1 Introduction

In this paper we consider an optimal consumption-investment problem on infinite time horizon for a general incomplete market model. The market model considered here consists of \(m+1\) securities, one of which is a risk-less asset and the other \(m\) assets are risky ones. The price of risk-less asset is governed by an ordinary differential equation, while the prices of risky ones are defined by the stochastic differential equations. We suppose that all coefficients appearing in those dynamics of the asset prices are affected by the economic factors. The dynamics of the economic factors are also formulated by the stochastic differential equations (cf. (2.1)–(2.3)). In such a general market model, the investor divides his (her) wealth among those \(m+1\) securities and decides the rate for consumption. The goal is to select consumption and investment strategies which maximize the total expected (discounted) power utility of consumption on a long run (cf. (2.5)).

Our approach is based on the dynamic programming principle, in which the H–J–B (Hamilton–Jacobi–Bellman) equation (cf. (2.10) and (2.11)) are derived relevant to the consumption-investment problem. One can construct an optimal consumption-investment strategy from the smooth solution to the equation. Indeed, the optimal consumption rate and optimal portfolio strategy can be explicitly expressed in terms of the smooth solution. In this approach, a pioneering work by Merton for the market with the risky asset price having constant volatility and return has been done in [14] and recent progress of the further studies in this direction is seen in [4–8] etc.. In [4–6] one dimensional H–J–B equations are considered, where the ordinary differential equations concern. On the other hand, in [7, 8], they prove existence of the smooth solution to the H–J–B equation in general dimension by employing a modification of the Leray-Schauder fixed point theorem. Then, constructing the optimal consumption-investment strategy by using the solution, the verification theorems are shown under certain incomplete market settings. In their proofs they have also the best possible discount factor. The current paper is motivated by these works [7, 8]. Following their methods we prove existence of the solution to the H–J–B equation (cf. (2.11)) by giving sub- and super- solutions under the general settings, while we obtain newly the uniqueness theorems on the solution, which is one of our main concern. In the current paper the definition of the set of admissible strategies is given by using the unique solution of the H–J–B equation and thus our uniqueness theorems have a crucial meaning. As a result our set of admissible strategies generalizes the one defined in [8] and the proofs of the verification theorems have become quite natural and simple in the current paper. Indeed, by introducing the new measure \({\overline{P}}^{\hat{h}}\) defined by (5.3) from the unique solution \(z(x)\) of H–J–B equation (2.11), the relevant criterion function to the optimal strategy \(({\hat{c}}_s,{\hat{h}}_s)\) turns out to be described as

$$\begin{aligned} e^{z(x)}{\overline{E}}^{\hat{h}}\left[ -\int \limits _0^{\infty }d\varphi _t\right] \end{aligned}$$

by using a multiplicative functional \(\varphi _t=e^{-\int _0^te^{-\frac{z(X_s)}{1-\gamma }}ds}=e^{-\int _0^t{\hat{c}}_s^{\gamma }e^{-z(X_s)}ds}\) as you see in (5.2) and (5.5). Thus, identification of the value function with the solution to the H–J–B equation can be done by showing \(\varphi _{\infty }=0, \) a.s. Comparing the value function with the criterion for strategy \((c_s,h_s)\) is also seen by looking at \(e^{z(x)}{\overline{E}}^{ h}[-\int _0^{\infty }d{\tilde{\varphi }}_t]\) with \({\tilde{\varphi }}_t=e^{-\int _0^tc_s^{\gamma }e^{-z(X_s)}ds}\) (cf. Proofs of Theorem 5.1 and 5.2).

There are slight difference between the market models discussed in [7, 8] and ours although the both are factor models. They treat a linear Gaussian model and another factor model with the boundedness assumptions, where all coefficients appearing in the dynamics for the asset prices and the factor process are bounded. In the current paper, the general factor model including linear Gaussian models is discussed without boundedness assumptions on the returns of the price process and the drift coefficient of the factor process. Such difference may cause certain technical difference to treat.

We mention some other works with different approach from ours. Approach using the martingale methods for a complete market model appears in [2, 11, 19]. Analytical solutions are given in [3, 12, 20]. More attentive introduction to the historical works can be seen in [7].

The paper is organized as follows. Derivation of the H–J–B equation and our assumptions are described in Sect. 2 under the setting of the factor model. We devote Sect. 3 to construction of sub- and super- solutions to the H–J–B equation. In Sect. 4, we present the existence and uniqueness theorems for the H–J–B equation and the proofs of uniqueness are given. The proofs of existence of the solution following the methods due to Hata and Sheu [7] is completed in Appendix 1. We give the proofs of the verification theorems in Sect.  5. Some notes on the useful gradient estimates for the proofs are given in Appendix 2.

2 Derivation of H–J–B Equations and Assumptions

Consider a market model with \(m+1\) securities and \(n\) factors, where the bond price is governed by ordinary differential equation

$$\begin{aligned} dS^0(t)=r(X_t)S^0(t)dt,\quad S^0(0)=s^0. \end{aligned}$$
(2.1)

The other security prices and factors are assumed to satisfy stochastic differential equations

$$\begin{aligned}&dS^i(t)=S^i(t)\{\alpha ^i(X_t)dt+\sum _{k=1}^{n+m}\sigma ^i_k(X_t) dW^k_t\},\nonumber \\&S^i(0)=s^i,\;\;i=1,\ldots ,m, \end{aligned}$$
(2.2)

and

$$\begin{aligned}&dX_t=\beta (X_t)dt +\lambda (X_t) dW_t,\nonumber \\&X(0)=x, \end{aligned}$$
(2.3)

where \(W_t=(W_t^k)_{k=1,\ldots ,(n+m)}\) is an \(m+n\)-dimensional standard Brownian motion process on a probability space \((\Omega ,{\mathcal F},P)\). Let \(N_t^i\) be the number of the shares of \(i\)th security. Then the total wealth that the investor possesses is defined as

$$\begin{aligned} V_t=\sum _{i=0}^m N_t^iS_t^i \end{aligned}$$

the portfolio proportion invested to \(i\)th security as

$$\begin{aligned} h_t^i=\frac{N_t^iS_t^i}{V_t},\quad i=0,1,2,\ldots ,m. \end{aligned}$$

We assume that self-financing condition holds for a consumption investment strategy \((h_t,C_t)\):

$$\begin{aligned} dV_t=\sum _{i=0}^mN_t^idS_t^i-C_t dt=\sum _{i=0}^m\frac{V_th_t^i}{S_t^i}dS_t^i-C_tdt, \end{aligned}$$

where \(C_t\) is the instantaneous nominal consumption. Setting \(C_t=c_tV_t\), we have

$$\begin{aligned} \frac{dV_t}{V_t}&= h_t^0r(X_t)dt+\sum _{i=1}^mh_t^i\{\alpha ^i(X_t)dt+\sum _{j=1}^{n+m}\sigma _j^i(X_t)dW_t^j\}-c_tdt\\&= \{r(X_t)-c_t\}dt+\sum _{i=1}^mh_t^i\{(\alpha ^i(X_t)-r(X_t))dt+\sum _{j=1}^{n+m}\sigma _j^i(X_t)dW_t^j\} \end{aligned}$$

Thus the equation describing the dynamics of wealth \(V_t=V(c,h)_t\) is given by

$$\begin{aligned} \frac{dV_t}{V_t}=\{r(X_t)+h_t^*(\alpha (X_t)-r(X_t){\mathbf 1})-c_t\}dt+h_t^*\sigma (X_t)dW_t. \end{aligned}$$
(2.4)

Here \(h^*\) stands for the transposed vector of \(h\) and \({\mathbf 1}=(1,1,\ldots ,1)^*\). As for the filtration to be satisfied by admissible investment strategies,

$$\begin{aligned} {\mathcal G}_t=\sigma (S(u),X(u),\;\;u\le t) \end{aligned}$$

is relevant in the present problem and we introduce the following definition.

Definition 2.1

\(h(t)_{0\le t\le T}\) is said an investment strategy if \(h(t)\) is an \(R^m\) valued \({\mathcal G}_t\)- progressively measurable stochastic process such that

$$\begin{aligned} P\left( \;\;\int \limits _0^T|h(s)|^2ds<\infty ,\;\; \forall T\right) =1. \end{aligned}$$

The set of all investment strategies is denoted by \({\mathcal H}(T)\). For a given \(h\in {\mathcal H}(T)\), the process \(V_t=V_t(h)\) representing the total wealth of the investor at time \(t\) is determined by the stochastic differential equation as was seen above. For \(\rho \ge 0\), let us consider the following problem

$$\begin{aligned} v(x):= \sup _{h., c_.} E\left[ \int \limits _0^{\infty }\frac{1}{\gamma }e^{-\rho t}\{c_tV_t(c,h)\}^{\gamma }dt\right] , \quad \gamma <1, \;\gamma \ne 0. \end{aligned}$$
(2.5)

The following equations are equivalent to (2.5)

$$\begin{aligned} v^*(x)&:= \sup _{h., c_.}E\left[ \int \limits _0^{\infty }e^{-\rho t}\{c_tV_t(c,h)\}^{\gamma }dt\right] , \quad 0<\gamma <1,\end{aligned}$$
(2.6)
$$\begin{aligned} v_*(x)&:= \inf _{h., c_.}E\left[ \int \limits _0^{\infty }e^{-\rho t}\{c_tV_t(c,h)\}^{\gamma }dt\right] , \quad \gamma <0, \end{aligned}$$
(2.7)

in each case of \(0<\gamma <1\) and \(\gamma <0\), respectively. It is easy to see that

$$\begin{aligned} V_t^{\gamma }=v_0^{\gamma }e^{\gamma \int _0^t\{\eta (X_s,h_s)-c_s\}ds+\gamma \int _0^th_s^*\sigma (X_s)dW_s-\frac{\gamma ^2}{2}\int _0^th_s^*\sigma \sigma ^*(X_s)h_sds}, \end{aligned}$$

where, \(v_0\) is the initial wealth and

$$\begin{aligned} \eta (x,h)=-\frac{1-\gamma }{2}h^*\sigma \sigma ^*(x)h+h^*{\hat{\alpha }}(x)+r(x),\;\;\; {\hat{\alpha }}(x)=\alpha (x)-r(x){\mathbf 1}. \end{aligned}$$

Therefore

$$\begin{aligned}&E\left[ \int \limits _0^{T}e^{-\rho t}\{c_tV_t(c,h)\}^{\gamma }dt\right] \\&\quad =v_0^{\gamma }\int \limits _0^{T}e^{-\rho t}E[c_t^{\gamma }e^{\gamma \int _0^t\{\eta (X_s,h_s)-c_s\}ds+\gamma \int _0^th_s^*\sigma (X_s)dW_s-\frac{\gamma ^2}{2}\int _0^th_s^*\sigma \sigma ^*(X_s)h_sds}]dt. \end{aligned}$$

Now let us assume that \((\Omega ,{\mathcal F})\) be a standard measurable space (cf. [18]). Then, if \(M^h_t\) defined by \(M^h_t=\gamma \int _0^th_s^*\sigma (X_s)dW_s\) satisfies

$$\begin{aligned} E[e^{M^h_T-\frac{1}{2}\langle M^h\rangle _T}]=1, \quad \forall T, \end{aligned}$$
(2.8)

then there is a probability measure \(P^h\) satisfying

$$\begin{aligned} \left. \frac{d P^h}{dP}\right| _{{\mathcal F}_T}=e^{M^h_T-\frac{1}{2}\langle M^h\rangle _T}, \quad \forall T. \end{aligned}$$

