2.1. The dynamics

Let $(\Omega,\mathcal{F},\mathbb{F},\mathbb{P})$ be a complete filtered probability space satisfying the usual conditions, where $\mathbb{F}=\{\mathcal{F}_t\}_{t\geq0}$ ; see [Reference Borodin and Salminen7, p. 2]. We assume that the underlying X is a strong Markov process defined on $(\Omega,\mathcal{F},\mathbb{F},\mathbb{P})$ and taking values in $E\subseteq\mathbb{R}^d$ for some $d\geq1$ with the initial state $x\in E$ . We further assume that $\mathbb{F}$ is generated by X. We take $E=(a_1,b_1)\times\dots\times(a_d,b_d)$ , where $- \infty \leq a_i < b_i\leq\infty$ for all $i=1,\dots,d$ . As usual, we augment the state space E with a topologically isolated element $\Delta$ if the process X is non-conservative. Then the process X can be made conservative on the augmented state space $E^{\Delta}\;:\!=\; E\cup\{\Delta\}$ ; see [Reference Rogers and Williams25]. In what follows, we drop the superscript $\Delta$ from the notation. By convention, we augment the definition of functions g on E with $g(\Delta)=0$ . Define the life-time of the process X as $\zeta=\inf\{t\geq0 \colon X_t=\Delta \}$ .

Let $\mathbb{P}_x$ denote the probability measure $\mathbb{P}$ conditioned on the initial state x and let $\mathbb{E}_x$ denote the expectation with respect to $\mathbb{P}_x$ . The process X is assumed to evolve under $\mathbb{P}_x$ and the sample paths are assumed to be right-continuous and left-continuous over stopping times, meaning the following: if the sequence of stopping times $\tau_n\uparrow \tau$ , then $X_{\tau_n}\rightarrow X_\tau$ $\mathbb{P}_x$ -almost surely as $n\rightarrow\infty$ . There is a well-established theory of optimal stopping for this class of processes; see [Reference Peskir and Shiryaev24].

For $r>0$ , we let $L_1^r$ denote the class of real-valued measurable functions f on E satisfying the integrability condition

\[\mathbb{E}_x\biggl[\int_0^\zeta \,{\textrm{e}}^{-rt} |f(X_t)| \,{\textrm{d}} t \biggr]<\infty \quad\text{for all $x\in E$.}\]

For a function $f\in L_1^r$ , the resolvent $R_rf\colon E\rightarrow\mathbb{R}$ is defined as

\[ (R_rf)(x)=\mathbb{E}_x \biggl[\int_0^\zeta \,{\textrm{e}}^{-rs} f(X_s) \,{\textrm{d}} s \biggr] \]

for all $x \in \mathbb{R}_+$ . It is well known that the family $(R_\lambda)_{\lambda\geq0}$ is a strongly continuous contraction resolvent and that it has the following connection to exponentially distributed random times: if $U\sim \mathrm{Exp}(\lambda)$ and independent of X, then $\lambda(R_{r+\lambda}g)(x)=\mathbb{E}_x[{\textrm{e}}^{-rU}g(X_U)]$ whenever $g\in L_1^r$ ; see [Reference Rogers and Williams25]. Finally, the function h is said to be r-harmonic for X if $h(x)=\mathbb{E}_x[{\textrm{e}}^{-r\tau}h(X_\tau)]$ for all $\mathbb{F}$ -stopping times $\tau$ .

2.2. The timing problem

We define the timing problem inductively and start by considering the case where the stopping decision cannot be reversed. Let $U \sim \mathrm{Exp}(\lambda)$ be independent of X, and define

(2.1) \begin{equation}\begin{split} V_a^0(x) &= \mathbb{E}_x[{\textrm{e}}^{-rU}g(X_U)] = \lambda(R_{r+\lambda}g)(x), \\[5pt] V_i^0(x) &= \sup\nolimits_{\tau} \mathbb{E}_x[ {\textrm{e}}^{-r\tau}(\lambda(R_{r+\lambda}g)(X_\tau) - K_1) ].\end{split}\end{equation}

Here the function g is the payoff function; we assume that this function satisfies the following.

1. (A1) The payoff $g\colon E\rightarrow\mathbb{R}$ is in $L^1_r$ , is lower-bounded, satisfies the condition $S^+\;:\!=\; \{x \colon g(x)>0 \}\neq\emptyset$ , and the process X reaches the set $S^+$ with positive probability for all initial states x,

2. (A2) There exists an r-harmonic function $h\colon E\rightarrow\mathbb{R}_+$ such that the function $x\mapsto{{g(x)}/{h(x)}}$ is bounded.

In (2.1), the function $V^0_a$ is the value of an active investment (i.e. the investment is initiated) when there is no possibility of calling the investment off. Note that the value function $V_i^0$ is (essentially) that of [Reference Lempa17]. This corresponds to the case where the investment opportunity is inactive (i.e. the investment is not initiated), the investment decision cannot be called off, and initiation of the investment incurs a cost of $K_1$ .

Proceeding inductively, for the case where a stopping decision can be reversed k times, we write

(2.2) \begin{equation} \begin{split} V_a^k(x) &= \sup\nolimits_{\tau}\mathbb{E}_x[ {\textrm{e}}^{-rU} g(X_U)\textbf{1}(U < \tau) + {\textrm{e}}^{-r\tau} (V_i^{k-1}(X_\tau) + K_2)\textbf{1}(U>\tau) ], \\[5pt] V_i^k(x) &= \sup\nolimits_{\tau} \mathbb{E}_x [ {\textrm{e}}^{-r\tau}(V_a^{k}(X_\tau) - K_1) ]. \end{split}\end{equation}

An alternative expression for the function $V_a$ above can be found. To find it, we first note that the following holds for any measurable function f:

\begin{align*} \mathbb{E}_{x}[{\textrm{e}}^{-r\tau}f(X_\tau)\textbf{1}(U > \tau)] &= \mathbb{E}_{x}[\mathbb{E}_{x}[{\textrm{e}}^{-r\tau}f(X_\tau)\textbf{1}(U > \tau) | \mathcal{F}_\tau]] \\[5pt] &= \mathbb{E}_{x}[{\textrm{e}}^{-r\tau}f(X_\tau)\mathbb{E}_{x}[\textbf{1}(U > \tau) | \mathcal{F}_\tau]] \\ &= \mathbb{E}_{x}[{\textrm{e}}^{-r\tau}f(X_\tau){\textrm{e}}^{-\lambda\tau}] \\[5pt] &= \mathbb{E}_{x}[{\textrm{e}}^{-(r+\lambda)\tau}f(X_\tau)].\end{align*}

Hence

\begin{align*}V_a^k(x) &= \sup\nolimits_{\tau}\mathbb{E}_x [ {\textrm{e}}^{-rU} g(X_U)\textbf{1}(U < \tau) + {\textrm{e}}^{-r\tau} (V_i^{k-1}(X_\tau) + K_2)\textbf{1}(U>\tau) ] \\[5pt] &= \sup\nolimits_{\tau}\mathbb{E}_x[ {\textrm{e}}^{-rU} g(X_U)) + {\textrm{e}}^{-r\tau} (V_i^{k-1}(X_\tau)- {\textrm{e}}^{-rU} g(X_U)) + K_2)\textbf{1}(U>\tau) ] \\&= \sup\nolimits_{\tau}\mathbb{E}_x [ {\textrm{e}}^{-rU} g(X_U)) + {\textrm{e}}^{-r\tau} (V_i^{k-1}(X_\tau)- \lambda (R_{r+\lambda}g) (X_\tau) + K_2)\textbf{1}(U>\tau) ] \\ &= \lambda(R_{r+\lambda}g)(x) + \sup\nolimits_{\tau}\mathbb{E}_x [{\textrm{e}}^{-(r+\lambda)\tau} (V_i^{k-1}(X_\tau) - \lambda(R_{r+\lambda}g)(X_\tau) + K_2) ].\end{align*}

Our main problem can then be written as the limiting case

(2.3) \begin{align} V_a(x) = V^{\infty}_a(x) = \lim_{k\rightarrow\infty} V^k_a(x), \quad V_i(x) = V^\infty_i(x) = \lim_{k\rightarrow\infty} V^k_i(x).\end{align}

The corresponding decision variables are then increasing sequences of stopping times denoted as $\bar{\tau}=(\tau_n)_{n\geq1}$ . The following proposition provides us with sufficient conditions for the main problem (2.3) to be well-defined.

Proof. We prove the result for $V_i$ ; the function $V_a$ is handled similarly. For the purpose of the argument, we write the value function $V_i^k = V_i^k(\cdot,K_1,K_2)$ ; here, $K_i$ are the cost parameters in the definition of the problem. Then it is straightforward to show by induction that $V_i^k(\cdot,K_1,K_2) \leq V_i^k(\cdot,0,0)$ .

Our task is to show that the function $V_i^k(\cdot,0,0)$ can be represented as the value of an optimal stopping problem. To this end, we first note that by [Reference Peskir and Shiryaev24, Theorem 1 and Corollary, p. 124], the optimal stopping problem $V(x) = \sup\nolimits_{\tau\in\mathcal{T}} \mathbb{E}_x[ {\textrm{e}}^{-r\tau} g(X_{\tau}) ]$ , where $\mathcal{T}$ is the set of $\mathbb{F}$ -stopping times, has a finite solution under assumptions (A1) and (A2). Now let N be a Poisson process, independent of X, with rate $\lambda$ . Furthermore, let $\mathbb{F}_N$ be the filtration generated by N and $(X_t)$ and let $\mathcal{T}_N$ be the set of $\mathbb{F}_N$ -stopping times. Let $V_N(x) = \sup\nolimits_{\tau\in\mathcal{T}_N} \mathbb{E}_x[ {\textrm{e}}^{-r\tau} g(X_{\tau}) ]$ . Since X is $\mathbb{F}_N$ -adapted and the payoff is independent of N, we find that $V=V_N$ .

