Subordinator – conditions for the Levy triple

Question

different books write Levy Khinchnin's formula in different ways and thus I have a problem with understanding the conditions that Levy's triple must satisfy for the process to be a subordinator. In my notation, Levy process with characteristic triple $(b,\sigma, \nu)$ has characteristic function:
$$\phi_{X_t}(u)=\exp\{t(ibu-\frac{1}{2}\sigma^2u^2+\int_{\mathbb{R}}e^{iux}-1-iux1_{|x|<1})\nu(dx))\}$$

What conditions, then, must be satisfy by the levy triple for the process to be subordinator?

Answer 1

Using your convention, the characteristic triple of a subordinator satisfies:

• $\sigma = 0$ (no Brownian part),
• $\nu$ has support in $(0,\infty)$ (positive jumps) and $\int_0^\infty (1\wedge x) \, \nu(dx) < \infty$ (bounded variation),
• $b - \int_0^1 x\,\nu(dx) \geqslant 0$ (positive drift).

This is Theorem 1.3.15 in Applebaum's Lévy Processes and Stochastic Calculus, see also Chapter 3 of Bertoin's Lévy Processes.

Notice that when working with subordinators, it is more convenient to use the Laplace exponent instead of the characteristic exponent. It is defined by $\mathbb{E}[\exp(-\lambda X_t)] = \exp(-t\psi(\lambda))$ and for subordinators the Lévy-Khintchine formula reads $$\psi(\lambda) = d \lambda + \int_0^\infty (1-e^{-\lambda x}) \, \nu(dx)$$ where $d\geqslant 0$ and $\nu$ a measure on $(0,\infty)$ satisfying $\int_0^\infty (1\wedge x) \, \nu(d x) <\infty$.