Optimal Investment and Consumption with Proportional Transaction Costs in Regime-Switching Model

Abstract

This paper is concerned with an infinite-horizon problem of optimal investment and consumption with proportional transaction costs in continuous-time regime-switching models. An investor distributes his/her wealth between a stock and a bond and consumes at a non-negative rate from the bond account. The market parameters (the interest rate, the appreciation rate, and the volatility rate of the stock) are assumed to depend on a continuous-time Markov chain with a finite number of states (also known as regimes). The objective of the optimization problem is to maximize the expected discounted total utility of consumption. We first show that for a class of hyperbolic absolute risk aversion utility functions, the value function is a viscosity solution of the Hamilton–Jacobi–Bellman equation associated with the optimization problem. We then treat a power utility function and generalize the existing results to the regime-switching case.

Appendix

Proof of Theorem 3.1

In view of Remark 2.1 and Corollary 3.1, it suffices to show that for each $i\in \mathcal {M}$, V(x,y,i) has limit 0 at any point (x ₀,y ₀)∈∂Π. Using (6), we obtain V≤K ₁ V _p where V _p denotes the value function for the power utility function U(c)=c ^γ/γ where γ is specified in (6) and K ₁ is some positive constant. In view of the upper bound property established in Proposition 5.3, we have

$$ V(x,y,i) \leq \left \{ \begin{array}{l@{\quad}l} K_2 [x+(1-\mu)y]^\gamma, & \mbox{if } (x,y) \in \overline{D}_0, \\ K_3 [x+(1+\lambda)y]^\gamma, & \mbox{if } (x,y) \in\overline{E}_0 \\ \end{array} \right . $$

(82)

for some constants K ₂,K ₃>0, where D ₀ and E ₀ are defined in (32) and (43), respectively.

Given (x ₀,y ₀)∈∂ ₁ Π and (x ₀,y ₀)≠(0,0). Consider an δ-neighborhood of (x ₀,y ₀): B _δ (x ₀,y ₀)={(x,y):|(x,y)−(x ₀,y ₀)|<δ} for some δ>0. Then for any point $(x,y)\in B_{\delta}(x_{0},y_{0})\cap\overline{D}_{0}$, we have 0≤V(x,y,i)≤K ₂[x+(1−μ)y]^γ. Sending (x,y)→(x ₀,y ₀), we obtain

$$ \lim_{(x,y)\rightarrow (x_0,y_0)} V(x,y,i)=0. $$

(83)

Similarly, we can show that (83) holds at any point (x ₀,y ₀)∈∂ ₂ Π∖{(0,0)}.

Finally, we consider the origin (0,0). For any point $(x,y)\in B_{\delta}(0, 0)\cap\overline{\varPi}$ where B _δ (0,0)={(x,y):|(x,y)|<δ} for some δ>0, we consider three cases:

(I) If $(x,y)\in\overline{D}_{0}$, then

$$ 0\leq V(x,y,i)\leq K_2 \bigl[x+(1-\mu )y \bigr]^\gamma; $$

(84)

(II) If $(x,y)\in\overline{E}_{0}$, then

$$ 0\leq V(x,y,i)\leq K_3 \bigl[x+(1+\lambda )y\bigr]^\gamma; $$

(85)

(III) If $(x,y)\in\varPi\setminus(\overline{D}_{0}\cup\overline{E}_{0})$, then in view of the homotheticity property (25) of the function V _p , we have

$$ V(x,y,i)\leq K_1 V_p(x,y,i)=K_1 \bigl(\sqrt{x^2+y^2} \bigr)^\gamma V_p \biggl(\frac{x}{\sqrt{x^2+y^2}}, \frac{y}{\sqrt{x^2+y^2}}, i \biggr). $$

(86)

Note that $( x/\sqrt{x^{2}+y^{2}}, y/\sqrt{x^{2}+y^{2}})$ is on the arc segment of the unit circle r=1, −arctan(1/ξ ₀)≤θ≤π−arctan(1/δ ₀), which is a closed subset in Π. In view of Corollary 3.1, V _p is continuous on this arc segment and therefore bounded. It follows that

$$ V(x,y,i)\leq K_4 \bigl(\sqrt {x^2+y^2} \bigr)^\gamma, \quad \forall(x,y)\in \varPi\setminus (\overline{D}_0\cup\overline{E}_0), $$

