Timeline for If $\mu$ is an infinitely divisible probability measure on $[0,\infty)$, then the Lévy measure of $\mu$ is the vague limit of $n\mu^{*1/n}$
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27 events
| when toggle format | what | by | license | comment | |
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| S Nov 3, 2020 at 20:24 | history | bounty ended | 0xbadf00d | ||
| S Nov 3, 2020 at 20:24 | history | notice removed | 0xbadf00d | ||
| Nov 3, 2020 at 20:24 | vote | accept | 0xbadf00d | ||
| Nov 1, 2020 at 0:02 | answer | added | Mateusz Kwaśnicki | timeline score: 1 | |
| Oct 31, 2020 at 11:14 | comment | added | 0xbadf00d | @MateuszKwaśnicki Thank you for your help so far. I've almost figured out how we can show the desired claim. Please take a look at my (partial) answer: mathoverflow.net/a/375322/91890. Can you help out? | |
| Oct 31, 2020 at 11:13 | answer | added | 0xbadf00d | timeline score: 0 | |
| Oct 30, 2020 at 11:34 | comment | added | Mateusz Kwaśnicki | (a) You are asking very basic questions that you can easily answer yourself, for example, by searching the web. See this question on Math.SE, for example. (b) Change $e^{-t}$ to $e^{-\alpha t}$ in my comment. | |
| Oct 30, 2020 at 11:23 | comment | added | 0xbadf00d | @MateuszKwaśnicki (a) You didn't respond to my equivalence of weak/vague convergence. am I missing something? (b) I will take a look at that section of Sato's book. Regarding what you wrote about $\alpha$: $\alpha$ is not occurring in what you wrote after "then"; so, I'm afraid, I don't understand what you mean. | |
| Oct 30, 2020 at 10:57 | comment | added | Mateusz Kwaśnicki | Regarding the support of $\mu$: I am sure this is answered in Section 24 of Sato's book, titled "Supports". (To be more constructive: if $\alpha$ is the left edge of the support of $\mu$, then $\mu_t(A) = \mu(A - t)$ is infinitely divisible and its support is still in $[0, \infty)$. And, of course, the Laplace transform of $\mu$ is $e^{-t}$ times the Laplace transform of $\mu_t$. Is this what you were asking about?) | |
| Oct 30, 2020 at 10:53 | comment | added | 0xbadf00d | @MateuszKwaśnicki No, I'm actually aware of the link. I'll have a second look at that. However, regarding the actual question, what I was really missing is that - since we are dealing with probability measures - the notion of vague and weak convergence should coincide, right? And do you know how we can prove or do you have a reference for the formula for $\alpha$? | |
| Oct 30, 2020 at 8:11 | comment | added | Mateusz Kwaśnicki | It looks like what you are missing is the link between weak convergence of probability distributions and pointwise convergence of the corresponding characteristic functions. This immediately proves (5) and uniqueness of $\mu^{*t}$ from what you wrote in your comment above. | |
| Oct 30, 2020 at 6:00 | comment | added | 0xbadf00d | @MateuszKwaśnicki BTW, I'm still searching for a reference for $(5)$ and for $\alpha=\sup\left\{x\ge0:\mu([0,x))=0\right\}$. Sato doesn't contain these claims and I didn't find them in other books. | |
| Oct 30, 2020 at 5:56 | comment | added | 0xbadf00d | @MateuszKwaśnicki Sato shows that for each infinitely divisible characteristic function $\varphi$ of a probability measure $\mu$, there is a unique continuous $\psi$ with$\psi(0)=0$ and $\varphi=e^\psi$. He then shows that for all $k$, the unique continuous $g$ with $g(0)=1$ and $\varphi=g^k$ is given by $g=e^{\frac\psi k}$. Now he defines $\varphi^t:=e^{t\psi}$ for $t\ge0$. So, for rational $t$, there is a unique probability measure $\nu$ with $\mu=v^{\ast1/t}$ and $\varphi_\nu=\varphi^t$. Now we define $\mu^{\ast t}:=\nu$. | |
| Oct 30, 2020 at 0:57 | comment | added | Mateusz Kwaśnicki | I am sorry, I am lost. I do not even know how you define rational convolutive powers of $\mu$, so I am afraid I cannot help. | |
