Consumption in incomplete markets

Abstract

We develop a method to find approximate solutions, and their accuracy, to consumption–investment problems with isoelastic preferences and infinite horizon, in incomplete markets where state variables follow a multivariate diffusion. We construct upper and lower contractions; these are fictitious complete markets in which state variables are fully hedgeable, but their dynamics is distorted. Such contractions yield pointwise upper and lower bounds for both the value function and the optimal consumption of the original incomplete market, and their optimal policies are explicit in typical models. Approximate consumption–investment policies coincide with the optimal one if the market is complete or utility is logarithmic.

Appendix: Proofs

We first recall a well-known duality property of wealth processes and stochastic discount factors. For any $\eta \in \mathcal{R}$ and $(\pi ,\ell )\in \mathcal{A}$,

$$ \frac{d(M^{\eta }_{t}X^{\pi ,\ell }_{t})}{M^{\eta }_{t}X^{\pi ,\ell }_{t}} = (\pi _{t}'-\mu '\Sigma ^{-1}-\eta _{t}'\Upsilon '\Sigma ^{-1}) \sigma dZ_{t} + \eta _{t}'a dW_{t}-\ell _{t}dt. $$

Thus as $X^{\pi ,\ell } \ge 0$ and $X^{\pi ,\ell }_{t} \ell _{t} = c_{t}$, it follows that $M^{\eta } X^{\pi ,\ell } + \int _{0}^{\cdot }M^{\eta }_{s}c_{s}ds$ is a nonnegative local martingale and therefore a supermartingale. Thus $\mathbb{E}[\int _{0}^{t}M^{\eta }_{s}c_{s}ds]\leq x$, and in the limit as $t\uparrow \infty$, $\mathbb{E}[\int _{0}^{\infty }M^{\eta }_{s}c_{s}ds]\leq x$. As this inequality holds true for all $\eta \in \mathcal{R}$, we get

$$ \sup _{\eta \in \mathcal{R}}\mathbb{E}\bigg[\int _{0}^{\infty }M^{ \eta }_{t}c_{t}dt\bigg]\leq x. $$

(A.1)

The next lemma establishes an upper bound, uniform for any policy $(\pi ,\ell ) \in \mathcal{A}$, for the expected utility from consumption up to a horizon $T$ (cf. [18, Lemma 5] for expected utility from terminal wealth). We refer to the left-hand side of (A.2) below as the primal bound and to the right-hand side as the dual bound. The limits of the primal and dual bounds as $T\uparrow \infty$ give the lower and upper bounds for the value function. If there exist $(\hat{\pi },\hat{\ell })\in \mathcal{A}$ and $\hat{\eta }\in \mathcal{R}$ such that (with $\hat{c} = \hat{\ell }X^{\hat{\pi },\hat{\ell }}$)

$$ \mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t} \frac{\hat{c}^{1-\gamma }_{t}}{1-\gamma }dt\bigg] = \frac{x^{1-\gamma }}{1-\gamma }\mathbb{E}\bigg[\int _{0}^{\infty }e^{- \frac{\beta }{\gamma }t}(M_{t}^{\hat{\eta }})^{ \frac{\gamma -1}{\gamma }}dt\bigg]^{\gamma }, $$

then $\hat{\pi }$, $\hat{\ell }$ and $\hat{\eta }$ are the optimal portfolio, consumption and market price of nontraded risk, respectively.

Lemma A.1

For any$(\pi ,\ell ) \in \mathcal{A}$, $\eta \in \mathcal{R}$and$T>0$,

$$ \mathbb{E}\bigg[\int _{0}^{T}e^{-\beta t} \frac{c^{1-\gamma }_{t}}{1-\gamma }dt\bigg] \leq \frac{x^{1-\gamma }}{1-\gamma }\mathbb{E}\bigg[\int _{0}^{T}e^{- \frac{\beta }{\gamma }t} (M_{t}^{\eta } )^{\frac{\gamma -1}{\gamma }}dt \bigg]^{\gamma }. $$

(A.2)

Proof

Recall that for any differentiable, strictly increasing, and strictly concave $f$ on $(0,\infty )$ and $z> 0$, we have $\sup _{x>0}(f(x) - xz) = f((f')^{-1}(z)) - (f')^{-1}(z)z$. Let $f(x) = e^{-\beta t}\frac{x^{1-\gamma }}{1-\gamma }$; then $(f')^{-1}(z) = e^{-\frac{\beta t}{\gamma }}z^{-\frac{1}{\gamma }}$. Replacing $x$ by $c_{t}$ and $z$ by $yM^{\eta }_{t}$ and setting $q = \frac{\gamma -1}{\gamma }$, it follows that

$$ e^{-\beta t}\frac{c^{1-\gamma }_{t}}{1-\gamma } \leq e^{- \frac{\beta }{\gamma }t}\frac{(yM^{\eta }_{t})^{q}}{1-\gamma } - e^{- \frac{\beta }{\gamma }t}(yM^{\eta }_{t})^{q} + yM^{\eta }_{t}c_{t} \qquad \text{for all }y>0 , $$

whence integrating and recalling (A.1) yields

$$ \mathbb{E}\bigg[\int _{0}^{T}e^{-\beta t} \frac{c^{1-\gamma }_{t}}{1-\gamma }dt\bigg] \leq y^{q} \frac{\gamma }{1-\gamma }\mathbb{E}\left [\int _{0}^{T}e^{- \frac{\beta }{\gamma }t}(M_{t}^{\eta })^{q}dt\right ] + y\mathbb{E} \left [\int _{0}^{T}M^{\eta }_{t}c_{t}dt\right ]. $$

The right-hand side reaches its minimum at $\hat{y} = {x^{-\gamma }}/{\mathbb{E}[\int _{0}^{T}e^{- \frac{\beta }{\gamma }t}(M_{t}^{\eta })^{q}dt]^{-\gamma }}$, and the claim follows by substituting this value. □

Proof of Lemma 3.1

Consider the differential equation

$$\begin{aligned} &\bigg(-\frac{\beta }{1-\gamma } + \pi \mu - \frac{\gamma \pi '\Sigma \pi }{2}- \ell + r \bigg)f \\ &+ \frac{(\nabla f)' b}{1-\gamma } + \pi '\Upsilon \nabla f + \frac{1}{2(1-\gamma )}\operatorname{tr} (A D^{2} f) = - \frac{ \ell ^{1-\gamma }}{1-\gamma }. \end{aligned}$$

(A.3)

From Gilbarg and Trudinger [16, Theorem 6.6.13], assumptions (i) and (iii) imply that (A.3) with the boundary condition $f_{n}(y) = u_{1}(y) = g_{1}(y)^{\gamma }$ on $\partial E_{n}$ has a solution $f_{n} \in C^{2,\alpha }(E_{n})$. By Itô’s lemma,

$$\begin{aligned} &d\bigg(e^{-\beta t}\frac{ X_{t}^{1-\gamma }}{1-\gamma }f_{n}(Y_{t}) \bigg) \\ &= e^{-\beta t} X^{1-\gamma }_{t}\bigg( \frac{(\nabla f_{n})' b}{1-\gamma } + \frac{1}{2(1-\gamma )}\operatorname{tr} (A D^{2} f_{n}) \\ &\qquad \qquad \qquad + \Big(-\frac{\beta }{1-\gamma } + \pi (Y_{t})\mu - \frac{\gamma \pi (Y_{t})'\Sigma \pi (Y_{t})}{2} - \ell (Y_{t}) + r\Big) f_{n} \bigg)dt \\ &\phantom{=:}+e^{-\beta t} X^{1-\gamma }_{t}\pi '\Upsilon \nabla f_{n} +e^{-\beta t} X^{1-\gamma }_{t}\frac{(\nabla f_{n})' a}{1-\gamma }dW_{t} + e^{- \beta t} X^{1-\gamma }_{t}\pi '\sigma dZ_{t}. \end{aligned}$$

For any initial value $x\in \mathbb{R}^{n}$ and $y\in E$, there exists $n$ such that $\frac{1}{n}< x< n$ and $y\in E_{n}$. Let $\tau _{n} = \inf \{t\geq 0: (X_{t},Y_{t}) \notin (\frac{1}{n},n) \times E_{n}\}$. Because $f_{n}$ satisfies (A.3) in the bounded domain $E_{n}$, we have

$$ \frac{x^{1-\gamma }}{1-\gamma }f_{n}(y) = \mathbb{E}\bigg[e^{-\beta \tau _{n}}\frac{ X_{\tau _{n}}^{1-\gamma }}{1-\gamma } u_{1}(Y_{\tau _{n}}) + \int _{0}^{\tau _{n}} \frac{e^{-\beta t} X_{t}^{1-\gamma } \ell (Y_{t})^{1-\gamma }}{1-\gamma }dt \bigg]. $$

The existence of a unique solution to the martingale problem from Assumption 2.1(ii) implies that $(X,Y)$ never explodes. Now the same argument as in the proof of Theorem 1 in Heath and Schweizer [24] together with the local Hölder-continuity of the model parameters implies that $(X,Y)$ is a strong Markov process. Thus for some finite $T>0$,

$$\begin{aligned} &\mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t} \frac{ ( \ell _{t} X_{t} )^{1-\gamma }}{1-\gamma }dt\bigg\rvert \mathcal{F}_{\tau _{n}\wedge T}\bigg] \\ &= \mathbb{E}\bigg[\int _{\tau _{n}\wedge T}^{\infty }e^{-\beta t} \frac{ ( \ell _{t} X_{t} )^{1-\gamma }}{1-\gamma }dt\bigg\rvert \mathcal{F}_{\tau _{n}\wedge T}\bigg] + \int _{0}^{\tau _{n}\wedge T}e^{- \beta t}\frac{\left ( \ell _{t} X_{t}\right )^{1-\gamma }}{1-\gamma }dt \\ &= e^{-\beta (\tau _{n}\wedge T)} \frac{X^{1-\gamma }_{\tau _{n}\wedge T}}{1-\gamma }u_{1}(Y_{\tau _{n} \wedge T}) + \int _{0}^{\tau _{n}\wedge T}e^{-\beta t} \frac{ ( \ell _{t} X_{t} )^{1-\gamma }}{1-\gamma }dt. \end{aligned}$$

The tower property of conditional expectations implies that

$$\begin{aligned} \frac{x^{1-\gamma }}{1-\gamma }u_{1}(y) &= \mathbb{E}\bigg[\int _{0}^{ \infty }e^{-\beta t} \frac{ ( \ell _{t} X_{t} )^{1-\gamma }}{1-\gamma }dt\bigg] \\ &=\mathbb{E}\bigg[e^{-\beta (\tau _{n}\wedge T)} \frac{X^{1-\gamma }_{\tau _{n}\wedge T}}{1-\gamma }u_{1}(Y_{\tau _{n} \wedge T}) + \int _{0}^{\tau _{n}\wedge T}e^{-\beta t} \frac{ ( \ell _{t} X_{t} )^{1-\gamma }}{1-\gamma }dt\bigg]. \end{aligned}$$

