2. Problem specification

We consider a Bayesian set-up where one observes a diffusion process X in continuous time, the drift of which depends on an unknown state $\theta$ that takes values 0 and 1 with probabilities $1-\pi$ and $\pi$ , respectively, with $\pi\in[0,1]$ . Given payoff functions g and h, the problem is to stop the process so as to maximize the expected reward in (1). Here the random horizon $\gamma$ has a state-dependent distribution but is independent of the noise of X.

The above set-up can be realized by considering a probability space $(\Omega, \mathcal F,{\mathbb P}_\pi)$ that hosts a standard Brownian motion W and an independent Bernoulli-distributed random variable $\theta$ with ${\mathbb P}_\pi(\theta=1)=\pi=1-{\mathbb P}_\pi(\theta=0)$ . Additionally, we let $\gamma$ be a random time (possible infinite) independent of W and with state-dependent survival distribution

\[{\mathbb P}_\pi(\gamma>t\mid \theta=i) = F_i(t),\]

where $F_i$ is continuous and non-increasing with $F_i(0)=1$ and $F_i(t)>0$ for all $t\geq 0$ , $i=0,1$ . We remark that we include the possibility that $F_i\equiv 1$ for some $i\in\{0,1\}$ (or for both), corresponding to an infinite horizon. We then have

(2) \begin{equation}{\mathbb P}_\pi=(1-\pi){\mathbb P}_0 + \pi{\mathbb P}_1,\end{equation}

where ${\mathbb P}_0({\cdot})={\mathbb P}_\pi(\cdot\mid \theta=0)$ and ${\mathbb P}_1({\cdot})={\mathbb P}_\pi(\cdot\mid \theta=1)$ .

Now consider the equation

(3) \begin{equation}{\mathrm{d}} X_t= \mu(X_t,\theta) \,{\mathrm{d}} t + \sigma(X_t) \,{\mathrm{d}} W_t.\end{equation}

Here $\mu(\cdot, \cdot)\,:\, {\mathbb R}\times\{0,1\}\to{\mathbb R}$ is a given function of the unknown state $\theta$ and the current value of the underlying process; we denote $\mu_0(x)=\mu(x,0)$ and $\mu_1(x)=\mu(x,1)$ . The diffusion coefficient $\sigma({\cdot})\,:\, {\mathbb R}\to (0,\infty)$ is a given function of x, independent of the unknown state $\theta$ . We assume that the functions $\mu_0,\mu_1$ and $\sigma$ satisfy standard Lipschitz conditions so that the existence and uniqueness of a strong solution X is guaranteed. We are also given two functions

\[g(\cdot,\cdot,\cdot)\,:\, [0,\infty)\times{\mathbb R}\times\{0,1\}\to{\mathbb R}\]

and

\[h(\cdot,\cdot,\cdot)\,:\, [0,\infty)\times{\mathbb R}\times\{0,1\}\to{\mathbb R},\]

which we refer to as the payoff functions. We assume that $g_i$ and $h_i$ are continuous for $i=0,1$ , where we use the notation $g_i(\cdot,\cdot)\,:\!=\, g(\cdot,\cdot,i)$ and $h_i(\cdot,\cdot)\,:\!=\, h(\cdot,\cdot,i)$ to denote the payoff functions on the event $\{\theta=i\}$ , $i=0,1$ .

Let $\mathcal F^X$ denote the smallest right-continuous filtration that makes X adapted, and let $\mathcal T^X$ be the set of $\mathcal F^X$ -stopping times. Similarly, let $\mathcal F^{X,\gamma}$ denote the smallest right-continuous filtration to which both X and the process $1_{\{\cdot\geq \gamma\}}$ are adapted, and let $\mathcal T^{X,\gamma}$ be the set of $\mathcal F^{X,\gamma}$ -stopping times.

We now consider the optimal stopping problem

(4) \begin{equation}V= \sup_{\tau\in\mathcal T^{X,\gamma}}{\mathbb E}_\pi\bigl[g(\tau,X_\tau,\theta)1_{\{\tau< \gamma\}}+ h(\gamma,X_\gamma,\theta)1_{\{ \gamma\leq\tau\}}\bigr].\end{equation}

In (4), and in similar expressions throughout the paper, we use the convention that $h(\tau,X_\tau,\theta)\,:\!=\, 0$ on the event $\{\tau=\gamma=\infty\}$ . We further assume that the integrability condition

\[{\mathbb E}_\pi\biggl[\sup_{t\geq0}\{ | g(t,X_t,\theta)| + | h(t,X_t,\theta)|\}\biggr]<\infty\]

holds.

More precisely, on the event $\{\theta=0\}$ , the drift of X is $\mu_0({\cdot})$ , the payoff functions are $g_0(\cdot,\cdot)$ and $h_0(\cdot,\cdot)$ , and the random horizon has survival distribution function $F_0({\cdot})$ ; on the event $\{\theta=1\}$ , the drift is $\mu_1$ , the payoff functions are $g_1(\cdot,\cdot)$ and $h_1(\cdot,\cdot)$ , and the random horizon has survival distribution function $F_1({\cdot})$ .

