bio | website | case.edu/artsci/math/mwmeckes |
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location | Toulouse, France | |
age | 37 | |
visits | member for | 5 years, 3 months |
seen | 7 hours ago | |
stats | profile views | 4,129 |
Assoc. Prof. at CWRU. My main research interests are in probability, geometry, and functional analysis.
Nov 14 |
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Best possible concentration inequality in high dimensions
Ah, or by $\|X_i\|_2 \le 1$ did you mean that the Euclidean norm of $X_i$ is almost surely at most 1? I thought you were referring to the $L^2$ norm of a random variable. In that case, you can e.g. use a vector-valued version of Talagrand's inequality to get the same type of result as the one dimensional case. |
Nov 14 |
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Best possible concentration inequality in high dimensions
Yes, in 1 dimension this is Hoeffding's inequality if you have boundedness, but that was never mentioned. |
Nov 12 |
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Best possible concentration inequality in high dimensions
You need additional assumptions, even in the one-dimensional case, for Hoeffding's inequality to hold. |
Oct 22 |
awarded | Yearling |
Sep 30 |
awarded | Explainer |
Jul 17 |
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Clusters of uniformly distributed random points
@ChristianRemling: for the most part, MathSciNet only lists citations from papers indexed in MathSciNet beginning sometime in the 1990s, so it misses almost all citations older than that and many newer ones as well. (It looks like Naus's paper has lots of citations from papers in other fields.) |
Jul 3 |
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Nonexistence of stable random variables
I think I really should just read Feller straight through one of these days. |
Jul 2 |
awarded | Curious |
Jul 2 |
accepted | Nonexistence of stable random variables |
Jul 2 |
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Functions that Calculate their $L_p$ Norm
How about $a=0$, $b=1$, $f(x) = 1$? |
Jul 2 |
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Nonexistence of stable random variables
Here's a purely probabilistic proof that if nonzero $\alpha$-stable variables with finite second moment exist then $\alpha = 2$. Suppose that $X$ is $\alpha$-stable with finite second moment, and assume without loss of generality that $X$ is symmetric, so that $\mathbb{E} X = 0$. Then $$ \mathbb{E} X^2 = \mathbb{E} \left(n^{-1/\alpha} \sum_{i=1}^n X_i \right)^2 = n^{-2/\alpha} \left(\sum_{i=1}^n \mathbb{E} X_i^2 + \sum_{i\neq j} \mathbb{E} X_i X_j \right) = n^{1-2/\alpha} \mathbb{E} X^2, $$ and so $\alpha = 2$. |
Jul 2 |
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Nonexistence of stable random variables
@BillJohnson: That's a nice reduction. It becomes even nicer if you can take $p=2$, in which case you could even replace the (easy) facts about cotype with just the (utterly trivial) parallelogram identity. |
Jul 2 |
revised |
Nonexistence of stable random variables
added 398 characters in body |
Jul 2 |
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Nonexistence of stable random variables
Now one could ask the question, how do you prove that a stable distribution (as I defined it) must be $\alpha$-stable for some $\alpha$? But here I'm not interested in that; I really want to start with $\alpha$-stable distributions as I defined them above. |
Jul 2 |
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Nonexistence of stable random variables
@wolfies: Good point. Definition: $X$ has an $\alpha$-stable distribution if, whenever $X_1, \ldots, X_n$ are independent copies of $X$, there is a real number $d$ such that $X_1 + \cdots + X_n$ has the same distribution as $n^{1/\alpha} X + d$. |
Jul 2 |
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Nonexistence of stable random variables
I would say this is the "real" definition of a stable distribution. Versions given in terms of characteristic functions are only stated that way because the cf is the usual technical tool used to study them. |
Jul 2 |
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Nonexistence of stable random variables
@wolfies: Definition: $X$ has a stable distribution if, whenever $Y$ is an independent copy of $X$ and $a$ and $b$ are real numbers, there are real numbers $c$ and $d$ such that $aX + bY$ has the same distribution as $cX + d$. No characteristic functions in sight. |
Jul 2 |
asked | Nonexistence of stable random variables |
Jun 25 |
awarded | Popular Question |
Jun 25 |
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Equivalent Norms for the Dual of Sobolev / Bessel Spaces
The point of my first comment is that for $s < 0$, you need to replace $L^2$ with $\mathcal{S}'$. For example, $H^{-1}(\mathbb{R})$ contains delta functions, which are not in $L^2$. As for your second question, this is well outside my expertise, so I'm just guessing but: probably you can do something like this when $s$ is an integer, but you may need to assume some kind of regularity of $A$. |