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I've been reading the literature and I am not sure whether this is a necessary condition or not...

Suppose the probability density function $f$ is infinitely divisible and generates a Levy process $X_t$.

Must the density function $f$ be continuous? Thanks

A reference would be great also thanks

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1 Answer 1

There are conditions such as the ones provided by Orey (On Continuity Properties of Infinitely Divisible Distribution Functions) that guarantee the existence of a smooth density for an infinitely divisible distribution. In Sato, chapter 5 pg 190 you can find: Let $X$ be a real-valued Levy process with characteristic triplet $(\gamma, \sigma, \nu)$. If there exists $\beta \in (0,2)$ satisfying \begin{equation} \liminf_{r\to 0} r^{-\beta}\int_{-r}^r |x|^2\, \nu(\mathrm{d} x) > 0 \end{equation}

then $X_t$ has a $\mathcal{C}^\infty$ density $p_t(\cdot)$ for every $t$. The message is that the existence of the density entirely depends on the behaviour of the density in a neighbourhood of the origin.

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