Equilibrium consumption and portfolio decisions with stochastic discount rate and time-varying utility functions

Abstract

This paper studies a multi-period investment–consumption optimization problem with a stochastic discount rate and a time-varying utility function, which are governed by a Markov-modulated regime switching model. The investment is dynamically reallocated between one risk-free asset and one risky asset. The problem is time inconsistent due to the stochastic discount rate. An analytical equilibrium solution is established by resorting to a game theoretical framework. Numerous sensitivity analyses and numerical examples are provided to demonstrate the effects of the stochastic discount rate and time-varying utility coefficients on the decision-maker’s investment–consumption behavior. Our results show that many properties which are satisfied in the classical models do not hold any more due to either the stochastic discount rate or the time-varying utility function.

Appendices

The proof of Lemma 3.2

Proof

We only give the proof of (16) for $\gamma <0$, as (15) can be proved in a similar way. When $\gamma <0$,

$$\begin{aligned}\mathop {\sup }\limits _{0 \le c \le w,\alpha } \left\{ \frac{c^\gamma }{\gamma } + \frac{(w - c)^\gamma }{\gamma } f(\alpha ) \right\} = \frac{1}{\gamma }\mathop {\inf }\limits _{0 \le c \le w,\alpha } \left\{ {c^\gamma } + {(w - c)^\gamma }f(\alpha )\right\} . \end{aligned}$$

Consider the following maximization problem:

$$\begin{aligned}\mathop {\inf }\limits _{0 \le c \le w} \left\{ {c^\gamma } + {(w - c)^\gamma }H\right\} ,\end{aligned}$$

where H is a positive constant. Since

$$\begin{aligned} \frac{d^2}{d{c^2}}\left[ {c^\gamma } + {(w - c)^\gamma }H\right] =\gamma (\gamma -1)\left( c^{\gamma -2}+(w-c)^{\gamma -2}H\right) >0, \end{aligned}$$

the minimal value of ${c^\gamma } + {(w - c)^\gamma }H$ is achieved at

$$\begin{aligned} {\hat{c}}=\frac{w}{1+H^{\frac{1}{1-\gamma }}}. \end{aligned}$$

(A.1)

As a consequence,

$$\begin{aligned}&\mathop {\sup }\limits _{0 \le c \le w } \left\{ \frac{c^\gamma }{\gamma } + \frac{(w - c)^\gamma }{\gamma }H \right\} \nonumber \\&\quad =\frac{1}{\gamma }\mathop {\inf }\limits _{0 \le c \le w} \left\{ {c^\gamma } + {(w - c)^\gamma }H \right\} \nonumber \\&\quad =\frac{1}{\gamma }w^{\gamma }\left[ 1+H^{1/(1-\gamma )}\right] ^{1-\gamma }, \end{aligned}$$

(A.2)

which is a decreasing function of H, and thus we have (16).

Proof of Theorem 3.1

Proof

The proof will be shown only for $0<\gamma <1$, as the results for $\gamma <0$ can be proved in a parallel way. Throughout the proof, we always keep the constraints $0\le c_n \le w_n$ and $0\le \alpha _n \le 1$ for all $n=0,1,\ldots ,T-1$. The proof will be achieved by an induction argument. To proceed, let

$$\begin{aligned} \begin{aligned} \varpi _n(i)&=\left( \frac{(A_n(i))^{\frac{1}{1-\gamma }}}{1+(A_n(i))^{\frac{1}{1-\gamma }}}\right) ^\gamma ,\\ \zeta _n(\alpha , i)&=\left[ (1-\alpha )r_f(i)+\alpha R_n(i)\right] ^\gamma ,~~i=1,2,\ldots ,L,~n=0,1,\ldots , T-1. \end{aligned}\end{aligned}$$

Then, according to Lemma 3.1, there exists a unique solution ${\hat{\alpha }}(i)$ such that

$$\begin{aligned} {\hat{\alpha }}(i)=\arg \max _{\alpha \in [0,1]} \mathrm{E}\left[ \zeta _n(\alpha , i)\right] , \end{aligned}$$

and according to (17), $Y(i)=\mathrm{E}\left[ \zeta _n(\hat{\alpha }(i), i)\right]$. It should be noted that both $\hat{\alpha }(i)$ and Y(i) depend on the market regime i only and are independent of time n. When $n=T-1$, it follows from (13) and (21) that

$$\begin{aligned} V_{T-1}(i,w_{T-1})= & {} \mathop {\sup }\limits _{{\pi _{T - 1}}} {\mathrm{E}_{T - 1,i,{w_{T - 1}}}}\left[ \begin{aligned}&\frac{{{\vartheta _{T - 1}}(i){{\left( {{c_{T - 1}}} \right) }^\gamma }}}{\gamma } + \frac{{{\vartheta _T}({\xi _T}){{\left( {W_T^{{\pi _{T - 1}}}} \right) }^\gamma }/\gamma }}{{1 + \rho (T -1,i)}} \end{aligned}\right] \nonumber \\= & {} \mathop {\sup }\limits _{{c_{T - 1}},{\alpha _{T - 1}}} \left[ \begin{aligned}&\frac{{{\vartheta _{T - 1}}(i){{\left( {{c_{T - 1}}} \right) }^\gamma }}}{\gamma }\\&+ {H_{T - 1}}(i)\frac{{{{({w_{T - 1}} - {c_{T - 1}})}^\gamma }}}{\gamma }\mathrm{E}\left[ \zeta _{T-1}(\alpha _{T-1}, i)\right] \end{aligned} \right] ,\nonumber \\ \end{aligned}$$

