# Optimal consumption of the stochastic Ramsey problem for non-Lipschitz diffusion

## Abstract

The stochastic Ramsey problem is considered in a growth model with the production function of a Cobb-Douglas form. The existence of a unique classical solution is proved for the Hamilton-Jacobi-Bellman equation associated with the optimization problem. A synthesis of the optimal consumption policy in terms of its solution is proposed.

MSC:49L20, 49L25, 91B62.

## 1 Introduction

We are concerned with the stochastic Ramsey problem in a growth model discussed by Merton . For recent contribution in this direction, we refer to . A firm produces goods according to the Cobb-Douglas production function $x γ$ for capital x, where $0<γ<1$ (cf. Barro and Sala-i-Martin ). The stock of capital $X t$ at time t is modeled by the stochastic differential equation

$d X t = X t γ dt+σ X t d B t ,t>0, X 0 =x>0,σ≠0,$

on a complete probability space $(Ω,F,P)$ carrying a standard Brownian motion ${ B t }$ endowed with the natural filtration $F t$ generated by $σ( B s ,s≤t)$.

The capital stock can be consumed and the flow of consumption at time t is assumed to be written as $c t X t$. The rate of consumption $c={ c t }$ per capital stock is called an admissible policy if ${ c t }$ is an ${ F t }$-progressively measurable process such that

(1.1)

We denote by the set of admissible policies. Given a policy $c∈A$, the capital stock process ${ X t }$ obeys the equation

$d X t = [ X t γ − c t X t ] dt+σ X t d B t , X 0 =x>0.$
(1.2)

The objective is to find an optimal policy $c ∗ ={ c t ∗ }$ so as to maximize the expected discounted utility of consumption

$J x (c)=E [ ∫ 0 ∞ e − α t U ( c t X t ) d t ]$
(1.3)

over $c∈A$, where $α>0$ is a discount rate and $U(x)$ is a utility function in $C 2 (0,∞)∩C[0,∞)$, which is assumed to have the following properties:

$U ′ (∞)=U(0)=0, U ′ (0+)=U(∞)=∞, U ″ <0.$
(1.4)

The Hamilton-Jacobi-Bellman (HJB for short) equation associated with this problem is given by

$αu(x)= 1 2 σ 2 x 2 u ″ (x)+ x γ u ′ (x)+ U ˜ ( x , u ′ ( x ) ) ,x>0,$
(1.5)

where

(1.6)

This kind of economic growth problem has been studied by Kamien and Schwartz  and Sethi and Thompson [, Chapter 11]. However, the optimization problem is unsolved. It is not guaranteed that (1.2) admits a unique positive solution ${ X t }$ and the value function is a viscosity solution of the HJB equation. The main difficulty stems from the fact that (1.5) is a degenerate nonlinear equation of elliptic type with the non-Lipschitz coefficient $x γ$ in $(0,∞)$. It is also analytically studied by , nevertheless in the finite time horizon. The resulting HJB equation is a parabolic partial differential equation (PDE, for short), which is very different from the elliptic PDE dealt with in the present paper.

In this paper, we provide the existence results on a unique positive solution ${ X t }$ to (1.2) and a classical solution u of (1.5) by the theory of viscosity solutions. For the detail of the theory of viscosity solutions, we mention the works [7, 8] and . An optimal policy is shown to exist in terms of u.

This paper is organized as follows. In Section 2, we show that (1.2) admits a unique positive solution. In Section 3, we show the existence of a viscosity solution u of the HJB equation (1.5). Section 4 is devoted to the $C 2$-regularity of its solution. In Section 5, we present a synthesis of the optimal consumption policy.

## 2 Preliminaries

In this section, we show the existence of a unique solution ${ X t }$ to (1.2).