Let $((\tau_i,\sigma_i))_{i=1}^k$ be a vector of pairs of $\mathbb{F}$ -stopping times such that $\tau_i\leq\sigma_{i}\leq\tau_{i+1}\leq\sigma_{i+1}$ for all $i=1,\dots,k-1$ with $\sigma_k=\infty$ . For such a vector, let $\tau$ be the first arrival of N such that its arrival time $T \in (\tau_i,\sigma_i)$ for some $i = 1,\dots,k$ ; if this does not occur, set $\tau=\infty$ . Then $\tau \in \mathcal{T}_N$ ; denote the set of such $\tau$ as $\mathcal{S}_k$ . Moreover, since $\tau \geq \tau_1$ , we find that

\begin{align*} \mathbb{E}_x[ {\textrm{e}}^{-r\tau} g(X_\tau) ] &= \mathbb{E}_x[ {\textrm{e}}^{-r\tau_1} \mathbb{E}_{X_{\tau_1}}[ {\textrm{e}}^{-r\tau} g(X_\tau) ] ] \\[5pt] &= \mathbb{E}_x[ {\textrm{e}}^{-r\tau_1} \mathbb{E}_{X_{\tau_1}}[ {\textrm{e}}^{-r\tau} g(X_\tau)\textbf{1}(\tau < \sigma_1) + {\textrm{e}}^{-r\tau} g(X_\tau)\textbf{1}(\tau > \sigma_1) ] ] \\[5pt] &= \mathbb{E}_x [ {\textrm{e}}^{-r\tau_1} \mathbb{E}_{X_{\tau_1}} [ {\textrm{e}}^{-rU} g(X_U)\textbf{1}(U < \sigma_1) + {\textrm{e}}^{-r\sigma_1}\mathbb{E}_{X_{\sigma_1}}[{\textrm{e}}^{-r\tau} g(X_\tau)]\textbf{1}(U > \sigma_1) ] ], \end{align*}

where $U\sim\mathrm{Exp}(\lambda)$ is independent of X. In the last equality, we used the memoryless property of the exponential distribution. Now we can proceed with the term $\mathbb{E}_{X_{\sigma_1}}[{\textrm{e}}^{-r\tau} g(X_\tau)]$ in the same way and eventually recover the objective functional in (2.2) with costs equal to zero. Finally, since the stopping times $\tau$ are indexed by the vectors $((\tau_i,\sigma_i))_{i=1}^k$ , we find that $V_i^k(x,0,0) = \sup\nolimits_{\tau \in \mathcal{S}_k}\mathbb{E}_x[ {\textrm{e}}^{-r\tau} g(X_{\tau}) ]$ . Thus

\[V_i^k(x)=V_i^k(x,K_1,K_2)\leq V_i^k(x,0,0)\leq V_N(x) = V(x)\quad\text{for all}\, \textit{k},\]

since $\mathcal{S}_k \subset \mathcal{T}_N$ .

To conclude, we observe that the sequence $(V_i^k)$ is increasing; indeed, we can augment any k-vector $((\tau_i,\sigma_i))_{i=1}^k$ to a $(k+1)$ -vector with $\tau_{k+1}=\sigma_{k+1}=\infty$ , which yields the same payoff as the original k-vector. Consequently the limit $V_i = \lim_{k\rightarrow \infty} V_i^k$ exists and is finite.

4.1. The problem specification

We assume that the process X follows a regular linear diffusion on the positive real line $\mathbb{R}_+$ . Furthermore, we assume that the boundaries of the state space are natural. Now, the evolution of X is completely determined by its scale function S and speed measure m inside $\mathbb{R}^+$ ; see [Reference Borodin and Salminen7, pp. 13–14]. Furthermore, we assume that the function S and the measure m are both absolutely continuous with respect to the Lebesgue measure, have smooth derivatives, and that S is twice continuously differentiable. Under these assumptions, we know that the infinitesimal generator $\mathcal{A}\colon \mathcal{D}(\mathcal{A})\rightarrow C_b(\mathbb{R}_+)$ of X can be expressed as

\[\mathcal{A}=\dfrac{1}{2}\sigma^2(x)\dfrac{{\textrm{d}}^2}{{\textrm{d}} x^2}+\mu(x)\dfrac{{\textrm{d}}}{{\textrm{d}} x},\]

where the functions $\sigma$ and $\mu$ are related to S and m via the formulæ

\[m'(x)=\dfrac{2}{\sigma^2(x)}\,{\textrm{e}}^{B(x)}\quad\text{and}\quad S'(x)= {\textrm{e}}^{-B(x)}\quad\text{for all $x \in \mathbb{R}_+$,}\]

where

\[B(x)\;:\!=\; \int^x \dfrac{2\mu(y)}{\sigma^2(y)}\,{\textrm{d}} y,\]

see [Reference Borodin and Salminen7, p. 17]. From these definitions we find that

\[\sigma^2(x)=\dfrac{2}{S'(x)m'(x)}\quad\text{and}\quad \mu(x)=-\dfrac{S''(x)}{{S'}^2(x)m'(x)}\quad\text{for all $x \in \mathbb{R}^+$.}\]

In what follows, we assume that the functions $\mu$ and $\sigma^2$ are continuous. The assumption that the state space is $\mathbb{R}^+$ is done for convenience. In fact we could assume that the state space is any interval $\mathcal{I}$ in $\mathbb{R}$ and the subsequent analysis would hold with obvious modifications.

Denote the hitting time for set S as $\tau_S$ and the hitting time for point y as $\tau_y$ . Then we call a state in $E= \mathbb{R}^+$ regular if $\mathbb{P}(\tau_{(0, x)} = 0) = \mathbb{P}(\tau_{(x, \infty)} = 0) = 1$ . Under our assumptions, the process X is a regular linear diffusion and the speed measure m is absolutely continuous with respect to the Lebesgue measure. Thus we see that by [Reference Borodin and Salminen7, p. 13], all states in $\mathbb{R}_+$ are regular.

Then we let $\psi_r$ and $\varphi_r$ , respectively, denote the increasing and the decreasing solution of the second-order linear ordinary differential equation $\mathcal{A}u=ru$ , where $r>0$ , defined on the domain of the characteristic operator of X. The functions $\psi_r$ and $\varphi_r$ can be identified as the minimal r-excessive functions $\psi_r$ and $\varphi_r$ of X; see [Reference Borodin and Salminen7, pp. 18–20]. Finally, it is well known (see [Reference Borodin and Salminen7, p. 19]) that for a given $f\in L_1^r$ the resolvent $R_rf$ can be expressed as

(4.1) \begin{equation}(R_rf)(x)=B_r^{-1}\varphi_r(x)\int_0^x \psi_r(y)f(y)m'(y)\,{\textrm{d}} y+B_r^{-1}\psi_r(x)\int_x^\infty \varphi_r(y)f(y)m'(y)\,{\textrm{d}} y\end{equation}

for all $x \in \mathbb{R}_+$ , where

\[B_r=\dfrac{\psi_r'(x)}{S'(x)}\varphi_r(x)-\dfrac{\varphi_r'(x)}{S'(x)}\psi_r(x)\]

denotes the Wronskian determinant.

Next, we propose the class of payoff functions for which we study the problem (2.3). In what follows, we use the notation

(4.2) \begin{align} g_l(x) = g(x) - K_l \dfrac{r+\lambda}{\lambda}, \quad l=1,2,\end{align}

for brevity.

Assumption 4.1 is fairly weak and easy to verify; similar assumptions appear frequently in optimal stopping. Roughly speaking, the assumption means that the payoff g should be continuous and non-decreasing, and satisfy suitable limiting conditions at the boundaries. Furthermore, the payoff can be unbounded but the rate of growth is constrained by item (3). The assumptions are similar to the irreversible problem with exercise lag studied in [Reference Lempa17].

We show that that the problem specification of this section satisfies Assumptions 3.2 and 3.3; Assumption 3.1 is assumed to hold. To this end, we recall the definition of the fine topology. A set A is finely open with respect to a process X if for each $x \in A$ there exists a nearly Borel set $B \subset A$ such that $\mathbb{P}_x(\tau_{B^c} > 0) = 1$ . We prove the following lemma.

Proof. First, let A be an open set. Thus, for each $x \in A$ , there exists a ball $B(x, \delta) \subset A$ for some $\delta > 0$ . Now, since the paths of X are almost surely continuous, it follows that for each $\omega \in \Omega$ , barring a zero-measured set of exceptions, there exists $t_\delta(\omega)$ such that $X_t(\omega) \in B(x, {{\delta}/{2}})$ for all $0 \leq t \leq t_\delta(\omega)$ . Consequently, $\tau_{B(x, \delta)^c}(\omega) \geq \tau_{B(x, {{\delta}/{2}})^c}(\omega) \geq t_\delta(\omega) > 0$ , yielding

\[\mathbb{P}_x(\tau_{b(x,\delta)^c} > 0) = 1.\]

Since this is true for all $x \in A$ , the set A is finely open by definition.

Next, assume that A is not open. Then there exists a sequence $x_i$ of points in $A^c$ that converges to some $x \in A$ . Either an infinite number of those points are on the left side of x, or an infinite number of them are on the right side of x. By symmetry, let us assume the left side of x has infinite number of points $x_i$ . Then those points form a subsequence that also converges to x.

Now, since x is a regular point, the stopping time $\tau_{(0, x)} = 0$ almost surely. Thus, if $t > 0$ , for each $\omega$ , barring a zero-measured set of exceptions, there exists a time index $u(\omega) < t$ such that $X_{u(\omega)}(\omega) < x$ . There exists a member of $x_i$ such that $X_{u(\omega)}(\omega) < x_{i(\omega)} < x$ , implying that

\[\tau_{B^c} \leq \tau_{A^c} \leq \tau_{x_i} \leq u < t\]

for chosen $\omega$ , where the inequality $\tau_{x_i} \leq u$ follows from the path of $X_t$ being almost surely continuous (the exceptions to this can also be disregarded as a null set). Since the claim is true for any nearly Borel $B \subset A$ and for any $t > 0$ , it follows that $\tau_{B^c} = 0$ almost surely for any nearly Borel $B \subset A$ , implying that A is not finely open, and completing the proof.

As an immediate consequence we get the following.

We have shown that fine continuity and continuity are equivalent in one-dimensional Markov processes satisfying the assumptions of Lemma 4.1. Thus the next result follows by [Reference Shiryaev28, Theorem 5, p. 135].

The process X does satisfy the assumptions of Lemma 4.1, so we only need to show that Assumption 3.4 holds. By [Reference Øksendal22, Lemma 8.1.4, p. 143] the function $u(t, x) = \mathbb{E}_x[f(X_t)]$ is continuous for all bounded and continuous functions f. Then, if $\varepsilon > 0$ and M is an upper bound for $|g(x)|$ , there exists a number $T > 0$ such that

\[\int_T^{\infty}\,{\textrm{e}}^{-rt} \,{\textrm{d}} t = \dfrac{1}{r}\,{\textrm{e}}^{-rT} < \dfrac{\varepsilon}{4M}.\]

Then, if $\delta > 0$ is such that for all $y \in B(x, \delta)$ we have $|g(x)-g(y)|<{{r\varepsilon}/{2}}$ , we get

\begin{align*} |(R_rg)(x)-(R_rg)(y)|&\leq \int_0^T \,{\textrm{e}}^{-rt}|u(t,x) - u(t, y)| \,{\textrm{d}} t\\[5pt] &\quad +\biggl|\mathbb{E}_x\biggl[\int_T^{\infty} \,{\textrm{e}}^{-rt}g(X_t) \,{\textrm{d}} t\biggr] - \mathbb{E}_y\biggl[\int_T^{\infty} \,{\textrm{e}}^{-rt}g(X_t) \,{\textrm{d}} t\biggr]\biggr| \\ &\leq \int_0^T \,{\textrm{e}}^{-rt}\dfrac{r\varepsilon}{2} \,{\textrm{d}} t + 2\int_T^{\infty} M\,{\textrm{e}}^{-rt} \,{\textrm{d}} t \\& = \dfrac{\varepsilon(1-{\textrm{e}}^{-rT})}{2}+\dfrac{\varepsilon}{2} \\[5pt] & < \varepsilon.\end{align*}

Thus the resolvent $(R_rg) (x)$ is also continuous when g is bounded, as also claimed by [Reference Øksendal22, Lemma 8.1.3, p. 143]. Now, if g is continuous and the ratio ${{g}/{h}}$ is bounded for some r-harmonic function h, we get

\begin{align*} (R_{r+\lambda}g)(x) = \int_0^{\infty}\mathbb{E}_x[{\textrm{e}}^{-(r+\lambda)t}g(X_t)] \,{\textrm{d}} t = \int_0^{\infty}\mathbb{E}_x^h\biggl[{\textrm{e}}^{-(\lambda)t}\dfrac{g(X_t)}{h(X_t)}\biggr] \,{\textrm{d}} t = R_{\lambda}^h\biggl(\dfrac{g}{h}\biggr)(x)\end{align*}

in other words, the resolvent can be represented in terms of another resolvent in the h-transformed space, where the argument is ${{g}/{h}}$ and the rate is $\lambda$ . That resolvent is continuous, since its argument is bounded and continuous. Thus we have shown the following proposition.