(B.1)

where, obviously, $\mathrm{E}\left[ \zeta _{T-1}(\alpha _{T-1}, i)\right] >0$. Consequently, applying Lemmas 3.1 and 3.2 to (B.1), we obtain

$$\begin{aligned} V_{T-1}(i,w_{T-1})= & {} \mathop {\sup }\limits _{{c_{T - 1}}} \left[ \frac{{{\vartheta _{T - 1}}(i){{\left( {{c_{T - 1}}} \right) }^\gamma }}}{\gamma } \right. \\&\left. \quad + {H_{T - 1}}(i)Y(i)\frac{{{{({w_{T - 1}} - {c_{T - 1}})}^\gamma }}}{\gamma } \right] \end{aligned}$$

and

$$\begin{aligned} {\hat{c}}_{T - 1} (i,w_{T - 1} ) =\frac{{{w_{T - 1}}}}{{1 + {{\left( {\frac{{Y(i){H_{T - 1}}(i)}}{{{\vartheta _{T - 1}}(i)}}} \right) }^{\frac{1}{{1 - \gamma }}}}}}= \frac{{w_{T - 1} }}{{1+\left( {A_{T -1} (i)} \right) ^{\frac{1}{{1 - \gamma }}}}}. \end{aligned}$$

Substituting ${\hat{c}}_{T-1}(i,w_{T - 1} )$ into $V_{T-1}$ yields

$$\begin{aligned} V_{T-1}(i,w_{T-1})= \frac{(w_{T - 1})^\gamma }{\gamma }\vartheta _{T - 1}(i){{\left( {1 + {{\left( {{A_{T - 1}}(i)} \right) }^{\frac{1}{{1 - \gamma }}}}} \right) }^{1 - \gamma }}. \end{aligned}$$

This implies that (22)–(24) hold for $n=T-1$. Generally, we assume that (22)–(24) hold for $T-1,T-2,\ldots ,n+1$, and we shall prove that they hold for n as well. From the given assumptions, we obtain

$$\begin{aligned} W_k^{{\pi _n},{{{\hat{\pi }} }_{n + 1}},\ldots ,{{{\hat{\pi }} }_{k - 1}}} = W_{n + 1}^{{\pi _n}}\prod \limits _{m = n + 1}^{k - 1} {\frac{{{{\left( {{A_m}({\xi _m})} \right) }^{\frac{1}{{1 - \gamma }}}}}}{{1 + {{\left( {{A_m}({\xi _m})} \right) }^{\frac{1}{{1 - \gamma }}}}}}\left( {(1 - {{{\hat{\alpha }} }_m}){r_f}({\xi _m}) + {{{\hat{\alpha }} }_m}{R_m}({\xi _m})} \right) } \end{aligned}$$

for $k=T-1,T-2,\ldots ,n+1$. This recursive equation implies

$$\begin{aligned} \frac{\left( W_k^{{\pi _n},{{{\hat{\pi }} }_{n + 1}},\ldots ,{{{\hat{\pi }} }_{k-1}}}\right) ^{\gamma }}{\gamma }=\frac{\left( W_{n+1}^{\pi _n}\right) ^\gamma }{\gamma }\prod \limits _{m=n+1}^{k-1}\left( \varpi _m(\xi _m) \zeta _m({\hat{\alpha }}_m,\xi _m)\right) , \end{aligned}$$

(B.2)

which along with (23) implies

$$\begin{aligned} \mathrm{E}_{n, i, w_n} [U_k({\hat{c}}_k )]= & {} \mathrm{E}_{n, i, w_n} \left[ {\frac{{\left( {W_k^{( \pi _n,{\hat{\pi }}_{n+1},\ldots ,{\hat{\pi }} _{k - 1} ) } } \right) ^\gamma }}{\gamma }\frac{\vartheta _k(\xi _k)}{{\left( {\left( {A_k (\xi _k )}\right) ^{\frac{1}{{1 -\gamma }}} + 1} \right) ^\gamma }}} \right] \nonumber \\= & {} \mathrm{E}_{n, i,w_n}\left[ \frac{\left( W_{n+1}^{\pi _n}\right) ^\gamma }{\gamma }\frac{\vartheta _k(\xi _k)\prod \limits _{m=n+1}^{k-1}\left( \varpi _m(\xi _m) \zeta _m({\hat{\alpha }}_m,\xi _m)\right) }{\left( {\left( {A_k (\xi _k )}\right) ^{\frac{1}{{1-\gamma }}} + 1} \right) ^\gamma }\right] .\qquad \quad \end{aligned}$$

(B.3)