Then we have

$$\begin{aligned}&E\left[ \int \limits _0^{T}e^{-\rho t}\{c_tV_t(c,h)\}^{\gamma }dt\right] \\&\quad =v_0^{\gamma }E^h\left[ \int \limits _0^{T}e^{-\rho t}c_t^{\gamma }e^{\gamma \int _0^t\{\eta (X_s,h_s)-c_s\}ds}dt\right] , \end{aligned}$$

and thus

$$\begin{aligned} v^*(x)= \sup _{h., c_.}v_0^{\gamma } E^h\left[ \int \limits _0^{\infty }e^{-\rho t}c_t^{\gamma }e^{\gamma \int _0^t\{\eta (X_s,h_s)-c_s\}ds}dt\right] ,\quad 0 <\gamma <1 \end{aligned}$$
(2.6')

and

$$\begin{aligned} v_*(x)= \inf _{h., c_.}v_0^{\gamma } E^h\left[ \int \limits _0^{\infty }e^{-\rho t}c_t^{\gamma }e^{\gamma \int _0^t\{\eta (X_s,h_s)-c_s\}ds}dt\right] ,\quad \gamma <0. \end{aligned}$$
(2.7')

Note that, under the probability measure \(P^h\),

$$\begin{aligned} W^h_t=W_t-\int \limits _0^t \gamma \sigma ^*(X_s)h_sds \end{aligned}$$

is a Brownian motion process and the stochastic differential equation for the economic factor \(X_t\) is written as

$$\begin{aligned} dX_t=\{\beta (X_t)+\gamma \lambda \sigma ^*(X_t)h_t\}dt+\lambda (X_t)dW^h_t,\quad X_0=x. \end{aligned}$$
(2.9)

When setting

$$\begin{aligned} v(x):=v_0^{-\gamma }v_*(x),\quad \gamma <0, \end{aligned}$$

the H–J–B equation for \(v(x)\) turns out to be

$$\begin{aligned} \rho v=\frac{1}{2}\text{ tr }[\lambda \lambda ^*D^2v]+\beta ^*Dv+\inf _{c\ge 0,h\in R^m}\{[\gamma \lambda \sigma ^*(x)h]^*Dv+\gamma (\eta (x,h)-c)v+c^{\gamma }\}. \end{aligned}$$

By taking a transformation \(z(x)=\log v(x)\), we obtain

$$\begin{aligned} \rho&= \frac{1}{2}\text{ tr }[\lambda \lambda ^*D^2z]+\frac{1}{2}(Dz)^*\lambda \lambda ^*Dz+\beta ^*Dz\nonumber \\&\quad +\inf _{c\ge 0,h\in R^m}\{[\gamma \lambda \sigma ^*(x)h]^*Dz+\gamma (\eta (x,h)-c)+c^{\gamma }e^{-z}\}, \end{aligned}$$
(2.10)

which can be written as

$$\begin{aligned} \rho =\frac{1}{2}\text{ tr }[\lambda \lambda ^*D^2z]+\beta _{\gamma }^*Dz+\frac{1}{2}(Dz)^*\lambda N_{\gamma }^{-1}\lambda ^*Dz+U_{\gamma }+(1-\gamma )e^{-\frac{z}{1-\gamma }}, \end{aligned}$$
(2.11)

where

$$\begin{aligned} \beta _{\gamma }&= \beta +\frac{\gamma }{1-\gamma }\lambda \sigma ^*(\sigma \sigma ^*)^{-1}{\hat{\alpha }},\;\;N_{\gamma }^{-1}=I+\frac{\gamma }{1-\gamma }\sigma ^*(\sigma \sigma ^*)^{-1}\sigma ,\;\;\\ U_{\gamma }&= \frac{\gamma }{2(1-\gamma )}{\hat{\alpha }}^*(\sigma \sigma ^*)^{-1}{\hat{\alpha }}+\gamma r. \end{aligned}$$

On the other hand, for \(1>\gamma >0\), we set

$$\begin{aligned} {\tilde{v}}(x):=v_0^{-\gamma }v^*(x). \end{aligned}$$

Then, the H–J–B equation of \({\tilde{v}}(x)\) is seen to be

$$\begin{aligned} \rho {\tilde{v}}=\frac{1}{2}\text{ tr }[\lambda \lambda ^*D^2{\tilde{v}}]+\beta ^*D{\tilde{v}}+\sup _{c\ge 0,h\in R^m}\{[\gamma \lambda \sigma ^*(x)h]^*D{\tilde{v}}+\gamma (\eta (x,h)-c)v+c^{\gamma }\} \end{aligned}$$

and, by taking a transformation \({\tilde{z}}(x)=\log {\tilde{v}}(x)\), we obtain

$$\begin{aligned} \rho&= \frac{1}{2}\text{ tr }[\lambda \lambda ^*D^2{\tilde{z}}]+\frac{1}{2}(D{\tilde{z}})^*\lambda \lambda ^*D{\tilde{z}}+\beta ^*D{\tilde{z}}\nonumber \\&\quad +\sup _{c\ge 0,h\in R^m}\{[\gamma \lambda \sigma ^*(x)h]^*D{\tilde{z}}+\gamma (\eta (x,h)-c)+c^{\gamma }e^{-{\tilde{z}}}\}, \end{aligned}$$
(2.12)

which turns out to be the same equation as (2.11).

When \(\gamma <0\), we assume that

$$\begin{aligned}&\displaystyle \lambda ,\;\beta ,\;\sigma ,\;\alpha \; \text{ and }\; r\;\;\text{ are } \text{ globally } \text{ Lipschitz, } \text{ smooth },\end{aligned}$$
(2.13)
$$\begin{aligned}&\displaystyle \left\{ \begin{array}{l} c_1|\xi |^2\le \xi ^*\lambda \lambda ^*(x)\xi \le c_2|\xi |^2,\;\;c_1,\;c_2>0,\;\; \xi \in R^n,\\ c_1|\zeta |^2\le \zeta ^*\sigma \sigma ^*(x)\zeta \le c_2|\zeta |^2,\;\;\; \zeta \in R^m, \end{array} \right. \end{aligned}$$
(2.14)
$$\begin{aligned}&\displaystyle r \;\hbox {is nonnegative }, \end{aligned}$$
(2.15)

and that, in the case of \(\rho =0\),

$$\begin{aligned} k_1\le {\hat{\alpha }}^*(\sigma \sigma ^*)^{-1}{\hat{\alpha }}(x)+r(x),\;\;\; k_1>0. \end{aligned}$$
(2.16)

On the other hand, when \(0<\gamma <1\), we assume (2.13)–(2.15),

$$\begin{aligned} {\hat{\alpha }}^*(\sigma \sigma ^*)^{-1}{\hat{\alpha }}+r \rightarrow \infty , \;\; |x|\rightarrow \infty , \end{aligned}$$
(2.16')

and the following condition

$$\begin{aligned} \left\{ \begin{array}{l} \text{ there } \text{ exists } \text{ a } \text{ function } z_0 \text{ bounded } \text{ below } \text{ such } \text{ that } \\ \frac{1}{2}\text{ tr }[\lambda \lambda ^*D^2z_0]+\beta _{\gamma }^*Dz_0+\frac{1}{2}(Dz_0)^*\lambda N_{\gamma }^{-1}\lambda ^*Dz_0+U_{\gamma }\rightarrow -\infty ,\\ \text{ as }\; |x|\rightarrow \infty ,\;\text{ and } \text{ that }\;\; |Dz_0(x)|\le C_0(|x|+1),\quad C_0>0. \end{array}\right. \end{aligned}$$
(2.17)

Note that

$$\begin{aligned}&\frac{1}{1-\gamma }I\le N_{\gamma }^{-1}\le I, \quad \gamma <0, \nonumber \\&I\le N_{\gamma }^{-1}\le \frac{1}{1-\gamma }I, \quad 0<\gamma <1 \end{aligned}$$
(2.18)

hold under these assumptions.

In considering (2.6) and (2.7) we formulate the set of strategies defined by

$$\begin{aligned} {\mathcal A}_0:=\{(c_s,h_s); c_s\ge 0 \text{ and }\; h_s\; \text{ are } \text{ progressively } \text{ measurable } \text{ and } h_s \text{ satisfies } \text{(2.9) }\}. \end{aligned}$$

Then, we confine the sets of admissible strategies in each case of \(0<\gamma <1\) and \(\gamma <0\) as follows. Set

$$\begin{aligned} {\tilde{M}}_t=\int \limits _0^t(Dz)^*\lambda (X_s)dW_s^h, \end{aligned}$$

where \(z(x)\) is the unique solution to H–J–B equation (2.11) (see Theorem 4.1 and Theorem 4.2 ), and consider strategies satisfying

$$\begin{aligned} E^h[e^{{\tilde{M}}_T-\frac{1}{2}\langle {\tilde{M}}\rangle _T}]=1, \quad \forall T. \end{aligned}$$
(2.19)

For such \(h_s\) we have a probability measure \({\overline{P}}^h\) on \((\Omega ,{\mathcal F})\) such that

$$\begin{aligned} \left. \frac{d {\overline{P}}^h}{dP^h}\right| _{{\mathcal F}_T}=e^{{\tilde{M}}_T-\frac{1}{2}\langle {\tilde{M}}\rangle _T}, \quad \forall T. \end{aligned}$$

The set \({\mathcal A}_1\) of admissible strategies is defined by

$$\begin{aligned} {\mathcal A}_1=\{(c_s,h_s)\in {\mathcal A}_0; h_s\; \text{ satisfies } \text{(2.20) } \}. \end{aligned}$$

3 Sub- and Super-Solution

3.1 Risk-Averse Case (\(\gamma <0\))

We first note that there exists a positive constants \(c_0\) and \(c\) such that

$$\begin{aligned} \frac{1}{2(1-\gamma )}{\hat{\alpha }}^*(\sigma \sigma ^*)^{-1}{\hat{\alpha }}+r-c\ge c_0>0 \end{aligned}$$
(3.1)

under assumption (2.16). We consider the following stochastic differential equation

$$\begin{aligned} d{\bar{X}}_t=\lambda ({\bar{X}}_t)dW_t+\beta _{\gamma }({\bar{X}}_t)dt,\quad {\bar{X}}_0=x \end{aligned}$$

and set

$$\begin{aligned} {\underline{z}}(x)= (1-\gamma )\log E_x\left[ \int \limits _0^{\infty }e^{\frac{1}{1-\gamma }\int _0^tU_{\gamma }({\bar{X}}_s)ds}dt\right] \end{aligned}$$
(3.2)

and

$$\begin{aligned} {\overline{z}}(x)=\log E_x\left[ \int \limits _0^{\infty }c^{\gamma }e^{\int _0^t\{U_{\gamma }({\bar{X}}_s)-\gamma c\}ds}dt\right] . \end{aligned}$$
(3.3)

Then, we have the following lemma.

