The optimal stopping problem revisited

Abstract

We consider an optimal stopping time problem, related with many models found in real options problems. We present analytical solutions for a broad class of gain functions, considering quite general assumptions over the model. Also, an extensive and general sensitivity analysis is provided.

Appendices

A Proofs

1.1 A.1 Preliminaries

We start by stating an auxiliary result that will be useful to prove the remaining results in the paper.

Proposition 6

Let X be defined as in (2) and $g:]0,\infty [\rightarrow [0,+\infty ]$ be a Borel measurable function, such that $\int _0^{\infty }g(x)dx>0$ (i.e., g is not almost everywhere zero). If $d_1=d_2$ or $d_1=\overline{d_2}\in \mathbb {C}{\setminus }\mathbb {R}$, then

$$\begin{aligned} E_x\left[ \int _0^{+\infty }e^{-rt}g(X(t))dt\right] =\infty . \end{aligned}$$

Proof

Using Fubini’s Theorem and equalities in (8), we have:

$$\begin{aligned} E_x\left[ \int _0^{+\infty }e^{-rt}g(X(t))dt\right] =&\int _0^{+\infty }e^{-rt}E_x\left[ g\left( X(t)\right) \right] dt\\ =&\int _0^{+\infty }e^{\frac{\sigma ^2}{2}d_1d_2t}\int _{-\infty }^{+\infty }g\left( xe^{-\frac{\sigma ^2}{2}\left( d_1+d_2\right) t+\sigma w}\right) \\&\quad \times \frac{e^{-\frac{w^2}{2t}}}{\sqrt{2\pi t}}dwdt. \end{aligned}$$

Using the change of variable $w=\frac{1}{\sigma }\log \frac{y}{x}+\sigma \frac{d_1+d_2}{2}t$, it is possible to obtain

$$\begin{aligned} E_x\left[ \int _0^{+\infty }e^{-rt}g(X(t))dt\right]&=\int _0^{+\infty }e^{-\frac{\sigma ^2}{2}\left( \frac{d_2-d_1}{2}\right) ^2t}\nonumber \\&\quad \times \int _{0}^{+\infty }\frac{g\left( y\right) }{\sigma y}e^{-\frac{1}{2t\sigma ^2}(\log \frac{y}{x})^2-\frac{d_1+d_2}{2}\log \frac{y}{x}}dydt. \end{aligned}$$

(29)

Additional calculations, allow us to obtain:

$$\begin{aligned} E_x\left[ \int _0^{+\infty }e^{-rt}g(X(t))dt\right] \ge A\int _1^{+\infty }e^{-\frac{\sigma ^2}{2}\left( \frac{d_1 - d_2}{2}\right) ^2}dt, \end{aligned}$$

where $A=\int _0^{+\infty }\frac{g(y)}{\sigma y}e^{-\frac{1}{2\sigma ^2}\left( \log \frac{y}{x}\right) ^2-\frac{d_1+d_2}{2}\log \left( \frac{y}{x}\right) }dy>0$. This proves the proposition, since $d_1=\overline{d_2}$ means that $(d_1-d_2)^2 \le 0$ and therefore $\int _1^{+\infty }\frac{1}{\sqrt{2\pi t}}e^{-\frac{\sigma ^2}{2}( d_1 - d_2)^2}dt = + \infty$. $\square$

1.2 A.2 Proof of Proposition 4

Fix a point $a \in ]0,+\infty [$, and consider the initial value problem defined by (11) and (14). Using the change of variable: $y=ln\left( \frac{x}{a} \right)$ and $u(y)=v(a e^y)$, the solution of the problem above can be obtained from the solution of the equation

$$\begin{aligned} -\tilde{{\mathcal {L}}}u(y)-\varPi (a e^y)=0, \end{aligned}$$

(30)

with initial conditions

$$\begin{aligned} u(0) = {\hat{v}}_1, \qquad u'(0) = a {\hat{v}}_2 . \end{aligned}$$

Here $\tilde{ {\mathcal {L}}} u(y)$ is defined as in Eq. (6). Defining the vector $w(y)=(u(y), u'(y))^T$, where the superscript $^T$ denotes the transpose, we may represent the ODE (30) as $w'(y)=Aw(y)+b(y)$, where $b:]0,\infty [\rightarrow \mathbb {R}^{2}$ is a vector function and A is a constant $2 \times 2$ matrix, defined as follows:

$$\begin{aligned} b(y)= \left( 0, -\frac{2}{\sigma ^2}\varPi (ae^y)\right) ^T; \quad A=\left( \! \begin{array}{cc} 0 &{} 1 \\ \frac{2r}{\sigma ^2}&{}-\frac{\sigma ^2-2\alpha }{\sigma ^2} \end{array} \!\right) =\left( \! \begin{array}{cc} 0 &{} 1 \\ -d_1d_2&{}d_1+d_2 \end{array} \!\right) . \end{aligned}$$

The last equality follows from the parametrization defined in (8). Furthermore, straightforward calculations lead to the fundamental matrix

$$\begin{aligned} e^{yA}=\left( \! \begin{array}{cc} \displaystyle \frac{d_2e^{yd_1}-d_1e^{yd_2}}{d_2-d_1} &{}\displaystyle \frac{e^{yd_2}-e^{yd_1}}{d_2-d_1} \\ \displaystyle -d_1d_2\frac{e^{yd_2}-e^{yd_1}}{d_2-d_1}&{}\displaystyle \frac{d_2e^{yd_2}-d_1e^{yd_1}}{d_2-d_1} \end{array} \!\right) . \end{aligned}$$

The solution for this system is given by $w(y)=e^{yA}w(0){+}\int _0^ye^{(y-s)A}b(s)ds$. Returning to the original variables, we get the expressions (16)–(17).

