6
votes
Accepted
Independent stationary increment process but with finite propagation speed
If by "an independent stationary increment process" you mean "a stochastic process with independent, stationary increments", that is a Lévy process, then the answer is: of course, ...
- 128k
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 ...
- 17.2k
4
votes
Accepted
Characteristic function and moments
The answers to your questions are no and no.
Question 1: Let $X$ have the standard double exponential distribution, so that the pdf $p_X$ of $X$ is given by $p_X(x)=\frac12\,e^{-|x|}$ for real $x$. ...
- 128k
4
votes
Accepted
A question about the proof of the Levy-Khintchine representation Theorem
$\newcommand\R{\mathbb R}\newcommand\ip[1]{\langle #1 \rangle}$In these notes, two related definitions of truncation functions are given. In Definition 5.6, a truncation function is defined as a ...
- 128k
4
votes
Accepted
Stationary Distribution of Langevin Dynamics driven by Lévy Process
There are several generalizations of the Brownian case results to general L'evy processes. For instance, under a classical log-concave assumption on $U$ we have a convergence to equilibrium in ...
- 2,703
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.'...
- 128k
3
votes
Accepted
Simulation of Lévy walk
Indeed, the Lévy flight is a random walk where the step increments $\nu$ are i.i.d. with a Lévy distribution, $p(\nu)\rightarrow 1/\nu^{1+\alpha}$ for $\nu\rightarrow\infty$, with exponent $0<\...
- 188k
3
votes
Accepted
translation invariance of expectation value of hit counting variable for Lévy process
No, this is not true is general. For instance, if $(X_t)$ is a Poisson process of intensity (say) $\lambda=1$, then
$$M_u(a,s)=\lfloor X_{u+s}/a\rfloor - \lceil X_u/a\rceil + 2.$$
So,
$$EM_{1/2}(2,2)=...
- 128k
2
votes
Accepted
Existence of unique convolution semigroups of probability measures on more general spaces then $\mathbb R^d$
A quick Google search on "infinitely divisible" and "Banach space" leads to Linde's Probability in Banach Spaces: Stable and Infinitely Divisible Distributions (John Wiley & ...
- 17.2k
2
votes
Accepted
Existence of the differential entropy for infinitely divisible laws
For real $t>0$, let
\begin{equation}
p_t:=e^{-t}e^{*tf}*g_t:=e^{-t}\sum_{n=0}^\infty\frac{t^n f^{*n}}{n!}*g_t, \tag{0}
\end{equation}
where $f$ is the (bounded by $c:=1/e$) pdf given by
\begin{...
- 128k
2
votes
Accepted
If $(\exp(\mu_n))_{n\in\mathbb N}$ is weakly convergent, is the normalized sequence convergent as well?
$\newcommand\R{\mathbb R}$The answer to the second question (and hence to the first one) is no.
Indeed, let $E:=\R$.
For all odd natural $n$, let $\mu_n:=\mu$, where $\mu$ is the uniform distribution ...
- 128k
2
votes
Accepted
Blumenthal 0-1 law
$\newcommand\F{\mathcal F}\newcommand\N{\mathbb N}$First, some preliminary remarks:
Your link to Bertoin's book is not very good. Here is a link with a better reference to the book.
It is Blumenthal,...
- 128k
2
votes
Accepted
A Levy process is a.s. continuous
$\newcommand{\R}{\mathbb R}\newcommand{\Q}{\mathbb Q}$This proof is insufficient (where did you take it from?).
Indeed, the logic at the end of the proof boils to down to
Proposition 1: Suppose that ...
- 128k
2
votes
Accepted
Second moment of stochastic integral wrt Levy Processes
For any random variable $Y$ with a finite second moment, one has
$$
\left.\frac{\partial^2}{\partial \lambda^2}\mathbb{E} e^{i\lambda Y}\right|_{\lambda=0} =\left. \mathbb{E}\frac{\partial^2}{\partial ...
- 8,992
2
votes
Accepted
Lévy measure and jump behaviour of the corresponding Lévy process
This follows immediately from (say) the following statements in Schilling - An Introduction to Lévy and Feller Processes:
parts b) and c) of Lemma 9.4, stating that $N(\cdot,A)$ is a Poisson process ...