Remark 2. In (4), the payoff on the event $\{\tau=\gamma\}$ is specified in terms of h. In some applications, however, one may want to use the alternative formulation

(5) \begin{equation}U\,:\!=\, \sup_{\tau\in\mathcal T^{X,\gamma}}{\mathbb E}_\pi\bigl[g(\tau,X_\tau,\theta)1_{\{\tau\leq \gamma\}}+ h(\gamma,X_\gamma,\theta)1_{\{ \gamma<\tau\}}\bigr].\end{equation}

If $g\geq h$ , then we have

\[ U = \sup_{\tau\in\mathcal T^{X,\gamma}}{\mathbb E}_\pi \bigl[g(\tau,X_{\tau},\theta)1_{\{\tau< \gamma\}}+ g(\gamma,X_\gamma,\theta)1_{\{ \gamma\leq\tau\}}\bigr]\]

since $\tau\wedge\gamma\in\mathcal T^{X,\gamma}$ for every $\tau\in\mathcal T^{X,\gamma}$ , and thus the formulation (5) is contained in the formulation (4).

Similarly, if $g\leq h$ , then

\begin{equation*}U=\sup_{\tau\in\mathcal T^{X,\gamma}}{\mathbb E}_\pi\bigl[g(\tau,X_\tau,\theta)1_{\{\tau< \gamma\}}+ h(\gamma,X_\gamma,\theta)1_{\{ \gamma\leq \tau\}}\bigr]=V,\end{equation*}

where the first equality uses that for a given stopping time $\tau$ , the time

\[\tau^{\prime}=\begin{cases}\tau & \text{on $\{\tau<\gamma\}$}\\\infty & \text{on $\{\tau\geq \gamma\}$}\end{cases}\]

is also a stopping time. In the general case where no ordering between g and h is given, however, the formulation (5) is more involved; such cases are not covered by the results of the current article.

3. A useful reformulation of the problem

In this section we rewrite the optimal stopping problem (4) with incomplete information as an optimal stopping problem with respect to stopping times in $\mathcal T^X$ and with complete information.

First consider the stopping problem

(6) \begin{equation}\hat V= \sup_{\tau\in\mathcal T^X}{\mathbb E}_\pi\bigl[g(\tau,X_\tau,\theta)1_{\{\tau< \gamma\}}+ h(\gamma,X_\gamma,\theta)1_{\{\gamma\leq\tau\}}\bigr],\end{equation}

where the supremum is taken over $\mathcal F^X$ -stopping times. Since $\mathcal T^X\subseteq\mathcal T^{X,\gamma}$ , we have $\hat V\leq V$ . On the other hand, by a standard argument (see [Reference Chakrabarty and Guo1] or [Reference Lempa and Matomäki17]), we also have the reverse inequality, so $\hat V=V$ . Indeed, first recall that for any $\tau\in\mathcal T^{X,\gamma}$ there exists $\tau^\prime\in\mathcal T^X$ such that $\tau\wedge\gamma = \tau^\prime\wedge\gamma$ ; see [Reference Protter20, p. 378]. Consequently, $\tau=\tau^{\prime}$ on $\{\tau<\gamma\}=\{\tau^{\prime}<\gamma\}$ and $\tau\wedge\gamma=\tau^{\prime}\wedge\gamma=\gamma$ on $\{\tau\geq \gamma\}=\{\tau^{\prime}\geq\gamma\}$ , so

\begin{equation*}{\mathbb E}_\pi\bigl[g(\tau,X_\tau,\theta)1_{\{\tau< \gamma\}}+ h(\gamma,X_\gamma,\theta)1_{\{\gamma\leq\tau\}}\bigr] = {\mathbb E}_\pi\bigl[g(\tau^\prime,X_{\tau^\prime},\theta)1_{\{\tau^\prime< \gamma\}} + h(\gamma,X_\gamma,\theta)1_{\{\gamma\leq\tau^{\prime}\}}\bigr],\end{equation*}

from which $\hat V=V$ follows. Moreover, if $\tau^{\prime}\in\mathcal T^X$ is optimal in (6), then it is also optimal in (4).

Remark 3. Since we assume that the survival distributions are continuous, we have

\[{\mathbb P}_\pi(\tau=\gamma<\infty)=0\]

for any $\tau\in\mathcal T^X$ . Consequently, we can alternatively write

\[\hat V=\sup_{\tau\in\mathcal T^X}{\mathbb E}_\pi\bigl[g(\tau,X_\tau,\theta)1_{\{\tau\leq \gamma\}}+h(\gamma,X_\gamma,\theta)1_{\{\gamma<\tau\}}\bigr].\]

To study the stopping problem (4), or equivalently the optimal stopping problem (6), we introduce the conditional probability process

\[\Pi_t\,:\!=\, {\mathbb P}_\pi\bigl(\theta=1\mid \mathcal F^X_t\bigr)\]

and the corresponding probability ratio process

(7) \begin{equation}\Phi_t\,:\!=\, \dfrac{\Pi_t}{1-\Pi_t}.\end{equation}

Note that $\Pi_0=\pi$ and $\Phi_0=\varphi\,:\!=\, \pi/(1-\pi)$ , ${\mathbb P}_\pi$ -a.s.

Proposition 1. We have

(8) \begin{align}V &= \sup_{\tau\in\mathcal T^X}{\mathbb E}_\pi\biggl[ g_0(\tau,X_\tau)(1-\Pi_\tau)F_0(\tau) + g_1(\tau,X_\tau)\Pi_\tau F_1(\tau)\\ \notag&\quad- \int_0^{\tau} h_0(t,X_t)(1-\Pi_t )\,{\mathrm{d}} F_0(t) -\int_0^{\tau}h_1(t,X_t)\Pi_t \,{\mathrm{d}} F_1(t)\biggr].\end{align}

Moreover, if $\tau\in\mathcal T^X$ is optimal in (8), then it is also optimal in (4).

