# The distribution of the jump diffusion process

In the Merton jump diffusion model the process of the share price can be expressed as $$S_{t}=S_{0}\cdot\exp\left\{ X_{t}\right\} ,$$ where $$X_{t}=\mu t+\sigma W_{t}+\sum_{i=1}^{N_{t}}Y_{i}.$$ Here $$\mu$$ and $$\sigma$$ are constants, $$W_{t}$$ is a Wiener process, $$N_{t}$$ is a Poisson process, and $$Y_{i}$$s are $$N\left(\mu_{A},\sigma_{A}\right)$$ $$iid$$ variables. $$W_{t}$$, $$N_{t}$$ and $$Y_{i}$$ are independent.

I know that we can express the distribution of $$X_{t}$$ and hence the distribution of $$S_{t}$$ as well, so we know the distribution of the process for a given $$t$$, but can we calculate the distribution of the (full) $$X$$ process? I mean do we know the joint distribution of $$X_{t_{1}},X_{t_{2}},...,X_{t_{n}}$$ for every $$n$$ and for every $$t_{1},t_{2},...,t_{n}$$ instants?

• Reading this question again, I have become a little bit unsure about my statement, that "I know we can express the distribution of X_t...". I changed my mind and now I don't even know how to find this margin distribution. If N_t was fixed, then the statement was right, and we could say something about the distribution of X_t, but since N_t is not fixed, but a random number I don't know how to find its distribution. We could find its Laplace-transform, but it would look like an exp(exp(something)), and I don't know how I would use any inversion formula for this expression. – Kapes Mate Mar 23 at 13:08