Time-consistent stopping under decreasing impatience

Abstract

Under non-exponential discounting, we develop a dynamic theory for stopping problems in continuous time. Our framework covers discount functions that induce decreasing impatience. Due to the inherent time inconsistency, we look for equilibrium stopping policies, formulated as fixed points of an operator. Under appropriate conditions, fixed-point iterations converge to equilibrium stopping policies. This iterative approach corresponds to the hierarchy of strategic reasoning in game theory and provides “agent-specific” results: it assigns one specific equilibrium stopping policy to each agent according to her initial behavior. In particular, it leads to a precise mathematical connection between the naive behavior and the sophisticated one. Our theory is illustrated in a real options model.

Appendices

Appendix A: Proofs for Sect. 3

Throughout this appendix, we constantly use the notation

$$ \tau_{n} := \Theta^{n} \tau,\qquad n\in\mathbb{N}, \hbox{for any}\ \tau\in{\mathcal {T}}(\mathbb{X}). $$

(A.1)

1.1 A.1 Proof of Proposition 3.11

Fix $(t,x)\in\mathbb{X}$. We deal with the two cases $\widetilde{\tau}(t,x) = 0$ and $\widetilde{\tau}(t,x) = 1$ separately. If $\widetilde{\tau}(t,x) = 0$, i.e., $\widetilde{\tau}_{t,x} = t$, then by (2.6),

$$g(x) =\sup_{\tau\in{\mathcal {T}}_{t}}\mathbb{E}^{t,x}[\delta(\tau-t) g(X_{\tau})] \ge\mathbb{E}^{t,x}\big[\delta\big({\mathcal {L}}^{*}\widetilde{\tau}(t,x)-t\big) g(X_{{\mathcal {L}}^{*}\widetilde{\tau}(t,x)})\big], $$

which implies $(t,x)\in S_{\widetilde{\tau}}\cup I_{\widetilde{\tau}}$. We then conclude from (3.6) that

$$\Theta\widetilde{\tau}(t,x)= \left\{ \textstyle\begin{array}{ll} 0, \quad & \text{ if }(t,x)\in S_{\widetilde{\tau}} \\ \widetilde{\tau}(t,x), \quad & \text{ if }(t,x)\in I_{\widetilde {\tau}} \end{array}\displaystyle \right\} \ = \widetilde{\tau}(t,x). $$

If $\widetilde{\tau}(t,x) =1$, then ${\mathcal {L}}^{*}\widetilde{\tau}(t,x) = {\mathcal {L}}\widetilde{\tau}(t,x) = \inf\{s\ge t :\widetilde{\tau}_{s,X^{t,x}_{s}}=s \}$. By (2.5) and (2.4), $\widetilde{\tau}_{s,X^{t,x}_{s}}=s$ means that

$$g\big(X^{t,x}_{s}(\omega)\big) = \mathop{\rm ess\, sup}_{\tau\in {\mathcal {T}}_{s}}\mathbb{E}^{s,X^{t,x}_{s}(\omega)}[\delta(\tau-s) g(X_{\tau})], $$

which is equivalent to

$$\begin{aligned} \delta(s-t) g\big(X^{t,x}_{s}(\omega)\big) &= \delta(s-t) \mathop {\rm ess\, sup}_{\tau\in{\mathcal {T}}_{s}}\mathbb {E}^{s,X^{t,x}_{s}(\omega)}[\delta(\tau-s) g(X_{\tau})]\\ &= \mathop{\rm ess\, sup}_{\tau\in{\mathcal {T}}_{s}}\mathbb {E}^{s,X^{t,x}_{s}(\omega)}[\delta(\tau-t) g(X_{\tau})] = Z^{t,x}_{s}(\omega), \end{aligned}$$

where the second equality follows from (2.7). As a result, we can conclude that ${\mathcal {L}}^{*}\widetilde{\tau}(t,x) = \inf\{s\ge t: \delta(s-t) g(X^{t,x}_{s})= Z^{t,x}_{s}\}=\widetilde{\tau}_{t,x}$. This together with (2.6) shows that

$$\begin{aligned} \mathbb{E}^{t,x}\big[\delta\big({\mathcal {L}}^{*}\widetilde{\tau}(t,x)-t\big) g(X_{{\mathcal {L}}^{*}\widetilde{\tau}(t,x)})\big] &= \mathbb{E}^{t,x}[\delta(\widetilde{\tau}_{t,x}-t) g(X_{\widetilde{\tau}_{t,x}})]\ge g(x), \end{aligned}$$

which implies $(t,x)\in I_{\widetilde{\tau}}\cup C_{\widetilde{\tau}}$. By (3.6), we have

$$\Theta\widetilde{\tau}(t,x)= \left\{ \textstyle\begin{array}{ll} \widetilde{\tau}(t,x), \quad & \text{ if }(t,x)\in I_{\widetilde {\tau}} \\ 1,\quad & \text{ if }(t,x)\in C_{\widetilde{\tau}} \end{array}\displaystyle \right\} \ = \widetilde{\tau}(t,x). $$

We therefore have $\Theta\widetilde{\tau}(t,x) = \widetilde{\tau}(t,x)$ for all $(t,x)\in\mathbb{X}$, i.e., $\widetilde{\tau}\in{\mathcal {E}}(\mathbb{X})$.