1.3 A.3 Proof of Proposition 2

Let v be a solution of the HJB equation. Then,

$$\begin{aligned} -\mathcal{L}v(x)=rv(x) - \alpha x v^\prime (x) - \frac{1}{2}\sigma ^2x^2 v^{\prime \prime }(x) \ge \varPi (x) \end{aligned}$$

for almost every $x \in ]0,+\infty [$.

Fix $t>0$, and let $\tau \in {\mathcal {S}}$ be any stopping time. Since the function $v'$ is continuous in $]0,\infty [$, then $\int _0^{t\wedge \tau }e^{-rs}v'(X(s))dW(s)$ is a martingale (we set $a \wedge b=\min (a,b)$). Consequently:

$$\begin{aligned} E_x\left[ \int _0^{t\wedge \tau }e^{-rs}v'(X(s))dW(s)\right] =0, \quad \text {for all }x>0, \tau \in {\mathcal {S}}. \end{aligned}$$

Since v is continuously differentiable with absolutely continuous derivative, it can be written as the difference of two convex functions. Therefore, the Itô-Tanaka formula holds (see Revuz and Yor (2013), Theorem VI.1.5), and therefore we obtain

$$\begin{aligned} 0&\le E_x \left[ e^{-r(t\wedge \tau )} v(X(t\wedge \tau )) \right] = v(x) + E_x \left[ \int _0^{t\wedge \tau } e^{-rs} {\mathcal {L}} v(X(s)) ds \right] \le \nonumber \\&\le v(x) - E_x \left[ \int _0^{t\wedge \tau } e^{-rs} \varPi (X(s)) ds \right] . \end{aligned}$$

(31)

The expressions $\int _0^{t\wedge \tau } e^{-rs} \varPi ^+(X(s)) ds$, $\int _0^{t\wedge \tau } e^{-rs} \varPi ^-(X(s)) ds$ are monotonically increasing with respect to t, and $E_x \left[ \int _0^{t\wedge \tau } e^{-rs} \varPi ^+(X(s)) ds \right] \le E_x \left[ \int _0^{+\infty } e^{-rs} \varPi ^+(X(s)) ds \right] <\infty$. Therefore, the monotonic convergence theorem guarantees that

$$\begin{aligned} \lim _{t \rightarrow +\infty } E_x \left[ \int _0^{t \wedge \tau } e^{-rs} \varPi (X(s)) ds \right] = E_x \left[ \int _0^\tau e^{-rs} \varPi (X(s)) ds \right] \in [-\infty , + \infty [ . \end{aligned}$$

Thus, the inequality (31) implies

$$\begin{aligned} v(x) \ge E_x \left[ \int _0^{\tau } e^{-rs} \varPi (X(s)) ds \right] =J(x,\tau ). \end{aligned}$$

Since this holds for arbitrary $\tau \in {\mathcal {S}}$, it shows that $v(x) \ge V(x)$.

1.4 A.4 Proof of Proposition 3

It can be checked that the argument in the proof of Proposition 2 holds in the present case and, therefore,

$$\begin{aligned} J(x,\tau ) \le v(x) \qquad \forall x \in ]a,b[, \ \tau \in {\mathcal {S}}_{[a,b]} . \end{aligned}$$

Let

$$\begin{aligned} \tau _0 = \inf \left\{ t\ge 0: v(X(t)) = 0 \right\} . \end{aligned}$$

It is clear that $\tau _0 \in {\mathcal {S}}_{[a,b]}$. To finish the proof, we only need to show that $J(x,\tau _0) = v(x)$ for every $x \in ]a,b[$. By definition of $\tau _0$, $e^{-r(t\wedge \tau _0)}v(X(t\wedge \tau _0))=0$ whenever $t\wedge \tau _0=\tau _0$, therefore:

$$\begin{aligned} E_x \left[ e^{-r(t \wedge \tau _0)} v(X(t \wedge \tau _0)) \right] = E_x \left[ e^{-rt} v(X(t)) {\mathcal {I}}_{\{ \tau _0>t\}} \right] \end{aligned}$$

Since $v(X(s))>0$ whenever $s < \tau _0$, the argument used in the proof of Proposition 2 shows that

$$\begin{aligned} E_x \left[ e^{-rt} v(X(t)) {\mathcal {I}}_{\{ \tau _0>t\}} \right] = v(x) - E_x \left[ \int _0^{t \wedge \tau _0} e^{-rs}\varPi (X(s)) ds \right] \end{aligned}$$

for every $t \in ]0,+\infty [$, and $\lim \limits _{t \rightarrow + \infty } E_x \left[ \int _0^{t \wedge \tau _0} e^{-rs}\varPi (X(s)) ds \right] = E_x \left[ \int _0^{\tau _0} e^{-rs}\varPi (X(s)) ds \right]$.