Lemma 3.1

Assume assumptions (2.13)–(2.16). Then, \({\underline{z}}(x)\) (respectively \({\overline{z}}(x)\)) is a sub-solution (resp. super-solution) to (2.11) for \(\rho =0\). Further

$$\begin{aligned} {\underline{z}}(x)\le {\overline{z}}(x) \end{aligned}$$

Proof

Set

$$\begin{aligned} \varphi _1(x)=E_x\left[ \int \limits _0^{\infty }e^{\frac{1}{1-\gamma }\int _0^tU_{\gamma }({\bar{X}}_s)ds}dt\right] . \end{aligned}$$

Then, it satisfies

$$\begin{aligned} \frac{1}{2}\text{ tr }[\lambda \lambda ^*D^2\varphi _1]+\beta _{\gamma }^*D\varphi _1+\frac{1}{1-\gamma }U_{\gamma }\varphi _1+1=0 \end{aligned}$$

and thus \({\underline{z}}\) does

$$\begin{aligned} \frac{1}{2}\text{ tr }[\lambda \lambda ^*D^2{\underline{z}}]+\beta _{\gamma }^*D{\underline{z}}+\frac{1}{2(1-\gamma )}(D{\underline{z}})^*\lambda \lambda ^*D{\underline{z}}+U_{\gamma } +(1-\gamma )e^{-\frac{\underline{z}}{1-\gamma }}=0. \end{aligned}$$

Since \(\frac{1}{1-\gamma }I\le N_{\gamma }^{-1}\), \({\underline{z}}\) turns out to be a sub-solution to (2.11).

On the other hand, set

$$\begin{aligned} \varphi _2(x)=E_x\left[ \int \limits _0^{\infty }c^{\gamma }e^{\int _0^t\{U_{\gamma }({\bar{X}}_s)-\gamma c\}ds}dt\right] . \end{aligned}$$

Then, it satisfies

$$\begin{aligned} \frac{1}{2}\text{ tr }[\lambda \lambda ^*D^2\varphi _2]+\beta _{\gamma }^*D\varphi _2+(U_{\gamma }-\gamma c)\varphi +c^{\gamma }=0. \end{aligned}$$

Therefore, \({\overline{z}}\) satisfies

$$\begin{aligned} \frac{1}{2}\text{ tr }[\lambda \lambda ^*D^2{\overline{z}}]+\beta _{\gamma }^*D{\overline{z}}+\frac{1}{2}(D{\overline{z}})^*\lambda \lambda ^*D{\overline{z}}+ (U_{\gamma }-\gamma c)+c^{\gamma }e^{-{\overline{z}}}=0. \end{aligned}$$

The left hand side is obtained when taking in the right hand of (2.12) \(h=\frac{1}{1-\gamma }(\sigma \sigma ^*)^{-1}{\hat{\alpha }}(x)\) and the constant \(c>0\) satisfying (3.1), and thus we see that \({\overline{z}}\) is a super-solution to (2.11) with \(\rho =0\).

It is a direct consequence from the following lemma that \({\underline{z}}(x)\le {\overline{z}}(x)\). \(\square \)

Lemma 3.2

The following inequality holds

$$\begin{aligned} E\left[ \int \limits _0^Te^{\frac{1}{1-\gamma }\int _0^tU_{\gamma }({\bar{X}}_s)ds}dt\right] ^{1-\gamma }\le E\left[ \int \limits _0^Tc^{\gamma }e^{\int _0^t\{U_{\gamma }({\bar{X}}_s)-\gamma c\}ds}dt\right] (1-e^{-cT})^{-\gamma }. \end{aligned}$$

Proof

Set

$$\begin{aligned} f_t&= e^{\frac{1}{1-\gamma }\int _0^tU_{\gamma }({\bar{X}}_s)ds}c^{\frac{\gamma }{1-\gamma }}e^{-\frac{\gamma }{1-\gamma }ct}, \\ g_t&= c^{-\frac{\gamma }{1-\gamma }}e^{\frac{\gamma }{1-\gamma }ct}. \end{aligned}$$

Then, we have

$$\begin{aligned} E\left[ \int \limits _0^Tf_tg_t dt\right]&\le E\left[ \int \limits _0^Tf_t^{1-\gamma }dt\right] ^{\frac{1}{1-\gamma }}E\left[ \int \limits _0^Tg_t^{-\frac{1-\gamma }{\gamma }}dt\right] ^{-\frac{\gamma }{1-\gamma }}\\&= E\left[ \int \limits _0^Tc^{\gamma }e^{\int _0^t\{U_{\gamma }({\bar{X}}_s)-\gamma c\}ds}dt\right] ^{\frac{1}{1-\gamma }}(1-e^{-cT})^{-\frac{\gamma }{1-\gamma }}. \end{aligned}$$

Hence, we obtain the present lemma. \(\square \)

Remark

When \(\rho >0\), we do not need to assume (2.16) because there exist positive constants \(c\) and \(c_0\) such that

$$\begin{aligned} \frac{1}{2(1-\gamma )}{\hat{\alpha }}^*(\sigma \sigma ^*)^{-1}{\hat{\alpha }}+r-\frac{\rho }{\gamma }-c\ge c_0>0 \end{aligned}$$

and thus, considering \({\tilde{U}}_{\gamma }:=\frac{\gamma }{2(1-\gamma )}{\hat{\alpha }}^*(\sigma \sigma ^*)^{-1}{\hat{\alpha }}+\gamma r-\rho \) in place of \(U_{\gamma }\) the parallel arguments to the above apply.

3.2 Risk Seeking Case (\(0<\gamma <1\))

It is not easy to construct a super-solution to (2.11) in risk seeking case, \(0<\gamma <1\). Let us start with considering the H–J–B equation of risk-sensitive portfolio optimization without consumption:

$$\begin{aligned} \chi =\frac{1}{2}\text{ tr }[\lambda \lambda ^*D^2z]+\beta _{\gamma }^*D z+\frac{1}{2}(D z)^*\lambda N_{\gamma }^{-1}\lambda ^*Dz+U_{\gamma }. \end{aligned}$$
(3.4)

To study the existence and uniqueness of the solution to (3.4), we introduce the discounted type equation:

$$\begin{aligned} \epsilon z_{\epsilon } =\frac{1}{2}\text{ tr }[\lambda \lambda ^*D^2z_{\epsilon }]+\beta _{\gamma }^*D z_{\epsilon }+\frac{1}{2}(D z_{\epsilon })^*\lambda N_{\gamma }^{-1}\lambda ^*Dz_{\epsilon }+U_{\gamma }. \end{aligned}$$
(3.5)

We first note that (3.5) can be written as

$$\begin{aligned} \epsilon z_{\epsilon }=L z_{\epsilon }+Q(x,Dz_{\epsilon }), \end{aligned}$$
(3.6)

where

$$\begin{aligned} L z:=\frac{1}{2}\text{ tr }[\lambda \lambda ^* D^2 z]+\beta ^*Dz \end{aligned}$$

and

$$\begin{aligned} Q(x,p)&:= \frac{1}{2}(\lambda ^*p+\gamma \sigma ^*(\sigma \sigma ^*)^{-1}{\hat{\alpha }})^*N_{\gamma }^{-1}(\lambda ^*p+\gamma \sigma ^*(\sigma \sigma ^*)^{-1}{\hat{\alpha }})\\&\quad +\frac{\gamma }{2}{\hat{\alpha }}^*(\sigma \sigma ^*)^{-1}{\hat{\alpha }}+\gamma r. \end{aligned}$$

We have the following lemma.

Lemma 3.3

Under assumptions (2.13)–(2.15) and (2.17), (3.5) has a solution \(z_{\epsilon }\in C^2(R^n)\) such that \(z_{\epsilon }-z_0\) is bounded above.

Proof

In light of assumption (2.17), we can assume that \(z_0\ge 0\) and also take \(R_0\) such that, for \(R\ge R_0\),

$$\begin{aligned} Lz_0+Q(x,Dz_0)<0,\quad x\in B_R^c. \end{aligned}$$

Set

$$\begin{aligned} \Phi _{\epsilon }(x)=\frac{M}{\epsilon }+z_0,\quad M:=\sup _{x\in B_{R_0}}|Lz_0+Q(x,Dz_0)|. \end{aligned}$$

Then, \(\Phi _{\epsilon }(x)\) turns out to be a super-solution to (3.5). In proving the existence of the solution to (3.5), we first consider the Dirichlet problem for \(R>R_0\):

$$\begin{aligned} \left\{ \begin{array}{l} \epsilon z_{\epsilon ,R}=Lz_{\epsilon ,R}+Q(x,Dz_{\epsilon ,R}), \quad x\in B_R\\ z_{\epsilon ,R}=\Phi _{\epsilon }, \quad x\in \partial B_R. \end{array}\right. \end{aligned}$$
(3.7)

Owing to Theorem 8.3 [13, Chapter 4], Dirichlet problem (3.7) has a solution \(z_{\epsilon ,R}\in C^{2,\mu }(R^n)\). We extend \(z_{\epsilon ,R}\) to the whole Euclidean space as

$$\begin{aligned} {\tilde{z}}_{\epsilon ,R}(x)=\left\{ \begin{array}{lcl} &{} z_{\epsilon ,R}(x) &{} x\in B_R,\\ &{} \Phi _{\epsilon }(x)&{} x\in B_R^c. \end{array} \right. \end{aligned}$$

Then, we can see that \({\tilde{z}}_{\epsilon ,R}(x)\) is non-increasing with respect to \(R\). Indeed, for \(R<R'\),

$$\begin{aligned}&\epsilon ({\tilde{z}}_{\epsilon ,R}-{\tilde{z}}_{\epsilon ,R'})=L({\tilde{z}}_{\epsilon ,R}-{\tilde{z}}_{\epsilon ,R'})\\&\quad + \frac{1}{2}\{\lambda ^*(D{\tilde{z}}_{\epsilon ,R}+D{\tilde{z}}_{\epsilon ,R'}+2\gamma \sigma ^*(\sigma \sigma ^*)^{-1}{\hat{\alpha }}\}^*N_{\gamma }^{-1}\lambda ^*({\tilde{z}}_{\epsilon ,R}-{\tilde{z}}_{\epsilon ,R'}). \end{aligned}$$

Therefore, from the maximum principle, we see that

$$\begin{aligned} {\tilde{z}}_{\epsilon ,R}(x)-{\tilde{z}}_{\epsilon ,R'}(x)\ge 0\quad \text{ on } \;\;B_{R'} \end{aligned}$$

since \({\tilde{z}}_{\epsilon ,R}(x)={\tilde{z}}_{\epsilon ,R'}(x),\; x\in \partial B_{R'}\). We further note that \({\tilde{z}}_{\epsilon ,R}(x)\ge 0\) for each \(R\) because \(z_1(x)\equiv 0\) is a sub-solution to (3.5) for \(\gamma >0\) and the maximum principle again applies.

