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Oct
23 |
awarded | Yearling |
Jun
25 |
awarded | Yearling |
Aug
11 |
revised |
How to compute KL-divergence when PMF contains 0s?
added 362 characters in body |
Aug
11 |
answered | How to compute KL-divergence when PMF contains 0s? |
Jul
13 |
comment |
Conic hulls and cones
BTW I believe there is a typo in your description: primal certificates mean no extreme point, and dual certificates mean some extreme point amongst the remaining input points. |
Jul
13 |
comment |
Conic hulls and cones
Welcome to MO! That is a very nice algorithm! Just a note--I can't figure out how to directly convert this to conic hulls, so it seems necessary to use a conic-to-convex reduction as remarked by Jean-Marc Schlenker in the comments above. (To be specific about the problem: it is possible that a non-extremal point has the largest projection onto the dual certificate, simply because it is "farthest down" that polyhedral face. I tried a couple quick fixes, but all failed. Anyway, the projection solution by Jean-Marc is fast and sufficient.) |
Jul
12 |
revised |
Conic hulls and cones
math parse error |
Jul
12 |
answered | Conic hulls and cones |
May
9 |
awarded | Nice Answer |
Apr
6 |
revised |
Existence of nonnegative solutions to an underdetermined system of linear equations
sampling for pos & neg |
Apr
6 |
answered | Existence of nonnegative solutions to an underdetermined system of linear equations |
Dec
17 |
awarded | Yearling |
Aug
22 |
comment |
Kernel width in Kernel density estimation
About using neighbors--although this seems like a reasonable idea, there is very little work in this direction. Luc Devroye has a few papers on this topic http://cg.scs.carleton.ca/~luc/devs.html , and you'll see that just looking at the closest is inadequate. In general, cross-validation is the standard technique. If you are interested in L1 density estimates, there is an excellent book by Devroye and Lugosi titled "Combinatorial Methods in Density Estimation". |
Aug
19 |
revised |
Why were matrix determinants once such a big deal?
added 227 characters in body |
Aug
19 |
answered | Why were matrix determinants once such a big deal? |
Jul
23 |
answered | Is quadratic programming still NP-hard if you have bounds and a feasible point? |
Mar
18 |
comment |
Are Bregman divergences quasi-convex?
(and set $x= (0,0)$.) in the bad example, since $f$ is linear along $x-y$ and $x-z$, then $b_x(y) = b_x(z) = 0$. On the other hand, since it is quadratic along $x-w$, the Bregman divergence is nonzero; in fact, it is $1/2$. I have an argument that $f$ is convex, but it is vague. I have to run, but tomorrow hopefully I can come back with something better. |
Mar
18 |
comment |
Are Bregman divergences quasi-convex?
thanks for clarifying. If you can verify the following example (in cylindrical coordinates) is convex, then the general case does not work. Set $\lambda(\theta) = \frac {4}{\pi}\left | \theta - \frac \pi 4\right|$, $S= [0,1] \times [0,\pi/2]$, and $f : S\to \mathbb{R}$ to $f(r,\theta) = \lambda(\theta)r + (1-\lambda(\theta))r^2$. Since $\lambda(\cdot)$ goes between 0 (at $\theta \in \{0,\pi/2\}$) and 1 (at $\theta = \pi/4$), $f$ interpolates (rotationally) between linear and quadratic. The bad choice is $y = (1,0)$, $z =(1,\pi/2)$, and $w = (y+z)/2 = (\sqrt{2}/2,\pi/4)$. |
Mar
17 |
answered | Are Bregman divergences quasi-convex? |
Feb
24 |
answered | Counting trailing zeros for factorials |