Proposition 2.1 There exists a unique positive solution ${ X t }={ X t x }$ to (1.2), for each $c∈A$, such that

$E[ X t ]≤ { ( 1 − γ ) t + x 1 − γ } 1 / ( 1 − γ ) ,$
(2.1)
$E [ X t 2 ] ≤ e σ 2 t { 2 ( 1 − λ ) t + x 2 ( 1 − λ ) } 1 / ( 1 − λ ) ,λ:=(1+γ)/2,$
(2.2)
(2.3)

Proof We set $x t = X t 1 − γ$. Then, by Ito’s formula and (1.2),

$d x t = ( 1 − γ ) X t − γ d X t + σ 2 2 ( 1 − γ ) ( − γ ) X t 1 − γ d t = ( 1 − γ ) [ 1 − ( c t + σ 2 2 γ ) x t ] d t + ( 1 − γ ) σ x t d B t , x 0 = x 1 − γ .$
(2.4)

By linearity, (2.4) has a unique solution ${ x t }$. Since

$d x ˆ t =(1−γ) [ − ( c t + σ 2 2 γ ) x ˆ t ] dt+(1−γ)σ x ˆ t d B t , x ˆ 0 = x 1 − γ$
(2.5)

has a positive solution ${ x ˆ t }$, we see by the comparison theorem [, Chapter 6, Theorem 1.1] that $x t ≥ x ˆ t >0$ holds almost surely (a.s.). Therefore, (1.2) admits a unique positive solution ${ X t }$ of the form $X t = x t 1 / ( 1 − γ )$, which satisfies $sup 0 ≤ t ≤ T E[ X t 4 ]<∞$ for each $T≥0$.

Let $θ t$ be the right-hand side of (2.1) and $ϕ t =E[ X t ]$. Obviously, we see that $θ t$ is a unique solution of

$d θ t = θ t γ dt, θ 0 =x>0.$

By (1.2) and Jensen’s inequality,

$d ϕ t =dE[ X t ]=E [ X t γ − c t X t ] dt≤ ϕ t γ dt.$

Since $θ 0 = ϕ 0 =x$, we deduce $ϕ t ≤ θ t$, which implies (2.1).

Similarly, let $Θ t$ be the right-hand side of (2.2) and $Φ t =E[ X t 2 ]$. By substitution, it is easy to see that $Θ ¯ t := e − σ 2 t Θ t$ is a unique solution of

$d Θ ¯ t =2 Θ ¯ t λ dt, Θ ¯ 0 = x 2 >0.$

Hence

$d Θ t = e σ 2 t ( 2 Θ ¯ t λ + σ 2 Θ ¯ t ) dt≥ ( 2 Θ t λ + σ 2 Θ t ) dt.$

Furthermore, by (1.2), Ito’s formula and Jensen’s inequality,

$d Φ t = d E [ X t 2 ] = E [ 2 X t 2 λ − 2 c t X t 2 + σ 2 X t 2 ] d t ≤ ( 2 Φ t λ + σ 2 Φ t ) d t .$

Thus, we deduce $Φ t ≤ Θ t$ and $Φ 0 = Θ 0$, which implies (2.2).

Next, let ${ Y t }$ denote the solution ${ X t y }$ of (1.2) with $Y 0 =y$ and $y t = Y t 1 − γ$. Then, by (2.4)

$d( x t − y t )=−(1−γ) ( c t + σ 2 2 γ ) ( x t − y t )dt+(1−γ)σ( x t − y t )d B t ,$

which implies

$x t − y t =( x 0 − y 0 )exp { − ( 1 − γ ) ( ∫ 0 t c s d s + σ 2 2 γ t ) + ( 1 − γ ) σ B t − σ 2 2 ( 1 − γ ) 2 t } .$

Setting $β=1/(1−γ)>1$, we have

$E [ | x t − y t | β ] ≤ | x 0 − y 0 | β E [ exp { σ B t − σ 2 2 t } ] = | x 1 − γ − y 1 − γ | 1 / ( 1 − γ ) ≤ | x − y | .$
(2.6)

By Young’s inequality , for any $ε 0 >0$,

$| x β − y β | ≤ β ( x β − 1 + y β − 1 ) | x − y | ≤ β [ 1 β ( 1 ε 0 ) β | x − y | β + β − 1 β { ε 0 ( x β − 1 + y β − 1 ) } β / ( β − 1 ) ] ≤ ( 1 ε 0 ) β | x − y | β + ( β − 1 ) ( 2 ε 0 ) β / ( β − 1 ) ( x β + y β ) , x , y ≥ 0 .$