Letting $T\rightarrow \infty$, by dominated convergence (for the first term in the the expectation) and monotone convergence (for the second term), the right-hand side converges to

$$ \mathbb{E}\bigg[e^{-\beta \tau _{n}} \frac{ X_{\tau _{n}}^{1-\gamma }}{1-\gamma } u_{1}(Y_{\tau _{n}}) + \int _{0}^{\tau _{n}} \frac{e^{-\beta t} X_{t}^{1-\gamma } \ell (Y_{t})^{1-\gamma }}{1-\gamma }dt \bigg] = \frac{x^{1-\gamma }}{1-\gamma }f_{n}(y). $$

As this equality holds for every $n$, it follows that $u_{1}$ is in $C^{2}(E)$ and solves (A.3) in $E$, whence $g_{1}= u_{1}^{\frac{1}{\gamma }}$ solves

$$\begin{aligned} &r +\pi \mu - \frac{\gamma \pi '\Sigma \pi }{2}+ \frac{\beta }{\gamma -1} + \frac{\gamma (\nabla g_{1})' b}{(1-\gamma )g_{1}} \\ &+\gamma \pi ' \Upsilon \frac{\nabla g_{1}}{g_{1}}+ \frac{\gamma \operatorname{tr} (AD^{2}g_{1})}{2(1-\gamma )g_{1}} - \frac{\gamma (\nabla g_{1})'A\nabla g_{1}}{2g_{1}^{2}} + \frac{g_{1}^{-\gamma }\ell ^{1-\gamma }}{1-\gamma } -\ell =0. \end{aligned}$$

Thus

$$\begin{aligned} &r + \frac{\beta }{\gamma -1} + \frac{\gamma (\nabla g_{1})' b}{(1-\gamma )g_{1}} + \frac{\gamma \operatorname{tr} (AD^{2}g_{1})}{2(1-\gamma )g_{1}} - \frac{\gamma (\nabla g_{1})'A\nabla g_{1}}{2g_{1}^{2}} \\ &+\sup _{\pi ,\ell }\bigg(\pi '\mu - \frac{\gamma }{2}\pi '\Sigma \pi + \gamma \pi ' \Upsilon \frac{\nabla g_{1}}{g_{1}}+ \frac{g_{1}^{-\gamma }\ell ^{1-\gamma }}{1-\gamma } -\ell \bigg) \geq 0. \end{aligned}$$

Then $\mathcal{H}(y,g_{1},\nabla g_{1}, D^{2} g_{1}) \geq 0$ for $0 < \gamma <1$ and $\mathcal{H}(y,g_{1},\nabla g_{1}g, D^{2}g_{1}) \leq 0$ for $\gamma >1$. On the other hand, consider the differential equation

$$\begin{aligned} &\bigg(-\frac{\beta }{\gamma -1}- r - \frac{1}{2\gamma }\mu '\Sigma ^{-1} \mu - \frac{1}{2\gamma }\eta 'A\eta + \frac{\eta '\Upsilon '\Sigma ^{-1}\Upsilon \eta }{2\gamma }\bigg)f \\ &+(\nabla f)'\bigg( \frac{\gamma b}{\gamma -1} - \Upsilon '\Sigma ^{-1} (\mu + \Upsilon \eta ) + A\eta \bigg) + \frac{\gamma \operatorname{tr} (A D^{2} f )}{2(\gamma -1)} = \frac{\gamma }{1-\gamma }. \end{aligned}$$

(A.4)

With similar arguments as above, assumptions (ii) and (iii) imply that there exists a unique solution $h_{n}$ in $E_{n}$ with the boundary condition $h_{n} = g_{2}$ on $\partial E_{n}$. By Itô’s lemma,

$$\begin{aligned} &d\bigg(e^{-\frac{\beta }{\gamma }t} (M^{\eta }_{t} )^{ \frac{\gamma -1}{\gamma }}h_{n}(Y_{t})\bigg) \\ &= \frac{\gamma -1}{\gamma }e^{-\frac{\beta }{\gamma } t} (M^{\eta }_{t} )^{\frac{\gamma -1}{\gamma }} \bigg(\Big(-\frac{\beta }{\gamma -1}- r - \frac{\mu '\Sigma ^{-1}\mu }{2\gamma } - \frac{\eta 'A\eta }{2\gamma } + \frac{\eta '\Upsilon '\Sigma ^{-1}\Upsilon \eta }{2\gamma }\Big)h_{n} \\ & \qquad \qquad \qquad \qquad \qquad \,\,\,+ \frac{\gamma \nabla h_{n}' b}{\gamma -1} + \frac{\gamma \operatorname{tr} (A D^{2} h_{n} )}{2(\gamma -1)}\bigg)dt \\ &\phantom{=:}+\frac{\gamma -1}{\gamma }e^{-\frac{\beta }{\gamma } t} (M^{\eta }_{t} )^{\frac{\gamma -1}{\gamma }}\big(- (\mu '\Sigma ^{-1}+\eta ' \Upsilon '\Sigma ^{-1} )\Upsilon + \eta 'A\big)\nabla h_{n} dt \\ &\phantom{=:}+ \frac{\gamma -1}{\gamma }e^{-\frac{\beta }{\gamma }t} (M^{\eta }_{t} )^{\frac{\gamma -1}{\gamma }}h_{n}\big(- (\mu '\Sigma ^{-1} + \eta ' \Upsilon '\Sigma ^{-1} )\sigma dZ_{t} + \eta 'adW_{t}\big) \\ &\phantom{=:}+e^{-\frac{\beta }{\gamma }t} (M^{\eta }_{t} )^{ \frac{\gamma -1}{\gamma }}(\nabla h_{n})' a dW_{t}. \end{aligned}$$

For any initial value $y\in E$, there exists $n$ such that $y\in E_{n}$. We therefore define $\tau _{n} = \inf \{t\geq 0: Y_{t} \notin E_{n}\}$. As $h_{n}$ satisfies (A.4) in the bounded domain $E_{n}$, we have

$$ h_{n}(y) = \mathbb{E}\bigg[e^{-\frac{\beta }{\gamma } \tau _{n}} (M^{ \eta }_{\tau _{n}} )^{\frac{\gamma -1}{\gamma }}g_{2}(Y_{\tau _{n}}) + \int _{0}^{\tau _{n}}e^{-\frac{\beta }{\gamma }t} (M^{\eta }_{t} )^{ \frac{\gamma -1}{\gamma }}dt\bigg]. $$

The fact that $Y$ never explodes together with the local Hölder-continuity of the model parameters implies that $Y$ is a strong Markov process. Thus similarly to the argument above for $u_{1}$, we get $h_{n} = g_{2}$ in $E_{n}$. Because this holds for every $n$, $g_{2}$ solves (A.4) in $E$ or, equivalently,

$$\begin{aligned} 0&=r + \frac{\beta }{\gamma -1} + \frac{\gamma (\nabla g_{2})' b}{(1-\gamma )g_{2}}+ \frac{\gamma \operatorname{tr} (A D^{2} g_{2} )}{2(1-\gamma )g_{2}} + \frac{\mu '\Sigma ^{-1}\mu }{2\gamma } + \frac{(\nabla g_{2})'\Upsilon '\Sigma ^{-1}\mu }{g_{2}} \\ &\phantom{=:}+ \frac{\gamma g_{2}^{-1}}{1-\gamma } + \frac{\eta 'A\eta }{2\gamma } - \frac{\eta '\Upsilon '\Sigma ^{-1}\Upsilon \eta }{2\gamma } + \frac{(\nabla g_{2})'}{g_{2}} ( \Upsilon '\Sigma ^{-1}\Upsilon - A ) \eta . \end{aligned}$$

Note also that

$$\begin{aligned} &\inf _{\eta }\bigg( \frac{\eta 'A\eta }{2\gamma } - \frac{\eta '\Upsilon '\Sigma ^{-1}\Upsilon \eta }{2\gamma } + \frac{(\nabla g_{2})'}{g_{2}} ( \Upsilon '\Sigma ^{-1}\Upsilon - A ) \eta \bigg) \\ & = - \frac{\gamma (\nabla g_{2})' (A - \Upsilon '\Sigma ^{-1}\Upsilon )\nabla g_{2}}{2g_{2}^{2}} , \end{aligned}$$

thus

$$\begin{aligned} &r + \frac{\beta }{\gamma -1} + \frac{\gamma (\nabla g_{2})' b}{(1-\gamma )g_{2}}+ \frac{\gamma \operatorname{tr} (A D^{2} g_{2} )}{2(1-\gamma )g_{2}}- \frac{\gamma (\nabla g_{2})'A\nabla g_{2}}{2g_{2}^{2}} \\ &+ \frac{\mu '\Sigma ^{-1}\mu }{2\gamma } + \frac{(\nabla g_{2})'\Upsilon '\Sigma ^{-1}\mu }{g_{2}}+ \frac{\gamma g_{2}^{-1}}{1-\gamma } + \frac{\gamma (\nabla g_{2})'\Upsilon '\Sigma ^{-1}\Upsilon \nabla g_{2}}{2g_{2}^{2}} \leq 0. \end{aligned}$$

Since $\sup _{\ell } (\frac{g_{2}^{-\gamma }\ell ^{1-\gamma }}{1-\gamma } - \ell ) = \frac{\gamma g_{2}^{-1}}{1-\gamma }$ and

$$\begin{aligned} &\sup _{\pi }\!\bigg(\!\pi '\!\mu - \frac{\gamma }{2}\pi '\Sigma \pi + \gamma \pi ' \Upsilon \frac{\nabla g_{2}}{g_{2}}\bigg) \\ &= \frac{\mu '\Sigma ^{-1}\mu }{2\gamma } + \frac{(\nabla g_{2})'\Upsilon '\Sigma ^{-1}\mu }{g_{2}}+ \frac{\gamma (\nabla g_{2})'\Upsilon '\Sigma ^{-1}\Upsilon \nabla g_{2}}{2g_{2}^{2}}, \end{aligned}$$

it follows that

$$\begin{aligned} &r + \frac{\beta }{\gamma -1} + \frac{\gamma (\nabla g_{2})' b}{(1-\gamma )g_{2}} + \frac{\gamma \operatorname{tr} (AD^{2}g_{2})}{2(1-\gamma )g_{2}} - \frac{\gamma (\nabla g_{2})'A\nabla g_{2}}{2g_{2}^{2}} \\ &+\sup _{\pi ,\ell }\bigg(\pi '\mu - \frac{\gamma }{2}\pi '\Sigma \pi + \gamma \pi ' \Upsilon \frac{\nabla g_{2}}{g_{2}}+ \frac{g_{2}^{-\gamma }\ell ^{1-\gamma }}{1-\gamma } -\ell \bigg) \leq 0. \end{aligned}$$

