= Laplace exponent of a subordinator
{c}
{title2=$\Phi(\lambda)$}
For a finite-valued <subordinator> $T$ starting at zero, the <Laplace exponent of a subordinator> is the <function> $\Phi$ determined by
$$
\mathbb E e^{-\lambda T_t}=e^{-t\Phi(\lambda)},\qquad t,\lambda\geq0.
$$
Its <Lévy–Khintchine formula> takes the form
$$
\Phi(\lambda)=d\lambda+\int_{(0,\infty)}(1-e^{-\lambda x})\,\nu(dx),\qquad
d\geq0,\quad\int_{(0,\infty)}(1\wedge x)\,\nu(dx)<\infty,
$$
where $d$ is the deterministic <drift coefficient> and $\nu$ is the <Lévy measure>. This representation separates steady motion of the clock from its positive jumps.<Topics in Probability: Lévy Processes by Davar Khoshnevisan>[Lecture 8, page 44, equation 2 and Definition 4]{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 $\Phi(0)=0$.
Back to article page
