# Meaning of “intensity measure” for Poisson processes

A compound Poisson process with jump intensity measure (Lévy measure) $\nu$ is a Lévy process $X_t$ on $\mathbb{R}^d$ such that $X_{t+s}-X_s$ has law $\pi_{t\nu}$, where $\pi_{t\nu}$ is the measure with characteristic function $p \mapsto\exp(-t\int (1-e^{ip\cdot y} \, \nu(dy))$. (Definition taken from the lecture notes I'm currently looking at).

That definition suggests that the intensity measure $\nu$ indicates the size of the jumps when they occur. However, in other contexts, it's seemingly also used to denote the measure governing how often jumps occur, which in the above example seems to be just the Lebesgue measure.

How can one separate the terminology here? "Intensity measure" seems to be quite an ambiguous concept.

• At first I thought you mean the characteristic function is $t\mapsto \exp(-t\int (1-e^{ip\cdot y} \, \nu(dy)),$ but on closer inspection it looks as if you must have meant $p\mapsto \exp(-t\int (1-e^{ip\cdot y} \, \nu(dy)).$ I'd be explicit about that in the question. $\qquad$ – Michael Hardy Jul 22 '18 at 17:09
• Your definition looks as if it may make sense on $\mathbb R^1$ but not for $\mathbb R^d$ for $d>1.$ For $t,s\in \mathbb R^1$ and t>0$,$ the intensity measure of the interval $(s,t+s)$ is the expected number of jumps in that interval. In higher dimensions, I'd expect to specify the measure of a set other than an interval, so the sum of the jump sizes would not be expressed as $X_{t+s}-X_s. \qquad$ – Michael Hardy Jul 22 '18 at 17:15
• So far it looks to me as if the intensity measure $\nu$ gives you the expected number of jumps within the interval $(s,t+s)$ (where $t>0$) and the measure $\pi_{t\nu}$ gives you the distribution of the size of a jump. $\qquad$ – Michael Hardy Jul 22 '18 at 17:17
• $\ldots\,$and the fact that not all jumps have size $1$ is the reason why the word "compound" is used. $\qquad$ – Michael Hardy Jul 22 '18 at 17:23
• The characteristic function of $\pi_{t\nu}$ would be the same in higher dimensions as in one dimension, but that thing about $X_{t+s} - X_s$ would be expressed differently. $\qquad$ – Michael Hardy Jul 22 '18 at 17:42