Then $\mathcal{H}(y,g_{2},\nabla g_{2}, D^{2} g_{2}) \leq 0$ if $0 < \gamma <1$ and $\mathcal{H}(y,g_{2},\nabla g_{2}, D^{2}g_{2}) \geq 0$ if $\gamma >1$. Finally, Lemma A.1 implies that $\frac{x^{1-\gamma }}{1-\gamma } g_{1}^{\gamma } \leq \frac{x^{1-\gamma }}{1-\gamma } g_{2}^{\gamma }$. Thus if $0 < \gamma <1$, we obtain $g_{1} \leq g_{2}$ and if $\gamma >1$, we have $g_{1} \geq g_{2}$. □

Lemma 3.1 is conceptually close to the result in Heath and Schweizer [24], where the equivalence is shown between a Feynman–Kac functional and the solution to a partial differential equation with a terminal condition. The difference is that (i) in the present setting the horizon is infinite, therefore the associated HJB equation does not have such a terminal condition, and (ii) the equivalence here is established for both the primal and dual bounds of the value function. In addition, the comparison between the primal and dual bounds is used for the existence result in Theorem 3.2, which relies on a method of sub- and supersolution akin to Hata and Sheu [20], Gilbarg and Trudinger [16, Chaps. 6 and 13], whereby solutions are established first locally and then globally. □

Proof of Theorem 3.2

With $u = \gamma \ln g$, we can first rewrite the HJB equation as $\mathcal{G}(y,u,\nabla u, D^{2} u) = 0$, where

$$\begin{aligned} \mathcal{G}(y,u,\nabla u, D^{2} u) &= \gamma e^{-\frac{u}{\gamma }} + ( \nabla u)'\bigg(b + \frac{(1-\gamma )}{\gamma }\Upsilon ' \Sigma ^{-1} \mu \bigg) + \frac{1}{2}\text{tr} (AD^{2}u ) \\ &\phantom{=:}+ \frac{1}{2}(\nabla u)'\bigg(A + \frac{(1-\gamma )}{\gamma } \Upsilon '\Sigma ^{-1}\Upsilon \bigg)\nabla u - \beta \\ &\phantom{=:}+ \frac{(1-\gamma )\mu '\Sigma ^{-1}\mu }{2\gamma } + (1-\gamma ) r, \end{aligned}$$

(A.5)

and $\underline{u} = \gamma \ln \underline{g}$ is a supersolution and $\bar{u} = \gamma \ln \bar{g}$ a subsolution. It suffices to show that a classical solution to $\mathcal{G}(y,u,\nabla u, D^{2} u) = 0$ exists.

For each $n \in \mathbb{N}$, since $A$ is positive definite and continuous, the eigenvalues of $A$ are bounded (away from 0) in $E_{n}$. Thus there exist $\underline{\lambda }_{n} < \bar{\lambda }_{n}$ such that for any $x\in \mathbb{R}^{k}$ and $y\in E_{n}$, we have $\underline{\lambda }_{n}\sum _{i=1}^{k} x_{i}^{2} \leq \sum _{i,j=1}^{k} A_{ij}(y)x_{i}x_{j} \leq \bar{\lambda }_{n}\sum _{i=1}^{k} x_{i}^{2}$. Then Lemma A.2 below implies that there exists a solution $u_{n}$ in $\bar{E}_{n}$ to the boundary value problem

$$\begin{aligned} &\mathcal{G}(y,u,\nabla u, D^{2} u) = 0, \qquad y\in E_{n}, \\ & u|_{\partial E_{n}} = \underline{u}|_{\partial E_{n}}. \end{aligned}$$

Since $\underline{u} \leq \bar{u}$, the comparison principle (cf. Gilbarg and Trudinger [16, Theorem 10.1]) yields $\underline{u} \leq u_{n} \leq \bar{u}$ in $E_{n}$. The same holds for every $m \geq n$, and thus $(u_{m})_{m\geq n}$ are uniformly bounded in $E_{n}$. Because $\bar{E}_{n} \subsetneq E_{n+1}$, Gilbarg and Trudinger [16, Theorem 13.6]) implies that for $m\geq n+1$, there exists $\alpha ' \in (0,1]$ such that $[\nabla u_{m}]_{\alpha ',E_{n}}$ is bounded above by a constant $C$, where $[f]_{\alpha ,\Omega } = \sup _{x,y\in \Omega ,x\neq y} \frac{|f(x)-f(y)|}{|x-y|^{\alpha }}$ and $C$, $\alpha '$ only depend on $\max _{E_{n+1}}|u_{m}|$, $\underline{\lambda }_{n+1}$, $\bar{\lambda }_{n+1}$ and are independent of $m$. Without loss of generality, assume $\alpha = \min (\alpha ,\alpha ')$ (otherwise reset $\alpha$ to the minimum). Then consider $u_{m}$ as solutions to the linear problem

$$ \mathcal{J}(y,u,\nabla u,D^{2} u) = f(y), $$

where

$$\begin{aligned} \mathcal{J}(y,u,\nabla u,D^{2} u) &= (\nabla u)'\bigg(b + \frac{(1-\gamma )}{\gamma }\Upsilon ' \Sigma ^{-1} \mu \bigg) + \text{tr} (AD^{2}u ), \\ f(y) &= -\gamma e^{-\frac{u}{\gamma }} - (\nabla u_{m})'\bigg(A + \frac{(1-\gamma )}{\gamma }\Upsilon '\Sigma ^{-1}\Upsilon \bigg) \nabla u_{m} \\ &\phantom{=:}+ \beta - \frac{(1-\gamma )\mu '\Sigma ^{-1}\mu }{2\gamma } - (1- \gamma ) r. \end{aligned}$$

Since $\nabla u_{m}$ is $\alpha$-Hölder-continuous in $E_{n}$, so is $f$, for every $m\geq n+1$. Then the Schauder interior estimates (see Gilbarg and Trudinger [16, Corollary 6.3]) imply that for $m\geq n+1$, with $d = \text{dist}(E_{n},\partial E_{n+1})$, we have

$$\begin{aligned} &d\max _{E_{n}} |\nabla u_{m}| + d^{2} \max _{E_{n}}|D^{2} u_{m}| + d^{2+ \alpha }[D^{2} u_{m}]_{\alpha ,E_{n}} \\ &\leq D\Big(\max _{E_{n+1}}|u_{m}| + \max _{E_{n+1}}|f| + [f]_{ \alpha ,E_{n+1}}\Big), \end{aligned}$$

where the constant $D$ is independent of $m$. Thus in any compact set $E_{n}$, for $m\geq n+1$, the $u_{m}$ are uniformly bounded in the essential supremum norm and the $\nabla u_{m}$ and $D^{2} u_{m}$ are equicontinuous. From the Arzelà–Ascoli theorem, $(u_{m})$ (up to a subsequence) converges locally uniformly to a function $u$ and on each $E_{n}$, $\nabla u_{m} \rightarrow \nabla u$ and $D^{2} u_{m} \rightarrow D^{2} u$ uniformly as $m \uparrow \infty$. Thus $u$ is a classical solution to (A.5), and $\underline{u}\leq u\leq \bar{u}$. □

Lemma A.2

There exists a solution to the boundary value problem

$$\begin{aligned} &\mathcal{G}(y,u,\nabla u, D^{2} u) = 0, \qquad y\in E_{n}, \\ & u|_{\partial E_{n}} = \underline{u}|_{\partial E_{n}}. \end{aligned}$$

Proof

This proof follows an idea similar to Hata and Sheu [20] and we discuss the case of $0<\gamma <1$. The case of $\gamma >1$ follows similarly. By Hata and Sheu [20, Theorem 3.4], it suffices to prove the boundedness, uniformly in $\tau \in [0,1]$, of the solutions to the two boundary value problems

$$ \begin{aligned} &\mathcal{G}^{\tau }(y,u,\nabla u, D^{2} u) = 0, \qquad y\in E_{n}, \\ & u|_{\partial E_{n}} = \tau \underline{u}|_{\partial E_{n}}, \end{aligned} $$

(A.6)

where $\mathcal{G}^{\tau }$ is defined by replacing $\gamma$ with $1-\tau (1-\gamma )$ in $\mathcal{G}$, and

$$ \begin{aligned} &\bar{\mathcal{G}}^{\tau }(y,u,\nabla u, D^{2} u) = 0, \qquad y\in E_{n}, \\ & u|_{\partial E_{n}} = 0, \end{aligned} $$

(A.7)

where $\bar{\mathcal{G}}^{\tau } = \tau e^{-u} + \tau (\nabla u)'b + \frac{1}{2}\text{tr}(AD^{2}u) + \frac{\tau }{2}(\nabla u)' A \nabla u - \tau \beta$. For (A.6), first note that the constant function $\underline{f}_{n} = -\sup _{y\in \partial E_{n}}|\underline{u}| - \ln \max (\frac{C_{n}}{\gamma },1)$, where

$$ C_{n} = \sup _{y\in \bar{E}_{n},\tau \in [0,1]}\left (\beta - \frac{\tau (1-\gamma )\mu '\Sigma ^{-1}\mu }{2(1-\tau (1-\gamma ))} - \tau (1-\gamma ) r\right ), $$

is a supersolution, and $\tau \underline{u} (y)\geq \underline{f}_{n}(y)$ for $y\in \partial E_{n}$. From the comparison principle, for any solution $u_{n,\tau }$ to (A.6), we get $u_{n,\tau } \geq \underline{f}_{n}$.

For an upper bound, consider the linear equation

$$\begin{aligned} \big(\nabla f(y)\big)'b + \frac{1}{2}\text{tr}\big(AD^{2}f(y)\big) - \beta f(y) & = 0 \qquad \text{for } y\in E_{n}, \\ f(y) & = 1 \qquad \text{for } y\in \partial E_{n}, \end{aligned}$$

which by Gilbarg and Trudinger [16, Theorem 8.34], has a solution in $C^{1,\alpha }$. By the Feynman–Kac formula, the solution is $\mathbb{E}_{y}[e^{-\beta \theta _{n}}]$, where $\theta _{n}$ is the hitting time of $\partial E_{n}$ by $(Y_{t})$ and $\mathbb{E}_{y}$ indicates the expectation for $Y_{0} = y$. Then the solution to the equation

$$\begin{aligned} \big(\nabla f(y)\big)'b + \frac{1}{2}\text{tr}\big(AD^{2}f(y)\big) - \beta f(y) +1 & = 0 \qquad \text{for } y\in E_{n}, \\ f(y) & = 1 \qquad \text{for } y\in \partial E_{n} \end{aligned}$$

is $\bar{f}_{n} = \frac{1}{\beta } + (1-\frac{1}{\beta })\mathbb{E}_{y}[e^{- \beta \theta _{n}}]$, and [20, Theorem 3.8] yields $e^{u_{n,\tau }} \leq \tau e^{\underline{u}} + (1-\tau )\bar{f}_{n}$, which is bounded from above.

