Timeline for Relationship between a logistic decision function and Gaussian Noise
Current License: CC BY-SA 4.0
22 events
| when toggle format | what | by | license | comment | |
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| S Aug 28, 2020 at 18:01 | history | bounty ended | CommunityBot | ||
| S Aug 28, 2020 at 18:01 | history | notice removed | CommunityBot | ||
| Aug 28, 2020 at 11:21 | vote | accept | monade | ||
| Aug 27, 2020 at 14:14 | comment | added | kjetil b halvorsen♦ | See also stats.stackexchange.com/questions/403575/… | |
| S Aug 27, 2020 at 13:57 | history | suggested | Gerardo Duran-Martin | CC BY-SA 4.0 |
Probit function has mean 0; remove "z" component from the distribution.
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| Aug 27, 2020 at 13:31 | review | Suggested edits | |||
| S Aug 27, 2020 at 13:57 | |||||
| Aug 27, 2020 at 13:22 | answer | added | Gerardo Duran-Martin | timeline score: 2 | |
| Aug 20, 2020 at 17:51 | comment | added | whuber♦ | The double exponential distribution is the location-scale family determined by the distribution with density function $f(x)=\exp(-|x|)/2.$ | |
| Aug 20, 2020 at 16:47 | comment | added | monade | Thanks, this question is an excellent resource. The double exponential error would be $\frac{e^{-x}}{(1 + e^{-x})^2}$ (using the notation of my question)? | |
| Aug 20, 2020 at 16:31 | comment | added | whuber♦ | If I follow you correctly, you are asking how logit and probit models might be related. Although the mathematical relationship is not simple, in practice they behave so similarly that they are considered interchangeable. Replacing the Gaussian error by a double exponential ("Laplacian") error makes the two approaches equivalent. Perhaps your questions are all satisfactorily addressed at stats.stackexchange.com/questions/20523/…? | |
| S Aug 20, 2020 at 16:26 | history | bounty started | monade | ||
| S Aug 20, 2020 at 16:26 | history | notice added | monade | Draw attention | |
| Aug 18, 2020 at 14:19 | history | edited | monade | CC BY-SA 4.0 |
added 28 characters in body
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| Aug 18, 2020 at 14:13 | comment | added | monade | I now realize your misunderstanding. $p_1$ and $p_2$ are two alternatives to achieve the same thing: modeling noisy perceptual choices. (your first question sounded like you thought that I add Gaussian noise p2 on top of the choice probabilities of p1, which of course would not make sense) | |
| Aug 18, 2020 at 14:01 | comment | added | monade | $p_2$ is the probability of choosing the positive stimulus category, given an observation $x$. The idea is that I fit both models ($p_1$ and $p_2$) to the choice data and obtain values for β and σ. One question would be whether the expected values of β and σ are related in terms of a formula. | |
| Aug 18, 2020 at 13:59 | comment | added | Tim | I'm not sure what you mean by $p_2$ in here. What exactly this ought to be? | |
| Aug 18, 2020 at 13:57 | comment | added | monade | I had a typo in $p_2(x)$, maybe things become more clear now. | |
| Aug 18, 2020 at 13:56 | comment | added | Tim | Than what you mean by "relation" between them? You could pick arbitrary $\sigma$ to generate $x$ and then multiply it by another arbitrary $\beta$ and use logistic transformation to generate such data, so both can be completely independent from each other. | |
| Aug 18, 2020 at 13:56 | history | edited | monade | CC BY-SA 4.0 |
added 8 characters in body
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| Aug 18, 2020 at 13:49 | comment | added | monade | Where did you see probability + Gaussian noise? My assumption is that there is Gaussian noise around each observation $x$, where $x$ is the stimulus variable, not a probability. | |
| Aug 18, 2020 at 13:47 | comment | added | Tim | Probabilities are bounded to $[0, 1]$ while normal distribution is unbounded $(-\infty, \infty)$, so probability + Gaussian noise is not probability any more since it goes outside the bounds... What exactly do you mean by their relation? | |
| Aug 18, 2020 at 13:43 | history | asked | monade | CC BY-SA 4.0 |