 Laplace exponent of a subordinator

For a finite-valued subordinator $T$ starting at zero, the Laplace exponent of a subordinator is the function $\Phi$ determined byIts Lévy–Khintchine formula takes the formwhere $d$ is the deterministic drift coefficient and $\nu$ is the Lévy measure. This representation separates steady motion of the clock from its positive jumps.^[ref]

For a deterministic clock $T_t=dt$, $\Phi (\lambda )=d\lambda$. For a Poisson process of rate $a$ with unit jumps, $\Phi (\lambda )=a(1-e^{-\lambda })$. For a sum of independent subordinators, the Laplace exponent of a subordinator is the sum of the individual functions, because the Laplace transforms multiply. The transform at $\lambda =0$ gives $\Phi (0)=0$.