Last-Passage American Cancelable Option in Lévy Models

Abstract

We derive the explicit price of the perpetual American put option canceled at the last-passage time of the underlying above some fixed level. We assume that the asset process is governed by a geometric spectrally negative Lévy process. We show that the optimal exercise time is the first moment when the asset price process drops below an optimal threshold. We perform numerical analysis considering classical Black–Scholes models and the model where the logarithm of the asset price has additional exponential downward shocks. The proof is based on some martingale arguments and the fluctuation theory of Lévy processes.

1. Introduction

The main goal of this paper is to find the closed-form formula for the price of the perpetual American put option canceled at the last-passage time of the underlying above some fixed level h. More formally, in this paper, we find the following value function

$$\overline{V}\left(s\right)=\underset{\tau \in \mathcal{T}}{sup}\mathbb{E}\left[e^{-r\tau }{(K-S_{\tau })}^{+}\mathbb{I}(\tau <\theta )\right|S_0=s],$$

(1)

for a family of stopping times $\mathcal{T}$, an underlying risky asset price process $S_t$, fixed strike price $K>0$ and the risk-free interest rate r, where

$$\theta =sup\{t\ge 0|S_t\ge h\},$$

(2)

for some fixed threshold $h>K$. We assume a Lévy market, that is, in our model, the asset price is described by a geometric spectrally negative Lévy process

$$S_t=e^{X_t},$$

(3)

where $X_t$ is a spectrally negative Lévy process and $S_0=s=e^x$ is an initial asset price. It is well known that a Lévy market allows for a more realistic representation of price dynamics, capturing certain features such as skewness and asymmetry, as well as a greater flexibility in calibrating the model to market prices; see, e.g., Cont (2001); Cont and Tankov (2004) and references therein.

In fact, we choose

$$X_t=x+\mu t+\sigma B_t-\sum _{k=1}^{N_t}{U}_k,$$

(4)

where $x=X_0=logs$ and $\sigma >0$. In (4), $B_t$ is a Brownian motion, $\mu$ is a fixed drift, $N_t$ is a homogeneous Poisson process (independent of $B_t$) with intensity $\lambda$ and ${\left\{U_k\right\}}_{\{k\in \mathbb{N}\}}$ is a sequence (independent of $B_t$ and $N_t$) of independent identically distributed exponential random variables with the expected value ${\rho }^{-1}$. We assume that all considered processes live in a common filtered probability space $(\Omega ,\mathcal{F},{\left\{{\mathcal{F}}_t\right\}}_{\{t\ge 0\}},\mathbb{P})$ with a natural filtration ${\left\{{\mathcal{F}}_t\right\}}_{\{t\ge 0\}}$ of $X_t$ satisfying the usual conditions. The above process (4) makes the asset price process $S_t$ a jump-diffusion model. Furthermore, we assume that dividends are not paid to the holders of the underlying asset and that the expectation in (1) is taken with respect to the martingale measure $\mathbb{P}$, that is, $e^{X_t-rt}$ is a $\mathbb{P}$ local martingale. In fact, as noted in (Cont and Tankov 2004, Table 1.1, p. 29), introducing jumps into the model implies a loss of completeness of the market, which results in the lack of uniqueness of an equivalent martingale measure. Still, we can choose one of these measures, and the price is the same regardless of the choice of the martingale measure.

In (4), we allow $\lambda =0$, which corresponds to the classical Black–Scholes (B-S) model with $\mu =r-\frac{{\sigma }^2}{2}$.

To find the value function (1), we will use the ‘guess and verify method’, in which we guess the candidate stopping rule and then verify that this is truly the optimal stopping rule using the verification Theorem 2. That is, we first guess the form of the stopping time as the first downward crossing time of some threshold $a>h$ and calculate the value function in terms of the scale functions using the fluctuation theory of Lévy processes. Then, we maximize it with respect to the exercise level a. In the last step, we prove that our guessed value function satisfies the Hamilton–Jacobi–Bellman equation (HJB) system and hence the verification step.

In the final part of the paper, we performed a comprehensive numerical analysis based on the known form of the scale functions for the process (4).

The pricing of American options has been investigated over the past four decades in various contexts; see Detemple (2006) for a review. This paper focuses mainly on adding a rather new cancellation feature built into the basic American contract. This canceling or recalling in the financial contracts can effectively mitigate undesirable positions in risky times or times when the markets are highly volatile. Therefore, we believe that this type of financial contract can be very attractive for many investors. Of course, these types of derivatives include a cancellation provision.