From (1), it is clear that $W_{n+1}^{\pi _n}$ depends on the random return $R_n(i)$ at time n. Moreover, the term $\prod \nolimits _{m=n+1}^{k-1}{\zeta _m({\hat{\alpha }}_m,\xi _m)}$ depends on the random return $R_{n+1}(\xi _{n+1}),\ldots ,R_{k-1}(\xi _{k-1})$, and $\varpi _k(\xi _k) \vartheta _k(\xi _k)$ and $A_k(\xi _k)$ are relevant to $\xi _k$. These variables are not independent of each other at time n because the market states $\xi _{n+1},\ldots ,\xi _k$ are interdependent. However, we recall that $\left\{ R_0(\cdot ),R_1(\cdot ),\ldots ,R_{T-1}(\cdot )\right\}$ are conditionally independent given the market states; hence, $W_{n+1}^{\pi _n}$, $\zeta _{n+1}({\hat{\alpha }}_{n+1},\xi _{n+1}),\ldots ,\zeta _{k-1}({\hat{\alpha }}_{k-1},\xi _{k-1})$ are independent for given values of $\xi _n,\xi _{n+1},\ldots ,\xi _{k-1}$. Moreover, by definition we conclude that $\prod \nolimits _{m=n+1}^{k-1}\varpi _{m}(\xi _m)$, $\vartheta _k(\xi _k)$ and $A_k(\xi _k)$ are all constant for given values of $\xi _{n+1},\ldots ,\xi _k$. Consequently, an application of the tower property of the expectation operator yields

$$\begin{aligned}&\mathrm{E}_{n, i, w_n} [U_k({\hat{c}}_k )]\\&\quad =\mathrm{E}_{n, i, w_n} \left[ \mathrm{E}_{\xi _n,\xi _{n+1},\ldots ,\xi _{k}}\left[ \frac{\left( W_{n+1}^{\pi _n}\right) ^\gamma }{\gamma }\frac{\vartheta _k(\xi _k)\prod \limits _{m=n+1}^{k-1}\left( \varpi _m(\xi _m) \zeta _m({\hat{\alpha }}_m,\xi _m)\right) }{\left( {\left( {A_k (\xi _k )}\right) ^{\frac{1}{{1-\gamma }}} + 1} \right) ^\gamma }\right] \right] , \end{aligned}$$

where

$$\begin{aligned} \begin{aligned}&\mathrm{E}_{\xi _n,\xi _{n+1},\ldots ,\xi _{k}}\left[ \frac{\left( W_{n+1}^{\pi _n}\right) ^\gamma }{\gamma }\frac{\vartheta _k(\xi _k)\prod \limits _{m=n+1}^{k-1}\left( \varpi _m(\xi _m) \zeta _m({\hat{\alpha }}_m,\xi _m)\right) }{\left( {\left( {A_k (\xi _k )}\right) ^{\frac{1}{{1-\gamma }}} + 1} \right) ^\gamma }\right] \\&\quad =\frac{\vartheta _k(\xi _k)}{\gamma \left( {\left( {A_k (\xi _k )}\right) ^{\frac{1}{{1-\gamma }}} + 1} \right) ^\gamma }\mathrm{E}_{\xi _n,\xi _{n+1},\ldots ,\xi _{k}}\left[ (W_{n+1}^{\pi _n})^\gamma \right] \\&\quad \quad \prod \limits _{m=n+1}^{k-1}{\varpi _m(\xi _m) \mathrm{E}_{\xi _n,\xi _{n+1},\ldots ,\xi _{k}}\left( \zeta _m({\hat{\alpha }}_m,\xi _m)\right) }\\&\qquad = \frac{\vartheta _k(\xi _k)}{\gamma \left( {\left( {A_k (\xi _k )}\right) ^{\frac{1}{{1-\gamma }}} + 1} \right) ^\gamma }\mathrm{E}_{\xi _n,\xi _{n+1},\ldots ,\xi _{k}}\left[ (W_{n+1}^{\pi _n})^\gamma \right] \prod \limits _{m=n+1}^{k-1}{\varpi _m(\xi _m) Y(\xi _m)}\\&\qquad =\mathrm{E}_{\xi _n,\xi _{n+1},\ldots ,\xi _{k}}\left[ \frac{\left( W_{n+1}^{\pi _n}\right) ^\gamma }{\gamma }\frac{\vartheta _k(\xi _k)\prod \limits _{m=n+1}^{k-1}\varpi _m(\xi _m) Y(\xi _m)}{\left( {\left( {A_k (\xi _k )}\right) ^{\frac{1}{{1-\gamma }}} + 1} \right) ^\gamma }\right] . \end{aligned}\end{aligned}$$

Therefore, we have

$$\begin{aligned}&\mathrm{E}_{n, i, w_n} [U_k({\hat{c}}_k )]\nonumber \\&\quad =\mathrm{E}_{n,i,w_n}\left[ \mathrm{E}_{\xi _n,\xi _{n+1},\ldots ,\xi _{k}}\left[ \frac{\left( W_{n+1}^{\pi _n}\right) ^\gamma }{\gamma }\frac{\vartheta _k(\xi _k)\prod \limits _{m=n+1}^{k-1}\varpi _m(\xi _m) Y(\xi _m)}{\left( {\left( {A_k (\xi _k )}\right) ^{\frac{1}{{1-\gamma }}} + 1} \right) ^\gamma }\right] \right] \nonumber \\&\quad =\mathrm{E}_{n, i, w_n} \left[ \frac{\left( W_{n+1}^{\pi _n}\right) ^\gamma }{\gamma }\frac{\vartheta _k(\xi _k)\prod \limits _{m=n+1}^{k-1}{\varpi _m(\xi _m) Y(\xi _m)}}{\left( {\left( {A_k (\xi _k )}\right) ^{\frac{1}{{1-\gamma }}} + 1} \right) ^\gamma }\right] . \end{aligned}$$