Similarly to the proof of Proposition 3.2 in [17], we have the following gradient estimate: for each \(R\), \(r<\frac{R}{2}\), and \(x\in B_r\),

$$\begin{aligned} |\nabla {\tilde{z}}_{\epsilon ,R}(x)|^2&\le C(|\nabla \Psi |_{2r}^2+\frac{1}{r^2}|\Psi |^2_{2r}+|\beta _{\gamma }|^2_{2r}+|\nabla \beta _{\gamma }|_{2r}\nonumber \\&\quad +\frac{|\beta _{\gamma }|_{2r}}{2r}+|U_{\gamma }|_{2r}+|\nabla U_{\gamma }|_{2r}+1)), \end{aligned}$$
(3.8)

where \(C\) is a positive constant independent of \(R\) and \(\epsilon \), \(|f|_r=|f|_{L^{\infty }(B_r(x))}\), and \(\Psi =\lambda N_{\gamma }^{-1}\lambda ^*\). Thus, by similar arguments to the proof of Lemma 2.8 in [9], we can see that \({\tilde{z}}_{\epsilon ,R}(x)\) converges \(H^1_{loc}\) strongly and uniformly on each compact set to the solution \(z_{\epsilon }\in C^2(R^n)\). Since \({\tilde{z}}_{\epsilon ,R}(x)\le \Phi _{\epsilon }(x)\) for each \(R>R_0\) we see that \(z_{\epsilon }\le \Phi _{\epsilon }\), and hence, \( z_{\epsilon }-z_0\) is bounded above. \(\square \)

Lemma 3.4

Assume assumptions (2.13)–(2.15) and (2.17). Then, the solution \(z_{\epsilon }\) to (3.5) such that \( z_{\epsilon }-z_0\) is bounded above satisfies

$$\begin{aligned} z_{\epsilon }(x)-z_0(x)\rightarrow -\infty ,\quad \text{ as }\;\; |x|\rightarrow \infty . \end{aligned}$$

Proof

Let \(z_{\epsilon }\) be a solution to (3.5) such that \(z_{\epsilon }-z_0\) is bounded above and set

$$\begin{aligned} V:=-\left\{ \frac{1}{2}\text{ tr }[\lambda \lambda ^*D^2z_0]+\beta _{\gamma }^*Dz_0+\frac{1}{2}(Dz_0)^*\lambda N_{\gamma }^{-1}\lambda ^*Dz_0+U_{\gamma }\right\} . \end{aligned}$$

Then, we have

$$\begin{aligned} \epsilon z_{\epsilon }+V&= \frac{1}{2}\text{ tr }[\lambda \lambda ^*D^2(z_{\epsilon }-z_0)]+\beta _{\gamma }^*D(z_{\epsilon }-z_0)+\frac{1}{2}(Dz_{\epsilon })^*\lambda N_{\gamma }^{-1}\lambda ^*Dz_{\epsilon }\nonumber \\&\quad -\frac{1}{2}(Dz_0)^*\lambda N_{\gamma }^{-1}\lambda ^*Dz_0=\frac{1}{2}\text{ tr }[\lambda \lambda ^*D^2(z_{\epsilon }-z_0)]\nonumber \\&\quad +{\tilde{\beta }}_{\gamma }^*D(z_{\epsilon }-z_0)+\frac{1}{2}D(z_{\epsilon }-z_0)^*\lambda N_{\gamma }^{-1}\lambda ^*D(z_{\epsilon }-z_0), \end{aligned}$$
(3.9)

where

$$\begin{aligned} {\tilde{\beta }}_{\gamma }=\beta _{\gamma }+\lambda N_{\gamma }^{-1}\lambda ^*Dz_0. \end{aligned}$$

Then, similarly to the proof of Lemma 2.1 in [16], we can see that \(z_{\epsilon }-z_0\rightarrow -\infty \) as \(|x|\rightarrow \infty \) since \(V(x)+\epsilon z_{\epsilon } \rightarrow \infty ,\;\;|x|\rightarrow \infty \). \(\square \)

Lemma 3.5

Under assumptions (2.13)–(2.15) and(2.17), (3.4) has a solution \(({\hat{\chi }}, {\hat{z}})\) such that \({\hat{z}}-z_0\) is bounded above and \({\hat{z}}\in C^2(R^n)\). Moreover, the solution \((\chi ,z)\) such that \( z-z_0\) is bounded above is unique, when admitting ambiguity of additive constants with respect to \(z\).

Proof

Let us first note that the same estimate as (3.8) holds for \(z_{\epsilon }\) for each \(\epsilon >0\). Owing to assumptions (2.13) - (2.15), estimate (3.8) implies that

$$\begin{aligned} |\nabla z_{\epsilon }(x)|^2\le C(|x|^2+1), \end{aligned}$$
(3.10)

where \(C\) is a positive constant independent of \(\epsilon \). According to Lemma 3.4, \(z_{\epsilon }(x)-z_0(x) \rightarrow -\infty \) as \(|x|\rightarrow \infty \). Therefore, we can take \(x_{\epsilon }\) such that

$$\begin{aligned} z_{\epsilon }(x_{\epsilon })-z_0(x_{\epsilon })=\sup _x\{z_{\epsilon }(x)-z_0(x)\}. \end{aligned}$$

Then, at \(x_{\epsilon }\)

$$\begin{aligned} D(z_{\epsilon }-z_0)=0,\quad \frac{1}{2}\text{ tr }[\lambda \lambda ^*D^2(z_{\epsilon }-z_0)]\le 0. \end{aligned}$$

Therefore, from (3.9), we have

$$\begin{aligned} V(x_{\epsilon })+\epsilon z_{\epsilon }(x_{\epsilon })\le 0, \end{aligned}$$

and thus, \(V(x_{\epsilon })\le 0\). Since \(V(x)\rightarrow \infty \) as \(|x|\rightarrow \infty \), there exists a compact set \(K\) such that \(x_{\epsilon }\in K\) for each \(\epsilon >0\). Therefore, we can take a subsequence \(\{ x_{\epsilon _n}\} \subset \{x_{\epsilon }\}\) and \({\hat{x}}\) such that \(x_{\epsilon _n}\rightarrow {\hat{x}}\), \(n\rightarrow \infty \). Once more again from (3.9) we see that

$$\begin{aligned} 0\le \varlimsup _{n\rightarrow \infty }\epsilon _nz_{\epsilon _n}(x_{\epsilon _n})\le -V({\hat{x}}). \end{aligned}$$

Thus, by taking a subsequence if necessary, \(\epsilon _nz_{\epsilon _n}(x_{\epsilon _n})\rightarrow {\hat{\chi }}\), \(n\rightarrow \infty \). On the other hand, by using (3.10) we can see that \(\{z_{\epsilon _n}(x)-z_{\epsilon _n}({\hat{x}})\}\) forms a sequence of uniformly bounded and equicontinuous functions on each compact set \(K' \) including \(K\). Thus, similarly to the proof of Theorem 3.1 in [9], we can see that it converges to \({\hat{z}}(x)\in C^2(R^n)\) in \(H^1_{loc}\) strongly and uniformly on each compact set, by using estimate (3.8) for \(z_{\epsilon }\), and that \(({\hat{\chi }}, {\hat{z}}(x))\) satisfies (3.4). Further, we can see that \(\epsilon _nz_{\epsilon _n}({\hat{x}})\rightarrow {\hat{\chi }}\). Note that

$$\begin{aligned} z_{\epsilon _n}(x)-z_{\epsilon _n}({\hat{x}})-z_0(x)\le z_{\epsilon _n}(x_{\epsilon _n})-z_{\epsilon _n}({\hat{x}})-z_0(x_{\epsilon _n}) \end{aligned}$$

and the left hand side converges to \({\hat{z}}(x)-z_0(x)\). Therefore we see that \({\hat{z}}(x)-z_0(x)\) is bounded above by the constant \(-z_0({\hat{x}})\) which is the limit of the right hand side.

Further, we can see that \({\hat{z}}(x)-z_0(x)\rightarrow -\infty \) as \(|x|\rightarrow \infty \) similarly to the proof of Lemma 3.4. Therefore, similarly to the proof of Lemma 3.2 in [16], we see that the solution to (3.4) such that \({\hat{z}}(x)-z_0(x)\) is bounded above is unique when admitting additive constant with respect to \(z\). \(\square \)

Let us define the operator by

$$\begin{aligned} {\hat{L}}\varphi :=\frac{1}{2}\text{ tr }[\lambda \lambda ^*D^2\varphi ]+\{\beta _{\gamma }^*+(\nabla {\hat{z}})^*\lambda N_{\gamma }^{-1}\lambda ^*\}D\varphi . \end{aligned}$$
(3.11)

Then,we have the following lemma.

Lemma 3.6

Under the assumptions of Lemma 3.3, the diffusion process with the generator \({\hat{L}}\) is ergodic.

Proof

We have shown that \(z_0(x)-{\hat{z}}(x)\rightarrow \infty \) as \(|x|\rightarrow \infty \) in the proof of Lemma 3.5. We moreover see that

$$\begin{aligned}&{\hat{L}}(z_0-{\hat{z}})=\frac{1}{2}\text{ tr }[\lambda \lambda ^*D^2(z_0-{\hat{z}})]+\beta ^*_{\gamma }D(z_0-{\hat{z}})+(D{\hat{z}})^*\lambda N_{\gamma }^{-1}\lambda ^*D(z_0-{\hat{z}})\\&\quad =\frac{1}{2}\text{ tr }[\lambda \lambda ^*D^2z_0]+\beta ^*_{\gamma }Dz_0+\frac{1}{2}(D z_0)^*\lambda N_{\gamma }^{-1}\lambda ^*Dz_0+U_{\gamma }\\&\qquad -\left\{ \frac{1}{2}\text{ tr }[\lambda \lambda ^*D^2{\hat{z}}]+\beta ^*_{\gamma }D{\hat{z}}+\frac{1}{2}(D {\hat{z}})^*\lambda N_{\gamma }^{-1}\lambda ^*D{\hat{z}}+U_{\gamma }\right\} \\&\qquad -\frac{1}{2}D(z_0-{\hat{z}})^*\lambda N_{\gamma }^{-1}\lambda ^*D(z_0-{\hat{z}})\\&\quad \le -V-{\hat{\chi }}. \end{aligned}$$

Since \(-V-{\hat{\chi }}\rightarrow -\infty \) as \(|x|\rightarrow \infty \), we see that the Has’minskii’s conditions are satisfied and that \({\hat{L}}\) is ergodic. \(\square \)

Lemma 3.7

Besides the assumptions of Lemma 3.3, we assume (2.17’). Then, \({\hat{z}}(x)\) is bounded below.