We extend the function v to the interval $]0,+\infty [$, by setting $v(x)=0$ for every $x \in ]0,a] \cup [b,+\infty [$. Thus, there is a constant $K<+\infty$ such that

$$\begin{aligned} {v(x) \le K x^{\frac{d_1+d_2}{2}}, \qquad \forall x \in ]0,+\infty [ .} \end{aligned}$$

Therefore,

$$\begin{aligned} E_x \left[ e^{-rt} v(X(t)) {\mathcal {I}}_{\{ \tau _0>t\}} \right]&< K E_x \left[ e^{-rt} X(t)^{\frac{d_1+d_2}{2}} \right] \\&= K x^{\frac{d_1+d_2}{2}} E \left[ e^{-rt} e^{ \left( (\alpha - \frac{\sigma ^2}{2}) t + \sigma W_t \right) \frac{d_1+d_2}{2}} \right] \\&= K x^{\frac{d_1+d_2}{2}} E \left[ e^{\frac{\sigma ^2}{2} \left( d_1d_2 - \frac{(d_1+d_2)^2}{4} \right) t} e^{ -\frac{1}{2} \sigma ^2 \left( \frac{d_1+d_2}{2} \right) ^2 + \sigma \frac{d_1+d_2}{2}W_t } \right] \\&= K x^{\frac{d_1+d_2}{2}} e^{-\frac{\sigma ^2}{2} \frac{(d_1-d_2)^2}{4} t} , \end{aligned}$$

where $\{W_t:t\ge 0\}$ is a standard Brownian motion. This shows that $\lim \limits _{t \rightarrow +\infty }E_x \left[ e^{-rt} v(X(t)) {\mathcal {I}}_{\{ \tau _0>t\}} \right] = 0$, and, therefore $v(x) = E_x \left[ \int _0^{ \tau _0} e^{-rs}\varPi (X(s)) ds \right]$, which concludes the proof.

1.5 A.5 Proof of Lemma 1

We prove assertions (i) and (iii) simultaneously.

If $x_{1l}=0$, then $\beta =0$, and Remark 3 implies that

$$\begin{aligned} \lim _{b \downarrow \beta } \inf \{ x>b : v_b(x)< 0 \} = x_{1r} < \gamma . \end{aligned}$$

Suppose now that $x_{1l}>0$, and, therefore, $\beta < x_{1l}$. For every $b \in ]\beta , x_{1l}[$, let $c(b) = \inf \{ x>b : v_b(x) < 0 \}$. By Remark 5, the function $b \mapsto c(b)$ is monotonically decreasing in $]b,x_{1l}[$ and therefore, $\lim \limits _{b \downarrow \beta } c(b)$ exists (possibly, infinite). If $\lim \limits _{b \downarrow \beta } c(b) \le x_{2r}$, then Remark 6 states that $\lim \limits _{b \downarrow \beta } c(b) < \gamma$. If $\lim \limits _{b \downarrow \beta } c(b) > x_{2r}$, then Remark 3 implies that for every $b \in ]\beta , x_{1l}[$, we have $v_b(c(b)) = 0$, $v_b'(c(b)) < 0$. Therefore, due to Remark 2, the inequality $v_{c(b)} (b) < v_b(b) =0$ must hold. This shows that $c(b) < \gamma$ and therefore, since b is arbitrary, $\lim \limits _{b \downarrow \beta } c(b) \le \gamma$.

Now, suppose that $\beta >0$. If $\lim \nolimits _{b \downarrow \beta } c(b) = + \infty$, then the equality in (i) holds trivially. Thus, we assume that $\lim \nolimits _{b \downarrow \beta } c(b) = c < + \infty$. Since $\lim \nolimits _{b \downarrow \beta } v_b(x) = v_\beta (x)$ uniformly with respect to x on compact subintervals of $]0,+\infty [$, it follows that $0 = \lim \nolimits _{b \downarrow \beta } v_b(c(b)) = v_\beta (c)$. Since $v_\beta$ is non-negative (Remark 6), it follows that $v_\beta '(c) = 0$, and therefore $v_\beta = v_c$. Notice that for every ${{\tilde{c}}} >c$, $v_{{\tilde{c}}}(x) > v_c(x)$ holds for every $x \in ]0,c]$ (Remark 5), and $v_{{\tilde{c}}}(x) >0$ for every $x \in [x_{2l},{\tilde{c}}[$ (Remark 3), it follows that $\gamma =c$.

The proof of assertion (ii) is analogous.

1.6 A.6 Proof of Lemma 2

To prove assertion (i):

Using the equalities (18)–(19), a simple computation shows that condition (25) is equivalent to

$$\begin{aligned} v_b(c) = v_b^\prime (c) = 0, \end{aligned}$$

i.e., $v_b \equiv v_c$. Therefore, Lemma 1 states that if $\beta >0$ and $\gamma < +\infty$, then $\beta$, $\gamma$ satisfy the conditions in (25).

Let $b \in ]0, x_{1l}[$ and $c\in ]x_{2r},+\infty [$ be constants satisfying the conditions in (25). Since $v_b(c) =v_c(b)= 0$, Remark 5 implies that $\beta \le b$ and $\gamma \ge c$.

Suppose that $\beta < b$. By Remarks 3 and 5, there must be some $x>b$ such that $v_b(x) < 0$. Due to Remark 3, x must lie in the interval $]x_{1r},x_{2l}[$, but this case is excluded by Remark 4, and therefore $\beta = b$. A similar argument shows that $\gamma = c$, and the proof of assertion (i) is complete.

To prove assertion (ii):

We can write equality (18) in the form:

$$\begin{aligned} v_a(x) = \frac{-2 x^{d_2}}{\sigma ^2 (d_2-d_1)} \int _a^x \left( 1 - \left( \frac{s}{x} \right) ^{d_2-d_1} \right) s^{-d_2-1} \varPi (s) ds . \end{aligned}$$

(32)

Breaking the interval $[a,+\infty [$ into intervals where $\varPi$ does not change sign and using the Lebesgue monotone convergence theorem, we see that

$$\begin{aligned} \lim _{x \rightarrow +\infty } \int _a^x \left( 1 - \left( \frac{s}{x} \right) ^{d_2-d_1} \right) s^{-d_2-1} \varPi (s) ds = \int _a^{+\infty } s^{-d_2-1} \varPi (s) ds \end{aligned}$$

(33)

for every $a \in ]0,+\infty [$.