1.2 A.2 Derivation of Proposition 3.13

To prove the technical result in Lemma A.1 below, we need to introduce shifted random variables as in Nutz [25]. Recall from Sect. 2 that $\Omega$ is the canonical path space. For any $t\ge0$ and $\omega\in\Omega$, we define the concatenation of $\omega$ and $\tilde{\omega}\in\Omega$ at time $t$ by

$$(\omega\otimes_{t}\tilde{\omega})_{s} := \omega_{s} \mathbf {1}_{[0,t)}(s) + \big(\tilde{\omega}_{s}-(\tilde{\omega}_{t} - \omega _{t})\big) \mathbf{1}_{[t,\infty)} (s),\quad s\ge0. $$

For any ${\mathcal {F}}_{\infty}$-measurable random variable $\xi:\Omega \to\mathbb{R}$, we define the shifted random variable $[\xi]_{t,\omega}:\Omega\to\mathbb{R}$, which is ${\mathcal {F}}^{t}_{\infty}$-measurable, by

$$[\xi]_{t,\omega} (\tilde{\omega}):= \xi(\omega\otimes_{t} \tilde{\omega})\quad\forall\tilde{\omega}\in\Omega. $$

Given $\tau\in{\mathcal {T}}$, we write $\omega\otimes_{\tau(\omega )}\tilde{\omega}$ as $\omega\otimes_{\tau}\tilde{\omega}$, and $[\xi]_{\tau(\omega ),\omega} (\tilde{\omega})$ as $[\xi]_{\tau,\omega} (\tilde{\omega})$. A detailed analysis of shifted random variables can be found in [4, Appendix A]; Proposition A.1 there implies that for fixed $(t,x)\in\mathbb{X}$, any $\theta\in{\mathcal {T}}_{t}$ and ${\mathcal {F}}^{t}_{\infty}$-measurable $\xi$ with $\mathbb {E}^{t,x}[|\xi|]<\infty$ satisfy

$$ \mathbb{E}^{t,x}[\xi\, |\, {\mathcal {F}}^{t}_{\theta}](\omega) = \mathbb {E}^{t,x}\big[[\xi]_{\theta,\omega}\big]\quad \hbox{for a.e. $\omega\in\Omega$}. $$

(A.2)

Lemma A.1

For any $\tau\in{\mathcal {T}}(\mathbb{X})$ and $(t,x)\in\mathbb {X}$, define $t_{0}:= {\mathcal {L}}^{*}\tau_{1}(t,x)\in{\mathcal {T}}_{t}$ and $s_{0} := {\mathcal {L}}^{*}\tau(t,x)\in{\mathcal {T}}_{t}$, with $\tau_{1}$ as in (A.1). If $t_{0}\le s_{0}$, then for a.e. $\omega\in\{t < t_{0}\}$,

$$g\big(X^{t,x}_{t_{0}}(\omega)\big) \le\mathbb{E}^{t,x}[\delta (s_{0}-t_{0})g(X_{s_{0}})\, |\,{\mathcal {F}}^{t}_{t_{0}}] (\omega). $$

Proof

For a.e. $\omega\in\{t< t_{0}\}\in{\mathcal {F}}_{t}$, we deduce from $t_{0} (\omega) = {\mathcal {L}}^{*}\tau_{1}(t,x) (\omega)>t$ that $\tau_{1}(s,X^{t,x}_{s}(\omega))=1$ for all $s\in(t,t_{0}(\omega))$. In view of (A.1) and (3.6), this implies $(s,X^{t,x}_{s}(\omega))\notin S_{\tau}$ for all $s\in (t,t_{0}(\omega))$. Thus,

$$\begin{aligned} g\big(X^{t,x}_{s}(\omega)\big) \le \mathbb{E}^{s,X^{t,x}_{s}(\omega)}\big[\delta\big({\mathcal {L}}^{*}\tau (s,X_{s})-s\big)g(X_{{\mathcal {L}}^{*}\tau(s,X_{s})})\big]\quad \forall s\in\big(t,t_{0}(\omega)\big). \end{aligned}$$

(A.3)

For any $s\in(t,t_{0}(\omega))$, note that

$$[t_{0}]_{s,\omega} (\tilde{\omega})= t_{0}(\omega\otimes_{s}\tilde{\omega}) = {\mathcal {L}}^{*}\tau_{1}(t,x)(\omega\otimes_{s}\tilde{\omega})= {\mathcal {L}}^{*}\tau_{1}\big(s,X^{t,x}_{s}(\omega)\big)(\tilde{\omega}) $$

for all $\tilde{\omega}\in\Omega$. As $t_{0}\le s_{0}$, a similar calculation gives

$$[s_{0}]_{s,\omega} (\tilde{\omega})= {\mathcal {L}}^{*}\tau\big(s,X^{t,x}_{s}(\omega)\big)(\tilde{\omega}). $$

We thus conclude from (A.3) that

$$\begin{aligned} g\big(X^{t,x}_{s}(\omega)\big) &\le\mathbb{E}^{s,X^{t,x}_{s}(\omega )}\big[\delta([s_{0}]_{s,\omega}-s)g([X_{s_{0}}]_{s,\omega})\big] \\ &\le\mathbb{E}^{s,X^{t,x}_{s}(\omega)}\big[\delta([s_{0}]_{s,\omega }-[t_{0}]_{s,\omega})g([X_{s_{0}}]_{s,\omega})\big]\quad\forall s\in \big(t,t_{0}(\omega)\big), \end{aligned}$$

(A.4)

where the second line holds because $\delta$ is decreasing and also $\delta$ and $g$ are both nonnegative. On the other hand, by (A.2), it holds a.s. that

$$\begin{aligned} & \mathbb{E}^{t,x}[\delta(s_{0}-t_{0}) g(X_{s_{0}})\, | \, {\mathcal {F}}^{t}_{s}](\omega) \\ &= \mathbb{E}^{t,x}\left[\delta([s_{0}]_{s,\omega}-[t_{0}]_{s,\omega}) g([X^{t,x}_{s_{0}}]_{s,\omega})\right]\quad \forall s\ge t, s\in \mathbb{Q}. \end{aligned}$$