Hence, for any $ε>0$, we choose $C ε >0$ such that

$| x β − y β |≤ C ε | x − y | β +ε ( 1 + x β + y β ) ,x,y≥0.$

Therefore, by (2.1) and (2.6), we have

$E [ | X t − Y t | ] = E [ | x t β − y t β | ] ≤ C ε E [ | x t − y t | β ] + ε E [ 1 + x t β + y t β ] ≤ C ε | x − y | + ε E [ 1 + X t + Y t ] ≤ C ε | x − y | + ε { 1 + 2 β ( t β + x ) + 2 β ( t β + y ) } ,$

which implies (2.3). □

Remark 2.1 The uniqueness for (1.2) is violated if $x=0$ and $c t$ is deterministic since 0 and the limit process $χ t := lim x → 0 + X t x$ satisfy (1.2) with $X 0 =0$, and

$E [ χ t 1 − γ ] =E [ ∫ 0 t ( 1 − γ ) { 1 − ( c s + σ 2 2 γ ) χ s 1 − γ } d s ] ≠0.$
(2.7)

## 3 Viscosity solutions of the HJB equation

Definition 3.1 Let $u∈C(0,∞)$. Then u is called a viscosity solution of (1.5) if the following relations are satisfied:

$α u ( x ) ≤ 1 2 σ 2 x 2 q + x γ p + U ˜ ( x , p ) , ∀ ( p , q ) ∈ J 2 , + u ( x ) , ∀ x > 0 , α u ( x ) ≥ 1 2 σ 2 x 2 q + x γ p + U ˜ ( x , p ) , ∀ ( p , q ) ∈ J 2 , − u ( x ) , ∀ x > 0 ,$

where $J 2 , + u(x)$ and $J 2 , − u(x)$ are the second-order superjets and subjets .

Define the value function u by

$u(x)= sup c ∈ A E [ ∫ 0 ∞ e − α t U ( c t X t ) d t ] ,$
(3.1)

where the supremum is taken over all systems $(Ω,F,P,{ F t };{ B t },{ c t })$.

In this section, we study the viscosity solution u of the HJB equation (1.5). Due to Proposition 2.1, we can show the value function u with the following properties.

Lemma 3.1 We assume (1.4). Then we have

$0≤u(x)≤ζ(x):=x+ ζ 0 ,x>0$
(3.2)

for some constant $ζ 0 >0$, and there exists $C ρ >0$ for any $ρ>0$ such that

$|u(x)−u(y)|≤ C ρ |x−y|+ρ(1+x+y),x,y>0.$
(3.3)

Proof Clearly, u is nonnegative. By concavity, there is $C ¯ >0$ such that

$U(x)≤α 2 − 1 / ( 1 − γ ) x+ C ¯ ,x≥0.$

By (1.1) and (2.1), we have

$E [ ∫ 0 ∞ e − α t U ( c t X t ) d t ] ≤ E [ ∫ 0 ∞ e − α t ( α 2 − 1 / ( 1 − γ ) X t + C ¯ ) d t ] ≤ ∫ 0 ∞ e − α t { α ( t 1 / ( 1 − γ ) + x ) + C ¯ } d t = x + α ∫ 0 ∞ e − α t t 1 / ( 1 − γ ) d t + C ¯ / α ,$

which implies (3.2).

Now, let $ρ>0$ be arbitrary. By (1.4), there is $δ>0$ such that $U(x)≤ρ$ for all $x∈[0,δ]$. Furthermore,

$|U(x)−U(y)|≤ U ′ (δ)|x−y|,x,y≥δ.$

Thus, we obtain a constant $C ρ >0$, depending on $ρ>0$, such that

$|U(x)−U(y)|≤ C ρ |x−y|+ρ,∀x,y≥0.$
(3.4)

By (1.1), (2.3) and (3.4), we get

$| u ( x ) − u ( y ) | ≤ sup c ∈ A E [ ∫ 0 ∞ e − α t | U ( c t X t ) − U ( c t Y t ) | d t ] ≤ sup c ∈ A E [ ∫ 0 ∞ e − α t { C ρ | X t − Y t | + ρ } d t ] ≤ C ρ ∫ 0 ∞ e − α t { C ε | x − y | + ε ( 1 + t 1 / ( 1 − γ ) + x + y ) } d t + ρ / α ≤ C { C ρ C ε | x − y | + ( ε + ρ ) ( 1 + x + y ) } , x , y > 0 ,$
(3.5)

where the constant $C>0$ is independent of ε, $ρ>0$. Replacing ρ by $ρ/2C$ and choosing sufficiently small $ε>0$, we deduce (3.3). □

Remark 3.1 The continuity of u is immediate from (3.3).