Suppose that $\beta >0$ and $\gamma = + \infty$. Since $v_\beta$ is nonnegative, the equalities (32), (33) imply that $\int _\beta ^{+\infty } s^{-d_2-1}\varPi (s) ds \le 0$. If $\int _\beta ^{+\infty } s^{-d_2-1}\varPi (s) ds < 0$, then there are constants $\varepsilon >0$, $c< +\infty$ such that $v_b(x) >0$ for every $x>c$, $b< \beta + \varepsilon$. However, Lemma 1 states that $\lim \limits _{b \downarrow \beta } \inf \{ x>b: v_b(x) < 0 \} = +\infty$. This shows that $\int _\beta ^{+\infty } s^{-d_2-1}\varPi (s) ds = 0$.

For each $b > \beta$, let $c(b) = \inf \{ x> b : v_b(x) < 0 \}$. By Remark 5, we have

$$\begin{aligned} 0&= v_b(b) > v_{c(b)}(b) =\frac{-2 b^{d_1}}{\sigma ^2 (d_2-d_1)} \left( \int _b^{c(b)} s^{-d_1-1} \varPi (s) ds - b^{d_2-d_1}\right. \\&\quad \left. \int _b^{c(b)} s^{-d_2-1} \varPi (s) ds \right) . \end{aligned}$$

Lemma 1 states that $\lim \limits _{b \downarrow \beta } c(b) = \gamma = + \infty$. Hence,

$$\begin{aligned} \int _\beta ^{+ \infty } s^{-d_1-1} \varPi (s) ds \ge \beta ^{d_2-d_1} \int _\beta ^{+ \infty } s^{-d_2-1} \varPi (s) ds = 0 . \end{aligned}$$

Now, let $b \in ]0, x_{1l}[$ be a constant satisfying (21). Since for every sufficiently small $\varepsilon >0$, we have

$$\begin{aligned} \int _{b+\varepsilon }^{+\infty } s^{-d_2-1}\varPi (s) ds >0 , \end{aligned}$$

the equalities (32) and (33) show that $v_{ b+\varepsilon }(x) <0$ for every sufficiently large x. Thus, $\beta \le b$. The equalities (32) and (33) also show that

$$\begin{aligned} \lim \limits _{x \rightarrow + \infty } \frac{v_b(x)}{x^{d_2}} = 0 . \end{aligned}$$

(34)

Therefore, if there is some $x \in ]b,+\infty [$ such that $v_b(x) <0$, the function $x \mapsto \frac{v_b(x)}{x^{d_2}}$ must have a global minimizer in $]b,+\infty [$. Using equalities (18)–(19), we see that

$$\begin{aligned} \frac{d}{dx} \left( \frac{v_b(x)}{x^{d_2}} \right) = \frac{-2}{\sigma ^2 x^{d_2-d_1+1}} \int _b^x s^{-d_1-1} \varPi (s) ds . \end{aligned}$$

Thus, Assumption 1 implies that if c is a minimizer of $\frac{v_b(x)}{x^{d_2}}$, then $c > x_{2r}$ and $\int _b^c s^{-d_1-1}\varPi (s) ds =0$. However, this implies that $\int _b^{+\infty } s^{-d_1-1} \varPi (s) ds < 0$. Hence, $v_b$ must be nonnegative and therefore $\beta = b$.

The proof of statement (iii) is analogous to the proof of statement (ii).

The statement (iv) is straightforward: by the statements (i) and (ii), $\beta >0$ implies $\int _0^{+\infty } s^{-d_2-1}\varPi (s) ds < 0$, and, by the statements (i) and (iii), $\gamma < + \infty$ implies $\int _0^{+\infty } s^{-d_1-1}\varPi (s) ds < 0$. Hence, $\beta = 0$ and $\gamma = +\infty$ imply (28). Conversely, if (28) holds, then none of the equalities in conditions (25), (21) or (23) may hold.

1.7 A.7 Proof of Theorem 1

Suppose that there are constants $b \in ]0, x_{1l}[$, $c \in ]x_{2r},+\infty [$ satisfying conditions in (25). By Lemma 2, $\beta = b$, $\gamma =c$, and the function V defined by (26) is a Carathéodory solution of the HJB equation (10). Therefore, Proposition 2 states that V is an upper bound for the value function (4).

For every $n \in {\mathbb {N}}$, let

$$\begin{aligned} b_n= \beta + \frac{1}{n} , \qquad c_n = \inf \{ x >b_n: v_{b_n}(x) <0 \} . \end{aligned}$$

By Proposition 3, the function

$$\begin{aligned} V_n(x) = \left\{ \begin{array}{ll} v_{b_n}(x) , &{} \text {for } x \in [b_n,c_n]\\ 0 , &{} \text {for } x \in ]0,b_n] \cup [c_n, +\infty [ \end{array} \right. , \end{aligned}$$

coincides with the function $V_{[b_n,c_n]}(x)= \sup \nolimits _{\tau \in {\mathcal {S}}_{[b_n,c_n]}} J(x,\tau )$. Since ${\mathcal {S}}_{[b_n,c_n]} \subset {\mathcal {S}}$, $V_n$ is a lower bound for the value function (4). By Lemma 1, $\lim V_n =V$ and therefore the assertion (iii) of the theorem holds. Notice that the argument above does not require $\gamma < +\infty$. Hence, it can be used verbatim to prove assertion (i). If we set

$$\begin{aligned} c_n = \gamma - \frac{1}{n}, \qquad b_n = \sup \{ x< c_n: v(x) <0 \} , \end{aligned}$$

then the argument above proves assertion (ii).