Note that we used the countability of ℚ to obtain the above almost sure statement. This together with (A.4) shows that it holds a.s. that

$$ g\big(X^{t,x}_{s}(\omega)\big)\ \mathbf{1}_{\{(t,t_{0}(\omega))\cap \mathbb{Q}\}}(s) \le\mathbb{E}^{t,x}[\delta(s_{0}-t_{0}) g(X_{s_{0}})\, | \,{\mathcal {F}}^{t}_{s}](\omega)\ \mathbf{1}_{\{(t,t_{0}(\omega))\cap \mathbb{Q}\}}(s). $$

(A.5)

Since our sample space $\Omega$ is the canonical space for Brownian motion with the right-continuous Brownian filtration $\mathbb{F}$, the martingale representation theorem holds under the current setting. This implies that every martingale has a continuous version. Let $(M_{s})_{s\ge t}$ be the continuous version of the martingale $(\mathbb{E}^{t,x}[\delta(s_{0}-t_{0}) g(X_{s_{0}})\, | \,{\mathcal {F}}^{t}_{s}])_{s\ge t}$. Then (A.5) immediately implies that it holds a.s. that

$$ g\big(X^{t,x}_{s}(\omega)\big) \mathbf{1}_{\{(t,t_{0}(\omega))\cap \mathbb{Q}\}}(s) \le M_{s}(\omega)\ \mathbf{1}_{\{(t,t_{0}(\omega))\cap \mathbb{Q}\}}(s). $$

(A.6)

Also, using the right-continuity of $M$ and (A.2), one can show that for any $\tau\in{\mathcal {T}}_{t}$, we have $M_{\tau}= \mathbb{E}^{t,x}[\delta(s_{0}-t_{0}) g(X_{s_{0}})\, | \, {\mathcal {F}}^{t}_{\tau}]$ a.s. Now we can take some $\Omega^{*}\in{\mathcal {F}}_{\infty}$ with $\mathbb{P}[\Omega^{*}] =1$ such that (A.6) holds true and $M_{t_{0}}(\omega) = \mathbb{E}^{t,x}[\delta(s_{0}-t_{0}) g(X_{s_{0}})\, | \,{\mathcal {F}}^{t}_{t_{0}}](\omega)$ for all $\omega\in\Omega^{*}$. For any $\omega\in\Omega^{*}\cap\{t< t_{0}\}$, take $({k_{n}})\subseteq\mathbb {Q}$ such that $k_{n} >t$ and $k_{n}\uparrow t_{0}(\omega)$. Then (A.6) implies that $g(X^{t,x}_{k_{n}}(\omega)) \le M_{k_{n}}(\omega),\ \forall n\in\mathbb{N}$. As $n\to\infty$, we obtain from the continuity of $s\mapsto X_{s}$ and $z\mapsto g(z)$ and the left-continuity of $s\mapsto M_{s}$ that $g(X^{t,x}_{t_{0}}(\omega)) \le M_{t_{0}}(\omega) = \mathbb{E}^{t,x}[\delta(s_{0}-t_{0}) g(X_{s_{0}})\, | \,{\mathcal {F}}^{t}_{t_{0}}](\omega)$. □

Now we are ready to prove Proposition 3.13.

Proof of Proposition 3.13

We prove (3.10) by induction. We know that the result holds for $n=0$ by (3.9). Now assume that (3.10) holds for $n=k\in\mathbb{N}\cup\{0\}$, and we intend to show that (3.10) also holds for $n=k+1$. Recall the notation in (A.1). Fix $(t,x)\in\ker(\tau_{k+1})$, i.e., $\tau_{k+1}(t,x)=0$. If ${\mathcal {L}}^{*}\tau_{k+1}(t,x) = t$, then $(t,x)$ belongs to $I_{\tau_{k+1}}$. By (3.6), we get $\tau _{k+2}(t,x) = \Theta\tau_{k+1}(t,x)= \tau_{k+1}(t,x) =0$, i.e., $(t,x)\in\ker(\tau_{k+2})$ as desired. We therefore assume below that ${\mathcal {L}}^{*}\tau_{k+1}(t,x)>t$.

By (3.6), $\tau_{k+1}(t,x)=0$ implies

$$ g(x)\ge\mathbb{E}^{t,x}\big[\delta\big({\mathcal {L}}^{*}\tau _{k}(t,x)-t\big) g(X_{{\mathcal {L}}^{*}\tau_{k}(t,x)})\big]. $$

(A.7)

Let $t_{0} := {\mathcal {L}}^{*}\tau_{k+1}(t,x)$ and $s_{0} := {\mathcal {L}}^{*}\tau_{k}(t,x)$. Under the induction hypothesis that $\ker(\tau_{k})\subseteq\ker(\tau_{k+1})$, we have $t_{0}\le s_{0}$ as $t_{0}$ and $s_{0}$ are hitting times to $\ker(\tau_{k+1})$ and $\ker(\tau_{k})$, respectively; see (3.5). Using (A.7), $t_{0}\le s_{0}$, Assumption 3.12 and $g$ being nonnegative, we obtain