To close the subsection, we present the following lemma without a proof, as it follows from the representation (4.1) and [Reference Lempa16, Lemma 2.1] by means of differentiation.

Lemma 4.2. Let $f\in L_1^r$ . Then

\begin{align*} \psi_r'(x)\varphi_{r+\lambda}(x) -\psi_r(x)\varphi_{r+\lambda}'(x) &= \lambda S'(x)(\Phi \psi_r)(x), \\[5pt] \lambda(R_{r+\lambda}f)'(x)\varphi_{r+\lambda}(x)-\lambda(R_{r+\lambda}f)(x)\varphi_{r+\lambda}'(x)& = \lambda S'(x)(\Phi f)(x), \\[5pt] \lambda(R_{r+\lambda}f)'(x)\psi_{r}(x)-\lambda(R_{r}f)(x)\psi_{r}'(x) &= \dfrac{\lambda^2S'(x)}{B_{r+\lambda}}( (\Phi f)(x)(\Psi \psi_r)(x) - (\Psi f)(x)(\Phi \psi_r)(x)), \end{align*}

where

\begin{align*} (\Phi f)(x) = \int_x^\infty \varphi_{r+\lambda}(y)f(y)m'(y)\,{\textrm{d}} y, \quad (\Psi f)(x) = \int_0^x \psi_{r+\lambda}(y)f(y)m'(y)\,{\textrm{d}} y. \end{align*}

4.2. The solution

We start by first deriving a candidate solution to our main problem, then show that this function is in the function space $\hat{\textbf{B}}$ (for the r-harmonic function $\psi_r$ ), and finally show that it satisfies the Bellman principle of Proposition 3.1.

4.2.1. Deriving the candidate

Since we are dealing with a time-homogeneous problem, we start with the working assumption that the optimal policy is of the following type: consider thresholds $y_1>y_2$ and the rule that

• • an inactive investor should engage the investment once the state variable X is above the threshold $y_1$ ,

• • an active investor should disengage the investment once the state variable X is below the threshold $y_2$ .

Denote the candidate solution for the inactive investor as $G_i$ . Then the rule described above can be expressed as

(4.3) \begin{align} G_i(x) = \psi_r(x)\mathbb{E}_x^{\psi_r}\biggl[ \biggl( \mathbb{E}_{X_{\tau_{y_1}}}^{\psi_r} \biggl[ \dfrac{g(X_U)}{\psi_r(X_U)}\textbf{1}(U < \sigma_{y_2}) + \dfrac{G_i(X_{\sigma_{y_2}})+K_2}{\psi_r(X_{\sigma_{y_2}})} \textbf{1}(U > \sigma_{y_2}) \biggr] - \dfrac{K_1}{\psi_r(X_{\tau_{y_1}})}\biggr) \biggr].\end{align}

This condition is expressed in terms of the $\psi_r$ -transform; we have already used this way of writing in the previous section. The reason for this is the same as above: it lends itself well to fixed point arguments. The following lemma tells us that it is reasonable to work with the condition (4.3) in the first place.

Lemma 4.3. There is a unique continuous function $G_i$ satisfying the condition (4.3) such that ${{G_i}/{\psi_r}}$ is bounded.

Proof. Recall the function space $\textbf{B}$ . For $y_1 > y_2$ , define the operator $\Theta\colon \textbf{B} \rightarrow \textbf{B}$ as $(\Theta f) = J(\cdot;\tau_{y_1},\sigma_{y_2},f)$ ; see the definition of the operator $\Lambda$ in (3.5). Take $f_1,f_2 \in \textbf{B}$ . Then

\begin{align*} \| \Theta f_1 - \Theta f_2 \|_\textbf{B} &= \sup\nolimits_x \biggl| \mathbb{E}_x^{\psi_r}\biggl[ \mathbb{E}_{y_1}^{\psi_r}\biggl[ \biggl(\dfrac{f_1(y_2)}{\psi_r(y_2)} - \dfrac{f_2(y_2)}{\psi_r(y_2)} \biggr) \textbf{1}(U > \sigma_{y_2}) \biggr] \biggr] \biggr| \\[5pt] & \leq \eta \|\, f_1 - f_2 \|_\textbf{B},\end{align*}

where

\begin{align*} \eta = \sup\nolimits_x \mathbb{E}_x^{\psi_r}\bigl[ \mathbb{E}_{y_1}^{\psi_r}[ \textbf{1}(U > \sigma_{y_2}) ] \bigr] < 1.\end{align*}

Thus there is a unique fixed point $G_i$ to the operator $\Theta$ .

By reversing the $\psi_r$ -transform, we rewrite the condition (4.3) as

(4.4) \begin{equation} G_i(x) = \mathbb{E}_x\bigl[ {\textrm{e}}^{-r\tau_{y_1}}(G_a(X_{\tau_{y_1}}) - K_1) \bigr],\\\end{equation}

where

\begin{equation*} G_a(x) = \mathbb{E}_{x} \bigl[ {\textrm{e}}^{-rU}g(X_U)\textbf{1}(U < \sigma_{y_2}) + {\textrm{e}}^{-r\sigma_{y_2}}(G_i(X_{\sigma_{y_2}})+K_2) \textbf{1}(U > \sigma_{y_2}) \bigr].\end{equation*}

Let $x < y_1$ and stopping time $\tau < \tau_{y_1}$ . Then the condition (4.4) and the strong Markov property yield

\begin{align*} G_i(x) = \mathbb{E}_x[ {\textrm{e}}^{-r\tau} G_i(X_\tau) ].\end{align*}

In other words, the function $G_i$ is r-harmonic for $x<y_1$ , so we can write $G_i(x) = A\psi_r(x) + A'\varphi_r(x)$ , where A and A ^′ are constants, for $x<y_1$ . Since ${{G_i}/{\psi_r}}$ is bounded, the constant $A' = 0$ . Moreover, since $G_i$ is continuous, we can write

\begin{equation*} G_i(x) = \begin{cases} G_a(x) - K_1, & x \geq y_1 , \\ A \psi_r(x), & x < y_1. \end{cases}\end{equation*}

Let $x \geq y_1$ . Then

\begin{align*} G_a(x) &= \mathbb{E}_{x} \bigl[ {\textrm{e}}^{-rU}g(X_U)\textbf{1}(U < \sigma_{y_2}) + {\textrm{e}}^{-r\sigma_{y_2}}(G_i(X_{\sigma_{y_2}})+K_2) \textbf{1}(U > \sigma_{y_2}) \bigr] \\[5pt] &= \lambda(R_{r+\lambda}g)(x) + \mathbb{E}_x\bigl[ {\textrm{e}}^{-(r+\lambda)\sigma_{y_2}}(G_i(y_2) + K_2 - \lambda(R_{r+\lambda}g)(y_2)) \bigr] \\[5pt] &= \lambda(R_{r+\lambda}g)(x) + ( A\psi_r(y_2) + K_2 - \lambda(R_{r+\lambda}g)(y_2) )\dfrac{\varphi_{r+\lambda}(x)}{\varphi_{r+\lambda}(y_2)}.\end{align*}

By continuity, this yields

\begin{align*} A\psi_r(y_1) = \lambda(R_{r+\lambda}g)(y_1) + ( A\psi_r(y_2) + K_2 - \lambda(R_{r+\lambda}g)(y_2) )\dfrac{\varphi_{r+\lambda}(y_1)}{\varphi_{r+\lambda}(y_2)} - K_1.\end{align*}

Solving for A, we find that

\begin{align*} A = A(y_1,y_2) = \dfrac{(K_2-\lambda(R_{r+\lambda}g)(y_2))\varphi_{r+\lambda}(y_1) - (K_1-\lambda(R_{r+\lambda}g)(y_1))\varphi_{r+\lambda}(y_2)}{\psi_r(y_1)\varphi_{r+\lambda}(y_2) - \psi_r(y_2)\varphi_{r+\lambda}(y_1)} .\end{align*}

Summarizing, we can write the candidate solutions as

(4.5) \begin{equation} G_i(x) = \begin{cases} \lambda(R_{r+\lambda}g)(x) + C(y_1,y_2) \varphi_{r+\lambda}(x) - K_1, & x \geq y_1, \\ A(y_1,y_2) \psi_r(x), & x > y_1, \end{cases}\end{equation}

and

(4.6) \begin{equation} G_a(x) = \begin{cases} \lambda(R_{r+\lambda}g)(x) + C(y_1,y_2) \varphi_{r+\lambda}(x), & x \geq y_2 ,\\ A(y_1,y_2) \psi_r(x) + K_2, & x > y_2, \end{cases}\end{equation}

where

\begin{align*} C(y_1,y_2) &= \dfrac{1}{\varphi_{r+\lambda}(y_2)} (A(y_1,y_2)\psi_r(y_2) + K_2 - \lambda(R_{r+\lambda}g)(y_2)) \\[5pt] &= \dfrac{(K_2-\lambda(R_{r+\lambda}g)(y_2))\psi_r(y_1) - (K_1-\lambda(R_{r+\lambda}g)(y_1))\psi_r(y_2)}{\psi_r(y_1)\varphi_{r+\lambda}(y_2) - \psi_r(y_2)\varphi_{r+\lambda}(y_1)}.\end{align*}

To find optimal thresholds, we impose the smooth-pasting condition: we assume that the candidate functions are continuously differentiable over their respective boundaries. This leads to the conditions

(4.7) \begin{equation}\begin{split} \lambda(R_{r+\lambda}g)'(y_1) + C(y_1,y_2) \varphi_{r+\lambda}'(y_1) &= A(y_1,y_2) \psi_r'(y_1), \\[5pt] \lambda(R_{r+\lambda}g)'(y_2) + C(y_1,y_2) \varphi_{r+\lambda}'(y_2) &= A(y_1,y_2) \psi_r'(y_2).\end{split}\end{equation}

A simple computation yields

\begin{align*} \lambda(R_{r+\lambda}g)(x) - K_l = \lambda (R_{r+\lambda}g_l)(x);\end{align*}

recall expression (4.2). Using this notation, we readily verify that the necessary conditions (4.7) can be rewritten as