To prove assertion (iv), we note that if there are no constants b, c as in (i)–(iii), then Lemma 2 states that $\beta = 0$ and $\gamma = +\infty$. Suppose, without loss of generality, that $x_{1l} >0$, and pick a sequence ${\tilde{a}}_n < x_{1l}$, converging to zero. Due to Equations (28), (32) and (33), for each $n \in {\mathbb {N}}$ there is some $c_n < +\infty$ such that $v_{{\tilde{a}}_n}(x) < 0$ for every $x \in {]}c_n,+\infty [$. Hence, there is a sequence $b_n$ such that $\lim b_n=+\infty$ and $v_{{\tilde{a}}_n}(b_n) < 0$. If, on the one hand, $v_{b_n}(x) >0$ for every $x \in ]{\tilde{a}}_n,b_n[$, then we set $a_n = \sup \{x<b_n : v_{b_n}(x) < 0 \} \le {\tilde{a}}_n$, and, consequently, $v_{[a_n,b_n]}(x) = v_{b_n}(x) >0$ holds for every $x \in ]a_n,b_n[$. If, on the other hand, there is some $x \in ]{\tilde{a}}_n, b_n[$ such that $v_{b_n}(x) \le 0$, then we set $a_n={\tilde{a}}_n$. Combining the expressions in Equation (20) and the inequalities $v_{[a_n,b_n]}(b_n)=0>v_{a_n}(b_n)$ and $v_{[a_n,b_n]}(a_n)=0>v_{b_n}(a_n)$, it follows that $v_{[a_n,b_n]}'(a_n) >0$ and $v_{[a_n,b_n]}'(b_n) <0$. Therefore, Remark 2 implies that

$$\begin{aligned}&v_{[a_n,b_n]}(x)> v_{a_n}(x) \qquad \text {for every } x > a_n, \end{aligned}$$

(35)

$$\begin{aligned}&v_{[a_n,b_n]}(x)> v_{b_n}(x) \qquad \text {for every } x < b_n. \end{aligned}$$

(36)

If $x_{2r}=+\infty$, then the inequality (36) implies that $v_{[a_n,b_n]}(x)>0$ for every $x \in [x_{2l},b_n[$ (if $b_n>x_{2l}$). If $x_{2r}<+\infty$, then we can chose $b_n>x_{2r}$ and the Remark 3 states that $v_{b_n}(x) >0$ for every $x \in [x_{2l},b_n[$. In both cases, the Remark 3 states that $v_{a_n}(x) >0$ for every $x \in ]a_n,x_{1r}]$. Hence, the Remark 4 guarantees that $v_{[a_n,b_n]}(x)>0$ for every $x \in ]x_{1r},x_{2l}[$, and therefore $v_{[a_n,b_n]}$ is strictly positive in $]a_n,b_n[$.

Fix a sequence $\left\{ [a_n,b_n] \right\} _{n \in {\mathbb {N}}}$ as above. Due to Remark 2, the sequence

$$\begin{aligned} v_n(x) = \left\{ \begin{array}{ll} v_{[a_n,b_n]}(x) , &{} \text {for } x \in [a_n,b_n] , \\ 0 , &{} \text {for } x \in ]0,a_n] \cup [b_n,+\infty [ \end{array} \right. \end{aligned}$$

is monotonically increasing. By Proposition 3, $v_n(x) = \sup \nolimits _{\tau \in {\mathcal {S}}_{[a_n,b_n]}}J(x,\tau ) < v_p^+(x)$. Therefore, Assumption 2 implies that $v_n(x)$ is a bounded monotonic sequence and hence it converges. Since solutions of Equation (11) depend continuously on boundary conditions, the convergence of $v_n(x)$ and $v'_n(x)$ is uniform on compact intervals. Therefore, $v(x) = \lim \limits _{n \rightarrow \infty } v_n(x)$ is a positive Carathéodoty solution of Equation (11) and therefore it is a solution of the HJB equation (10), which, by Proposition 2, is an upper bound for the value function. Since $v_n(x) = \sup \nolimits _{\tau \in {\mathcal {S}}_{[a_n,b_n]}}J(x,\tau ) \le \sup \nolimits _{\tau \in {\mathcal {S}}}J(x,\tau )$, v must coincide with the value function.

To see that v coincides with the function defined in Equation (27), use Equation (16) and the boundary conditions $v_{[a_n,b_n]}(a_n) = v_{[a_n,b_n]}(b_n) = 0$ to obtain

$$\begin{aligned} v_{[a_n,b_n]}'(a_n) = \frac{2}{\sigma ^2} \frac{\int _{a_n}^{b_n} (b^{d_2} s^{-d_2-1} - b^{d_1} s^{-d_1-1}) \varPi (s) ds}{a_n \left( \left( \frac{b_n}{a_n} \right) ^{d_2} - \left( \frac{b_n}{a_n} \right) ^{d_1} \right) } . \end{aligned}$$