$$\begin{aligned} g(x) &\ge \mathbb{E}^{t,x}[\delta(s_{0}-t) g(X_{s_{0}})] \\ & \ge\mathbb{E}^{t,x}[\delta(t_{0}-t)\delta(s_{0}-t_{0}) g(X_{s_{0}})]\\ &=\mathbb{E}^{t,x}\big[\delta(t_{0}-t)\mathbb{E}^{t,x}[\delta (s_{0}-t_{0})g(X_{s_{0}})\ |\ {\mathcal {F}}^{t}_{t_{0}}]\big]\\ &\ge\mathbb{E}^{t,x}[\delta(t_{0}-t)g(X_{t_{0}})], \end{aligned}$$

where the third line follows from the tower property of conditional expectations and the fourth is due to Lemma A.1. This implies $(t,x)\notin C_{\tau_{k+1}}$ and thus

$$\tau_{k+2}(t,x) = \left\{ \textstyle\begin{array}{ll} 0 \quad & \text{ for }(t,x)\in S_{\tau_{1}} \\ \tau_{k+1} (t,x) \quad & \text{ for }(t,x)\in I_{\tau_{1}} \end{array}\displaystyle \right\} = 0. $$

That is, $(t,x)\in\ker(\tau_{k+2})$. Thus, we conclude that $\ker (\tau_{k+1})\subseteq\ker(\tau_{k+2})$ as desired.

It remains to show that $\tau_{0}$ defined in (3.7) is a stopping policy. Observe that for any $(t,x)\in\mathbb{X}$, $\tau_{0}(t,x) =0$ if and only if $\Theta^{n}\tau (t,x)=0$, i.e., $(t,x)\in\ker(\Theta^{n}\tau)$, for $n$ large enough. This together with (3.10) implies that

$$\{(t,x)\in\mathbb{X}: \tau_{0}(t,x)=0 \} = \bigcup_{n\in\mathbb {N}}\ker(\Theta^{n}\tau)\in{\mathcal {B}}(\mathbb{X}). $$

Hence $\tau_{0}:\mathbb{X}\to\{0,1\}$ is Borel-measurable and thus an element in ${\mathcal {T}}(\mathbb{X})$. □

1.3 A.3 Proof of Proposition 3.15

Fix $(t,x)\in\ker(\widetilde{\tau})$. Since $\widetilde{\tau}(t,x)=0$, i.e., $\widetilde{\tau}_{t,x}=t$, (2.5), (2.4) and (2.6) imply

$$g(x) = \sup_{\tau\in{\mathcal {T}}_{t}} \mathbb{E}^{t,x}[\delta(\tau -t) g(X_{\tau})] \ge\mathbb{E}^{t,x} \big[\delta\big({\mathcal {L}}^{*}\widetilde{\tau}(t,x)-t\big) g(X_{{\mathcal {L}}^{*}\widetilde{\tau}(t,x)})\big]. $$

This shows that $(t,x)\in S_{\widetilde{\tau}}\cup I_{\widetilde{\tau}}$. Thus we have $\ker(\widetilde{\tau})\subseteq S_{\widetilde{\tau}}\cup I_{\widetilde{\tau}}$. It follows that

$$\ker(\widetilde{\tau}) = \big(\ker(\widetilde{\tau})\cap S_{\widetilde{\tau}}\big) \cup \big(\ker(\widetilde{\tau})\cap I_{\widetilde{\tau}}\big) \subseteq S_{\widetilde{\tau}} \cup \big(\ker(\widetilde{\tau})\cap I_{\widetilde{\tau}}\big) = \ker (\Theta\widetilde{\tau}), $$

where the last equality follows from (3.6).

1.4 A.4 Derivation of Theorem 3.16

Lemma A.2

Suppose Assumption  3.12 holds and $\tau\in{\mathcal {T}}(\mathbb{X})$ satisfies (3.9). Then $\tau_{0}$ defined in (3.7) satisfies

$${\mathcal {L}}^{*}\tau_{0}(t,x) = \lim_{n\to\infty} {\mathcal {L}}^{*}\Theta ^{n}\tau(t,x)\quad\forall(t,x)\in\mathbb{X}. $$

Proof

We use the notation in (A.1). Recall that we have $\ker(\tau _{n})\subseteq\ker(\tau_{n+1})$ for all $n\in\mathbb{N}$ and $\ker(\tau_{0}) = \bigcup_{n\in\mathbb {N}}\ker(\tau_{n})$ from Proposition 3.13. By (3.5), this implies that $({\mathcal {L}}^{*}\tau_{n}(t,x))_{n\in \mathbb{N}}$ is a nonincreasing sequence of stopping times and

$${\mathcal {L}}^{*}\tau_{0}(t,x) \le t_{0}:=\lim_{n\to\infty} {\mathcal {L}}^{*}\tau_{n}(t,x). $$

It remains to show that ${\mathcal {L}}^{*}\tau_{0}(t,x) \ge t_{0}$. We deal with the following two cases.

(i) On $\{\omega\in\Omega: {\mathcal {L}}^{*}\tau_{0}(t,x)(\omega)=t\}$: By (3.5), there exists a sequence $(t_{m})_{m\in\mathbb{N}}$ in $\mathbb{R}_{+}$, depending on $\omega\in \Omega$, such that $t_{m}\downarrow t$ and $\tau_{0}(t_{m}, X^{t,x}_{t_{m}}(\omega)) = 0$ for all $m\in\mathbb{N}$. For each $m\in \mathbb{N}$, by the definition of $\tau_{0}$ in (3.7), there exists $n^{*}\in\mathbb{N}$ large enough such that $\tau_{n^{*}}(t_{m}, X^{t,x}_{t_{m}}(\omega)) = 0$, which implies ${\mathcal {L}}^{*}\tau_{n^{*}}(t,x)(\omega)\le t_{m}$. Since $({\mathcal {L}}^{*}\tau_{n}(t,x))_{n\in\mathbb{N}}$ is nonincreasing, we have $t_{0}(\omega)\le{\mathcal {L}}^{*}\tau_{n^{*}}(t,x)(\omega)\le t_{m}$. With $m\to\infty$, we obtain $t_{0}(\omega)\le t={\mathcal {L}}^{*}\tau_{0}(t,x)(\omega)$.

