I'm working in double exponential jump-diffusion model known as the Kou model with following form, under the physical probability measure $P$.

$$ ‎\frac{dS(t)}{S(t-)}=\mu‎‏ ‎dt+\sigma ‎dW(‎t)+d(\sum_{i=1}^{N(t)}(V_i-1)) $$

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

$$ f_Y(y)=p.‎\eta_1 e^{-‎\eta_{1}y‎‎}‎\upharpoonleft_{y‎\geq 0‎}+q.‎\eta_2 e^{‎\eta_2 y‎} \upharpoonleft_{y<0},\eta_{1}>1,\eta_{2}>0 ‎‎‎$$

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:

$$ S(t)=S(0)‎\exp‎\{(\mu- ‎\frac{1}{2}‎\sigma‎^2‎)t+‎\sigma ‎W(t)‎\} ‎\prod‎_{i=1}^{N(t)}V_i ‎‎$$

and also The stock price process, $(S_t)_{t ‎\geq 0‎} $‎‎, driven by these model, is given by:

$$ S_{t}=S_{0}e^{L_t} $$

is the stock price at time zero and $L_t$ is defined by:

$$ ‎L_t:=‎\gamma‎_{c}t+‎\sigma ‎W_t‎‎+\sum_{i=1}^{N_i}Y_i‎‎ $$

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 by $ (‎\gamma‎_{c},‎\sigma‎,‎‎\nu‎‎) $, where

$$ \begin{aligned} ‎\gamma‎_{c} & = r-q-‎\frac{1}{2}‎\sigma‎^2+ ‎\int_{‎\mathbb{R}‎} ‎(e^x-1) ‎\nu(dx) ‎\\‎ &=‎ ‎‎r-q-‎\frac{1}{2}‎\sigma‎^2+ ‎\lambda ‎\eta‎ \end{aligned}$$

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:

$$ f(x):=p.‎\eta_1 e^{-‎\eta_{1}x‎‎}‎\upharpoonleft_{x‎\geq 0‎}+(1-p).‎\eta_2 e^{‎\eta_2 x‎} ‎‎‎\upharpoonleft_{x<0},\eta_1>1,\eta_2>0 $$

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:

$$ \eta=‏ ‎‎\frac{p‏ ‎‎\alpha‎‎_1}{‎‎\alpha‎_{1}-1}+‎\frac{(1-p)‎\alpha‎_2}{‎\alpha‎_2+1}-1‎‎ $$

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 partial integro-differential equation (PIDE) in the non-exercise region:

$$ \begin{aligned} ‎\partial_‎\tau\, u(x,‎\tau‎)‏ &‎ = ‎‎\frac{1}{2}‎\sigma‎^2‏ ‎‎\partial‎_{x}^2‎ u‎ ‎+(r-q-‎\frac{1}{2}‎\sigma‎^2‎-\lambda \eta)‎\partial‎_x ‎u-(r+‎\lambda‎)u \\ ‎&+ ‎‎\lambda ‎\int ‎_{‎\mathbb{R}‎} ‎u(x+y,‎\tau‎) ‎f(y) ‎dy‎ \end{aligned} $$

with initial value

$$ u(x,0)=g(x):‎=G(e^x)= \begin{cases} \max\{e^x-k,0\}, & \text{call option} \\ \max\{k-e^x,0\}, & \text{put option} \end{cases}$$

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 & Tankov book and also I've searched in this book but I could not find the Exactly above PIDE.

  • $\begingroup$ Should $‎\partial_‎\tau(x,‎\tau‎)$ be $‎\partial_‎\tau \, u(x,‎\tau‎)$? $\endgroup$
    – Gordon
    Jul 10, 2015 at 20:36

2 Answers 2


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*}


Let's start with the main idea, I hope you can finish the computations yourself. Whenever you want to derive a pricing equation, try the following approach: discounted value of portfolio/option/derivative must be a martingale for non-arbitrage reasons. Since you have a Markovian dynamics in variables $t$ and $S$, you assume that the price is some function $V(t,S)$. What you need to do know is to compute the Ito differential $$ \mathrm d\left(\mathrm e^{-rt}V(t,S_t)\right) = (\dots)\mathrm dt + (\dots)\mathrm dw_t + (\dots)\mathrm d\bar N_t $$ where $\bar N_t$ is the compensated Poisson process. The last two terms are martingales, so the term in the first bracket must be zero: that gives you the pricing equation. For the computation of this differential, and $\bar N_t$ see e.g. Shreve's II volume.

  • $\begingroup$ The original poster may just want to know how to derive the first term. $\endgroup$
    – Gordon
    Jul 9, 2015 at 13:08

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