(B.4)

Substituting (B.2) and (B.4) into (13) implies

$$\begin{aligned} \begin{aligned} V_n(i,w_n):=&\mathop {\max }\limits _{c_n ,\alpha _n }\left[ {U_n}\left( c_n\right) +I_1({n, i, w_n})-I_2({n, i, w_n})+\frac{{{{(W_{n + 1}^{{\pi _n}})}^\gamma }}}{\gamma }{H_n}(i)\right] , \end{aligned}\nonumber \\ \end{aligned}$$

(B.5)

where

$$\begin{aligned} I_1({n, i, w_n}) =\mathrm{E}_{n,i,w_n}\left[ \begin{aligned}&\frac{{{{(W_{n + 1}^{{\pi _n}})}^\gamma }}}{\gamma }\frac{{{{\left( {1 + \rho (n + 1,{\xi _{n + 1}})} \right) }^{T - n - 1}}}}{{{{\left( {1 + \rho (n,i)} \right) }^{T - n}}}}{\vartheta _{n + 1}({\xi _{n + 1}})}\\&\times \left( {1 + {{\left( {{A_{n + 1}}({\xi _{n + 1}})} \right) }^{\frac{1}{{1 - \gamma }}}}} \right) ^{1 - \gamma } \end{aligned}\right] , \end{aligned}$$

and

$$\begin{aligned}&I_2(n, i, w_n) \\&\quad =\mathrm{E}_{n,i,w_n}\left[ \frac{{{{(W_{n + 1}^{{\pi _n}})}^\gamma }}}{\gamma }\sum \limits _{k = n + 1}^T {\left( \begin{aligned}&\frac{{{{\left( {1 + \rho (n + 1,{\xi _{n + 1}})} \right) }^{T - k}}}}{{\left( {1 + \rho (n,i)} \right) }^{T - n}}\\&\quad \times \frac{\vartheta _k(\xi _k)}{\left( {1 + {\left( {{A_k}({\xi _k})} \right) }^{\frac{1}{{1 - \gamma }}}}\right) ^\gamma } \prod \limits _{m = n + 1}^{k - 1} {\varpi _m(\xi _m)Y({\xi _m})} \end{aligned}\right) }\right] . \end{aligned}$$

