An Analytic Approach for Stochastic Differential Utility for Endowment and Production Economies

Abstract

In this paper, an analytic approach is applied to approximate stochastic differential utility for both endowment and production economies. This utility function satisfies a second order nonlinear differential equation and is analytic near the stationary mean of the state variable. The radius of convergence of its power series representation is at lease two standard deviations of the stationary distribution of the state variable around its mean. The main contribution of this paper is to use a higher order Taylor polynomial to accurately approximate the utility function, which confirms the empirical properties of the pricing kernel and equity premium on wealth.

Appendix

1.1 Derivation of Differential Equation for (14)

The lifetime utility $V_t$ for a representative investor satisfies the stochastic differential equation (1)

$$\begin{aligned} dV_t = -f(C_t,V_t) \,dt+\sigma _{V}(t)\,d\omega _t \ \hbox { subject to } V_T=\frac{\xi ^{\frac{\alpha }{\rho }}}{\alpha }C_T^\alpha , \end{aligned}$$

(47)

where the aggregator is given by

$$\begin{aligned} f(C,V) =\frac{\beta }{\rho }\frac{C^{\rho }-(\alpha V)^{\frac{\rho }{\alpha }}}{(\alpha V)^{\frac{\rho }{\alpha }-1}}. \end{aligned}$$

(48)

The lifetime utility function satisfies the differential equation

$$\begin{aligned} E_t[dV_t]=-f(C_t,V_t)\,dt. \end{aligned}$$

(49)

Duffie and Epstein (1992b) and Schroder and Skiadas (1999, 2003) demonstrate that the solution to the differential equation (49) is of the separable form

$$\begin{aligned} V(C,x)=J(C,x) = \frac{1}{\alpha }C^\alpha v(x). \end{aligned}$$

(50)

Following an insight from Fisher and Gilles (1998), the differential equation can be simplified further by introducing the change of variable

$$\begin{aligned} v(x)= g(x)^{\frac{\alpha }{\rho }}. \end{aligned}$$

(51)

By the stochastic process (8) and Ito’s lemma, we have

$$\begin{aligned} \frac{dC_t}{C_t}=\left[ x_t+\bar{x}+\frac{\sigma (x_t)^2}{2}\right] \,dt +\sigma (x_t)\,d\tilde{\omega }_{1,t} \end{aligned}$$

(52)

where $\bar{C}=e^{\bar{c}}$ is the steady state level for consumption $C_t$. The long-run risk variable $x$ follows the stochastic process (9). Based on (9), (52), and the change of variable which leads to the equivalent aggregator (48), the lifetime utility follows the backward stochastic differential equation (49).

Duffie and Lions (1992) assume that $V_t=V(C_t,x_t)$, where $x$ is included in the lifetime utility function because it is a driving force of consumption.

$$\begin{aligned} dV(C_t,x_t)&= \frac{\partial V}{\partial t}(C_t,x_t)\,dt +\frac{\partial V}{\partial C}(C_t,x_t)\,dC_t +\frac{\partial V}{\partial x}(C_t,x_t)\,dx_t \nonumber \\&+\,\frac{1}{2}\frac{\partial ^2 V}{\partial C_t^2}(C_t,x_t)(dC_t)^2 +\frac{\partial ^2 V}{\partial C\partial x}(C_t,x_t)(dC_t)(dx_t)\nonumber \\&+\,\frac{1}{2}\frac{\partial ^2 V}{\partial x^2}(C_t,x_t)(dx_t)^2 \end{aligned}$$

(53)

By (49), we get the differential equation

$$\begin{aligned} 0&= f(C,V)+\left[ x+\bar{x}+\frac{\sigma (x)^2}{2}\right] C \frac{\partial V}{\partial C}-\kappa x \frac{\partial V}{\partial x} \nonumber \\&+\frac{\sigma (x)^2}{2}C^2\frac{\partial ^2 V}{\partial C^2} +\varphi _e\rho _{xc}\sigma (x)^2C\frac{\partial ^2 V}{\partial C\partial x} +\frac{\varphi _e^2\sigma (x)^2}{2}\frac{\partial ^2 V}{\partial x^2} \end{aligned}$$

(54)

Note that $V(C,x)=(1/\alpha )C^\alpha v(x)$.

$$\begin{aligned}&\displaystyle \frac{\partial V}{\partial C}=C^{\alpha -1} v(x), \quad \frac{\partial V}{\partial x}=\frac{1}{\alpha }C^\alpha v'(x)\end{aligned}$$

(55)

$$\begin{aligned}&\displaystyle \frac{\partial ^2 V}{\partial C^2}=(\alpha -1)C^{\alpha -2}v(x), \quad \frac{\partial ^2 V}{\partial C\partial x}=C^{\alpha -1}v'(x), \quad \frac{\partial ^2 V}{\partial x^2}=\frac{1}{\alpha }C^\alpha v''(x)\nonumber \\ \end{aligned}$$

(56)

The differential equation (54) is equivalent to

$$\begin{aligned} 0&= \frac{\beta }{\rho }v(x)^{1-\frac{\rho }{\alpha }} +\left[ \alpha \frac{\sigma (x)^2}{2}+x+\bar{x}-\frac{\beta }{\rho }\right] v(x) +\left[ \varphi _e\rho _{xc}\sigma (x)^2- \frac{\kappa x}{\alpha }\right] v'(x)\nonumber \\&+\,\frac{\varphi _e^2\sigma ^2(x)}{2\alpha }v''(x). \end{aligned}$$

(57)

Use the change of variable (51).

$$\begin{aligned} v'(x)=\frac{\alpha }{\rho } g(x)^{\frac{\alpha }{\rho }-1} g'(x) \end{aligned}$$

(58)

and

$$\begin{aligned} v''(x)=\frac{\alpha (\alpha -\rho )}{\rho ^2}g(x)^{\frac{\alpha }{\rho }-2}g'(x)^2 +\frac{\alpha }{\rho } g(x)^{\frac{\alpha }{\rho }-1} g''(x). \end{aligned}$$

(59)

Substitute these results into differential equation (57) to get

$$\begin{aligned} 0&= \frac{\beta }{g(x)}+\left[ \alpha \rho \frac{\sigma (x)^2}{2}+\rho x+\rho \bar{x}-\beta \right] +\left[ \alpha \varphi _e\rho _{xc}\sigma (x)^2- \kappa x\right] \frac{g'(x)}{g(x)} \nonumber \\&+\left( \frac{\alpha }{\rho }-1\right) \frac{\varphi _e^2\sigma (x)^2}{2} \left( \frac{g'(x)}{g(x)}\right) ^2 +\frac{\varphi _e^2\sigma (x)^2}{2}\frac{g''(x)}{g(x)}. \end{aligned}$$

(60)

After the change of variable leads to the ODE (27) its coefficients are

• (i) $A_0 =(\rho -\alpha )/\rho$,

• (ii) $B =\epsilon b\sigma _0\varphi _e$,

• (iii) $B_0=-2\epsilon \alpha \sigma _0\rho _{xc}$,

• (iv) $B_1=2\epsilon ^2a\kappa /(a+1), B_2=2\epsilon ^2\kappa /(a+1)$,

• (v) $C_0=2\epsilon ^2a(\beta -\rho \bar{x})/(a+1)-\epsilon ^2\alpha \rho \sigma _0^2$,

• (vi) $C_1=-2\epsilon ^3a\rho \sigma _0\varphi _e/(a+1)$,

• (vii) $C_2=2\epsilon ^2(\beta -\rho \bar{x})/(a+1)$,

• (viii) $C_3=-2\epsilon ^3\rho \sigma _0\varphi _e$,

• (ix) $D_0=-2\epsilon ^2a\beta /(a+1)$, and

• (x) $D_1=-2\epsilon ^2\beta /(a+1)$.

