Contraction Options and Optimal Multiple-Stopping in Spectrally Negative Lévy Models

Abstract

This paper studies the optimal multiple-stopping problem arising in the context of the timing option to withdraw from a project in stages. The profits are driven by a general spectrally negative Lévy process. This allows the model to incorporate sudden declines of the project values, generalizing greatly the classical geometric Brownian motion model. We solve the one-stage case as well as the extension to the multiple-stage case. The optimal stopping times are of threshold-type and the value function admits an expression in terms of the scale function. A series of numerical experiments are conducted to verify the optimality and to evaluate the efficiency of the algorithm.

Appendix: Proofs

1.1 Proof of Lemma 2.1

Let $g_1(x) := -bx$, $x \in \mathbb {R}$. We have

$$\begin{aligned} \rho _{g_1,A}^{(r)}&= -b \int _{(0,\infty )}\Pi (\mathrm{d}u) \int _0^{u} e^{-\Phi _r z} (z-u) \mathrm{d}z = -\frac{b}{\Phi _r^2}\int _{(0,\infty )} \Pi (\mathrm{d}u) (1- e^{-\Phi _r u} - \Phi _r u), \end{aligned}$$

and hence

$$\begin{aligned} \Lambda (A; 0, g_1)&:= -\frac{r}{\Phi _r} g_1(A) - \frac{\sigma ^2}{2} g_1'(A) + \rho _{g_1,A}^{(r)}\\&= b \Big [ \frac{r}{\Phi _r} A + \frac{\sigma ^2}{2} - \frac{1}{\Phi _r^2} \int _{(0,\infty )}\Pi (\mathrm{d}u) (1- e^{-\Phi _r u} - \Phi _r u) \Big ] \\&= \frac{b}{\Phi _r^2}\Big [ c \Phi _r +\frac{\sigma ^2}{2} \Phi _r^2 +\int _{( 0,\infty )}(e^{- \Phi _r u}-1+ \Phi _r u 1_{\{0 <u<1\}})\,\Pi (\mathrm{d}u) \Big ] \\&\quad - \frac{b}{\Phi _r} \Big [ c - \int _{[1,\infty )} u \Pi (\mathrm{d}u) - r A \Big ] \\&= b \Big [ \frac{r}{\Phi _r^2}- \frac{\psi '(0+) -r A }{\Phi _r} \Big ]. \end{aligned}$$

Let $g_2(x) = -\sum _{i=1}^N c_i e^{a_i x}$, $x \in \mathbb {R}$. Then,

$$\begin{aligned} \rho _{g_2,A}^{(r)} = - \sum _{i=1}^N c_i e^{a_i A} \int _{(0,\infty )} \Pi (\mathrm{d}u) \int _0^u e^{-\Phi _r z} (e^{a_i(z-u)} - 1) \mathrm{d}z. \end{aligned}$$

(6.1)

(Case 1) First suppose $a_i \ne \Phi _r$ for all $1 \le i \le N$. Simple algebra gives

$$\begin{aligned} \Lambda (A)= -\frac{r}{\Phi _r}K + b \Big ( \frac{r}{\Phi _r^2}+ \frac{r A - \psi '(0+)}{\Phi _r} \Big ) + \sum _{i=1}^N c_i e^{a_i A} \frac{M_r^{(a_i)}}{\Phi _r} + \Psi _f(A) \end{aligned}$$

(6.2)

where

$$\begin{aligned} M_r^{(a)}&:= r + \frac{a \sigma ^2}{2} \Phi _r+ \int _{(0,\infty )} \Pi (\mathrm{d}u) \Big [ (1 - e^{-\Phi _ru})\\&\qquad - e^{-a u} (1 - e^{-(\Phi _r-a) u}) \frac{\Phi _r}{\Phi _r-a}\Big ], \quad a \in \mathbb {R}\backslash \{\Phi _r \}. \end{aligned}$$