For (A.7), let $u^{0}_{n}$ be a solution. Note that $u_{1} = -\ln \beta$ is a solution with boundary condition $-\ln \beta$ on $\partial E_{n}$. When $0<\beta \leq 1$, we have $-\ln \beta \geq 0$ and hence $u^{0}_{n} \leq u_{1}$ by the comparison principle. Similarly, $u_{2} = 0$ is a supersolution, while $u_{2} \leq u^{0}_{n}$. When $\beta >1$, we have $-\ln \beta < 0$, and $u_{1}\leq u^{0}_{n} \leq u_{2}$. □

Proof of Theorem 3.3

First we prove the equalities

$$\begin{aligned} \mathbb{E}\bigg[\int _{0}^{T}e^{-\beta t} \frac{\hat{c}^{1-\gamma }_{t}}{1-\gamma }dt\bigg] & = \frac{x^{1-\gamma }}{1-\gamma }g(y)^{\gamma }\big(1-\mathbb{E}_{ \hat{\mathbb{P}}}\big[e^{-\int _{0}^{T}g(Y_{s})^{-1}ds}\big]\big), \end{aligned}$$

(A.8)

$$\begin{aligned} \frac{x^{1-\gamma }}{1-\gamma }\mathbb{E}\bigg[\int _{0}^{T}e^{- \frac{\beta }{\gamma }t} (M^{\hat{\eta }}_{t} )^{ \frac{\gamma -1}{\gamma }}dt\bigg]^{\gamma } & = \frac{x^{1-\gamma }}{1-\gamma }g(y)^{\gamma }\big(1 - \mathbb{E}_{ \hat{\mathbb{P}}}\big[e^{-\int _{0}^{T}g(Y_{s})^{-1}ds}\big]\big)^{ \gamma }. \end{aligned}$$

(A.9)

Since $X^{\pi ,\ell }_{t} = xe^{\int _{0}^{t}((r + \pi '_{s}\mu - \frac{\pi '_{s}\Sigma \pi _{s}}{2}-\ell _{s})ds + \pi '_{s}\sigma dZ_{s})}$, we get

$$ e^{-\beta t}\frac{c^{1-\gamma }_{t}}{1-\gamma } = \frac{x^{1-\gamma }}{1-\gamma }\ell ^{1-\gamma }_{t}e^{(1-\gamma ) \int _{0}^{t}((r + \frac{\beta }{\gamma -1} + \pi '_{s}\mu - \frac{\pi _{s}'\Sigma \pi _{s}}{2}-\ell _{s})ds + \pi '_{s}\sigma dZ_{s})}. $$

(A.10)

Then substituting $\pi = \frac{\Sigma ^{-1}\mu }{\gamma } + \Sigma ^{-1}\Upsilon \frac{\nabla g}{g}$ and $\ell = g^{-1}$, the integral in the last exponential function above becomes

$$\begin{aligned} &(1-\gamma )\int _{0}^{t}\bigg(\Big(r + \frac{\beta }{\gamma -1} + \pi '_{s}\mu - \frac{\pi '_{s}\Sigma \pi _{s}}{2}-\ell _{s}\Big)ds + \pi '_{s}\sigma dZ_{s}\bigg) \\ & = -\int _{0}^{t}g^{-1}ds - \gamma \int _{0}^{t}\left ( \frac{(\nabla g)'b}{g}+\frac{\operatorname{tr} (AD^{2}g)}{2g} - \frac{(\nabla g)'A\nabla g}{2g^{2}}\right )ds \\ &\phantom{=:}- \gamma \int _{0}^{t} \frac{(\nabla g)' a}{g}dW_{s}+ \gamma \int _{0}^{t} \mathcal{H}(Y_{s},g,\nabla g,D^{2} g)ds +\ln D_{t}, \end{aligned}$$

where

$$\begin{aligned} D_{t} & = \mathcal{E}\bigg(\int _{0}^{\cdot }\Big( \frac{1-\gamma }{\gamma }\Sigma ^{-1}\mu + \frac{(1-\gamma )\Sigma ^{-1}\Upsilon \nabla g}{g}\Big)'\sigma \bar{\rho }dB_{s} \bigg)_{t} \\ &\phantom{=:}\times \mathcal{E}\bigg(\int _{0}^{\cdot }\Big( \frac{(1-\gamma )}{\gamma }\Upsilon '\Sigma ^{-1}\mu + \big(\gamma A+(1- \gamma )\Upsilon '\Sigma ^{-1}\Upsilon \big)\frac{\nabla g}{g}\Big)'(a')^{-1}dW_{s} \bigg)_{t}. \end{aligned}$$

From Lemma A.3 below, $D$ is an $(\mathbb{F},\mathbb{P})$-martingale and $\hat{\mathbb{P}}|_{\mathcal{F}_{t}} = D_{t} \mathbb{P}|_{ \mathcal{F}_{t}}$.

Since $g$ solves the HJB equation, $\mathcal{H}(y,g,\nabla g,D^{2} g) = 0$. Also, note that by Itô’s formula,

$$ \int _{0}^{t}\!\bigg(\frac{(\nabla g)'b}{g}+\frac{\operatorname{tr} (AD^{2}g)}{2g} - \frac{(\nabla g)'A\nabla g}{2g^{2}}\bigg)ds + \int _{0}^{t} \frac{(\nabla g)' a}{g}dW_{s} = \ln g(Y_{t}) - \ln g(y). $$

Hence (A.10) equals $\frac{x^{1-\gamma }}{1-\gamma }g(y)^{\gamma }g(Y_{t})^{-1}e^{- \int _{0}^{t}g^{-1}(Y_{s})ds}D_{t}$. Then with the candidate portfolio and consumption in (3.6), we obtain

$$\begin{aligned} \mathbb{E}\bigg[\int _{0}^{T}e^{-\beta t} \frac{c^{1-\gamma }_{t}}{1-\gamma }dt\bigg] & = \frac{x^{1-\gamma }}{1-\gamma }g(y)^{\gamma }\mathbb{E}\bigg[\int _{0}^{T}g(Y_{t})^{-1}e^{- \int _{0}^{t}g(Y_{s})^{-1}ds}D_{t}dt\bigg] \\ & = \frac{x^{1-\gamma }}{1-\gamma }g(y)^{\gamma }\big(1 - \mathbb{E}_{ \hat{\mathbb{P}}}\big[e^{-\int _{0}^{T}g(Y_{s})^{-1}ds}\big]\big), \end{aligned}$$

where $\mathbb{E}_{\hat{\mathbb{P}}}$ indicates the expectation under $\hat{\mathbb{P}}$, and the equality (A.8) is proved.

On the other hand, plugging in the candidate $\eta = \frac{\gamma \nabla g}{g}$ and following similar calculations with $q = \frac{\gamma -1}{\gamma }$, we get

$$\begin{aligned} &e^{-\frac{\beta }{\gamma }t}(M_{t}^{\eta })^{q} = \frac{g(y)}{g(Y_{t})}e^{-\int _{0}^{t}g(Y_{s})^{-1}ds}D_{t}. \end{aligned}$$

Thus

$$\begin{aligned} \frac{x^{1-\gamma }}{1-\gamma }\mathbb{E}\bigg[\int _{0}^{T}e^{- \frac{\beta }{\gamma }t} (M_{t}^{\eta } )^{\frac{\gamma -1}{\gamma }}dt \bigg]^{\gamma } & = \frac{x^{1-\gamma }}{1-\gamma }\mathbb{E}\bigg[ \int _{0}^{T}\frac{g(y)}{g(Y_{t})}e^{-\int _{0}^{t}g(Y_{s})^{-1}ds}D_{t}dt \bigg]^{\gamma } \\ & = \frac{x^{1-\gamma }}{1-\gamma }g(y)^{\gamma }\big(1 - \mathbb{E}_{ \hat{\mathbb{P}}}\big[e^{-\int _{0}^{T}g(Y_{s})^{-1}ds}\big]\big)^{ \gamma }, \end{aligned}$$

which concludes the proof of the equality (A.9). Now by monotone convergence,

$$ \mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t} \frac{c^{1-\gamma }_{t}}{1-\gamma }dt\bigg] = \lim _{T\rightarrow \infty }\mathbb{E}\bigg[\int _{0}^{T}e^{-\beta t} \frac{c^{1-\gamma }_{t}}{1-\gamma }dt\bigg] $$

for any $(\pi ,\ell )\in \mathcal{A}$. Thus with $(\hat{\pi },\hat{\ell },\hat{\eta })$ in (3.6), we have

$$\begin{aligned} &\mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t} \frac{\hat{c}^{1-\gamma }_{t}}{1-\gamma }dt\bigg] = \frac{x^{1-\gamma }}{1-\gamma }g(y)^{\gamma }\lim _{T\rightarrow \infty }\big(1 - \mathbb{E}_{\hat{\mathbb{P}}}\big[e^{-\int _{0}^{T}g(Y_{s})^{-1}ds} \big]\big) \\ &\leq \frac{x^{1-\gamma }}{1-\gamma }g(y)^{\gamma }\lim _{T \rightarrow \infty }\big(1 - \mathbb{E}_{\hat{\mathbb{P}}}\big[e^{- \int _{0}^{T}g(Y_{s})^{-1}ds}\big]\big)^{\gamma } \\ &= \frac{x^{1-\gamma }}{1-\gamma }\mathbb{E}\bigg[\int _{0}^{\infty }e^{- \frac{\beta t}{\gamma }} (M^{\hat{\eta }}_{t} )^{q}dt\bigg]^{\gamma }, \end{aligned}$$

(A.11)

and equality holds if and only if $\lim _{T\rightarrow \infty }(1 - \mathbb{E}_{\hat{\mathbb{P}}}[e^{- \int _{0}^{T}g(Y_{s})^{-1}ds}]) = 1$. But the term $1 - e^{-\int _{0}^{T}g(Y_{s})^{-1}ds}$ is nonnegative and increasing in $T$, and so by monotone convergence,

$$ \lim _{T\rightarrow \infty }\big(1 - \mathbb{E}_{\hat{\mathbb{P}}} \big[e^{-\int _{0}^{T}g(Y_{s})^{-1}ds}\big]\big) = 1 - \mathbb{E}_{ \hat{\mathbb{P}}}\big[e^{-\int _{0}^{\infty }g(Y_{s})^{-1}ds}\big] . $$

Thus the inequality (A.11) becomes an equality, i.e., $(\hat{\pi },\hat{\ell })$ in (3.6) is optimal, exactly when $\mathbb{E}_{\hat{\mathbb{P}}}[e^{-\int _{0}^{\infty } g(Y_{s})^{-1}ds}] = 0$, which is equivalent to $\int _{0}^{\infty }g(Y_{s})^{-1}ds = \infty \text{ } \hat{\mathbb{P}}\text{-a.s.}$, and in this case, both sides of (A.11) are equal to $\frac{x^{1-\gamma }}{1-\gamma }g(y)^{\gamma }$. □