Following the procedure in Chen et al. (2010), the power series representations of $g(x), g'(x)$, and $g''(x)$ are substituted into (27). After equating the coefficients of the terms of degree $n$, we find the recursive formula for $g_k$, where $k \ge 2$.

$$\begin{aligned} g_0 g_2 = A_0 g_1^2 + B_0 g_0 g_1 + (C_2 + C_0) g_0^2 + ( D_1 + D_0) g_0, \end{aligned}$$

(61)

and if $n \ge 1$, then

$$\begin{aligned}&(n+1)(n+2)g_0g_{n+2} =-\sum _{k=0}^{n-1}(k+1)(k+2)g_{k+2}g_{n-k}\nonumber \\&\quad +\,A_0\sum _{k=0}^n (k+1)(n\!-\!k\!+\!1)g_{k+1}g_{n-k+1} \!+\!B_2\sum _{k=0}^{n-1}\sum _{r=0}^k (r+1)g_{r+1}g_{k-r}\frac{B^{n-k-1}}{(n-k-1)!}\nonumber \\&\quad +B_1\sum _{k=0}^{n-1} (k+1)g_{k+1}g_{n-k-1} +B_0\sum _{k=0}^n (k+1)g_{k+1}g_{n-k} \nonumber \\&\quad +\,C_3\sum _{k=0}^{n-1}\sum _{r=0}^k g_rg_{k-r}\frac{B^{n-k-1}}{(n-k-1)!} +C_2\sum _{k=0}^n\sum _{r=0}^k g_rg_{k-r}\frac{B^{n-k}}{(n-k)!} \nonumber \\&\quad +\,C_1\sum _{k=0}^{n-1} g_kg_{n-k-1} +C_0\sum _{k=0}^n g_kg_{n-k} +D_1\sum _{k=0}^n g_k\frac{B^{n-k}}{(n-k)!} +D_0g_n. \end{aligned}$$

(62)

1.2 Derivation of Differential Equation (20) from Ai (2010)

Ai’s differential equation is (IA.3) in the appendix of Ai (2010).

$$\begin{aligned} 0&= \frac{\beta ^\psi }{\psi -1} H(m)^{\frac{1-\psi }{1-\gamma }} +\left( m-\frac{\beta \psi }{\psi -1}-\gamma \frac{\sigma _K^2}{2}\right) \nonumber \\&+\left( \rho \sigma _K\sigma _\theta +Q+\frac{a\bar{\theta }}{1-\gamma } -\frac{a}{1-\gamma }m\right) \frac{H'(m)}{H(m)}+\frac{1}{1-\gamma } \frac{\sigma _m^2}{2}\frac{H''(m)}{H(m)} \end{aligned}$$

(63)

Multiply the above equation by $H(m)$.

$$\begin{aligned} 0&= \frac{\beta ^\psi }{\psi -1} H(m)^{\frac{1-\psi }{1-\gamma }+1} +\left( m-\frac{\beta \psi }{\psi -1}-\gamma \frac{\sigma _K^2}{2}\right) H(m)\nonumber \\&+\left( \rho \sigma _K\sigma _\theta +Q+\frac{a\bar{\theta }}{1-\gamma } -\frac{a}{1-\gamma }m\right) H'(m) +\frac{1}{1-\gamma } \frac{\sigma _m^2}{2} H''(m) \end{aligned}$$

(64)

To make it similar to (14), we introduce the change of variable

$$\begin{aligned} H(m)=h(m)^{-\frac{1-\gamma }{1-\psi }}. \end{aligned}$$

(65)

Then

$$\begin{aligned} H'(m)=-\frac{1-\gamma }{1-\psi } h(m)^{-\frac{1-\gamma }{1-\psi }-1}h'(m) \end{aligned}$$

(66)

and

$$\begin{aligned} H''(m)=\frac{1-\gamma }{1-\psi }\left( \frac{1-\gamma }{1-\psi } +1\right) h(m)^{-\frac{1-\gamma }{1-\psi }-2}h'(m) -\frac{1-\gamma }{1-\psi } h(m)^{-\frac{1-\gamma }{1-\psi }-1}h''(m).\nonumber \\ \end{aligned}$$

(67)

The differential equation (64) becomes

$$\begin{aligned} 0&= \beta ^\psi \frac{1}{h(m)} +\left[ (\psi -1)m-\beta \psi -\gamma (\psi -1)\frac{\sigma _K^2}{2}\right] \nonumber \\&+\,\left[ a\bar{\theta }+(1-\gamma )(\rho \sigma _K\sigma _\theta +Q)-am\right] \frac{h'(m)}{h(m)} \nonumber \\&-\left( \frac{1-\gamma }{1-\psi }+1\right) \frac{\sigma _m^2}{2} \left( \frac{h'(m)}{h(m)}\right) ^2 +\frac{\sigma _m^2}{2}\frac{h''(m)}{h(m)} \end{aligned}$$

(68)

This is the differential equation (20) when $m = x + \bar{\theta }$.

Now introduce the change of variable $m=\epsilon \sigma _m x+\bar{\theta }$ and set $g(x)=h(m)$ so that the differential equation is in the form which is simulated.

$$\begin{aligned} g(x)g''(x)&= \!\left( \frac{1-\gamma }{1-\psi }\!+\!1\right) g'(x)^2 \!+\!\left[ 2\epsilon ^2ax-\frac{2\epsilon (1-\gamma ) (\rho \sigma _K\sigma _\theta \!+\!Q)}{\sigma _m}\right] g(x)g'(x)\nonumber \\&-\,\left[ 2\epsilon ^3(\psi \!-\!1)\sigma _mx\!+\!2\epsilon ^2(\psi -1)\bar{\theta } \!-\!2\epsilon ^2\beta \psi -\epsilon ^2\gamma (\psi -1)\sigma _K^2\right] g(x)^2\nonumber \\&-\,2\epsilon ^2\beta ^\psi g(x) \end{aligned}$$

(69)

Define the constants

$$\begin{aligned} A_0&= \frac{1-\gamma }{1-\psi }+1, \nonumber \\ B_0&= -\frac{2\epsilon (1-\gamma )(\rho \sigma _K\sigma _\theta +Q)}{\sigma _m}, \nonumber \\ B_1&= 2\epsilon ^2a, \nonumber \\ B_2&= 0, \nonumber \\ C_0&= -2\epsilon ^2(\psi -1)\bar{\theta }+2\epsilon ^2\beta \psi +\epsilon ^2\gamma (\psi -1)\sigma _K^2, \nonumber \\ C_1&= -2\epsilon ^3(\psi -1)\sigma _m, \nonumber \\ C_2&= 0, \nonumber \\ C_3&= 0, \nonumber \\ D_0&= -2\epsilon ^2\beta ^\psi , \ \hbox { and } \nonumber \\ D_1&= 0. \end{aligned}$$

(70)

So

$$\begin{aligned} g(x)g''(x)=A_0g'(x)^2+(B_1x+B_0)g(x)g'(x)+(C_1x+C_0)g(x)^2+D_0g(x).\nonumber \\ \end{aligned}$$

(71)

1.3 Proof of Theorem 3.1

Define $\tilde{g_n}=n^2L(g_n/g_0)$ for $n \ge 1$, where $L \ge 1$ is a constant to be determined later. For all $n \ge 3$,

$$\begin{aligned} \tilde{g}_{n+2}&= \frac{n+2}{n+1} \left\{ -\frac{1}{L}\sum _{k=0}^{n-1} \frac{(k+1)\tilde{g}_{k+2}\tilde{g}_{n-k}}{(k+2)(n-k)^2} +\frac{A_0}{L}\sum _{k=0}^n \frac{\tilde{g}_{k+1}\tilde{g}_{n-k+1}}{(k+1)(n-k+1)} \right. \nonumber \\&+\,\frac{B_2}{L} \sum _{k=1}^{n-1}\sum _{r=0}^{k-1} \frac{\tilde{g}_{r+1}\tilde{g}_{k-r}}{(r+1)(k-r)^2} \frac{B^{n-k-1}}{(n-k-1)!} +\frac{B_1}{L} \sum _{k=0}^{n-2} \frac{\tilde{g}_{k+1}\tilde{g}_{n-k-1}}{(k+1)(n-k-1)^2} \nonumber \\&+\,\frac{B_0}{L} \sum _{k=0}^{n-1} \frac{\tilde{g}_{k+1}\tilde{g}_{n-k}}{(k+1)(n-k)^2} +\frac{C_3}{L} \sum _{k=2}^{n-1} \sum _{r=1}^{k-1} \frac{\tilde{g}_r\tilde{g}_{k-r}}{r^2(k-r)^2}\frac{B^{n-k-1}}{(n-k-1)!}\nonumber \\&\!+\,\frac{C_2}{L} \sum _{k=2}^{n} \sum _{r=1}^{k-1} \frac{\tilde{g}_r\tilde{g}_{k-r}}{r^2(k-r)^2} \frac{B^{n-k}}{(n-k)!} \!+\!\frac{C_1}{L} \sum _{k=1}^{n-2} \frac{\tilde{g}_k\tilde{g}_{n-k-1}}{k^2(n-k-1)^2} \!+\!\frac{C_0}{L} \sum _{k=1}^{n-1} \frac{\tilde{g}_k\tilde{g}_{n-k}}{k^2(n-k)^2}\nonumber \\&+\,\frac{D_1}{g_0}\sum _{k=1}^n \frac{\tilde{g}_k}{k^2} \frac{B^{n-k}}{(n-k)!} \!+\!\frac{D_0}{g_0}\frac{\tilde{g}_n}{n^2} \!+\!B_2\sum _{k=0}^{n-1} \frac{\tilde{g}_{k+1}}{k+1}\frac{B^{n-k-1}}{(n-k-1)!} +B_1\frac{\tilde{g}_n}{n}+B_0\frac{\tilde{g}_{n+1}}{n+1}\nonumber \\&+\,C_3L\frac{B^{n-1}}{(n-1)!} +2C_3\sum _{k=1}^{n-1} \frac{\tilde{g}_k}{k^2} \frac{B^{n-k-1}}{(n-k-1)!} +C_2L\frac{B^n}{n!} +2C_2\sum _{k=1}^n \frac{\tilde{g}_k}{k^2}\frac{B^{n-k}}{(n-k)!} \nonumber \\&\left. +\,2C_1\frac{\tilde{g}_{n-1}}{(n-1)^2} +2C_0\frac{\tilde{g}_n}{n^2} +\frac{D_1L}{g_0}\frac{B^n}{n!} \right\} . \end{aligned}$$