(4.8) \begin{equation}\begin{split} \lambda(R_{r+\lambda}g_2)(y_2) G(y_1) + F_2(y_1) \varphi_{r+\lambda}(y_2) &= F_1(y_1) \psi_r(y_2), \\[5pt] \lambda(R_{r+\lambda}g_1)(y_1) I(y_2) + J_2(y_2) \varphi_{r+\lambda}(y_1) &= J_1(y_2) \psi_r(y_1),\end{split}\end{equation}

where

\begin{align*} F_1(x)&=\lambda(R_{r+\lambda}g_1)(x)\varphi_{r+\lambda}'(x) - \lambda(R_{r+\lambda}g_1)'(x)\varphi_{r+\lambda}(x), \\[5pt] F_2(x)&=\lambda(R_{r+\lambda}g_1)(x)\psi_r'(x) - \lambda(R_{r+\lambda}g_1)'(x)\psi_r(x), \\ G(x) &= -I(x) = \psi_r'(x)\varphi_{r+\lambda}(x)-\psi_r(x)\varphi_{r+\lambda}'(x), \\ J_1(x)&=\lambda(R_{r+\lambda}g_2)'(x)\varphi_{r+\lambda}(x) - \lambda(R_{r+\lambda}g_2)(x)\varphi_{r+\lambda}'(x), \\[5pt] J_2(x)&=\lambda(R_{r+\lambda}g_2)'(x)\psi_r(x) - \lambda(R_{r+\lambda}g_2)(x)\psi_r'(x).\end{align*}

We observe that the conditions (4.8) can be further expressed as

\begin{align*} \lambda(R_{r+\lambda}g)(y_2)+\dfrac{F_2(y_1)}{G(y_1)}\varphi_{r+\lambda}(y_2) &= \dfrac{F_1(y_1)}{G(y_1)}\psi_r(y_2) + K_2, \\[5pt] \lambda(R_{r+\lambda}g)(y_1)+\dfrac{J_2(y_2)}{I(y_2)}\varphi_{r+\lambda}(y_1) - K_1& = \dfrac{J_1(y_2)}{I(y_2)}\psi_r(y_1).\end{align*}

By coupling these with expressions (4.5) and (4.6), we obtain the conditions

\begin{equation*}\begin{split} \dfrac{F_2(y_1)}{G(y_1)} &= \dfrac{J_2(y_2)}{I(y_2)}, \\[5pt] \dfrac{F_1(y_1)}{G(y_1)} &= \dfrac{J_1(y_2)}{I(y_2)}.\end{split}\end{equation*}

Using Lemma 4.2, these can be further expressed as

(4.9) \begin{equation} \begin{split} H_1(y_1) &= H_2(y_2), \\[5pt] R_1(y_1) &= R_2(y_2), \end{split}\end{equation}

where

\begin{align*} H_l(x)&= \dfrac{(\Phi g_l)(y_l)}{(\Phi \psi_r)(y_l)}, \\[5pt] R_l(x) &= (\Phi g_l)(y_l)\dfrac{(\Psi \psi_r)(y_l)}{(\Phi \psi_r)(y_l)} - (\Psi g_l)(y_l),\end{align*}

for $l=1,2$ .

To analyse the solvability of the pair (4.9), we prove some auxiliary results. The next lemma shows that the fractions

\[\dfrac{(\Phi g_l)(x)}{(\Phi \psi_r)(x)}\quad\text{and}\quad \dfrac{(\Psi g_l)(x)}{(\Psi \psi_r)(x)}\]

have properties very similar to those of

\[\dfrac{g_l(x)}{\varphi_r(x)}\quad\text{and}\quad \dfrac{g_l(x)}{\psi_r(x)}\]

in Assumption 4.1.

Lemma 4.4. There exist unique states $\hat{x}_l < x^*_l$ and $\check{x}_l > x^*_l$ such that

\[\hat{x}_l=\mathrm{argmax} \biggl\{ \dfrac{\Phi g_l}{\Phi \psi_r} \biggr\}\quad{and}\quad\check{x}_l=\mathrm{argmax} \biggl\{ \dfrac{\Psi g_l}{\Psi \psi_r} \biggr\}\]

and the functions

\[x \mapsto \dfrac{(\Phi g_l)(x)}{(\Phi \psi_r)(x)}\quad{and}\quad x \mapsto \dfrac{(\Psi g_l)(x)}{(\Psi \psi_r)(x)}\]

are non-decreasing on $(0, \hat{x}_l)$ and $(0, \check{x}_l)$ , and non-increasing on $(\hat{x}_l,\infty)$ and $(\check{x}_l,\infty)$ . Furthermore, $\hat{x}_2 < \hat{x}_1$ and $\check{x}_2 < \check{x}_1$ .

Proof. The main claim is [Reference Lempa17, Lemma 3.4]. Thus we show that $\hat{x}_2 < \hat{x}_1$ , and then $\check{x}_2 < \check{x}_1$ follows similarly.

Let $\hat{x}_1$ be the unique maximum of the function ${{(\Phi g_1)}/{(\Phi \psi_r)}}$ , so that it satisfies the equation

\[\psi_r(\hat{x}_1) (\Phi g_1)(\hat{x}_1) = g_1(\hat{x}_1)(\Phi \psi_r)(\hat{x}_1).\]

Then, because $g_2 = g_1 + \Omega$ , where $\Omega = (K_1 - K_2) {{(r+\lambda)}/{\lambda}}$ , we find that

\begin{align*}& \psi_r(\hat{x}_1) (\Phi g_2)(\hat{x}_1) - g_2(\hat{x}_1)(\Phi \psi_r)(\hat{x}_1) \\[5pt] &\quad < \psi_r(\hat{x}_1) (\Phi g_1)(\hat{x}_1) - g_1(\hat{x}_1)(\Phi \psi_r)(\hat{x}_1) + \Omega (\psi_r(\hat{x}_1) (\Phi 1)(\hat{x}_1) - (\Phi \psi_r)(\hat{x}_1)) \\[5pt] &\quad < 0,\end{align*}

where the last inequality follows because $\psi_r$ is increasing. Hence the point $\hat{x}_1$ is on the part where

\[x \mapsto \dfrac{(\Phi g_2)(x)}{(\Phi \psi_r)(x)}\]

is decreasing, which implies that $\hat{x}_2 < \hat{x}_1$ .

The following lemma summarizes the needed properties of the functions $H_l$ and $R_l$ , $l=1,2$ .

Lemma 4.5. Let $\hat{x}_1$ and $\hat{x}_2$ be as in Lemma 4.4. Then

1. (1) $H_l'(x) = -\dfrac{\varphi_{r+\lambda}(x)m'(x)}{(\Phi \psi_r)(x)^2} (\psi_r(x)(\Phi g_l)(x)-(\Phi \psi_r)(x) g_l(x))$ ,

2. (2) $R_l'(x) = \dfrac{B_{r+\lambda } \psi_r(x) m'(x)}{\lambda (\Phi \psi_r)(x)^2} (\psi_r(x)(\Phi g_l)(x)-(\Phi \psi_r)(x) g_l(x))$ ,

3. (3) $H_2(\hat{x}_2) < H_1(\hat{x}_1)$ ,

4. (4) $R_2(\hat{x}_2) < R_1(\hat{x}_1)$ ,

5. (5) $\lim_{x \to 0} H_2(x) > 0$ and $\lim_{x \to 0} R_2(x) < 0$ .

Proof. Parts (1) and (2) follow by straightforward differentiation and the formula (4.1).

For parts (3) and (4), since $g_1(x) < g_2(x)$ we find that $H_2(x) < H_1(x)$ for all x. By Lemma 4.4 we have $\hat{x}_2 < \hat{x}_1$ and thus $H_2(\hat{x}_2) < H_2(\hat{x}_1)$ . We find that

\[ H_2(\hat{x}_2) < H_2(\hat{x}_1) < H_1(\hat{x}_1).\]

The proof of part (4) is analogous.

For part (5), let $x < x_0$ . Then we find that

\begin{equation*} (\Phi g_2)(x) = \int_x^{x_0} g_2(z) \varphi_{r+\lambda}(z) m'(z) \,{\textrm{d}} z + \int_{x^0}^{\infty } g_2(z) \varphi_{r+\lambda}(z) m'(z) \,{\textrm{d}} z.\end{equation*}

The last integral is finite since $g_2 \in L_1^r$ , and for the first we find by the mean value theorem that

\begin{equation*} \int_x^{x_0} g_2(z) \varphi_{r+\lambda}(z) m'(z) \,{\textrm{d}} z = g_2(\xi) \int_x^{x_0} \varphi_{r+\lambda}(z) m'(z) \,{\textrm{d}} z = g_2(\xi) \bigg( \dfrac{\varphi_{r+\lambda}'(x_0)}{S'(x_0)} -\dfrac{\varphi_{r+\lambda}'(x)}{S'(x)} \bigg),\end{equation*}

where $\xi \in [x, x_0]$ . As $g_2(\xi)<0$ and

\[\lim_{x \to 0} -\dfrac{\varphi_{r+\lambda}'(x)}{S'(x)} = \infty\]

(the lower boundary is natural), we find that $\lim_{x \to 0 } (\Phi g_2)(x) = - \infty$ . As $\psi_r$ is increasing, we find by similar calculations that $\lim_{x \to 0 } (\Phi \psi_r)(x) = \infty$ . Thus L’Hôpital’s rule with part (4) of Assumption 4.1 yields

\begin{equation*} \lim_{x \to 0} H_2 (x) = \lim_{x \to 0} - \dfrac{g_2(x) \varphi_{r+\lambda}(x)m'(x)}{\psi_r(x)\varphi_{r+\lambda}(x)m'(x)} = - \lim_{x \to 0} \dfrac{g_2(x)}{\psi_r(x)} \geq 0.\end{equation*}

The proof for the limit of $R_1(x)$ is analogous.

We are now in a position to prove that the pair (4.9) has a unique solution under our assumptions. With the next result, we can continue our analysis with a unique candidate function $G_i$ .

Proposition 4.2. Let Assumption 4.1 hold. Then there exists a unique solution to the pair of equations given by (4.9).