Lemma A.3

Assume that for some$f\in C^{1}(E;\mathbb{R})$, there exists a unique solution$\hat{\mathbb{P}}$to the martingale problem on$\mathbb{R}^{n}\times E \ni x = (z,y)$for

$$\begin{aligned} \hat{L} &= \frac{1}{2}\sum _{i,j=1}^{n+k}\tilde{A}_{i,j}(y) \frac{\partial ^{2}}{\partial x_{i}\partial x_{j}} + \sum _{i=1}^{n+k} \hat{b}_{i}(y)\frac{\partial }{\partial x_{i}}, \\ \tilde{A}(y) &= \bigg( \textstyle\begin{array}{c@{\quad }c} \Sigma (y) & \Upsilon (y) \\ \Upsilon '(y) & A(y) \\ \end{array}\displaystyle \bigg), \\ \hat{b} &= \bigg( \textstyle\begin{array}{c} \frac{\mu }{\gamma } + \Upsilon \frac{\nabla f}{f} \\ b + \frac{(1-\gamma )\Upsilon '\Sigma ^{-1}\mu }{\gamma } + (\gamma A + (1-\gamma )\Upsilon '\Sigma ^{-1}\Upsilon )\frac{\nabla f}{f} \\ \end{array}\displaystyle \bigg). \end{aligned}$$

Then the process

$$\begin{aligned} D_{t} &= \mathcal{E}\bigg(\int _{0}^{\cdot }\Big( \frac{1-\gamma }{\gamma }\Sigma ^{-1}\mu + \frac{(1-\gamma )\Sigma ^{-1}\Upsilon \nabla f}{f}\Big)'\sigma \bar{\rho }dB_{s} \bigg)_{t} \\ &\phantom{=:}\times \mathcal{E}\bigg(\int _{0}^{\cdot }\Big( \frac{(1-\gamma )}{\gamma }\Upsilon '\Sigma ^{-1}\mu + \big(\gamma A+(1- \gamma )\Upsilon '\Sigma ^{-1}\Upsilon \big)\frac{\nabla f}{f}\Big)'(a')^{-1}dW_{s} \bigg)_{t}, \end{aligned}$$

$t \geq 0$, is an$(\mathbb{F},\mathbb{P})$-martingale. Furthermore, for any$t<\infty$, $\hat{\mathbb{P}}|_{\mathcal{F}_{t}} = D_{t} \mathbb{P}|_{ \mathcal{F}_{t}}$.

Proof

Since Assumption 2.1 holds and $\frac{\nabla f}{f}$ is locally bounded, Cheridito et al. [7, Theorem 2.4 and Remark 2.5] imply that there exists a $(\mathbb{B}, \mathbb{P})$-martingale $\hat{D}$ such that $\hat{\mathbb{P}}|_{\mathcal{B}_{\tau }} = \hat{D}_{\tau } \mathbb{P}|_{\mathcal{B}_{\tau }}$ for any finite stopping time $\tau$ (with respect to $\mathbb{B}$). Note that from Revuz and Yor [36, Theorem II.2.8], $\hat{D}$ is also an $(\mathbb{F},\mathbb{P})$-martingale. Furthermore, from [36, Proposition VII.2.4 and Theorem VII.2.7], there exist Brownian motions $Z$ and $W$ adapted to $\mathbb{F}$ such that (2.1) and (2.2) hold (cf. footnote 5).

By definition, $\mathbb{F}$ is the right-continuous envelope of the filtration generated by $(R,Y)$. On the other hand, we have

$$\begin{aligned} dR_{t} & = \mu (Y_{t})dt + \sigma (Y_{t})dZ_{t}, \\ dY_{t} & = b(Y_{t})dt + a(Y_{t})dW_{t}. \end{aligned}$$

Thus $\mathbb{F}$ coincides with the right-continuous envelope of the filtration generated by $(Z,W)$. (Recall that $\Sigma (y)$ and $A(y)$ are positive definite for all $y \in E$ by Assumption 2.1). Because $d\langle Z,W\rangle _{t} = \rho (Y_{t})dt$, where $\rho (y) = \sigma ^{-1}(y)\Upsilon (y){(a')}^{-1}(y)$, it follows that the Brownian motion $Z$ admits the decomposition

$$ dZ_{t} = \rho (Y_{t}) dW_{t} + \bar{\rho }(Y_{t}) dB_{t}, $$

where $B$ is a Brownian motion independent of $W$ and $\bar{\rho }$ is the unique positive definite, symmetric matrix such that $\rho \rho '+\bar{\rho }\bar{\rho }'= I_{n}$, with $I_{n}$ denoting the identity matrix of dimension $n$. Thus by the martingale representation theorem with respect to the filtration $\mathbb{F}$ above,

$$ \hat{D}= \mathcal{E}\bigg(\int _{0}^{\cdot } (d'_{1s}dB_{s} + d'_{2s}dW_{s} )\bigg). $$

for some adapted processes $d_{1}$ and $d_{2}$. Then by Girsanov’s theorem,

$$ \hat{B}_{t} = B_{t} - \int _{0}^{t}d_{1s}ds \qquad \text{and}\qquad \hat{W}_{t} = W_{t} -\int _{0}^{s}d_{2s}ds $$

are Brownian motions under $\hat{\mathbb{P}}$, and the dynamics of $(R,Y)$ under $\hat{\mathbb{P}}$ are

$$\begin{aligned} dR_{t} & = \left (\mu + \sigma \bar{\rho }d_{1t} + \sigma \rho d_{2t} \right )dt + \sigma d\hat{Z}_{t}, \\ dY_{t} & = \left (b + ad_{2t}\right )dt + ad\hat{W}_{t}. \end{aligned}$$

On the other hand, the infinitesimal generator for $(R,Y)$ under $\hat{\mathbb{P}}$ is $\hat{L}$. Thus we have

$$ \left ( \textstyle\begin{array}{c} \mu + \sigma \bar{\rho }d_{1} + \sigma \rho d_{2} \\ b + ad_{2} \\ \end{array}\displaystyle \right ) = \hat{b} , $$

which implies that

$$\begin{aligned} d_{1} & = \bar{\rho }\sigma \left (\frac{1-\gamma }{\gamma }\Sigma ^{-1} \mu + \frac{(1-\gamma )\Sigma ^{-1}\Upsilon \nabla f}{f}\right ), \\ d_{2} & = a^{-1}\left ( \frac{(1-\gamma )\Upsilon '\Sigma ^{-1}\mu }{\gamma } + \left ( \gamma A + (1-\gamma )\Upsilon '\Sigma ^{-1}\Upsilon \right ) \frac{\nabla f}{f}\right ). \end{aligned}$$

Thus $D = \hat{D}$, and $D$ is an $(\mathbb{F},\mathbb{P})$-martingale. Finally, for $t< T < \infty$ and every $A\in \mathcal{F}_{t}\subseteq \mathcal{B}_{T}$, using $\hat{\mathbb{P}}|_{\mathcal{B}_{T}} =D_{T} \mathbb{P}|_{ \mathcal{B}_{T}}$ implies $\hat{\mathbb{P}}|_{\mathcal{F}_{t}} =D_{t} \mathbb{P}|_{ \mathcal{F}_{t}}$. □

Proof of Theorem 4.1

Consider the stochastic discount factor $M^{\eta }$ with $\eta = \frac{\gamma \nabla g^{d}}{g^{d}}$. Similar calculations as for the dual bound in Theorem 3.3 (with $q = \frac{\gamma -1 }{\gamma }$) yield

$$\begin{aligned} -\frac{\beta t}{\gamma } + \ln \big((M^{\eta }_{t})^{q}\big) &= \int _{0}^{t}-d \ln g^{d}(Y_{s}) + \ln \bar{D}_{t} + \int _{0}^{t}g^{d}(Y_{s})^{-1}ds \\ &\phantom{=:}+\int _{0}^{t}\bigg(\mathcal{H}^{d}(Y_{s},g^{d},\nabla g^{d}, D^{2} g^{d}) \\ &\phantom{=:}\qquad \qquad + \frac{(\gamma -1) (\nabla g^{d} )' (A-\Upsilon '\Sigma ^{-1}\Upsilon )\nabla g^{d}}{2 (g^{d} )^{2}} \bigg)ds, \end{aligned}$$

where

$$\begin{aligned} \bar{D}& = \mathcal{E}\bigg(\int _{0}^{\cdot }\Big( \frac{(1-\gamma )\Sigma ^{-1}\mu }{\gamma } + \frac{(1-\gamma )\Sigma ^{-1}\Upsilon \nabla g^{d}}{g^{d}}\Big)' \sigma \bar{\rho }dB\bigg) \\ &\phantom{=:}\times \mathcal{E}\bigg(\int _{0}^{\cdot }\Big( \frac{(1-\gamma )\Upsilon '\Sigma ^{-1}\mu }{\gamma } + \frac{ (\gamma A + (1-\gamma )\Upsilon '\Sigma ^{-1}\Upsilon )\nabla g^{d}}{g^{d}} \Big)' (a' )^{-1}dW \bigg). \end{aligned}$$

(A.12)

Now $\mathcal{H}^{d}(Y_{s},g^{d},\nabla g^{d}, D^{2} g^{d}) = 0$ implies that for any $T < \infty$,

$$\begin{aligned} &\mathbb{E}\bigg[\int _{0}^{T}e^{-\frac{\beta t}{\gamma }} (M^{\eta }_{t} )^{q}dt\bigg]^{\gamma } \\ &=g^{d}(y)^{\gamma }\mathbb{E}\bigg[\int _{0}^{T}e^{\int _{0}^{t} (-g^{d}(Y_{s})^{-1} + \frac{(\gamma -1) (\nabla g^{d} )' (A-\Upsilon '\Sigma ^{-1}\Upsilon )\nabla g^{d}}{2 (g^{d} )^{2}})ds}g^{d}(Y_{t})^{-1} \bar{D}_{t}dt\bigg]^{\gamma }. \end{aligned}$$

(A.13)

Since the martingale problem for $\bar{L}^{d}$ has a unique solution, $\bar{D}$ is an $(\mathbb{F},\mathbb{P})$-martingale by Lemma A.3, and $\bar{\mathbb{P}}^{d}|_{\mathcal{F}_{T}}=\bar{D}_{T}\mathbb{P}|_{ \mathcal{F}_{T}}$. Thus (A.13) is equal to

$$ g^{d}(y)^{\gamma }\mathbb{E}_{\bar{\mathbb{P}}^{d}}\bigg[\int _{0}^{T}e^{ \int _{0}^{t}( -g^{d}(Y_{s})^{-1}+ \frac{(\gamma -1) (\nabla g^{d} )' (A-\Upsilon '\Sigma ^{-1}\Upsilon )\nabla g^{d}}{2 (g^{d} )^{2}})ds}g^{d}(Y_{t})^{-1}dt \bigg]^{\gamma }. $$

(A.14)

Since $A-\Upsilon '\Sigma ^{-1}\Upsilon$ is nonnegative definite, (A.14) is for $\gamma > 1$$(\text{resp.} < 1)$ greater (resp. less) than or equal to

$$\begin{aligned} &g^{d}(y)^{\gamma }\mathbb{E}_{\bar{\mathbb{P}}^{d}}\bigg[\int _{0}^{T}g^{d}(Y_{s})^{-1}e^{- \int _{0}^{t} g^{d}(Y_{s})^{-1}ds }dt\bigg]^{\gamma } \\ &=g^{d}(y)^{\gamma }\big(1 - \mathbb{E}_{\bar{\mathbb{P}}^{d}}\big[e^{- \int _{0}^{T} g^{d}(Y_{s})^{-1}ds}\big]\big)^{\gamma }. \end{aligned}$$

