t)=\frac{1}{(1-\beta )\int _t^{\infty }e^{-\lambda (x^\gamma -t^\gamma )}e^{-\rho (x-t)} dx+\frac{\beta }{\rho }}\rightarrow \frac{\rho }{\beta }$ as $t\rightarrow \infty$ and $\gamma >1$. From Eq. (26), we have

$$\begin{aligned} \frac{\partial {\hat{c}}(t)}{\partial t}=\frac{(1-\beta )\left[ 1-(\lambda \gamma t^{\gamma -1}+\rho )\int _t^{\infty }(e^{-\lambda (x^\gamma -t^\gamma )}e^{-\rho (x-t)})dx\right] }{\left( (1-\beta )\int _t^{\infty } e^{-\lambda (x^\gamma -t^\gamma )}e^{-\rho (x-t)}dx+\frac{\beta }{\rho }\right) ^2} \end{aligned}$$

(H1)

By defining $x=t+y$, from Eq. (H1), we have

$$\begin{aligned} \frac{\partial {\hat{c}}(t)}{\partial t}= & {} \frac{(1-\beta )\left[ 1-(\lambda \gamma t^{\gamma -1}+\rho )\int _0^{\infty }e^{-\lambda ((t+y)^\gamma -t^\gamma )}e^{-\rho y}dy\right] }{\left( (1-\beta )\int _t^{\infty } e^{-\lambda (x^\gamma -t^\gamma )}e^{-\rho (x-t)} dx+\frac{\beta }{\rho }\right) ^2}\\> & {} \frac{(1-\beta )\left[ 1-(\lambda \gamma t^{\gamma -1}+\rho )\int _0^{\infty }e^{-\lambda \gamma t^{\gamma -1}y}e^{-\rho y}dy \right] }{\left( (1-\beta )\int _t^{\infty }[ e^{-\lambda (x^\gamma -t^\gamma )}e^{-\rho (x-t)}] dx+\frac{\beta }{\rho }\right) ^2}\\= & {} 0 \end{aligned}$$

The second strict inequality holds because $(t+y)^\gamma -\tau ^\gamma > \gamma t^{\gamma -1}y$ for $\gamma >1$, $y>0$ and $t>0$.

We second show ${\hat{c}}(t)\rightarrow \frac{\rho }{\beta }$. Define $x=t+y$, then

$$\begin{aligned}&\int _t^{\infty }[(1-\beta )e^{-\lambda (x^\gamma -t^\gamma )}e^{-\rho (x-t)}] dx+\frac{\beta }{\rho }\\&\quad =\int _ 0^{\infty }[(1-\beta )e^{-\lambda ((t+y)^\gamma -t^\gamma )}e^{-\rho y}] dy+\frac{\beta }{\rho }\\&\quad <\int _0^{\infty }[(1-\beta )e^{-\lambda \gamma t^{\gamma -1}y}e^{-\rho y}] dy+\frac{\beta }{\rho }\\&\quad =\frac{1-\beta }{\lambda \gamma t^{\gamma -1}+\rho }+\frac{\beta }{\rho } \end{aligned}$$

where the first inequality holds because $(t+y)^\gamma -t^\gamma > \gamma t^{\gamma -1}y$ for $\gamma >1$, $y>0$ and $t>0$. Since $\frac{1-\beta }{\lambda \gamma t^{\gamma -1}+\rho }+\frac{\beta }{\rho }\rightarrow \frac{\beta }{\rho }$ as $t\rightarrow \infty$ and $\gamma >1$, we have

$$\begin{aligned} {\hat{c}}(t)\rightarrow \frac{\rho }{\beta }\quad \mathrm{as}\quad t\rightarrow \infty \quad \mathrm{and}\quad \gamma >1 \end{aligned}$$

Appendix I

From Eqs. (E1), (E2) and (E10) in “Appendix E”, the average budget constraint equation is

$$\begin{aligned} \frac{{\bar{w}}(t)}{w(t)}=R-{\hat{c}}(t)=R-\frac{1}{(1-\beta )\int _t^{\infty } [e^{-\lambda (x^{\gamma }-t^{\gamma })}e^{-\rho (x-t)}] dx+\frac{\beta }{\rho }} \end{aligned}$$

(I1)

where ${\bar{w}}(t)=\frac{E_t[dw]}{dt}$ and $R=\frac{a^2}{{\bar{\sigma }}^2}+r$.

Differentiating Eq. (I1) and from Eq. (34), we obtain

$$\begin{aligned} \frac{\partial }{\partial t}\left[ \frac{{\bar{w}}(t)}{w(t)}\right] =-\frac{\partial {\hat{c}}(t)}{\partial t}<0 \end{aligned}$$

(I2)

Equation (I2) shows that the expected growth rate of wealth is a decreasing function of time. Therefore, if $R<{\hat{c}}(0)$, the individual plans to disinvest, i.e., she plans to consume more than her expected income.