(87)

for some constant K ₄>0. Sending (x,y)→(0,0) in (84), (85), and (87), we obtain

$$ \lim_{(x,y)\rightarrow(0,0)} V(x,y,i)=0. $$

(88)

This completes the proof. □

Proof of the ≤ part of Proposition 5.4

We only show the inequality for x≤0 since that for x≥0 can be proved analogously and therefore omitted. For this purpose, we introduce a sequence of closed subsets in Π. For n=1,2,… , define

$$ \varPi_n= \biggl\{(x,y): x+(1-\mu)y\geq \frac{1}{n}, x+(1+\lambda)y\geq\frac{1}{n} \biggr\} \subset\varPi. $$

(89)

Given (x ₀,y ₀)∈Π, $i\in \mathcal {M}$, and $(C,M,N)\in \mathcal {A}(x_{0},y_{0},i)$, define

$$ \nu_n=\inf\bigl\{ t\geq0: \bigl(x(t),y(t)\bigr)\notin\varPi_n\bigr\} $$

(90)

and

$$ \nu=\inf\bigl\{ t\geq0: \bigl(x(t),y(t) \bigr)=(0,0) \bigr\} $$

(91)

where (x(t),y(t)) are given by (3). It can be shown as in [3, Lemma 5.6] that ν _n ↑ν a.s. as n→∞.

We consider φ defined in (55). It is readily seen that φ(x,y,i)=V(x,y,i), ∀(x,y)∈∂D, $i\in \mathcal {M}$. We extend the definition of φ to $\overline{\varPi}$ by letting φ=V on $\overline{\varPi }\setminus \overline{D}$.

Next, we introduce a sequence of closed subsets in $\overline{D}$. For n=1,2,… , define

$$ D_n= \bigl\{(x,y)\in \overline{D}: y\leq n \bigr\} \subset\overline{D}. $$

(92)

Let (x ₀,y ₀)∈D. Define

$$ \tau_n=\inf \bigl\{t\geq0: \bigl(x(t),y(t)\bigr)\notin D_n\bigr\} $$

(93)

and

$$ \tau=\inf\bigl\{ t\geq0: \bigl(x(t),y(t)\bigr) \notin\overline{D} \bigr\}. $$

(94)

Then τ _n ↑τ a.s. as n→∞. Moreover, similar to [3, Proposition 5.7], we have

$$ \{ \tau<\infty\} =\bigcup _{n=1}^{\infty}\{\nu_n\wedge \tau_n=\tau\leq n\}. $$

(95)

Now consider 0≤t<n∧ν _n ∧τ _n . Note that (57) implies that for each $i\in \mathcal {M}$, the function φ _i is non-increasing in the jump directions (1−μ,−1) and (−(1+λ),1) at any point in D. Thus,

$$ \varphi \bigl(x(t),y(t),i\bigr)\leq \varphi\bigl(x(t-),y(t-),i\bigr), \quad 0\leq t <n \wedge\nu_n\wedge \tau_n, i\in \mathcal {M}. $$

(96)

Applying the Itô formula to e ^−βt φ(x(t),y(t),α _t ) for 0≤t≤n∧ν _n ∧τ _n , noting that φ(x(t),y(t),α _t ) and y(t)φ _y (x(t),y(t),α _t ) are bounded on [0,n∧ν _n ∧τ _n [, we obtain

$$\begin{aligned} &E \bigl[ e^{-\beta(n \wedge\nu_n\wedge\tau_n )} \varphi\bigl(x({n \wedge\nu_n\wedge\tau_n } ),y({n \wedge\nu _n\wedge \tau_n }),\alpha_{n \wedge\nu_n\wedge\tau_n }\bigr) \bigr] \\ &\quad \leq \varphi (x_0,y_0,i) - E \biggl[ \int_0^{n \wedge\nu_n\wedge\tau_n } e^{-\beta t} \bigl[ \mathcal {L}_{\alpha_t}\varphi\bigl(x(t),y(t),\alpha_t\bigr) \\ &\quad\quad{} +C(t) \varphi_x\bigl(x(t),y(t),\alpha_t\bigr) -Q \varphi\bigl(x(t), y(t), \cdot\bigr) ( \alpha_t) \bigr]\,dt \biggr]. \end{aligned}$$