Proof. Let $\tau$ denote a stopping time in $\mathcal T^X$ . Using the tower property, we find that

\begin{align*}{\mathbb E}_\pi\bigl[g(\tau,X_\tau,\theta)1_{\{\tau< \gamma\}}\bigr]&= {\mathbb E}_\pi\bigl[{\mathbb E}_\pi\bigl[ g_i(\tau,X_\tau) 1_{\{\tau< \gamma\}}\mid \mathcal F^X_\tau\bigr]\bigr] \\ &={\mathbb E}_\pi\bigl[ {\mathbb E}_\pi\bigl[g(\tau,X_\tau,\theta)1_{\{\tau< \gamma\}}\mid \mathcal F^X_\tau,\theta=0\bigr] (1-\Pi_\tau) \\ & \quad + {\mathbb E}_\pi\bigl[g(\tau,X_\tau,\theta)1_{\{\tau< \gamma\}}\mid \mathcal F^X_\tau,\theta=1\bigr] \Pi_\tau\bigr]\\&={\mathbb E}_\pi\bigl[g_0(\tau,X_\tau){\mathbb P}_\pi\bigl(\tau< \gamma \mid \mathcal F^X_\tau,\theta=0\bigr) (1-\Pi_\tau) \\ & \quad + g_1(\tau,X_\tau) {\mathbb P}_\pi\bigl(\tau< \gamma\mid \mathcal F^X_\tau,\theta=1\bigr) \Pi_\tau\bigr]\\ &={\mathbb E}_\pi\bigl[ g_0(\tau,X_\tau)(1-\Pi_\tau)F_0(\tau) + g_1(\tau,X_\tau)\Pi_\tau F_1(\tau)\bigr],\end{align*}

where for the last equality we recall that $\gamma$ and $\tau$ are independent on the event $\{\theta = i\}$ , $i=0,1$ .

Similarly, for the second term we use

\[{\mathbb P}_\pi\bigl(\theta=0, \gamma\geq t\mid \mathcal F^X_\tau\bigr)=(1-\Pi_\tau) F_0(t),\quad{\mathbb P}_\pi\bigl(\theta=1, \gamma\geq t\mid \mathcal F^X_\tau\bigr)=\Pi_\tau F_1(t),\]

so that

\begin{align*} {\mathbb E}_\pi\bigl[ h(\gamma,X_\gamma,\theta)1_{\{\gamma\leq\tau\}}\bigr] &= {\mathbb E}_\pi\bigl[ {\mathbb E}_\pi\bigl[h(\gamma,X_\gamma,\theta) 1_{\{\gamma\leq\tau\}}\mid \mathcal F^{X}_\tau\bigr] \bigr]\\ &= -{\mathbb E}_\pi\biggl[ \int_0^\infty {\mathbb E}_\pi\bigl[h(\gamma,X_\gamma,\theta) 1_{\{\gamma\leq\tau\}}\mid \mathcal F^{X}_\tau, \theta = 0, \gamma=t \bigr](1-\Pi_\tau) \,{\mathrm{d}} F_0(t) \\ & \quad + \int_0^\infty {\mathbb E}_\pi\bigl[h(\gamma,X_\gamma,\theta) 1_{\{\gamma\leq\tau\}}\mid \mathcal F^{X}_\tau, \theta = 1, \gamma=t \bigr] \Pi_\tau \,{\mathrm{d}} F_1(t)\biggr]\\&= -{\mathbb E}_\pi\biggl[ \int_0^\infty h_0(t,X_t) 1_{\{t\leq\tau\}}(1-\Pi_\tau) \,{\mathrm{d}} F_0(t) \\ & \quad +\int_0^\infty h_1(t,X_t) 1_{\{t\leq\tau\}} \Pi_\tau \,{\mathrm{d}} F_1(t)\biggr]\\ &= -{\mathbb E}_\pi\biggl[\int_0^{\tau} h_0(t,X_t)(1-\Pi_t )\,{\mathrm{d}} F_0(t) +\int_0^{\tau}h_1(t,X_t)\Pi_t \,{\mathrm{d}} F_1(t)\biggr],\end{align*}

where the last equality holds because $\Pi$ is a martingale. The optimal stopping problem (4) therefore coincides with the stopping problem

\begin{align*}& \sup_{\tau\in\mathcal T^X}{\mathbb E}_\pi\biggl[ g_0(\tau,X_\tau)(1-\Pi_\tau)F_0(\tau) + g_1(\tau,X_\tau)\Pi_\tau F_1(\tau) \\ &\quad - \int_0^{\tau} h_0(t,X_t)(1-\Pi_t )\,{\mathrm{d}} F_0(t) -\int_0^{\tau}h_1(t,X_t)\Pi_t \,{\mathrm{d}} F_1(t)\biggr].\end{align*}