If ${\hat{c}}(0)<R<\frac{\rho }{\beta }$, she plans to increase her wealth for $0<t<{\bar{t}}$ and then disinvests her wealth at an expected rate $R-{\hat{c}}(t)$ for $t>{\bar{t}}$, where ${\bar{t}}$ is defined as the solution to

$$\begin{aligned} R=\frac{1}{(1-\beta )\int _t^{\infty }[e^{-\lambda (x^{\gamma }-t^{\gamma })}e^{-\rho (x-t)}] dx+\frac{\beta }{\rho }} \end{aligned}$$

If $R>\frac{\rho }{\beta }$, she plans to grow her wealth.

Appendix J

From Eq. (E2) in “Appendix E”,

$$\begin{aligned} E_t[d{\hat{C}}(t)]={\hat{c}}(t)E_t[dw(t)]={\hat{c}}(t)w(t)\left[ \frac{a^2}{{\bar{\sigma }}^2}+r-\frac{1}{A(t)}\right] dt \end{aligned}$$

(J1)

Therefore, from Eq. (J1), the average budget equation for ${\hat{C}}(t)$ is

$$\begin{aligned} \frac{\hat{C} (t)}{\bar{\hat{C}}(t)}=\left[ \frac{a^2}{{\bar{\sigma }}^2}+r-\frac{1}{A(t)}\right] =R-{\hat{c}}(t) \end{aligned}$$

(J2)

where $\bar{{\hat{C}}}(t)=\frac{E_t[d\bar{{\hat{C}}}(t)]}{dt}$.

From “Appendix I”, if $R<{\hat{c}}(0)$, the individual plans to decrease her expected consumption amount. If ${\hat{c}}(0)<R<\frac{\rho }{\beta }$, she plans to increase her expected consumption amount for $0<t<{\bar{t}}$, and then decreases her expected consumption amount at an expected rate $R-{\hat{c}}(t)$ for $t>{\bar{t}}$, where ${\bar{t}}$ is defined as the solution to

$$\begin{aligned} R=\frac{1}{(1-\beta )\int _t^{\infty }[e^{-\lambda (x^{\gamma }-t^{\gamma })}e^{-\rho (x-t)}] dx+\frac{\beta }{\rho }} \end{aligned}$$

If $R>\frac{\rho }{\beta }$, she plans to increase her expected consumption amount.

Appendix K

Define $\{c_t^*(\tau ),\alpha _t^*(\tau )\}$, $0\le t\le \tau$, as the optimal consumption rate and the portion of risky asset from the perspective of the naive individual at time t with the initial condition $(\tau ,w(\tau ))$. The value function is therefore, $Q(t,\tau ,w(\tau ))$. By the same analysis as in “Appendix B”, the corresponding HJB equation is

$$\begin{aligned}&Q_\tau +\left( r+\alpha _t^*(\tau )a-c_t^*(\tau )\right) w(\tau )Q_w+\frac{1}{2}(\alpha _t^*(\tau ))^2{\bar{\sigma }}^2w^2Q_{ww}\nonumber \\&\quad +D(t,\tau )u(c_t^*(\tau )w(\tau ))=0 \end{aligned}$$

(K1)

We conjecture the value function to be $Q(t,\tau ,w(\tau ))=\psi (t,\tau )\frac{w^{1-b}}{1-b}$. From Eq. (K1), we can see that

$$\begin{aligned} c^*_t(\tau )=(D(t,\tau ))^{\frac{1}{b}}\frac{1}{\psi (t,\tau )}, \quad \alpha ^*_t(\tau )=\frac{a}{b{\bar{\sigma }}^2} \end{aligned}$$

(K2)

where

$$\begin{aligned} \psi (t,\tau )=\int _\tau ^{\infty }(D(t,x))^{\frac{1}{b}}e^{q(x-\tau )}dx \end{aligned}$$

(K3)

and $q=\frac{1-b}{b}(r+\frac{a^2}{2b{\bar{\sigma }}^2})$. Notice that $\lim _{\tau \rightarrow \infty }\psi (t,\tau )=0$, which means that the expected discounted utility generated from the wealth in the far future $w(\tau )$ is zero from the perspective of the present because of discounting. Another boundary condition is $\lim _{\tau \rightarrow \infty }\psi (\tau ,\tau )=\beta \xi ^b$. Footnote (22) shows the $\xi$ and $\lim _{\tau \rightarrow \infty }\psi (\tau ,\tau )=\beta \xi ^b$.