Our article continues the research conducted by Gapeev et al. (2020), where the authors also evaluate the American-style option (1), but they carry this out by solving an appropriate HJB system of equations. In Gapeev et al. (2020), the underlying asset price was described by geometric Brownian motion, for which the above approach is very natural due to the locality of the diffusive generator of the asset price process $S_t$. Still, in the context of nonlocal generators, a ‘guess-and-verify’ method used in this paper seems to be more efficient. To show the relationship of these approaches, we also provide a connection of our results to the seminal HJB equations.

There are other works strongly related to our paper. Most of them reduce the original optimal stopping problems to the associated free boundary problems. Wu and Li (2022) valuate the American put option in a very similar model as Gapeev et al. (2020) did, that is, any stopping rules must be before the last exit time of the price of the underlying assets at any fixed level. This paper focuses only on the Black–Scholes market as well. The difference is that Wu and Li (2022) analyze the finite-time maturity. Gapeev and Motairi (2022) consider the perpetual American cancelable dividend-paying put and call option in the Black–Merton–Scholes market, where the cancellation times are assumed to occur when the underlying risky asset price process hits some unobservable random thresholds. They proved that the optimal stopping times are the first times at which the asset price reaches stochastic boundaries depending on the current values of its running maximum and minimum processes. Other related articles are (Gapeev and Li 2022a, 2022b), who study perpetual American standard and lookback options terminated by the writers at the last times at which the underlying stock reaches its running maximum or minimum. Then, the linear and fractional recovery amounts are paid to the holders. Other important papers are Dumitrescu et al. (2018); Gapeev and Motairi (2018); Gapeev and Rodosthenous (2014); Linetsky (2006); Szimayer (2005) (see also references therein).

The paper is organized as follows. Our main result is given in Section 2. The proof of the main result is contained in Section 3. At the end of the paper, in Section 4, we describe some numerical analyses.

2. Main Result

To present the main result of this paper, we introduce required notations. Let $\mathbb{P}$ be a martingale measure and $\mathbb{E}$ be the expectation with respect to $\mathbb{P}$ with the convention ${\mathbb{E}}_{logs}[·]=\mathbb{E}[·|S_0=s]=\mathbb{E}[·|X_0=logs]$. We skip the subindex in expectation when $X_0=0$ (hence $S_0=1$).

We define a Laplace exponent of the process $X_t$ through

$$\Psi \left(\theta \right)=\frac{1}{t}log\mathbb{E}{e}^{\theta X_t}.$$

For the process $X_t$ defined in (4), we have

$$\Psi \left(\theta \right)=\mu \theta +\frac{{\sigma }^2{\theta }^2}{2}-\frac{\lambda \theta }{\theta +\rho }.$$

(5)

Since under the risk-neutral measure $\mathbb{P}$ the discounted asset price process $e^{-rt}{S}_t$ is a martingale, we assume throughout this paper that

$$\Psi \left(1\right)=r.$$

(6)

In other words, we take

$$\mu =r-\frac{{\sigma }^2}{2}+\frac{\lambda }{1+\rho }.$$

(7)

For $r\ge 0$, the so-called scale function is defined as a continuous function $W^{\left(r\right)}:[0,\infty )\to [0,\infty )$ such that:

$${\int }_0^{\infty }{e}^{-\beta x}{W}^{\left(r\right)}\left(x\right)dx=\frac{1}{\Psi \left(\beta \right)-r},\mathrm{for}\beta >\Phi \left(r\right).$$

(8)

With the first scale function, we associate the second one given by

$$Z^{\left(r\right)}\left(x\right)=1+r{\int }_0^x{W}^{\left(r\right)}\left(y\right)dy.$$

Lemma 1.

The scale function for process X defined by (4) is given by

$$W^{\left(r\right)}\left(x\right)=\sum _{i=1}^3{C}_i{e}^{{\eta }_{i}x},$$

(9)

where:

$$\begin{array}{cc}\hfill {\eta }_1=1,{\eta }_{2/3}& =\frac{-1}{2(\rho {\sigma }^2+{\sigma }^2)}\left(2\lambda +2r+{\rho }^2{\sigma }^2+\rho {\sigma }^2+2r\rho \pm 2\sqrt{\omega }\right)\hfill \end{array}$$

(10)

and

$$\omega ={\lambda }^2+\lambda (\rho +1)(2r+{\sigma }^2)+{(\rho +1)}^2{\left(r-\frac{1}{2}\rho {\sigma }^2\right)}^2.$$

(11)