(A.15)

Therefore, since $\int _{0}^{\infty }g^{d}(Y_{t})^{-1}dt = \infty$$\bar{\mathbb{P}}^{d}$-a.s., we get

$$\begin{aligned} \frac{x^{1-\gamma }}{1-\gamma }\mathbb{E}\bigg[\int _{0}^{\infty }e^{- \frac{\beta }{\gamma }} (M^{\eta }_{t} )^{q}dt\bigg]^{\gamma } &= \lim _{T\rightarrow \infty }\frac{x^{1-\gamma }}{1-\gamma }\mathbb{E} \bigg[\int _{0}^{T}e^{-\frac{\beta }{\gamma }} (M^{\eta }_{t} )^{q}dt \bigg]^{\gamma } \\ &\leq \lim _{T\rightarrow \infty }\frac{x^{1-\gamma }}{1-\gamma }g^{d}(y)^{ \gamma }\big(1 - \mathbb{E}_{\bar{\mathbb{P}}^{d}}\big[e^{-\int _{0}^{T} g^{d}(Y_{s})^{-1}ds }\big]\big)^{\gamma } \\ &=\frac{x^{1-\gamma }}{1-\gamma }g^{d}(y)^{\gamma }. \end{aligned}$$

Finally, since $A -\Upsilon '\Sigma ^{-1}\Upsilon$ is nonnegative definite and $g^{d}$ solves

$$ \mathcal{H}\big(y,g(y),\nabla g, D^{2} g\big) - \frac{(\gamma -1)(\nabla g)'\left (A -\Upsilon '\Sigma ^{-1}\Upsilon \right )\nabla g}{2g^{2}} = 0, $$

we obtain $\mathcal{H}(y,g^{d}(y),\nabla g^{d}, D^{2} g^{d}) \leq (\geq ) \, 0$ when $0<\gamma <1$ (resp. $\gamma >1$), and so $g^{d}$ is a sub(resp. super)solution. □

Proof of Theorem 4.2

With $\ell = (g^{p})^{-1}$ and $\pi = \frac{\Sigma ^{-1}\mu }{\gamma } + \frac{\Sigma ^{-1}\Upsilon \nabla g^{p}}{g^{p}}$, similar calculations as for the primal bound in Theorem 3.3 yield for every $T<\infty$ that

$$\begin{aligned} &\mathbb{E}\bigg[\int _{0}^{T}e^{-\beta t} \frac{c_{t}^{1-\gamma }}{1-\gamma }dt\bigg] \\ &=\frac{x^{1-\gamma }}{1-\gamma }g^{p}(y)^{\gamma } \\ &\phantom{=:}\times \mathbb{E}\bigg[\int _{0}^{T}e^{\int _{0}^{t}(1-\gamma ) ( \phi (Y_{s}) - \frac{\gamma (\nabla g^{p} )' (A-\Upsilon '\Sigma ^{-1}\Upsilon )\nabla g^{p}}{2 (g^{p} )^{2}} )ds}g^{p}(Y_{t})^{-1}e^{-\int _{0}^{t}g^{p}(Y_{s})^{-1}ds}\bar{D}_{t}dt \bigg], \end{aligned}$$

where $\bar{D}$ is defined in (A.12) but with $g^{d}$ replaced by $g^{p}$. Since the martingale problem for $\bar{L}^{p}$ has a unique solution, Lemma A.3 implies that $\bar{D}$ is an $(\mathbb{F},\mathbb{P})$-martingale, and $\bar{\mathbb{P}}^{p}|_{\mathcal{F}_{T}} = \bar{D}_{T} \mathbb{P}|_{ \mathcal{F}_{T}}$. Thus we obtain

$$\begin{aligned} &\mathbb{E}\bigg[\int _{0}^{T}e^{-\beta t} \frac{c_{t}^{1-\gamma }}{1-\gamma }dt\bigg] \\ &=\frac{x^{1-\gamma }}{1-\gamma }g^{p}(y)^{\gamma } \\ &\phantom{=:}\times \mathbb{E}_{\bar{\mathbb{P}}^{p}}\bigg[\int _{0}^{T}e^{\int _{0}^{t}(1- \gamma ) (\phi (Y_{s}) - \frac{\gamma (\nabla g^{p} )' (A-\Upsilon '\Sigma ^{-1}\Upsilon )\nabla g^{p}}{2 (g^{p} )^{2}} )ds}g^{p}(Y_{t})^{-1}e^{-\int _{0}^{t} g^{p}(Y_{s})^{-1}ds}dt\bigg]. \end{aligned}$$

(A.16)

Since $\phi \geq \frac{\gamma (\nabla g^{p})'(A-\Upsilon '\Sigma ^{-1}\Upsilon )\nabla g^{p}}{2(g^{p})^{2}}$, the above is greater than or equal to

$$\begin{aligned} &\frac{x^{1-\gamma }}{1-\gamma }g^{p}(y)^{\gamma }\mathbb{E}_{ \bar{\mathbb{P}}^{p}}\bigg[\int _{0}^{T}g^{p}(Y_{t})^{-1}e^{-\int _{0}^{t} g^{p}(Y_{s})^{-1}ds}dt\bigg] \\ &=\frac{x^{1-\gamma }g^{p}(y)^{\gamma }}{1-\gamma }\big(1-\mathbb{E}_{ \bar{\mathbb{P}}^{p}}\big[e^{- \int _{0}^{T} g^{p}(Y_{s})^{-1}ds} \big]\big). \end{aligned}$$

(A.17)

Since $\int _{0}^{\infty }g^{p}(Y_{t})^{-1}dt = \infty$$\bar{\mathbb{P}}^{p}$-a.s., we get

$$\begin{aligned} \mathbb{E}\bigg[\int _{0}^{\infty }e^{-\beta t} \frac{c_{t}^{1-\gamma }}{1-\gamma }dt\bigg] &= \lim _{T\rightarrow \infty }\mathbb{E}\bigg[\int _{0}^{T}e^{-\beta t} \frac{c_{t}^{1-\gamma }}{1-\gamma }dt\bigg] \\ & \geq \frac{x^{1-\gamma }}{1-\gamma }g^{p}(y)^{\gamma }\lim _{T \rightarrow \infty }\big(1-\mathbb{E}_{\bar{\mathbb{P}}^{p}}\big[e^{- \int _{0}^{T}g^{p}(Y_{s})^{-1}ds}\big]\big) \\ &= \frac{x^{1-\gamma }g^{p}(y)^{\gamma }}{1-\gamma }. \end{aligned}$$

If $\frac{\gamma (\nabla g^{p})' (A-\Upsilon '\Sigma ^{-1}\Upsilon )\nabla g^{p}}{2(g^{p})^{2}} = \phi$, then by (A.16), the above inequality becomes an equality. Note that $g^{p}$ solves

$$ \mathcal{H}(y,g,\nabla g, D^{2} g) -\frac{(1-\gamma )}{\gamma } \bigg( \phi - \frac{\gamma (\nabla g)' (A -\Upsilon '\Sigma ^{-1}\Upsilon )\nabla g}{2g^{2}} \bigg) = 0. $$

Thus $\mathcal{H}(y,g^{p}(y),\nabla g^{p}, D^{2} g^{p}) \geq (\leq ) \, 0$ if $0<\gamma <1$ (resp. $\gamma >1$), and so $g^{p}$ is a super(resp. sub)solution. Finally we have $\frac{x^{1-\gamma }}{1-\gamma }g^{p}(y)^{\gamma } \leq \frac{x^{1-\gamma }}{1-\gamma }g^{d}(y)^{\gamma }$ by Lemma A.1 and Theorem 4.1. Thus we get $g^{p} \leq g^{d}$ ($g^{p} \geq g^{d}$) when $0<\gamma <1$ (resp. $\gamma >1$). □

Proof of Proposition 4.4

By Lemma A.1 and Theorem 4.1

$$ V(x,y) \leq \inf _{\eta \in \mathcal{R}} \frac{x^{1-\gamma }}{1-\gamma }\mathbb{E}\bigg[\int _{0}^{\infty }e^{- \frac{\beta }{\gamma }} (M^{\eta }_{t} )^{q}dt\bigg]^{\gamma } \leq \frac{x^{1-\gamma }}{1-\gamma }g^{d}(y)^{\gamma }. $$

On the other hand, since $V(x,y)$ is homogeneous in $x$, the definition of the $\operatorname{CEL}$ and Theorem 4.2 give

$$\begin{aligned} \big(1-\operatorname{CEL} (\pi ,\ell )\big)^{1-\gamma }V(x,y) &= V\Big(x\big(1- \operatorname{CEL} (\pi ,\ell )\big),y\Big) \\ &=\int _{0}^{\infty }e^{-\beta t} \frac{ (\ell _{t}X^{\pi ,\ell }_{t} )^{1-\gamma }}{1-\gamma }dt \geq \frac{x^{1-\gamma }}{1-\gamma }g^{p}(y)^{\gamma }. \end{aligned}$$

Thus if $\gamma >1$, then $1\leq (1-\operatorname{CEL} (\pi ,\ell ))^{1-\gamma } \leq ( \frac{g^{p}(y)}{g^{d}(y)})^{\gamma }$, and if $0<\gamma <1$, the inequalities are reversed. Therefore $0\leq \operatorname{CEL} (\pi ,\ell )\leq 1- (\frac{g^{p}(y)}{g^{d}(y)})^{ \frac{\gamma }{1-\gamma }}$. □

Proof of Proposition 4.6

First consider the well-posedness of the martingale problem for $L^{Y}$, where $L^{Y}$ is the operator associated to $Y$ (with $u = \gamma \ln g$) given by

$$\begin{aligned} L^{Y} &= \frac{1}{2}\sum _{i,j=1}^{k}A_{i,j}(x) \frac{\partial ^{2}}{\partial x_{i}\partial x_{j}} \\ &\phantom{=:}+ \sum _{i=1}^{k}\bigg(b + \frac{(1-\gamma )\Upsilon '\Sigma ^{-1} \mu }{\gamma } + \Big(A + \frac{(1-\gamma )}{\gamma }\Upsilon '\Sigma ^{-1}\Upsilon \Big) \nabla u\bigg)_{i}\frac{\partial }{\partial x_{i}}. \end{aligned}$$