It is well known (see e.g. [Reference Shiryaev22, pp. 180--181] or [Reference Gapeev and Shiryaev10, p. 522]) that the posterior probability is given by

\[\Pi_t=\dfrac{\frac{\pi}{1-\pi}L_t}{1+\frac{\pi}{1-\pi}L_t},\]

where

\[L_t\,:\!=\, \exp\biggl\{\int_0^t\dfrac{\mu_1(X_s)-\mu_0(X_s)}{\sigma^2(X_s)}\,{\mathrm{d}} X_s-\dfrac{1}{2}\int_0^t\dfrac{\mu^2_1(X_s)-\mu^2_0(X_s)}{\sigma^2(X_s)}\,{\mathrm{d}} s\biggr\}.\]

Therefore, by an application of Itô’s formula, the pair $(X,\Pi)$ satisfies

(9) \begin{equation}\begin{cases}{\mathrm{d}} X_t=(\mu_0(X_t)+ (\mu_1(X_t)-\mu_0(X_t))\Pi_t)\,{\mathrm{d}} t + \sigma(X_t) \,{\mathrm{d}}\hat W_t,\\{\mathrm{d}} \Pi_t=\omega(X_t) \Pi_t(1-\Pi_t)\,{\mathrm{d}}\hat W_t,\end{cases}\end{equation}

where $\omega(x)\,:\!=\, (\mu_1(x)-\mu_0(x))/\sigma(x)$ is the signal-to-noise ratio and

\[\hat W_t\,:\!=\, \int_0^t\dfrac{{\mathrm{d}} X_t}{\sigma(X_s)}-\int_0^t\dfrac{1}{\sigma(X_t)}(\mu_0 (X_s)+(\mu_1(X_s)-\mu_0(X_s))\Pi_s)\,{\mathrm{d}} s\]

is the so-called innovation process; by P. Lévy’s theorem, $\hat W$ is a ${\mathbb P}_\pi$ -Brownian motion. By our non-degeneracy and Lipschitz assumptions on the coefficients, there exists a unique strong solution to the system (9), and the pair $(X,\Pi)$ is a strong Markov process. Moreover, using Itô’s formula, it is straightforward to check that the likelihood ratio process $\Phi$ in (7) satisfies

(10) \begin{equation}{\mathrm{d}}\Phi_t=\dfrac{\mu_1(X_t)-\mu_0(X_t)}{\sigma^2(X_t)}\Phi_t({\mathrm{d}} X_t-\mu_0(X_t)\,{\mathrm{d}} t).\end{equation}

4. A measure change

In the current section we provide a measure change that decouples X from $\Pi$ . This specific measure change technique was first used in [Reference Klein15] and has since been used by several authors (see [Reference De Angelis2], [Reference Ekström and Lu5], [Reference Ekström and Vannestål7], [Reference Johnson and Peskir14]).

Lemma 1. For $t\in[0,\infty)$ , let ${\mathbb P}_{\pi,t}$ denote the measure ${\mathbb P}_\pi$ restricted to $\mathcal F^X_t$ . We then have

\[\dfrac{{\mathrm{d}}{\mathbb P}_{0,t}}{{\mathrm{d}}{\mathbb P}_{\pi,t}}=\dfrac{1+\varphi}{1+\Phi_t}.\]

Proof. For any $A\in\mathcal F^X_t$ we have

\[{\mathbb E}_\pi[(1-\Pi_t) 1_{A}]={\mathbb E}_\pi[1_{\{\theta=0\}} 1_A]=(1-\pi){\mathbb E}_0[ 1_A]\]

by (2). Consequently,

\begin{equation*} \dfrac{{\mathrm{d}}{\mathbb P}_{0,t}}{{\mathrm{d}}{\mathbb P}_{\pi,t}}=\dfrac{1-\Pi_t}{1-\pi}= \dfrac{1+\varphi}{1+\Phi_t}.\end{equation*}

Since $1-\Pi_\tau=1/(1+\Phi_t)$ and $\Pi_t={{\Phi_t}/{(1+\Phi_t)}}$ , it is now clear that the stopping problem (4) (or equivalently problem (8)) can be written as

\begin{align*}V &= \dfrac{1}{1+\varphi} \sup_{\tau\in\mathcal T^X}{\mathbb E}_0\biggl[g_0(\tau,X_\tau)F_0(\tau) + g_1(\tau,X_\tau)\Phi_\tau F_1(\tau) \\ &\quad-\int_0^\tau h_0(t,X_t)\,{\mathrm{d}} F_0(t) - \int_0^\tau h_1(t,X_t)\Phi_t \,{\mathrm{d}} F_1(t)\biggr],\end{align*}

where the expected value is with respect to ${\mathbb P}_0$ , under which the process $(X,\Phi)$ is strong Markov and satisfies

(11) \begin{equation}\begin{cases}{\mathrm{d}} X_t=\mu_0(X_t)\,{\mathrm{d}} t + \sigma(X_t) \,{\mathrm{d}} W_t\\{\mathrm{d}}\Phi_t=\omega(X_t)\Phi_t \,{\mathrm{d}} W_t\end{cases}\end{equation}

(see (3) and (10)).

