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Mar
24 |
awarded | Benefactor |
Mar
19 |
accepted | Are gaussians with different moments far in total variation distance? |
Mar
19 |
comment |
Are gaussians with different moments far in total variation distance?
Thanks! I'll give this the bounty unless a significantly cleaner solution crops up (unless there is some etiquette about this I don't know). Minor (I think) issue: the covariance matrices can have eigenvalues more than 1. But you only need to apply this bound for sparse vectors, so it looks like you are good. |
Mar
18 |
comment |
Are gaussians with different moments far in total variation distance?
Suvrit: it's a big integral, of the absolute value of the difference between the Gaussian densities. I don't really see how to simplify it usefully, and writing it out is a mess. |
Mar
18 |
revised |
Are gaussians with different moments far in total variation distance?
Added remark about 2 dimensional case. |
Mar
18 |
comment |
Are gaussians with different moments far in total variation distance?
I certainly should have made that observation... I don't see how to do this nicely even in the case of 2 dimensional Gaussians, but upon consideration it does seem like that should be much easier. |
Mar
18 |
awarded | Nice Question |
Mar
17 |
awarded | Promoter |
Mar
15 |
comment |
Are gaussians with different moments far in total variation distance?
No; together with the strong concavity of the entropy it might give an alternative and conceptually clearer proof of this claim, in which I am once again interested. What suggests this might be a homework problem? |
Mar
14 |
revised |
Are gaussians with different moments far in total variation distance?
edited body; edited title |
Mar
14 |
asked | Are gaussians with different moments far in total variation distance? |
Nov
8 |
awarded | Self-Learner |
Oct
30 |
comment |
An approximate infinite-dimensional fixed point theorem
This is surely too much to ask, but do you know what happens if every coordinate of $f$ is continuous in the product topology, except for one of them? Intuitively there are two kinds of obstructions from infinite dimension, and it seems like this eliminates one of them (we no longer have to intersect infinitely many sets in a non-compact space). I don't understand the counterexamples to the approximate fixed point property well enough to see whether they work in this setting. (For my purposes, this case would be almost as good as the whole thing.) |
Oct
29 |
accepted | An approximate infinite-dimensional fixed point theorem |
Oct
28 |
awarded | Teacher |
Oct
27 |
answered | An approximate infinite-dimensional fixed point theorem |
Oct
27 |
accepted | Infinite-dimensional hex |
Oct
27 |
comment |
Infinite-dimensional hex
Yes, if the players alternate turns, then everyone only plays once. Calling it a "game" is a bit of a stretch. |
Oct
26 |
answered | Infinite-dimensional hex |
Oct
25 |
revised |
Infinite-dimensional hex
Added the case $k = 3$. |