bio  website  

location  Rehovot, Israel  
age  23  
visits  member for  2 years 
seen  11 hours ago  
stats  profile views  584 
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? Pseudoarc 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
deleted 109 characters in body; edited tags 
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 
Localtoglobal inequalities for measures: BrunnMinkowski, AhlswedeDaykin, what else?
edited tags 
Mar 16 
revised 
Localtoglobal inequalities for measures: BrunnMinkowski, AhlswedeDaykin, what else?
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