Skip to main content
3 votes
Accepted

Can we show that the characteristic function of an infinitely divisible probability measure has no zeros

$\newcommand\vpi\varphi\newcommand\R{\mathbb R}$ The approach involving (4) will not work, because Lévy's continuity theorem will guarantee that the pointwise limit of characteristic functions (c.f.'...
Iosif Pinelis's user avatar
2 votes
Accepted

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

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_{...
JGWang's user avatar
  • 724
2 votes
Accepted

Existence of unique convolution semigroups of probability measures on more general spaces then $\mathbb R^d$

A quick Google search on "infinitely divisible" and "Banach space" leads to Linde's Probability in Banach Spaces: Stable and Infinitely Divisible Distributions (John Wiley & ...
Mateusz Kwaśnicki's user avatar
1 vote
Accepted

If $\mu$ is an infinitely divisible probability measure on $[0,\infty)$, then the Lévy measure of $\mu$ is the vague limit of $n\mu^{*1/n}$

There are at least three ways to show that $n \mu^{*1/n}$ converges to $\nu$ vaguely in $\mathbb{R} \setminus \{0\}$. Let $X_t$ be the Lévy process such that $X_1$ has distribution $\mu$, and let $f$ ...
Mateusz Kwaśnicki's user avatar
1 vote
Accepted

How can we show this estimate for the convolution of two probability measures?

$\newcommand\ep\varepsilon\newcommand\de\delta$Let $\mu:=\mu_n$, $\nu:=\nu_n$, $B:=B_n$, $$C_k:=\{x\in E:\mu(K_k-x)>1-\de_k\},$$ so that $$B=\bigcap_k C_k.$$ We have $$1-\ep_k<(\mu*\nu)(K_k)=\...
Iosif Pinelis's user avatar
1 vote

Characterization of the generator of a Lévy process using martingale problems

I think you can do it by computing the expectation of the Laplace transform of both the function f(X_t) and the integral (the one with the martingale vanishes) and then identity the Laplace transform ...
AntoineL's user avatar
  • 131
1 vote

Is this statement of the Lévy–Khintchine formula ill-posed?

Apologies if this is an inappropriate place to post a comment (I am a newbie). But I wanted to mention that an easier to read treatment of infinite divisibility is Chapter 3 of the classic book by ...
AndreyF's user avatar
  • 171

Only top scored, non community-wiki answers of a minimum length are eligible