(72)

Lemma A.1

If $a, b, c$, and $d$ are integers such that $a \ge 0, b \ge 0, a+c>0, b+d>0$, and $c+d \ge 0$, then

$$\begin{aligned} \sum _{k=a}^{n-b} \frac{1}{(k+c)(n-k+d)} \le U_{n+c+d}, \end{aligned}$$

(73)

where $U_k=2[1+\ln (k+1)]/k$ for $k=1,2,3,\ldots .$

Proof

Note that $\sum _{k=1}^n(1/k) \le 1+\ln (n+1)$ for $n \ge 1$.

$$\begin{aligned} \sum _{k=a}^{n-b} \frac{1}{(k+c)(n-k+d)}&= \frac{1}{n+c+d} \sum _{k=a}^{n-b}\left( \frac{1}{k+c}+\frac{1}{n-k+d}\right) \\&\le \frac{1}{n+c+d}\left( \sum _{k=1}^{n-b+c}\frac{1}{k}+\sum _{k=1}^{b+d}\frac{1}{k}\right) \\&\le \frac{2}{n+c+d}\sum _{k=1}^{n+c+d}\frac{1}{k} \\&\le \frac{2[1+\ln (n+c+d)]}{n+c+d} \\&= U_{n+c+d} \end{aligned}$$

$\square$

Lemma A.2

Let $B \ge 0$ be a real number. If $b$ and $d$ are integers such that $b \ge 0, d \ge 0$, and $b+d>0$, then

$$\begin{aligned} \sum _{k=b}^{n} \frac{1}{k+d} \cdot \frac{B^{n-k}}{(n-k)!} \le U^B_{n+d}, \end{aligned}$$

(74)

where $U^B_k=[(1+2Be^B)+2Be^B\ln (k+1)]/k$ for $k=1,2,3,\ldots .$

Proof

By Lemma A.1, we have

$$\begin{aligned} \sum _{k=b}^n \frac{1}{k+d}\frac{B^{n-k}}{(n-k)!}&= \frac{1}{n+d}+B\sum _{k=b}^{n-1} \frac{1}{(k+d)(n-k)} \cdot \frac{B^{n-k-1}}{(n-k-1)!} \\&\le \frac{1}{n+d}+Be^B\sum _{k=b}^{n-1} \frac{1}{(k+d)(n-k)} \\&\le \frac{1}{n+d}+Be^BU_{n+d} \\&= U^B_{n+d}. \end{aligned}$$

$\square$

There exist a real number $L \ge 1$ and an integer $N \ge 3$ such that for all $n \ge N$,

$$\begin{aligned}&\frac{n+2}{n+1} \left\{ \frac{1}{L}\sum _{k=0}^{n-1} \frac{k+1}{(k+2)(n-k)^2} +\frac{|A_0|}{L}\sum _{k=0}^n \frac{1}{(k+1)(n-k+1)} \right. \nonumber \\&\quad +\,\frac{|B_2|}{L} \sum _{k=1}^{n-1}\sum _{r=0}^{k-1} \frac{1}{(r+1)(k-r)^2} \frac{|B|^{n-k-1}}{(n-k-1)!} +\frac{|B_1|}{L} \sum _{k=0}^{n-2} \frac{1}{(k+1)(n-k-1)^2}\nonumber \\&\quad +\,\frac{|B_0|}{L} \sum _{k=0}^{n-1} \frac{1}{(k+1)(n-k)^2} +\frac{|C_3|}{L} \sum _{k=2}^{n-1} \sum _{r=1}^{k-1} \frac{1}{r^2(k-r)^2}\frac{|B|^{n-k-1}}{(n-k-1)!}\nonumber \\&\quad +\,\frac{|C_2|}{L} \sum _{k=2}^{n} \sum _{r=1}^{k-1} \frac{1}{r^2(k-r)^2} \frac{|B|^{n-k}}{(n-k)!} \!+\!\frac{|C_1|}{L} \sum _{k=1}^{n-2} \frac{1}{k^2(n-k-1)^2} \!+\!\frac{|C_0|}{L} \sum _{k=1}^{n-1} \frac{1}{k^2(n-k)^2}\nonumber \\&\quad +\,\frac{|D_1|}{|g_0|}\sum _{k=1}^n \frac{1}{k^2} \frac{|B|^{n-k}}{(n-k)!} +\frac{|D_0|}{|g_0|}\frac{1}{n^2} +|B_2|\sum _{k=0}^{n-1} \frac{1}{k+1}\frac{|B|^{n-k-1}}{(n-k-1)!} +\frac{|B_1|}{n}+\frac{|B_0|}{n+1}\nonumber \\&\quad +\,|C_3|L\frac{|B|^{n-1}}{(n-1)!} +2|C_3|\sum _{k=1}^{n-1} \frac{1}{k^2} \frac{|B|^{n-k-1}}{(n-k-1)!} +|C_2|L\frac{|B|^n}{n!} +2|C_2|\sum _{k=1}^n \frac{1}{k^2}\frac{|B|^{n-k}}{(n-k)!}\nonumber \\&\quad \left. +\,\frac{2|C_1|}{(n-1)^2} +\frac{2|C_0|}{n^2} +\frac{|D_1|L}{|g_0|}\frac{|B|^n}{n!} \right\} \le \mathcal B (N,L)<1, \end{aligned}$$

(75)

where

$$\begin{aligned} \mathcal B (N,L)&= \frac{N+2}{N+1} \bigg [\frac{\pi ^2}{6L}+\frac{|A_0|}{L}U_{N+2}+\frac{|B_0|}{L}U_{N+1} +\frac{|B_1|+|C_1|+|C_0|}{L}U_N \nonumber \\&+\,\frac{|B_2|+|C_3|+|C_2|}{L}e^{|B|}U_2+\frac{|C_1|}{L(N-2)^2}\nonumber \\&+\,(|B_2|+2|C_3|+2|C_2|+|D_1/g_0|)U^{|B|}_N\nonumber \\&+\,2|C_3|\frac{|B|^{N-2}}{(N-2)!}+\frac{2|C_0|+|D_0/g_0|}{N^2} +\frac{2|C_1|}{(N-1)^2}+\frac{|B_0|}{N+1}+\frac{|B_1|}{N} \nonumber \\&+\,L(|C_2|+|D_1/g_0|)\frac{|B|^N}{N!}+L|C_3|\frac{|B|^{N-1}}{(N-1)!}\bigg ]. \end{aligned}$$

(76)

Pick a real number $M_g \ge 1$ such that $|\tilde{g}_n| \le M_g^n$ for $1 \le n \le N+1$. Apply the following algorithm to construct a sequence $\{G_n\}$ of nonnegative real numbers.

• (i) Use the recurrence relation (62) and $g_0,\,g_1$ to calculate $g_n$, where $2 \le n \le N+1$.

• (ii) Calculate $G_n=n^2L|g_n/g_0|$ for $1 \le n \le N+1$.