Let $\psi = \gamma \ln \bar{g}$ and $\tilde{u} = \psi - u$. Then with $\mathcal{G}$ defined in (A.5), we have

$$\begin{aligned} L^{Y} \tilde{u} &= L^{Y} \psi - L^{Y} u \\ &= L^{Y} \psi -\mathcal{G}(y,u,\nabla u,D^{2} u) \\ & \phantom{=:}- \frac{1}{2}(\nabla u)'\bigg(A + \frac{(1-\gamma )}{\gamma } \Upsilon '\Sigma ^{-1}\Upsilon \bigg)\nabla u + \gamma e^{- \frac{u}{\gamma }} - \beta \\ & \phantom{=:} + \frac{(1-\gamma )\mu '\Sigma ^{-1}\mu }{2\gamma } + (1-\gamma ) r \\ &= \mathcal{G}(y,\psi , \nabla \psi ,D^{2}\psi ) + (\nabla u)'\left (A + \frac{(1-\gamma )}{\gamma }\Upsilon '\Sigma ^{-1}\Upsilon \right ) \nabla \psi \\ &\phantom{=:}-\frac{1}{2}(\nabla \psi )'\bigg(A + \frac{(1-\gamma )}{\gamma } \Upsilon '\Sigma ^{-1}\Upsilon \bigg)\nabla \psi \\ & \phantom{=:}-\frac{1}{2}(\nabla u)'\bigg(A + \frac{(1-\gamma )}{\gamma }\Upsilon ' \Sigma ^{-1}\Upsilon \bigg)\nabla u + \gamma e^{-\frac{u}{\gamma }}- \gamma e^{-\frac{\psi }{\gamma }} \\ &= \mathcal{G}(y,\psi , \nabla \psi ,D^{2}\psi ) \\ &\phantom{=:}- \frac{1}{2}(\nabla u-\nabla \psi )'\bigg(A + \frac{(1-\gamma )}{\gamma }\Upsilon '\Sigma ^{-1}\Upsilon \bigg)( \nabla u-\nabla \psi ) + \gamma e^{-\frac{u}{\gamma }}-\gamma e^{- \frac{\psi }{\gamma }} \\ &\leq \mathcal{G}(y,\psi , \nabla \psi ,D^{2}\psi ) + \gamma e^{- \frac{u}{\gamma }}- \gamma e^{-\frac{\phi }{\gamma }} \\ &= \gamma \big(\mathcal{H}(y,\bar{g}, \nabla \bar{g},D^{2}\bar{g}) + g^{-1} - \bar{g}^{-1} \big), \end{aligned}$$

where the last inequality holds because $A + \frac{(1-\gamma )}{\gamma }\Upsilon '\Sigma ^{-1}\Upsilon$ is nonnegative definite. Because $\bar{g}$ is a subsolution to $\mathcal{H}(y, g, \nabla g,D^{2} g) = 0$ and

$$ \sup _{y\in E}\big( g(y)^{-1} - \bar{g}(y)^{-1}\big) < \sup _{y\in E} \big(\underline{g}(y)^{-1} - \bar{g}(y)^{-1}\big)< \infty $$

as assumed in (i), we get $L^{Y} \tilde{u} < C$ in $E$ for some constant $C$. Furthermore, Theorem 4.1 implies that $\tilde{u} = \psi -u = \gamma (\ln \bar{g} - \ln g) \geq 0$. Thus for a positive constant $\lambda \geq C$, we have $L^{Y} (\tilde{u} + 1) = L^{Y} \tilde{u} \leq C \leq \lambda (\tilde{u} +1)$ and

$$ \lim _{n\uparrow \infty }\inf _{y\in E\setminus E_{n}}\tilde{u} (y) \geq \lim _{n\uparrow \infty }\inf _{y\in E\setminus E_{n}} \big(\ln \bar{g}(y)- \ln \underline{g}(y)\big) = \infty . $$

Then Stroock and Varadhan [38, Theorem 10.2.1] implies that the martingale problem for $L^{Y}$ is well posed, and there exists a Brownian motion $W$ such that [36, Proposition VII.2.4 and Theorem VII.2.7]

$$ dY_{t} = \bigg(b + \frac{(1-\gamma )\Upsilon '\Sigma ^{-1} \mu }{\gamma } + \Big(A + \frac{(1-\gamma )}{\gamma }\Upsilon '\Sigma ^{-1}\Upsilon \Big) \nabla u\bigg)dt + adW_{t}. $$

For the martingale problem associated to $\hat{L}$, expand the probability space to make it support a Brownian motion $B$ independent of $W$. Write $Z = \rho W + \bar{\rho }B$, where $\rho \rho ' + \bar{\rho }\bar{\rho }' = I_{n}$, and define $R$ by $dR_{t} = (\frac{\mu }{\gamma } + \Upsilon \frac{\nabla g}{g})dt + \sigma dZ_{t}$. Then $(R,Y)$ is the unique weak solution to the stochastic differential equation corresponding to $\hat{L}$ (i.e., the SDE with drift $\hat{b}$ and diffusion matrix $\tilde{A}$) and the martingale problem for $\hat{L}$ has a unique solution. □

Proof of Lemma 5.1

Let $g(y) = \int _{0}^{\infty }h(y,t)dt$ and suppose that

$$ g_{y} = \int _{0}^{\infty }h_{y}dt\qquad \text{and} \qquad g_{yy} = \int _{0}^{\infty }h_{yy}dt. $$

(A.18)

In order to prove that $g$ solves (5.3), it suffices to prove that $h(t,y)$ solves

$$ 1 + \left (-ky + \lambda \right )\int _{0}^{\infty }h_{y}dt + \frac{Ay\int _{0}^{\infty }h_{yy}dt}{2} - \left (cy + K\right )\int _{0}^{ \infty }hdt=0. $$

The above equation holds true if $h(y,t)$ satisfies

$$ -h_{t} + \left (-ky + \lambda \right )h_{y} + \frac{Ay}{2}h_{yy} - \left (cy+K\right )h = 0 $$

(A.19)

with the boundary condition $h(y,\infty ) - h(y,0) = -1$ for every $y \in E$, because such an $h$ guarantees that

$$\begin{aligned} &1 + (-ky + \lambda )\int _{0}^{\infty }h_{y}dt+ \frac{Ay\int _{0}^{\infty }h_{yy}dt}{2} - (cy + K )\int _{0}^{\infty }hdt \\ &=1+\int _{0}^{\infty }h_{t}dt = 1 + h(y,\infty ) - h(y,0)= 0. \end{aligned}$$

Taking derivatives of $h(t,y) = e^{C(t) - B(t)y}$, with $C(t)$ and $B(t)$ defined in (5.4) and (5.5), shows that $h(t,y)$ satisfies (A.19). Moreover, $C(0) = B(0) = 0$, $C(\infty ) = -\infty$ and $B(\infty ) = \frac{2c}{k+a}$ implies $h(y,0) = 1$ and $h(y,\infty ) = 0$, and so the boundary condition for $h$ holds for any $y > 0$. Finally, since $B'(t) = \frac{4c\alpha ^{2}e^{\alpha t}}{(e^{\alpha t}(k + \alpha ) - k + \alpha )^{2}} >0$, we have

$$ 0=B(0)\leq B(t)\leq B(\infty ) = \frac{2c}{k+\alpha } \qquad \text{for all } t\geq 0, $$

and so (A.18) holds by Lemma A.4 below. □

Lemma A.4

For the functions$g(y) = \int _{0}^{\infty }h(y,t)dt$and$h(y,t) = e^{C(t)-B(t)y}$, if$B(t) \geq 0$and is bounded, then$g_{y} = \int _{0}^{\infty }h_{y}dt$and$g_{yy} = \int _{0}^{\infty }h_{yy}dt$.

Proof

By definition,

$$\begin{aligned} g_{y} &= \lim _{\epsilon \rightarrow 0}\int _{0}^{\infty } \frac{h(t,y+\epsilon )-h(t,y)}{\epsilon }dt \\ & =\lim _{\epsilon \rightarrow 0}\int _{0}^{\infty } \frac{e^{C(t)-B(t)y-B(t)\epsilon } - e^{C(t)-B(t)y}}{\epsilon }dt \\ &= \lim _{\epsilon \rightarrow 0}\int _{0}^{\infty } \frac{e^{C(t)-B(t)y} (e^{-B(t)\epsilon } - 1 )}{\epsilon }dt. \end{aligned}$$

Now for $\epsilon >0$, we have $e^{-B(t)\epsilon } \geq 1 - B(t)\epsilon$, hence

$$ \bigg|\frac{e^{-B(t)\epsilon }-1}{\epsilon }\bigg| = \frac{1 - e^{-B(t)\epsilon }}{\epsilon } \leq \frac{1 - (1-B(t)\epsilon )}{\epsilon } = B(t), $$

and when $\epsilon <0$, we have $1 \geq e^{-B(t)\epsilon } + B(t)\epsilon e^{-B(t)\epsilon }$ so that

$$ \bigg|\frac{e^{-B(t)\epsilon }-1}{\epsilon }\bigg| = \frac{1 - e^{-B(t)\epsilon }}{\epsilon } \leq \frac{e^{-B(t)\epsilon }B(t)\epsilon }{\epsilon } = B(t) e^{-B(t) \epsilon }. $$

Since $B(t)$ is bounded, $|\frac{e^{-B(t)\epsilon } - 1}{\epsilon }|$ is bounded for $\epsilon$ close to 0. Thus dominated convergence gives

$$\begin{aligned} g_{y} &= \lim _{\epsilon \rightarrow 0}\int _{0}^{\infty } \frac{h(t,y+\epsilon )-h(t,y)}{\epsilon }dt \\ &=\int _{0}^{\infty }\lim _{\epsilon \rightarrow 0} \frac{h(t,y+\epsilon )-h(t,y)}{\epsilon }dt = \int _{0}^{\infty }h_{y}dt, \end{aligned}$$

and $g_{yy} = \int _{0}^{\infty }h_{yy}dt$ follows by a similar argument. □

Proof of Proposition 5.2

Assumptions 2.1 (i), 2.2 and 2.3 are straightforward, by inspection of the model (5.1) and (5.2), and we omit the proof.

To check Assumption 2.1 (ii), similarly to the argument in Proposition 4.6, it suffices to check the well-posedness of the martingale problem for the operator $L^{Y} = b(\theta - y) \frac{\partial }{\partial y} + \frac{a^{2}y}{2} \frac{\partial ^{2}}{\partial y^{2}}$, the generator of the state variable $Y$. By Karatzas and Shreve [29, Corollary 5.4.9], this is equivalent to the uniqueness of the weak solution to $dY_{t} = b(\theta - Y_{t})dt + a\sqrt{Y_{t}}dW_{t}$. Since $b(\theta - y)$ and $a\sqrt{y}$ are Lipschitz-continuous on $(\epsilon ,\infty )$ for any $\epsilon >0$, there exists a unique weak solution $Y$ on $(\epsilon ,\infty )$. Then, since $Y$ is a CIR process satisfying the parameter restriction $b\theta \geq \frac{A}{2}$ under ℙ, it never reaches 0, and so there exists a unique solution on $(0,\infty )$.

