1

I will take the domain to be $\mathbb{R}$ for simplicity, in which case a Levy measure $\Pi$ is defined as a $\sigma$-finite measure such that $$\int_{\mathbb{R}\setminus \{0\}} 1 \wedge |x|^2 \Pi(dx) < \infty.$$ Is there any reason why this condition is a part of the definition? Why is zero excluded from the domain?

Levy measures naturally arise in the Levy-Khintchine representation of Levy processes, and one other mathSE question said that this definition is so that the integral appearing in that representation is integrable but didn't give any details. I also have read this question but the answer/comments do not explain why this condition is needed.

CBBAM
  • 7,149
  • The integrand in the Levy-Khintchine representation behaves like $x^{2}$ near $0$ and it is bounded outside a neighborhood of $0$. – Kavi Rama Murthy Nov 28 '24 at 05:21
  • @geetha290krm Thanks, so I guess this condition is just a technical one so that the integral converges? – CBBAM Nov 28 '24 at 05:23

2 Answers2

2

The Levy -Kintchine formula is a great conquest of probability of the thirties. For understanding it, better is to assume that $\Pi$ is concentrated on $ (0,\infty)$ (thus using rather Laplace transform, a bit easier to manipulate) and to consider 3 cases:

A) Type zero: $\Pi$ is a bounded measure of mass $\lambda$: you probably know that then we get a so called compound Poisson, namely the law of $$Y=X_1+\cdots+X_N$$ where $X_i\sim \Pi/\lambda$ and $N$ is Poisson of mean $\lambda$ with $N, X_1,\ldots, X_n,\ldots $independent. $$E(e^{-sY})=\exp(\int_0^{\infty}(e^{-sx}-1)\Pi(dx).$$

B) Type one: $\Pi$ is unbounded but $$\int_0^{\infty}\min (1,x)\Pi(dx)<\infty.$$ Then $E(e^{-sY})=\exp(\int_0^{\infty}(e^{-sx}-1)\Pi(dx).$ still holds. To understand $Y$, consider the bounded $\Pi_n,$ which is the restriction of $\Pi$ to $(1/n,\infty)$ and the corresponding $Y_n.$ Type 1 implies that $Y_n\rightarrow Y$ in law - just compute. Note that $Y\geq 0.$

C) Type two, the most difficult to understand. We have $\int_0^{\infty}\min (1,x)\Pi(dx)=\infty$ but $$\int_0^{\infty}\min (1,x^2)\Pi(dx)<\infty.$$ Unfortunately $\int_0^{\infty}(e^{-sx}-1)\Pi(dx)$ diverges and we have to add a compensatory term like $\tau(x)=\sin x$ or $x/(1+x^2)$ or else (depending of the authors) such that $$\int_0^{\infty}(e^{-sx}-1-s\tau(x))\Pi(dx)$$ converges. With this choice $E(e^{-sY})$ does exist but behold! $Y$ is not positive. To see this, use again our good $Y_n$, which is not positive anymore because of $\tau$ (again, just compute). We have again $Y_n\rightarrow Y$ in law.

Letac Gérard
  • 3,820
  • Thank you! Perhaps I have misunderstood, but your answer describes the consequences of these three cases. But why would we consider the type two case in the first place? Why is it a natural condition to impose? – CBBAM Nov 29 '24 at 21:36
  • If you impose only $\int_0^{\infty}\min (1,x^2)\Pi(dx)<\infty$ without $\int_0^{\infty}\min (1,x)\Pi(dx)=\infty$ then you are considering the three cases altogether in one shot. But as you surely notice, $\tau$ is useless for types 0 and 1. If you keep it in types 0 and 1, you carry on a useless translation factor in the distribution of $Y.$ – Letac Gérard Nov 30 '24 at 07:17
  • The condition $\int_0^{\infty}\min(1,x^2)\Pi(dx)<\infty$ is not only a sufficient condition for having $Y$ infinitely divisible but also a necessary condition: this is the difficult part of the Levy Kintchine theorem. This is why it is 'natural.' – Letac Gérard Nov 30 '24 at 07:24
  • Thank you for your help! – CBBAM Dec 02 '24 at 06:56
1

Here's one (perhaps intuitive) way to think about the necessity of the condition $\int_{\Bbb R\setminus\{0\}}(1\wedge |x|^2)\,\Pi(dx)<\infty$. Suppose $X=(X_t)_{t\ge 0}$ is a (one-dimensional) Lévy process with Lévy measure $\Pi$. The jumps $\{(t,X_t-X_{t-}): t>0\}$ form a Poisson point process in $(0,\infty)\times(\Bbb R\setminus\{0\})$ with intensity measure $dt\otimes\Pi(dx)$.

The process $X$ is necessarily a semimartingale; as such the sum $\sum_{0<s\le t}(X_s-X_{s-})^2$ is a.s. finite for each $t>0$. From the well-known formula for the Laplace transform of a Poisson point process, this is equivalent to the condition that $$ \int_{[0,\infty)\times(\Bbb R\setminus \{0\})}(1-\exp(-1_{[0,t]}(s)|x|^2)\,ds\,\Pi(dx)<\infty. $$ This is in turn equivalent to $$ \int_{\Bbb R\setminus \{0\}}(1-e^{-|x|^2})\,\Pi(dx)<\infty, $$ which is equivalent to $$ \int_{\Bbb R\setminus \{0\}}(1\wedge |x|^2)\,\Pi(dx)<\infty. $$

John Dawkins
  • 29,845
  • 1
  • 23
  • 39