• (iii) Calculate the remaining terms $G_{n+2}$, with $n \ge N$, by the recurrence relation:

$$\begin{aligned} G_{n+2}&= \frac{n+2}{n+1} \left\{ \frac{1}{L}\sum _{k=0}^{n-1} \frac{(k+1)G_{k+2}G_{n-k}}{(k+2)(n-k)^2} +\frac{|A_0|}{L}\sum _{k=0}^n \frac{G_{k+1}G_{n-k+1}}{(k+1)(n-k+1)} \right. \nonumber \\&+\,\frac{|B_2|}{L} \sum _{k=1}^{n-1}\sum _{r=0}^{k-1} \frac{G_{r+1}G_{k-r}}{(r\!+\!1)(k-r)^2} \frac{|B|^{n-k-1}}{(n-k-1)!} \!+\!\frac{|B_1|}{L} \sum _{k=0}^{n-2} \frac{G_{k+1}G_{n-k-1}}{(k\!+\!1)(n-k-1)^2} \nonumber \\&+\,\frac{|B_0|}{L} \sum _{k=0}^{n-1} \frac{G_{k+1}G_{n-k}}{(k\!+\!1)(n-k)^2} +\frac{|C_3|}{L} \sum _{k=2}^{n-1} \sum _{r=1}^{k-1} \frac{G_rG_{k-r}}{r^2(k-r)^2}\frac{|B|^{n-k-1}}{(n-k-1)!}\nonumber \\&+\,\frac{|C_2|}{L} \sum _{k=2}^{n} \sum _{r=1}^{k-1} \frac{G_rG_{k-r}}{r^2(k-r)^2} \frac{|B|^{n-k}}{(n-k)!} +\frac{|C_1|}{L} \sum _{k=1}^{n-2} \frac{G_kG_{n-k-1}}{k^2(n-k-1)^2}\nonumber \\&+\,\frac{|C_0|}{L} \sum _{k=1}^{n-1} \frac{G_kG_{n-k}}{k^2(n-k)^2} \nonumber \\&+\,\frac{|D_1|}{|g_0|}\sum _{k=1}^n \frac{G_k}{k^2} \frac{|B|^{n-k}}{(n-k)!} +\frac{|D_0|}{|g_0|}\frac{G_n}{n^2} +|B_2|\sum _{k=0}^{n-1} \frac{G_{k+1}}{k+1}\frac{|B|^{n-k-1}}{(n-k-1)!}\nonumber \\&+\,|B_1|\frac{G_n}{n}+|B_0|\frac{G_{n+1}}{n+1} \nonumber \\&+\,|C_3|L\frac{|B|^{n-1}}{(n-1)!} +2|C_3|\sum _{k=1}^{n-1} \frac{G_k}{k^2} \frac{|B|^{n-k-1}}{(n-k-1)!} +|C_2|L\frac{|B|^n}{n!}\nonumber \\&+\,2|C_2|\sum _{k=1}^n \frac{G_k}{k^2}\frac{|B|^{n-k}}{(n-k)!} \left. +2|C_1|\frac{G_{n-1}}{(n-1)^2} +2|C_0|\frac{G_n}{n^2} +\frac{|D_1|L}{|g_0|}\frac{|B|^n}{n!} \right\} . \end{aligned}$$

(77)

Lemma A.3

Let $M_g=\max \left\{ 1,\root n \of {n^2L|g_n/g_0|} \mid 1 \le n \le N+1\right\}$ and $r_g=\epsilon \sigma _0\varphi _e/M_g$. For $n \ge 1$,

$$\begin{aligned} |g_n| \le \frac{|g_0|}{L} \frac{M_g^n}{n^2}. \end{aligned}$$

(78)

Proof

Note that for any integer $n \ge 3$, we have

$$\begin{aligned} \sum _{k=1}^{n-2} \frac{1}{k^2(n-k-1)^2}&= \frac{1}{(n-2)^2}+\sum _{k=1}^{n-3}\frac{1}{k^2(n-k-2)^2} \nonumber \\&\le \frac{1}{(n-2)^2}+\sum _{k=1}^{n-3}\frac{1}{k(n-k)} \nonumber \\&\le \frac{1}{(n-2)^2}+U_n \end{aligned}$$

and

$$\begin{aligned} \sum _{k=1}^{n-1} \frac{1}{k^2} \frac{|B|^{n-k-1}}{(n-k-1)!}&= \sum _{k=2}^n \frac{1}{(k-1)^2} \frac{|B|^{n-k}}{(n-k)!} \nonumber \\&= \frac{|B|^{n-2}}{(n-2)!}+\sum _{k=3}^n\frac{1}{(k-1)^2} \frac{|B|^{n-k}}{(n-k)!} \nonumber \\&\le \frac{|B|^{n-2}}{(n-2)!}+\sum _{k=3}^n\frac{1}{k} \frac{|B|^{n-k}}{(n-k)!} \nonumber \\&\le \frac{|B|^{n-2}}{(n-2)!}+U^{|B|}_n. \end{aligned}$$

The inequality follows from mathematical induction on $n$ and the facts that $M_g \ge 1$ and $\mathcal B (n,L)<1$ for $n \ge N$. $\square$

Applying the root test for convergence, the lifetime utility can be represented as a convergent power series in Theorem 3.1 is true. Suppose that $|x| \le r_g$.

$$\begin{aligned} \left| g(x)-T_{g,n}(x)\right|&\le \sum _{k=n+1}^\infty \frac{|g_k|}{(\epsilon \sigma _0\varphi _e)^k}r_g^k \nonumber \\&\le \sum _{k=n+1}^\infty \frac{|g_0|}{Lk^2} \nonumber \\&\le \frac{|g_0|}{L}\int \limits _n^\infty \frac{d\lambda }{\lambda ^2} \nonumber \\&= \frac{|g_0|}{Ln} \end{aligned}$$

Suppose that $|x| \leqslant \nu r_g$.

$$\begin{aligned} \left| g'(x)-T_{g,n}'(x)\right|&\le \sum _{k=n+1}^\infty \frac{(k+1)|g_{k+1}|}{(\epsilon \sigma _0\varphi _e)^k}(\nu r_g)^k\nonumber \\&\le \sum _{k=n+1}^\infty \frac{M_g|g_0|}{\epsilon \sigma _0\varphi _e L(k+1)}\nu ^k \nonumber \\&= \frac{M_g|g_0|\nu ^n}{\epsilon \sigma _0\varphi _eL(n+1)(1-\nu )} \end{aligned}$$

The proof is complete.

1.4 Derivation of the Stochastic Differential Equation for the State Price (44)

Given the lifetime utility of a representative investor, the stochastic process for the state price $\Lambda _t$ can be found by using Duffie and Epstein (1992b) and Schroder and Skiadas (2003).

$$\begin{aligned} \frac{d\Lambda _t}{\Lambda _t}=f_V(C_t,V_t)\,dt+\frac{df_C(C_t,V_t)}{f_C(C_t,V_t)} \end{aligned}$$

(79)

where $d f_C(C_t, V_t)$ can be determined by using Ito’s Lemma. Then $\Lambda _t$ can be computed by integrating (79).

$$\begin{aligned} \Lambda _t=f_C(C_t,V_t)\exp \left( \int \limits _0^tf_V(C_s,V_s)\,ds\right) \end{aligned}$$

(80)

where it is assumed that $f_C(C_0, V_0)=\Lambda _0$.³¹

The stochastic differential equation (44) follows from (79). Recall that

$$\begin{aligned} f(C,V)=\frac{\beta }{\rho }\frac{C^{\rho }-(\alpha V)^{\frac{\rho }{\alpha }}}{(\alpha V)^{\frac{\rho }{\alpha }-1}}. \end{aligned}$$

(81)

Then

$$\begin{aligned} f_C(C,V)=\beta \frac{C^{\rho -1}}{(\alpha V)^{\frac{\rho }{\alpha }-1}} \end{aligned}$$

(82)

and

$$\begin{aligned} f_V(C,V)=-\frac{\beta }{\rho }\left[ (\rho -\alpha )\frac{C^\rho }{(\alpha V)^{\frac{\rho }{\alpha }}}+\alpha \right] . \end{aligned}$$

(83)

Now take the partial derivatives of $f_C(C,V)$ which will be used in the computation of $d\Lambda$, and then evaluate using the functional form $V(C,x)=C^\alpha v(x)$. The first order partial derivatives of $f_C(C,V)$ are

$$\begin{aligned} \frac{\partial f_C}{\partial C}(C,V)&= \beta (\rho -1)\frac{C^{\rho -2}}{(\alpha V)^{\frac{\rho }{\alpha }-1}}=\frac{\rho -1}{C}f_C(C,V), \nonumber \\ \frac{\partial f_C}{\partial V}(C,V)&= \beta (\alpha -\rho )\frac{C^{\rho -1}}{(\alpha V)^{\frac{\rho }{\alpha }}}=\frac{\alpha -\rho }{\alpha V}f_C(C,V). \end{aligned}$$

(84)

The second order partial derivatives of $f_C(C,V)$ are

$$\begin{aligned} \frac{\partial ^2 f_C}{\partial C^2}(C,V)&= \frac{(\rho -1)(\rho -2)}{C^2}f_C(C,V), \nonumber \\ \frac{\partial ^2 f_C}{\partial C\partial V}(C,V)&= \frac{(\alpha -\rho )(\rho -1)}{\alpha CV}f_C(C,V),\nonumber \\ \frac{\partial ^2 f_C}{\partial V^2}(C,V)&= \frac{\rho (\rho -\alpha )}{\alpha ^2V^2}f_C(C,V). \end{aligned}$$