Next we introduce the process

(12) \begin{equation}\Phi^\circ_t\,:\!=\, \dfrac{F_1(t)}{F_0(t)}\Phi_t,\end{equation}

so that

\begin{align*} V &= \dfrac{1}{1+\varphi}\sup_{\tau\in\mathcal T^X}{\mathbb E}_0\biggl[ F_0(\tau)\bigl( g_0(\tau,X_\tau) + g_1(\tau,X_\tau)\Phi^\circ_\tau \bigr) \notag\\ &\quad-\int_0^\tau h_0(t,X_t) \,{\mathrm{d}} F_0(t) -\int_0^\tau \dfrac{F_0(t)}{F_1(t)} h_1(t,X_t)\Phi^\circ_t \,{\mathrm{d}} F_1(t)\biggr].\end{align*}

Note that the process $\Phi^\circ$ satisfies

\[{\mathrm{d}}\Phi^\circ_t= \dfrac{1}{f(t)}\Phi^\circ_t\,{\mathrm{d}} f(t) + \omega(X_t)\Phi^\circ_t \,{\mathrm{d}} W_t\]

under ${\mathbb P}_0$ , where $f(t)=F_1(t)/F_0(t)$ .

Remark 4. The process $\Phi^\circ$ is the likelihood ratio given observations of the processes X and $1_{\{\cdot\geq \gamma\}}$ on the event $\{ \gamma>t\}$ . Indeed, for $t\leq T$ , defining

\[\Pi^\circ_t \,:\!=\, {\mathbb P}_\pi\bigl(\theta=1\mid \mathcal F^X_t, \gamma>t\bigr)=\dfrac{{\mathbb P}_\pi\bigl(\theta=1,\gamma>t\mid \mathcal F^X_t\bigr)}{{\mathbb P}_\pi\bigl(\gamma>t\mid \mathcal F^X_t\bigr)}= \dfrac{\Pi_tF_1(t)}{\Pi_tF_1(t) + (1-\Pi_t)F_0(t)},\]

we have

\[\Pi^\circ_t=\dfrac{\Phi^\circ_t}{\Phi^\circ_t+1}.\]

We summarize our theoretical findings in the following theorem.

Theorem 1. Let

(13) \begin{align}v &= \sup_{\tau\in\mathcal T^X}{\mathbb E}_0\biggl[F_0(\tau)\bigl(g_0(\tau,X_\tau)+g_1(\tau,X_\tau)\Phi^\circ_\tau \bigr) \notag \\ &\quad -\int_0^\tau h_0(t,X_t) \,{\mathrm{d}} F_0(t) -\int_0^\tau \dfrac{F_0(t)}{F_1(t)} h_1(t,X_t)\Phi^\circ_t \,{\mathrm{d}} F_1(t) \biggr],\end{align}

where $(X,\Phi^\circ)$ is given by (11) and (12). Then $V=v/(1+\varphi)$ , where $\varphi=\pi/(1-\pi)$ . Moreover, if $\tau\in\mathcal T^X$ is an optimal stopping in (13), then it is also optimal in the original problem (4).

5. An example: a hiring problem

In this section we consider a (simplistic) version of a hiring problem. To describe this, consider a situation where a company tries to decide whether or not to employ a certain candidate, where there is considerable uncertainty about the candidate’s ability. The candidate is either of a ‘strong type’ or of a ‘weak type’, and during the employment procedure, tests are performed to find out which is the true state. At the same time, the candidate is potentially lost for the company as he/she may receive other offers. Moreover, the rate at which such offers are presented may depend on the ability of the candidate; for example, a candidate of the strong type could be more likely to be recruited to other companies than a candidate of the weak type.

To model the above hiring problem, we let $h_i\equiv 0$ , $i=0,1$ , and

\[g(t,x,\theta)=\begin{cases}-{\mathrm{e}}^{-rt}c & \text{if $\theta=0$,}\\{\mathrm{e}}^{-rt}d & \text{if $\theta=1$,}\end{cases}\]

where c and d are positive constants representing the overall cost and benefit of hiring the candidate, respectively, and $r> 0$ is a constant discount rate. To learn about the unknown state $\theta$ , tests are performed and represented as a Brownian motion

\[X_t=\mu (\theta) t+\sigma W_t\]

with state-dependent drift

\[\mu(\theta)=\begin{cases}\mu_0 & \text{if $\theta=0$,}\\\mu_1 & \text{if $\theta=1$,}\end{cases}\]

where $\mu_0<\mu_1$ . We further assume that the survival probabilities $F_0$ and $F_1$ decay exponentially in time, that is,

\[F_0(t) = {\mathrm{e}}^{-\lambda_0 t},\quad F_1(t) = {\mathrm{e}}^{-\lambda_1 t},\]

where $\lambda_0, \lambda_1\geq 0$ are known constants. The stopping problem (4) under consideration is thus

\[V=\sup_{\tau\in\mathcal T^{X,\gamma}}{\mathbb E}_\pi\bigl[ {\mathrm{e}}^{-r\tau}\bigl(d1_{\{\theta=1\}}-c1_{\{\theta=0\}}\bigr)1_{\{\tau<\gamma\}}\bigr],\]

where $\pi={\mathbb P}_{\pi}(\theta=1)$ .