Substituting in Equation (20) and rearranging, we obtain

$$\begin{aligned} v_{[a_n,b_n]}(x) =&\frac{\left( \frac{x}{a_n} \right) ^{d_2} - \left( \frac{x}{a_n} \right) ^{d_1}}{d_2-d_1} \frac{2}{\sigma ^2} \frac{\int _{a_n}^{b_n} (b^{d_2} s^{-d_2-1} - b^{d_1} s^{-d_1-1}) \varPi (s) ds}{\left( \frac{b_n}{a_n} \right) ^{d_2} - \left( \frac{b_n}{a_n} \right) ^{d_1} }\\&- \frac{2}{(d_2-d_1)\sigma ^2} \int _{a_n}^x \left( x^{d_2}s^{-d_2-1} - x^{d_1}s^{-d_1-1} \right) \varPi (s) ds \\ =&\frac{2}{(d_2-d_1) \sigma ^2} \frac{1- \left( \frac{x}{b_n} \right) ^{d_2-d_1}}{1- \left( \frac{a_n}{b_n} \right) ^{d_2-d_1}} \Bigg ( x^{d_1} \int _{a_n}^x \left( 1- \left( \frac{a_n}{s} \right) ^{d_2-d_1}\right) s^{-d_1-1} \varPi (s) ds \\&+ x^{d_2} \int _x^{b_n} \left( 1- \left( \frac{s}{b_n} \right) ^{d_2-d_1}\right) s^{-d_2-1} \varPi (s) ds \Bigg ), \end{aligned}$$

for every $x \in [a_n,b_n]$. For every $x \in ]0,+\infty [$ (fixed), the Lebesgue monotone convergence theorem shows that

$$\begin{aligned} \lim _{n \rightarrow \infty } \int _{a_n}^x \left( 1- \left( \frac{a_n}{s} \right) ^{d_2-d_1}\right) s^{-d_1-1} \varPi ^+(s) ds&= \int _0^x s^{-d_1-1} \varPi ^+(s) ds ,\\ \lim _{n \rightarrow \infty } \int _x^{b_n} \left( 1- \left( \frac{s}{b_n} \right) ^{d_2-d_1}\right) s^{-d_2-1} \varPi ^+(s) ds&= \int _x^{+\infty } s^{-d_2-1} \varPi ^+(s) ds , \end{aligned}$$

and the same holds for $\varPi ^-$. Assumption 2 guarantees that the integrals with $\varPi ^+$ are finite. Hence,

$$\begin{aligned}&\lim _{n \rightarrow \infty } v_{[a_n,b_n]}(x)\\&\quad = \frac{2}{(d_2-d_1) \sigma ^2} \left( x^{d_1} \int _0^x s^{-d_1-1} \varPi (s) ds + x^{d_2} \int _x^{+ \infty } s^{-d_2-1} \varPi (s) ds \right) , \end{aligned}$$

and the proof is complete.

1.8 A.8 Proof of Proposition 1

Since Proposition 6, states that if $v_p^+(x)<+\infty$, for any x, then $d_1<d_2$, the argument used to prove assertion (iv) of Theorem 1 can be easily adapted to prove Proposition 1.

1.9 A.9 Proof of Lemma 3

For each a, b, d such that $0<a<b < +\infty$ and $d \ne 0$, let

$$\begin{aligned} f_{a,b,d}(x) = \ln \frac{x}{a} - \frac{\ln b - \ln a}{b^d - a^d} \left( x^d - a^d \right) . \end{aligned}$$

Notice that $f_{a,b,d}$ is continuous, and for $d>0$:

$$\begin{aligned}&f_{a,b,d}(x) < 0 \qquad \text {for } x \in ]0,a[ \cup ]b,+\infty [, \\&f_{a,b,d}(x) > 0 \qquad \text {for } x \in ]a,b[ . \end{aligned}$$

For $d<0$, $f_{a,b,d}$ takes the oposites of the signs above.

Suppose that the assumptions of statement (i) hold.

Since $\int _b^c s^{-d_2-1} \varPi (s) ds = 0$, we see that $\int _b^c s^{-d_2-1} \varPi (s) \ln s \, ds = \int _b^c s^{-d_2-1} \varPi (s) \left( \ln s +C \right) \, ds$, for every constant $C \in {\mathbb {R}}$. If $c \le x_{2r}$, this implies

$$\begin{aligned} \int _b^c s^{-d_2-1} \varPi (s) \ln s \, ds = \int _b^c s^{-d_2-1} \varPi (s) \ln \frac{s}{x_{1l}} \, ds > 0 , \end{aligned}$$

since the integrand of the right-hand side integral is non-negative and is not zero on a set of positive measure. If $c > x_{2r}$, then

$$\begin{aligned}&\int _b^c s^{-d_2-1} \varPi (s) \ln s \, ds \\&\quad = \int _b^c s^{-d_2-1} \varPi (s) \left( f_{x_{1l},x_{2r},d_2-d_1} (s) + \frac{\ln x_{2r} - \ln x_{1l}}{x_{2r}^{d_2-d_1} - x_{1l}^{d_2-d_1}}s^{d_2-d_1} \right) \, ds\\&\quad = \int _b^c s^{-d_2-1} \varPi (s) f_{x_{1l},x_{2r},d_2-d_1} (s) \, ds + \frac{\ln x_{2r} - \ln x_{1l}}{x_{2r}^{d_2-d_1} - x_{1l}^{d_2-d_1}} \int _b^c s^{-d_1-1} \varPi (s) \, ds\\&\quad \ge \int _b^c s^{-d_2-1} \varPi (s) f_{x_{1l},x_{2r},d_2-d_1} (s) \, ds , \end{aligned}$$

and again the integrand in the last expression is a non-negative function which differs from zero on a set of positive measure.

The proof of statement (ii) is analogous.