Proof

Note that

$$\begin{aligned} Q(z,p)\ge \frac{\gamma }{2}{\hat{\alpha }}^*(\sigma \sigma ^*)^{-1}{\hat{\alpha }}+\gamma r\equiv f_{\gamma }(x). \end{aligned}$$

Then, from (3.6), it follows that

$$\begin{aligned} (\epsilon -L)z_{\epsilon }\ge f_{\gamma }(x). \end{aligned}$$

Set

$$\begin{aligned} {\bar{z}}_{\epsilon }:=z_{\epsilon }({\hat{x}}),\;\; {\tilde{z}}_{\epsilon }:=z_{\epsilon }- {\bar{z}}_{\epsilon }. \end{aligned}$$

Then, \(|\epsilon {\bar{z}}_{\epsilon }|<M\) for some positive constant \(M\) and we can take \(R>0\) such that

$$\begin{aligned} f_{\gamma }(x)>M,\quad x\in B_R^c \end{aligned}$$

under assumption (2.17’). Therefore, by using Itô’s formula, we have

$$\begin{aligned} {\tilde{z}}_{\epsilon }(x)&\ge E_x\left[ \int \limits _0^{\sigma \wedge T}e^{-\epsilon t}(f_{\gamma }-\epsilon {\bar{z}}_{\epsilon })(Y_t)dt+e^{-\epsilon \sigma \wedge T}{\tilde{z}}_{\epsilon }(Y_{\sigma })\right] \\&\ge E_x[e^{-\epsilon \sigma \wedge T}{\tilde{z}}_{\epsilon }(Y_{\sigma \wedge T})]\\&\ge E_x[e^{-\epsilon \sigma }{\tilde{z}}_{\epsilon }(Y_{\sigma }); \sigma <T]- {\bar{z}_{\epsilon }}e^{-\epsilon T}P_x(T\le \sigma ) \end{aligned}$$

since \(z_{\epsilon }(x)\ge 0\), where \((Y_t,P_x)\) is the diffusion process with the generator \(L\) and

$$\begin{aligned} \sigma =\left\{ \begin{array}{l@{\quad }l} \inf \{t; Y_t\in B_R\},&{} \text{ if } \{t; Y_t\in B_R\}\ne \phi \\ \infty , &{} \text{ if } \{t; Y_t\in B_R\}= \phi . \end{array} \right. \end{aligned}$$

Therefore, as \(T\rightarrow \infty \) we see that

$$\begin{aligned} {\tilde{z}}_{\epsilon }(x) \ge \inf _{|x|=R}{\tilde{z}}_{\epsilon }(x), \quad x\in B_R^c. \end{aligned}$$

Since \({\tilde{z}}_{\epsilon }\) converges to \({\hat{z}}\) uniformly on each compact set we have \(\inf _{|x|=R}{\tilde{z}}_{\epsilon }(x)\ge -K,\;\; K>0\) and hence obtain \({\hat{z}}(x)\ge -K\). \(\square \)

Now, we consider H–J–B equation (2.11) for \(0<\gamma <1\).

Proposition 3.1

Assume assumptions (2.13)–(2.15), 2.17’ and (2.17). Then, when taking \(C\) to be sufficiently large, \({\overline{z}}(x)={\hat{z}}(x)+ C\) is a super-solution to (2.11) with \(\rho >{\hat{\chi }}\). Moreover, \({\underline{z}}(x)\equiv -C'\) is a sub-solution to (2.11) if \(C'>0\) is sufficiently large.

Proof

Take \(\epsilon >0\) such that \(\rho -{\hat{\chi }}>\epsilon \). Then, we can see that

$$\begin{aligned} \frac{1}{1-\gamma }e^{-\frac{{\hat{z}}(x)+ C}{1-\gamma }}\le \frac{1}{1-\gamma }e^{-\frac{C+\inf {\hat{z}}}{1-\gamma }}<\epsilon , \end{aligned}$$

by taking \(C\) to be sufficiently large because of Lemma 3.7. Since \(({\hat{\chi }},{\hat{z}})\) is a solution to (3.4), \({\overline{z}}={\hat{z}}(x)+ C\) turns out to be a super-solution to (2.11).

It is easy to see that \({\underline{z}}(x)\) is a sub-solution to (2.11). \(\square \)

4 Existence and Uniqueness

We first prepare the following lemma.

Lemma 4.1

Assume assumptions (2.13)–(2.16) and \(\gamma <0\). Then, the bounded above solution to H–J–B equation (2.11) is unique.

Proof

Note that each smooth solution \(z\) to (2.11) satisfies the estimate

$$\begin{aligned} |\nabla z(x)|\le C(1+|x|) \end{aligned}$$

for a positive constant \(C>0\) under our assumptions (cf. Appendix 2). Let \(z\) and \(z_1\) be bounded above solutions to (2.11) and set

$$\begin{aligned} \phi (x)=e^{\frac{1}{1-\gamma }(z-z_1)(x)}. \end{aligned}$$

Then, we have

$$\begin{aligned}&\frac{1}{2}\text{ tr }[\lambda \lambda ^*D^2\phi ]+(\beta _{\gamma }+\lambda N_{\gamma }^{-1}\lambda ^*Dz_1)^*D\phi \nonumber \\&\quad =\frac{1}{1-\gamma }\Bigg [\frac{1}{2}\text{ tr }[\lambda \lambda ^*D^2z] +(\beta _{\gamma }+\lambda N_{\gamma }^{-1}\lambda ^*Dz_1)^*Dz\nonumber \\&\qquad -\Bigg \{\frac{1}{2}\text{ tr }[\lambda \lambda ^*D^2z_1] +(\beta _{\gamma }+\lambda N_{\gamma }^{-1}\lambda ^*Dz_1)^*Dz_1\Bigg \}\nonumber \\&\qquad +\frac{1}{2(1-\gamma )}D(z-z_1)^*\lambda \lambda ^*D(z-z_1)\Bigg ]\phi \nonumber \\&\quad =\frac{1}{1-\gamma }\Bigg \{-(1-\gamma )e^{-\frac{z}{1-\gamma }}+(1-\gamma )e^{-\frac{z_1}{1-\gamma }}\nonumber \\&\qquad -\frac{1}{2}D(z-z_1)^*\lambda (N_{\gamma }^{-1}-\frac{1}{1-\gamma }I)\lambda ^*D(z-z_1)\Bigg \}\phi \nonumber \\&\quad \le e^{-\frac{z}{1-\gamma }}(e^{\frac{z-z_1}{1-\gamma }}-1)\phi =e^{-\frac{z}{1-\gamma }}(\phi -1)\phi \end{aligned}$$
(4.1)

since \(\frac{1}{1-\gamma }I\le N_{\gamma }^{-1}\) for \(\gamma <0\).

Let \(Y_t\) be a solution of the stochastic differential equation

$$\begin{aligned} dY_t=\lambda (Y_t)dW_t+\{\beta _{\gamma }(Y_t)+\lambda N_{\gamma }^{-1}\lambda ^*Dz_1(Y_t)\}dt,\quad Y_0=x \end{aligned}$$

and set

$$\begin{aligned}f(x)=-e^{-\frac{z}{1-\gamma }}\phi =-e^{-\frac{z_1}{1-\gamma }}.\end{aligned}$$

Then, from (4.1) it follows that

$$\begin{aligned} d\{(\phi (Y_t)-1)e^{\int _0^tf(Y_s)ds}\}&= f(Y_t)(\phi (Y_t)-1)e^{\int _0^tf(Y_s)ds}dt+e^{\int _0^tf(Y_s)ds}d\phi (Y_t)\\&\le e^{\int _0^tf(Y_s)ds}D\phi (Y_t)^*\lambda (Y_t)dW_t. \end{aligned}$$

Setting \(t=\tau _G\wedge T\), where

$$\begin{aligned} \tau _G=\inf \{t;z(Y_t)\ge z_1(Y_t)\}, \end{aligned}$$

we have

$$\begin{aligned} \phi (x)-1&\ge E[(\phi (Y_{\tau _G\wedge T})-1)e^{\int _0^{\tau _G\wedge T}f(Y_s)ds}]\\&= E[(\phi (Y_{\tau _G})-1)e^{\int _0^{\tau _G}f(Y_s)ds}; \tau _G\le T]\\&\quad +E[(\phi (Y_{T})-1)e^{\int _0^{T}f(Y_s)ds}; T<\tau _G]\\&\ge E[(\phi (Y_{T})-1)e^{\int _0^{T}f(Y_s)ds}; T<\tau _G]. \end{aligned}$$

Note that

$$\begin{aligned} (\phi (Y_{T})-1)e^{\int _0^{T}f(Y_s)ds}\ge -e^{-\int _0^Te^{-\frac{1}{1-\gamma }z_1(Y_s)}ds}\ge -e^{-\frac{T}{K}},, \end{aligned}$$

where \(K\) is a positive constant such that \(e^{\frac{1}{1-\gamma }z_1(x)}\le K\). Thus, we see that

$$\begin{aligned} \phi (x)-1 \ge -e^{-\frac{T}{K}}. \end{aligned}$$

Sending \(T\) to \(\infty \) we have \(\phi (x)-1\ge 0\) and so \(z(x)\ge z_1(x)\).

Exchanging a role of \(z_1\) by \(z\), we obtain converse inequality \(z_1(x)\ge z(x)\). \(\square \)

Let \( {\underline{z}}(x)\) and \({\overline{z}}(x)\) be, respectively sub- and super- solution to (2.11) with \(\rho =0\) obtained in Lemma 3.1. Then, we have the following theorem.

Theorem 4.1

For \(\gamma <0\), we assume assumptions (2.13)–(2.16). Then, for \(\rho = 0\), (2.11) has a solution \(z\) such that \({\underline{z}}(x)\le z(x)\le {\overline{z}}(x)\). Moreover, the bounded above solution to (2.11) is unique.

Proof

Since we have a sub- and a super- solution as was seen in Lemma 3.1, the existence of a solution can be shown in a similar manner to the proof of Theorem 3.5 in [7]. We complete the proof in Appendix 1.

Let us prove uniqueness. Let \(z(x)\) be the solutions to (2.11) with \(\rho =0\) such that \({\underline{z}}(x)\le z(x)\le {\overline{z}}(x)\). Under our assumptions we can see that

$$\begin{aligned} {\overline{z}}(x)\le \gamma \log c+\log \left( \frac{-1}{\gamma c_0}\right) \end{aligned}$$

holds and so \(z(x)\) is bounded above. Further, owing to Lemma 4.1 we have the uniqueness of the bounded above solution to (2.11). \(\square \)

Remark

It is to be noted that even in the case of \(\rho =0\), H–J–B equation (2.11) has the unique solution without ambiguity of additive constants with respect to \(z(x)\). Moreover, considering (2.11) with \(\rho >0\) and without assumption (2.16) can be reduced to the case of \(\rho =0\) with assumption (2.16) (cf. Remark in Sect. 3).

Theorem 4.2

For \(0<\gamma <1\), we assume assumptions (2.13)–(2.15), 2.17’ and (2.17), and let \({\underline{z}}(x)\) and \({\overline{z}}(x)\) be, respectively sub- and super- solutions to (2.11) appeared in Proposition 3.1. Then, for each \(\rho > {\hat{\chi }}\), (2.11) has a solution \(z\) such that \({\underline{z}}(x)\le z(x)\le {\overline{z}}(x)\) and that it satisfies

$$\begin{aligned} z(x)-z_0(x) \rightarrow -\infty ,\;\; \text{ as }\;\; |x|\rightarrow \infty . \end{aligned}$$
(4.2)

Moreover, the solution satisfying (4.2) is unique.