(ii) On $\{\omega\in\Omega: {\mathcal {L}}^{*}\tau_{0}(t,x)(\omega)>t\}$: Set $s_{0}:={\mathcal {L}}^{*}\tau_{0}(t,x)$ and focus on the value of $\tau_{0}(s_{0}(\omega), X^{t,x}_{s_{0}}(\omega))$. If $\tau _{0}(s_{0}(\omega), X^{t,x}_{s_{0}}(\omega))=0$, then by (3.7) there exists $n^{*}\in\mathbb{N}$ large enough such that $\tau_{n^{*}}(s_{0}(\omega), X^{t,x}_{s_{0}}(\omega))=0$. Since $({\mathcal {L}}^{*}\tau_{n}(t,x))_{n\in \mathbb{N}}$ is nonincreasing, $t_{0}(\omega)\le {\mathcal {L}}^{*}\tau_{n^{*}}(t,x)(\omega)\le s_{0}(\omega)$ as desired. If $\tau_{0}(s_{0}(\omega), X^{t,x}_{s_{0}}(\omega))=1$, then by (3.5), there exists a sequence $(t_{m})_{m\in\mathbb{N}}$ in $\mathbb{R}_{+}$, depending on $\omega\in\Omega$, such that $t_{m}\downarrow s_{0}(\omega)$ and $\tau_{0}(t_{m}, X^{t,x}_{t_{m}}(\omega)) = 0$ for all $m\in\mathbb{N}$. Then we can argue as in case (i) to show that $t_{0}(\omega)\le s_{0}(\omega)$ as desired. □

Now we are ready to prove Theorem 3.16.

Proof of Theorem 3.16

By Proposition 3.13, $\tau_{0}\in{\mathcal {T}}(\mathbb{X})$ is well defined. For simplicity, we use the notation in (A.1). Fix $(t,x)\in\mathbb{X}$. If $\tau_{0}(t,x)=0$, then (3.7) gives $\tau_{n}(t,x)=0$ for $n$ large enough. Since $\tau_{n}(t,x) = \Theta\tau_{n-1}(t,x)$, we deduce from “$\tau_{n}(t,x)=0$ for $n$ large enough” and (3.6) that $(t,x)\in S_{\tau_{n-1}}\cup I_{\tau _{n-1}}$ for $n$ large enough. That is, $g(x)\ge\mathbb{E}^{t,x}[\delta({\mathcal {L}}^{*}\tau_{n-1}(t,x)-t) g(X_{{\mathcal {L}}^{*}\tau_{n-1}(t,x)})]\ \hbox{for $n$ large enough}$. With $n\to\infty$, the dominated convergence theorem and Lemma A.2 yield

$$g(x)\ge\mathbb{E}^{t,x}\big[\delta\big({\mathcal {L}}^{*}\tau _{0}(t,x)-t\big) g(X_{{\mathcal {L}}^{*}\tau_{0}(t,x)})\big], $$

which shows that $(t,x)\in S_{\tau_{0}}\cup I_{\tau_{0}}$. We then deduce from (3.6) and $\tau_{0}(t,x)=0$ that $\Theta\tau_{0}(t,x) = \tau_{0}(t,x)$. On the other hand, if $\tau _{0}(t,x)=1$, then (3.7) gives $\tau_{n}(t,x)=1$ for $n$ large enough. Since $\tau_{n}(t,x) = \Theta \tau_{n-1}(t,x)$, we deduce from “$\tau_{n}(t,x)=1$ for $n$ large enough” and (3.6) that $(t,x)\in C_{\tau_{n-1}}\cup I_{\tau_{n-1}}$ for $n$ large enough. That is, $g(x)\le\mathbb{E}^{t,x}[\delta ({\mathcal {L}}^{*}\tau_{n-1}(t,x)-t) g(X_{{\mathcal {L}}^{*}\tau_{n-1}(t,x)})]\ \hbox{for }n\hbox{ large enough}$. With $n\to\infty$, the dominated convergence theorem and Lemma A.2 yield

$$g(x)\le\mathbb{E}^{t,x}\big[\delta\big({\mathcal {L}}^{*}\tau _{0}(t,x)-t\big) g(X_{{\mathcal {L}}^{*}\tau_{0}(t,x)})\big], $$

which shows that $(t,x)\in C_{\tau_{0}}\cup I_{\tau_{0}}$. We then deduce from (3.6) and $\tau_{0}(t,x)=1$ that $\Theta\tau_{0}(t,x) = \tau_{0}(t,x)$. We therefore conclude that $\tau_{0}\in{\mathcal {E}}(\mathbb{X})$. □

Appendix B: Proofs for Sect. 4

2.1 B.1 Derivation of Proposition 4.1

In the classical case of exponential discounting, (2.7) ensures that for all $s\ge0$,

$$ \delta(s)v(X_{s}^{x})=\sup_{\tau\in{\mathcal {T}}}\mathbb {E}^{X^{x}_{s}}[\delta(s+\tau) g(X_{\tau})]= \sup_{\tau\in{\mathcal {T}}_{s}}\mathbb{E}^{x}[\delta(\tau) g(X_{\tau}) \, | \, {\mathcal {F}}_{s}], $$

(B.1)

which shows that $(\delta(s)v(X_{s}^{x}))_{s\ge0}$ is a supermartingale. Under hyperbolic discounting (4.1), since $\delta(r_{1})\delta(r_{2}) \leq\delta (r_{1}+r_{2})$ for all $r_{1},r_{2}\ge0$, $(\delta(s)v(X_{s}^{x}))_{s\ge t}$ need no longer be a supermartingale as the first equality in the above equation fails.