(97)

In view of (14) and (58), we have, for t∈]0,n∧ν _n ∧τ _n [,

$$ \mathcal {L}_{\alpha_t}\varphi \bigl(x(t),y(t), \alpha_t\bigr) +C(t)\varphi_x\bigl(x(t),y(t), \alpha_t\bigr) - Q \varphi\bigl(x(t), y(t), \cdot \bigr) ( \alpha_t) \geq U\bigl(C(t)\bigr). $$

(98)

Using (98) and the fact that φ=V on $\overline {\varPi}\setminus\overline{D}$, we get from (97)

$$\begin{aligned} \varphi(x_0,y_0,i) \geq& E \bigl[ e^{-\beta(n \wedge \nu_n\wedge\tau_n )}\varphi\bigl(x({n \wedge\nu_n\wedge \tau_n } ),y({n \wedge\nu_n\wedge\tau_n }), \alpha_{n \wedge\nu_n\wedge\tau_n }\bigr) \bigr] \\ &{} + E \biggl[ \int_0^{n \wedge\nu_n\wedge\tau_n } e^{-\beta t} U \bigl(C(t)\bigr)\,dt \biggr] \\ \geq& E \bigl[I_{\{\nu_n\wedge\tau_n=\tau\leq n\}} e^{-\beta\tau }\varphi\bigl(x({ \tau} ),y({ \tau}),\alpha_{ \tau}\bigr) \bigr] \\ &{}+ E \biggl[ \int_0^{n \wedge\nu_n\wedge\tau_n } e^{-\beta t} U\bigl(C(t)\bigr)\,dt \biggr] \\ =& E \bigl[I_{\{\nu_n\wedge\tau_n=\tau\leq n\}} e^{-\beta\tau }V\bigl(x({ \tau} ),y({ \tau}), \alpha_{ \tau}\bigr) \bigr] \\ &{}+ E \biggl[ \int_0^{n \wedge\nu_n\wedge\tau_n } e^{-\beta t } U\bigl(C(t)\bigr)\,dt \biggr]. \end{aligned}$$

(99)

Note that n∧ν _n ∧τ _n →ν∧τ as n→∞. We have two cases. (I) If τ(ω)<∞, then we must have τ(ω)<ν(ω). Otherwise the state (x(t,ω),y(t,ω)) would reach the origin (0,0) and stay there forever and hence never exit $\overline{D}$, resulting in τ(ω)=∞, a contradiction! (II) If τ(ω)=∞, then ν(ω)∧τ(ω)=ν(ω). For this case, we have

$$ \int_0^{\nu(\omega) } e^{-\beta t} U\bigl(C(t,\omega) \bigr)\,dt = \int_0^{\infty} e^{-\beta t} U \bigl(C(t,\omega)\bigr)\,dt = \int_0^{\tau(\omega) } e^{-\beta t} U\bigl(C(t, \omega)\bigr)\,dt, $$

where the first equality is due to the fact that C(t,ω)=0 for t>ν(ω). It follows by using the monotone convergence theorem that

$$ \begin{aligned} \lim_{n\rightarrow\infty} E \biggl[ \int _0^{n \wedge \nu _n\wedge\tau_n } e^{-\beta t } U\bigl(C(t)\bigr)\,dt \biggr] =E \biggl[ \int_0^{ \tau} e^{-\beta t } U\bigl(C(t)\bigr)\,dt \biggr]. \end{aligned} $$