Proof

As in the proof of the previous theorem, the existence of a solution is given in a similar manner to the proof of Theorem 3.5 in [7] (cf. Appendix 1). We give the proof of unique existence of the solution satisfying (4.2). First note that the solution \(z(x)\) to (2.11) necessarily satisfies (4.2) under our assumptions. Indeed, we have seen that \({\hat{z}}(x)-z_0(x)\rightarrow -\infty \) as \(|x|\rightarrow \infty \) in the proof of Lemma 3.5, and thus, (4.2) follows from \(z(x)\le {\hat{z}}(x)+C\). Let us prove uniqueness. Set

$$\begin{aligned} \psi =z-z_0 \end{aligned}$$

for a solution \(z\) to (2.11). Then,

$$\begin{aligned} \rho&= \frac{1}{2}\text{ tr }[\lambda \lambda ^*D^2\psi ]+\frac{1}{2}\text{ tr }[\lambda \lambda ^*D^2z_0] +\beta _{\gamma }^*D\psi +\beta _{\gamma }^*Dz_0+U_{\gamma }\\&\quad +\frac{1}{2}D(\psi +z_0)^*\lambda N_{\gamma }^{-1}\lambda ^*D(\psi +z_0)+(1-\gamma )e^{-\frac{\psi +z_0}{1-\gamma }}. \end{aligned}$$

Therefore,

$$\begin{aligned} \rho&= \frac{1}{2}\text{ tr }[\lambda \lambda ^*D^2\psi ]+{\tilde{\beta }}_{\gamma }^*D\psi +\frac{1}{2}(D\psi )^*\lambda N_{\gamma }^{-1}\lambda ^*D\psi \nonumber \\&\quad -V+(1-\gamma )e^{-\frac{\psi +z_0}{1-\gamma }}, \end{aligned}$$
(4.3)

where

$$\begin{aligned} V=-\left\{ \frac{1}{2}\text{ tr }[\lambda \lambda ^*D^2z_0]+\beta _{\gamma }^*Dz_0+\frac{1}{2}(Dz_0)^*\lambda N_{\gamma }^{-1}\lambda ^*Dz_0+U_{\gamma }\right\} \end{aligned}$$
(4.4)

and

$$\begin{aligned} {\tilde{\beta }}_{\gamma }=\beta _{\gamma }+\lambda N_{\gamma }^{-1}\lambda ^*Dz_0. \end{aligned}$$

Let \(z_1\) and \(z_2\) be solutions to (2.11) satisfying (4.2) and set \(\psi _i= z_i-z_0\), \(i=1,2\). Assume that there exists \(x_0\) such that \(z_2(x_0)>z_1(x_0)\). Then, \(\psi _2(x_0)>\psi _1(x_0)\) and \(\psi _i(x)\rightarrow -\infty \), as \(|x|\rightarrow \infty \). Therefore, for each \(\epsilon >0\) there exists \(x_\epsilon \) such that

$$\begin{aligned} \psi _{\epsilon }(x):=\sup _x\{e^{\epsilon \psi _2(x)}-e^{\epsilon \psi _1(x)}\}=e^{\epsilon \psi _2(x_{\epsilon })} -e^{\epsilon \psi _1(x_{\epsilon })}\ge e^{\epsilon \psi _2(x_0)} -e^{\epsilon \psi _1(x_0)}>0. \end{aligned}$$
(4.5)

At \(x_{\epsilon }\) we have

$$\begin{aligned} 0&\le -\frac{1}{2}\text{ tr }[\lambda \lambda ^* D^2\psi _{\epsilon }]-{\tilde{\beta }}_{\gamma }^*D\psi _{\epsilon }\\&= -\frac{\epsilon }{2}e^{\epsilon \psi _2}\text{ tr }[\lambda \lambda ^*D^2 \psi _2]-\frac{\epsilon ^2}{2}e^{\epsilon \psi _2}(D \psi _2)^*\lambda \lambda ^*DZ_2\\&\quad +\frac{\epsilon }{2}e^{\epsilon \psi _1}\text{ tr }[\lambda \lambda ^*D^2 \psi _1]+\frac{\epsilon ^2}{2}e^{\epsilon \psi _1}(D \psi _1)^*\lambda \lambda ^*D\psi _1\\&\quad -\epsilon e^{\epsilon \psi _2}{\tilde{\beta }}_{\gamma }^*D\psi _2+\epsilon e^{\epsilon \psi _1}{\tilde{\beta }}_{\gamma }^*D\psi _1\\&= \epsilon e^{\epsilon \psi _2}\Bigg \{\frac{1}{2}(D\psi _2)^*\lambda N_{\gamma }^{-1}\lambda ^* D \psi _2-V+(1-\gamma )e^{-\frac{\psi _2+z_0}{1-\gamma }}-\rho \Bigg \}\\&\quad -\epsilon e^{\epsilon \psi _1}\Bigg \{\frac{1}{2}(D\psi _1)^*\lambda N_{\gamma }^{-1}\lambda ^* D \psi _1-V+(1-\gamma )e^{-\frac{\psi _1+z_0}{1-\gamma }}-\rho \Bigg \}\\&\quad -\frac{\epsilon ^2}{2}e^{\epsilon \psi _2}(D\psi _2)^*\lambda \lambda ^* D \psi _2+\frac{\epsilon ^2}{2}e^{\epsilon \psi _1}(D\psi _1)^*\lambda \lambda ^* D \psi _1\\&= \epsilon e^{\epsilon \psi _2}\Bigg \{\frac{1}{2}(D\psi _2)^*\lambda (N_{\gamma }^{-1}-\epsilon I)\lambda ^* D \psi _2-V+(1-\gamma )e^{-\frac{\psi _2+z_0}{1-\gamma }}-\rho \Bigg \}\\&\quad -\epsilon e^{\epsilon \psi _1}\Bigg \{\frac{1}{2}(D\psi _1)^*\lambda (N_{\gamma }^{-1}-\epsilon I)\lambda ^* D \psi _1-V+(1-\gamma )e^{-\frac{\psi _1+z_0}{1-\gamma }}-\rho \Bigg \}. \end{aligned}$$

Therefore,

$$\begin{aligned}&\frac{1}{2}e^{\epsilon \psi _2}(D\psi _2)^*\lambda (N_{\gamma }^{-1}-\epsilon I)\lambda ^* D \psi _2-\frac{1}{2}e^{\epsilon \psi _1}(D\psi _1)^*\lambda (N_{\gamma }^{-1}-\epsilon I)\lambda ^* D \psi _1\nonumber \\&\quad \ge e^{\epsilon \psi _2}(V+\rho -(1-\gamma )e^{-\frac{\psi _2+z_0}{1-\gamma }})- e^{\epsilon \psi _1}(V+\rho -(1-\gamma )e^{-\frac{\psi _1+z_0}{1-\gamma }}) \end{aligned}$$
(4.6)

On the other hand, \(D\psi _2=D\psi _1e^{\epsilon \psi _1-\epsilon \psi _2}\) because \(D\psi _{\epsilon }=0\) at \(x_\epsilon \), and thus,

$$\begin{aligned}&\frac{1}{2}e^{\epsilon \psi _2}(D\psi _2)^*\lambda (N_{\gamma }^{-1}-\epsilon I)\lambda ^* D \psi _2-\frac{1}{2}e^{\epsilon \psi _1}(D\psi _1)^*\lambda (N_{\gamma }^{-1}-\epsilon I)\lambda ^* D \psi _1\\&\quad =\frac{1}{2}e^{2\epsilon \psi _1-\epsilon \psi _2}(D\psi _1)^*\lambda (N_{\gamma }^{-1}-\epsilon I)\lambda ^* D \psi _1-\frac{1}{2}e^{\epsilon \psi _1}(D\psi _1)^*\lambda (N_{\gamma }^{-1}-\epsilon I)\lambda ^* D \psi _1\\&\quad =\frac{1}{2}e^{\epsilon \psi _1-\epsilon \psi _2}(e^{\epsilon \psi _1}-e^{\epsilon \psi _2})(D\psi _1)^*\lambda (N_{\gamma }^{-1}-\epsilon I)\lambda ^* D \psi _1\\&\quad \le \frac{1-\epsilon }{2}e^{\epsilon \psi _1-\epsilon \psi _2}(e^{\epsilon \psi _1}-e^{\epsilon \psi _2})(D\psi _1)^*\lambda \lambda ^* D \psi _1\le 0. \end{aligned}$$

Then, from (4.6), it follows that

$$\begin{aligned} (e^{\epsilon \psi _2}-e^{\epsilon \psi _1})(V+\rho )- (1-\gamma )e^{-\frac{z_0}{1-\gamma }}(e^{(\epsilon -\frac{1}{1-\gamma })\psi _2}-e^{(\epsilon -\frac{1}{1-\gamma })\psi _1})\le 0, \end{aligned}$$

and so,

$$\begin{aligned} (e^{\epsilon \psi _2}-e^{\epsilon \psi _1})(V+\rho )\le (1-\gamma )e^{-\frac{z_0}{1-\gamma }}(e^{(\epsilon -\frac{1}{1-\gamma })\psi _2}-e^{(\epsilon -\frac{1}{1-\gamma })\psi _1})<0 \end{aligned}$$
(4.7)

by taking \(\epsilon \) such that \(\epsilon <\frac{1}{1-\gamma }\). Thus, we obtain

$$\begin{aligned} V(x_{\epsilon })<-\rho \end{aligned}$$

from (4.7). Since \(V(x)\rightarrow \infty \) as \(|x|\rightarrow \infty \), there exists \(R>0\) independent of \(\epsilon \) such that \(x_{\epsilon }\in B_R\) and \(\epsilon _n\) such that \(x_{\epsilon _n}\rightarrow {\hat{x}}\in {\overline{B}_R}\). From (4.5), we have

$$\begin{aligned} \frac{1}{\epsilon _n}(e^{\epsilon _n \psi _2(x_{\epsilon _n})} -e^{\epsilon _n \psi _1(x_{\epsilon _n})})\ge \frac{1}{\epsilon _n}( e^{\epsilon _n \psi _2(x_0)} -e^{\epsilon _n \psi _1(x_0)})>0 \end{aligned}$$

and, by letting \(\epsilon _n\rightarrow 0\), we obtain

$$\begin{aligned} \psi _2({\hat{x}})-\psi _1({\hat{x}})\ge \psi _2( x_0)-\psi _1( x_0)>0. \end{aligned}$$

From (4.7), at \(x_{\epsilon }\), we have

$$\begin{aligned} -K_0\le V+\rho \le \frac{(1-\gamma )e^{-\frac{z_0}{1-\gamma }}(e^{(\epsilon -\frac{1}{1-\gamma })\psi _2}-e^{(\epsilon -\frac{1}{1-\gamma })\psi _1})}{e^{\epsilon \psi _2}-e^{\epsilon \psi _1}} \end{aligned}$$

and letting \(\epsilon _n\rightarrow 0\), the right-hand side tends to \(-\infty \), which is a contradiction. Therefore \(\psi _2(x)\le \psi _1(x)\) for each \(x\). In the same way, we have the converse inequality, and hence, we proved uniqueness of the solution to (2.11). \(\square \)

Let us set the operator \({\overline{L}}\) by

$$\begin{aligned} {\overline{L}}g:=\frac{1}{2}\text{ tr }[\lambda \lambda ^* D^2g]+\beta _{\gamma }^*Dg+(Dz)^*\lambda N_{\gamma }^{-1}\lambda ^*Dg. \end{aligned}$$
(4.8)

Then, inspired by Lemma 3.6 we have the following proposition useful in the proof of the verification theorem.

Proposition 4.1

Under the assumptions of Theorem 4.2, the diffusion process with the generator \({\overline{L}}\) is ergodic.