Furthermore,

$$\begin{array}{c}\hfill C_1=\frac{2({\eta }_1+\rho )}{{\sigma }^2({\eta }_1-{\eta }_2)({\eta }_1-{\eta }_3)},C_2=\frac{2({\eta }_2+\rho )}{{\sigma }^2({\eta }_2-{\eta }_1)({\eta }_2-{\eta }_3)},\end{array}$$

(12)

$$\begin{array}{c}\hfill C_3=\frac{2({\eta }_3+\rho )}{{\sigma }^2({\eta }_3-{\eta }_1)({\eta }_3-{\eta }_2)}.\end{array}$$

(13)

The proof of this lemma is given in Appendix A. We show that the optimal exercise time is of the form

$${\tau }_a=inf\{t\ge 0:S_t\le a\}=inf\{t\ge 0:X_t\le log\left(a\right)\}.$$

(14)

The threshold a needs to be lower than the strike price K (and hence of the canceling threshold h), so that exercising the option can be profitable to the holder. We take

$$0<a<K.$$

(15)

We denote

$$G\left(s\right)={(K-s)}^{+}\left({\left(\frac{h}{s}\right)}^{\alpha }\wedge 1\right).$$

(16)

The main result of this paper is as follows.

Theorem 1.

The price of the perpetual American cancelable put option defined in (1) equals

$$\begin{array}{cc}\hfill \overline{V}\left(s\right)& =\frac{{\sigma }^2}{2}\left[W^{{\left(r\right)}^{\prime }}\left(log\frac{s}{a^{*}}\right)-\Phi \left(r\right)W^{\left(r\right)}\left(log\frac{s}{a^{*}}\right)\right]G\left(a^{*}\right)\hfill \\ & +\left[Z^{\left(r\right)}\left(log\frac{s}{a^{*}}\right)-\frac{{\sigma }^2}{2}{W}^{{\left(r\right)}^{\prime }}\left(log\frac{s}{a^{*}}\right)-W^{\left(r\right)}\left(log\frac{s}{a^{*}}\right)\left(\frac{r}{\Phi \left(r\right)}-\frac{\Phi \left(r\right){\sigma }^2}{2}\right)\right]\hfill \\ \hfill & \times {\int }_0^{\infty }\rho e^{-\rho y}G\left(a^{*}{e}^{-y}\right)dy\hfill \end{array}$$

and ${\tau }_{a^{*}}$ defined in (14) is the optimal stopping rule for the optimal stopping threshold

$$a^{*}=\frac{K\left(\frac{{\sigma }^2}{2}{\sum }_{i=2}^3{C}_i{\eta }_i({\eta }_i-1)+\alpha +\frac{\rho }{\rho -\alpha }{\sum }_{i=2}^3{C}_i{\eta }_i\left[r\left(\frac{1}{{\eta }_i}-1\right)-\frac{{\sigma }^2}{2}\left({\eta }_i-1\right)\right]\right)}{\frac{{\sigma }^2}{2}{\sum }_{i=2}^3{C}_i{\eta }_i({\eta }_i-1)-(1-\alpha )+\frac{\rho }{\rho -\alpha +1}{\sum }_{i=2}^3{C}_i{\eta }_i\left[r\left(\frac{1}{{\eta }_i}-1\right)-\frac{{\sigma }^2}{2}\left({\eta }_i-1\right)\right]}.$$

(17)

3. Proof of the Main Result

To prove Theorem 1 we start by transforming the value function $\overline{V}\left(s\right)$. Let $Z_t=\mathbb{P}(\theta >t|{\mathcal{F}}_t)$ be the conditional survival process. Additionally, let us introduce a parameter $\alpha$ solving

$$\Phi (-\alpha )=0.$$

(18)

Equation (18) has three solutions. The only one that can be negative is

$$\alpha =\frac{\rho }{2}+\frac{\mu }{{\sigma }^2}-\sqrt{{\left(\frac{\rho }{2}-\frac{\mu }{{\sigma }^2}\right)}^2+\frac{2\lambda }{{\sigma }^2}}.$$

(19)

Observe that $-1<\alpha <0$. Then,

$$Z_t=\left\{\begin{array}{cc}{\left(\frac{h}{S_t}\right)}^{\alpha }\wedge 1,\hfill & \mathrm{if}\alpha <0,\hfill \\ 1,\hfill & \mathrm{if}\alpha \ge 0\hfill \end{array}\right.$$

and

$${\mathbb{E}}_{logs}[e^{-r\tau }{(K-S_{\tau })}^{+};\tau <\theta ]={\mathbb{E}}_{logs}\left[e^{-r\tau }{(K-S_{\tau })}^{+}\left({\left(\frac{h}{S_{\tau }}\right)}^{\alpha }\wedge 1\right)\right].$$