Proof. Let $\hat{\cdot}$ and $\check{\cdot}$ denote the restrictions to the domains $(0, \hat{x}_2)$ and $(\hat{x}_1,\infty)$ respectively. Define the function $K\colon (0, \hat{x}_2) \to (0, \hat{x}_2)$ as

\begin{equation*} K(x) = \bigl(\hat{H}_2^{-1} \circ \check{H}_1 \circ \check{R}_1^{-1} \circ \hat{R}_2\bigr)(x).\end{equation*}

We notice that $y_2$ is the fixed point of K if and only if the pair $(y_2, y_1)$ , where $y_1 = \bigl(\check{R}_1^{-1} \circ \hat{R}_2\bigr)(x)$ , solves the pair of equations (4.9). By Lemma 4.5 the function K is well-defined. A direct differentiation and parts (1) and (2) of Lemma 4.5 yield

\begin{equation*} K'(x) = \bigl(\hat{H}_2^{-1 '} \circ \check{H}_1 \circ \check{R}_1^{-1} \circ \hat{R}_2\bigr)(x) \cdot \bigl(\check{H}_1' \circ \check{R}_1^{-1} \circ \check{H}_2\bigr)(x) \cdot \bigl(\check{R}_1^{-1'} \circ \hat{R}_2\bigr)(x) \cdot \hat{R}_2'(x) > 0,\end{equation*}

showing that K is increasing. Thus K is a monotonic mapping from an interval to its open subset and must have a fixed point. We denote this fixed point by $y_2$ . Consequently, the pair $(y_2, y_1)$ , where $y_1 = (\check{R}_1^{-1} \circ \hat{R}_2)(x)$ , gives a solution to the pair of equations (4.9).

To prove uniqueness we use the fixed point property of $y_2$ and find by Lemma 4.5 parts (1) and (2) that

\begin{equation*} K'(y_2) = \dfrac{H_1'(y_1)}{R_1'(y_1)} \dfrac{R_2'(y_2)}{H_2'(y_2)} = \dfrac{\psi_r(y_2)}{\psi_r(y_1)} \dfrac{\varphi_{r+\lambda}(y_1)}{\varphi_{r+\lambda}(y_2)} < 1.\end{equation*}

Thus, whenever K intersects the diagonal of $\mathbb{R}_+$ , the intersection must be from above.

4.2.2. Candidate belongs to $\hat{\mathbf{B}}$

To use Proposition 3.1 to show that the candidate value is the actual value of our main problem, we need to show that $G_i \in \hat{\textbf{B}}$ . To this end, we first prove the following lemma.

Now we are ready to prove the following proposition.

Proposition 4.3. The candidate function $G_i$ belongs to the space $\hat{\textbf{B}}$ .

Proof. To complete the proof, we have to show that the following claims are true:

1. (1) $G_i$ is continuous,

2. (2) ${{G_i}/{h}}$ is bounded,

3. (3) $\Lambda G_i \geq G_i$ .

Claim (1) is true since $G_i$ is readily known to be continuous. Claim (3) is also trivially true, since it is known that $\Lambda G_i = G_i$ . It is left to show that claim (2) holds.

First, let $x \geq y_1$ . Then, for some constants a and b, we have

\begin{align*} \dfrac{G_i(x)}{h(x)} &= \dfrac{\lambda( R_{r+\lambda}g_1)(x) + C\varphi_{r+\lambda}(x)}{a\psi_r(x) + b\varphi_r(x)} \\[5pt] &\leq \dfrac{\lambda (R_{r+\lambda}g_1)(x)}{a \psi_r(x)} + \dfrac{C\varphi_{r+\lambda}(x)}{a\psi_r(x)} \\[5pt] &= \dfrac{\lambda (R_{r+\lambda}g_1)(x)}{a \psi_r(x)} + \dfrac{C\varphi_{r+\lambda}(y_1)}{a\psi_r(y_1)}.\end{align*}

Now, since ${{\lambda (R_{r+\lambda}g_1)}/{(a \psi_r)}}$ is bounded by Lemma 3.4, the rightmost function is also bounded. Thus ${{G_i(x)}/{h(x)}}$ is bounded for $x \geq y_1$ .

Then let $x \leq y_1$ . Now

\begin{align*} \dfrac{G_i(x)}{h(x)} = \dfrac{A \psi_r(x)}{a\psi_r(x) + b\varphi_r(x)} \leq \dfrac{A \psi_r(x)}{a \psi_r(x)} &= \dfrac{A}{a}\end{align*}

and thus ${{G_i(x)}/{h(x)}}$ is also bounded for $x \leq y_1$ . This completes the proof.

4.2.3. Candidate satisfies the Bellman principle

Knowing that $G_i \in \hat{\textbf{B}}$ , we show next that it satisfies the Bellman principle. We start by proving a series of lemmas.

Lemma 4.6. Define the functions $f_{\psi}\colon \mathbb{R}_+ \to \mathbb{R}$ and $f_{\varphi}\colon \mathbb{R}_+ \to \mathbb{R}$ as

\begin{equation*} f_{\psi}(x) = \dfrac{\lambda(R_{r+\lambda}g_1)(x)-C(y_1, y_2)\varphi_{r+\lambda}(x)}{\psi_r(x)}, \quad f_{\varphi}(x) = \dfrac{H_1(x) - C(y_1, y_2)\varphi_{r+\lambda}(x)}{\varphi_r(x)}. \end{equation*}

Then the following claims are true:

1. (i) $f_{\psi}$ is non-increasing at $x \geq y_1$ ,

2. (ii) $f_{\psi}$ is non-decreasing at x for $y_2 \leq x \leq y_1$ ,

3. (iii) $f_{\varphi}$ is non-decreasing at x for $x \geq y_1$ .

Proof. Define the function

\begin{equation*} \hat{C}(x) = \dfrac{\lambda(R_{r+\lambda}g_1)'(x)\psi_r(x)-\lambda(R_{r+\lambda}g_1)(x)\psi_r'(x)}{\varphi_{r+\lambda}'(x) \psi_r(x)-\varphi_{r+\lambda}(x) \psi_r'(x)}. \end{equation*}

Then, differentiating and rearranging the terms, we find that the inequality $f_{\psi}'(x) \geq 0$ is equivalent to

(4.10) \begin{equation} \hat{C}(x) \leq \hat{C}(y_1). \end{equation}

This inequality holds if the function $\hat{C}$ is increasing on the interval $[x,y_1]$ . In addition, if $\hat{C}(x)$ is also increasing in $[y_1, \infty)$ , the converse of (4.10) holds for $x \geq y_1$ , which would complete the proof of claim (i). Proceeding as in the derivation of (4.9), we can re-express $\hat{C}(x)$ as

(4.11) \begin{equation} \hat{C}(x) = - \dfrac{\lambda}{B_{r+\lambda}} R_1(x). \end{equation}

Thus, by Lemma 4.5 (2), we find that

(4.12) \begin{equation} \hat{C}'(x) \geq 0 \quad \text{if and only if}\quad \dfrac{g_1(x)}{\psi_r(x)} \geq \dfrac{(\Phi g_1)(x)}{(\Phi\psi_r)(x)}. \end{equation}

Since by the lemma (4.4) it holds that $\hat{x}_1 < y_1$ , we find that inequality (4.12) holds when $x \in [\hat{x}_1, \infty)$ . The same calculation also shows that $\hat{C}(x)$ is decreasing when $x \in [y_2, \hat{x}_1]$ . This implies that (i) is proved, and for (ii) it remains to be shown that (4.10) is true at $x = y_2$ .

Using the expression (4.11) for $\hat{C}(y_2)$ and the second equation in (4.9), we find that we need to show that

\begin{equation*} (\Psi g_1)(y_2) -\dfrac{ (\Phi g_1)(y_2)(\Psi \psi_r)(y_2)}{(\Phi \psi_r)(y_2)} \leq (\Psi g_2)(y_2) - \dfrac{ (\Psi g_2)(y_2)(\Phi \psi_r)(y_2)}{(\Phi \psi_r)(y_2)}. \end{equation*}

Since the boundaries of the state space are natural, we have

\begin{equation*} (\Psi 1)(y_2) = \dfrac{\psi_{r+\lambda}'(y_2)}{S'(y_2)}, \quad (\Phi 1)(y_2) = -\dfrac{\varphi_{r+\lambda}'(y_2)}{S'(y_2)}, \end{equation*}

which implies, together with the application of Lemma 4.2 to function G defined in (4.8), that

\begin{align*} & (\Psi g_1)(y_2) -\dfrac{ (\Phi g_1)(y_2)(\Psi \psi_r)(y_2)}{(\Phi \psi_r)(y_2)} - (\Psi g_2)(y_2) + \dfrac{ (\Phi g_2)(y_2)(\Psi \psi_r)(y_2)}{(\Phi \psi_r)(y_2)} \\[5pt] & \quad = -\Omega (\Psi 1)(y_2) + \dfrac{(\Psi \psi_r)(y_2)}{(\Phi \psi_r)(y_2)} \Omega (\Phi 1)(y_2) \\[5pt] & \quad = \dfrac{\Omega}{S'(y_2)} \bigg( -\psi_{r+\lambda}'(y_2) -\dfrac{\psi_{r+\lambda}(y_2)\psi_r'(y_2)-\psi_{r+\lambda}'(y_2)\psi_r(y_2)}{\varphi_{r+\lambda}'(y_2)\psi_r(y_2)-\varphi_{r+\lambda}(y_2)\psi_r'(y_2)} \varphi_{r+\lambda}'(y_2) \bigg), \end{align*}

where

\[\Omega = \dfrac{r+\lambda}{r}(K_1+K_2) = g_2(x)-g_1(x).\]

Because $\varphi_{r+\lambda}'(y_2)\psi_r(y_2)-\varphi_{r+\lambda}(y_2)\psi_r'(y_2) < 0$ , S ^′ and $\Omega$ are always positive, we find that the above expression is negative if and only if

\begin{align*} & -\psi_{r+\lambda}'(y_2)[\varphi_{r+\lambda}'(y_2)\psi_r(y_2)-\varphi_{r+\lambda}(y_2)\psi_r'(y_2) ]\\[5pt] &\quad -\varphi_{r+\lambda}'(y_2)[\psi_{r+\lambda}(y_2)\psi_r'(y_2)-\psi_{r+\lambda}'(y_2)\psi_r(y_2)] > 0. \end{align*}

Finally, cancelling the terms, we get the equivalent condition

\begin{equation*} \psi_{r+\lambda}'(y_2)\varphi_{r+\lambda}(y_2)\psi_r'(y_2) -\varphi_{r+\lambda}'(y_2) \psi_{r+\lambda}(y_2)\psi_r'(y_2) = B_{r+\lambda} \psi_r'(y_2) > 0, \end{equation*}

which holds since $\psi_r$ is increasing. This completes the proof for (ii).

