Timeline for If $\mu$ is an infinitely divisible probability measure on $[0,\infty)$, then the Lévy measure of $\mu$ is the vague limit of $n\mu^{*1/n}$
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 11, 2021 at 17:24 | comment | added | Mateusz Kwaśnicki | I do not really remember the question, but (a) $t^{-1} \mu^{*t} f$ converges to infinity if $f(0) > 0$, even if the Lévy measure is zero; (b) for vague convergence of non-negative measures, it is sufficient to consider smooth functions: every continuous one can be squeezed between two smooth ones with arbitrary precision; (c) $t^{-1}\mu^{*t}$ converges to the Lévy measure vaguely on $\mathbb R^d\setminus\{0\}$, not on $\mathbb R^d$. | |
| Oct 11, 2021 at 14:53 | comment | added | 0xbadf00d | However, by definition of vague convergence, we need to show $\frac1t\mu^{\ast t}\xrightarrow{t\to0+}f\xrightarrow{t\to0+}\lambda f$ for all $f\in C_c(\mathbb R^d)$. Since $\lambda(\{0\}=0$, it's clearly legitimate to pick $f\in C_c(\mathbb R^d\setminus\{0\})$ instead, but why is it even sufficient to assume $f$ is smooth? | |
| Oct 11, 2021 at 14:51 | comment | added | 0xbadf00d | Sorry for replying to this old post, but there is some issue in your reasoning in (3.) which I'm not able to resolve. It's clear to me that $\frac1t\mu^{\ast t}\xrightarrow{t\to0+}f\xrightarrow{t\to0+}\lambda f$ (I write $\eta f:=\int f\:{\rm d}\eta$ whenever $\eta$ is a measure) for all $f\in C_c^\infty(\mathbb R^d\setminus\{0\})$ (yes, we're dealing with $d=1$ in the question, but what I wrote before holds for an arbitrary infinitely divisible probaiblity measure $\mu$ on $\mathcal B(\mathbb R^d)$, where $\lambda$ is the Lévy measure from the canonical decomposition of $\mu$). | |
| Nov 5, 2020 at 12:40 | comment | added | Mateusz Kwaśnicki | Section 36 "Laws of large numbers" in Sato's book has Theorem 36.5, which is exactly the SLLN that you ask for in your previous comment. Regarding ergodicity in your last comment, a natural counterpart would be the Lévy–Itô decomposition, which in particular asserts that $Z_t = \sum_{s \leqslant t} f(X_s - X_{s-})$ is a Lévy process with Lévy measure $\nu \circ f^{-1}$, and hence $t^{-1} Z_t$ converges a.s. as $t \to \infty$ to $\mathbb{E} Z_1 = \int f d\nu$, of course given that $f$ is sufficiently "nice". | |
| Nov 4, 2020 at 5:43 | comment | added | 0xbadf00d | Yes, sorry, my intend was a strong law of large numbers. We easily observe that $$\frac1n\sum_{i=1}^nf(Y_i)\xrightarrow{n\to\infty}\int f\:{\rm d}\nu\;\;\;\text{a.s. for all }f\in\mathcal L^p(\nu)\text{ and }p\ge1,$$ since $$Y_n:=X_n-X_{n-1}\;\;\;\text{for }n\in\mathbb N$$ is an independent identically $\nu$-distributed process. I wondered whether there is a generalization of this observation. | |
| Nov 3, 2020 at 21:07 | comment | added | Mateusz Kwaśnicki | You mean: as $t \to \infty$, I suppose? This is a variant of the strong law of large numbers, I suppose. On the other hand, as $t \to 0^+$, I believe the random variables $t^{-1} X_t$ converge a.s. to the drift coefficient in the finite variation case. | |
| Nov 3, 2020 at 20:25 | comment | added | 0xbadf00d | Assuming that $\nu:=\mathcal L(X_1)$ has a finite first moment, is there a result yielding $\frac1tX_t\xrightarrow{t\to0+}\operatorname E[X_1]$ a.s.? | |
| Nov 3, 2020 at 20:24 | history | bounty ended | 0xbadf00d | ||