According to the expressions of $I_1(n,i,w_n)$ and $I_2(n,i,w_n)$, we have

$$\begin{aligned}&I_1(n,i,w_n)-I_2(n,i,w_n) \\&\quad = \mathrm{E}_{n,i,w_n}\left[ \begin{aligned}&\frac{{{{(W_{n + 1}^{{\pi _n}})}^\gamma }}}{\gamma }\frac{{{{\left( {1 + \rho (n + 1,{\xi _{n + 1}})} \right) }^{T - n - 1}}}}{{{{\left( {1 + \rho (n,i)} \right) }^{T - n}}}}{\vartheta _{n + 1}({\xi _{n + 1}})}\left( {1 + {{\left( {{A_{n + 1}}({\xi _{n + 1}})} \right) }^{\frac{1}{{1 - \gamma }}}}} \right) ^{1 - \gamma } \end{aligned}\right] \\&\qquad -\,\mathrm{E}_{n,i,w_n}\left[ \frac{{{{(W_{n + 1}^{{\pi _n}})}^\gamma }}}{\gamma }\sum \limits _{{k = n + 1}}^T {\left( \begin{aligned}&\frac{{{{\left( {1 + \rho (n + 1,{\xi _{n + 1}})} \right) }^{T - k}}}}{{\left( {1 + \rho (n,i)} \right) }^{T - n}}\\&\times \frac{\vartheta _k(\xi _k)}{\left( {1 + {\left( {{A_k}({\xi _k})} \right) }^{\frac{1}{{1 - \gamma }}}}\right) ^\gamma } \prod \limits _{m = n + 1}^{k - 1} {{\varpi _m(\xi _m)}Y({\xi _m})} \end{aligned}\right) }\right] \\&\quad =\mathrm{E}_{n,i,w_n}\left[ \begin{aligned}&\frac{{{{(W_{n + 1}^{{\pi _n}})}^\gamma }}}{\gamma }\frac{{{{\left( {1 + \rho (n + 1,{\xi _{n + 1}})} \right) }^{T - n - 1}}}}{{{{\left( {1 + \rho (n,i)} \right) }^{T - n}}}}{\vartheta _{n + 1}({\xi _{n + 1}})}\\&\times \left[ \left( {1 + {{\left( {{A_{n + 1}}({\xi _{n + 1}})} \right) }^{\frac{1}{{1 - \gamma }}}}} \right) ^{1 - \gamma }-\left( {1 + {{\left( {{A_{n + 1}}({\xi _{n + 1}})} \right) }^{\frac{1}{{1 - \gamma }}}}} \right) ^{ - \gamma }\right] \end{aligned}\right] \\&\qquad -\,\mathrm{E}_{n,i,w_n}\left[ \frac{{{{(W_{n + 1}^{{\pi _n}})}^\gamma }}}{\gamma }\sum \limits _{{k = n + 2}}^T {\left( \begin{aligned}&\frac{{{{\left( {1 + \rho (n + 1,{\xi _{n + 1}})} \right) }^{T - k}}}}{{\left( {1 + \rho (n,i)} \right) }^{T - n}}\\&\times \frac{\vartheta _k(\xi _k)}{\left( {1 + {\left( {{A_k}({\xi _k})} \right) }^{\frac{1}{{1 - \gamma }}}}\right) ^\gamma } \prod \limits _{m = n + 1}^{k - 1} {\varpi _m(\xi _m)Y({\xi _m})} \end{aligned}\right) }\right] \\&\quad =\mathrm{E}_{n,i,w_n}\left[ \begin{aligned}&\frac{{{{(W_{n + 1}^{{\pi _n}})}^\gamma }}}{\gamma }\frac{{{{\left( {1 + \rho (n + 1,{\xi _{n + 1}})} \right) }^{T - n - 1}}}}{{{{\left( {1 + \rho (n,i)} \right) }^{T - n}}}}{\vartheta _{n + 1}}({\xi _{n + 1}})\frac{{{{\left( {{A_{n + 1}}({\xi _{n + 1}})} \right) }^{\frac{1}{{1 - \gamma }}}}}}{{{{\left( {1 + {{\left( {{A_{n + 1}}({\xi _{n + 1}})} \right) }^{\frac{1}{{1 - \gamma }}}}} \right) }^\gamma }}}\end{aligned}\right] \nonumber \\&\qquad -\,\mathrm{E}_{n,i,w_n}\left[ \frac{{{{(W_{n + 1}^{{\pi _n}})}^\gamma }}}{\gamma }\sum \limits _{{k = n +2}}^T {\left( \begin{aligned}&\frac{{{{\left( {1 + \rho (n + 1,{\xi _{n + 1}})} \right) }^{T - k}}}}{{\left( {1 + \rho (n,i)} \right) }^{T - n}}\\&\times \frac{{\vartheta _k}({\xi _k})}{\left( {1 + {\left( {{A_k}({\xi _k})} \right) }^{\frac{1}{{1 - \gamma }}}}\right) ^\gamma } \prod \limits _{m = n + 1}^{k - 1} {{\varpi _m(\xi _m)}Y({\xi _m})} \end{aligned}\right) }\right] \nonumber \\&\quad =:\bigtriangleup _1(n,i,w_n)-\bigtriangleup _2(n,i,w_n). \end{aligned}$$

In view of

$$\begin{aligned} A_k(\xi _k)=\frac{{H_k}(\xi _k)Y(\xi _k)}{\vartheta _k(\xi _k)}~~\mathrm{{and}}~~\varpi _k(\xi _k)=\left( \frac{(A_k(\xi _k))^{\frac{1}{1-\gamma }}}{1+(A_k(\xi _k))^{\frac{1}{1-\gamma }}}\right) ^\gamma , \end{aligned}$$

we first obtain

$$\begin{aligned} \bigtriangleup _1(n,i,w_n)= & {} \mathrm{E}_{n,i,w_n}\left[ \frac{{{{(W_{n + 1}^{{\pi _n}})}^\gamma }}}{\gamma }\frac{{{{\left( {1 + \rho (n + 1,{\xi _{n + 1}})} \right) }^{T - n - 1}}}}{{{{\left( {1 + \rho (n,i)} \right) }^{T - n}}}}\right. \\&\left. Y(\xi _{n + 1})\varpi _{n+1}(\xi _{n+1}){H_{n + 1}}({\xi _{n + 1}})\right] .\end{aligned}$$

Further referring to (21), we have

$$\begin{aligned}&\bigtriangleup _1(n,i,w_n)\\&\quad =\mathrm{E}_{n,i,w_n}\left[ \begin{aligned}&\frac{{{{(W_{n + 1}^{{\pi _n}})}^\gamma }}}{\gamma }\frac{{{{\left( {1 + \rho (n + 1,{\xi _{n + 1}})} \right) }^{T - n - 1}}}}{{{{\left( {1 + \rho (n,i)} \right) }^{T - n}}}}Y({\xi _{n + 1}})\varpi _{n+1}(\xi _{n+1})\\&\times \mathrm{E}_{n + 1,{\xi _{n + 1}}}\left[ \sum \limits _{k = n + 2}^T {\left( \begin{aligned}&\frac{1}{\left( {1 + \rho (n + 1,{\xi _{n + 1}})} \right) ^{k - n - 1}}\\&\times \frac{\vartheta _k(\xi _k)}{\left( {1 + {{\left( {{A_k}({\xi _k})} \right) }^{\frac{1}{{1 - \gamma }}}}} \right) ^\gamma }\prod \limits _{m = n + 2}^{k - 1} {\varpi _m(\xi _{m})Y(\xi _m)} \end{aligned}\right) } \right] \end{aligned}\right] \\&\quad =\mathrm{E}_{n,i,w_n}\left[ \frac{{{{(W_{n + 1}^{{\pi _n}})}^\gamma }}}{\gamma }\sum \limits _{k = n +2}^T {\left( \begin{aligned}&\frac{{{{\left( {1 + \rho (n + 1,{\xi _{n + 1}})} \right) }^{T - k}}}}{{\left( {1 + \rho (n,i)} \right) }^{T - n}}\\&\times \frac{\vartheta _k({\xi _k})}{\left( {1 + {\left( {{A_k}({\xi _k})} \right) }^{\frac{1}{{1 - \gamma }}}}\right) ^\gamma }\prod \limits _{m = n + 1}^{k - 1} {\varpi _m(\xi _{m})Y({\xi _m})} \end{aligned}\right) } \right] \\&\quad =\bigtriangleup _2(n,i,w_n), \end{aligned}$$