By [Reference Saarinen26, Lemma 3.6], the mapping

\[x \mapsto \dfrac{\varphi_{r+\lambda}(x)}{\varphi_{r}(x)}\]

is decreasing, which implies that

\begin{align*} \varphi_{r+\lambda}'(x)\varphi_{r}(x) - \varphi_{r+\lambda}(x)\varphi_{r}'(x) &= \varphi_r^2(x) \biggl( \dfrac{\varphi_{r+\lambda}(x)}{\varphi_{r}(x)}\biggr)' < 0. \end{align*}

To show that $f_{\varphi}(x)$ is non-decreasing at $x \geq y_1$ , we first find by differentiation that

\begin{align*} f_{\varphi}'(x) &= \dfrac{(H_1'(x)\varphi_r(x) - H_1(x)\varphi_r'(x)) - C(y_1, y_2)((\varphi_{r+\lambda}'(x)\varphi_r(x) - \varphi_{r+\lambda}(x)\varphi_r'(x)))}{\varphi_r^2(x)}. \end{align*}

Thus the claim is equivalent to

\begin{align*} (H_1'(x)\varphi_r(x) - H_1(x)\varphi_r'(x)) - C(y_1, y_2)((\varphi_{r+\lambda}'(x)\varphi_r(x) - \varphi_{r+\lambda}(x)\varphi_r'(x))) &\geq 0. \end{align*}

Noting that $C(y_1, y_2) = \hat{C}(y_1)$ , we find after rearrangement that the above is equivalent to

\begin{align*} \dfrac{\varphi_r'(x)H_1(x) - \varphi_r(x)H_1'(x)}{\varphi_r'(x)\varphi_{r+\lambda}(x) - \varphi_r(x)\varphi_{r+\lambda}'(x)} &\leq \dfrac{\psi_r'(y_1)H_1(y_1) - \varphi_r(y_1)H_1'(y_1)}{\varphi_r'(y_1)\varphi_{r+\lambda}(y_1) - \varphi_r(y_1)\varphi_{r+\lambda}'(y_1)}. \end{align*}

Similar to (4.11), we can re-express the inequality as

(4.13) \begin{align} \dfrac{(\Psi\varphi_r)(x)}{(\Phi \varphi_r)(x)}(\Phi g_1)(x) - (\Psi g_1)(x) &\geq \dfrac{(\Psi\psi_r)(y_1)}{(\Phi \psi_r)(y_1)}(\Phi g_1)(y_1) - (\Psi g_1)(y_1). \end{align}

We now show that (4.13) holds at $x=y_1$ and that the left-hand side is non-decreasing with respect to x. At $x = y_1$ it suffices to show that

\begin{equation*} \dfrac{(\Psi\varphi_r)(y_1)}{(\Phi \varphi_r)(y_1)} \geq \dfrac{(\Psi\psi_r)(y_1)}{(\Phi \psi_r)(y_1)}. \end{equation*}

Now, since $\varphi_r$ is strictly decreasing and $\psi_r$ strictly increasing, we have

\[\dfrac{\varphi(z)}{\varphi(y_1)} \geq 1 \geq \dfrac{\psi_r(z)}{\psi_r(y_1)}\]

whenever $z \leq y_1$ and

\[\dfrac{\varphi(z)}{\varphi(y_1)} \leq 1 \leq \dfrac{\psi_r(z)}{\psi_r(y_1)}\]

when $z \geq y_1$ . Thus

\begin{align*} \dfrac{(\Psi\varphi_r)(y_1)}{(\Phi \varphi_r)(y_1)} &= \dfrac{\int_0^{y_1}\varphi_r(z)\psi_{r+\lambda}(z)m'(z)\,{\textrm{d}} z}{\int_{y_1}^{\infty}\varphi_r(z)\varphi_{r+\lambda}(z)m'(z)\,{\textrm{d}} z} \\[5pt] &= \dfrac{\int_0^{y_1}\frac{\varphi(z)}{\varphi(y_1)}\psi_{r+\lambda}(z)m'(z)\,{\textrm{d}} z}{\int_{y_1}^{\infty}\frac{\varphi(z)}{\varphi(y_1)}\varphi_{r+\lambda}(z)m'(z)\,{\textrm{d}} z} \\ &\geq \dfrac{\int_0^{y_1}\frac{\psi_r(z)}{\psi_r(y_1)}\psi_{r+\lambda}(z)m'(z)\,{\textrm{d}} z}{\int_{y_1}^{\infty}\frac{\psi_r(z)}{\psi_r(y_1)}\varphi_{r+\lambda}(z)m'(z)\,{\textrm{d}} z}\\& = \dfrac{(\Psi\psi_r)(y_1)}{(\Phi \psi_r)(y_1)}. \end{align*}

Denote the left-hand side of equation (4.13) by u, that is,

\begin{equation*} u(x) = \dfrac{(\Psi\varphi_r)(x)}{(\Phi \varphi_r)(x)}(\Phi g_1)(x) - (\Psi g_1)(x). \end{equation*}

The derivative of u can be expressed as

\begin{align*} u'(x) &= \dfrac{(\Phi g_1)'(x)(\Psi \varphi_r)(x)(\Phi \varphi_r)(x) + (\Phi g_1)(x)(\Psi \varphi_r)'(x)(\Phi \varphi_r)(x)}{(\Phi \varphi_r(x))^2} \\[5pt] &\quad - \dfrac{(\Phi g_1)(x)(\Psi \varphi_r)(x)(\Phi \varphi_r)'(x) + (\Psi g_1)'(x)(\Phi \varphi_r)^2(x)}{(\Phi \varphi_r(x))^2} \\[5pt] &= \dfrac{m'(x)}{(\Phi \varphi_r)^2(x)}[\varphi_{r+\lambda}(\Psi\varphi_r)(x) + \psi_{r+\lambda}(x)(\Phi \varphi_r)(x)]\, [\varphi_r(x)(\Phi g_1)(x) - g_1(x)(\Phi \varphi_r(x))(x) ]. \end{align*}

Since the first two factors are positive, it now suffices to prove

\begin{align*} \dfrac{(\Phi g_1)(x)}{(\Phi \varphi_r(x))(x)} &\geq \dfrac{g_1(x)}{\varphi_r(x)}. \end{align*}

Now, since ${{g_1(x)}/{\varphi_r(x)}}$ is non-decreasing (by item (4) of Assumption 4.1),

\begin{align*} (\Phi g_1)(x) &= \int_{x}^{\infty}(g_1(z)\varphi_{r+\lambda}(z)m'(z)\,{\textrm{d}} z) \\[5pt] &= \int_{x}^{\infty}\biggl(\dfrac{g_1(z)}{\varphi_r(z)}\varphi_r(z)\varphi_{r+\lambda}(z)m'(z)\,{\textrm{d}} z\biggr) \\ &\geq \int_{x}^{\infty}\biggl(\dfrac{g_1(x)}{\varphi_r(x)}\varphi_r(z)\varphi_{r+\lambda}(z)m'(z)\,{\textrm{d}} z\biggr) \\&= \dfrac{g_1(x)}{\varphi_r(x)}\int_{x}^{\infty}(\varphi_r(z)\varphi_{r+\lambda}(z)m'(z)\,{\textrm{d}} z) \\[5pt] &= \dfrac{g_1(x)}{\varphi_r(x)}(\Phi \varphi_r(x))(x), \end{align*}

which concludes the proof.

The following lemma gives a useful ordering of the candidate functions $G_i$ and $G_a$ .

Lemma 4.7. The candidate value functions $G_i$ and $G_a$ satisfy

\begin{equation*} G_a(x) - K_2 \geq G_i(x) \geq G_a(x)-K_1 \end{equation*}

for all $x \in \mathbb{R}_+.$

Proof. For $x \in (y_1, \infty)$ and for $x \in (0, y_2)$ , the claim holds by construction. Thus we assume that $x \in [y_2, y_1]$ for the rest of this proof.

By a direct calculation we find for $x \in [y_2, y_1]$ that

\begin{align*} G_i(x) - G_a(x) + K_1 &= \dfrac{\psi_r(x)}{\psi_r(y_1)} ( \lambda(R_{r+\lambda}g_1)(y_1) -C(y_1, y_2)\varphi_{r+\lambda}(y_1) ) \\[5pt] &\quad - \lambda(R_{r+\lambda}g_1)(x) + C(y_1, y_2)\varphi_{r+\lambda}(x). \end{align*}

Reorganizing the terms above, we find that the proposed inequality is equivalent to

\begin{equation*} f_{\psi}(x) \leq f_{\psi}(y_1), \end{equation*}

where

\begin{equation*} f_{\psi}(x) = \dfrac{\lambda(R_{r+\lambda}g_1)(x)-C(y_1, y_2)\varphi_{r+\lambda}(x)}{\psi_r(x)}. \end{equation*}

The claim then follows from part (ii) of Lemma 4.6.

Next, we show that $G_i$ is r-excessive and $G_a - \lambda (R_{r+\lambda}g_2)$ is $(r+\lambda)$ -excessive, in order to use the inequalities

\begin{equation*} G_i(x) \geq \mathbb{E}_x[{\textrm{e}}^{-r\tau}G_i(X_{\tau})]\end{equation*}

and

\begin{equation*} G_a(x) - \lambda (R_{r+\lambda}g_2) (x) \geq \mathbb{E}_x[{\textrm{e}}^{-(r+\lambda)\tau}(G_a(X_{\tau}) - \lambda (R_{r+\lambda}g_2) (X_{\tau}))]\end{equation*}

for all stopping times $\tau$ .

Lemma 4.8. The function $G_i$ is r-excessive.

Proof. Since $\lim_{x \to \infty} f_{\psi}(x) \geq 0$ , we find that that $G_i(x) \geq 0$ for all $x \in \mathbb{R}_+$ . We now show that

\[ \mathbb{E}_x[{\textrm{e}}^{-r \tau'} G_i(X_{\tau'})] \leq G_i(x), \]

where $\tau' = \inf\{t \geq 0 \mid X_t \in [c,d] \}$ , from which the r-excessivity follows by [Reference Borodin and Salminen7, p. 32]. When $x \in [c,d]$ , this condition trivially holds, so we consider the remaining cases (1) $x < c$ and (2) $x>d$ . We split these cases further depending on where $y_1$ is located. Let $x < c$ . Then for $c < y_1$ we find by Lemma 4.6 that

\begin{align*} \mathbb{E}_x[{\textrm{e}}^{-r \tau_c} G_i(X_{\tau_c})] & = \dfrac{\psi_r(x)}{\psi_r(c)} G_i(c) \\[5pt] & = \dfrac{\psi_r(x)}{\psi_r(c)} \psi_r(c) f_{\psi}(y_1) \\ & = \psi_r(x) f_{\psi}(y_1) \\[5pt] &= G_i(x), \end{align*}

and for $y_1 < c$ ,

\begin{align*} \mathbb{E}_x[{\textrm{e}}^{-r \tau_c} G_i(X_{\tau_c})] & = \dfrac{\psi_r(x)}{\psi_r(c)} G_i(c) \\[5pt] & = \dfrac{\psi_r(x)}{\psi_r(c)} \psi_r(c) f_{\psi}(\max\{y_1, c\}) \\ & = \psi_r(x) f_{\psi}(c) \\ & \leq \psi_r(x) f_{\psi}(\max\{x, y_1\}) \\[5pt] &= G_i(x). \end{align*}

The case $x>d$ is proved similarly by using part (iii) of Lemma 4.6.

Lemma 4.9. The function $G_a - \lambda (R_{r+\lambda}g)$ is $(r+\lambda)$ -excessive.