Hence,

$$\overline{V}\left(s\right)=\underset{\tau \in \mathcal{T}}{sup}{\mathbb{E}}_{logs}\left[e^{-r\tau }G\left(S_{\tau }\right)\right],$$

(20)

where the function G is defined in (16). Representation (20) is very convenient from the point of view of the general optimal stopping theory. However, we can still modify this representation. Observe that by (Ken-Iti 1999, Thm. 31.5, p. 208)

$$\mathcal{L}f\left(s\right)=\tilde{\mu }s{f}^{\prime }\left(s\right)+\frac{1}{2}{\sigma }^2{s}^2{f}^{″}\left(s\right)+\lambda \rho {\int }_0^{\infty }\left(f\left(s{e}^{-y}\right)-f\left(s\right)\right)e^{-\rho y}dy$$

(21)

is the infinitesimal generator of the process $X_t$ acting on ${\mathcal{C}}_0^2\left(\mathbb{R}\right)$, where

$$\tilde{\mu }=r+\frac{\lambda }{1+\rho }.$$

(22)

Due to the localization procedure, $\mathcal{L}$ is an extended generator that also acts on ${\mathcal{C}}^2\left(\mathbb{R}\right)$. For $s<K$, we denote

$$H\left(s\right)=\mathcal{L}G\left(s\right)-rG\left(s\right)={\left(\frac{h}{s}\right)}^{\alpha }(\delta s-rK)\mathbb{I}(s<h)-rK\mathbb{I}(s\ge h),$$

(23)

where

$$\delta =\alpha {\sigma }^2-\frac{\lambda }{1+\rho }+\frac{\lambda \rho }{(\rho -\alpha )(1+\rho -\alpha )}$$

(24)

and $\mathbb{I}\left(C\right)$ denotes the indicator of an event C. Let us also introduce the local time $l_t^{log\left(h\right)}\left(X\right)$ of the process X at the point $log\left(h\right)$ (see, e.g., Peskir (2007)):

$$l_t^{log\left(h\right)}\left(X\right)=\mathbb{P}-\underset{\epsilon \downarrow 0}{lim}\frac{1}{2\epsilon }{\int }_0^t\mathbb{I}(log\left(h\right)-\epsilon <X_u<log\left(h\right)+\epsilon )d<X{>}_u.$$

(25)

The key representation of $\overline{V}\left(s\right)$ is given in the next lemma, for which proof is given in Appendix A.

Lemma 2.

The following holds true:

$$\overline{V}\left(s\right)=G\left(s\right)+V^{*}\left(s\right),$$

(26)

where

$$\begin{array}{cc}\hfill V^{*}\left(s\right)& =\underset{\tau \in \mathcal{T}}{sup}{\mathbb{E}}_{logs}\left[{\int }_0^{\tau }{e}^{-r\tau }H\left(S_u\right)du\right.\hfill \\ \hfill & \left.+{\int }_0^{\tau }{e}^{-r\tau }h\left(G^{\prime }(h+)-G^{\prime }(h-)\right)\mathbb{I}(S_u=h)d{l}_t^{log\left(h\right)}\left(X\right)\right].\hfill \end{array}$$

(27)

The next step is a verification theorem that allows us to identify $V^{*}\left(s\right)$. Its proof is also given in Appendix B.

Theorem 2.

Suppose that function $V\in {\mathcal{C}}^2\left(\mathbb{R}\right)$, except the point h and a point a, where it is of class ${\mathcal{C}}^1\left(\mathbb{R}\right)$. Assume that V satisfies the following HJB system of equations

$$\begin{array}{cc}\hfill (\mathcal{L}V-rV)\left(s\right)& \le -H\left(s\right),\hfill \end{array}$$

(28)

$$\begin{array}{cc}\hfill V^{\prime }(h+)-V^{\prime }(h-)& =-(G^{\prime }(h+)-G^{\prime }(h-)).\hfill \end{array}$$

(29)

Then, $V\left(s\right)\ge V^{*}\left(s\right)$.