By Theorem 1, we have

\[V = \dfrac{1}{1+\varphi}\sup_{\tau\in\mathcal T^X}{\mathbb E}_0\bigl[ {\mathrm{e}}^{-(r+\lambda_0)\tau}\bigl( \Phi^\circ_\tau d -c\bigr) \bigr],\]

where the underlying process $\Phi^{\circ}$ is a geometric Brownian motion satisfying

\[{\mathrm{d}}\Phi_t^\circ = {-(\lambda_1-\lambda_0)\Phi_t^\circ\,{\mathrm{d}} t + \omega\Phi^\circ_t\,{\mathrm{d}} W_t.}\]

Clearly, the value of the stopping problem is

\[V = \dfrac{d}{1+\varphi}\sup_{\tau\in\mathcal T^X}{\mathbb E}_0\biggl[ {\mathrm{e}}^{-(r+\lambda_0)\tau}\biggl( \Phi^\circ_\tau-\dfrac{c}{d} \biggr) \biggr]= \dfrac{d}{1+\varphi} V^{Am}(\varphi),\]

where $V^{Am}$ is the value of the American call option with underlying $\Phi^{\circ}$ starting at $\Phi^\circ_0=\varphi$ and with strike ${{c}/{d}}$ . Standard stopping theory gives that the corresponding value is

\[V =\begin{cases} \dfrac{d b^{1-\eta}}{\eta(1+\varphi)}{\varphi}^\eta, & \varphi<b,\\[9pt] \dfrac{d}{1+\varphi}\biggl(\varphi-\dfrac{c}{d}\biggr), & \varphi\geq b,\end{cases}\]

where $\eta>1$ is the positive solution of the quadratic equation

\[\dfrac{\omega^2}{2}\eta(\eta-1)+(\lambda_0-\lambda_1)\eta-(r+\lambda_0)=0,\]

and $b = {{c \eta}/{(d(\eta-1))}}$ . Furthermore,

\[\tau\,:\!=\, \inf\{t\geq 0\,:\, \Phi^\circ_t\geq b\}\]

is an optimal stopping time. More explicitly, in terms of the process X, we have

\[\tau=\inf\biggl\{t\geq 0\,:\, X_t\geq x + \dfrac{\sigma}{\omega}\biggl(\ln\biggl(\dfrac{b}{\varphi}\biggr) +(\lambda_1-\lambda_0)t\biggr)+\dfrac{\mu_0+\mu_1}{2}t\biggr\},\]

where $\omega\,:\!=\, (\mu_1-\mu_0)/\sigma$ .

6. An example: closing a short position

In this section we study an example of optimal closing of a short position under recall risk; see [Reference Glover and Hulley11]. We consider a short position in an underlying stock with unknown drift, where the random horizon corresponds to a time point at which the counterparty recalls the position. Naturally, the counterparty favours a large drift, and we thus assume that the risk of recall is greater in the state with a small drift. A similar model (but with no recall risk) was studied in [Reference Ekström and Lu5].

Let the stock price be modelled by geometric Brownian motion with dynamics

\[{\mathrm{d}} X_t = \mu(\theta) X_t\,{\mathrm{d}} t+ \sigma X_t \,{\mathrm{d}} W_t,\]

where the drift is state-dependent with $\mu(0)=\mu_0 <\mu_1=\mu(1)$ , and $\sigma$ is a known constant. We let $g(t,x,\theta)=h(t,x,\theta)=x\,{\mathrm{e}}^{-rt}$ and consider the stopping problem

\[V=\inf_{\tau\in\mathcal T^{X,\gamma}}{\mathbb E}_\pi[{\mathrm{e}}^{-r\tau\wedge\gamma} X_{\tau\wedge\gamma}],\]

where

\[F_i(t)\,:\!=\, {\mathbb P}_\pi(\gamma>t\mid \theta=i) = {\mathrm{e}}^{-\lambda_i t},\]

with $\lambda_0>0=\lambda_1$ and

\[{\mathbb P}_\pi(\theta= 1)=\pi =1-{\mathbb P}_\pi(\theta = 0) .\]

Here r is a constant discount rate; to avoid degenerate cases, we assume that $r\in(\mu_0,\mu_1)$ .

Then the value function can be written as $V=v/(1+\varphi)$ , where

(14) \begin{equation}v = \inf_{\tau\in\mathcal T^X}{\mathbb E}_0\biggl[ {\mathrm{e}}^{-r\tau}F_0(\tau)X_{\tau}\bigl( 1+ \Phi^\circ_\tau \bigr) + \lambda_0\int_0^\tau \,{\mathrm{e}}^{-rt}F_0(t) X_{t}\bigl( 1+ \Phi^\circ_t \bigr)\,{\mathrm{d}} t\biggr],\end{equation}

with ${\mathrm{d}}\Phi^\circ_t = \lambda_0\Phi^\circ_t \,{\mathrm{d}} t + \omega \Phi^\circ_t\,{\mathrm{d}} W_t$ and $\Phi^\circ_0 = \varphi$ . Here $\omega = {{(\mu_1-\mu_0)}/{\sigma}}$ .

Another change of measure will remove the occurrences of X in (14). In fact, let $\tilde {\mathbb {P}}$ be a measure with

\[\dfrac{{\mathrm{d}}\tilde {\mathbb P}}{{\mathrm{d}}{\mathbb P}_0}\bigg\vert_{\mathcal F_t} = {\mathrm{e}}^{-({{\sigma^2}/{2}})t + \sigma W_t},\]

so that $\tilde W_t=-\sigma t+W_t$ is a $\tilde{\mathbb P}$ -Brownian motion. Then

(15) \begin{equation}v=x\inf_{\tau\in\mathcal T^X}\tilde{\mathbb {E}}\biggl[ {\mathrm{e}}^{-(r+\lambda_0-\mu_0)\tau}\bigl( 1+ \Phi^\circ_\tau \bigr) + \lambda_0\int_0^\tau \,{\mathrm{e}}^{-(r+\lambda_0-\mu_0)t} \bigl( 1+ \Phi^\circ_t \bigr)\,{\mathrm{d}} t\biggr],\end{equation}

with

\[{\mathrm{d}}\Phi^0_t=(\lambda_0 +\sigma\omega)\Phi^0_t\,{\mathrm{d}} t + \omega \Phi^0_t\,{\mathrm{d}}\tilde W.\]