On the other hand, we can show that $I_{\{\nu_{n}\wedge\tau_{n}=\tau \leq n\}} \uparrow I_{\{\tau<\infty\}}$ as n→∞. Again by the monotone convergence theorem we have

$$ \lim_{n\rightarrow\infty} E \bigl[I_{\{\nu _n\wedge\tau _n=\tau\leq n\}} e^{-\beta\tau}V\bigl(x({ \tau} ),y({ \tau}),\alpha_{ \tau}\bigr) \bigr] = E \bigl[I_{\{\tau<\infty\}} e^{-\beta\tau}V\bigl(x({ \tau} ),y({ \tau}),\alpha_{ \tau}\bigr) \bigr]. $$

Therefore we obtain

$$ \begin{aligned} \varphi(x_0,y_0,i) & \geq E \bigl[I_{\{\tau<\infty\}} e^{-\beta\tau}V\bigl(x({ \tau} ),y({ \tau}), \alpha_{ \tau}\bigr) \bigr] + E \biggl[ \int_0^{ \tau} e^{-\beta t } U\bigl(C(t)\bigr)\,dt \biggr]. \end{aligned} $$

(100)

Taking the supremum of the RHS of (100) over $(C, M, N)\in \mathcal {A}(x_{0}, y_{0}, i)$ and using (19), we obtain φ(x ₀,y ₀,i)≥V(x ₀,y ₀,i). □

Partition of the Region Π by Convex Analysis

For each (x,y)∈Π and each $i\in \mathcal {M}$, define the sub-differential of V by

$$\begin{aligned} \partial V(x,y,i) =&\bigl \{ \bigl(\delta^i_x, \delta^i_y \bigr)\in \mathbb{R}^2: V(\tilde{x}, \tilde{y},i)\leq V(x,y,i)+ \delta^i_x(\tilde {x}-x)+\delta ^i_y( \tilde{y}-y), \\ & \ \forall(\tilde{x}, \tilde{y})\in\varPi\bigr\}. \end{aligned}$$

(101)

Since V _i (x,y) is concave and finite on Π, ∂V(x,y,i) is a nonempty, compact and convex subset of $\mathbb{R}^{2}$. The function V _i (x,y) is differentiable at the point (x,y)∈Π iff ∂V(x,y,i) is a singleton. In this case we have ∂V(x,y,i)={V _x (x,y,i),V _y (x,y,i)}.

The following two results can be established by applying [3, Lemma 6.1 and Proposition 6.2] to each function V _i , $i\in \mathcal {M}$.

Lemma A.1

Given $i\in \mathcal {M}$. Let {(x _n ,y _n ),n≥1} be a sequence of points in Π with limit (x ₀,y ₀)∈Π. If $(\delta^{i,n}_{x}, \delta^{i,n}_{y})\in \partial V(x_{n},y_{n},i)$ for every n≥1, then the sequence $\{ (\delta ^{i,n}_{x}, \delta^{i,n}_{y}), n\geq1\}$ is bounded and its every limit point is in ∂V(x ₀,y ₀,i).

Proposition A.1

Let O be an open subset of Π. Then V _i is of C ¹ in O if and only if ∂V(x,y,i) is a singleton for every (x,y)∈O.

Given $i\in \mathcal {M}$, (x,y)∈Π, and $(\delta^{i}_{x}, \delta^{i}_{y})\in \partial V(x,y,i)$. We define function $\phi(\tilde{x}, \tilde{y},i)= V(x,y,i)+\delta^{i}_{x}(\tilde{x}-x)+\delta^{i}_{y}(\tilde{y}-y), (\tilde {x}, \tilde{y})\in\varPi$. Then $\phi(\tilde{x}, \tilde{y},i)\geq V(\tilde{x}, \tilde{y},i), \forall(\tilde{x}, \tilde{y})\in\varPi$. For each h>0 satisfying (x−(1−μ)h,y+h)∈Π from which (x,y) can be reached by a transaction, by Proposition 3.3, we have

$$ \begin{aligned} \phi(x,y,i)& = V(x,y,i) \leq V\bigl(x-(1-\mu)h,y+h,i \bigr) \\ & \leq\phi\bigl(x-(1-\mu)h,y+h,i\bigr) = \phi(x,y,i) + h\bigl[-(1-\mu)\delta ^i_x+\delta ^i_y\bigr]. \end{aligned} $$