Now, the main idea of the proof of the main result is to find the value function

$${\overline{V}}_a\left(s\right)={\mathbb{E}}_s[e^{-r{\tau }_a}(K-S_{{\tau }_a});{\tau }_a<\theta ],s>a,$$

when the exercise time is the first downward passage time ${\tau }_a$ of the asset price defined in (14). We let

$$V\left(s\right)={\overline{V}}_{a^{*}}\left(s\right)-G\left(s\right)\mathrm{for}s>a\mathrm{and}0\mathrm{otherwise}$$

(30)

for the unique $0<a^{*}<K$ (see assumption (15)) solving $V_{a^{*}}\left(s\right){|}_{s=a^{*}+}=0$ and $V_{a^{*}}^{\prime }\left(s\right){|}_{s=a^{*}+}=0$.

In the final step, we show that $V\left(s\right)$ satisfies all the conditions of the verification theorem (Theorem 2), and therefore we obtain the assertion of the main Theorem 1.

We now prove the following proposition, which is interesting in itself, and its proof is presented in Appendix C.

Proposition 1.

The value ${\overline{V}}_a\left(s\right)$ is equal to

$$\begin{array}{cc}\hfill {\overline{V}}_a\left(s\right)& =\frac{{\sigma }^2}{2}\left[W^{{\left(r\right)}^{\prime }}\left(log\frac{s}{a}\right)-\Phi \left(r\right)W^{\left(r\right)}\left(log\frac{s}{a}\right)\right]G\left(a\right)\hfill \\ & \hfill +\left[Z^{\left(r\right)}\left(log\frac{s}{a}\right)-\frac{{\sigma }^2}{2}{W}^{{\left(r\right)}^{\prime }}\left(log\frac{s}{a}\right)-W^{\left(r\right)}\left(log\frac{s}{a}\right)\left(\frac{r}{\Phi \left(r\right)}-\frac{\Phi \left(r\right){\sigma }^2}{2}\right)\right]\\ \hfill & \times {\int }_0^{\infty }\rho e^{-\rho y}G\left(a{e}^{-y}\right)dy.\hfill \end{array}$$

(31)

We are now ready to give the proof of the main result of this paper.

Proof of Theorem 1.

We recall that

$$V\left(s\right)=\left\{\begin{array}{cc}{\overline{V}}_{a^{*}}\left(s\right)-G\left(s\right),\hfill & \mathrm{if}s>a\hfill \\ 0,\hfill & \mathrm{if}0<s\le a\hfill \end{array}\right.$$

(32)

for ${\overline{V}}_a\left(s\right)$ defined in (31) and $a^{*}$ solving $V_{a^{*}}^{\prime }\left(s\right){|}_{s=a^{*}+}=0$. We verify that the inequality (28) and the Equation (29) formulated in Theorem 2 are satisfied.

By (Kuznetsov 2012, Thm. 3.10), both scale functions $W^{\left(r\right)}$ and $Z^{\left(r\right)}$ belong to ${\mathcal{C}}^2\left(\mathbb{R}\right)$. Hence, by (31), ${\overline{V}}_{a^{*}}\left(s\right)\in {\mathcal{C}}^2(\mathbb{R}\backslash \{a^{*},h\})$ and of class ${\mathcal{C}}^1\left(\mathbb{R}\right)$ at $a^{*}$ by the choice of $a^{*}$. Moreover,

$$V^{\prime }(h+)-V^{\prime }(h-)={\overline{V}}^{\prime }(h+)-{\overline{V}}^{\prime }(h-)-(G^{\prime }(h+)-G^{\prime }(h-))=-(G^{\prime }(h+)-G^{\prime }(h-))$$

(33)

and hence (29) is satisfied.

Observe that the only candidate for $0<a<K$ that satisfies condition

$V_a^{\prime }\left(s\right){|}_{s=a+}=0$ is given as a solution of the following equation

$$\begin{array}{c}\frac{1}{a}{\left(\frac{h}{a}\right)}^{\alpha }\left[\frac{{\sigma }^2}{2}(K-a)\left[C_2{\eta }_2({\eta }_2-1)+C_3{\eta }_3({\eta }_3-1)\right](1-\alpha )a+\alpha K\right.\hfill \\ \hfill \left.+\rho \left(\frac{K}{\rho -\alpha }+\frac{a}{\rho -\alpha +1}\right)\times \sum _{i=2}^3{C}_i{\eta }_i\left(r\left(\frac{1}{{\eta }_i}-1\right)-\frac{{\sigma }^2}{2}\left({\eta }_i-1\right)\right)\right]=0\end{array}$$

and hence is given in (17). We still have to verify if $0<a^{*}<K$. To do so, we rewrite the representation (17) of $a^{*}$ as follows:

