The Levy measure of the compound Poisson distribution

Question

The compound Poisson distribution is defined as(see Levy processes and infinitely divisible distributions page: 18):

Let $c>0$ and $\sigma$ be a measure on $\mathbb{R}$ with $\sigma(\{0\})=0$, a compound distribution $\mu$ on $\mathbb{R}$ is called a compound Poisson if its characteristic function is

$$ \Phi_{\mu}(t) = e^{c(\Phi_\sigma - 1)} $$

My question is: Why we need the condition $\sigma(\{0\})=0$?

Similarly, in the book Probability: a comprehensive course (page: 333) stated the following theorem:

A probability measure $\mu$ on $\mathbb{R}$ is infinitely divisible if and only if there exists a sequence of finite measures $(\nu_n)_{n\in\mathbb{N}}$ on $\mathbb{R} - \{0\}$ such that $CPoi_{\nu_n} \rightarrow \mu(n\rightarrow\infty)$, where $CPoi_{\nu_n}$ denote the compound distribution with intensity $\nu_n$.

My question is: Why $\nu_n$ is defined on $\mathbb{R}-\{0\}$ other than $\mathbb{R}$?

Answer 1

This requirement is not necessary, but convenient.

The measure $\sigma$ is the distribution of jumps. If $\sigma(\{0\})>0$, then the jump is $0$ with positive probability, meaning that there is in fact no jump.

With the restriction $\sigma(\{0\})=0$, the parameters $\sigma$ and $c$ are uniquely determined by the distribution of process. But if we allow jumps to be equal to zero, we can increase intensity of jumps, at the same time allowing some of them to be zero, which would lead to the same process.

Another point is that when analyzing the behavior of the process you need to know when it changes value. With $\sigma(\{0\})>0$ it would be possible that the underlying counting measure is positive, however, there is no jump, since it is equal to zero.

I hope this is enough to convince you that $\sigma(\{0\}) = 0$ is a convenient assumption.

EDIT I've just noticed Jean Duchon's answer. He is right, the last paragraph was some mess. I can't remember what I was going to say, so I removed it.

Answer 2

Although I do agree with most of zhoraster's answer, I wish to make a few points, as complements at least.

First, the formula $\Phi_\mu=e^{c(\Phi_\sigma-1)}$ defines the (compound Poisson) probability distribution $\mu$ only if $\sigma$ itself is a probability ($\Phi_\sigma(0)=1$), not just a finite measure.

Second, this same formula makes sense with $\sigma=\delta_0$ (then $\mu=\delta_0$). Whether this is considered compound Poisson or not is somewhat arbitrary. If one defines $\mu$ as the distribution of $\sum_1^N X_j$ with i.i.d. $X_j$ of law $\sigma$ and $N$ Poisson (with $P[N=n]=e^{-c}c^n/n!$, and independent from the $X_j$), then obviously $\delta_0$ is a valid, albeit degenerate, example of a compound Poisson distribution. The drawback of this unrestricted definition is that the one-to-one correspondence between $\mu$ and $(c,\sigma)$ is lost. Whence the condition $\sigma\{0\}=0$, that restores this one-to-one correspondence, at the price of excluding some degenerate cases from the definition.

Third, the notion of a Lévy measure is best applied to Lévy processes rather than to compound Poisson distributions. It defines the jump rate, $\nu$, which, together with a drift rate and a diffusion rate, characterize the process (in law). The condition $\nu\{0\}=0$ then reflects the fact that a jump is nonzero by definition (as already said). Naturally, compound Poisson processes are a class of Lévy processes, those with a.s. finitely many jumps per time unit, occurring at the instants of a Poisson process of intensity $c\ dt$, independently, with distribution $\sigma$ ($\nu=c\sigma$ ; and of course no drift and no diffusion).

Fourth, it is true that every infinitely divisible distribution, including the degenerate $\delta_0$ (which is compound Poisson in the unrestricted sense) or $\delta_x$ (which is not compound Poisson), is a limit of compound Poisson distributions (in the unrestricted sense, or in the restricted sense as well, i.e. with or without the condition $\nu\{0\}=0$). The quoted statement is correct, then (this is the only point in zhoraster's answer that seems unclear to me).