Let $\chi(\xi) := \mathbb{E}e^{i \xi X}$, $\xi \in \mathbb{R}^d$, be the characteristic function of a random variable $X$. It is widely known that $\chi$ admits a holomorphic extension to the strip $\{z \in \mathbb{C}; |\text{Im} \, z| < m\}$ whenever

$$\mathbb{E}e^{m |X|}<\infty.$$

Are there other, more general, statements on the existence of (holomorphic) extensions of characteristic functions to subsets of the complex plane which do not contain such a strip, e.g. to sets of the form

$$\mathbb{C} \backslash (i \mathbb{R} \backslash \{0\}):= \mathbb{C} \backslash \{z \in \mathbb{C}; \text{Re} \, z =0, \text{Im} \, z \neq 0\}$$


$$\{z \in \mathbb{C}; \left| \frac{\text{Re} z}{\text{Im} z} \right| \leq c\}$$

More specifically, I'm interested in characteristic functions of Lévy processes; that is characteristic functions of the form $e^{-\psi}$ where $\psi$ is a continuous negative definite function. (In this case, the question is under which conditions continuous negative definite functions have such holomorphic extensions.)

Motivation: The characteristic function of a symmetric $\alpha$-stable random variable equals $e^{- c|\xi|^{\alpha}}$. It is obvious that this function does not admit a holomorphic extension to a strip containing $0$, but it has a holomorphic extension to the (interior) of $\mathbb{C}\backslash (i \mathbb{R} \backslash \{0\}$. There are quite a few examples where similar things happen (because of missing regularity we cannot expect a holomorphic extension to a neighborhood of $0$, but away from $0$ it's fine).

There is a result (due to Cuppens and Lukacs) stating that for any open subset $\Omega \subseteq \mathbb{C}$ which is symmetric with respect to to the imaginary axis and whose closure contains $ia$ and $-ib$ for some $a,b > 0$, there exists a characteristic function $\chi$ which cannot extended (analytically) beyond $\Omega$. That means that we cannot expect, in general, that "nice" holomorphic extensions exist. However, I'm interested in sufficient conditions for the existence of such extensions.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.