1.10 A.10 Proof of Lemma 4

Let $f_{x_1,x_2,d}$ be as in the proof of Lemma 3. Using the argument in the proof of Lemma 3, we see that for any constant $k \in ]0,+\infty [$:

$$\begin{aligned}&\int _b^c s^{-d_1-1} \varPi (s) \ln s \, ds + k^{d_2-d_1} \int _b^c s^{-d_2-1} \varPi (s) \ln s \, ds \\&\quad = \int _b^c s^{-d_1-1} \varPi (s) \left( f_{x_1,x_2,d_1-d_2}(s) + \frac{k^{d_2-d_1}}{s^{d_2-d_1}} f_{x_1,x_2,d_2-d_1}(s) \right) \, ds . \end{aligned}$$

A simple computation shows that for $d>0$:

$$\begin{aligned}&f_{x_1,x_2,-d}(s) + \frac{k^d}{s^d} f_{x_1,x_2,d}(s) \\&\quad = \frac{1}{d s^d} \left( (s^d + k^d) \ln \frac{s^d}{x_1^d} - (x_2^d + k^d) \frac{\ln x_2^d - \ln x_1^d}{x_2^d - x_1^d}(s^d - x_ 1^d) \right) . \end{aligned}$$

This last expression is equal to zero if $s=x_1$ or $s= x_2$. It is a strictly concave function of $s^d$ for $s \in ]0,k[$, and strictly convex for $s \in ]k,+\infty [$. It follows that

$$\begin{aligned}&f_{x_1,x_2,d_1-d_2}(s) + \frac{c^{d_2-d_1}}{s^{d_2-d_1}} f_{x_1,x_2,d_2-d_1}(s)>0 \qquad \text {for } s \in ]x_1,x_2[, \\&f_{x_1,x_2,d_1-d_2}(s) + \frac{c^{d_2-d_1}}{s^{d_2-d_1}} f_{x_1,x_2,d_2-d_1}(s)<0 \qquad \text {for } s \in [b,x_1[ \cup ]x_2,c], \\&f_{x_1,x_2,d_1-d_2}(s) + \frac{b^{d_2-d_1}}{s^{d_2-d_1}} f_{x_1,x_2,d_2-d_1}(s) <0 \qquad \text {for } s \in ]x_1,x_2[, \\&f_{x_1,x_2,d_1-d_2}(s) + \frac{b^{d_2-d_1}}{s^{d_2-d_1}} f_{x_1,x_2,d_2-d_1}(s) >0 \qquad \text {for } s \in [b,x_1[ \cup ]x_2,c], \end{aligned}$$

and the result follows.

$\square$

B Auxiliary results for the sensitivity analysis

In this section we present some trivial calculations in order to obtain the monotonicity of the parameters $\beta$ and $\gamma$ in Sects. 4.1, 4.2 and 4.3.

In Sect. 4.1, we note that the threshold $\beta$ is the unique solution of equation (21). Excluding singular cases, we assume that $\int _\beta ^{+\infty } s^{-d_1-1} \varPi (s) ds >0$, and, therefore, in view of the Implicit Function Theorem and the chain rule, it follows that:

$$\begin{aligned} \frac{\partial \beta }{\partial \alpha } =&\frac{\partial \beta }{\partial d_2} \frac{\partial d_2}{\partial \alpha } = \frac{\partial \beta }{\partial d_2} \frac{-d_2}{\sqrt{\left( \frac{\sigma ^2}{2} - \alpha \right) ^2 + 2 \sigma ^2 r}},\\ \frac{\partial \beta }{\partial \sigma ^2} =&\frac{\partial \beta }{\partial d_2} \frac{\partial d_2}{\partial \sigma ^2} = \frac{\partial \beta }{\partial d_2} \frac{\alpha d_2-r}{\sigma ^2 \sqrt{\left( \frac{\sigma ^2}{2} - \alpha \right) ^2 + 2 \sigma ^2 r}},\\ \frac{\partial \beta }{\partial r} =&\frac{\partial \beta }{\partial d_2} \frac{\partial d_2}{\partial r} = \frac{\partial \beta }{\partial d_2} \frac{1}{\sqrt{\left( \frac{\sigma ^2}{2} - \alpha \right) ^2 + 2 \sigma ^2 r}}, \end{aligned}$$

where $\frac{\partial \beta }{\partial d_2} = - \frac{\int _\beta ^{+\infty } s^{-d_2-1} \varPi (s) \ln s \, ds}{\beta ^{-d_2-1} \varPi (\beta )}$. By Lemma 3, this expression is strictly positive, and therefore the derivatives of $\beta$ with respect to the parameters have the signs:

$$\begin{aligned} \mathrm {sg}\left( \frac{\partial \beta }{\partial \alpha } \right) = - \mathrm {sg}(d_2), \qquad \mathrm {sg}\left( \frac{\partial \beta }{\partial \sigma ^2} \right) = \mathrm {sg}(\alpha d_2 - r), \qquad \frac{\partial \beta }{\partial r} >0. \end{aligned}$$

(37)

In Sect. 4.2, we can se that the threshold $\gamma$ is the unique solution of equation (23). Excluding singular cases, we assume that $\int _0^\gamma s^{-d_2-1} \varPi (s) ds >0$. Therefore, the argument above shows that

$$\begin{aligned} \frac{\partial \gamma }{\partial \alpha } =&\frac{\int _0^\gamma s^{-d_1-1} \varPi (s) \ln s \, ds}{\gamma ^{-d_1-1} \varPi (\gamma )} \frac{d_1}{\sqrt{\left( \frac{\sigma ^2}{2} - \alpha \right) ^2 + 2 \sigma ^2 r}} ;\\ \frac{\partial \gamma }{\partial \sigma ^2} =&\frac{\int _0^\gamma s^{-d_1-1} \varPi (s) \ln s \, ds}{\gamma ^{-d_1-1} \varPi (\gamma )} \frac{r - \alpha d_1}{\sigma ^2 \sqrt{\left( \frac{\sigma ^2}{2} - \alpha \right) ^2 + 2 \sigma ^2 r}} ;\\ \frac{\partial \gamma }{\partial r} =&\frac{\int _0^\gamma s^{-d_1-1} \varPi (s) \ln s \, ds}{\gamma ^{-d_1-1} \varPi (\gamma )} \frac{-1}{\sqrt{\left( \frac{\sigma ^2}{2} - \alpha \right) ^2 + 2 \sigma ^2 r}} ; \end{aligned}$$