It follows that

$$ -(1-\mu )\delta^i_x+ \delta ^i_y\geq0, \quad \forall\bigl( \delta^i_x, \delta^i_y\bigr)\in \partial V(x,y,i), \forall(x,y)\in\varPi. $$

(102)

Similarly, we can show

$$ (1+\lambda )\delta ^i_x- \delta ^i_y\geq0, \quad \forall\bigl( \delta^i_x, \delta^i_y\bigr)\in \partial V(x,y,i), \forall(x,y)\in\varPi. $$

(103)

Since for (x,y)∈Π, x+(1−μ)y>0 and x+(1+λ)y>0, we have V(x,y,i)>0 in view of Proposition 5.2. It follows from Corollary 5.1 that $\delta^{i}_{x}=\phi_{x}(x,y)>0$, and then from (102) that $\delta^{i}_{y}>0$. To summarize, we have

$$ \delta ^i_x>0,\quad \quad \delta^i_y>0, \quad\forall\bigl(\delta^i_x, \delta^i_y\bigr)\in\partial V(x,y,i), \forall (x,y)\in \varPi. $$

(104)

For θ sufficiently close to 1, using the homotheticity property (25), we have

$$\begin{aligned} \bigl( \theta^\gamma-1\bigr)V(x,y,i) =&V(\theta x,\theta y, i)-V(x,y,i) \\ \leq&\phi(\theta x,\theta y, i)-V(x,y,i)=(\theta -1) \bigl(x \delta^i_x + y\delta^i_y\bigr). \end{aligned}$$

(105)

Dividing both sides of (105) by θ−1, sending θ→1+ and θ→1−, respectively, we obtain

$$ x\delta ^i_x + y \delta ^i_y=\gamma V(x,y,i), \quad \forall\bigl( \delta^i_x, \delta^i_y\bigr)\in \partial V(x,y,i), \forall(x,y)\in\varPi. $$

(106)

Now, for $(\tilde{x}, \tilde{y})\in\varPi$, θ>0, we have

$$ \begin{aligned} V(\tilde{x}, \tilde{y},i) &= \theta^\gamma V \biggl(\frac{\tilde{x}}{\theta}, \frac{\tilde{y}}{\theta},i \biggr)\leq \theta ^\gamma\phi \biggl(\frac{\tilde{x}}{\theta}, \frac{\tilde {y}}{\theta },i \biggr) \\ &= \theta^\gamma \biggl[ V(x,y,i)+ \delta^i_x \biggl(\frac{\tilde {x}}{\theta}-x \biggr) +\delta^i_y \biggl( \frac{\tilde{y}}{\theta }-y \biggr) \biggr] \\ &= V(\theta x, \theta y, i) +\bigl(\theta^{\gamma -1}\delta^i_x \bigr) (\tilde{x}-\theta x) + \bigl(\theta^{\gamma-1}\delta^i_y \bigr) (\tilde{y}-\theta y). \end{aligned} $$

It follows that $(\theta^{\gamma-1}\delta^{i}_{x}, \theta^{\gamma -1}\delta ^{i}_{y}) \in\partial V(\theta x, \theta y, i)$. Thus

$$\theta^{\gamma-1} \partial V(x,y,i) \subset\partial V(\theta x, \theta y, i).$$

The reverse set containment can be established by simply replacing θ by $\frac{1}{\theta}$, x by θx and y by θy. Therefore we have

$$ \theta ^{\gamma-1} \partial V(x,y,i) = \partial V(\theta x, \theta y, i), \quad \forall(x,y)\in\varPi, \forall\theta>0, \forall i\in \mathcal {M}. $$

(107)

Using the sub-differentials, we can partition the solvency region Π into three convex cones. First, for $i\in \mathcal {M}$, (x,y)∈Π, $(\bar {x},\bar{y})\in\varPi$, $(\delta^{i}_{x}, \delta^{i}_{y}) \in\partial V(x,y,i)$ and $(\delta^{i}_{\bar{x}}, \delta^{i}_{\bar{y}}) \in\partial V(\bar {x},\bar{y},i)$, in view of (101), we have

$$ \begin{aligned} V(\bar{x},\bar{y},i) &\leq V(x,y,i) + \delta ^i_{x} (\bar{x}-x)+ \delta^i_{y} ( \bar{y}-y) \\ &\leq V(\bar {x},\bar {y},i) + \delta^i_{\bar{x}} (x-\bar{x})+ \delta^i_{\bar{y}} (y-\bar{y}) + \delta^i_{x} (\bar{x}-x)+ \delta^i_{y} (\bar{y}-y). \end{aligned} $$