$$\begin{array}{cc}\hfill a^{*}& =K+K\frac{1+\frac{\rho }{(\rho -\alpha )(\rho -\alpha +1)}{\sum }_{i=2}^3{C}_i{\eta }_i\left[r\left(\frac{1}{{\eta }_i}-1\right)-\frac{{\sigma }^2}{2}\left({\eta }_i-1\right)\right]}{(\alpha -1)+{\sum }_{i=2}^3{C}_i{\eta }_i\left[\frac{r\rho }{\rho -\alpha +1}\left(\frac{1}{{\eta }_i}-1\right)+\frac{{\sigma }^2}{2}({\eta }_i-1)\left(1-\frac{\rho }{\rho -\alpha +1}\right)\right]}.\hfill \end{array}$$

(34)

Furthermore,

$$C_2{\eta }_2({\eta }_2-1)=\frac{\rho +1}{\omega }({\eta }_2+\rho ),C_3{\eta }_3({\eta }_3-1)=-\frac{\rho +1}{\omega }({\eta }_3+\rho )$$

(35)

and

$$-{\eta }_2<\rho <{\eta }_3;$$

(36)

see also Yamazaki (2017). Therefore, we can see that the right-hand sides of equations in (35) are positive, which means that also the left-hand sides are also positive. As both ${\eta }_2$ and ${\eta }_3$ are negative, it means that both $C_2$ and $C_3$ are also negative. By virtue of the fact that ${\sum }_{i=2}^3{C}_i\left[r\left(\frac{1}{{\eta }_i}-1\right)-\frac{{\sigma }^2}{2}\left({\eta }_i-1\right)\right]=1$, we can see that:

$$C_2\left[r\left(\frac{1}{{\eta }_2}-1\right)-\frac{{\sigma }^2}{2}\left({\eta }_2-1\right)\right]=-C_3\left[r\left(\frac{1}{{\eta }_3}-1\right)-\frac{{\sigma }^2}{2}\left({\eta }_3-1\right)\right].$$

(37)

Now, by (36)

$$C_2{\eta }_2\left[r\left(\frac{1}{{\eta }_2}-1\right)-\frac{{\sigma }^2}{2}\left({\eta }_2-1\right)\right]>-C_3{\eta }_3\left[r\left(\frac{1}{{\eta }_3}-1\right)-\frac{{\sigma }^2}{2}\left({\eta }_3-1\right)\right].$$

(38)

which gives

$$\sum _{i=2}^3{C}_i{\eta }_i\left[r\left(\frac{1}{{\eta }_i}-1\right)-\frac{{\sigma }^2}{2}\left({\eta }_i-1\right)\right]>0.$$

(39)

This leads to the conclusion that the numerator of the rhs of (34) is strictly positive. Moreover, since ${\eta }_1$ and ${\eta }_2$ are negative, and $-1<\alpha <0$, we know that its denominator is negative. This immediately gives $a^{*}<K$. To show that $a^{*}>0$, we need to verify that the numerator plus the denominator is smaller than 0, that is, that

$$\begin{array}{cc}\hfill & 1+\frac{\rho }{(\rho -\alpha )(\rho -\alpha +1)}\sum _{i=2}^3{C}_i{\eta }_i\left[r\left(\frac{1}{{\eta }_i}-1\right)-\frac{{\sigma }^2}{2}\left({\eta }_i-1\right)\right]\hfill \\ \hfill & +(\alpha -1)+\sum _{i=2}^3{C}_i{\eta }_i\left[\frac{r\rho }{\rho -\alpha +1}\left(\frac{1}{{\eta }_i}-1\right)+\frac{{\sigma }^2}{2}({\eta }_i-1)\left(1-\frac{\rho }{\rho -\alpha +1}\right)\right]\hfill \\ \hfill & =\alpha +\sum _{i=2}^3{C}_i{\eta }_i\left[\frac{\rho }{\rho -\alpha }r\left(\frac{1}{{\eta }_i}-1\right)+\frac{{\sigma }^2}{2}\left({\eta }_i-1\right)\left(1-\frac{\rho }{\rho -\alpha }\right)\right]<0.\hfill \end{array}$$

This follows from the fact that $\alpha ,C_2,C_3,{\eta }_2,{\eta }_3$ are all strictly negative.