which leads to $I_1(n,i,w_n)-I_2(n,i,w_n)=0$. Consequently, (B.5) is equivalent to

$$\begin{aligned} V_n(i,w_n)= & {} \mathop {\max }\limits _{c_n ,\alpha _n }\left[ {U_n}\left( {{c_n}} \right) +\frac{{{{(W_{n + 1}^{{\pi _n}})}^\gamma }}}{\gamma }{H_n}(i)\right] \nonumber \\= & {} \mathop {\max }\limits _{{c_n},{\alpha _n}} \left[ {{\vartheta _n}(i)\frac{{{{\left( {{c_n}} \right) }^\gamma }}}{\gamma } + \frac{{{{({w_n} - {c_n})}^\gamma }\mathrm{E}(\zeta _n(i))}}{\gamma }{H_n}(i)} \right] . \end{aligned}$$

(B.6)

Since $\mathrm{E}(\zeta _n(i))H_n(i)>0$ and is a function of $\alpha _n$, we can apply Lemma 3.2 to obtain

$$\begin{aligned} V_n(i,w_n)=\mathop {\sup }\limits _{{c_n}} \left[ {{\vartheta _n}(i)\frac{{{{\left( {{c_n}} \right) }^\gamma }}}{\gamma } + \frac{{{{({w_n} - {c_n})}^\gamma }Y(i)}}{\gamma }{H_n}(i)} \right] , \end{aligned}$$

where the supremum of the value function is attained at

$$\begin{aligned} {{\hat{c}}_n}\left( {i,{w_n}} \right) = \frac{{{w_n}}}{{1 + {{\left( {\frac{{Y(i){H_n}(i)}}{{{\vartheta _n}(i)}}} \right) }^{\frac{1}{{1 - \gamma }}}}}} = \frac{{{w_n}}}{{1 + {{\left( {{A_n}(i)} \right) }^{\frac{1}{{1 - \gamma }}}}}}.\end{aligned}$$

Substituting ${\hat{c}}_n(i,w_n)$ into $V_n(i,w_n)$ and rearranging yield

$$\begin{aligned}V_n(i,w_n)=\frac{{{{\left( {{w_n}} \right) }^\gamma }}}{\gamma } {\vartheta _n}(i){{\left( {1 + {{\left( {{A_n}(i)} \right) }^{\frac{1}{{1 - \gamma }}}}} \right) }^{1 - \gamma }}.\end{aligned}$$

This implies that the desired results also hold for n, and thus, by induction principle, the proof is complete.

Proof of Proposition 4.1

Proof

It follows from (21) that $H_n(i)$ does not depend on $\vartheta _n(i)$, which along with (23) immediately yields that the equilibrium consumption proportion ${\hat{\phi }}_n(i)$ is strictly increasing in $\vartheta _n(i)$. Further note that ${\hat{\eta }}_n(i)=(1-{\hat{\phi }}_n(i)){\hat{\alpha }}_n(i)$ and ${\hat{\alpha }}_n(i)$ is irrelevant to $\vartheta _n(i)$. Thus, ${\hat{\eta }}_n(i)$ is strictly decreasing with $\vartheta _n(i)$.

Proof of Proposition 4.2

Proof

When the discount rates are a constant, (21) can be written as

$$\begin{aligned} H_n(i)=\frac{1}{1+\rho }\mathrm{E}_{n,i}\left[ {\vartheta _{n + 1}}({\xi _{n + 1}}){{\left( {1 + {{\left( \frac{{{H_{n+ 1}}(\xi _{n + 1})Y(\xi _{n + 1})}}{{{\vartheta _{n+1}}(\xi _{n + 1})}} \right) }^{\frac{1}{{1 - \gamma }}}}} \right) }^{1 - \gamma }} \right] .\nonumber \\ \end{aligned}$$

(D.1)

By (18), $H_T(\xi _T)=0$ and thus (D.1) implies

$$\begin{aligned} {H_{T - 1}}(i)= & {} {\mathrm{E}_{T - 1,i}}\left[ \frac{\vartheta _T(\xi _T)}{1 + \rho }\right] \\= & {} \frac{\sum \limits _{j=1}^L{Q(i,j)\vartheta _T(j)}}{1+\rho }, \end{aligned}$$

which is obviously increasing in ${\varvec{\vartheta }}_T$ for each $i=1,\ldots , L$. Consequently, it follows from (20) and (27) that ${\hat{\phi }}_{T-1}(i)$ is decreasing in ${\varvec{\vartheta }}_T$ for each $i=1,\ldots , L$.

