If you combine the compound Poisson process with the Brownian motion you obtain the simplest case of a Jump-diffusion. Let’s define $$X_t = \mu t + \sigma W_t + J_t$$ where $W_t$ is a Wiener process and $J_t$ is a compound Poisson process. In what sense is possible to write an SDE to represent the dynamics of $X_t$? what is the meaning of $dJ_t$?


2 Answers 2


$dJ_{t}$ can be understood as a Steljes measure , when you want to define jumps using bounded variation function , but you can simply understand it as $J_{t}-J_{t-}$

Those processes belong to a more general class of process called Levy processes through Lévy–Khintchine representation where you can define clearly the jump part, you can find better expanding of Ito's formula and exponential form based on Doleans-Dade forumula.

There are also more complex jumps models like Bates model or double exponential Kou model

  • $\begingroup$ Hello @Kupoc allahoui. Thank for your help. $\endgroup$
    – RedZoro
    Commented Jul 3, 2020 at 14:49

Let $$ J_t = \sum_{i=1}^{N_t} Z_i$$ be a compound Poisson process, with $(T_n)_{n\geq 1}$ being the jump times for Poisson process $(N_t)_{t\geq 0}$ and $(Z_i)_{i\geq 1}$ sequence of i.i.d. variables independent of $(N_t)_{t\geq 0}$.

We need the stochastic integral against $dJ_t$ in order to make sense of $dJ_t$.

For discrete jump size we have $$\delta J_t = J_t-J_{t^-} = Z_{N_t}(N_t - N_{t^-}) = Z_{N_t}\delta N_t$$

Then for a process $(u_s)_{s\geq 0}$ we have:

$$ \int_0^t u_s dJ_s = \int_0^t u_s Z_{N_s}dN_s = \sum_{i=1}^{N_t} u_{T_i} Z_i$$

In particular, for $u$ set to constant $1$, we have:

$$ \int_0^t dJ_s = J_t$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.