Proof

Let us set

$$\begin{aligned} {\bar{\psi }}(x):= z_0(x)-z(x). \end{aligned}$$

Then, as was seen in the proof of Theorem 4.2, \({\bar{\psi }}(x)\rightarrow \infty \), as \(|x|\rightarrow \infty \). Further, from (4.3) it follows that

$$\begin{aligned} -\rho&= \frac{1}{2}\text{ tr }[\lambda \lambda ^*D^2{\bar{\psi }}]+ \beta _{\gamma }^*D{\bar{\psi }}+(Dz_0)^*\lambda N_{\gamma }^{-1}\lambda ^*D{\bar{\psi }}\\&\quad -\frac{1}{2}(D{\bar{\psi }})^*\lambda N_{\gamma }^{-1}\lambda ^*D{\bar{\psi }}+V-(1-\gamma )e^{-\frac{z}{1-\gamma }}. \end{aligned}$$

Since \(z_0=z+{\bar{\psi }}\) we have

$$\begin{aligned}&\frac{1}{2}\text{ tr }[\lambda \lambda ^*D^2{\bar{\psi }}]+ \beta _{\gamma }^*D{\bar{\psi }}+(Dz)^*\lambda N_{\gamma }^{-1}\lambda ^*D{\bar{\psi }}\\&\quad =-\frac{1}{2}(D{\bar{\psi }})^*\lambda N_{\gamma }^{-1}\lambda ^*D{\bar{\psi }}-V+(1-\gamma )e^{-\frac{z}{1-\gamma }}-\rho \end{aligned}$$

Thus, we can see that \({\overline{L}}{\bar{\psi }}(x)\rightarrow -\infty \) as \(|x|\rightarrow \infty \) and hence the proof is complete. \(\square \)

5 Verification Theorems

Let us set the value functions

$$\begin{aligned} {\hat{v}}(x)= \inf _{(h., c_.)\in {\mathcal A}_1} E^h\left[ \int \limits _0^{\infty }e^{-\rho t}c_t^{\gamma }e^{\gamma \int _0^t\{\eta (X_s,h_s)-c_s\}ds}dt\right] , \end{aligned}$$

for \(\gamma <0\), and

$$\begin{aligned} {\check{v}}(x)= \sup _{(h., c_.)\in {\mathcal A}_1} E^h\left[ \int \limits _0^{\infty }e^{-\rho t}c_t^{\gamma }e^{\gamma \int _0^t\{\eta (X_s,h_s)-c_s\}ds}dt\right] , \end{aligned}$$

for \(0<\gamma <1\), where \({\mathcal A}_1\) is the sets of admissible strategies defined in the end of Sect. 2. For a solution \(z(x)\) to (2.11), we shall prove that \(e^{z(x)}={\hat{v}}(x)\), for \(\gamma <0\), under assumptions (2.13)-(2.15) (resp. (2.13)-(2.16)) for \(\rho >0\) (resp. \(\rho =0\)), and that \(e^{z(x)}={\check{v}}(x)\), for \(0<\gamma <1\), under the assumptions of Theorem 4.2. For the solution \(z(x)\) to (2.11), define a function \({\hat{h}}(x)\) by

$$\begin{aligned} {\hat{h}}(x)=\frac{1}{1-\gamma }(\sigma \sigma ^*)^{-1}(\sigma \lambda ^* Dz+{\hat{\alpha }})(x) \end{aligned}$$

and

$$\begin{aligned} {\hat{c}}(x)=e^{-\frac{z(x)}{1-\gamma }}. \end{aligned}$$

We define also

$$\begin{aligned} {\hat{h}}_s={\hat{h}}(X_s),\quad {\hat{c}}_s={\hat{c}}(X_s). \end{aligned}$$

Let us prepare the following lemma for the proof of the verification theorem.

Lemma 5.1

Let \(Y_t\) be a solution to the stochastic differential equation

$$\begin{aligned} dY_t=\sigma (t,Y_t)dW_t+\mu (t,Y_t)dt, \quad Y_0=x, \end{aligned}$$

where \(\sigma \) and \(\mu \) are locally Lipschitz continuous with respect to \(x\) and continuous in \(t\). We moreover assume that \( |\mu (t,x)|\le C(1+|x|) \) and \(\sigma \) is bounded. For a given continuous function \(H(t,x)\) satisfying \( |H(t,x)|\le C(1+|x|), \) define \(\rho _t\) by

$$\begin{aligned} \rho _t:=e^{\int _0^tH(s,Y_s)^*dW_s-\frac{1}{2}\int _0^t|H(s,X_s)|^2ds}. \end{aligned}$$

Then, we have

$$\begin{aligned} E[\rho _t]=1,\quad \forall t. \end{aligned}$$

The proof of this lemma is similar to that of Lemma 4.1.1 in [1] and we omit the proof.

We have the following theorem.

Theorem 5.1

For \(\rho =0\) (resp. \(\rho >0\)), assume assumptions (2.13)–(2.16) (resp. (2.13)–(2.15)). Then, for a solution \(z(x)\) to (2.11), we have

$$\begin{aligned} e^{z(x)}&= \inf _{{h., c_.}\in {\mathcal A}_1} E^h\left[ \int \limits _0^{\infty }e^{-\rho t}c_t^{\gamma }e^{\gamma \int _0^t\{\eta (X_s,h_s)-c_s\}ds}dt\right] \nonumber \\&= E^{\hat{h}}\left[ \int \limits _0^{\infty }e^{-\rho t}{\hat{c}}_t^{\gamma }e^{\gamma \int _0^t\{\eta (X_s,{\hat{h}}_s)-{\hat{c}}_s\}ds}dt\right] \end{aligned}$$
(5.1)

with \(\gamma <0\).

Proof

Let us first note that \({\hat{h}}_s\) satisfies (2.8). Indeed, because of the gradient estimates for the solution \(z\) given in (7.1) in Appendix 2, we can see it by using the above lemma. Thus, we have a probability measure \(P^{\hat{h}}\) under which \(X_t\) satisfies the stochastic differential equation

$$\begin{aligned} dX_t=\lambda (X_t) dW_t^{{\hat{h}}}+\{\beta _{\gamma }(X_t)+\frac{\gamma }{1-\gamma }\lambda \sigma ^*(\sigma \sigma ^*)^{-1}\lambda ^*Dz(X_t)\}dt, \end{aligned}$$

where

$$\begin{aligned} \beta _{\gamma }(x)=\beta (x)+\frac{\gamma }{1-\gamma }\lambda \sigma ^*(\sigma \sigma ^*)^{-1}{\hat{\alpha }}(x). \end{aligned}$$

Set

$$\begin{aligned} L^{\gamma }f:=\frac{1}{2}\text{ tr }[\lambda \lambda ^*D^2f]+\beta _{\gamma }^*Df+\frac{\gamma }{1-\gamma }(Dz)^*\lambda \sigma ^*(\sigma \sigma ^*)^{-1}\sigma \lambda ^*Df. \end{aligned}$$

Then, by Itô’s formula, we have

$$\begin{aligned}&z(X_t)-z(X_0)= \int \limits _0^tL^{\gamma }z(x_s)ds+\int \limits _0^t(Dz)^*\lambda (X_s)dW_s^{{\hat{h}}}\\&\quad =\int \limits _0^t\Bigg \{\frac{1}{2}\text{ tr }[\lambda \lambda ^*D^2z]+\beta _{\gamma }^*Dz+\frac{1}{2}(Dz)^*\lambda N_{\gamma }^{-1}\lambda ^*Dz\\&\qquad +\frac{\gamma }{2(1-\gamma )}(Dz)^*\lambda \sigma ^*(\sigma \sigma ^*)^{-1}\sigma \lambda ^*Dz\Bigg \}(X_s)ds\\&\qquad +\int \limits _0^t(Dz)^*\lambda (X_s)dW_s^{{\hat{h}}}-\frac{1}{2}\int \limits _0^t(Dz)^*\lambda \lambda ^*Dz(X_s)ds\\&\quad =\int \limits _0^t\Bigg \{-U_{\gamma }(X_s)-(1-\gamma )e^{-\frac{z(X_s)}{1-\gamma }}+\rho \\&\qquad +\frac{\gamma }{2(1-\gamma )}(Dz)^*\lambda \sigma ^*(\sigma \sigma ^*)^{-1}\sigma \lambda ^*Dz\Bigg \}(X_s)ds\\&\qquad +\int \limits _0^t(Dz)^*\lambda (X_s)dW_s^{{\hat{h}}}-\frac{1}{2}\int \limits _0^t(Dz)^*\lambda \lambda ^*Dz(X_s)ds. \end{aligned}$$

Note that

$$\begin{aligned} \eta (x,{\hat{h}}(x))=-\frac{1}{2(1-\gamma )}(Dz)^*\lambda \sigma ^*(\sigma \sigma ^*)^{-1}\sigma \lambda ^*Dz+\frac{1}{2(1-\gamma )}{\hat{\alpha }}^*(\sigma \sigma ^*)^{-1}{\hat{\alpha }}+r, \end{aligned}$$

and hence,

$$\begin{aligned} \gamma \eta (x,{\hat{h}}(x))=U_{\gamma }-\frac{\gamma }{2(1-\gamma )}(Dz)^*\lambda \sigma ^*(\sigma \sigma ^*)^{-1}\sigma \lambda ^*Dz. \end{aligned}$$

Therefore,

$$\begin{aligned} z(X_t)-z(X_0)=-\gamma \int \limits _0^t(\eta (X_s,{\hat{h}}_s)-{\hat{c}}_s)ds+\rho t-\int \limits _0^te^{-\frac{z(X_s)}{1-\gamma }}ds+ M_t-\frac{1}{2}\langle M\rangle _t \end{aligned}$$

where

$$\begin{aligned} M_t=\int \limits _0^t(Dz)^*\lambda (X_s)dW_s^{{\hat{h}}}. \end{aligned}$$

Once again, owing to the above lemma, we see that \(M_t\) satisfies (2.19) with respect to \(P^{{\hat{h}}}\) and therefore \(({\hat{c}},{\hat{h}})\in {\mathcal A}_1\). Thus, we obtain

$$\begin{aligned}&E^{\hat{h}}\left[ \int \limits _0^{\infty }e^{-\rho t}{\hat{c}}_t^{\gamma }e^{\gamma \int _0^t(\eta (X_s,{\hat{h}}_s)-{\hat{c}}_s)ds}dt\right] \nonumber \\&\quad =E^{\hat{h}}\left[ \int \limits _0^{\infty }e^{-\frac{\gamma }{1-\gamma }z(X_t)-z(X_t)+z(X_0)-\int _0^te^{-\frac{z(X_s)}{1-\gamma }}ds +M_t-\frac{1}{2}\langle M\rangle _t}dt\right] \nonumber \\&\quad =e^{z(x)}E^{\hat{h}}\left[ \int \limits _0^{\infty }e^{-\frac{1}{1-\gamma }z(X_t)-\int _0^te^{-\frac{z(X_s)}{1-\gamma }}ds +M_t-\frac{1}{2}\langle M\rangle _t}dt\right] \nonumber \\&\quad =e^{z(x)}{\overline{E}}^{{\hat{h}}}\left[ \int \limits _0^{\infty }e^{-\frac{1}{1-\gamma }z(X_t)-\int _0^te^{-\frac{z(X_s)}{1-\gamma }}ds}dt\right] , \end{aligned}$$
(5.2)

since

$$\begin{aligned} \left. \frac{d {\overline{P}}^{{\hat{h}}}}{d P^{\hat{h}}}\right| _{{\mathcal F}_t}=e^{M_t-\frac{1}{2}\langle M\rangle _t}. \end{aligned}$$
(5.3)

When setting \(\varphi _t=e^{-\int _0^t{\hat{c}}_s^{\gamma }e^{-z(X_s)}ds}=e^{-\int _0^te^{-\frac{z(X_s)}{1-\gamma }}ds}\), we have

$$\begin{aligned} -d\varphi _t=e^{-\frac{1}{1-\gamma }z(X_t)-\int _0^te^{-\frac{z(X_s)}{1-\gamma }}ds}dt. \end{aligned}$$