For $n=T-2$, (D.1) implies

$$\begin{aligned} H_{T-2}(i)=\frac{1}{1+\rho } \sum \limits _{j=1}^L{Q(i,j){\vartheta _{T- 1}}(j){{\left( {1 + {{\left( \frac{{{H_{T- 1}}(j)Y(j)}}{{{\vartheta _{T-1}}(j)}} \right) }^{\frac{1}{{1 - \gamma }}}}} \right) }^{1 - \gamma }}}. \end{aligned}$$

(D.2)

Thus, for each $i=1,\ldots , L$, $H_{T-2}(i)$ is also increasing in ${\varvec{\vartheta }}_T$ as a property inherited from $H_{T-1}(i)$, and consequently, ${\hat{\phi }}_{T-2}(i)$ is decreasing in ${\varvec{\vartheta }}_T$ due to (20) and (27).

In general, we assume that $H_{n+1}(i)$ is increasing in ${\varvec{\vartheta }}_T$ for each $i=1,\ldots , L$, and from (D.1) we can obtain a similar equation for $H_n(\cdot )$ in terms of $H_{n+1}(\cdot )$ similar to (D.2) to conclude that $H_n(i)$ is indeed increasing in ${\varvec{\vartheta }}_T$. Then, by (20) and (27), ${\hat{\phi }}_n(i)$ is decreasing in ${\varvec{\vartheta }}_T$, and therefore, by the induction principle, we complete the proof w.r.t. the monotonic property of ${\hat{\phi }}_n$ in the general case.

The results for ${\hat{\eta }}_n(i)$ follow straightforward from the equation ${\hat{\eta }}_n(i)=(1-{\hat{\phi }}_n(i)){\hat{\alpha }}_n(i)$ and the fact that ${\hat{\alpha }}_n(i)$ is irrelevant to $\vartheta _n(i)$. The proof for results with all elements of Q being positive can be obtained in a complete parallel way.

Proof of Proposition 4.3

Proof

When the discount rates are constant, (21) can be equivalently written as Eq. (D.1), from which it follows for $k=n, \ldots , T-1$ that

$$\begin{aligned} H_k(i)=\frac{1}{1+\rho }\sum _{j=1}^LQ(i,j) \vartheta _{k + 1}(j)\left( 1 + {{\left( \frac{{{H_{k+ 1}}(j)Y(j)}}{{{\vartheta _{k+1}}(j)}} \right) }^{\frac{1}{{1 - \gamma }}}} \right) ^{1 - \gamma }, \end{aligned}$$

(E.1)

where $H_{k+1}(j)$ does not depend on ${\varvec{\vartheta }}_{k+1}$. Therefore, it is straightforward to verify that, for any $i,j=1,\ldots ,L$ and $k=n,\ldots , T-1$,

$$\begin{aligned} \frac{\partial }{\partial \vartheta _{k+1}(j)}H_k(i) =\frac{Q(i,j)}{1+\rho } \left( {1 + {{\left( {\frac{{{H_{k + 1}}(j)Y(j)}}{{{\vartheta _{k + 1}}(j)}}} \right) }^{\frac{1}{{1 - \gamma }}}}} \right) ^{ - \gamma } \ge 0, \ \, \end{aligned}$$

(E.2)

which implies that $H_k(i)$ is increasing in ${\varvec{\vartheta }}_{k+1}$. On the other hand, by (20) and (27), ${\hat{\phi }}_n(i)$ admits the following expression

$$\begin{aligned} \hat{\phi }_n(i)=\frac{1}{1+\left( \frac{H_n(i)Y(i)}{\vartheta _n(i)}\right) ^{\frac{1}{1-\gamma }}}, \end{aligned}$$

(E.3)

and hence it is decreasing in ${\varvec{\vartheta }}_{n+1}$. Moreover, (E.1) and (E.3) together imply that $\hat{\phi }_n(i)$ depends on ${\varvec{\vartheta }}_{n+2}$ only via $H_{n+1}(\cdot )$. Since (E.2) indicates that $H_{n+1}(j)$ is increasing in ${\varvec{\vartheta }}_{n+2}$, $\hat{\phi }_n(i)$ is also decreasing in ${\varvec{\vartheta }}_{n+2}$. Repeatedly using recursion (E.1) and the positiveness of the derivative in (E.2), we can conclude by the induction principle that the equilibrium consumption proportion ${\hat{\phi }}_n(i)$ is decreasing in ${{\varvec{\vartheta }}}_k$ for each $k=n+1,n+2,\ldots ,T-1$.

The increasing property of ${\hat{\eta }}_n(i)$ with respect to ${\varvec{\vartheta }}_k$ for each $k=n+1,n+2,\ldots ,T-1$ follows trivially from the equation ${\hat{\eta }}_n(i)=(1-{\hat{\phi }}_n(i)){\hat{\alpha }}_n(i)$ and the fact that ${\hat{\alpha }}_n(i)$ is irrelevant to $\vartheta _n(i)$. The results with all elements Q being positive can be obtained in a complete parallel way.

