# Questions about Levy measure in the canonical representation of infinitely divisible distributions

Let $X$ be a random variable with infinitely divisible and symmetric distribution $F$ distributed on $\mathbb{R}$.

It is well known that the characteristic function of $X$ has a canonical representation of the form: \begin{align} \phi(t)=e^{ -\frac{\sigma^2}{2}t^2-\int_{-\infty}^\infty (1-cos(tx) ) dV(x)} \end{align} where $V$ is a non-negative measure such that $V(\{0\})=0$ and $\int_{-\infty}^\infty \min(1,x^2) dV(x)<\infty$.

The measure $V$ is called Levy measure and here we are interested in its properties.

My questions are:

1. Under what condition on $F$ is $V$ an absolutely continuous measure? For example, is absolute continuity of $F$ enough?
2. Can we say anything about $V$ based on the tail behavior of $F$?
3. Can we say anything about $V$ based on the tail behavior of $\phi(t)$?
4. I know it is generally difficult to determine $V$ but what can generally be said about $V$ from basic properties of $F$ and $\phi(t)$?

Also, any good reference would be appreciated.

I would discuss your questions partially under the following condition: $$\int_{-\infty}^\infty x^2\,dF(x)<\infty, \tag{1}$$ or equivalently $\varphi^{\prime\prime}(0)$ exists and finite or $\int_{-\infty}^\infty y^2\,dV(y)<\infty$. Let $K(x)\triangleq\int_{-\infty}^x y^2\,dV(y)<\infty$, it is easy to obtain the following expressions: \begin{gather} -(\log\varphi(t))^{\prime\prime}=\sigma^2+\int_{-\infty}^{\infty}\cos(tx)dK(x),\\ -\lim_{T\to\infty}\frac1{2T}\int_{-T}^T (\log\varphi(t))^{\prime\prime} dt =\sigma^2,\\ \psi(t)\triangleq -(\log\varphi(t))^{\prime\prime}-\sigma^2=\int_{-\infty}^\infty \cos(tx)\,dK(x).\tag{2} \end{gather} Now from the inversion formula of characteristic functions we have following conclusion: Under (1), \begin{gather} \int_{-\infty}^\infty|\psi(t)|\,dt<\infty \quad\Rightarrow\quad dK(x)\ll d\lambda\quad \Rightarrow \quad dV(x)\ll d\lambda\;(\text{$V$ has density})\\ \begin{aligned}\int x^{2n}dF(x)<\infty \quad &\Rightarrow \quad \varphi^{(2n)}(0)\; \text{exists and finite}\\ &\Rightarrow \quad \psi^{(2n)}(0)\; \text{exists and finite}\\ &\Rightarrow \quad \int x^{2n}\,dV(x)<\infty, \qquad n\in\mathbb{N} \end{aligned} \end{gather}
• Hi. I finally got this book from the library. Unfortunately, I couldn't find the exact theorem that would say under what condition the measure $V$ is absolutely continuous. Can you point this out to me? – Boby May 30 '17 at 20:25
• @Boby， Sorry， I confused your questions. You concerned is the sufficient conditions of $V$ is absolute continuous, and the book given is the the conditions of $F$ is absolute continuous. I am sorry to bother you. – JGWang May 31 '17 at 3:49
• In your proof $\psi(t)$ is a characteristic function of the distribution $K(x)$, correct? Or is it only a cosine transform? Of course, cosine transform and characteristic function are equivalent if the distribution $K(x)$ is symmetric. Do we know here that $K(x)$ is symmetric? Does symmetry of $F(x)$ imply symmetry of $K(x)$? Thanks. – Boby May 31 '17 at 15:46
• @Boby As a measure, $dK(x)$ is symmetry. $K(x)$ and $K(x)+c$ define same measure. Distribution $F$ is symmetry($F(-x-)=1-F(x)$) $\Leftrightarrow$ $\varphi(-t)=\varphi(t)$ $\Leftrightarrow$ $\psi(-t)=\psi(t)$ $\Leftrightarrow$ $dK(x)$ is symmetry. – JGWang Jun 1 '17 at 1:21
• After equation (1) when you say: "...or equivalently $\phi^{"}(0)$ exists and is finite or $\int_{-\infty}^\infty y^2 dV(y)$ ." This sentence seems incomplete. What did you mean to say about $\int_{-\infty}^\infty y^2 dV(y)$?? Did you mean to say that $\int_{-\infty}^\infty y^2 dV(y)$?\$ is finite? – Boby Jun 2 '17 at 15:59