# Given a Levy Exponent find the jump-measure and drift

A Levy subordinator is an finite variation Levy process with non-negative drift and positive jumps. The Levy exponent is given by

$\phi(\lambda) = \gamma \lambda + \int_0^\infty $$1 - e^{-\lambda s}$$ \nu(ds)$

where $\gamma>0$ is the drift of the subordinator and $\nu$ is the jump measure (Levy measure). If the jumps are a compound Poisson process with (net) jump intensity $\alpha$ and jump-size distribution $\mu$ then $\nu = \alpha \mu$ and the levy exponent becomes

$\phi(\lambda) = \gamma \lambda + \alpha $$1 - \widehat{\mu}( \lambda )$$$

where

$\widehat{\mu}( \lambda ) = \int_0^\infty e^{- \lambda s} \mu(ds)$

My Questions are as follows:

1. Given a function $\phi(\lambda)$, how do I know if there is a $\nu$ and a $\gamma$ that generates it?

2. For a given $\phi$ is the pair that generates it $(\gamma,\nu)$ unique?

3. Assume the jumps are a compound Poisson process. If you are given $\phi(\lambda)$ can you find $\alpha$ and $\gamma$? Finding $\alpha$ and $\gamma$ would uniquely determine $\widehat{\mu}( \lambda )$ and allow us to reconstruct $\mu(ds)$ from the inverse Laplace transform. Then $\nu(ds) = \alpha \mu(ds)$.

4. More generally, given $\phi(\lambda)$, can you find $\nu$ and $\gamma$.

The reason for these questions is that I am going to numerically construct $\phi(\lambda)$ from data. Ideally, I would like to then construct $\gamma$ and $\nu$ (or $\alpha$ for a Poisson process) as well. At this point, it isn't clear to me that I actually need $\gamma$ and $\nu$ for my calculations. It may be that $\phi(\lambda)$ is enough (this project is in its nascent stage at the moment). But, even if I don't need $\nu$ and $\gamma$ I am curious to see if I can construct them. And an existance and uniqueness result would definitely strengthen my paper.

....

So, I have a partial answer to the construction of $(\gamma,\nu)$ from $\phi$. Clearly

$\gamma = \lim_{\lambda \to \infty} \phi(\lambda)/\lambda$

Still looking for a construction of $\nu$ at the moment.

-

Once you get $\gamma$, you can calculate
$\frac{\phi'(\lambda)-\gamma}{\lambda}=\int_0^\infty e^{-\lambda s}\nu(ds)$
and $\nu$ can be obtained by Laplace inversion.