The optimal stopping problem (15) is a one-dimensional time-homogeneous problem, and is thus straightforward to analyse using standard stopping theory. Indeed, setting

\[\bar v\,:\!=\, \dfrac{v}{x}= \inf_{\tau\in\mathcal T^X}\tilde{\mathbb E}\biggl[ {\mathrm{e}}^{-(r+\lambda_0-\mu_0)\tau}\bigl( 1+ \Phi^\circ_\tau \bigr) + \lambda_0\int_0^\tau \,{\mathrm{e}}^{-(r+\lambda_0-\mu_0)t} \bigl( 1+ \Phi^\circ_t \bigr)\,{\mathrm{d}} t\biggr],\]

the associated free-boundary problem is to find $(\bar v,B)$ such that

(16) \begin{equation}\begin{cases}\dfrac{\omega^2\varphi^2}{2}\bar v_{\varphi\varphi} + (\lambda_0+\mu_1-\mu_0)\varphi\bar v_\varphi-(r+\lambda_0-\mu_0)\bar v + \lambda_0(1+\varphi)=0, & \varphi<B,\\[7pt]\bar v(\varphi)=1+\varphi, &\varphi\geq B,\\\bar v_\varphi(B)=1,\end{cases}\end{equation}

and such that $\bar v\leq 1+\varphi$ . Solving the free-boundary problem (16) gives

\[B=\dfrac{\eta(r-\mu_0)(\mu_1-r)}{(1-\eta)(r+\lambda_0-\mu_0)(\lambda_0+\mu_1-r)}\]

and

\[\bar v(\varphi)=\begin{cases}\dfrac{r-\mu_0}{(1-\eta)(r+\lambda_0-\mu_0)}\biggl(\dfrac{\varphi}{B}\biggr)^\eta-\dfrac{\lambda_0}{\mu_1-r}\varphi +\dfrac{\lambda_0}{r+\lambda_0-\mu_0}, & \varphi <B,\\[7pt]1+\varphi, & \varphi\geq B,\end{cases}\]

where $\eta<1$ is the positive solution of the quadratic equation

\[\dfrac{\omega^2}{2}\eta(\eta-1)+(\lambda_0+\mu_1-\mu_0)\eta-(r+\lambda_0-\mu_0)=0.\]

Note that $\bar v$ is concave since it satisfies the smooth-fit condition at B and since $\eta<1$ , so $\bar v(\varphi)\leq 1+\varphi$ . A standard verification argument then gives that

\[V=\frac{x}{1+\varphi}\bar v(\varphi),\]

and

\[\tau_B\,:\!=\, \inf\{t\geq 0\,:\, \Phi^\circ_t\geq B\}= \inf\{t\geq 0\,:\, \Phi_t\geq B\,{\mathrm{e}}^{-\lambda_0t}\}\]

is optimal in (14).

7. An example: a sequential testing problem with a random horizon

Consider the sequential testing problem for a Wiener process, i.e. the problem of determining as quickly, and accurately, an unknown drift $\theta$ from observations of the process

\[X_t=\theta t + \sigma W_t.\]

Similar to the classical version (see [Reference Shiryaev21]), we assume that $\theta$ is Bernoulli-distributed with ${\mathbb P}_\pi(\theta=1)=\pi=1-{\mathbb P}_\pi(\theta=0)$ , where $\pi\in(0,1)$ . In [Reference Novikov and Palacios-Soto19], the sequential testing problem has been studied under a random horizon. Here we consider an instance of a testing problem which further extends the set-up by allowing the distribution of the random horizon to depend on the unknown state.

More specifically, we assume that when $\theta = 1$ , then the horizon $\gamma$ is infinite, i.e. $F_1(t) = 1$ for all t; and when $\theta = 0$ , the time horizon is exponentially distributed with rate $\lambda$ , i.e. $F_0(t) = {\mathrm{e}}^{-\lambda t}$ . Mimicking the classical formulation of the problem, we study the problem of minimizing

\[{\mathbb P}_\pi(\theta\not= d) + c{\mathbb E}_\pi[\tau]\]

over all stopping times $\tau\in\mathcal T^{X,\gamma}$ and $\mathcal F^{X,\gamma}_\tau$ -measurable decision rules d with values in $\{0,1\}$ . By standard methods, the above optimization problem reduces to a stopping problem

\[V=\inf_{\tau\in\mathcal T^{X,\gamma}}{\mathbb E}_\pi\bigl[\hat\Pi_\tau\wedge (1-\hat\Pi_\tau) + c\tau\bigr],\]

where

\[\hat\Pi_t\,:\!=\, {\mathbb P}_\pi\bigl(\theta=1\mid \mathcal F^{X,\gamma}_t\bigr).\]