Now, note that $e^{-rt\wedge {\tau }_{a^{*}}}V\left(S_{t\wedge {\tau }_{a^{*}}}\right)$ is a martingale. Indeed, from (Avram et al. 2004, Rem. 5), we know that $e^{-rt\wedge {\tau }_{a^{*}}\wedge {\tau }_b^{+}}{W}^{\left(r\right)}\left(S_{t\wedge {\tau }_{a^{*}}\wedge {\tau }_b^{+}}\right)$ and $e^{-rt\wedge {\tau }_{a^{*}}\wedge {\tau }_b^{+}}{Z}^{\left(r\right)}\left(S_{t\wedge {\tau }_{a^{*}}\wedge {\tau }_b^{+}}\right)$ are martingales, where ${\tau }_b^{+}inf\{t\ge 0:S_t\ge b\}$. Furthermore, by (A10), the process $e^{-rt\wedge {\tau }_{a^{*}}\wedge {\tau }_b^{+}}{W}^{{\left(r\right)}^{\prime }}\left(S_{t\wedge {\tau }_{a^{*}}}\right)$ is a martingale as well, since $e^{-rt\wedge {\tau }_{a^{*}}\wedge {\tau }_b^{+}}A\left(S_{t\wedge {\tau }_{a^{*}}}\right)$ is martingale, where

$$A\left(s\right)=\frac{{\sigma }^2}{2}\left[W^{{\left(r\right)}^{\prime }}\left(log\frac{s}{a}\right)-\Phi \left(r\right)W^{\left(r\right)}\left(log\frac{s}{a}\right)\right]$$

is the right-hand side of (A10). To show this, observe that by Markov property of $S_t$, we have

$$\begin{array}{cc}\hfill & {\mathbb{E}}_{logs}\left[e^{-r{\tau }_{a^{*}}}\mathbb{I}({\tau }_{a^{*}}<\infty ,S_{{\tau }_{a^{*}}}=a^{*})\right|{\mathcal{F}}_t]\hfill \\ \hfill & =\mathbb{I}({\tau }_{a^{*}}>t){\mathbb{E}}_{log{S}_t}\left[e^{-r{\tau }_{a^{*}}}\mathbb{I}({\tau }_{a^{*}}<\infty ,S_{{\tau }_{a^{*}}}=a^{*})\right]\hfill \\ & +I({\tau }_{a^{*}}\le t)e^{-r{\tau }_{a^{*}}}\mathbb{I}({\tau }_{a^{*}}<\infty ,S_{{\tau }_{a^{*}}}=a^{*})=e^{-r{\tau }_{a^{*}}\wedge t}A\left(S_{t\wedge {\tau }_{a^{*}}}\right),\hfill \end{array}$$

where we used the fact that $A\left(s\right)=0$ for $s<a$ and $A\left(a^{*}\right)=\frac{{\sigma }^2}{2}{W}^{{\left(r\right)}^{\prime }}\left(0\right)=1$ because $W^{\left(r\right)}\left(0\right)=0$ due to the assumption that $\sigma >0$ (see (Kuznetsov 2012, Lem. 3.1 and Lem. 3.2)). Since b appearing in ${\tau }_b^{+}$ above is general, hence

$$(\mathcal{L}{\overline{V}}_{a^{*}}-r{\overline{V}}_{a^{*}})\left(s\right)=0\mathrm{for}s>a$$

and thus for $s>a$

$$(\mathcal{L}V-rV)\left(s\right)=-(\mathcal{L}G-rG)\left(s\right)=-H\left(s\right)$$

(40)

by definition (23) of $H\left(s\right)$.

Now, for $s<a$, $V=0$ and $(\mathcal{L}V-rV)\left(s\right)=0$. To prove (28), one need thus prove that $H\left(s\right)=\mathcal{L}G\left(s\right)-rG\left(s\right)\le 0$ for $s<a$. Using the fact that $a<K<h$, we can write $G\left(s\right)$ for $s<a$ in the following form:

$$G\left(s\right)=(K-s){\left(\frac{h}{s}\right)}^{\alpha }.$$

(41)

Then,

$$\begin{array}{c}\hfill H\left(s\right)=\tilde{\mu }s{G}^{\prime }\left(s\right)+\frac{{\sigma }^2}{2}{s}^2{G}^{″}\left(s\right)+\lambda {\int }_0^{\infty }\left(G\left(s{e}^{-y}\right)-G\left(s\right)\right)d{F}_U\left(y\right)-rG\left(s\right)\\ \hfill ={\left(\frac{h}{s}\right)}^{\alpha }\left[s\left(\alpha -1\right)\left(r+\frac{\lambda }{1+\rho }-\frac{{\sigma }^2}{2}\alpha \right)+s\lambda +sr-\frac{s\rho \lambda }{\rho +1-\alpha }\right.\\ \hfill \left.-K(\alpha +1)\left(r-\frac{\lambda \alpha }{(1+\rho )(\rho -\alpha )}-\frac{{\sigma }^2}{2}\alpha \right)\right].\end{array}$$

