bio | website | perimeterinstitute.ca/… |
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visits | member for | 2 years, 9 months |
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stats | profile views | 367 |
Jun 24 |
comment |
A probability distribution in n dimensional space which its projection on any line is a uniform distribution?
Addendum to my comments: now I understand that Will achieves something equivalent by considering a random projection. Clever! |
Jun 24 |
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A probability distribution in n dimensional space which its projection on any line is a uniform distribution?
Hence we work wlog with a measure of the form $f(r)\, d\Omega\, dr$, where $d\Omega$ is the volume element of the sphere (and $f$ may be distributional), i.e. an integral over the uniform measures on spheres for varying radius of the sphere. For these measures, it should be a straightforward calculation to determine whether there exists some $f$ for which $f(r) \, d\Omega\, dr$ satisfies the OP's requirement. |
Jun 24 |
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A probability distribution in n dimensional space which its projection on any line is a uniform distribution?
Using Will's observation on mean and covariance, one can also assume the measure to be symmetric under rotations without loss of generality. The reason is that having projection uniform on $[-1,1]$ is a property which is invariant under taking mixtures of measures. Hence if some measure has this property, then so does its convolution with a random rotation (chosen from the Haar measure on $O(n)$). |
Jun 24 |
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A probability distribution in n dimensional space which its projection on any line is a uniform distribution?
@Salavati: do you know the answer for the two-dimensional case? A positive answer in three dimensions would imply a positive answer in two dimensions by projecting the measure onto the plane. So I think that it will help to consider the two-dimensional case first. |
May 29 |
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Reference for an unbiased definition of a symmetric monoidal category
I had asked this question on the nForum about a year ago: nforum.ncatlab.org/discussion/3101/symmetric-monoidal-category/… |
Mar 24 |
revised |
Dissolution of Tensors
fixed tag, typo and latex |
Mar 24 |
suggested | approved edit on Dissolution of Tensors |
Mar 9 |
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Numbers greater than Skewes's whose existence can be found in number theoretic proofs
This seems close to the "eventual counterexamples" question: mathoverflow.net/questions/15444/… |
Mar 6 |
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Abstract connectedness
@MikeShulman: yes, I strongly agree about the minimality of the separoid axioms! It feels like they're missing something important... |
Mar 6 |
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Abstract connectedness
The notion of "separoid" may be related: en.wikipedia.org/wiki/Separoid "Given a topological space, we can define a separoid saying that two subsets are separated if there exist two disjoint open sets which contains them (one for each of them)." But possibly it captures something slightly different from what you're looking for; I haven't thought about it. |
Mar 3 |
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Base of a cone in a vector space: can one always choose a convex base?
@WillieWong: sure, done! |
Mar 3 |
revised |
Base of a cone in a vector space: can one always choose a convex base?
added example due to request in the comments |
Mar 1 |
revised |
Base of a cone in a vector space: can one always choose a convex base?
small improvements |
Feb 28 |
revised |
Base of a cone in a vector space: can one always choose a convex base?
yet another correction |
Feb 28 |
revised |
Base of a cone in a vector space: can one always choose a convex base?
this is final, sorry for the many edits ;) |
Feb 28 |
answered | Base of a cone in a vector space: can one always choose a convex base? |
Feb 27 |
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Base of a cone in a vector space: can one always choose a convex base?
@AlexDegtyarev: that's a good question, but actually the answer is negative. Here's a simple example of a pointed convex cone which does not have a convex base. In $\mathbb{R}^2$, define $(x,y)$ to be in $C$ if $x>0$, or if $x=0$ and $y\geq 0$. (This cone is a simple counterexample also to many other elementary conjectures that one might have about convex cones, by the way.) |
Feb 27 |
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Base of a cone in a vector space: can one always choose a convex base?
@WillieWong: thanks, that's exactly right. I'm not sure whether a convex base always exists; if it does, then the question is uninteresting. So I assume that the OP knows examples in which a convex base does not exist. In these cases, my argument shows that a non-convex base still exists, for trivial reasons. This shows that what the OP is trying to do is not possible. |
Feb 27 |
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Base of a cone in a vector space: can one always choose a convex base?
Every cone has a base in your sense, and this follows from the axiom of choice. In $C\setminus\{0\}$, put $x\sim y$ whenever $x$ and $y$ are positive scalar multiples of each other. This is an equivalence relation, and the axiom of choice lets you pick one representative of each equivalence class. These representatives form a base in your sense. |
Feb 26 |
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Symplectic (contact) structure on $M_{n}(\mathbb{R})$
Concerning the second sentence: the derivative of the Hamiltonian indeed vanishes in certain directions, but how do you conclude that the associated vector field vanishes as well? (A generic Hamiltonian has vanishing directional derivatives in a subspace of codimension 1 of the tangent space, but that doesn't mean that the flow is zero!) |