It follows that

$$ \bigl(\delta ^i_{x}- \delta ^i_{\bar {x}}\bigr) (x-\bar{x})+\bigl(\delta^i_{y}- \delta^i_{\bar{y}}\bigr) (y-\bar{y}) \leq0. $$

(108)

For $i\in \mathcal {M}$, (x,y)∈Π, define

$$ \begin{aligned} \vartheta^+(x,y,i)&= \max\bigl\{-(1-\mu)\delta ^i_{x}+\delta ^i_{y}: \bigl(\delta^i_x, \delta^i_y\bigr) \in\partial V(x,y,i)\bigr\}, \\ \vartheta^-(x,y,i)&=\min\bigl\{-(1-\mu)\delta^i_{x}+ \delta^i_{y}: \bigl(\delta ^i_x, \delta^i_y\bigr) \in\partial V(x,y,i)\bigr\}. \end{aligned} $$

(109)

Note that the maxima and minima in (109) are attained since ∂V(x,y,i) is compact. In addition, ϑ ⁺(x,y,i)≥ϑ ⁻(x,y,i)≥0 due to (102).

Consider the parametric equations of the half line L ₁ originating at the point (1+λ,−1) on ∂ ₂ Π and parallel to ∂ ₁ Π, defined by

$$ L_1: \left \{ \begin{aligned} x(\rho)&=1+\lambda-(1-\mu)\rho, \\ y(\rho )&=-1+\rho, \end{aligned} \quad \forall\rho\geq0. \right . $$

(110)

Define

$$ \rho_0^i=\inf\bigl \{ \rho>0: \vartheta ^-\bigl(x(\rho),y(\rho),i\bigr)=0\bigr\}, $$

(111)

where $\rho_{0}^{i}=\infty$ if the above set is empty.

Lemma A.2

Given $i\in \mathcal {M}$. For $0<\rho<\bar{\rho}<\infty$,

$$ \vartheta^+\bigl(x(\bar {\rho}),y( \bar{\rho}),i\bigr)\leq\vartheta^-\bigl(x(\rho),y(\rho),i\bigr). $$

(112)

If $0<\rho_{0}^{i}<\infty$, then $\vartheta^{-}(x(\rho_{0}^{i}),y(\rho _{0}^{i}),i)=0$ and

$$ \vartheta^+\bigl(x(\rho ),y( \rho),i\bigr)=0, \quad \forall\rho>\rho_0^i. $$

(113)

Let $i\in \mathcal {M}$ be fixed. We partition Π into two open, convex (possibly empty) cones as follows:

$$ \begin{aligned} SA^i&=\bigl\{ (x,y)\in\varPi: (\theta x, \theta y)=\bigl(x(\rho ),y(\rho)\bigr) \mbox{ for some $\theta>0$} \\ & \quad\quad \mbox{and some $ \rho$ with $\rho_{0}^{i}<\rho<\infty$} \bigr\}, \\ \varPi\setminus\overline{SA^i} &=\bigl\{ (x,y)\in\varPi: (\theta x, \theta y)=\bigl(x(\rho),y(\rho)\bigr) \mbox{ for some $\theta>0$} \\ & \quad\quad\mbox{and some $\rho$ with $0<\rho<\rho_{0}^{i}$ }\bigr \}. \end{aligned} $$

Proposition A.2

Given $i\in \mathcal {M}$. We have

$$ -(1-\mu)\delta ^{i}_{x}+ \delta^{i}_{y}=0, \quad \forall\bigl(\delta^i_x, \delta^i_y\bigr)\in\partial V(x,y,i), \forall (x,y)\in SA^i. $$

(114)

