I defined such a model for stock price

(1).... $$dS = \mu\ S\ dt + \sigma\ S\ dW + \rho\ S(dH - \mu) $$

, where $H$ is a so-called "resettable poisson process" defined as

(2).... $$dH(t) = dN_{\lambda}(t) - H(t-)dN_{\eta}(t) $$

, and $\mu := \frac{\lambda}{\eta}$.

Is it possible to derive some analytic results similar to Black-Scholes equation (3)?

(3).... $$ \frac{\partial V}{\partial t} + r\ S \frac{\partial V}{\partial S} + \frac{\sigma^2S^2}{2}\frac{\partial^2 V}{\partial S^2} - r\ V = 0$$

Even better, could we derive something similar to Black-Scholes formula for call/put option prices?

I tried but failed.

In classic GBM model, to get Black-Scholes euqation (3), the essential steps are:

By Ito's lemma,

(4)... $$df = (f_x+\mu f_x+\sigma^2/2\cdot f_{xx})dt + \sigma f_x dW$$

Based on GBM stock price model (5),

(5)... $$ds = \mu S dt + \sigma S dW$$

We have

(6).... $$dV = \left( \frac{\partial V}{\partial t} +\mu S \frac{\partial V}{\partial S}+\frac{\sigma^2 S^2}{2}\frac{\partial^2 V}{\partial S^2}\right)dt + \sigma S \frac{\partial V}{\partial S}dW$$

Putting (5) in (6) again we have

(7).... $$dV - \sigma S \frac{\partial V}{\partial S}dS = \left( \frac{\partial V}{\partial t} +\frac{\sigma^2 S^2}{2}\frac{\partial^2 V}{\partial S^2}\right)dt $$

then we can define (8)... $$\Pi = V - \frac{\partial V}{\partial S} S$$ , so the LHS of (7) is just $d\Pi$ and it's not related to any random effect, so we have

(9)... $$d\Pi = r \Pi dt$$ then we get (3).

After I introduce the "resettable Poisson process" $H_{\lambda, \eta}(t)$ in the model, I couldn't find a way to cancel both the $dW$ and $dN$....

Do you know how to solve this?

Any suggestions are appreciated, I'm stuck here...

  • 4
    $\begingroup$ do you means that you would like to price under semi-martingale assumptions (and especially jumps)? you should look at Financial modelling with Jump Processes ( Chapman & Hall / CRC Press) by Rama CONT and Peter TANKOV cmap.polytechnique.fr/~rama/Jumps . $\endgroup$
    – lehalle
    Commented Jun 10, 2012 at 8:06
  • $\begingroup$ It is possible, do you know how to calculate the generator of (S,H)? The answer will have a local term - i.e. A term involving integral, you will get a PIDE rather than a PDE. $\endgroup$
    – Lost1
    Commented Dec 29, 2013 at 11:05
  • 1
    $\begingroup$ What is this model called? $\endgroup$
    – Lost1
    Commented Dec 29, 2013 at 11:07
  • $\begingroup$ Lehalle has the best answers. I followed a course by A. Popier wich was based on this book. These are famous teachers, you may find their slides online. $\endgroup$ Commented Sep 2, 2014 at 13:54

2 Answers 2


I didn't work out the explicit details, but you can reproduce Black&Scholes methodology using the Ito's formula for Jump Diffusions. See for example, the sectio about Poisson jump processes in http://en.wikipedia.org/wiki/Itō's_lemma

In general every Markov process admits some kind of Ito's formula, known as Dynkin formula, which says that for a markov process $X$ with generator $\mathcal{L}$, and every sufficiently smooth $f$,$$ M^f_t = f( X_t ) - f( X_0 ) - \int_0^t \mathcal{L}f(S_s)ds$$ is a martingale.


Yes, one can derive an analytic call pricing formula for this model. It is a close analogue of the Merton jump-diffusion, and the same techniques apply. The final distribution is the superposition of lognormal distributions, with a common variance but differing means depending on the final level of the resetting Poisson process.

Lets find the probability p that H(t) = 0. I will assume H(0)=0; the other case is similar. Note that p'(t) = -lambda p(t) + eta (1-p(t)), reflecting the lambda arrival rate of jumps out of the state H=0 and the eta arrival rate of jumps back to that state. This is a first order seperable ODE with solution p(t) = eta/(eta+lambda) + lambda exp(-(eta+lambda)t)/(eta + lambda).

The probability q that H(t)=1 satisfies q'(t) = lambda p(t) - (lambda + eta) q(t). Again this is first order seperable; integrating we get q(t)=lambda eta/(eta+lambda)^2 + exp(-(eta+lambda)t)(lambda^2(eta+1)t + lambda eta)/(eta+lambda)^2.

This is getting messy but not intractible. The form is constant + exponential * polynomial. Generalizing, for each n the probability p_n(t) that H(t)=n satisfies p_n'(t) = lambda p_n(t) - (lambda + eta) p_{n-1}(t) and will have a solution of the form p_n(t) = lambda^n eta/(eta + lambda)^(n+1) + exp(-(eta+lambda)t)*f_n(t) where f_n(t) is a polynomial of degree n. I don't see how to write coefficients of f_n explicitly, but we can calculate them recursively from coefficients of f_{n-1}; if anyone really wants to see the details ask.

Finally let BS(S,K,T,r,mu,sigma) be the Black-Scholes call pricing formula. The price of a call under the resetting poisson jump-diffusion is sum_{n=0}^infinity p_n(T) BS(S,K,T,r,mu-rho lambda/eta + n rho/T,sigma).

In practise the sum should converge quickly, so we can truncate after a few terms.


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