Timeline for How sensitive are probability distributions to noise?
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 31, 2019 at 16:30 | vote | accept | Alfred | ||
| S Jan 31, 2019 at 16:30 | history | bounty ended | Alfred | ||
| S Jan 31, 2019 at 16:30 | history | notice removed | Alfred | ||
| Jan 30, 2019 at 17:08 | vote | accept | Alfred | ||
| Jan 30, 2019 at 17:09 | |||||
| Jan 30, 2019 at 10:07 | answer | added | RaphaelB4 | timeline score: 1 | |
| Jan 26, 2019 at 0:05 | comment | added | user44143 | You can see what I did in WolframAlpha and redo it with $x=(0.6,0.8)$ if you like | |
| Jan 25, 2019 at 21:23 | comment | added | Alfred | There is one thing I don't understand from your example: $x = (2,3)$ doesn't have norm one. Did you take care of it in the Wolfram alpha experimentation? Thanks again | |
| Jan 25, 2019 at 20:03 | comment | added | user44143 | My next step would be numerical experimentation. | |
| Jan 25, 2019 at 18:10 | comment | added | Alfred | Thank you! do you know a way to show the same thing in more dimensions? I'm trying to get an answer but I have no idea how to keep going. Thanks again! | |
| Jan 25, 2019 at 16:05 | history | edited | Alfred | CC BY-SA 4.0 |
better explanation of the problem
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| S Jan 24, 2019 at 16:19 | history | bounty started | Alfred | ||
| S Jan 24, 2019 at 16:19 | history | notice added | Alfred | Draw attention | |
| Jan 23, 2019 at 15:22 | history | edited | Alfred | CC BY-SA 4.0 |
[Edit removed during grace period]
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| Jan 22, 2019 at 19:18 | history | edited | user44143 | CC BY-SA 4.0 |
distinguished domain and range of softmax
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| Jan 22, 2019 at 18:48 | history | edited | Alfred | CC BY-SA 4.0 |
fixed grammar
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| Jan 22, 2019 at 17:31 | comment | added | user44143 | In 2 dimensions this can be calculated numerically and visualized, e.g. as below with $x=(2,3)$, $k=10$, $E=0.025$. wolframalpha.com/input/… If $f(\eta)$ is the norm, then $f=0$ when $\eta$ is proportional to $(1,1)$ or to $(-1,-1)$. So $f$ can be approximated by two parabolas, and $E[f]\simeq (1/3)(f((1,-1)/k\sqrt{2})+f((-1,1)/k\sqrt{2}))$. In the example, that gives $0.026$. | |
| Jan 22, 2019 at 16:13 | history | asked | Alfred | CC BY-SA 4.0 |