From Theorem 4.1, we have

$$\begin{aligned} 0\le \varphi _t\le e^{-\int _0^te^{-\frac{{\overline{z}}(X_s)}{1-\gamma }}ds}\le e^{-K_{\gamma }t} \end{aligned}$$

with \(K_{\gamma }=c^{-\frac{\gamma }{1-\gamma }}(\frac{-1}{\gamma c_0})^{-\frac{1}{1-\gamma }}\) and thus, \( \lim _{T\rightarrow \infty }\varphi _T=\varphi _{\infty }=0\). Therefore we see that

$$\begin{aligned} {\overline{E}}^{{\hat{h}}}\Bigg [\int _0^{\infty }e^{-\frac{1}{1-\gamma }z(X_t)-\int _0^te^{-\frac{z(X_s)}{1-\gamma }}ds}dt\Bigg ]={\overline{E}}^{{\hat{h}}}[\varphi _0-\varphi _{\infty }]=1. \end{aligned}$$

Hence,

$$\begin{aligned} E^{\hat{h}}\left[ \int \limits _0^{\infty }e^{-\rho t}{\hat{c}}_t^{\gamma }e^{\gamma \int _0^t(\eta (X_s,{\hat{h}}_s)-{\hat{c}}_s)ds}dt\right] =e^{z(x)}. \end{aligned}$$

Now, we shall prove that

$$\begin{aligned} e^{z(x)}\le E^{ h}\left[ \int \limits _0^{\infty }e^{-\rho t}c_t^{\gamma }e^{\gamma \int _0^t(\eta (X_s, h_s)- c_s)ds}dt\right] , \quad \forall (c,h)\in {\mathcal A}_1. \end{aligned}$$
(5.4)

For the controlled process defined by

$$\begin{aligned} dX_t=\lambda (X_t)dW_t^h+\{\beta (X_t)+\gamma \lambda ^*\sigma (X_t)h_t\}dt,\quad X_0=x, \;\; (c, h)\in {\mathcal A}_1, \end{aligned}$$

we have

$$\begin{aligned} z(X_t)-z(X_0)&= \int \limits _0^t(Dz)^*\lambda (X_s)dW^h_s\\&\quad +\int \limits _0^t\Bigg \{\frac{1}{2}\text{ tr }[\lambda \lambda ^*D^2z]+(\beta +\gamma \lambda \sigma ^*h_s)^*Dz\Bigg \}(X_s)ds\\&\ge \int \limits _0^t\{\rho -\gamma (\eta (X_s,h_s)-c_s)-c_s^{\gamma }e^{-z(X_s)}\}ds\\&\quad +\int \limits _0^t(Dz)^*\lambda (X_s)dW_s^h-\frac{1}{2}\int \limits _0^t(Dz)^*\lambda \lambda ^*Dz(X_s)ds, \end{aligned}$$

from H–J–B equation (2.11). Therefore,

$$\begin{aligned} -\rho t+\gamma \int \limits _0^t(\eta (X_s,h_s)-c_s)ds\ge z(X_0)-z(X_t)-\int \limits _0^tc_s^{\gamma }e^{-z(X_s)}ds+{\tilde{M}}_t-\frac{1}{2}\langle {\tilde{M}}\rangle _t, \end{aligned}$$

where

$$\begin{aligned} {\tilde{M}}_t=\int \limits _0^t(Dz)^*\lambda (X_s)dW_s^h. \end{aligned}$$

Thus, we obtain

$$\begin{aligned}&E^{h}\left[ \int \limits _0^{\infty }e^{-\rho t} c_t^{\gamma }e^{\gamma \int _0^t(\eta (X_s,h_s)- c_s)ds}dt\right] \\&\quad \ge e^{z(x)}E^h\left[ \int \limits _0^{\infty }c_t^{\gamma }e^{-z(X_t)-\int _0^tc_s^{\gamma }e^{-z(X_s)}ds}e^{{\tilde{M}}_t-\frac{1}{2}\langle {\tilde{M}}\rangle _t}dt\right] \\&\quad =e^{z(x)}{\overline{E}}^h\left[ \int \limits _0^{\infty }c_t^{\gamma }e^{-z(X_t)-\int _0^tc_s^{\gamma }e^{-z(X_s)}ds}dt\right] , \end{aligned}$$

where \({\overline{P}}^{h}\) is a probability measure defined by

$$\begin{aligned} \left. \frac{d {\overline{P}}^h}{d P^{ h}}\right| _{{\mathcal F}_t}=e^{{\tilde{M}}_t-\frac{1}{2}\langle {\tilde{M}}\rangle _t}. \end{aligned}$$

Let us first assume that \(c_t\le K\), \(\forall t\), for some positive constant \(K>0\). In this case,

$$\begin{aligned} \int \limits _0^t c_s^{\gamma }e^{-z(X_s)}ds\ge K^{\gamma }e^{-C}t \end{aligned}$$

holds since \(z(x)\) is bounded above by a constant \(C\). Then we have

$$\begin{aligned} {\tilde{\varphi }}_t:=e^{-\int _0^tc_s^{\gamma }e^{-z(X_s)}ds}\le e^{-K^{\gamma }e^{-C}t} \end{aligned}$$

and thus obtain

$$\begin{aligned} {\overline{E}}^h\left[ \int \limits _0^{\infty }c_t^{\gamma }e^{-z(X_t)-\int _0^tc_s^{\gamma }e^{-z(X_s)}ds}dt\right] ={\overline{E}}^h[{\tilde{\varphi }}_0-{\tilde{\varphi }}_{\infty }]=1. \end{aligned}$$

Hence (5.4) holds.

For general \((c,h)\in {\mathcal A}_1\) we set \(c_s^{(n)}:=\min \{c_s,n\}\). Then, we have

$$\begin{aligned} E^{ h}\left[ \int \limits _0^{\infty }e^{-\rho t}(c_t^{(n)})^{\gamma }e^{\gamma \int _0^t(\eta (X_s, h_s)- c_s^{(n)})ds}dt\right] \!\le \! E^{ h}\left[ \int \limits _0^{\infty }e^{-\rho t}(c_t^{(n)})^{\gamma }e^{\gamma \int _0^t(\eta (X_s, h_s)- c_s)ds}dt\right] \end{aligned}$$

and therefore

$$\begin{aligned} e^{z(x)} \le E^{ h}\left[ \int \limits _0^{\infty }e^{-\rho t}(c_t^{(n)})^{\gamma }e^{\gamma \int _0^t(\eta (X_s, h_s)- c_s)ds}dt\right] . \end{aligned}$$

Hence, by monotone convergence theorem we have (5.4). \(\square \)

Theorem 5.2

Under the assumptions of Theorem 4.2, for a solution \(z(x)\) to (2.11), we have

$$\begin{aligned} e^{z(x)}&= E^{\hat{h}}\left[ \int \limits _0^{\infty }e^{-\rho t}{\hat{c}}_t^{\gamma }e^{\gamma \int _0^t\{\eta (X_s,{\hat{h}}_s)-{\hat{c}}_s\}ds}dt\right] \\&= \sup _{{h., c_.}\in {\mathcal A}_1} E^h \left[ \int \limits _0^{\infty }e^{-\rho t}c_t^{\gamma }e^{\gamma \int _0^t\{\eta (X_s,h_s)-c_s\}ds}dt\right] \end{aligned}$$

Proof

Similarly to the proof of Theorem 5.1, we see that \(({\hat{c}}_s,{\hat{h}}_s)\in {\mathcal A}_1\) and have

$$\begin{aligned} E^{\hat{h}}\left[ \int \limits _0^{\infty }e^{-\rho t}{\hat{c}}_t^{\gamma }e^{\gamma \int _0^t(\eta (X_s,{\hat{h}}_s)-{\hat{c}}_s)ds}dt\right]&= e^{z(x)}{\overline{E}}^{{\hat{h}}} \left[ \int \limits _0^{\infty }e^{-\frac{1}{1-\gamma }z(X_t)-\int _0^te^{-\frac{z(X_s)}{1-\gamma }}ds}dt\right] \nonumber \\&= e^{z(x)}{\overline{E}}^{{\hat{h}}}\left[ -\int \limits _0^{\infty }d\varphi _t\right] , \end{aligned}$$
(5.5)

where

$$\begin{aligned} \varphi _t=e^{-\int _0^te^{-\frac{z(X_s)}{1-\gamma }}ds}. \end{aligned}$$

We first note that \((X_t,{\overline{P}}^{{\hat{h}}})\) is the ergodic diffusion process with the generator \({\overline{L}}\) defined by (4.8) according to Proposition 4.1. Moreover,

$$\begin{aligned} z(x)\le {\hat{z}}(x)+C \end{aligned}$$

and, for each \(R>0\), there exists a positive constant \(C_R\) such that

$$\begin{aligned} e^{-C_R}\le e^{-\frac{{\hat{z}}(x)+C}{1-\gamma }},\quad x\in B_R. \end{aligned}$$

Therefore,

$$\begin{aligned} \frac{1}{T}\int \limits _0^Te^{-\frac{{\hat{z}}(X_s)+C}{1-\gamma }}ds\ge \frac{1}{T}e^{-C_R}\int \limits _0^T{\mathbf 1}_{B_R}(X_s)ds\rightarrow e^{-C_R}m(B_R) \end{aligned}$$

as \(T\rightarrow \infty \), where \(m(dx)\) is the invariant measure of \((X_t,{\overline{P}}^{{\hat{h}}})\). Therefore, we see that

$$\begin{aligned} \int \limits _0^Te^{-\frac{{\hat{z}}(X_s)+C}{1-\gamma }}ds\rightarrow \infty ,\quad {\overline{P}}^{\hat{h}} \; \text{ a.s. } \end{aligned}$$

as \(T\rightarrow \infty \), and thus we have \(\varphi _T\rightarrow 0,\;\;{\overline{P}}^{\hat{h}} \; \text{ a.s. } \) since

$$\begin{aligned} e^{-\int _0^Te^{-\frac{{\hat{z}}(X_s)+C}{1-\gamma }}ds}\ge e^{-\int _0^Te^{-\frac{z(X_s)}{1-\gamma }}ds}=\varphi _T=e^{-\int _0^T{\hat{c}}_t^{\gamma }e^{-z(X_t)}dt}. \end{aligned}$$

Thus, we have the first equality.

To prove

$$\begin{aligned} e^{z(x)}\ge E^{ h}\left[ \int \limits _0^{\infty }e^{-\rho t}c_t^{\gamma }e^{\gamma \int _0^t(\eta (X_s, h_s)- c_s)ds}dt\right] , \quad \forall (c,h)\in {\mathcal A}_1, \end{aligned}$$

we use H–J–B equation (2.12) similarly to the above, and then we arrive at

$$\begin{aligned} E^{h}\left[ \int \limits _0^{\infty }e^{-\rho t} c_t^{\gamma }e^{\gamma \int _0^t(\eta (X_s,h_s)- c_s)ds}dt\right]&\le e^{z(x)}{\overline{E}}^h\left[ \int \limits _0^{\infty }c_t^{\gamma }e^{-z(X_t)-\int _0^tc_s^{\gamma }e^{-z(X_s)}ds}dt\right] \\&= e^{z(x)}\lim _{T\rightarrow \infty } {\overline{E}}^h[{\tilde{\varphi }}_0-{\tilde{\varphi }}_T]\\&\le e^{z(x)}. \end{aligned}$$

\(\square \)