Timeline for A seeming paradox regarding estimation of the number of buttons
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
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| Feb 26 at 5:29 | history | edited | Feri | CC BY-SA 4.0 |
edited body
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| Feb 26 at 2:53 | vote | accept | Feri | ||
| S Feb 26 at 2:53 | history | bounty ended | Feri | ||
| S Feb 26 at 2:53 | history | notice removed | Feri | ||
| Feb 26 at 2:53 | history | edited | Feri | CC BY-SA 4.0 |
adding conclusion
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| Feb 25 at 20:00 | answer | added | Eoin | timeline score: 4 | |
| Feb 25 at 10:07 | answer | added | Spätzle | timeline score: 3 | |
| Feb 25 at 8:00 | comment | added | Spätzle | Have you got any additional assumptions regarding the distro of $\mu_i$ values, or is it just $U[1,10^6]$? That is, are they far enough from each other for us to distinguish? Because if for example $\mu_j=1, \mu_k = 1.05, \mu_l = 1.1$ it's gonna be really difficult to tell these apart. This also has a great effect on the solution strategy and the answer to @Flounderer 's comment. | |
| Feb 23 at 4:49 | answer | added | Cliff AB | timeline score: 4 | |
| Feb 23 at 3:12 | comment | added | Feri | @Flounderer why? | |
| Feb 23 at 2:30 | comment | added | Flounderer | "If I decide to bug the person in the room one more time, I know beforehand that my best estimate would be $N=k+1$" is not true. It depends on the value of $d_{k+1}$. For example, if you observed $d_1 = 1.5$, $d_2 = 123.1$ then your best estimate for $N$ is $2$. But what if, say, $d_3 = 123.2$?. Now I think your best estimate for $N$ will still be $2$. | |
| Feb 22 at 18:11 | answer | added | Sextus Empiricus | timeline score: 3 | |
| S Feb 22 at 16:02 | history | bounty started | Feri | ||
| S Feb 22 at 16:02 | history | notice added | Feri | Draw attention | |
| Jan 24 at 22:49 | history | edited | Feri | CC BY-SA 4.0 |
added 207 characters in body
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| Jan 24 at 22:06 | comment | added | Feri | Thanks. But even if the $d_i$s are close, still increasing the number of estimated $N$ would give a larger posterior than not (the prior regarding smaller $N$s is so small that effectively only comes into play when the likelihood for two $N$s is equal) | |
| Jan 24 at 21:59 | comment | added | Jarle Tufto | Given that $\sigma$ is very small relative to the range of the different $\mu_i$'s of which there are at most $100$, you can to a good approximation treat $d_i$'s that are sufficiently close as if generated by the same button. Hence, you can estimate $N$ using en.wikipedia.org/wiki/Mark_and_recapture methods. | |
| Jan 24 at 21:11 | history | edited | Feri | CC BY-SA 4.0 |
edited title
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| Jan 24 at 20:20 | history | asked | Feri | CC BY-SA 4.0 |