By the definition of $\psi$ and $\Phi _r$, we rewrite $M_r^{(a)}$ as

$$\begin{aligned}&r + \frac{a \sigma ^2}{2} \Phi _r+ \int _{(0,\infty )} \Pi (\mathrm{d}u) \left[ (1 - e^{-\Phi _ru} - \Phi _ru 1_{\{ 0 < u < 1 \}})\right. \\&\qquad \left. - e^{-au} (1 - e^{-(\Phi _r-a) u}) \frac{\Phi _r}{\Phi _r-a} + \Phi _ru 1_{\{ 0 < u < 1\}}\right] \\&= \Big ( c + a \sigma ^2 - \int _{(0,1)} u (e^{-a u}-1) \Pi (\mathrm{d}u) \Big ) \Phi _r+ \frac{\sigma ^2}{2} \Phi _r(\Phi _r- a ) \\&\qquad - \frac{\Phi _r}{\Phi _r-a} \int _{(0,\infty )} \Pi (\mathrm{d}u) e^{-a u} \left( 1 - e^{-(\Phi _r-a) u} - (\Phi _r-a) u 1_{\{0 < u <1 \}} \right) \\&= \frac{\Phi _r}{\Phi _r- a} \psi _a(\Phi _r-a), \end{aligned}$$

where the last equality holds by (2.6). On the other hand, $\psi _a(\Phi _r-a) = \psi (\Phi _r) - \psi (a) = r - \psi (a)$; see page 213 of [24]. Hence

$$\begin{aligned} \Phi _r\varpi _r(a) = \frac{\Phi _r}{\Phi _r- a} (r- \psi (a))&= \frac{\Phi _r}{\Phi _r- a} \psi _a(\Phi _r-a), \end{aligned}$$

which shows for the case $a_i \ne \Phi _r$ for all $1 \le i \le N$.

(Case 2) Suppose $a_j = \Phi _r$ for some $1 \le j \le N$ (with $a_i \ne a_j$ for $i \ne j$ by assumption). Take a sequence of (strictly) increasing sequence $a_j^{(m)} \uparrow a_j = \Phi _r$. Then a modification of (2.18) with $a_j$ replaced with $a_j^{(m)}$ is by Case 1

$$\begin{aligned} \Lambda ^{(m)}(A)&= -\frac{r}{\Phi _r}K + b \Big ( \frac{r}{\Phi _r^2}+ \frac{r A - \psi '(0+)}{\Phi _r} \Big )\\&\qquad + \sum _{1 \le i \le N, i \ne j} c_i e^{a_i A} \varpi _r(a_i) + c_j e^{a_j^{(m)} A} \varpi _r(a_j^{(m)}) + \Psi _f(A). \end{aligned}$$

By the definition of $\varpi _r$ as in (2.23), we have

$$\begin{aligned} \lim _{m \uparrow \infty }\Lambda ^{(m)}(A)= -\frac{r}{\Phi _r}K + b \Big ( \frac{r}{\Phi _r^2}+ \frac{r A - \psi '(0+)}{\Phi _r} \Big ) + \sum _{i=1}^N c_i e^{a_i A} \varpi _r(a_i) + \Psi _f(A). \end{aligned}$$

On the other hand, in view of (6.1), its integrand is monotone in $a$. Hence by the monotone convergence theorem and because $g$ and $g'$ are continuous in $a$, $\lim _{m \uparrow \infty }\Lambda ^{(m)}(A) = \Lambda (A)$, and the proof is complete for Case 2.

1.2 Proof of Lemma 2.2

Because $g(x)$ is infinitely differentiable, the results are clear for $x \in (-\infty , A^*)$. Hence we show for $x \in (A^*, \infty )$. Because $W^{(r)}(y)$ is differentiable on $y > 0$ as in Remark 2.1(1), $K Z^{(r)} (x-A^*) - b \Big [ \overline{Z}^{(r)}(x-A^*) + \big (A^*- \frac{\psi '(0+)}{r} \big )Z^{(r)}(x-A^*) + \frac{\psi '(0+)}{r} \Big ] - \sum _{i=1}^N c_i e^{a_i x} Z_{a_i}^{(r - \psi (a_i))} (x-A^*)$ is twice differentiable.

Regarding $\Theta _f(x;A^*)$, integration by parts thanks to the continuity of $f$ gives (with $\overline{W}^{(r)}(x) := \int _0^x W^{(r)}(y) \mathrm{d}y$, $x \in \mathbb {R}$)

$$\begin{aligned} \Theta _f(x;A^*) = f(A^*) \overline{W}^{(r)} (x-A^*) + \int _{A^*}^x f'(y) \overline{W}^{(r)}(x-y) \mathrm{d}y. \end{aligned}$$

It is differentiable with

$$\begin{aligned} \Theta _f'(x;A^*) = f(A^*) {W^{(r)}(x-A^*)} + \int _{A^*}^x f'(y) {W^{(r)}(x-y)} \mathrm{d}y. \end{aligned}$$