Theorem 3.1 We assume (1.4). Then the value function u is a viscosity solution of (1.5).

Proof According to , the viscosity property of u follows from the dynamic programming principle for u, that is,

$u(x)= sup c ∈ A E [ ∫ 0 τ e − α t U ( c t X t ) d t + e − α τ u ( X τ ) ] ,x>0$
(3.6)

for any bounded stopping time $τ≥0$, where the supremum is taken over all systems $(Ω,F,P,{ F t };{ B t },{ c t })$. Let $u ¯ (x)$ be the right-hand side of (3.6). Let $X ˜ t = X t + r$ and $B ˜ t = B t + r − B r$, when $τ=r$ is non-random. Then we have

$d X ˜ t = [ X ˜ t γ − c ˜ t X ˜ t ] dt+σ X ˜ t d B ˜ t , X ˜ 0 = X r$

for the shifted process $c ˜ ={ c ˜ t }$ of c by r, i.e., $c ˜ t = c t + r$. It is easy to see that

$e α r E [ ∫ r ∞ e − α t U ( c t X t ) d t | F r ] =E [ ∫ 0 ∞ e − α t U ( c ˜ t X ˜ t ) d t | F r ] = J X r ( c ˜ )a.s.$

with respect to the conditional probability $P(⋅| F r )$. We take $ζ 1 >0$ such that $x γ ≤αx+ ζ 1$ and sufficiently large $ζ 0 >0$ to obtain

$−αζ+ 1 2 σ 2 x 2 ζ ″ + x γ ζ ′ ≤−α ζ 0 + ζ 1 ≤0.$

By (3.2) in Lemma 3.1, Ito’s formula and Doob’s inequality, we have

$E [ sup 0 ≤ t ≤ T e − α t J X t ( c ˜ ) ] ≤E [ sup 0 ≤ t ≤ T e − α t ζ ( X t ) ] ≤ζ(x)+C,T>0$

for some constant $C>0$. Hence, approximating τ by a sequence of countably valued stopping times, we see that

$E [ e − α τ J X τ ( c ˜ ) ] =E [ ∫ τ ∞ e − α t U ( c t X t ) d t ] .$

Thus

$J x ( c ) = E [ ∫ 0 τ e − α t U ( c t X t ) d t + ∫ τ ∞ e − α t U ( c t X t ) d t ] ≤ E [ ∫ 0 τ e − α t U ( c t X t ) d t + e − α τ u ( X τ ) ] .$

Taking the supremum, we deduce $u≤ u ¯$.

We shall show the reverse inequality in the case that $τ=r$ is constant. For any $ε>0$, we consider a sequence ${ S j :j=1,…,n+1}$ of disjoint subsets of $(0,∞)$ such that

$diam( S j )<δ, ⋃ j = 1 n S j =(0,R)and S n + 1 =[R,∞)$

for $δ,R>0$ chosen later. We take $x j ∈ S j$ and $c ( j ) ∈A$ such that

$u( x j )−ε≤ J x j ( c ( j ) ) ,j=1,…,n+1.$
(3.7)

By the same argument as (3.5), we note that

$| J x ( c ( j ) ) − J y ( c ( j ) ) |+|u(x)−u(y)|≤ C ε |x−y|+ ε 4 (1+x+y),x,y>0$

for some constant $C ε >0$. We choose $0<δ<1$ such that $C ε δ<ε/2$. Then we have

$| J x ( c ( j ) ) − J y ( c ( j ) ) |+|u(x)−u(y)|≤ε(1+x),x,y∈ S j ,j=1,2,…,n.$
(3.8)

Hence, by (3.7) and (3.8),

(3.9)