(85)

We can apply Ito’s lemma to compute

$$\begin{aligned} \frac{df_C(C_t,V_t)}{f_C(C_t,V_t)}&= (\rho -1)\frac{dC_t}{C_t}+\frac{\alpha -\rho }{\alpha }\frac{dV_t}{V_t} +\frac{(\rho -1)(\rho -2)}{2}\left( \frac{dC_t}{C_t}\right) ^2 \nonumber \\&+\,\frac{(\rho -1)(\alpha -\rho )}{\alpha }\frac{dC_t}{C_t}\frac{dV_t}{V_t} +\frac{\rho (\rho -\alpha )}{2\alpha ^2}\left( \frac{dV_t}{V_t}\right) ^2. \end{aligned}$$

(86)

Note that

$$\begin{aligned} V=\frac{1}{\alpha }C^\alpha g(x)^{\frac{\alpha }{\rho }} \end{aligned}$$

(87)

The first and second partial derivatives of $V$ are

$$\begin{aligned} \frac{\partial V }{\partial C}&= \alpha \frac{V}{C}, \nonumber \\ \frac{\partial V }{\partial x}&= \frac{\alpha }{\rho } V \frac{g'(x)}{g(x)}, \nonumber \\ \frac{\partial ^2 V }{\partial C^2}&= \alpha (\alpha -1) \frac{V}{C^2}, \nonumber \\ \frac{\partial ^2 V }{\partial C \partial x}&= \frac{\alpha ^2}{\rho } \frac{V}{C} \frac{g'(x)}{g(x)} \nonumber \\ \frac{\partial ^2 V }{\partial x^2}&= V\left[ \frac{\alpha (\alpha -\rho )}{\rho ^2}\left( \frac{g'(x)}{g(x)}\right) ^2+\frac{\alpha }{\rho } \frac{g''(x)}{g(x)}\right] . \end{aligned}$$

(88)

By Ito’s Lemma, we get

$$\begin{aligned} \frac{dV_t}{V_t}&= \alpha \frac{dC_t}{C_t}+\frac{\alpha }{\rho }\frac{g'(x_t)}{g(x_t)}\,dx_t +\frac{\alpha (\alpha -1)}{2}\left( \frac{dC_t}{C_t}\right) ^2 +\frac{\alpha ^2}{\rho }\frac{g'(x_t)}{g(x_t)}\frac{dC_t}{C_t}dx_t \nonumber \\&+\frac{1}{2}\left[ \frac{\alpha }{\rho }\frac{g''(x_t)}{g(x_t)} +\frac{\alpha (\alpha -\rho )}{\rho ^2}\left( \frac{g'(x_t)}{g(x_t)}\right) ^2\right] (dx_t)^2 \end{aligned}$$

(89)

By the stochastic processes (9) and (52), we get

$$\begin{aligned} \frac{dV_t}{V_t}&= \left\{ \left[ \frac{\alpha ^2}{2}+\frac{\alpha ^2\varphi _e\rho _{xc}}{\rho }\frac{g'(x_t)}{g(x_t)} +\frac{\alpha \varphi _e^2}{2\rho }\frac{g''(x_t)}{g(x_t)} +\frac{\alpha (\alpha -\rho )\varphi _e^2}{2\rho ^2} \left( \frac{g'(x_t)}{g(x_t)}\right) ^2\right] \sigma (x_t)^2\right. \nonumber \\&+\left. \alpha \bar{x}+\alpha x_t-\frac{\alpha \kappa }{\rho }x_t\frac{g'(x_t)}{g(x_t)} \right\} \,dt +\alpha \sigma (x_t)\,d\tilde{\omega }_{1,t}+\frac{\alpha \varphi _e}{\rho }\sigma (x_t)\frac{g'(x_t)}{g(x_t)}\,d\tilde{\omega }_{2,t}.\nonumber \\ \end{aligned}$$

(90)

Then (86) becomes

$$\begin{aligned} \frac{df_C(C_t,V_t)}{f_C(C_t,V_t)}&= \left\{ (\alpha -1)(x_t+\bar{x})+\frac{(\rho -\alpha )\kappa }{\rho } x_t \frac{g'(x_t)}{g(x_t)}\right. \nonumber \\&+\left[ \frac{(\alpha -1)^2}{2}+\frac{(\alpha -\rho )(\alpha -1)\varphi _e\rho _{xc}}{\rho } \frac{g'(x_t)}{g(x_t)} \right. \nonumber \\&+\left. \left. \frac{(\alpha -\rho )\varphi _e^2}{2\rho } \frac{g''(x_t)}{g(x_t)} +\frac{(\alpha -\rho )(\alpha -2\rho )\varphi _e^2}{2\rho ^2} \left( \frac{g'(x_t)}{g(x_t)}\right) ^2\right] \right\} \,dt\nonumber \\&+\,(\alpha )\sigma (x_t)\,d\tilde{\omega }_{1,t} +\frac{(\alpha -\rho )\varphi _e}{\rho }\sigma (x_t)\frac{g'(x_t)}{g(x_t)}\, d\tilde{\omega }_{2,t}. \end{aligned}$$

(91)

Substitute (87) into (83) to yield

$$\begin{aligned} f_V(C,V) =\frac{\beta (\alpha -\rho )}{\rho }\frac{1}{g(x)} -\frac{\alpha \beta }{\rho } \end{aligned}$$

(92)

Finally, substitute (91) and (92) into (79).

$$\begin{aligned} \frac{d\Lambda _t}{\Lambda _t}&= \bigg \{ (\alpha -1)(x_t+\bar{x})-\frac{(\alpha -\rho )\kappa }{\rho } x_t \frac{g'(x_t)}{g(x_t)} +\frac{(\alpha -\rho )\beta }{\rho }\frac{1}{g(x_t)}-\frac{\alpha \beta }{\rho }\nonumber \\&+\,\bigg [\frac{(\alpha -1)^2}{2}+\frac{(\alpha -\rho )(\alpha -1)\varphi _e\rho _{xc}}{\rho } \frac{g'(x_t)}{g(x_t)}+\frac{(\alpha -\rho )\varphi _e^2}{2\rho } \frac{g''(x_t)}{g(x_t)}\nonumber \\&+\,\frac{(\alpha -\rho )(\alpha -2\rho )\varphi _e^2}{2\rho ^2} \left( \frac{g'(x_t)}{g(x_t)}\right) ^2\bigg ]\sigma (x_t)^2\bigg \}\,dt\nonumber \\&+\,(\alpha -1)\sigma (x_t)\,d\tilde{\omega }_{1,t} +\frac{(\alpha -\rho )\varphi _e}{\rho }\sigma (x_t)\frac{g'(x_t)}{g(x_t)}\,d\tilde{\omega }_{2,t} \end{aligned}$$

(93)

The differential equation (14) is used to remove $g''(x_t)/g(x_t)$ from $\mu _{\Lambda }(x_t)$.

$$\begin{aligned} \frac{(\alpha -\rho )\varphi _e^2\sigma (x)^2}{2\rho } \frac{g''(x)}{g(x)}&= -\frac{(\alpha -\rho )\beta }{\rho } \frac{1}{g(x)} -\left[ (\alpha -\rho )(x+\bar{x})\right. \nonumber \\&\left. +\,\frac{\alpha (\alpha -\rho )\sigma (x)^2}{2} -\frac{(\alpha -\rho )\beta }{\rho }\right] \nonumber \\&+\,\left[ \frac{(\alpha -\rho )\kappa }{\rho }x -\frac{\alpha (\alpha -\rho )\varphi _e\rho _{xc}\sigma (x)^2}{\rho }\right] \frac{g'(x)}{g(x)}\nonumber \\&-\,\frac{(\alpha -\rho )^2\varphi _e^2\sigma (x)^2}{2\rho ^2}\left( \frac{g'(x)}{g(x)}\right) ^2 \end{aligned}$$

(94)

Substitute this result into (93) to get a new expression for the instantaneous mean of $d\Lambda _t/\Lambda _t$.

$$\begin{aligned} \mu _\Lambda (x)&= (\rho -1)(x+\bar{x})-\beta +\left[ \frac{\alpha \rho -2\alpha +1}{2}\right. \nonumber \\&\quad \left. -\frac{(\alpha -\rho )\varphi _e\rho _{xc}}{\rho } \frac{g'(x)}{g(x)} -\frac{(\alpha -\rho )\varphi _e^2}{2\rho } \left( \frac{g'(x)}{g(x)}\right) ^2\right] \sigma (x)^2 \end{aligned}$$

(95)

The rate of return for the risk-free bonds is determined by

$$\begin{aligned} r(x_t) \,dt=-E_{t}\left[ d\Lambda _t/\Lambda _t \right] =-\mu _{\Lambda }(x_t)\,dt. \end{aligned}$$

(96)

This completes the derivation of the stochastic differential equation for the state price (44) and its mean (45).