Proof of Proposition 4.4

Proof

When the assumptions in Proposition 4.4 hold, by (18) and (21), $H_n$ can be simplified as

$$\begin{aligned} H_n= & {} \sum \limits _{k = n + \mathrm{{1}}}^T {\frac{{\prod \limits _{m = n + 1}^{k - 1} {{{\left( {\frac{{{{({A_m})}^{\frac{1}{{1 - \gamma }}}}}}{{1 + {{({A_m})}^{\frac{1}{{1 - \gamma }}}}}}} \right) }^\gamma }Y} }}{{{{\left( {1 + \rho } \right) }^{k - n}}}}\frac{{{\vartheta _k}}}{{{{\left( {1 + {{({A_k})}^{\frac{1}{{1 - \gamma }}}}} \right) }^\gamma }}}}\nonumber \\= & {} \frac{\mathrm{{1}}}{{1 + \rho }}\frac{{{\vartheta _{n + 1}}}}{{{{\left( {1 + {{({A_{n + 1}})}^{\frac{1}{{1 - \gamma }}}}} \right) }^\gamma }}}+ \frac{\mathrm{{1}}}{{1 + \rho }}{\left( {\frac{{{{({A_{n + 1}})}^{\frac{1}{{1 - \gamma }}}}}}{{1 + {{({A_{n + 1}})}^{\frac{1}{{1 - \gamma }}}}}}} \right) ^\gamma }YH_{n + 1}, \end{aligned}$$

where $Y=\mathrm{E}\left[ \left( (1-{\hat{\alpha }}_n)r_f+{\hat{\alpha }}_nR_n\right) ^{\gamma }\right] >0$. According to (20), substituting $Y{H_n} = \vartheta _n{A_n}$ into the above equation yields

$$\begin{aligned} \frac{\vartheta _nA_n}{Y}= & {} \frac{{{\vartheta _{n + 1}}}}{{1 + \rho }}\frac{1}{{{{\left( {1 + {{({A_{n + 1}})}^{\frac{1}{{1 - \gamma }}}}} \right) }^\gamma }}}+ \frac{{{\vartheta _{n + 1}}}}{{1 + \rho }}{\left( {\frac{{{{({A_{n + 1}})}^{\frac{1}{{1 - \gamma }}}}}}{{1 + {{({A_{n + 1}})}^{\frac{1}{{1 - \gamma }}}}}}} \right) ^\gamma }{A_{n + 1}}\\= & {} \frac{{{\vartheta _{n + 1}}}}{{1 + \rho }}{\left( {1 + {{({A_{n + 1}})}^{\frac{1}{{1 - \gamma }}}}} \right) ^{1 - \gamma }}. \end{aligned}$$

Therefore, we have

$$\begin{aligned} {A_n} = \frac{{{\vartheta _{n + 1}}Y}}{{{\vartheta _n}(1 + \rho )}}{\left( {1 + {{({A_{n + 1}})}^{\frac{1}{{1 - \gamma }}}}} \right) ^{1 - \gamma }},~~n=0,1,\ldots ,T-1, \end{aligned}$$

(F.1)

with $A_T=0$. Next, under the assumptions in Proposition 4.4, we show that $A_n>A_{n+1}~(n=0,1,\ldots , T-1)$ by using induction principle. To this end, by (20), we first note that

$$\begin{aligned} A_{T-1}=0,~~\mathrm{{and}}~~{A_{T - 2}} = \frac{{{\vartheta _{T - 1}}Y}}{{{\vartheta _{T - 2}}(1 + \rho )}} \ge {A_{T - 1}}=0. \end{aligned}$$

In general, with the assumption $A_{n+1}\ge A_{n+2}$, we obtain

$$\begin{aligned} \begin{aligned} {A_n} =&\frac{{{\vartheta _{n + 1}}Y}}{{{\vartheta _n}(1 + \rho )}}{\left( {1 + {{({A_{n + 1}})}^{\frac{1}{{1 - \gamma }}}}} \right) ^{1 - \gamma }}\\ \ge&\frac{{{\vartheta _{n + 1}}Y}}{{{\vartheta _n}(1 + \rho )}}{\left( {1 + {{({A_{n + 2}})}^{\frac{1}{{1 - \gamma }}}}} \right) ^{1 - \gamma }}\\ =&\frac{{{{({\vartheta _{n + 1}})}^2}}}{{{\vartheta _n}{\vartheta _{n + 2}}}}\frac{{{\vartheta _{n + 2}}Y}}{{{\vartheta _{n + 1}}(1 + \rho )}}{\left( {1 + {{({A_{n + 2}})}^{\frac{1}{{1 - \gamma }}}}} \right) ^{1 - \gamma }} = \frac{{{{({\vartheta _{n + 1}})}^2}}}{{{\vartheta _n}{\vartheta _{n + 2}}}}{A_{n + 1}}. \end{aligned} \end{aligned}$$

If $\frac{{{\vartheta _n}}}{{{\vartheta _{n + 1}}}} \le \frac{{{\vartheta _{n + 1}}}}{{{\vartheta _{n + 2}}}},~~n=0,1,\ldots ,T-3$, then $A_n\ge A_{n+1}$. Consequently, by (29), we obtain

$$\begin{aligned} \Lambda _{n+1}-\Lambda _n=Z\left[ \frac{1}{1+(A_n)^{\frac{1}{1-\gamma }}}-\frac{1}{1+(A_{n+1})^{\frac{1}{1-\gamma }}}\right] \le 0. \end{aligned}$$