By definition, we find $c∈A$ such that

$u ¯ (x)−ε≤E [ ∫ 0 r e − α t U ( c t X t ) d t + e − α r u ( X r ) ] .$

In view of [, Chapter 6, Theorem 1.1], we can take c, $c ( j )$ on the same probability space. Define

where $1 { ⋅ }$ denotes the indicator function. Then ${ c t r }$ belongs to . Let ${ X t r }$ be the solution of

$d X t r = [ ( X t r ) γ − c t r X t r ] dt+σ X t r d B t , X 0 r =x>0.$

Clearly, $X t r = X t$ a.s. if $t. Further, for each $j=1,…,n+1$, we have on ${ X r ∈ S j }$

$X t + r r = X r + ∫ r t + r [ ( X s r ) γ − c s r X s r ] d s + ∫ r t + r σ X s r d B s = X r + ∫ 0 t [ ( X s + r r ) γ − c s ( j ) X s + r r ] d s + ∫ 0 t σ X s + r r d B ˜ s a.s.$

Hence, $X t + r r$ coincides with the solution $X t ( j )$ of (1.2) for $( Ω ˜ , F ˜ , P ˜ ,{ F ˜ t };{ B ˜ t },{ c t ( j ) })$ a.s. on ${ X r ∈ S j }$ with $X 0 ( j ) = X r$. Thus, we get

(3.10)

where $E P ˜$ denotes the expectation with respect to $P ˜$.

Now, we fix $x>0$ and choose $R>0$, by (2.1), (2.2) and (3.1), such that

$sup c ∈ A E [ u ( X r ) 1 { X r ≥ R } ] ≤ sup c ∈ A E [ ζ ( X r ) 1 { X r ≥ R } ] ≤ sup c ∈ A 1 R E [ X r 2 + ζ 0 X r ] ≤ C 0 R ( 1 + x + x 2 ) < ε ,$
(3.11)

where the constant $C 0 >0$ depends only on r, $ζ 0$. By (3.9), (3.10) and (3.11), we have

$E [ ∫ r ∞ e − α t U ( c t r X t r ) d t ] = E [ E [ ∫ r ∞ e − α t U ( c t r X t r ) d t | F r ] ] = E [ e − α r J X r ( c ˜ r ) ] = E [ ∑ j = 1 n + 1 e − α r J X r ( c ( j ) ) 1 { X r ∈ S j } ] ≥ E [ ∑ j = 1 n e − α r { u ( X r ) − 3 ε ( 1 + X r ) } 1 { X r ∈ S j } ] ≥ E [ e − α r { u ( X r ) − u ( X r ) 1 { X r ≥ R } } ] − 3 ε E [ 1 + X r ] ≥ E [ e − α r u ( X r ) ] − ε − 3 ε C ( 1 + x )$

for some constant $C>0$ independent of ε. Thus

$u ( x ) ≥ E [ ∫ 0 r e − α t U ( c t r X t r ) d t + ∫ r ∞ e − α t U ( c t r X t r ) d t ] ≥ E [ ∫ 0 r e − α t U ( c t X t ) d t + e − α r u ( X r ) ] − ε − 3 ε C ( 1 + x ) ≥ u ¯ ( x ) − 2 ε − 3 ε C ( 1 + x ) .$

Letting $ε→0$, we get $u ¯ ≤u$.

In the general case, by the above argument, we note that

Hence ${ e − α s u( X s )+ ∫ 0 s e − α t U( c t X t )dt}$ is a supermartingale. By the optional sampling theorem,

$u( X 0 )≥E [ ∫ 0 τ e − α t U ( c t X t ) d t + e − α τ u ( X τ ) | F 0 ] a.s.$

Taking the expectation and then the supremum over , we conclude that $u ¯ ≤u$. Noting the continuity of u, we obtain (3.6). □

## 4 Classical solutions

In this section, using the viscosity solutions technique, we show the $C 2$-regularity of the viscosity solution u of (1.5). For any fixed $0, we consider the boundary value problem

(4.1)

with boundary condition

$w(a)=u(a),w(b)=u(b),$
(4.2)

given by u.