| Nov 3, 2020 at 20:24 | vote | accept | 0xbadf00d | ||
| Nov 1, 2020 at 22:18 | history | edited | Mateusz Kwaśnicki | CC BY-SA 4.0 |
deleted 180 characters in body
|
| Nov 1, 2020 at 22:17 | comment | added | Mateusz Kwaśnicki | (1) Your approach seems to be similar to item 1. in the above list, and I believe one can make it complete, but I had no time to think about it. (2) You are absolutely right that I overcomplicated item 3., I will edit the answer momentarily. (3) Indeed, transition semigroups of Lévy processes are strongly continuous on $U_b$, but in general the formula requires all of $f$, $f'$ and $f''$ to be $U_b$. (4) However, all you need here is that $P_t$ (that you call $\kappa_t$) is a Feller semigroup (i.e. strongly continuous on $C_0$, continuous functions vanishing at infinity). | |
| Nov 1, 2020 at 15:23 | comment | added | 0xbadf00d | (a) Is this correct? (b) Am I missing something or is strong continuity of the semigroup not needed. (b) Everything should generalizes to arbitrary Hilbert spaces, but do you have a reference for the form of $L$ for the one-dimensional case considered here? (c) In any case, I'd still be interested on how we can fill the gaps in the approach described in my answer and would highly appreciate if you could comment on that as well. | |
| Nov 1, 2020 at 15:23 | comment | added | 0xbadf00d | My guess is that we can show that $(\kappa_t)_{t\ge0}$ is strongly continuous on the space $U_b(\mathbb R)$ of uniformly continuous functions on $\mathbb R$ and with that choice for $C$ it holds $$(Af)(x)=af''(x)+bf'(x)+cf(x)+\int f(y)-f(x)-1_{B_1(x)}(y-x)f'(x)\:\nu({\rm d}y)$$ for all $f\in U_b(\mathbb R)\cap C^2_b(\mathbb R)$ (maybe derivatives need to be uniformly continuous as well) for some $a,b,c\in\mathbb R$. | |
| Nov 1, 2020 at 15:22 | comment | added | 0xbadf00d | Let $L$ denote the $C$-generator of $(\kappa_t)_{t\ge0}$. Note that for all $f\in\mathcal D(L)$ it holds $$\lim_{t\to0+}\frac{\kappa_tf-f}t=\kappa_0Af=Af$$ and since the convergence is wrt the supremum norm, it particularly holds $$\lim_{t\to0+}\frac{(\kappa_tf)(0)-f(0)}t=(Af)(0).$$ So, you don't need to consider the integral and argue with the dominated convergence theorem, unless I'm missing something. Now, the question is how we need to choose $C$. | |
| Nov 1, 2020 at 15:22 | comment | added | 0xbadf00d | I really like the third approach, since it is very easy to establish the desired result, once we know the form of the generator $L$. However, I need your help to understand some minor things. Let $\mu_t:=\mathcal L(X_t)$ for $t\ge0$. I know that the transition semigroup $(\kappa_t)_{t\ge0}$ of $X$ is given by $\kappa_t(x,B)=\mu_t(B-x)$ for all $(x,B)\in\mathbb R\times\mathcal B(E)$. Moreover, (as this is true for any Markov semigroup), $(\kappa_t)_{t\ge0}$ is contractive on any closed subspace $C$ of the space of bounded Borel measurable functions equipped with the supremum norm. | |
| Nov 1, 2020 at 6:01 | comment | added | 0xbadf00d | Thank you for your answer. I will check these approaches. But what do you think about the approach I've described in my answer? Does it work and can we will the gaps? | |
| Nov 1, 2020 at 0:42 | history | edited | Mateusz Kwaśnicki | CC BY-SA 4.0 |
added 170 characters in body
|
| Nov 1, 2020 at 0:02 | history | answered | Mateusz Kwaśnicki | CC BY-SA 4.0 |