For the additional assumptions in Theorem 4.1, in the model (5.1) and (5.2), the ODE $\mathcal{H}^{d}(y,g^{d},\nabla g^{d},D^{2} g^{d}) = 0$ becomes

$$\begin{aligned} 0 &= (g^{d} )^{-1}+ \bigg(-by + b\theta + \frac{(1-\gamma )\rho a \mu y}{\gamma \sigma }\bigg) \frac{g^{d}_{y}}{g^{d}} + \frac{Ayg^{d}_{yy}}{2g^{d}} \\ &\phantom{=:}+ \frac{(1-\gamma )\mu ^{2} y}{2\gamma ^{2}\Sigma } - \frac{\beta }{\gamma } + \frac{(1-\gamma )r}{\gamma }. \end{aligned}$$

Since $\gamma >1$, $\frac{\beta }{\gamma } + \frac{(\gamma -1)r}{\gamma }>0$ and $\frac{(\gamma -1)\mu ^{2} }{2\gamma ^{2}\Sigma } > 0$, Lemma 5.1 implies that $g^{d}(y)$ defined in Proposition 5.2 is the solution to the above ODE.

For the martingale problem for $\bar{L}^{d}$ in assumption (i), the corresponding SDE for $Y$ is

$$ dY_{t} = (b\theta -\phi ^{d} Y_{t} )dt + a\sqrt{Y_{t}}d\bar{W}_{t}, $$

(A.20)

where $\phi ^{d} = b - \frac{(1-\gamma )\rho a \mu }{\gamma \sigma } - ( \gamma + (1-\gamma )\rho ^{2}) A \frac{g^{d}_{y}}{g^{d}}>0$ because $0\leq B(t) \leq \frac{2c}{k+\alpha }$ and $0\geq \frac{g^{d}_{y}}{g^{d}} = \frac{\int _{0}^{\infty }-B(t)h(y,t)dt}{\int _{0}^{\infty }h(y,t)dt} \geq - \frac{2c}{k+\alpha }$. Thus $b\theta - \phi y$ and $a\sqrt{y}$ are Lipschitz-continuous, and (A.20) has a unique weak solution, on $(\epsilon ,\infty )$ for any $\epsilon >0$. Then similarly to the argument for $L$, Lemma A.5 below shows that $Y$ in (A.20) never hits 0 or $\infty$, and so the solution to the martingale problem for $\bar{L}^{d}$ has a unique solution.

For assumption (ii), let $G(t) = \ln ((k+\alpha )e^{\alpha t} - k+\alpha ) - \ln 2\alpha - \frac{1}{2}(k+\alpha )t$. Since $G(0) = 0$ and

$$\begin{aligned} G'(t) &= \frac{(k+\alpha )\alpha e^{at}}{(k+\alpha )e^{\alpha t}-k+\alpha } - \frac{1}{2}(k+\alpha ) \\ &= (\alpha -k)\left (\frac{1}{2} - \frac{\alpha }{(k+\alpha )e^{\alpha t}-k+\alpha }\right )\geq 0, \end{aligned}$$

we obtain

$$ C(t) = -\frac{2b\theta G(t)}{A} - Kt\leq -Kt = - \frac{\beta + (\gamma -1)r}{\gamma }t. $$

Therefore, as $B(t) \geq 0$ and $y > 0$,

$$ g^{d}(y)< \int _{0}^{\infty }e^{C(t)}dt < \int _{0}^{\infty }e^{- \frac{\beta + (\gamma -1)r}{\gamma }t}dt = \bigg( \frac{\beta + (\gamma -1)r}{\gamma }\bigg)^{-1} . $$

Thus $(g^{d})^{-1}$ is bounded from below and $\int _{0}^{\infty }g^{d}(Y_{t})^{-1}dt = \infty$$\bar{\mathbb{P}}^{d}$-a.s.

For the additional assumptions in Theorem 4.2, first, from Lemma A.6 below, there exists a constant $Q$ such that $\bar{c} = \frac{(\gamma -1)\mu ^{2} }{2\gamma ^{2}\Sigma } - ( \gamma -1)(1-\rho ^{2})AQ>0$ and $\frac{2\bar{c}^{2}}{(k+\bar{\alpha })^{2}} = Q$. For $\phi (y) = \gamma (1-\rho ^{2})AQy$, the ODE $\mathcal{H}^{d}(y,g^{p},\nabla g^{p},D^{2} g^{p}) - \frac{(1-\gamma )\phi }{\gamma }=0$ becomes

$$ 0= g^{p}(y)^{-1}+ \left (-by + b\theta + \frac{(1-\gamma )\rho a\mu y}{\gamma \sigma }\right ) \frac{g^{p}_{y}}{g^{p}} + \frac{Ayg^{p}_{yy}}{2g^{p}} -\bar{c}y - \frac{\beta }{\gamma } + \frac{(1-\gamma )r}{\gamma } . $$

Then since $A>0$, $\frac{\beta + (\gamma -1)r}{\gamma } > 0$ and $\bar{c}>0$, Lemma 5.1 implies that $g^{p}(y)$ defined in Proposition 5.2 is the solution of the above ODE.

Similarly to $B(t)$, we get $0\leq \bar{B}(t) \leq \frac{2\bar{c}}{k + \bar{\alpha }}$ and $0\geq \frac{g^{p}_{y}}{g^{p}} \geq - \frac{2\bar{c}}{k + \bar{\alpha }}$. Then similarly to the argument for $\bar{L}^{d}$ in assumption (i) in Theorem 4.1, Lemma A.5 below implies that the diffusion $Y$ which follows $dY_{t} = (b\theta -\phi ^{p} Y_{t}) dt + a\sqrt{Y_{t}}d\bar{W}_{t}$ with $\phi ^{p} = b - \frac{(1-\gamma )\rho a \mu }{\gamma \sigma } - ( \gamma + (1-\gamma )\rho ^{2})A \frac{g^{p}_{y}}{g^{p}}$ never reaches 0 or $\infty$. Thus $Y$ has a unique weak solution and assumption (ii) holds.

Note that

$$ \phi (y) - \gamma \frac{ (\nabla g^{p} )' (A- \Upsilon '\Sigma ^{-1}\Upsilon )\nabla g^{p}}{2 (g^{p} )^{2} }y =\gamma (1 - \rho ^{2} )A\bigg(Q - \frac{1}{2}\Big( \frac{g^{p}_{y}}{g^{p}}\Big)^{2}\bigg)y\geq 0 , $$

and we already know that assumption (i) holds. Finally, similarly to $C(t)$, we obtain $\bar{C}(t)\leq -\frac{\beta +(\gamma -1)r}{\gamma }t$ which implies $\int _{0}^{\infty }g^{p}(Y_{t})^{-1}dt = \infty$$\bar{\mathbb{P}}^{p}$-a.s., and so assumption (iii) holds. □

Lemma A.5

If$b_{1}\leq b_{t} \leq b_{2}$for all$t\geq 0$for two constants$b_{1}$and$b_{2}>0$, $\theta \geq \frac{a^{2}}{2}$and the stochastic process$Y$satisfies$dY_{t} = (\theta -b_{t} Y_{t})dt + a\sqrt{Y_{t}}dW_{t}$with$Y_{0}>0$, then$Y$never explodes to 0 or$\infty$.

Proof

For nonexplosion to 0, by the comparison principle, $Y$ is bounded below by $Y_{2}$ which follows $dY_{2t} = (\theta -b_{2}Y_{2t})dt + a\sqrt{Y_{2t}}dW_{t}$. Since $\theta \geq \frac{a^{2}}{2}$, $Y_{2}$ never reaches 0, and thus neither does $Y$. For nonexplosion to $\infty$, consider $n$ Ornstein–Uhlenbeck processes $X^{i}$, $i=1,\dots ,n$, given by

$$ dX^{i}_{t} = -\frac{b_{1}}{2} X^{i}_{t}dt + \frac{a}{2}dW^{i}_{t}, $$

where $W^{i}$, $i=1,\dots ,n$, are $n$ independent Brownian motions. Let $\tilde{Y} = \sum _{i=1}^{n}(X^{i})^{2}$; then

$$ d\tilde{Y}_{t} = \left (\frac{nA}{4} - b_{1} \tilde{Y}_{t}\right )dt + a\sqrt{\tilde{Y}_{t}}\sum _{i=1}^{n} \frac{X^{i}_{t}}{\sqrt{\tilde{Y}_{t}}}dW^{i}_{t}. $$

Note that $\int _{0}^{\cdot }\sum _{i=1}^{n}({X^{i}_{s}}/{\sqrt{\tilde{Y}_{s}}})dW^{i}_{s}$ is a continuous local martingale starting from 0, and since $\sum _{i=1}^{n}{(X^{i}_{t})^{2}}/{\tilde{Y}_{t}} = 1$, its quadratic variation is $t$. Thus by Lévy’s theorem, it is a Brownian motion. Now let $n$ be large enough such that $\frac{nA}{4}\geq \theta$. Then by the comparison principle, $\tilde{Y}$ with dynamics

$$ d\tilde{Y}_{t} = \left (\frac{nA}{4} - b_{1} \tilde{Y}_{t}\right )dt + a\sqrt{\tilde{Y}_{t}}dW_{t} $$

dominates $Y_{1}$ satisfying

$$ dY_{1t} = \left (\theta -b_{1}Y_{1t}\right )dt + a\sqrt{Y_{1t}}dW_{t}, $$

which in turn dominates $Y$. Since $\tilde{Y}$ is the sum of $n$ independent squared Ornstein–Uhlenbeck processes which are Gaussian, $\tilde{Y}$ never explodes to $\infty$, and so neither does $Y$. □

Lemma A.6

For the model (5.1) and (5.2), there exists a constant$\hat{Q} > 0$such that$\bar{c} >0$and$\frac{2\bar{c}^{2}}{(k+\bar{\alpha })^{2}} = \hat{Q}$, where$\bar{c}$, $k$and$\bar{\alpha }$are defined in Proposition 5.2.

Proof

Consider $U(Q) = \frac{2\bar{c}^{2}}{(k+\bar{\alpha })^{2}} - Q$ and note from Proposition 5.2 that the term $\bar{c}= \frac{(\gamma -1)\mu ^{2}}{2\gamma ^{2}\Sigma }- (\gamma -1)A(1-\rho ^{2})Q$ depends on $Q$. For $Q = 0$, $\bar{c} = \frac{(\gamma -1)\mu ^{2}}{2\gamma ^{2}\Sigma } > 0$ implies that $U(0) = \frac{2\bar{c}^{2}}{(k+\bar{\alpha })^{2}}>0$, and when $Q = \frac{\mu ^{2}}{2(1-\rho ^{2})\gamma ^{2}\Sigma A}> 0$, then $\bar{c}=0$ and $U(Q) = -Q < 0$. Since $Q \mapsto U(Q)$ is continuous, there exists a constant $\hat{Q}$ between 0 and $\frac{\mu ^{2}}{2(1-\rho ^{2})\gamma ^{2}\Sigma A}$ such that $U(\hat{Q}) = 0$, and since $\bar{c}$ is strictly decreasing in $Q$, we have $\bar{c}(\hat{Q})>0$. □