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Apr
8
awarded  Yearling
Apr
3
asked L^1 maximal inequalities for the Ornstein-Uhlenbeck semigroup in infinite dimension
Mar
9
comment 'Nonclassical' abstract Wiener space
My not-so-educated guess is that this is impossible because the embedding $C^\alpha \to C$ should only be $2$-summing for $\alpha > d/2$.
Mar
9
comment 'Nonclassical' abstract Wiener space
What you are asking is equivalent to squeezing Hilbert spaces $H$ in between $C^{0,\frac{1}{2}}(\Omega)$ and $C(\Omega^\prime)$ for subsets $\Omega^\prime$ of arbitrarily large measure. Indeed, whenever a Gaussian measure lives in $L^0$ its Cameron-Martin space has to consist of functions that are continuous on sets of large measure, and conversely, for any such Cameron-Martin space, the canonical Gaussian measure lives in $W := L^1(\mu^\prime)$ for some equivalent measure $\mu^\prime$.
Feb
10
awarded  Custodian
Feb
10
reviewed Approve Is there a proof that OEIS-A002387 is $[ e^{n-\gamma} ]$?
Feb
4
answered Divergence of general random series and a special case
Dec
21
comment A measure of how “spread out” a probability measure is
I just posted my calculation as an answer.
Dec
21
answered A measure of how “spread out” a probability measure is
Dec
21
comment A measure of how “spread out” a probability measure is
Did you try to play with Young's convolution inequalities with optimal constants, and its relatives - entropy power, Brascamp-Lieb, etc.? I don't know of a fully general answer to your question, but, e.g., if the distribution of $X_1$ has bounded density then one using Young's inequality with the optimal constant I can get the bound $\mathsf{P}\{|X| \le 1\} = O(1/\sqrt n)$.
Nov
17
comment What is the optimal growth of the constant in BDG?
The time change is the quadratic variation. I thought you were talking about the case when it's bounded.
Nov
13
comment What is the optimal growth of the constant in BDG?
The question was about continuous martingales. A continuous martingale is a time-changed Brownian motion, and the maximum of a Brownian motion over a bounded interval is sub-Gaussian.
Oct
14
comment Exponential of approximate quadratic variation of Brownian motion
For the Brownian motion $B$ with bounded drift in Euclidean space and the same BM $B^\prime$ without drift one can use some trivial bound like $\Vert B_t - B_s \Vert^2 \le (1+\varepsilon) \Vert B_t^\prime - B_s^\prime \Vert^2 + \mathrm{const}_\varepsilon \cdot |t-s|^2$ to compare them.
Oct
14
revised Exponential of approximate quadratic variation of Brownian motion
added 2 characters in body
Oct
14
comment Exponential of approximate quadratic variation of Brownian motion
I used the first edition of Kallenberg. In the second edition it's Theorem 23.5.
Oct
14
comment Exponential of approximate quadratic variation of Brownian motion
BTW, correct me if I'm wrong, but it seems like at the cut locus the singular part of the drift of $d_p(X_t)$ is negative anyway, so whatever happens there, it happens in the right direction. So maybe we don't need to introduce this ugly $\theta$ stopping after all.
Oct
14
revised Exponential of approximate quadratic variation of Brownian motion
added 27 characters in body
Oct
14
answered Exponential of approximate quadratic variation of Brownian motion
Oct
13
comment Exponential of approximate quadratic variation of Brownian motion
No, and I'm probably wrong about that on short scales, but what I had in mind is that there are comparison theorems like Theorem 1.1 here: cvgmt.sns.it/media/doc/paper/2113/HessBV.pdf, which imply that if you have a Ricci curvature bound from below then the distance is stochastically dominated by that for the hyperbolic space. The same can be done for each increment conditionally on the past, so basically you only need to check your hypothesis on the hyperbolic space, for which you can calculate everything explicitly.
Oct
13
comment Exponential of approximate quadratic variation of Brownian motion
Do you assume bounded curvature? If not, I find it surprising how you get any a priori tail bounds on $d(X_0, X_1)$ at all, simply because in regions of arbitrarily largely negative curvature one can have arbitrarily strong drift of the distance, hence arbitrarily heavy tails. And conversely, bounded curvature implies that $d(X_0, X_t)$ differs from its Euclidean counterpart by at most $ct$, where $c$ depends on the curvature bound. I haven't checked the details though, so I might be saying something stupid.