To overcome this, we introduce an auxiliary value function: for $(s,x)\in\mathbb{R}^{2}_{+}$,

$$\begin{aligned} V(s,x)&:=\sup_{\tau\in{\mathcal {T}}}\mathbb{E}^{x}[\delta(s+\tau) g(X_{\tau})] =\sup_{\tau\in{\mathcal {T}}}\mathbb{E}^{x}\bigg[\frac {X_{\tau}}{1+\beta(s+\tau)}\bigg]. \end{aligned}$$

By definition, $V(0,x)=v(x)$, and $(V(s,X^{x}_{s}) )_{s\ge0}$ is a supermartingale as $V(s,X^{x}_{s})$ is equal to the right-hand side of (B.1).

Proof of Proposition 4.1

Recall that $X_{s}=|W_{s}|$ for a one-dimensional Brownian motion $W$. Let $y\in\mathbb{R}$ be the initial value of $W$ and define $\bar{V}(s,y) := V(s,|y|)$. The associated variational inequality for $\bar{V}(s,y)$ is the following: for $(s,y)\in[0,\infty)\times\mathbb{R}$,

$$ \min\left\{w_{s}(s,y)+\frac{1}{2}w_{yy}(s,y),\ w(s,y)-\frac {|y|}{1+\beta s}\right\} =0. $$

(B.2)

Taking $s\mapsto b(s)$ as the free boundary to be determined, we can rewrite (B.2) as

$$ \textstyle\begin{cases} \hbox{$w_{s}(s,y) + \frac{1}{2}w_{yy}(s,y)=0$ and $ w(s,y)>\frac {|y|}{1+\beta s}$} \quad & \hbox{for}\ |y|< b(s),\\ w(s,y)=\frac{|y|}{1+\beta s} \quad &\hbox{for}\ |y|\ge b(s). \end{cases} $$

(B.3)

Following [27], we propose the ansatz $w(s,y)=\frac{1}{\sqrt{1+\beta s}} h(\frac{y}{\sqrt{1+\beta s}})$. Equation (B.3) then becomes a one-dimensional free boundary problem, namely

$$ \left\{ \textstyle\begin{array}{ll} \hbox{$-\beta zh'(z)+h''(z)=\beta h(z)$ and $h(z)>|z|$} \quad \,& \hbox{for}\ |z|< \frac{b(s)}{\sqrt{1+\beta s}},\\ h(z)=|z| \quad \,& \hbox{for} \ |z|\ge\frac{b(s)}{\sqrt{1+\beta s}}. \end{array}\displaystyle \right. $$

(B.4)

As the variable $s$ does not appear in the above ODE, take $b(s) = \alpha\sqrt{1+ \beta s}$ for some $\alpha\ge0$. The general solution of the differential equation in the first line of (B.4) is

$$h(z) = e^{\frac{\beta}{2}z^{2}}\bigg(c_{1}+c_{2} \sqrt {\frac{2}{\beta}}\int_{0}^{\sqrt{{\beta}/{2}}\ z} e^{-u^{2}}du\bigg),\quad (c_{1}, c_{2}) \in\mathbb{R}^{2}\; . $$

We then have

$$w(s,y) = \textstyle\begin{cases} \frac{e^{\frac{\beta y^{2}}{2(1+\beta s)}}}{\sqrt{1+\beta s}}\bigg(c_{1} + c_{2} \sqrt{\frac{2}{\beta}}\int_{0}^{\frac{\sqrt{{\beta }/{2}} }{\sqrt{1+\beta s}}y} e^{-u^{2}}du\bigg), \quad & |y|< \alpha\sqrt{1+\beta s},\\ \frac{|y|}{1+ \beta s}, & |y|\ge\alpha\sqrt{1+\beta s}. \end{cases} $$

To find the parameters $c_{1}, c_{2}$ and $\alpha$, we equate the values of $w(s,y)$ and its partial derivatives on both sides of the free boundary. This yields the equations

$$ \alpha= e^{\frac{\beta}{2}\alpha^{2}}\bigg(c_{1} \pm c_{2}\sqrt{\frac {2}{\beta}}\int_{0}^{\sqrt{{\beta}/{2}}\ \alpha}e^{-u^{2}}du\bigg) \qquad \text{and}\qquad \alpha^{2}\beta+ c_{2} =1. $$

The first equation implies $c_{2}=0$. Then, these equations together yield $\alpha= 1/\sqrt{\beta}$ and $c_{1} = \alpha e^{-1/2}$. Thus we obtain

$$ w(s,y) = \textstyle\begin{cases} \frac{1}{\sqrt{\beta}\sqrt{1+\beta s}}\exp(\frac{1}{2}(\frac {\beta y^{2}}{1+\beta s}-1)), \quad & |y|< \sqrt{1/\beta+s},\\ \frac{|y|}{1+ \beta s}, & |y|\ge\sqrt{1/\beta+s}. \end{cases} $$

(B.5)