1.5 Derivation of the Solution (43)

The equivalent differential equation in our model is Eq. (57).

$$\begin{aligned} 0&= \frac{\beta }{\rho }v(x)^{1-\frac{\rho }{\alpha }} +\left[ \frac{\alpha \sigma (x)^2}{2}+x+\bar{x}-\frac{\beta }{\rho }\right] v(x) +\left[ \varphi _e\rho _{xc}\sigma (x)^2-\kappa x\right] v'(x)\nonumber \\&+\,\frac{\varphi _e^2\sigma ^2(x)}{2}v''(x). \end{aligned}$$

(97)

Let the function be chosen to agree with Ai’s.

$$\begin{aligned} y(x)=\frac{1}{\beta } v(x)^{-\frac{\alpha }{\rho }}, \end{aligned}$$

(98)

which is related to the wealth-consumption ratio. Find first order Taylor approximation of

$$\begin{aligned} e^{-\ln y(x)}=\beta v(x)^{-\frac{\rho }{\alpha }}. \end{aligned}$$

(99)

Set

$$\begin{aligned} z=\ln y(x) \quad \hbox { and } \quad \bar{z} = \ln y(0). \end{aligned}$$

(100)

The first order approximation of (99) is

$$\begin{aligned} e^{-z} \approx e^{-\bar{z}}-e^{-\bar{z}}(z-\bar{z}) =\frac{1}{\ln y(0)}-\frac{1}{\ln y(0)} [\ln y(x)-\ln y(0)]. \end{aligned}$$

(101)

Substitute it back into the definition of $z$.

$$\begin{aligned} \frac{1}{y(x)}=e^{-\ln y(x)} \approx \frac{1}{y(0)}[1+\ln y(0)]-\frac{\ln y(x)}{y(0)}. \end{aligned}$$

(102)

Express (99) as

$$\begin{aligned} -\ln y(x)&= \ln \left[ \beta v(x))^{-\frac{\rho }{\alpha }}\right] = \ln \beta -\frac{\rho }{\alpha } \ln v(x).\end{aligned}$$

(103)

$$\begin{aligned} \frac{1}{y(x)}&\approx \frac{1}{y(0)}[1+\ln y(0)]+\frac{1}{y(0)}\left[ \ln \beta -\frac{\rho }{\alpha } \ln v(x)\right] . \end{aligned}$$

(104)

Set $\kappa _1=1/y(0)$. In Ai (2010), he takes $\kappa _1=\beta$ and $\kappa _0=\kappa _1(1-\ln \kappa _1)$. Use (98) and (104) to yield

$$\begin{aligned} \beta v(x)^{-\frac{\rho }{\alpha }} \approx \kappa _0+\kappa _1\ln \beta -\frac{\kappa _1 \rho }{\alpha } \ln v(x) \end{aligned}$$

(105)

The differential equation (97) becomes

$$\begin{aligned} 0&= \frac{1}{\rho } \left[ \kappa _0+\kappa _1\ln \beta -\frac{\kappa _1 \rho }{\alpha } \ln v(x)\right] v(x) +\left[ \frac{\alpha \sigma (x)^2}{2}+x+\bar{x}-\frac{\beta }{\rho }\right] v(x) \nonumber \\&+\left[ \varphi _e\rho _{xc}\sigma (x)^2-\kappa x\right] v'(x) +\frac{\varphi _e^2\sigma (x)^2}{2} v''(x). \end{aligned}$$

(106)

Following Ai (2010), we guess the solution to be of the form

$$\begin{aligned} v(x)=\exp (A+Bx). \end{aligned}$$

(107)

Then

$$\begin{aligned} \frac{v'(x)}{v(x)}=B \quad \hbox { and } \quad \frac{v''(x)}{v(x)}=B^2. \end{aligned}$$

(108)

The differential equation (106) under this guess is

$$\begin{aligned} 0&= \left( \bar{x}+\frac{\kappa _0-\beta +\kappa _1\ln \beta }{\rho }-\frac{\kappa _1}{\alpha }A\right) +\left[ 1+\left( \frac{\kappa _1}{\alpha }-\kappa \right) B\right] x \nonumber \\&+\left( \frac{\alpha }{2}+\varphi _e\rho _{xc}B +\frac{\varphi _e^2}{2}B^2\right) \sigma (x)^2 \end{aligned}$$

(109)

Set

$$\begin{aligned} B=\frac{\alpha }{\alpha \kappa -\kappa _1}. \end{aligned}$$

(110)

If $\sigma (x)=\sigma$ is constant, then

$$\begin{aligned} A=\left( \frac{\alpha }{2}+\varphi _e\rho _{xc}B +\frac{\varphi _e^2}{2}B^2\right) \frac{\alpha \sigma ^2}{\kappa _1}+\frac{\alpha (\kappa _0-\beta +\kappa _1\ln \beta )}{\rho \kappa _1}. \end{aligned}$$

(111)

In this case, we have

$$\begin{aligned} g(x)=v(x)^{\frac{\rho }{\alpha }}=\exp \left( \frac{\rho }{\alpha } A+\frac{\rho }{\alpha } Bx\right) . \end{aligned}$$

(112)

So the initial conditions are

$$\begin{aligned} g(0)=\exp \left( \frac{\rho }{\alpha } A\right) \ \hbox { and } \ g'(0)=\frac{\rho }{\alpha } g(0). \end{aligned}$$

(113)

1.6 Solution to the Bansal and Yaron Model When $\psi =1$

To set the coefficients in (105), Ai uses the solution for the case when $\psi = 1$. In this case, Ai uses

$$\begin{aligned} f(C,V)=\beta (\alpha V)\left[ \ln C-\frac{1}{\alpha } \ln (\alpha V)\right] . \end{aligned}$$

(114)

Let

$$\begin{aligned} V=\frac{1}{\alpha } C^\alpha v(x). \end{aligned}$$

(115)

Then

$$\begin{aligned} f(C,V)=-\frac{\beta }{\alpha } C^\alpha v(x) \ln v(x). \end{aligned}$$

(116)

Recall that the differential equation (54) is

$$\begin{aligned} 0&= f(C,V)+\left[ x+\bar{x}+\frac{\sigma (x)^2}{2} \right] C\frac{\partial V}{\partial C}-\kappa x\frac{\partial V}{\partial x} \nonumber \\&+\,\frac{\sigma (x)^2}{2}C^2\frac{\partial ^2 V}{\partial C^2} +\varphi _e\rho _{xc}\sigma (x)^2\frac{\partial ^2 V}{\partial C\partial x} +\frac{\varphi _e^2 \sigma (x)^2}{2} \frac{\partial ^2 V}{\partial x^2}. \end{aligned}$$

(117)

With the guess (115), we have

$$\begin{aligned} \frac{\partial V}{\partial C}&= C^{\alpha -1} v(x), \quad \frac{\partial V}{\partial x}=\frac{1}{\alpha } C^\alpha v'(x),\end{aligned}$$

(118)

$$\begin{aligned} \frac{\partial ^2 V}{\partial C^2}&= (\alpha -1) C^{\alpha -2} v(x), \quad \frac{\partial ^2 V}{\partial C \partial x}=C^{\alpha -1} v'(x), \quad \frac{\partial ^2 V}{\partial x^2}=\frac{1}{\alpha } C^\alpha v''(x).\qquad \end{aligned}$$

(119)

The differential equation becomes

$$\begin{aligned} 0&= -\frac{\beta }{\alpha } v(x)\ln v(x)+\left[ x+\bar{x}+\frac{\alpha \sigma (x)^2}{2}\right] v(x)-\frac{\kappa }{\alpha } xv'(x) \nonumber \\&+\,\varphi _e \rho _{xc} \sigma (x)^2 v'(x)+\frac{\varphi _e^2 \sigma (x)^2}{2\alpha } v''(x). \end{aligned}$$

(120)

Suppose that the differential equation has a solution of the form

$$\begin{aligned} v(x)=\exp (A+Bx). \end{aligned}$$

(121)

Substituting it into the differential equation yields

$$\begin{aligned} 0&= -\frac{\beta }{\alpha }A+\bar{x}+\left( \frac{\alpha }{2} +\varphi _e\rho _{xc}B+\frac{\varphi _e^2}{2\alpha }B^2\right) \sigma (x)^2 +\left( 1-\frac{\beta +\kappa }{\alpha }B\right) x.\qquad \end{aligned}$$

(122)

Set

$$\begin{aligned} B=\frac{\alpha }{\beta +\kappa }. \end{aligned}$$

(123)

If $\sigma (x)=\sigma$, is constant, then

$$\begin{aligned} A=\frac{\alpha }{\beta } \bar{x}+\left( \frac{\alpha }{2}+\varphi _e\rho _{xc}B +\frac{\varphi _e^2}{2\alpha }B^2\right) \frac{\alpha \sigma ^2}{\beta }. \end{aligned}$$

(124)

This agrees with Ai (2010), since $\kappa _0+\kappa _1\ln \beta =\beta$, when $\kappa _1=\beta$.

