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?

  • $\begingroup$ 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. $\endgroup$ – Kapes Mate Mar 23 at 13:08

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