Note that $w(s,y)> \frac{|y|}{1+ \beta s}$ for $|y|<\sqrt{1/\beta +s}$. Indeed, by defining the function

$$h(y) := \frac{1}{\sqrt{\beta}\sqrt{1+\beta s}}\exp\bigg(\frac {1}{2}\Big(\frac{\beta y^{2}}{1+\beta s}-1\Big)\bigg)-\frac {y}{1+\beta s} $$

and observing that $h(0)>0$, $h(\sqrt{1/\beta+s})=0$ and $h'(y)<\frac {1}{1+\beta s} -\frac{1}{1+\beta s}=0$ for all $y\in(0,\sqrt{1/\beta+s})$, we conclude that $h(y)>0$ for all $y\in[0,\sqrt{1/\beta+s})$, i.e., $w(s,y)> \frac{|y|}{1+ \beta s}$ for $|y|<\sqrt{1/\beta+s}$. Also note that $w$ is ${\mathcal {C}}^{1,1}$ on $[0,\infty)\times\mathbb{R}$ and ${\mathcal {C}}^{1,2}$ on the domain $\{(s,y)\in[0,\infty)\times\mathbb{R} : |y|<\sqrt{1/\beta+s }\}$. Moreover, by (B.5), $w_{s}(s,y) + \frac {1}{2}w_{yy}(s,y)<0$ for $|y|>\sqrt{1/\beta+ s}$. We then conclude from a standard verification theorem (see e.g. [26, Theorem 3.2]) that $\bar{V}(s,y) = w(s,y)$ is a smooth solution of (B.3). This implies that $(\bar{V}(s,W^{y}_{s}))_{s\ge0}$ is a supermartingale, and $(\bar{V}(s\wedge\tau^{*}_{y},W^{y}_{s\wedge \tau^{*}_{y}}))_{s\ge0}$, with $\tau^{*}_{y} := \inf\{s\ge0: |W^{y}_{s}|\ge\sqrt{1/\beta+s}\}$, is a true martingale.

It then follows from standard arguments that $\tau^{*}_{y}$ is the smallest optimal stopping time for $\bar{V}(0,y)$. As a consequence, $\hat{\tau}_{x}:=\inf\{s\ge0: X^{x}_{s}\ge \sqrt{1/\beta+s}\}$ is the smallest optimal stopping time for (4.2). In view of Proposition 2.2, $\widetilde{\tau}_{x} = \hat{\tau}_{x}$. □

Remark B.1

With $X$ being reflected at the origin, it is expected that the variational inequality of the value function $V(s,x)$ should admit a Neumann boundary condition at $x=0$. This is not explicitly seen in (B.2) because of the change of variable $\bar{V}(s,y) := V(s,|y|)$ in the second line of the proof above, which shifts our analysis to a Brownian motion with no reflection at the origin. In fact, one may check directly from (B.5) that $V(s,x) = \bar{V}(s,x) = w(s,x)$ indeed satisfies the Neumann boundary condition $V_{x}(s,0+)=0$ for all $s\ge0$.

2.2 B.2 Proof of Lemma 4.3

First, we prove that $E$ is totally disconnected. If $\ker(\tau)=[a,\infty)$, then $E=\emptyset$ and there is nothing to prove. Assume that there exists $x^{*}> a$ such that $x^{*}\notin\ker(\tau)$. Define

$$ \ell:= \sup\{{b\in\ker(\tau) : b< x^{*}}\}\quad \;\mbox{ and }\; \quad u:=\inf\{{b\in\ker(\tau) : b>x^{*}}\}. $$

We claim that $\ell=u=x^{*}$. Assume to the contrary that $\ell< u$. Then $\tau(x)=1$ for all $x\in(\ell,u)$. Thus, given $y\in(\ell,u)$, ${\mathcal {L}}^{*}\tau(y) = T^{y} :=\inf\{ s\ge0 : X^{y}_{s} \notin(\ell,u)\}>0$ and

$$\begin{aligned} J\big(y; {\mathcal {L}}^{*}\tau(y)\big) = \mathbb{E}^{y}\left[\frac {X_{T^{y}}}{1+\beta T^{y}}\right] < \mathbb{E}^{y}[X_{T^{y}}] = \ell\mathbb{P}[X_{T^{y}}=\ell]+ u\mathbb {P}[X_{T^{y}}=u]. \end{aligned}$$

(B.6)

Since $X_{s}=|W_{s}|$ for a one-dimensional Brownian motion $W$ and $0<\ell <y<u$, by the optional sampling theorem, $\mathbb{P}[X_{T^{y}}=\ell] = \mathbb{P}[W^{y}_{s}\ \hbox{hits }\ell\hbox{ before hitting }u] = \frac{u-y}{u-\ell}$ and $\mathbb{P}[X_{T^{y}}=u]=\mathbb{P}[W^{y}_{s}\ \hbox{hits }u\hbox{ before hitting }\ell] =\frac{y-\ell}{u-\ell}$. Alternatively, one may evaluate $\mathbb{P}[X_{T^{y}}=\ell]$ and $\mathbb{P}[X_{T^{y}}= u]$ directly by using the fact that the scale function of a one-dimensional Bessel process is the identity mapping (see e.g. [8, Part I, Chap. 6, Sect. 15]). This together with (B.6) gives $J(y; {\mathcal {L}}^{*}\tau(y)) < y$. This implies $y\in S_{\tau}$, and thus $\Theta\tau(y)=0$ by (3.12). Then $\Theta\tau(y)\neq\tau(y)$, a contradiction to $\tau\in{\mathcal {E}}(\mathbb{R}_{+})$. This already implies that $E$ is totally disconnected, and thus $\overline{\ker(\tau)}=[a,\infty)$. The rest of the proof follows from Lemma 4.2.

