I'm working in double exponential Jump-diffusion model (we know as kou model) with following form , under the physical probability measure P: \begin{equation} \frac{dS(t)}{S(t-)}=\mu dt+\sigma dW(t)+d(\sum_{i=1}^{N(t)}(V_i-1)) \end{equation} where W(t) is a standard Brownian motion, N(t) is a Poisson process with rate \lambda , and \{V_i\} is a sequence of independent identically distributed (i.i.d.) non negative random variables such that Y = log(V) has an asymmetric double exponential distribution with the density \begin{equation} f_Y(y)=p.\eta_1 e^{-\eta_{1}y}\upharpoonleft_{y\geq 0}+q.\eta_2 e^{\eta_2 y} \upharpoonleft_{y<0},\eta_{1}>1,\eta_{2}>0 \end{equation} where p, q \ge 0, p+q = 1, represent the probabilities of upward and downward jumps.
Solving the stochastic differential equation gives the dynamics of the asset price: \begin{equation} S(t)=S(0)\exp\{(\mu- \frac{1}{2}\sigma^2)t+\sigma W(t)\} \prod_{i=1}^{N(t)}V_i \end{equation} and also The stock price process, (S_t)_{t \geq 0} , driven by these model, is given by: \begin{equation} S_{t}=S_{0}e^{L_t} \end{equation} where S_0 is the stock price at time zero and L_t is defined by: \begin{equation} L_t:=\gamma_{c}t+\sigma W_t+\sum_{i=1}^{N_i}Y_i \end{equation} here,\gamma_{c} is a drift term , \sigma is a volatility, W_t is a Brownian motion, N_t is a Possion process with intensity \lambda , Y_i is an i.i.d. sequence of random variables.Since \sigma>0 in up equation, there exists a risk-neutral probability measure Q such that the discounted process \{e^{-(r-q)} S_t\}_{t \geq 0} becomes a martingale,where r is the interest rate and q is the dividend rate.Then under this new measure Q , the risk-neutral Levy triplet of L_t can be described as follows: \begin{equation*} (\gamma_{c},\sigma,\nu) \end{equation*} where \begin{align*} \gamma_{c} & = r-q-\frac{1}{2}\sigma^2+ \int_{\mathbb{R}} (e^x-1) \nu(dx) \\ & = r-q-\frac{1}{2}\sigma^2+ \lambda \eta \end{align*} Here we focus on the case where the Levy measure is associated to the pure-jump component and hence the Levy measure \nu(dx) can be written as \lambda f(x) dx , where the weight function f(x) can take the following form: \begin{equation} f(x):=p.\eta_1 e^{-\eta_{1}x}\upharpoonleft_{x\geq 0}+(1-p).\eta_2 e^{\eta_2 x} \upharpoonleft_{x<0},\eta_1>1,\eta_2>0 \end{equation}
Also in \eta = \int_{\mathbb{R}}(e^x-1)f(x) dx represents the expected relative price change due to a jump. Since we have defined the Levy density function f(x) for double exponential Jump-diffusion model, \eta can be computed as: \begin{equation} \eta= \frac{p \alpha_1}{\alpha_{1}-1}+\frac{(1-p)\alpha_2}{\alpha_2+1}-1 \end{equation}
This is found by integrating e^x over the real line by setting \alpha_1 >1 and \alpha_{2}>0 .
We let \tau=T-t, the time-to-maturity, where T is the maturity of the financial option under consideration and we introduce x = log S_t, the underlying asset's log-price. If u(x; \tau ) denotes the values of some (American and European) contingent claim on S_t when log St = x and \tau = T - t, then it is well-known, see for example, (Cont and Tankov, 2004) that u satisfies the following PIDE in the non-exercise region: \begin{align*} \partial_\tau\, u(x,\tau) & = \frac{1}{2}\sigma^2 \partial_{x}^2 u +(r-q-\frac{1}{2}\sigma^2-\lambda \eta)\partial_x u-(r+\lambda)u \\ &+ \lambda \int _{\mathbb{R}} u(x+y,\tau) f(y) dy \end{align*} with initial value \begin{equation} u(x,0)=g(x):=G(e^x)= \begin{cases} max\{e^x-k,0\}, & \text{call option} \\ max\{k-e^x,0\}, & \text{put option} \end{cases} \end{equation}
my question is how we can derive the above PIDE I've searched a lot of article but most of them only mention the PIDE and we said you can find in Cont and Tankov Book and also I've searched in this book but I could not find the Exactly above PIDE.
thanks for help.