Moreover, the process $\hat\Pi$ satisfies

\[\hat\Pi_t=\begin{cases} \Pi^\circ_t, & t<\gamma,\\0, & t\geq \gamma,\end{cases}\]

where

\[\Pi^\circ_t=\dfrac{\Pi_t}{\Pi_t + (1-\Pi_t)\,{\mathrm{e}}^{-\lambda t}}=\dfrac{{\mathbb P}_\pi\bigl(\theta=1\mid \mathcal F^{X}_t\bigr)}{{\mathbb P}_\pi\bigl(\theta=1\mid \mathcal F^{X}_t\bigr) + \bigl(1-{\mathbb P}_\pi\bigl(\theta=1\mid \mathcal F^{X}_t\bigr)\bigr)\,{\mathrm{e}}^{-\lambda t}},\]

and it follows that

\begin{align*}V &= \inf_{\tau\in\mathcal T^{X,\gamma}}{\mathbb E}_\pi\bigl[\hat\Pi_\tau\wedge (1-\hat\Pi_\tau) + c\tau\bigr]\\ &= \inf_{\tau\in\mathcal T^{X,\gamma}}{\mathbb E}_\pi\bigl[\bigl(\Pi^\circ_\tau\wedge \bigl(1-\Pi_\tau^\circ\bigr)\bigr)1_{\{\tau<\gamma\}} + c(\tau\wedge\gamma) \bigr]\\ &=\inf_{\tau\in\mathcal T^{X,\gamma}}{\mathbb E}_\pi\bigl[\bigl(\Pi^\circ_\tau\wedge \bigl(1-\Pi_\tau^\circ\bigr)+ c\tau\bigr)1_{\{\tau<\gamma\}} + c\gamma 1_{\{\gamma\leq \tau\}}\bigr].\end{align*}

In other words, $g_i(t,\pi) = \pi \wedge (1-\pi) +ct$ and $h_i(t,\pi) = ct$ for $i \in\{0,1\}$ . Following the general methodology leading up to Theorem 1, we find that

\begin{align*} V &= \inf_{\tau\in\mathcal T^{X}}{\mathbb E}_\pi\biggl[\bigl(\Pi^\circ_\tau\wedge \bigl(1-\Pi_\tau^\circ\bigr)+c\tau\bigr)((1-\Pi_\tau)F_0(\tau) +\Pi_\tau)- c\int_0^\tau t(1-\Pi_t)\,{\mathrm{d}} F_0(t)\biggr]\\ &= \inf_{\tau\in\mathcal T^{X}} {\mathbb E}_\pi\biggl[\bigl(\Pi^\circ_\tau\wedge \bigl(1-\Pi_\tau^\circ\bigr)+c\tau\bigr)((1-\Pi_\tau)F_0(\tau) +\Pi_\tau)- c\tau(1-\Pi_{\tau}) F_0(\tau) \\ &\quad - c\int_0^\tau F_0(t)\,{\mathrm{d}}(t(1-\Pi_t))\biggr]\\&= \inf_{\tau\in\mathcal T^{X}}{\mathbb E}_\pi\biggl[\bigl(\Pi^\circ_\tau\wedge \bigl(1-\Pi_\tau^\circ\bigr)\bigr)((1-\Pi_\tau)F_0(\tau) +\Pi_\tau)+ c\int_0^\tau((1-\Pi_t)F_0(t) +\Pi_t)\,{\mathrm{d}} t\biggr]\\ &= \dfrac{1}{1+\varphi} \inf_{\tau\in\mathcal T^{X}}{\mathbb E}_0\biggl[F_0(\tau)\bigl(\Phi^\circ_\tau\wedge 1\bigr)+ c\int_0^\tau F_0(t)\bigl(1 +\Phi^\circ_t\bigr) \,{\mathrm{d}} t\biggr].\end{align*}

Here $\Phi^\circ\,:\!=\, \Pi^\circ/(1-\Pi^\circ)$ satisfies

\[{\mathrm{d}}\Phi^\circ_t=\lambda \Phi^\circ_t\,{\mathrm{d}} t + \omega \Phi^\circ_t \,{\mathrm{d}} W_t,\]

where $\omega = {{1}/{\sigma}}$ .

Standard stopping theory can now be applied to solve the sequential testing problem with a random horizon. Setting

\[v(\varphi)\,:\!=\,\inf_{\tau\in\mathcal T^{X}}{\mathbb E}_0\biggl[F_0(\tau)\bigl(\Phi^\circ_\tau\wedge 1\bigr)+ c\int_0^\tau F_0(t)\bigl(1 +\Phi^\circ_t\bigr) \,{\mathrm{d}} t\biggr],\]

where $\Phi^\circ_0=\varphi$ , one expects a two-sided stopping region $(0,A]\cup[B,\infty)$ , and v to satisfy

\[\begin{cases}\dfrac{1}{2}\omega^2\varphi^2 v_{\varphi\varphi}+\lambda \varphi v_\varphi -\lambda v+c(1+\varphi) =0, & \varphi \in (A,B),\\v(A) =A,\\v_\varphi(A) =1,\\v(B) =1,\\v_\varphi(B) =0,\end{cases}\]

for some constants A, B with $0<A<1<B$ . The general solution of the ODE is easily seen to be

\[v(\varphi)=C_1 \varphi^{-{{2\lambda}/{\omega^2}}}+C_2 \varphi+\dfrac{c}{\lambda} - \dfrac{c}{\lambda+\frac{1}{2}\omega^2}\varphi\ln(\varphi),\]

where $C_1, C_2$ are arbitrary constants. Since the stopping region is two-sided, explicit solutions are not expected. Instead, using the four boundary conditions, equations for the unknowns $C_1$ , $C_2$ , A, and B can be derived using standard methods; we omit the details.