Proposition 4.1 Let $w i ∈C[a,b]$, $i=1,2$, be two viscosity solutions of (3.1), (4.2). Then, under (1.4), we have

$w 1 = w 2 .$

Proof It is sufficient to show that $w 1 ≤ w 2$. Suppose that there exists $x 0 ∈[a,b]$ such that $w 1 ( x 0 )− w 2 ( x 0 )>0$. Clearly, by (4.2), $x 0 ≠a,b$, and we find $x ¯ ∈(a,b)$ such that

$ϱ:= sup x ∈ [ a , b ] { w 1 ( x ) − w 2 ( x ) } = w 1 ( x ¯ )− w 2 ( x ¯ )>0.$

Define

$Ψ k (x,y)= w 1 (x)− w 2 (y)− k 2 | x − y | 2$

for $k>0$. Then there exists $( x k , y k )∈ [ a , b ] 2$ such that

$Ψ k ( x k , y k )= sup ( x , y ) ∈ [ a , b ] 2 Ψ k (x,y)≥ Ψ k ( x ¯ , x ¯ )=ϱ,$
(4.3)

from which

$k 2 | x k − y k | 2 < w 1 ( x k )− w 2 ( y k ).$

Thus

(4.4)

Furthermore, by the definition of $( x k , y k )$,

$Ψ k ( x k , y k )≥ Ψ k ( x k , x k ).$

Hence, by uniform continuity

(4.5)

By (4.3), (4.4) and (4.5), extracting a subsequence, we have

(4.6)

Now, we may consider that $( x k , y k )∈ ( a , b ) 2$ for sufficiently large k. Applying Ishii’s lemma  to $Ψ k (x,y)$, we obtain $X,Y∈R$ such that

$( k ( x k − y k ) , X ) ∈ J ¯ 2 , + w 1 ( x k ) , ( k ( x k − y k ) , Y ) ∈ J ¯ 2 , − w 2 ( y k ) , ( X 0 0 − Y ) ≤ 3 k ( 1 − 1 − 1 1 ) .$
(4.7)

By Definition 3.1,

$α w 1 ( x k ) ≤ 1 2 σ 2 x k 2 X + x k γ μ + U ˜ ( x k , μ ) , α w 2 ( y k ) ≥ 1 2 σ 2 y k 2 Y + y k γ μ + U ˜ ( y k , μ ) ,$

where $μ=k( x k − y k )$. Putting these inequalities together, we get

$α { w 1 ( x k ) − w 2 ( y k ) } ≤ 1 2 σ 2 ( x k 2 X − y k 2 Y ) + ( x k γ − y k γ ) μ + { U ˜ ( x k , μ ) − U ˜ ( y k , μ ) } ≡ I 1 + I 2 + I 3 , say .$

By (4.5) and (4.7), it is clear that

Also, by (4.5)

By (1.6), (3.4), (4.5) and (4.6), we have

Consequently, by (4.6), we deduce that

$αϱ≤α { w 1 ( x ˜ ) − w 2 ( x ˜ ) } ≤0,$

Theorem 4.1 We assume (1.4). Then there exists a solution $u∈ C 2 (0,∞)$ of (1.5).

Proof For any $0, we recall the boundary value problem (4.1), (4.2). Since

$U(0)≤ U ′ (x)(0−x)+U(x),x>0,$

we have

$K 0 := sup 0 < x ≤ a x U ′ (x)<∞.$

Hence, by (1.4)

$| U ( c x 1 ) − U ( c x 2 ) | ≤ c U ′ ( c a ) | x 1 − x 2 | ≤ K 0 a | x 1 − x 2 | , x 1 , x 2 ∈ [ a , b ] , 0 ≤ c ≤ 1 .$

Also, by (1.6)

$| U ˜ ( x 1 , y 1 ) − U ˜ ( x 2 , y 2 ) | ≤ max 0 ≤ c ≤ 1 | U ( c x 1 ) − U ( c x 2 ) | + | x 1 y 1 − x 2 y 2 | ≤ K 0 a | x 1 − x 2 | + | x 1 − x 2 | | y 1 | + b | y 1 − y 2 | , y 1 , y 2 > 0 .$

