TITLE: Given a Levy Exponent find the jump-measure and drift
QUESTION [5 upvotes]: 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:

Given a function $\phi(\lambda)$, how do I know if there is a $\nu$ and a $\gamma$ that generates it?
For a given $\phi$ is the pair that generates it $(\gamma,\nu)$ unique?
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 $\nu(ds) = \alpha \mu(ds)$.
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 $\gamma = \lim\limits_{\lambda \to \infty} \phi(\lambda)/\lambda$.
Still looking for a construction of $\nu$ at the moment.

REPLY [2 votes]: 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.