Search Results
Search type | Search syntax |
---|---|
Tags | [tag] |
Exact | "words here" |
Author |
user:1234 user:me (yours) |
Score |
score:3 (3+) score:0 (none) |
Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
Views | views:250 |
Code | code:"if (foo != bar)" |
Sections |
title:apples body:"apples oranges" |
URL | url:"*.example.com" |
Saves | in:saves |
Status |
closed:yes duplicate:no migrated:no wiki:no |
Types |
is:question is:answer |
Exclude |
-[tag] -apples |
For more details on advanced search visit our help page |
Theory and applications of Lévy processes (stochastic processes with stationary and independent increments): e.g. path properties, stochastic differential equations driven by jump-type processes, fluctuation theory of Lévy processes, queuing theory.
2
votes
Accepted
Existence of unique convolution semigroups of probability measures on more general spaces th...
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 & Sons, 1986). There we find …
1
vote
Accepted
If $\mu$ is an infinitely divisible probability measure on $[0,\infty)$, then the Lévy measu...
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 $f$ …
1
vote
Quantiles of a Levy process
First of all, $X$ being non-lattice is not enough for $X_t$ being absolutely continuous. A simple counter-example is $$X_t = \sum_{n=1}^\infty \frac{N_t^{(n)}}{n!} \, ,$$ where $N_t^{(n)}$ are indepen …
1
vote
Probability of exiting on the boundary for a monotone Lévy-type process
(i,ii) Not quite. A trivial counterexample is $\ell(x, y) \equiv 0$. A simple non-trivial counterexample would be something like $$\ell(x,y) = \begin{cases} 1 / (x^{1/2} y^{3/2}) & \text{if $0 < x \le …
5
votes
Besov regularity of càdlàg functions?
(This is not a complete answer, someone more experience in Besov spaces is welcome to improve it).
I do not think there is much relation between these two concepts.
Obviously, $D(\mathbb{R})$ only c …
1
vote
Accepted
Identity for stable Lévy subordinator
It is of course possible to show that the left-hand side is continuous (e.g. using bounds for the derivative of $p_\beta(s, x)$), but there is a more direct way. Observe that $$p_\beta(s, x) = s^{-1/\ …
1
vote
escape points of Levy processes
This does not seem to be true: take $X_t$ to be a symmetric stable process in $\mathbf{R}$ with index $\alpha > 1$, and $D = (-1, 1) \setminus \{0\}$. Then $X_t$ hits $0$ with positive probability.
T …