When $X$ is of unbounded variation, because $W^{(r)}(0) = 0$ as in Remark 2.1(2), $\Theta _f(x;A^*)$ is twice-differentiable with

$$\begin{aligned} \Theta _f''(x;A^*) = f(A^*) {W^{(r)'}(x-A^*)}+ \int _{A^*}^x f'(y) {W^{(r)'}(x-y)} \mathrm{d}y. \end{aligned}$$

1.3 Proof of Proposition 2.1

(i) Suppose $-\infty < A^* \le \infty$. By directly using the results of [17] (Lemma 3.7 and Proposition 3.4), we obtain

$$\begin{aligned}&(\mathcal {L}-r) u_{A^*}(x) + f(x) = 0, \quad x \in (A^*, \infty ), \nonumber \\&u_{A^*}(x) \ge g(x), \quad x \in \mathbb {R}, \end{aligned}$$

(6.3)

where $\mathcal {L}$ is the infinitesimal generator of $X$ applied to a sufficiently smooth function $h$, i.e.,

$$\begin{aligned} \mathcal {L} h(x) = c h'(x) + \frac{1}{2} \sigma ^2 h''(x) + \int _{(0,\infty )} [h(x-z) - h(x) + h'(x) z 1_{\{0 < z < 1\}}] \Pi (\mathrm{d}z). \end{aligned}$$

Lemma 5.1

If $-\infty < A^* \le \infty$, we have $(\mathcal {L}-r) u_{A^*}(x) + f(x) \le 0$ on $x \in (-\infty , A^*)$.

Proof

Fix $-\infty < A^* < \infty$. First, if we define $g_l(x) := x$, $x \in \mathbb {R}$,

$$\begin{aligned} (\mathcal {L}-r) g_l(x) = \psi '(0+)- rx. \end{aligned}$$

By the definition of $\psi$, if we define $g_e (x) := e^{ax}$, $x \in \mathbb {R}$,

$$\begin{aligned} \mathcal {L} g_e(x) = e^{ax} \Big [ ca + \frac{1}{2} \sigma ^2 a^2+ \int _{(0,\infty )} (e^{-az} - 1 + a z 1_{\{0 < z < 1\}} ) \Pi (\mathrm{d}z) \Big ] = e^{ax} \psi (a) \end{aligned}$$

for any $a > 0$ and hence we have

$$\begin{aligned} (\mathcal {L}-r) g(x) + f(x) = - rK -b(\psi '(0+) -rx) + \sum _{i=1}^N c_i e^{a_i x} (r - \psi (a_i)) + f(x). \end{aligned}$$

(6.4)

By how $A^*$ is chosen,

$$\begin{aligned} 0 = -r K + b \Big (\frac{r}{\Phi _r}- {(\psi '(0+) -r A^*)}\Big ) + \sum _{i=1}^N c_i e^{a_i A^*} {\Phi _r} \varpi _r(a_i) + \Phi _r\Psi _f(A^*). \end{aligned}$$

(6.5)

Because $f$ is increasing and $x < A^*$

$$\begin{aligned} \Phi _r\Psi _f(A^*) \ge \Phi _r\int _0^\infty e^{-\Phi _ry} f(x) \mathrm{d}y = f(x). \end{aligned}$$

(6.6)

By $A^* \ge x$ and $\Phi _r > 0$,

$$\begin{aligned} b \Big (\frac{r}{\Phi _r}- {(\psi '(0+) -r A^*)}\Big ) \ge -b(\psi '(0+) -rx). \end{aligned}$$

It is also easy to see that

$$\begin{aligned} e^{a_i A^*}{\Phi _r} \varpi _r(a_i) \ge e^{a_i x} (r - \psi (a_i)), \quad 1 \le i \le N. \end{aligned}$$

(6.7)

Indeed, for the case $r - \psi (a_i) > 0$, we must have $\Phi _r-a_i > 0$ and hence (6.7) holds by $A^* > x$; for the case $r - \psi (a_i) < 0$, the left-hand side is positive while the right-hand side is negative in (6.7); for the case $r - \psi (a_i) = 0$, the left-hand side is positive because $\psi '(\Phi _r)$ is, while the right-hand side is zero. Hence, by (6.5)–(6.7), $(\mathcal {L}-r) u_{A^*}(x) + f(x) \le 0$ holds.