2.3 B.3 Proof of Lemma 4.4

(i) Given $a\ge0$, it is obvious from the definition that $\eta (0,a)\in(0,a)$ and $\eta(a,a)=a$. Fix $x\in(0,a)$ and let $f^{x}_{a}$ denote the density of $T^{x}_{a}$. We obtain

$$\begin{aligned} \mathbb{E}^{x}\left[\frac{1}{1+\beta T^{x}_{a}}\right] &= \int _{0}^{\infty}\frac{1}{1+\beta t}f^{x}_{a}(t)dt = \int_{0}^{\infty}\int _{0}^{\infty}e^{-(1+\beta t)s}f^{x}_{a}(t)ds\ dt \\ &= \int_{0}^{\infty}e^{-s}\bigg(\int_{0}^{\infty}e^{-\beta st}f^{x}_{a}(t)dt\bigg)\ ds \\ & = \int_{0}^{\infty}e^{-s} \mathbb{E}^{x}[e^{-\beta sT^{x}_{a}}] ds. \end{aligned}$$

(B.7)

Since $T^{x}_{a}$ is the first hitting time of a one-dimensional Bessel process, we compute its Laplace transform by using [19, Theorem 3.1] (or [8, Part II, Sect. 3, Formula 2.0.1]), as

$$ \mathbb{E}^{x}[e^{-\frac{\lambda^{2}}{2} T^{x}_{a}}] = \frac{\sqrt{x} I_{-\frac{1}{2}}(x\lambda)}{\sqrt{a} I_{-\frac{1}{2}}(a\lambda)}= \cosh(x\lambda)\operatorname {sech}(a\lambda)\qquad \hbox{for}\ x\le a. $$

Here, $I_{\nu}$ denotes the modified Bessel function of the first kind. Thanks to the above formula with $\lambda=\sqrt{2\beta s}$, we obtain from (B.7) that

$$ \eta(x,a) = a \int_{0}^{\infty}e^{-s} \cosh(x\sqrt{2\beta s})\operatorname{sech}(a\sqrt{2\beta s}) ds. $$

(B.8)

It is then obvious that $x\mapsto\eta(x,a)$ is strictly increasing. Moreover,

$$\eta_{xx}(x,a) = 2a\beta^{2} \int_{0}^{\infty}e^{-s} s\cosh(x\sqrt {2\beta s})\operatorname{sech}(a\sqrt{2 \beta s}) ds>0 \qquad \hbox{for}\ x\in[0,a], $$

which shows the strict convexity.

(ii) This follows from (B.8) and the dominated convergence theorem.

(iii) We first prove the desired result with $x^{*}(a)\in(0,a)$, and then upgrade it to $x^{*}(a)\in(0,a^{*})$. Fix $a\ge0$. In view of the properties in (i), we observe that the two curves $y=\eta(x,a)$ and $y=x$ intersect at some $x^{*}(a)\in(0,a)$ if and only if $\eta_{x}(a,a)>1$. Define $k(a):=\eta_{x}(a,a)$. By (B.8),

$$ k(a)=a\int_{0}^{\infty}e^{-s} \sqrt{2\beta s}\tanh(a\sqrt{2\beta s}) ds. $$

(B.9)

Thus we see that $k(0)=0$ and $k(a)$ is strictly increasing on $(0,1)$ since for any $a>0$,

$$k'(a)=\int_{0}^{\infty}e^{-s} \sqrt{2s}\bigg(\tanh(a\sqrt{2s})+\frac {a\sqrt{2s}}{\cosh^{2}(a\sqrt{2s})}\bigg) ds >0. $$

By numerical computation, $k(1/\sqrt{\beta}) =\int_{0}^{\infty}e^{-s} \sqrt{2s}\tanh(\sqrt{2s}) ds \approx 1.07461 >1$. It follows that there must exist $a^{*}\in(0,1/\sqrt{\beta })$ such that $k(a^{*})=\eta_{x}(a^{*},a^{*})=1$. Monotonicity of $k(a)$ then gives the desired result.

Now, for any $a> a^{*}$, we intend to upgrade the previous result to $x^{*}(a)\in(0,a^{*})$. Fix $x\ge0$. By the definition of $\eta$ and (ii), on the domain $a\in[x,\infty )$, the mapping $a\mapsto \eta(x,a)$ must either first increase and then decrease to 0, or directly decrease to 0. From (B.8), we have

$$\eta_{a}(x,x) = 1-x\int_{0}^{\infty}e^{-s} \sqrt{2\beta s}\tanh(x\sqrt {2\beta s}) ds = 1-k(x), $$

with $k$ as in (B.9). Recalling $k(a^{*})=1$, we have $\eta _{a}(a^{*},a^{*})=0$. Notice that

$$\begin{aligned} \eta_{aa}(a^{*},a^{*}) &= -\frac{2}{a^{*}}k(a^{*}) -2 \beta a^{*} + a^{*}\int _{0}^{\infty}4\beta s e^{-s}\tanh^{2}(a^{*}\sqrt{2\beta s}) ds\\ &\le -\frac{2}{a^{*}} + 2\beta a^{*} < 0, \end{aligned}$$

where the second line follows from $\tanh(x)\le1$ for $x\ge0$ and $a^{*}\in(0,1/\sqrt{\beta})$. Since $\eta_{a}(a^{*},a^{*})=0$ and $\eta_{aa}(a^{*},a^{*})<0$, we conclude that on the domain $a\in[a^{*},\infty)$, the mapping $a\mapsto\eta(a^{*},a)$ decreases to 0. On the other hand, for any $a>a^{*}$, since $\eta(a^{*},a) < \eta(a^{*},a^{*})= a^{*}$, we must have $x^{*}(a)< a^{*}$.