1,558 reputation
413
bio website
location Rehovot, Israel
age 23
visits member for 2 years
seen 11 hours ago

I'm a PhD student in Weizmann Institute, specializing in Probability.


11h
comment How to define the distributional Hessian for a convex function on a $C^0$ Riemannian manifold?
My comment would only apply if the manifold and the metric were smooth, in which case $f \mapsto \nabla^2 f$ would be a smooth differential operator. In your nonsmooth case I don't know how to make sense of this.
23h
comment How to define the distributional Hessian for a convex function on a $C^0$ Riemannian manifold?
Any differential operator with smooth coefficients acts on distributions,
Apr
11
comment When does topological homogeneity imply algebraic homogeneity? Pseudo-arc and Hilbert cube
If I remember correctly, the pseudoarc has the property that it admits no interesting binary operations at all: namely, every continuous map $X \times X \to X$ factors through the projection onto one of the coordinates.
Apr
8
awarded  Yearling
Apr
3
comment Finiteness of “novel variance” from a kernel on a compact space
Then I don't have a guess what you are computing in the case of more than 2 vectors.
Apr
3
answered Finiteness of “novel variance” from a kernel on a compact space
Apr
3
comment Finiteness of “novel variance” from a kernel on a compact space
On the other hand, it does depend on the order in which you take the $i_n$'s. For instance, even for 2 vectors, say, $i_1$ and $i_2$, your sum of squared norms will be $\Vert i_1 \Vert^2 + \Vert i_2 \Vert^2 - \frac{\langle i_1,i_2 \rangle}{\Vert i_1 \Vert^2}$, which is not symmetric in $i_1,i_2$.
Apr
3
comment Finiteness of “novel variance” from a kernel on a compact space
What do you mean by "doesn't depend on the choice of basis"? It seems that you are not choosing a basis.
Mar
23
revised An inequality for positive definite matrices
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Mar
23
comment An inequality for positive definite matrices
Actually, my original motivation was to avoid Gaussian tricks - and instead we arrived at special properties of exponential...
Mar
23
accepted An inequality for positive definite matrices
Mar
23
comment An inequality for positive definite matrices
Very nice, thanks!
Mar
23
comment An inequality for positive definite matrices
Actually, it is true that $\exp K$ is completely positive for every positive definite $K$. And guess what, this has a probabilistic reason: $\exp K_{ij} = \mathsf{E} \exp[X_i - \frac{1}{2} K_{ii}] \exp[X_j - \frac{1}{2} K_{jj}]$, where $X$ is the Gaussian with covariance $K$.
Mar
23
revised An inequality for positive definite matrices
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Mar
23
revised An inequality for positive definite matrices
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Mar
23
comment An inequality for positive definite matrices
@Suvrit: That's the whole point. :)
Mar
23
comment An inequality for positive definite matrices
@Suvrit: Actually, I don't see how to show even that unless the difference $K^\prime - K$ is positive definite.
Mar
23
asked An inequality for positive definite matrices
Mar
17
revised Local-to-global inequalities for measures: Brunn-Minkowski, Ahlswede-Daykin, what else?
edited tags
Mar
16
revised Local-to-global inequalities for measures: Brunn-Minkowski, Ahlswede-Daykin, what else?
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