| S Oct 29, 2020 at 20:29 | history | bounty started | 0xbadf00d | ||
| S Oct 29, 2020 at 20:29 | history | notice added | 0xbadf00d | Canonical answer required | |
| Oct 29, 2020 at 14:29 | comment | added | 0xbadf00d | @MateuszKwaśnicki Does $\mathcal L(X_t)=\mu^{\ast t}$ now simply follow from $\varphi_\mu^{t_n}=\varphi_{\mu^{\ast t_n}}=\varphi_{X_{t_n}}\xrightarrow{n\to\infty}\varphi_{X_t}$ and the former definition of $\mu^\ast$ in terms of characteristic functions? | |
| Oct 29, 2020 at 14:29 | comment | added | 0xbadf00d | @MateuszKwaśnicki How do we see that $\mu^{\ast t}$ is equivalently uniquely defined as the weak limit? For example, suppose $X$ is Lévy in law and $\mu:=\mathcal L(X_1)$. It then clearly holds $\mathcal L(X_t)=\mu^{\ast t}$ for all $t\in\mathbb Q\cap[0,\infty)$. Now taking $t>0$ and $(t_n)_{n\in\mathbb N}\subseteq\mathbb Q\cap[0,\infty)$ with $t_n\xrightarrow{n\to\infty}t$, we clearly have $X_{t_n}\xrightarrow{n\to\infty}X_t$ in distribution (assuming $X$ is at least continuous in probability). | |
| Oct 29, 2020 at 11:07 | comment | added | Mateusz Kwaśnicki | My intuition breaks in infinite dimensions, but I think the answer is "yes". Dave Applebaum likely discusses this (or at least gives some references) in his review article Lévy processes and stochastic integrals in Banach spaces, available at the PMS website. | |
| Oct 29, 2020 at 10:12 | comment | added | 0xbadf00d | @MateuszKwaśnicki Thank you for your comment. Do these results generalize to Banach spaces? | |
| Oct 29, 2020 at 10:11 | comment | added | Mateusz Kwaśnicki | I do not have access to Sato at the moment, so I am not sure how he defines $\mu^{*t}$, but in any case it is unique. Either uniquely determined by the characteristic function, or uniquely defined as the weak* limit of approximations by rational powers. | |
| Oct 29, 2020 at 9:34 | comment | added | 0xbadf00d | @MateuszKwaśnicki I've taken a look at Sato. Is the measure $\mu^\ast t$ uniquely determined by $\mu$ (assuming $\mu$ is infinitely divisible) for irrational $t$? Or to state this question differently: Is $\mu^{\ast t}$ the unique probability measure on $\mathbb R^d$ such that $$\varphi_{\mu^{\ast t}}=\varphi_\mu^t$$ for all $t\ge0$? The uniqueness is clear to me for rational $t$. | |
| Oct 27, 2020 at 20:39 | comment | added | Mateusz Kwaśnicki | In the same way as $\mu^{*1/n}$ in your question. For example, the characteristic function of $\mu$ is $e^{\phi(z)}$ for a negative definite $\phi$, and the characteristic function of $\mu^{*t}$ is $e^{t \phi(z)}$ for $t \geqslant 0$. This is all very standard, you may have a look at, say, Sato's Lévy processes and infinitely divisible distributions, Sections 7 and 8. | |
| Oct 27, 2020 at 19:43 | comment | added | 0xbadf00d | @MateuszKwaśnicki How do you define $\mu^{\ast t}$ for non-integer $t$? And with generator, do you mean the $C_c^\infty$-generator of $p_t$? | |
| Oct 27, 2020 at 19:21 | comment | added | Mateusz Kwaśnicki | This follows from the general fact that if $\mu_t = \mu^{*t}$ for $t > 0$, then $t^{-1} \mu_t$ converges vaguely in $\mathbb{R} \setminus \{0\}$ to the Lévy measure $\nu$. This, in turn, is closely related to the fact that if $p_t f = f * \mu_t$, then $t^{-1} (p_t f - f)$ converges to the generator applied to $f$ whenever $f$ is $C^2$. You should be able to find these facts in most books on Lévy processes, such as the one by Sato, but I do not have an exact reference at hand. | |
| Oct 27, 2020 at 19:08 | history | edited | LSpice | CC BY-SA 4.0 |
Title Unicode -> MathJax; other proofreading
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| Oct 27, 2020 at 18:40 | history | asked | 0xbadf00d | CC BY-SA 4.0 |