Proof. Recall that $G_a$ is defined as

\begin{equation*} G_a(x) = \begin{cases} \lambda (R_{r+\lambda}g)(x) + C(y_1, y_2)\varphi_{r+\lambda}(x), &\text{ if } x \geq y_2, \\ A\psi_r(x) + K_2, &\text{otherwise.} \end{cases}\end{equation*}

Then differentiation and rearranging yields

\begin{align*} (\mathcal{A} - (r+\lambda))(G_a(x) - \lambda R_{r+\lambda}g(x)) = \begin{cases} 0, &\text{ if }x \geq y_2, \\ -\lambda A \psi_r(x) + \lambda^2 g_2(x), &\text{ else. } \end{cases} \end{align*}

Thus we need to prove that $-\lambda A \psi_r(x) + \lambda^2 g_2(x) \leq 0$ for all $x \leq y_2$ .

First, by Lemma 4.4 and the pair of equations (4.9), we have

\begin{align*} \dfrac{g_2(x)}{\psi_r(x)}< \dfrac{(\Phi g_2)(x)}{(\Phi \psi_r)(x)} < \dfrac{(\Phi g_2)(y_2)}{(\Phi \psi_r)(y_2)} = \dfrac{(\Phi g_1)(y_1)}{(\Phi \psi_r)(y_1)}.\end{align*}

Then, similarly to the derivation of (4.9), we can rewrite the above as

\begin{equation*} \lambda \dfrac{g_2(x)}{\psi_r(x)} < f_{\psi}(y_1).\end{equation*}

Multiplying by $\psi_r(x)$ yields

\begin{align*} \lambda g_2(x) < \dfrac{\psi_r(x)}{\psi_r(y_1)}(H_1(y_1) - C(y_1, y_2)\phi_{r+\lambda}(y_1) ).\end{align*}

Consequently,

\begin{align*} -\lambda A\psi_r(x) + \lambda^2 g_2(x) &= -\lambda \dfrac{\psi_r(x)}{\psi_r(y_1)}(H_1(y_1) - C(y_1, y_2)\phi_{r+\lambda}(y_1)) + \lambda^2 g_2(x) < 0,\end{align*}

which concludes the proof.

Having the auxiliary results at our disposal, we can now show that $G_i$ satisfies the Bellman principle of Proposition 3.1. First, since the function $G_i$ satisfies the condition (4.3), we have

\begin{align*} G_i(x)&=\mathbb{E}_x\bigl[ {\textrm{e}}^{-r\tau_{y_1}}(G_a(X_{\tau_{y_1}})-K_1) \bigr], \\[5pt] G_a(x)&=\lambda(R_{r+\lambda}g)(x)+\mathbb{E}_x\bigl[{\textrm{e}}^{-(r+\lambda)\sigma_{y_2}}(G_i(X_{\sigma_{y_2}})-\lambda(R_{r+\lambda}g)(X_{\sigma_{y_2}})+K_2)\bigr].\end{align*}

On the other hand, by Lemmas 4.7–4.9 we find that

(4.14) \begin{align} G_i(x) \geq \mathbb{E}_x[{\textrm{e}}^{-r\tau}G_i(X_\tau)] \geq \mathbb{E}_x[{\textrm{e}}^{-r\tau}(G_a(X_\tau) - K_1)] \end{align}

and

\begin{align*} G_a(x) - \lambda (R_{r+\lambda}g)(x) &\geq \mathbb{E}_x[{\textrm{e}}^{-(r+\lambda)\tau}(G_a(X_\tau) - \lambda (R_{r+\lambda}g)(X_\tau))] \\[5pt] &\geq \mathbb{E}_x[{\textrm{e}}^{-(r+\lambda)\tau}(G_i(X_\tau) - \lambda (R_{r+\lambda}g)(X_\tau) + K_2)],\end{align*}

which implies

(4.15) \begin{align} G_a(x) &\geq \lambda (R_{r+\lambda}g)(x) + \mathbb{E}_x[{\textrm{e}}^{-(r+\lambda)\tau}(G_a(X_\tau) - \lambda (R_{r+\lambda}g)(X_\tau))]. \end{align}

Since inequalities (4.14) and (4.15) are true for all stopping times $\tau$ , it follows that

\begin{align*} G_i(x) &\geq \sup\nolimits_{\tau} \mathbb{E}_x[{\textrm{e}}^{-r\tau}(G_a(X_\tau) - K_1)], \\[5pt] G_a(x) &\geq \sup\nolimits_{\sigma} (\lambda R_{r+\lambda}g(x) + \mathbb{E}_x[{\textrm{e}}^{-(r+\lambda)\sigma}(G_a(X_\sigma) - \lambda (R_{r+\lambda}g)(X_\sigma))]).\end{align*}

Then, since for stopping times $\tau^* = \tau_{y_1}$ and $\sigma^* = \sigma_{y_2}$ we have

\begin{align*} G_i(x) &= \mathbb{E}_x[{\textrm{e}}^{-r\tau^*}(G_a(X_{\tau^*}) - K_1)], \\[5pt] G_a(x) &= (\lambda (R_{r+\lambda}g)(x) + \mathbb{E}_x[{\textrm{e}}^{-(r+\lambda){\sigma^*}}(G_a(X_{\sigma^*}) - \lambda (R_{r+\lambda}g)(X_{\sigma^*}))]),\end{align*}

it follows that

(4.16) \begin{align} G_i(x) &= \sup\nolimits_{\tau} \mathbb{E}_x[{\textrm{e}}^{-r\tau}(G_a(X_\tau) - K_1)], \end{align}

(4.17) \begin{align} G_a(x) &= \sup\nolimits_{\sigma} \bigl(\lambda (R_{r+\lambda}g)(x) + \mathbb{E}_x[{\textrm{e}}^{-(r+\lambda)\sigma}(G_a(X_\sigma) - \lambda (R_{r+\lambda}g)(X_\sigma))]\bigr). \end{align}

Using these, we obtain that (i) for all pairs of stopping times $(\tau,\sigma)$ , we can use equations (4.16) and (4.17) to conclude that

\begin{align*} G_i(x) &\geq \mathbb{E}_x[{\textrm{e}}^{-r\tau}(G_a(X_{\tau})-K_1)] \\[5pt] &\geq \mathbb{E}_x[{\textrm{e}}^{-r\tau}(\mathbb{E}_{X_\tau}[{\textrm{e}}^{-rU} g(X_U)\textbf{1}(U < \sigma) + {\textrm{e}}^{-r\sigma}(G_i(X_{\sigma})+K_2)\textbf{1}(U > \sigma)]-K_1)].\end{align*}

(ii) For the pair $(\tau^*,\sigma^*)$ , we have

\begin{align*} G_i(x) &= \mathbb{E}_x[{\textrm{e}}^{-r\tau_{y_1}}(G_a(X_{\tau_{y_1}})-K_1)] \\[5pt] &= \mathbb{E}_x[{\textrm{e}}^{-r\tau_{y_1}}(\mathbb{E}_{X_{\tau_{y_1}}}[{\textrm{e}}^{-rU} g(X_U)\textbf{1}(U < \sigma_{y_2})\\[5pt] &\quad + {\textrm{e}}^{-r\sigma_{y_2}}(G_i(X_{\sigma_{y_2}})+K_2)\textbf{1}(U > \sigma_{y_2})]-K_1)].\end{align*}

Thus $G_i$ is the solution to the problem (2.3). Summarizing, we have proved the following result.

Theorem 4.1. Let Assumption 4.1 hold. Then

\begin{equation*} V_i(x) = \begin{cases} \lambda(R_{r+\lambda}g)(x) + C(y_1,y_2) \varphi_{r+\lambda}(x) - K_1, & x \geq y_1, \\ A(y_1,y_2) \psi_r(x), & x > y_1, \end{cases}\end{equation*}

and

\begin{equation*} V_a(x) = \begin{cases} \lambda(R_{r+\lambda}g)(x) + C(y_1,y_2) \varphi_{r+\lambda}(x), & x \geq y_2, \\ A(y_1,y_2) \psi_r(x) + K_2, & x > y_2, \end{cases}\end{equation*}

where

\begin{align*} A(y_1,y_2) &= \dfrac{(K_2-\lambda(R_{r+\lambda}g)(y_2))\varphi_{r+\lambda}(y_1) - (K_1-\lambda(R_{r+\lambda}g)(y_1))\varphi_{r+\lambda}(y_2)}{\psi_r(y_1)\varphi_{r+\lambda}(y_2) - \psi_r(y_2)\varphi_{r+\lambda}(y_1)}, \\[5pt] C(y_1,y_2) &= \dfrac{(K_2-\lambda(R_{r+\lambda}g)(y_2))\psi_r(y_1) - (K_1-\lambda(R_{r+\lambda}g)(y_1))\psi_r(y_2)}{\psi_r(y_1)\varphi_{r+\lambda}(y_2) - \psi_r(y_2)\varphi_{r+\lambda}(y_1)}. \end{align*}

Here, the thresholds are uniquely given by the conditions

\begin{equation*} \begin{split} H_1(y_1) &= H_2(y_2), \\[5pt] R_1(y_1) &= R_2(y_2), \end{split}\end{equation*}

where

\begin{align*} H_l(x)&= \dfrac{(\Phi g_l)(y_l)}{(\Phi \psi_r)(y_l)}, \\[5pt] R_l(x) &= (\Phi g_l)(y_l)\dfrac{(\Psi \psi_r)(y_l)}{(\Phi \psi_r)(y_l)} - (\Psi g_l)(y_l)\end{align*}

for $l=1,2$ .

We have shown that in the diffusion case the optimal rule is to activate the investment once revenue process X is above the threshold $y_1$ and abandon an active investment if the revenue process reaches the level $y_2$ before the project is completed. Figure 1 shows an example of a realization of using this kind of stopping strategy. The agent starts as inactive (the path for the inactive agent is plotted in black), and when the process hits the threshold $y_1$ the agent invests and his/her status changes to active (the path for the active agent is plotted in grey). When the status is changed to active a Poisson process with intensity $\lambda$ is immediately started. This starting time is plotted as a dashed vertical line and marked as $T_0$ . In this realization of the path the process hits the threshold $y_2$ before the first jump of the Poisson process (dashed vertical line $T_1$ ). Thus the agent abandons the project and goes back to inactive at $y_2$ so that she can wait for a better opportunity. Then the agent again invests and activates when the process hits $y_1$ , but this time the Poisson process jumps before the process hits $y_2$ , so the agent receives the payoff $g(X_{T_4})$ .

Figure 1.

An illustration of a possible realization of the underlying process and the usage of the optimal policy given by Theorem 4.1.

