First, let us consider time period *dt*, and let *S* satisfy the following stochastic differential equation:

$$\begin{aligned}& dS=\mu (t)S\,dt+\sigma (t)S\,dW+\bigl(q(t)-1\bigr)S\,dN, \end{aligned}$$

(1)

where \(\mu (t)=r(t)-dt-\lambda (t)\kappa (t)\) is the drift rate, \(\sigma (t)\) is the volatility, *dW* is the increment of a continuous-time stochastic process, called a standard Brownian process, *dN* is a Poisson process, and *r* is the continuously compounded *risk-free interest rate*. \(\kappa (t)=E[q(t)-1]\) that depends on *t* is identically independent distributed random variable representing the expected relative jump size. Actually, any \(\kappa (t)\) that belongs to an interval \((-1, \infty )\) for all *t* and \(q(t) -1\) is an impulse function producing a jump from *S* to \(Sq(t)\). It is important that \(dN = 0\) with probability \(1-\lambda \,dt\) and \(dN=1\) with probability \(\lambda \,dt\), where *λ* is the Poisson arrival intensity, which is the expected number of “events” or “arrivals” that occur per unit time. For more information, we refer to [17].

Now, let us consider a case when \(dN = 0\) in (1), then the given equation will be equivalent to the usual stochastic process of “geometric Brownian motion” assumed in the Black–Scholes models. If the Poisson event occurs, then equation (1) can be written as

$$\begin{aligned} &\frac{dS}{S}\simeq q(t)-1. \end{aligned}$$

(2)

In this case the function \(q(t)-1\) is an impulse function producing a jump from *S* to \(Sq(t)\). After that, we consider \(V (S, t)\) as the contingent claim depending on the asset price *S* and time *t*. Let in equation (1) \(dt = 0\). Then the following backward-in-time partial integro-differential equation may be solved to determine \(v(x, \tau )=V(S, \tau )\):

$$\begin{aligned} v_{\tau } =&\frac{1}{2}\sigma ^{2}v_{xx} +\biggl(r-\lambda \kappa - \frac{1}{2}\sigma ^{2}\biggr)v_{x}+\lambda \int _{-\infty }^{\infty }v(z, \tau )f(z-x)\,dz \\ &{}-(\lambda +r)v, \quad (\tau , x)\in [0, T]\times (-\infty , \infty ), \end{aligned}$$

(3)

where \(x=\log (S)\), \(\tau = T -t \), and *T* is the expiry time of the contingent claim. In the Merton model the density function *f* of a normal distribution is given by

$$\begin{aligned} f(x)=\frac{1}{\sqrt{2\pi \eta ^{2}}}e^{-\frac{(x-\nu )^{2}}{2\eta ^{2}}}, \end{aligned}$$

where *ν* is the mean and \(\eta ^{2}\) is the variance of the jump size probability distribution. It is possible to present the expectation operator in the form \(E[q(t)] = \exp (\nu +\frac{\eta ^{2}}{2})\), i.e., \(\kappa (\tau )=E[q(\tau )-1]=\exp (\nu +\frac{\eta ^{2}}{2})-1\). Also we can divide the integral term into two parts as follows:

$$\begin{aligned} \int _{\mathbb{R}}v(z, \tau )f(z-x)\,dz= \int _{[-b, b]}v(z, \tau )f(z-x)\,dz+ \int _{\mathbb{R} \backslash [-b, b]}v(z, \tau )f(z-x)\,dz. \end{aligned}$$

Let us define \(\varPhi (\tau , x, b)= \int _{\mathbb{R} \backslash [-b, b]}v(z, \tau )f(z-x)\,dz\). In the case of the Merton model,

$$\begin{aligned} \varPhi (\tau , x, b)=Se^{x+\nu +\frac{\eta ^{2}}{2}}N\biggl(\frac{x-b+\nu +\eta ^{2}}{\eta } \biggr)-Ke^{-r\tau }N\biggl(\frac{x-b+\nu }{\eta }\biggr), \end{aligned}$$

where \(N(y)=\frac{1}{\sqrt{2\pi }}\int _{-\infty }^{y} e^{-\frac{x ^{2}}{2}}\,dx\) and *K* is a strike price. In the case of European options, the payoff functions of the call and put options are as follows:

$$\begin{aligned}& C_{E}=\max \bigl(0, Se^{x} -K\bigr), \qquad P_{E}=\max \bigl(0, K-Se^{x}\bigr). \end{aligned}$$

The asymptotic behavior of the European option when \(S = 0\) or \(S \rightarrow \infty \) is similar to the Black–Scholes PDE.

American options have the important additional feature that early exercise is permitted at any time during the life of the option. American options can be exercised at any time before expiry. Formally, the value of an American call option is

$$\begin{aligned}& V(0,K) = \max \bigl\{ {e^{ - r\tau }} {\bigl({e^{x}S_{\tau }} - K\bigr)^{+} } : 0 \le \tau \le T\bigr\} \end{aligned}$$

since the payoff at \(0 \le \tau \le T\) is \(\max (0,{({e^{x}S_{\tau }} - K) }) \). Also we have

$$\begin{aligned}& C_{A}(\tau ,K)\geq C_{E}(\tau ,K), \qquad P_{A}(\tau ,K)\geq P_{E}(\tau ,K). \end{aligned}$$

By the chain rule and the following new constants

$$\begin{aligned}& \alpha =-\frac{r-\lambda \kappa -\frac{1}{2}\sigma ^{2}}{\sigma ^{2}}, \qquad \beta =-\frac{1}{2}\sigma ^{2}\alpha ^{2}-(\lambda +r), \end{aligned}$$

the transformed Eq. (3) turns into a simpler form

$$\begin{aligned}& u_{\tau }=\frac{1}{2}\sigma ^{2}u_{xx}+\lambda \int _{-\infty }^{\infty }u(z, \tau )g(z-x)\,dz, \quad (\tau , x)\in [0, T]\times (-\infty , \infty ), \end{aligned}$$

(4)

where \(u(x, \tau )=e^{-\alpha x-\beta \tau }v(x, \tau )\) and \(g(x)=e^{\alpha x}f(x)\).