Answer
Let \{P_t \mid t \geq 0\} be a compound Poisson process, where \begin{align*} P_t = \sum_{i=1}^{N_t} (V_i -1), \end{align*} and N_t is a Poisson process with intensity \lambda and jump times \tau_i, i = 1, \ldots, \infty. Let Y_i=\ln V_i and f(x) be the density function. Then \begin{align*} P_t - \lambda t E(V_1) &= P_t - \lambda t \int_{\mathbb{R}}(e^x-1)f(x) dx \end{align*} is a martingale. We denote by \eta = \int_{\mathbb{R}}(e^x-1)f(x) dx. Moreover, we assume that the equity price process \{S_t \mid t \geq 0\} satisfies the SDE \begin{align*} \frac{dS_t}{S_t} = (r-q-\lambda \eta)dt + \sigma dW_t + dP_t, \end{align*} where \{W_t \mid t \geq 0\} is a standard Brownian motion. Then \begin{align*} S_t = S_0 \exp\Big(\big(r-q-\frac{1}{2}\sigma^2 - \lambda \eta \big)t + \sigma W_t + \sum_{i=1}^{N_t} Y_i \Big). \end{align*} That is, \begin{align*} d \ln S_t = (r-q-\frac{1}{2}\sigma^2-\lambda \eta)dt + \sigma dW_t + d\sum_{i=1}^{N_t} Y_i. \end{align*}
Let X_t = \ln S_t, and u(X_t, t) be the option price at time t, where 0 \leq t \leq T. Then, by Ito's formula, \begin{align*} u(X_t, t) &= u(X_0, 0) + \int_0^t\partial_t u(X_s, s) ds + \int_0^t\partial_x u(X_{s-}, s) dX_s + \frac{1}{2}\sigma^2 \int_0^t \partial_{xx} u(X_s, s)ds\\ & \qquad +\sum_{s \leq t}\big[u(X_s, s) - u(X_{s-}, s) - \partial_x u(X_{s-}, s)\Delta X_s\big] \quad (\mbox{where } \Delta X_s=X_s - X_{s-})\\ &= u(X_0, 0) + \int_0^t\partial_t u(X_s, s) ds + \int_0^t\partial_x u(X_{s}, s) dX_s^c + \frac{1}{2}\sigma^2 \int_0^t \partial_{xx} u(X_s, s)ds\\ & \qquad +\sum_{s \leq t}\big[u(X_t, t) - u(X_{t-}, t) \big] \quad (\mbox{where } X_t^c = \big(r-q-\frac{1}{2}\sigma^2 - \lambda \eta \big)t + \sigma W_t)\\ &= u(X_0, 0) + \int_0^t\partial_t u(X_s, s) ds + \int_0^t\partial_x u(X_{s}, s) dX_s^c + \frac{1}{2}\sigma^2 \int_0^t \partial_{xx} u(X_s, s)ds\\ & \qquad +\int_0^t \int_{\mathbb{R}}\big[ u(X_{s-} + y, s) - u(X_{s-}, s))\big]\mu(ds, dy) \quad (\mbox{where } \mu = \sum_{i=1}^{\infty} \delta_{\tau_i, Y_i})\\ &= u(X_0, 0) + \int_0^t\partial_t u(X_s, s) ds + \int_0^t\partial_x u(X_{s}, s) dX_s^c + \frac{1}{2}\sigma^2 \int_0^t \partial_{xx} u(X_s, s)ds\\ &\qquad +\int_0^t \int_{\mathbb{R}}\big[ u(X_{s-} + y, s) - u(X_{s-}, s))\big](\mu(ds, dy) - ds v(dy)) \\ &\qquad +\int_0^t ds\int_{\mathbb{R}}\big[ u(X_{s} + y, s) - u(X_{s}, s))\big]\lambda f(y)dy, \end{align*} where v(dy) = \lambda f(y)dy. Here \begin{align*} M_t = \int_0^t \int_{\mathbb{R}}\big[ u(X_{s-} + y, s) - u(X_{s-}, s))\big](\mu(ds, dy) - ds v(dy)) \end{align*} is a martingale. Since u(X_t, t) e^{-rt} is a martingale, and \begin{align*} d\big(u(X_t, t) e^{-rt}\big) &= e^{-rt}\big[-r u dt + du\big], \end{align*} we obtain that \begin{align*} &-ru(X_t, t) + \partial_t u(X_t, t) + \big(r-q-\frac{1}{2}\sigma^2 - \lambda \eta \big)\partial_x u(X_{s}, s) + \frac{1}{2}\sigma^2 \partial_{xx} u(X_t, t) \\ & \qquad\qquad + \int_{\mathbb{R}}\big[ u(X_{t} + y, t) - u(X_{t}, t))\big]\lambda f(y)dy = 0. \end{align*} That is, \begin{align*} & \partial_t u(X_t, t) + \big(r-q-\frac{1}{2}\sigma^2 - \lambda \eta \big)\partial_x u(X_{s}, s) + \frac{1}{2}\sigma^2 \partial_{xx} u(X_t, t) -(r+\lambda)u(X_t, t)\\ & \qquad\qquad + \lambda \int_{\mathbb{R}} u(X_{t} + y, t) f(y)dy = 0. \end{align*}
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