Thus the nonlinear term of (4.1) is Lipschitz. By uniform ellipticity, a standard theory of nonlinear elliptic equations yields that there exists a unique solution $w∈ C 2 (a,b)∩C[a,b]$ of (4.1), (4.2). For details, we refer to [, Theorem 17.18] and [, Chapter 5, Theorem 3.7]. Clearly, by Theorem 3.1, u is a viscosity solution of (4.1), (4.2). Therefore, by Proposition 4.1, we have $w=u$ and u is smooth. Since a, b are arbitrary, we obtain the assertion. □

## 5 Optimal consumption

In this section, we give a synthesis of the optimal policy $c ∗ ={ c t ∗ }$ for the optimization problem (1.4) subject to (1.2). We consider the stochastic differential equation

$d X t ∗ = [ ( X t ∗ ) γ − η ( X t ∗ ) X t ∗ ] dt+σ X t ∗ d B t , X 0 ∗ =x>0,$
(5.1)

where $η(x)=I(x, u ′ (x))$ and $I(x,y)$ denotes the maximizer of (1.6) for $x,y>0$, i.e.,

(5.2)

Our objective is to prove the following.

Theorem 5.1 We assume (1.4). Then the optimal consumption policy ${ c t ∗ }$ is given by

$c t ∗ =η ( X t ∗ ) .$
(5.3)

To obtain the optimal consumption policy ${ c t ∗ }$, we should study the properties of the value function u and the existence of strong solution ${ X t ∗ }$ of (5.1). We need the following lemmas.

Lemma 5.1 Under (1.4), the value function u is concave. In addition, we have

(5.4)
$u ′ (0+)=∞.$
(5.5)

Proof Let $x i >0$, $i=1,2$. For any $ε>0$, there exists $c ( i ) ∈A$ such that

$u( x i )−ε

where ${ X t ( i ) }$ is the solution of (1.2) corresponding to $c ( i )$ with $X 0 ( i ) = x i$. Let $0≤ξ≤1$, and we set

$c ¯ t = ξ c t ( 1 ) X t ( 1 ) + ( 1 − ξ ) c t ( 2 ) X t ( 2 ) ξ X t ( 1 ) + ( 1 − ξ ) X t ( 2 ) ,$

which belongs to . Define ${ X ¯ t }$ and ${ X ˜ t }$ by

$d X ¯ t = [ ( X ¯ t ) γ − c ¯ t X ¯ t ] d t + σ X ¯ t d B t , X ¯ 0 = ξ x 1 + ( 1 − ξ ) x 2 , X ˜ t = ξ X t ( 1 ) + ( 1 − ξ ) X t ( 2 ) .$

By concavity,

$X ˜ t ≤ξ x 1 +(1−ξ) x 2 + ∫ 0 t [ ( X ˜ s ) γ − c ¯ s X ˜ s ] ds+ ∫ 0 t σ X ˜ s d B s a.s.$

By the comparison theorem, we have

Thus, by (1.4)

$u ( ξ x 1 + ( 1 − ξ ) x 2 ) ≥ E [ ∫ 0 ∞ e − α t U ( c ¯ t X ¯ t ) d t ] ≥ E [ ∫ 0 ∞ e − α t U ( c ¯ t X ˜ t ) d t ] = E [ ∫ 0 ∞ e − α t U ( ξ c t ( 1 ) X t ( 1 ) + ( 1 − ξ ) c t ( 2 ) X t ( 2 ) ) d t ] ≥ ξ E [ ∫ 0 ∞ e − α t U ( c t ( 1 ) X t ( 1 ) ) d t ] + ( 1 − ξ ) E [ ∫ 0 ∞ e − α t U ( c t ( 2 ) X t ( 2 ) ) d t ] ≥ ξ u ( x 1 ) + ( 1 − ξ ) u ( x 2 ) − ε .$

Therefore, letting $ε→0$, we obtain the concavity of u.

To prove (5.4), by Theorem 4.1, we recall that u is smooth. Furthermore, we get $u ′ (x)≥0$ for $x>0$. If not, then $u ′ ( a 0 )<0$ for some $a 0 >0$. By concavity,

which is a contradiction. Suppose that $u ′ (z)=0$ for some $z>0$. Then, by concavity, we have $u ′ (x)=0$ for all $x≥z$. Hence, by (1.5) and (1.6),

$αu(z)=αu(x)= U ˜ (x,0)=U(x),x≥z.$

This is contrary to (1.4). Thus, we obtain (5.4).