1.7 Solution to Ai (2010) Under no Uncertainty

Consider the following first order linear differential equation for $y=y(x)$

$$\begin{aligned} xy'+(\nu + \zeta x)y=\chi , \end{aligned}$$

(125)

where $\nu , \zeta$ and $\chi$ are real constants with

$$\begin{aligned} \nu >0. \end{aligned}$$

(126)

Note that the point $x=0$ is regular singular for this equation. Here, we shall prove the following result for Eq. (125).

Theorem A.4

If $\nu >0$, then Eq. (125) has the following general solution

$$\begin{aligned} y_c(x) = c x^{-\nu } e^{-\zeta x} + \chi x^{-\nu } e^{-\zeta x} \int \limits _0^x t^{\nu -1} e^{\zeta t}dt, \quad \text {if} \quad x>0, \end{aligned}$$

(127)

and

$$\begin{aligned} y_c(x) = c (-x)^{-\nu } e^{-\zeta x} + \chi (-x)^{-\nu } e^{-\zeta x} \int \limits _x^0(-t)^{\nu -1} e^{\zeta t}dt, \quad \text {if} \quad x<0, \end{aligned}$$

(128)

where $c$ is any real number. Furthermore, if $\chi \ne 0$ then letting $c=0$ we get the following particular solution $y(x)=y_0(x)$

$$\begin{aligned} y(x) = \chi x^{-\nu } e^{-\zeta x} \int \limits _0^x t^{\nu -1} e^{\zeta t}dt, \quad \text {if} \quad x>0, \end{aligned}$$

(129)

and

$$\begin{aligned} y(x) = \chi (-x)^{-\nu } e^{-\zeta x} \int \limits _x^0(-t)^{\nu -1} e^{\zeta t}dt, \quad \text {if} \quad x<0. \end{aligned}$$

(130)

Moreover, setting

$$\begin{aligned} y(0) = \frac{\chi }{\nu }, \end{aligned}$$

(131)

makes the function $y(x)$ defined by (129) and (130) continuous at $x=0$.

Finally, setting

$$\begin{aligned} y'(0) = \chi \frac{\zeta }{\nu (\nu +1)} \end{aligned}$$

(132)

makes the function $y(x)$ continuously differentiable at $x=0$.

1.7.1 Solving Differential Equation (125)

Since equation (125) is singular at $x=0$, we shall solve it for $x>0$ and for $x<0$ separately.

Solution for $x>0$. Dividing both sides of (125) by $x$ we write this equation in the normal form

$$\begin{aligned} y'+\Big (\frac{\nu }{x} + \zeta \Big )y=\frac{\chi }{x}. \end{aligned}$$

(133)

Thus we find the integrating factor

$$\begin{aligned} \mu (x)=e^{\int (\frac{\nu }{x} + \zeta )dx} = e^{\nu \ln x +\zeta x} = x^\nu e^{\zeta x}. \end{aligned}$$

(134)

Then multiplying both sides of (133) by the the integrating factor (134) we obtain the equation

$$\begin{aligned} \frac{d}{dx} \Big ( x^\nu e^{\zeta x} y \Big ) = \chi x^{\nu -1} e^{\zeta x}, \end{aligned}$$

which integrated gives

$$\begin{aligned} x^\nu e^{\zeta x} y = c + \chi \int \limits _0^x t^{\nu -1} e^{\zeta t}dt. \end{aligned}$$

Thus, in this case the general solution of equation (125) is

$$\begin{aligned} y_c(x) = c x^{-\nu } e^{-\zeta x} + \chi x^{-\nu } e^{-\zeta x} \int \limits _0^x t^{\nu -1} e^{\zeta t}dt, \quad x>0, \end{aligned}$$

(135)

where $c$ is any real constant. Finally, setting $c=0$ we obtain the following particular solutions.

$$\begin{aligned} y(x) = \chi x^{-\nu } e^{-\zeta x} \int \limits _0^x t^{\nu -1} e^{\zeta t}dt, \quad x>0. \end{aligned}$$

(136)

Note, integral $\int _0^x t^{\nu -1} e^{\zeta t}dt$ is finite for any $x>0$ thanks to our assumption $\nu >0$.

Solution for $x<0$. Replacing $x$ with $-x$ changes our Eq. (125) for the function $y=y(x), x<0$, into the following equation for the function $u(x)=y(-x)$

$$\begin{aligned} (-x)(-u')+(\nu -\zeta x)u=\chi , \quad x>0, \end{aligned}$$

which has the normal form

$$\begin{aligned} y'+\Big (\frac{\nu }{x} - \zeta \Big )y=\frac{\chi }{x}, \quad x>0. \end{aligned}$$

(137)

Now, we observe that Eq. (137) can be obtained from Eq. (133) by replacing $\zeta$ with $-\zeta$. Thus, its general solution is obtained from (35) by replacing $\zeta$ with $-\zeta$. So, we get

$$\begin{aligned} u_c(x) = c x^{-\nu } e^{\zeta x} + \chi x^{-\nu } e^{\zeta x} \int \limits _0^x t^{\nu -1} e^{-\zeta t}dt, \quad x>0, \end{aligned}$$

and therefore

$$\begin{aligned} y_c(x)=u_c(-x) = c (-x)^{-\nu } e^{-\zeta x} + \chi (-x)^{-\nu } e^{-\zeta x} \int \limits _0^{-x} t^{\nu -1} e^{-\zeta t}dt, \quad x<0, \end{aligned}$$

Furtheremore, making the change of variables $t\mapsto -t$ gives

$$\begin{aligned} y_c(x) = c (-x)^{-\nu } e^{-\zeta x} + \chi (-x)^{-\nu } e^{-\zeta x} \int \limits _x^0(-t)^{\nu -1} e^{\zeta t}dt, \quad x<0, \end{aligned}$$

(138)

which is the general solution of Eq. (125) for $x<0$. Finally, setting $c=0$ gives

$$\begin{aligned} y(x) = \chi (-x)^{-\nu } e^{-\zeta x} \int \limits _x^0(-t)^{\nu -1} e^{\zeta t}dt, \quad x<0, \end{aligned}$$

(139)

which is a particular solution of Eq. (125) for $x<0$.

1.7.2 Proof of Conditions (39) and (132)

For $x>0$ using the definition (129) and making the change of variables $u=t/x$ we have

$$\begin{aligned} y(x)&= \chi x^{-\nu } e^{-\zeta x} \int _0^x t^{\nu -1} e^{\zeta t}dt\\&= \chi e^{-\zeta x}\int \limits _0^x \Big ( \frac{t}{x}\Big )^{\nu -1} e^{\zeta t}\frac{dt}{x}\\&= \chi e^{-\zeta x} \int \limits _0^1u^{\nu -1} e^{\zeta x u} du. \end{aligned}$$

Now, we observe that as $x\rightarrow 0^+$ the integrand $u^{\nu -1} e^{\zeta x u}$ converges to $u^{\nu -1}$. Therefore applying the dominated convergence theorem we get that

$$\begin{aligned} \lim _{x\rightarrow 0^+}y(x) = \chi \int \limits _0^1u^{\nu -1} du = \frac{\chi }{\nu }, \end{aligned}$$

which is the desired limit. Similarly, for $x<0$ using formula (130) we have

$$\begin{aligned} y(x)&= \chi (-x)^{-\nu } e^{-\zeta x} \int \limits _x^0(-t)^{\nu -1} e^{\zeta t}dt\\&= \chi e^{-\zeta x} \int \limits _x^0 \Big ( \frac{t}{x}\Big )^{\nu -1} e^{\zeta t}\frac{dt}{-x} = \chi e^{-\zeta x} \int \limits _0^1u^{\nu -1} e^{\zeta x u} du, \end{aligned}$$

which goes to ${\chi }/{\nu }$ as $x\rightarrow 0^-$.

This completes the proof of continuity condition (131).