In a similar way, we introduce the parametric equations of the half line L ₂ originating at the point (−(1−μ),1) on ∂ ₁ Π and parallel to ∂ ₂ Π:

$$ L_2: \left \{ \begin{aligned} \tilde{x}(\rho)&=-(1-\mu)+ (1+\lambda)\rho, \\ \tilde {y}(\rho)&= 1-\rho, \end{aligned} \quad \forall\rho\geq0. \right . $$

(115)

Define

$$ \tilde{\rho }_0^i= \inf\bigl\{ \rho>0: (1+\lambda)\delta_x^i- \delta_y^i=0 \mbox{ for some $\bigl(\delta_{x}^{i}, \delta _{y}^{i}\bigr)\in\partial V\bigl(\tilde{x}(\rho), \tilde{y}(\rho), i\bigr)$} \bigr\}. $$

(116)

Let $i\in \mathcal {M}$ be fixed. We partition Π into two open, convex (possibly empty) cones as follows:

$$ \begin{aligned} BU^i &=\bigl\{ (x,y)\in\varPi: (\theta x, \theta y)=\bigl(\tilde {x}(\rho),\tilde{y}(\rho)\bigr) \mbox{ for some $ \theta>0$} \\ &\quad\quad \mbox{and some $\rho$ with $\tilde{\rho}_{0}^{i}<\rho< \infty$} \bigr\}, \\ \varPi \setminus \overline{BU^i}& =\bigl\{ (x,y)\in\varPi: (\theta x, \theta y)=\bigl(\tilde {x}(\rho ),\tilde{y}(\rho)\bigr) \mbox{ for some $ \theta>0$} \\ & \quad\quad\mbox{and some $\rho $ with $0<\rho<\tilde{\rho}_{0}^{i}$ }\bigr\}. \end{aligned} $$

Proposition A.3

Given $i\in \mathcal {M}$. We have

$$ (1+\lambda)\delta ^{i}_{x}-\delta ^{i}_{y}=0, \quad \forall\bigl(\delta^i_x, \delta^i_y\bigr)\in\partial V(x,y,i), \forall(x,y)\in BU^i. $$

(117)

Corollary A.1

SA ⁱ∩BU ⁱ=∅, $\forall i \in \mathcal {M}$.

Corollary A.2

For each $i \in \mathcal {M}$, the value function V _i is C ¹ in SA ⁱ∪BU ⁱ.

Note that Corollary A.2 implies that −(1−μ)V _x (x,y,i)+V _y (x,y,i)=0 for (x,y)∈SA ⁱ and (1+λ)V _x (x,y,i)−V _y (x,y,i)=0 for (x,y)∈BU ⁱ. In view of Proposition 5.4, we conclude that both SA ⁱ and BU ⁱ are nonempty, since the two equations are satisfied by V in D and E, respectively, where D and E are defined by (54) and (60), respectively. Clearly, SA ⁱ⊃D and BU ⁱ⊃E. Moreover, it follows from these equations, the homotheticity of the value function, and the functions presented in Proposition 5.4 that

$$ V(x,y,i) = \left \{ \begin{array}{l@{\quad}l} \frac{1}{\gamma} A_i^{\gamma-1} [x+(1-\mu )y]^\gamma, & \forall(x,y) \in SA^i, \\ \frac{1}{\gamma} B_i^{\gamma-1} [x+(1+\lambda)y]^\gamma, & \forall(x,y) \in BU^i, \\ \end{array} \right . $$

(118)

where A _i , B _i are given by (56) and (62), respectively.

For each $i\in \mathcal {M}$, let $NT^{i}=\varPi\setminus\overline{SA^{i}\cup BU^{i}}$. Then we have

Proposition A.4

Given $i\in \mathcal {M}$. we have

$$ \left \{ \begin{aligned} -(1-\mu) \delta^{i}_{x}+\delta^{i}_{y}&>0, \\ (1+\lambda)\delta^{i}_{x}-\delta^{i}_{y}&>0, \end{aligned} \quad\forall\bigl(\delta^i_x, \delta^i_y\bigr)\in\partial V(x,y,i), \forall(x,y)\in NT^i. \right . $$

(119)