(42)

It can easily be seen that the term $K(\alpha +1)\left(r-\frac{\lambda \alpha }{(1+\rho )(\rho -\alpha )}-\frac{{\sigma }^2}{2}\alpha \right)$ is strictly positive, as $-1<\alpha <0$. Additionally, $\left(\alpha -1\right)\left(r+\frac{\lambda }{1+\rho }-\frac{{\sigma }^2}{2}\alpha \right)+\lambda +r-\frac{\rho \lambda }{\rho +1-\alpha }$ is strictly negative. As both h and s are positive, hence $H\left(s\right)<0$. This completes the proof. □

4. Numerical Analysis

4.1. Geometric Brownian Motion

As the first case in the numerical analysis, we consider the underlying asset price described by the geometric Brownian motion. We set the intensity $\lambda$ of the $N_t$ process from Equation (4) equal to zero, and thus $X_t$ becomes the arithmetic Brownian motion with drift parameter $\mu =r-\frac{{\sigma }^2}{2}$. This example corresponds to the option evaluated in Gapeev et al. (2020). The scope of the numerical analysis here is to find the optimal exercise level $a^{*}$ and the fair price $\overline{V}\left(s\right)$ of the option. The parameters are chosen as follows: the strike price $K=100$, the threshold $h=120$ so that $h>K$, risk-free rate $r=5\%$ and the volatility of the underlying asset ${\sigma }^2=0.2$. Additionally, the initial price of the underlying asset $s=S_0$ is set to 110. We first start with calculation of $a^{*}$. Using formula (17), we obtain $a^{*}=50$. This value fits well to the assumption (15). Furthermore, when we use formula (43) from Gapeev et al. (2020), we obtain the same result:

$$a^{*}=\frac{{\eta }_2+\alpha }{{\eta }_2+\alpha -1}K=100\frac{-0.5-0.5}{-0.5-0.5-1}=50.$$

(43)

Now, by using Theorem 1, we obtain the price of the cancelable option $\overline{V}\left(s\right)\approx 21.76$. Again, when we use the formula (37) proposed in Gapeev et al. (2020), we obtain the same result:

$$\overline{V}\left(s\right)=(K-a^{*}){\left(\frac{s}{a^{*}}\right)}^{{\eta }_2}{\left(\frac{h}{a^{*}}\right)}^{\alpha }=\frac{250}{\sqrt{132}}\approx 21.76.$$

(44)

Now, according to (Wilmott 2006, Chap. 9.2), the price of the standard perpetual American option for the no-dividend case is given by:

$$\overline{V}\left(s\right)=B{s}^{{\alpha }^{-}},$$

(45)

where $B=-\frac{1}{{\alpha }^{-}}{\left(\frac{K}{1-1/{\alpha }^{-}}\right)}^{1-{\alpha }^{-}}$ and ${\alpha }^{-}=-\frac{2r}{{\sigma }^2}$. The price of this option with parameters chosen as previously in this section is equal to 36.70. This price is significantly higher than the price of the corresponding cancelable option due to the higher risk the issuer of this contract has to deal with.

Finally, in Figure 1, we demonstrate the dependence of the cancelable option price on the initial price of the underlying instrument. Note that the price and payoff function fit smoothly at $s=a^{*}$.

4.2. Geometric Spectrally Negative Lévy Process

Here, we perform a similar calculation as in Section 4.1, but now we set a fixed $\lambda >0$. We keep the other parameters unchanged, i.e., $h=120,s=S_0=110,K=100,r=5\%,{\sigma }^2=0.2$, and additionally set $\lambda =5$ and $\rho =2$. Again, we start by finding the optimal threshold $a^{*}$. With formula (17), we obtain $a^{*}\approx 63.18$. One more time, we use Theorem 1 to find the fair price of the cancelable option, and we obtain $\overline{V}\left(s\right)\approx 18.99$. The price is smaller, although it cannot be predicted without numerical analysis. Indeed, although all the jumps of the underlying are downward, the drift is still greater, which is a consequence of applying the martingale measure in a pricing formula.

As the final step, in Figure 2, we show the behavior of the payoff and price functions of the cancelable put option, depending on the initial underlying asset price. Again, the smooth fit of the aforementioned functions is clearly visible for $S=a^{*}$.