7
$\begingroup$

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

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

or

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

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.

$\endgroup$

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.