This result also holds for the case $A^* = \infty$. In this case, $0 > -r K + \sum _{i=1}^N c_i e^{a_i \widehat{A}} \Phi _r\varpi _r (a_i) + b \big (\frac{r}{\Phi _r}- {(\psi '(0+) -r \widehat{A})}\big ) + \Phi _r\Psi _f(\widehat{A})$, for any $\widehat{A}\in \mathbb {R}$. Therefore $(\mathcal {L}-r) g(x) + f(x) < 0$ holds by the same reasoning as in (1) by simply replacing $A^*$ with $\widehat{A}$ for any $\widehat{A} > x$. $\square$

We are now ready to verify the optimality of $u_{A^*}(x)$ for the case $A^* \in (-\infty , \infty ]$. Thanks to Lemma 2.2 and the continuous/smooth fit condition as in Remark 2.4, a version of Meyer–Ito’s formula as in Theorem IV.71 of [32] (see also Theorem 2.1 of [30]) implies

$$\begin{aligned} e^{-r t}u_{A^*}(X_t)-u_{A^*}(X_0)&= \int _0^t e^{-rs}(\mathcal {L}-r)u_{A^*}(X_{s-})\mathrm{d}s+ M_t, \end{aligned}$$

with the local martingale part

$$\begin{aligned} M_t&:= \int _0^t \sigma e^{-rs} u_{A^*}'(X_{s-}) \mathrm{d}B_s \!-\!\int _{(0,t]} \int _{(0,1)} e^{-rs}u_{A^*}'(X_{s-})y (N(\mathrm{d}s\times \mathrm{d}y)\!-\!\Pi (\mathrm{d}y) \mathrm{d}s)\\&\quad +\int _{(0,t]} \int _{(0, \infty )}e^{-rs}(u_{A^*}(X_{s-}\!-\!y)\!-\!u_{A^*}(X_{s-})\!+\!u_{A^*}'(X_{s-})y1_{\{0 < y < 1\}})(N(\mathrm{d}s\!\times \! \mathrm{d}y)\\&\quad -\Pi (\mathrm{d}y) \mathrm{d}s), \end{aligned}$$

where $N(\mathrm{d}s \times \mathrm{d}x)$ is the Poisson random measure associated with the dual process $-X$.

Fix any stopping time $\tau$, and define for each $m\in {\mathbb {N}}$ the stopping time $T_m$ as

$$\begin{aligned} T_m :=\inf \{t>0\; : \;|X_t|> m\}, \end{aligned}$$

and the martingale process $M=\{M_{t\wedge \tau \wedge T_m}: t\ge 0\}$, with $M_0=0$. By optional sampling and because $(\mathcal {L} -r) u_{A^*} + f \le 0$ via (6.3) and Lemma 5.1,

$$\begin{aligned} \mathbb {E}^x \Big [ e^{-r (t \wedge \tau \wedge T_m)} u_{A^*}(X_{t \wedge \tau \wedge T_m}) + \int _0^{t \wedge \tau \wedge T_m} e^{-rs} f(X_s) \mathrm{d}s \Big ] \le u_{A^*}(x). \end{aligned}$$

When $x < A^*$, $u_{A^*}(x) = g(x) \ge g(A^*) > -\infty$ (the same result holds for $A^* = \infty$ by Remark 2.3). On the other hand, if $A^* \in (-\infty , \infty )$ and $x > A^*$, because $\partial u_A(x) / \partial A > 0$ on $(-\infty , A^*)$, the value $u_{A^*}(x)$ is bounded from below by a limit:

$$\begin{aligned} u_{-\infty }(x) := \lim _{A \downarrow -\infty } u_A(x) = \lim _{A \downarrow -\infty } \mathbb {E}^x \left[ \int _0^{\tau _A} e^{-rt} f(X_t) \mathrm{d}t + e^{-q \tau _A} g(X_{\tau _A}) 1_{\{ \tau _A < \infty \}}\right] . \end{aligned}$$

Hence,

$$\begin{aligned} (u_{A^*})_- (x) \le (u_{-\infty })_- (x)&\le \! \limsup _{A \downarrow -\infty } \mathbb {E}^x \left[ \int _0^{\tau _A} e^{-rt} f_-(X_t) \mathrm{d}t \!+\! e^{-q \tau _A} g_-(X_{\tau _A}) 1_{\{ \tau _A < \infty \}}\right] \\&= \mathbb {E}^x \left[ \int _0^{\infty } e^{-rt} f_-(X_t) \mathrm{d}t \right] , \end{aligned}$$

where the last equality holds by monotone convergence applied to the $f_-$ term and because $g_-(X_{\tau _A}) \le -g(A) \vee 0$ on $\{ \tau _A < \infty \}$, which is bounded because $g$ is decreasing. Because $f$ is increasing, the expectation on the right hand side decreases in $x$ and hence

$$\begin{aligned} (u_{A^*})_- (x) \le \mathbb {E}^{A^*} \left[ \int _0^{\infty } e^{-rt} f_-(X_t) \mathrm{d}t \right] , \quad x > A^*. \end{aligned}$$