and the derivatives of the threshold $\gamma$ with respect to the parameters have the signs:

$$\begin{aligned} \mathrm {sg}\left( \frac{\partial \gamma }{\partial \alpha } \right) = \mathrm {sg}(d_1), \qquad \mathrm {sg}\left( \frac{\partial \gamma }{\partial \sigma ^2} \right) = \mathrm {sg}(r - \alpha d_1), \qquad \frac{\partial \gamma }{\partial r} <0. \end{aligned}$$

(38)

Finally, in Sect. 4.3 the thresholds $\beta \in ]0,x_{1l}[$, $\gamma \in ]x_{2r},+\infty [$ are the unique solutions of Eq. (25), and, therefore, we can use the Implicit Function Theorem and the chain rule to derive:

$$\begin{aligned} \frac{\partial \beta }{\partial r}&= \frac{-\beta ^{d_2+1} \left( I_1 + \gamma ^{d_2-d_1} I_2 \right) }{\left( \gamma ^{d_2-d_1} - \beta ^{d_2-d_1} \right) \varPi (\beta ) \sqrt{\left( \frac{\sigma ^2}{2} - \alpha \right) ^2 + 2 \sigma ^2 r}};\\ \frac{\partial \gamma }{\partial r}&= \frac{-\gamma ^{d_2+1} \left( I_1 + \beta ^{d_2-d_1} I_2 \right) }{\left( \gamma ^{d_2-d_1} - \beta ^{d_2-d_1} \right) \varPi (\gamma ) \sqrt{\left( \frac{\sigma ^2}{2} - \alpha \right) ^2 + 2 \sigma ^2 r}}; \\ \frac{\partial \beta }{\partial \alpha }&= \frac{\beta ^{d_2+1} \left( I_1 d_1 + \gamma ^{d_2-d_1} I_2 d_2 \right) }{\left( \gamma ^{d_2-d_1} - \beta ^{d_2-d_1} \right) \varPi (\beta ) \sqrt{\left( \frac{\sigma ^2}{2} - \alpha \right) ^2 + 2 \sigma ^2 r}};\\ \frac{\partial \gamma }{\partial \alpha }&= \frac{\gamma ^{d_2+1} \left( I_1 d_1 + \beta ^{d_2-d_1} I_2 d_2 \right) }{\left( \gamma ^{d_2-d_1} - \beta ^{d_2-d_1} \right) \varPi (\gamma ) \sqrt{\left( \frac{\sigma ^2}{2} - \alpha \right) ^2 + 2 \sigma ^2 r}};\\ \frac{\partial \beta }{\partial \sigma ^2}&= \frac{\beta ^{d_2+1} \left( I_1 (r-\alpha d_1) + \gamma ^{d_2-d_1} I_2 (r - \alpha d_2 ) \right) }{\left( \gamma ^{d_2-d_1} - \beta ^{d_2-d_1} \right) \varPi (\beta ) \sqrt{\left( \frac{\sigma ^2}{2} - \alpha \right) ^2 + 2 \sigma ^2 r}}; \\ \frac{\partial \gamma }{\partial \sigma ^2}&= \frac{\gamma ^{d_2+1} \left( I_1 (r-\alpha d_1) + \beta ^{d_2-d_1} I_2 (r - \alpha d_2 ) \right) }{\left( \gamma ^{d_2-d_1} - \beta ^{d_2-d_1} \right) \varPi (\gamma ) \sqrt{\left( \frac{\sigma ^2}{2} - \alpha \right) ^2 + 2 \sigma ^2 r}}, \end{aligned}$$

where $I_i = \int _\beta ^\gamma s^{-d_i-1} \varPi (s) \ln s \, ds$, $i=1,2$. Thus, Lemma 4 shows that $\frac{\partial \beta }{\partial r} >0$ and $\frac{\partial \gamma }{\partial r} <0$, that is, an increase of the discount rate leads to an earlier exercise of the option.

Concerning the effect of changes in the parameters $\alpha$, $\sigma ^2$, notice that

$$\begin{aligned} \mathrm {sg}\left( \frac{\partial \beta }{\partial \alpha } \right)&= -\mathrm {sg}\left( I_1 d_1 + \gamma ^{d_2-d_1} I_2 d_2\right) , \\ \mathrm {sg}\left( \frac{\partial \beta }{\partial \sigma ^2} \right)&= -\mathrm {sg}\left( I_1 (r - \alpha d_1) + \gamma ^{d_2-d_1} I_2 (r - \alpha d_2) \right) ,\\ \mathrm {sg}\left( \frac{\partial \gamma }{\partial \alpha } \right)&= -\mathrm {sg}\left( I_1 d_1 + \beta ^{d_2-d_1} I_2 d_2\right) ,\\ \mathrm {sg}\left( \frac{\partial \gamma }{\partial \sigma ^2} \right)&= -\mathrm {sg}\left( I_1 (r - \alpha d_1) + \beta ^{d_2-d_1} I_2 (r - \alpha d_2) \right) . \end{aligned}$$