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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