Next, we prove condition (132). For $x>0$ differentiating $y(x)$ given by formula (129) gives

$$\begin{aligned} \frac{y'(x)}{\chi }&= x^{-\nu } e^{-\zeta x} x^{\nu -1} e^{\zeta x} + \Big (-\nu x^{-\nu -1} - \zeta x^{-\nu } \Big )e^{-\zeta x} \int \limits _0^x t^{\nu -1} e^{\zeta t}dt\\&= \frac{1}{x} - \Big ( \frac{\nu }{x}+\zeta \Big ) x^{-\nu }e^{-\zeta x} \int \limits _0^x t^{\nu -1} e^{\zeta t}dt\\&= \frac{1}{x} - \Big ( \frac{\nu }{x}+\zeta \Big ) e^{-\zeta x} \int \limits _0^x \Big ( \frac{t}{x}\Big )^{\nu -1} e^{\zeta t}\frac{dt}{x}\\&= \frac{1}{x} - \Big ( \frac{\nu }{x}+\zeta \Big ) \int \limits _0^1u^{\nu -1} e^{\zeta x (u-1)} du\\&= \frac{1}{x} - \Big ( \frac{\nu }{x}+\zeta \Big ) \int \limits _0^1u^{\nu -1} \Big ( 1+\zeta x (u-1) +\frac{1}{2!}[\zeta x (u-1)]^2+ \cdots \Big ) du\\&= \frac{1}{x} - \Big ( \frac{\nu }{x}+\zeta \Big )\frac{1}{\nu } - \Big ( \frac{\nu }{x}+\zeta \Big )\zeta x \Big ( \frac{1}{\nu +1}- \frac{1}{\alpha } \Big )\\&- \Big ( \frac{\nu }{x}+\zeta \Big ) \int \limits _0^1u^{\nu -1} \Big ( \frac{1}{2!}[\zeta x (u-1)]^2+ \cdots \Big ) du \end{aligned}$$

Now, letting $x\rightarrow 0^+$ we see that the last term goes to zero and therefore we get

$$\begin{aligned} \lim _{x\rightarrow 0^+} \frac{y'(x)}{\chi } = - \frac{\zeta }{\nu } - \nu \zeta \Big ( \frac{1}{\nu +1}-\frac{1}{\alpha } \Big ) = \frac{\beta }{\alpha (\alpha +1)}, \end{aligned}$$

which gives the desired limit (132). Similarly working, for $x<0$ we have

$$\begin{aligned} \frac{y'(x)}{\chi }&= - (-x)^{-\nu } e^{-\zeta x} (-x)^{\nu -1} e^{\zeta x} - \Big (-\nu (-x)^{-\nu -1} (-1)- \zeta x^{-\nu } \Big )e^{-\zeta x}\nonumber \\&\times \int \limits _0^x (-x)^{\nu -1} e^{\zeta t}dt\\&= \frac{1}{x} + \Big ( \frac{\nu }{x}+\zeta \Big ) (-x)^{-\nu }e^{-\zeta x} \int \limits _0^x (-t)^{\nu -1} e^{\zeta t}dt\\&= \frac{1}{x} - \Big ( \frac{\nu }{x}+\zeta \Big ) e^{-\zeta x} \int \limits _0^x \Big ( \frac{t}{x}\Big )^{\nu -1} e^{\zeta t}\frac{dt}{x}\\&= \frac{1}{x} - \Big ( \frac{\nu }{x}+\zeta \Big ) \int \limits _0^1u^{\nu -1} e^{\zeta x (u-1)} du. \end{aligned}$$

Now, proceeding as in the case $x>0$ we obtain the desired limit (132). This completes the proof of the Theorem A.4.

Remark

Also, Eq. (125) can be solved by letting $y(x)=|x|^ru(x)$ and reduce it to a simpler equation for $u$. Finally, Eq. (125) can be solved by using power series. Since $x=0$ is a regular singular point we look for solutions to Eq. (125) of the form

$$\begin{aligned} y(x) = |x|^r \left( \sum _{n=0}^{\infty }c_nx^n \right) , \quad x>0 \quad \text {or} \quad x<0, \end{aligned}$$

where $c_0\ne 0$. We do this first for $x>0$ and then for $x<0$. In both case we substitute the above series into Eq. (125) and derive an equation for $r$ and the corresponding recurrence relation for the coefficients $c_n$.

1.8 The Equity Premium and its Variance from Ai (2010)

Considering wealth as an asset which pays the consumption level each period, its value is given by

$$\begin{aligned} K = C \frac{g(x)}{\beta ^\psi }. \end{aligned}$$

Applying Ito’s lemma the change in wealth is given by

$$\begin{aligned} \frac{dK}{K} = \frac{dC}{C} + \frac{g'(x)}{g(x)} dx + \frac{g'(x)}{g(x)} \frac{dC}{C} dx + \frac{1}{2} \left( \frac{g'(x)}{g(x)} dx \right) ^2. \end{aligned}$$

By (30) of Ai (2010) the stochastic process for $x$ is

$$\begin{aligned} dx = -\alpha dt + \sigma _m d \widetilde{B}_{m,t}. \end{aligned}$$

By (37) of Ai the stochastic process for consumption is

$$\begin{aligned} \frac{dC}{C} = \mu _C(x)dt + \sigma _K d \widetilde{B}_{K,t} + \sigma _m \frac{g'(x)}{g(x)} d \widetilde{B}_{m,t}. \end{aligned}$$

Here $\mu _C(x)$ is an increasing function of $x$.

Using (26) for the covariance between wealth and the long run risk variable, the variance of the return on wealth and its equity premium is given by

$$\begin{aligned} Var_t \left( \frac{dK}{K}\right) = \left[ \frac{g'(x)}{g(x)} + \left( \frac{g'(x)}{g(x)}\right) ^2 \right] \sigma _m^2 + \sigma _K^2 + \rho _{K,\theta } \sigma _\theta + Q. \end{aligned}$$

1.9 The Equity Premium on Wealth from Ai (2010)

We can calculate the return on wealth as on p. 1346 of Ai (2010). We already have the risk free interest rate

$$\begin{aligned} r(x_t)\,dt=-E_{t}\left[ d\Lambda _t/\Lambda _t \right] . \end{aligned}$$

(140)

where

$$\begin{aligned} -r(x)&= (\rho -1)(x+\bar{x})-\beta +\left[ \frac{\alpha \rho -2\alpha +1}{2} -\frac{(\alpha -\rho )\varphi _e\rho _{xc}}{\rho }\frac{g'(x)}{g(x)}\right. \nonumber \\&\left. -\frac{(\alpha -\rho )\varphi _e^2}{2\rho } \left( \frac{g'(x)}{g(x)}\right) ^2\right] \sigma (x)^2.\end{aligned}$$

(141)

$$\begin{aligned} \frac{d\Lambda _t}{\Lambda _t}-\mu _{\Lambda }(x_t)\,dt&= (\alpha -1)\sigma (x_t)\,d\tilde{\omega }_{1,t}+\frac{(\alpha -\rho )\varphi _e}{\rho }\sigma (x_t)\frac{g'(x_t)}{g(x_t)}\,d\tilde{\omega }_{2,t}.\qquad \end{aligned}$$

(142)

Because we have an endowment economy, the wealth is a claim on the consumption stream so that the changes in wealth and consumption are identical.

$$\begin{aligned} \frac{dW_t}{W_t} =\frac{dC_t}{C_t}=\left( x_t+\bar{x}+\frac{\sigma (x_t)^2}{2}\right) \,dt +\sigma (x_t)\,d\tilde{\omega }_{1,t} \end{aligned}$$

(143)

Following Cochrane (2005) Eq. (1.36) the expected return on an asset relative to the risk free rate is given by the covariance between the change in asset’s price, $P_t$, and the pricing kernel.

$$\begin{aligned} \mu _W(x) - r(x) = - E_t \left[ \frac{d\Lambda _t}{\Lambda _t} \frac{dP_t}{P_t} \right] . \end{aligned}$$

(144)

Since wealth is the price of the claim on consumption, as in (34) of Duffie and Epstein (1992b), its change is given by (143), while the change in the pricing kernel is given by (142). As a result, the risk premium on aggregate wealth is

$$\begin{aligned} \mu _W(x) - r(x)&= \gamma (\sigma (x_t))^2 -\frac{(\alpha -\rho )\varphi _e}{\rho } (\sigma (x_t))^2 \rho _{xc}\frac{g'(x_t)}{g(x_t)} \nonumber \\&= \gamma (\sigma (x_t))^2 - \frac{(\alpha -\rho )\varphi _e}{\rho } (\sigma (x_t))^2 \rho _{xc} \frac{\rho }{\alpha } \frac{v'(x_t)}{v(x_t)} \nonumber \\&= \gamma (\sigma (x_t))^2 - \frac{(\alpha -\rho )\varphi _e}{\rho } (\sigma (x_t))^2 \rho _{xc}\frac{v'(x_t)}{v(x_t)}. \end{aligned}$$

(145)

This corresponds to (48) in Ai’s case but with endowment model for consumption.