In sum, $(u_{A^*})_-$ is bounded from above. Recall also Remark 2.2. Hence, Fatou’s lemma gives upon $t \uparrow \infty$ and $m \uparrow \infty$

$$\begin{aligned} \mathbb {E}^x \Big [ e^{-r \tau } u_{A^*}(X_{\tau }) 1_{\{\tau < \infty \}} + \int _0^{\tau } e^{-rs} f(X_s) \mathrm{d}s \Big ] \le u_{A^*}(x). \end{aligned}$$

Finally, (6.3) shows the result for $A^* \in (-\infty , \infty ]$.

(ii) It is now left to show for the case $A^* = -\infty$. Because $\partial u_A(x) / \partial A < 0$ for any $A \in \mathbb {R}$ as in (2.20), there again exists $u_{-\infty }(x) := \lim _{A \downarrow -\infty } u_A(x)$. Assumption 2.2 guarantees that $\lim _{A \downarrow -\infty }\mathbb {E}^x [e^{-q \tau _A}|X_{\tau _A}| 1_{\{ \tau _A < \infty \}}] = 0$; see e.g., [34]. Because the slope of $g_-(x)$ is bounded on the half-line, this shows that

$$\begin{aligned} \lim _{A \downarrow -\infty }\mathbb {E}^x [e^{-q \tau _A}g(X_{\tau _A}) 1_{\{ \tau _A < \infty \}}] = 0. \end{aligned}$$

This together with Remark 2.2 shows that $u_{-\infty }(x)$ has the desired expression.

Because $\partial u_A(x) / \partial A < 0$ for any $A \in \mathbb {R}$, $u_{-\infty }(x) > g(x)$ for any $x \in \mathbb {R}$ (hence the stopping region is an empty set). Moreover, because $u_{-\infty }$ is attained by $\tau ^*=\infty$, we have the claim.

1.4 Proof of Proposition 2.2

(1,2) We first suppose $x < A^*$. Then by definition $u_{A^*}(x) = g(x)$. Because $u_A(x) = g(x)$ for any $A \ge x$, it is sufficient to show $u_A(x) \le g(x)$ for $A < x$. This is indeed so because by (2.17), (2.20) and $\Lambda (x) < 0$ due to $x < A^*$,

$$\begin{aligned} u_A(x) \le u_x(x+) = g(x) + W^{(r)}(0) \Lambda (x) \le g(x) = u_{A^*}(x), \quad A < x < A^*. \end{aligned}$$

This proves (2). For (1), we additionally show for the case $x \ge A^*$. By (2.20), $u_{A^*}(x) \ge u_A(x)$ for any $A \le x$. For $A \ge x$, $u_A(x) = g(x)$ and by (2.17), (2.20) and $\Lambda (x) > 0$ due to $x > A^*$,

$$\begin{aligned} u_{A^*}(x) \ge u_x(x+) = g(x) + W^{(r)}(0) \Lambda (x) \ge g(x) = u_A(x), \quad A \ge x > A^*. \end{aligned}$$

Therefore $u_{A^*}(x) \ge u_A(x)$ uniformly in $x \in \mathbb {R}$, as desired.

The corresponding value function (for (1)) can be expressed as the sum of (2.15):

$$\begin{aligned} \widetilde{u}(x) v&= g(A^*) Z^{(r)}(x-A^*) + W^{(r)} (x-A^*) \left( - \frac{r}{\Phi _r} g(A^*) + \rho _{g,A^*}^{(r)} + \Psi _f(A^*) \right) \\&\qquad - \varphi _{g,A^*}^{(r)}(x) - \Theta _f(x; A^*). \end{aligned}$$

From the definition of $A^*$ that makes (2.18) vanish, it is simplified to

$$\begin{aligned} \widetilde{u}(x) = g(A^*) Z^{(r)}(x-A^*) + W^{(r)} (x-A^*) \frac{\sigma ^2}{2} g'(A^*) - \varphi _{g,A^*}^{(r)}(x) - \Theta _f(x; A^*), \end{aligned}$$

as desired.

(3) The proof is the same as that of Proposition 2.1(3).