The optimal consumption policies $(c^*_t(\tau ),\alpha ^*_t(\tau ))$ will not be implemented at time $\tau$ and the actual consumption and portfolio rules are

$$\begin{aligned} c^*_\tau (\tau )=\frac{1}{\psi (\tau ,\tau )}, \quad \alpha ^*_\tau (\tau )=\frac{a}{b{\bar{\sigma }}^2} \end{aligned}$$

(K4)

where

$$\begin{aligned} \psi (\tau ,\tau )=\int _\tau ^{\infty }(D(\tau ,x))^{\frac{1}{b}}e^{q(x-\tau )}dx \end{aligned}$$

(K5)

Following the method in Yong’s [67] and Wei et al. [64], we derive the actual expected discounted utility ${\bar{Q}}(t,\tau ,w(\tau ))$ following the consumption policies given by Eqs. (K4)–(K5),

$$\begin{aligned} {\bar{Q}}(t,\tau ,w(\tau ))=E_t\left[ \int _\tau ^{\infty }D(t,x)u(c^*_x(x)w(x))dx\right] \end{aligned}$$

(K6)

Equation (K6) can be written as

$$\begin{aligned} {\bar{Q}}(t,\tau ,w(\tau ))=D(t,\tau )u(c_\tau ^*(\tau )w(\tau ))d\tau +{\bar{Q}}(t,\tau +d\tau ,w(\tau +d\tau )) \end{aligned}$$

(K7)

By the Taylor expansion, we have

$$\begin{aligned} {\bar{Q}}(t,\tau +d\tau ,w(\tau +d\tau ))= & {} {\bar{Q}}(t,\tau ,w(\tau ))+{\bar{Q}}_\tau d\tau +{\bar{Q}}_wE_t[dw]+\frac{1}{2}{\bar{Q}}_{ww}E_t[(dw)^2]\nonumber \\= & {} {\bar{Q}}(t,\tau ,w(\tau ))+{\bar{Q}}_\tau d\tau +[r+\alpha _\tau ^*(\tau )a-c_\tau ^*(\tau )]w(\tau ){\bar{Q}}_wd\tau \nonumber \\&+\frac{1}{2}(\alpha _\tau ^*(\tau ))^2\sigma ^2w^2(\tau ){\bar{Q}}_{ww}d\tau \end{aligned}$$

(K8)

From Eqs. (K7)–(K8), we have

$$\begin{aligned}&{\bar{Q}}_\tau +[r+\alpha _\tau ^*(\tau )a-c_\tau ^*(\tau )]w(\tau ){\bar{Q}}_w+\frac{1}{2}(\alpha _\tau ^*(\tau ))^2{\bar{\sigma }}^2w^2(\tau ) {\bar{Q}}{ww}\nonumber \\&\quad +\, D(t,\tau )u(c_\tau ^*(\tau )w(\tau ))=0 \end{aligned}$$

(K9)

We conjecture the expected discounted utility to be ${\bar{Q}}(t,\tau ,w(\tau ))=\phi (t,\tau )\frac{w^{1-b}}{1-b}$ and therefore

$$\begin{aligned} {\bar{Q}}_\tau =\frac{\partial \phi (t,\tau )}{\partial \tau }\frac{w^{1-b}}{1-b},\quad {\bar{Q}}_w=\phi (t,\tau )w^{-b},\quad {\bar{Q}}_{ww}=(-b)\phi (t,\tau )w^{-1-b} \end{aligned}$$

(K10)

From Eqs. (K4), (K9) and (K10), we have

$$\begin{aligned} \frac{\partial \phi (t,\tau )}{\partial \tau }+\left( qb-\frac{1-b}{\psi (\tau ,\tau )}\right) \phi (t,\tau )+D(t,\tau )(\psi (\tau ,\tau ))^{b-1}=0 \end{aligned}$$

(K11)

Thus,

$$\begin{aligned} \phi (t,\tau )=e^{-\int _t^\tau (qb-\frac{1-b}{\psi (s,s)})ds}\times \int _\tau ^{\infty }D(t,x)(\psi (x,x))^{b-1}e^{\int _t^x(qb-\frac{1-b}{\psi (s,s)})ds}dx \end{aligned}$$

(K12)

and

$$\begin{aligned} \phi (\tau ,\tau )=\int _\tau ^{\infty }D(\tau ,x)(\psi (x,x))^{b-1}e^{\int _\tau ^x(qb-\frac{1-b}{\psi (s,s)})ds}dx \end{aligned}$$

(K13)

and $\lim _{\tau \rightarrow \infty }\phi (t,\tau )=0$ and $\lim _{\tau \rightarrow \infty }\phi (\tau ,\tau )=\beta \xi ^b$.