- 128k
2
votes
Accepted
Approximation of a random sum of random variables (infinitely divisible distribution) by a triangular array
$\newcommand\la\lambda$Let
$$S_n:=\sum_{j=1}^n\xi_j X_{j,n},$$
where the $\xi_j$'s are iid random variables (r.v.'s) and, for each $n$, the $X_{j,n}$'s are iid r.v.'s independent of $\xi_j$'s and such ...
- 128k
2
votes
Accepted
How to show that $\int x \,d\nu = 0$ using a pseudo-weak convergence of measures?
$\newcommand\de\delta$A counterexample is given by $p=1$, $\nu(dx):=|x|^{-5/2}\,1(0<|x|<1)\,dx$, and $\nu_n(dx):=|x|^{-5/2}\,1(1/n<|x|<1)\,dx$.
The OP has added certain conditions. The ...
- 128k
1
vote
A Lévy process is a semimartingale proof
The statement that cadlag Levy is semimartingale is proved here in theorem 4. Namely they prove that a cadlag Lévy process $X$ decomposes as $X_t=bt+W+Y$ where $Y$ is a semimartingale and $W$ is a ...
- 5,474
1
vote
Accepted
Exceedance distribution of Levy process
First. if you were only interested in the one-sided problem, then you look in at the jump times and you have the same problem for a random walk which is the difference of a uniform and an exponential. ...
- 1,172
1
vote
Accepted
Compound poisson processes (Construction)
$\newcommand{\N}{\mathbb N}\newcommand\ep\varepsilon$Apparently, the book "Levy processes and infinitely divisible distributions" you mentioned is the one by Sato.
According to the ...
- 128k
1
vote
Accepted
Expectation of killed subordinator at first-passage time
Because $f(0)=g(0)=0$, the passage over $x$ is being counted in $\Bbb E[f(X_{\tau_x^+}-x)g(x-X_{\tau_x^+-})]$ iff $X$ jumps at the crossing time. And
$$
\eqalign{
1_{\{\Delta X_{\tau_x^+}>0\}}f(X_{\...
- 1,989
1
vote
Accepted
Poisson point process in polar coordinates
This kind of thing is studied at length in Random Measures, Point Processes, and Stochastic Geometry by Baccelli, Blaszczyszyn, and Karray. The book is made freely available in pdf form by the authors,...
- 282
1
vote
Accepted
How can we show this estimate for the convolution of two probability measures?
$\newcommand\ep\varepsilon\newcommand\de\delta$Let $\mu:=\mu_n$, $\nu:=\nu_n$, $B:=B_n$,
$$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)=\...
- 128k
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 ...
- 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 ...
- 171
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 ...
- 17.2k
1
vote
Existence of a distinguished continuous version of the logarithm of a continuous function
The result (1) follows from the lifting property of covering spaces: Consider $f=\varphi/|\varphi|: E\to S^1$ and the standard covering map $p:\mathbb R\to S^1$, $t\mapsto e^{it}$. Since Banach space ...
- 16.4k
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 $f$ ...
- 17.2k
1
vote
Is the distribution of a Banach space valued Lévy process uniquely determined by its characteristic function?
If the Banach space $E$ is separable, then, by Lemma 2.1, there does exist a unique continuous $f\colon E'\to\mathbb C$ with $f(0)=0$ such that your condition (1) holds. Moreover, then the function $f$...
- 128k
1
vote
If $L_t=\sum_{i=1}^{N_t}Y_i$ is a compound Poisson process, then $\left|\left\{s\in[0,t]:\Delta L_s\in B\right\}\right|=\sum_{i=1}^{N_t}1_B(Y_i)$
$\newcommand{\De}{\Delta}$
This is not really a probability problem, since the equality
\begin{equation}\label{1}\tag{1}
l_t:=|\{s\in[0,t]\colon\De L_s\in B\}|=\sum_{i=1}^{N_t}1_B(Y_i)=:r_t
\end{...
- 128k
Only top scored, non community-wiki answers of a minimum length are eligible