4.3. An illustration

Let X be the diffusion with initial state $X_0=x$ and the infinitesimal generator

\begin{equation*}\mathcal{A} = \mu x\dfrac{{\textrm{d}}}{{\textrm{d}} x} + \dfrac{1}{2}\sigma^2 x^2\dfrac{{\textrm{d}}^2}{{\textrm{d}} x^2},\end{equation*}

where $\mu \in \mathbb{R}$ and $\sigma>0$ . This diffusion process is called a geometric Brownian motion. The state space of the process is $\mathbb{R}_+$ and the endpoints of the state space are natural. We further assume that $\mu < r$ and that $\mu - \tfrac{1}{2} \sigma^2 >0,$ so that $X_t \to \infty$ almost surely as $t \to \infty$ . The scale density and the density of the speed measure read as

\[S'(x) = x^{-{{2 \mu}/{\sigma^2}}}, \quad m'(x) = \dfrac{2}{\sigma^2} x^{{{2 \mu}/{\sigma^2}}-2}.\]

We fix the constants $r, \lambda > 0$ and denote

\begin{align*}& \beta_{\lambda}= \dfrac{1}{2}-\dfrac{\mu}{\sigma^2}+\sqrt{\bigg(\dfrac{1}{2}-\dfrac{\mu}{\sigma^2} \bigg)^2+\dfrac{2(r+\lambda)}{\sigma^2}} >1, \\[5pt] & \alpha_{\lambda} = \dfrac{1}{2}-\dfrac{\mu}{\sigma^2} - \sqrt{\bigg(\dfrac{1}{2}-\dfrac{\mu}{\sigma^2} \bigg)^2+\dfrac{2(r+\lambda)}{\sigma^2}} < 0.\end{align*}

Then the minimal r-excessive functions for X are

\[\psi_{r}(x) = x^{\beta_0}, \quad \varphi_{r}(x) = x^{\alpha_0}.\]

Further, we consider the linear payoff $g(x) = x^\theta - \eta$ , where $\theta \in (0,1]$ and $\eta > K_2 {{(r+\lambda)}/{\lambda}}$ . We notice now that our assumptions are satisfied.

By straightforward integration we find that

\begin{alignat*}{3} (\Phi g)(x) & = \dfrac{2}{\sigma^2} \dfrac{x^{\theta-\beta_{\lambda}}}{\beta_{\lambda}-\theta} - \eta \dfrac{2}{\sigma^2} \dfrac{x^{-\beta_{\lambda}}}{\beta_{\lambda}}, &\quad (\Psi g)(x) & = \dfrac{2}{\sigma^2} \dfrac{x^{\theta-\alpha_{\lambda}}}{\theta-\alpha_{\lambda}} + \eta \dfrac{2}{\sigma^2} \dfrac{x^{-\alpha_{\lambda}}}{\alpha_{\lambda}}, \\[5pt] (\Phi \psi_r)(x) & = \dfrac{2}{\sigma^2} \dfrac{x^{\beta_0 - \beta_{\lambda}}}{\beta_{\lambda}-\beta_0}, & \quad (\Psi \psi_r)(x) & = \dfrac{2}{\sigma^2} \dfrac{x^{\beta_0-\alpha_{\lambda}}}{\beta_0-\alpha_{\lambda}} .\end{alignat*}

Using the above calculations, we find using the representation (4.1) that

\begin{equation*} \lambda(R_{r+\lambda}g)(x) = \dfrac{2 \lambda}{\sigma^2} \dfrac{x^{\theta}}{(\beta_{\lambda}-\theta)(\theta-\alpha_{\lambda})} - \dfrac{\eta \lambda}{r+\lambda}.\end{equation*}

We note that analogous calculations also hold for $g_1$ and $g_2$ instead of g. Using these calculations, we first find that the solutions to the classical stopping problem $(V_c(x), x^*)$ and the problem with exercise lag but without reversibility (16) $(V_r(x), y^*)$ are given by

\begin{align*} V_c(x) &= \begin{cases} x^{\theta}-\eta-K_1, & x \geq x^*, \\ \dfrac{{x^*}^{\theta}-\eta-K_1}{{x^*}^{\beta_0}} x^{\beta_0}, & x < x^*, \end{cases} \\ V_r(x) &= \begin{cases} \dfrac{2 \lambda}{\sigma^2} \dfrac{x^{\theta}}{(\beta_{\lambda}-\theta)(\theta-\alpha_{\lambda})} - \dfrac{(\eta +K_1)\lambda}{r+\lambda}, & x \geq y^*,\\ \biggl(\dfrac{2 \lambda}{\sigma^2} \dfrac{{y^*}^{\theta-\beta_0}}{(\beta_{\lambda}-\theta)(\theta-\alpha_{\lambda})} - \dfrac{(\eta + K_1)\lambda {y^*}^{-\beta_0}}{r+\lambda}\biggr) x^{\beta_0}, & x < y^*, \end{cases}\end{align*}

where $x^*$ and $y^*$ are given by

\begin{align*} {x^*} &= \bigg(\dfrac{(\eta + K_1)\beta_0}{\beta_0 - \theta}\bigg)^{1/\theta}, \\[5pt] {y^*} &= \bigg(\dfrac{(\eta +K_1)\beta_0 (\beta_{\lambda} - \theta)(\theta - \alpha_{\lambda})}{\alpha_{\lambda} \beta_{\lambda} (\theta-\beta_0)}\bigg)^{1/\theta} = x^* \bigg( \dfrac{(\theta-\beta_{\lambda})(\theta - \alpha_{\lambda})}{\alpha_{\lambda} \beta_{\lambda} }\bigg)^{1/\theta}.\end{align*}

Similar calculations show that the solution to the reversible problem studied in previous sections is characterized by the value functions

(4.18) \begin{equation} V_i(x) = \begin{cases} \dfrac{2 \lambda}{\sigma^2} \dfrac{x^{\theta}}{(\beta_{\lambda}-\theta)(\theta-\alpha_{\lambda})} - \dfrac{\eta \lambda}{r+\lambda} + C(y_1, y_2) x^{\alpha_{\lambda}} - K_1, & x \geq y_1, \\ A(y_1, y_2) x^{\beta_0}, & x \leq y_1, \end{cases}\end{equation}

and

(4.19) \begin{equation} V_a(x) = \begin{cases} \dfrac{2 \lambda}{\sigma^2} \dfrac{x^{\theta}}{(\beta_{\lambda}-\theta)(\theta-\alpha_{\lambda})} - \dfrac{\eta \lambda}{r+\lambda} + C(y_1, y_2) x^{\alpha_{\lambda}}, & x \geq y_2, \\ A(y_1, y_2) x^{\beta_0} - K_2, & x \leq y_2, \end{cases}\end{equation}

where

\begin{align*} & C(y_1, y_2) = \dfrac{2 \lambda}{\sigma^2} \dfrac{\bigl( \frac{y_1^{\theta}}{(\beta_{\lambda}-\theta)(\theta-\alpha_{\lambda})} - \frac{\eta \lambda}{r+\lambda} - K_1 \bigr) y_2^{\beta_0}-\bigl( \frac{y_2^{\theta}}{(\beta_{\lambda}-\theta)(\theta-\alpha_{\lambda})} - \frac{\eta \lambda}{r+\lambda} - K_2 \bigr) y_1^{\beta_0}}{y_1^{\beta_0}y_2^{\alpha_{\lambda}}-y_1^{\alpha_{\lambda}} y_2^{\beta_0}}, \\[5pt] & A(y_1, y_2) = \dfrac{2 \lambda}{\sigma^2} \dfrac{\bigl( \frac{y_1^{\theta}}{(\beta_{\lambda}-\theta)(\theta-\alpha_{\lambda})} - \frac{\eta \lambda}{r+\lambda} - K_1 \bigr) y_2^{\alpha_{\lambda}}-\bigl( \frac{y_2^{\theta}}{(\beta_{\lambda}-\theta)(\theta-\alpha_{\lambda})} - \frac{\eta \lambda}{r+\lambda} - K_2 \bigr) y_1^{\alpha_{\lambda}}}{y_1^{\beta_0}y_2^{\alpha_{\lambda}}-y_1^{\alpha_{\lambda}} y_2^{\beta_0}}.\end{align*}

Figure 2.

Optimal thresholds when the rate $\lambda$ changes.

Further, the thresholds $y_1$ and $y_2$ in (4.18) and (4.19) are given as the unique solution to the pair of equations

\begin{align*} & y_1^{- \beta_0} \biggl( \dfrac{y_1^{\theta}}{\beta_{\lambda}-\theta} - \dfrac{\eta + K_1 \frac{r+\lambda}{\lambda}}{\beta_{\lambda}} \biggr) = y_2^{- \beta_0} \biggl( \dfrac{y_2^{\theta}}{\beta_{\lambda}-\theta} - \dfrac{\eta + K_2 \frac{r+\lambda}{\lambda}}{\beta_{\lambda}} \biggr), \\[5pt] & y_1^{-\alpha_{\lambda}} \biggl[ \dfrac{\beta_{\lambda}-\beta_0}{\beta_0-\alpha_{\lambda}} \biggl( \dfrac{y_1^{\theta}}{\beta_{\lambda}-\theta} - \dfrac{\eta + K_1 \frac{r+\lambda}{\lambda}}{\beta_{\lambda}} \biggr) - \dfrac{y_1^{\theta}}{\theta-\alpha_{\lambda}} - \dfrac{\eta + K_1 \frac{r+\lambda}{\lambda}}{\alpha_{\lambda}} \biggr] \\[5pt] &\quad = y_2^{-\alpha_{\lambda}} \biggl[ \dfrac{\beta_{\lambda}-\beta_0}{\beta_0-\alpha_{\lambda}} \biggl( \dfrac{y_2^{\theta}}{\beta_{\lambda}-\theta} - \dfrac{\eta + K_2 \frac{r+\lambda}{\lambda}}{\beta_{\lambda}} \biggr) - \dfrac{y_2^{\theta}}{\theta-\alpha_{\lambda}} - \dfrac{\eta + K_2 \frac{r+\lambda}{\lambda}}{\alpha_{\lambda}} \biggr]. \end{align*}

Figure 3.

Optimal thresholds when the payoff/cost $K_2$ changes.

Figure 4.

Limiting relations between the problems.

Since it seems that it is not possible to solve the pair of equations explicitly, we illustrate the results numerically. We select the parameters $\mu= 0.2$ , $\sigma= 0.5$ , $r= 0.25$ , $\lambda= 1.0$ , $\theta= 1.0$ , $\eta= 1.0$ , $K_1= 0.05$ , and $K_2= 0.04$ . Using these parameters we find that $y_2 \approx 4.46$ and $y_1 \approx 6.25$ . If we let $\lambda$ vary but keep the other parameters fixed, we find that the solution approaches a classical stopping problem, in the sense that $y_1 \to x^*$ , as $\lambda \to \infty$ ; see Figure 2. This result is intuitive since in the limit $\lambda \to \infty$ there are no changes to reverse the investment and thus the payoff is immediately realized. Similarly, when $K_2 \to -\infty$ (so that it is never optimal to reverse the investment because the cost is too high), we find that $y_2 \to 0$ and $y_1 \to y^*$ , and hence the problem reduces to the stopping problem with time-to-build considered in [Reference Lempa17]; see Figure 3. These observations are also collected in Figure 4.

Lastly, in Figure 3 we also notice that $y_2 \to y_1$ , when $K_2 \to K_1$ . Interestingly, in this case it is reasonable to assume that the decision maker effectively follows a Poisson process and at each jump time makes the decision to either continue or stop and receive the payoff. Consequently, we conjecture that in this limiting case our considered problem could be represented as a Poisson stopping problem, as in [Reference Dupuis and Wang12, Reference Lempa16], for example. Unfortunately, the proper treatment of these considerations is beyond the scope of the present study and therefore left for future research.