Next, by definition, we have

$00,$

where ${ X ˇ t }$ is the solution of (1.2) corresponding to $c t =1$. As in (2.7), the limit process $χ ˇ t := lim x → 0 + X ˇ t$ is different from 0. Hence

$0

Suppose that $u ′ (0+)<∞$. By (1.5) and concavity, we get $u(0+)=0$, which is a contradiction. This implies (5.5). □

Lemma 5.2 Under (1.4), there exists a unique positive strong solution ${ X t ∗ }$ of (5.1).

Proof Let ${ N t }$ be the solution of (1.2) corresponding to $c t =0$. We can take the Brownian motion ${ B t }$ on the canonical probability space [, p.71]. Since $0≤η≤1$, the probability measure $P ˆ$ is defined by

$d P ˆ /dP=exp { − ∫ 0 t η ( N s ) / σ d B s − 1 2 ∫ 0 t ( η ( N s ) / σ ) 2 d s }$

for every $t≥0$. Girsanov’s theorem yields that

Hence

Thus, (5.1) admits a weak solution.

Now, by (5.2), we have

$η(x)x=min { ( U ′ ) − 1 ∘ u ′ ( x ) , x } .$

Hence, by (1.4) and concavity,

$d d x ( U ′ ) − 1 ∘ u ′ (x)= u ″ ( x ) U ″ ∘ ( U ′ ) − 1 ∘ u ′ ( x ) ≥0.$

Thus, $η(x)x$ is nondecreasing on $(0,∞)$. We rewrite (5.1) as the form of (2.4) to obtain $X t ∗ >0$ a.s. Then we see that the pathwise uniqueness holds for (5.1). Therefore, by the Yamada-Watanabe theorem , we deduce that (5.1) admits a unique strong solution ${ X t ∗ }$. □

Proof of Theorem 5.1 Since ${ c t ∗ }$ satisfies (1.1), it belongs to . By Lemma 5.2, we note that

$0< u ′ (x)x≤u(x)−u(0+)0.$

Hence, by (2.2) and (3.2),

$E [ ∫ 0 t { e − α s u ′ ( X s ∗ ) X s ∗ } 2 d s ] ≤ E [ ∫ 0 t { e − α s u ( X s ∗ ) } 2 d s ] ≤ E [ ∫ 0 t e − α s ζ ( X s ∗ ) 2 d s ] < ∞ .$

This yields that ${ ∫ 0 t e − α s u ′ ( X s ∗ ) X s ∗ d B s }$ is a martingale. By (1.6), (5.3) and Ito’s formula,

$E [ e − α t u ( X t ∗ ) ] = u ( x ) + E [ ∫ 0 t e − α s { − α u ( X s ∗ ) + ( X s ∗ ) γ u ′ ( X s ∗ ) − c s ∗ X s ∗ u ′ ( X s ∗ ) + 1 2 σ 2 ( X s ∗ ) 2 u ″ ( X s ∗ ) } d s ] = u ( x ) − E [ ∫ 0 t e − α s U ( c s ∗ X s ∗ ) d s ] .$

By (2.1) and (3.2), it is clear that

Letting $t→∞$, we deduce

$E [ ∫ 0 ∞ e − α t U ( c t ∗ X t ∗ ) d t ] =u(x).$

By the same calculation as above, we obtain

$E [ ∫ 0 ∞ e − α t U ( c t X t ) d t ] ≤u(x)$

for any $c∈A$. The proof is complete. □

Remark 5.1 From the proof of Theorem 5.1, it follows that the solution u of the HJB equation (1.5) coincides with the value function. This implies that the uniqueness holds for (1.5).

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## Acknowledgements

I would like to thank Professor H Morimoto for his useful help. The research was supported by the National Natural Science Foundation of China (11171275) and the Fundamental Research Funds for the Central Universities (XDJK2012C045).

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Correspondence to Chuandi Liu. 