Timeline for Sequences with integral variances
Current License: CC BY-SA 3.0
19 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Nov 5, 2013 at 15:23 | comment | added | Joseph O'Rourke | (@Lucia: Sorry, cannot reply now...) | |
| Nov 5, 2013 at 15:11 | comment | added | Lucia | @JosephO'Rourke: Is it the case that this sequence always matches the sequence of means until it terminates? If that is always the case, then it would follow that the sequence of variances must terminate. | |
| Nov 5, 2013 at 11:56 | comment | added | Joseph O'Rourke | @BenjaminDickman: Very nice! E.g., $27722, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2$: length $12$. And for each of those, start with $n-1$: $27721, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1$. | |
| Nov 5, 2013 at 7:09 | answer | added | Greg Martin | timeline score: 3 | |
| Nov 5, 2013 at 3:46 | comment | added | Benjamin Dickman | @JosephO'Rourke How about the sequences arising from the elements in oeis.org/A174554? I believe this will answer your second question (there is no upper-bound as one can produce a long string of $2$s). | |
| Nov 5, 2013 at 1:32 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
Typo noticed by Ben.
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| Nov 5, 2013 at 1:00 | comment | added | Benjamin Dickman | Small typo: Your $(32)^2$ should be $(-3)^2$. | |
| Nov 4, 2013 at 17:33 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
Strengthened the "I believe" statement to an assertion (as I verified it in that one instance).
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| Nov 4, 2013 at 13:08 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
A few more data points.
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| Nov 4, 2013 at 11:38 | comment | added | Joseph O'Rourke | @RobertIsrael & JohnBentin & GerryMyerson: You are right; I have changed the variance to divide by $n$. I came upon this problem from a sampling context, but that context is left so far behind in the problem formulation that it doesn't make sense to divide by $n{-}1$. Thanks. | |
| Nov 4, 2013 at 11:01 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
Changed variance definition to divide by n.
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| Nov 4, 2013 at 9:13 | comment | added | John Bentin | Division by $n-1$ is used for an unbiased estimate of the population variance---but what is the "population" here? Division by $n$ is more natural in this case. (If we do have a "population" in mind, division by $n$ gives the maximum-likelihood estimate for its variance, also a natural quantity.) | |
| Nov 4, 2013 at 0:06 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
Highlighted n-1 vs n, as per the comments.
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| Nov 3, 2013 at 23:05 | comment | added | Robert Israel | Since this is not a situation of sampling from a distribution, I don't see what an "unbiased estimate" has to do with it. You're free to use whatever formula you want, but you should at least specify it. | |
| Nov 3, 2013 at 22:40 | comment | added | Joseph O'Rourke | Another unimpeachable source: the Kahn Academy! :-) Review and intuition why we divide by n-1 for the unbiased sample variance. | |
| Nov 3, 2013 at 22:27 | comment | added | Joseph O'Rourke | @GerryMyerson: Yes, dividing by $n-1$. Of course one could define the variance differently, I was just following the typical unbiased estimate of the variance... Perhaps for number-theoretic interest, dividing by $n$ would be more natural? | |
| Nov 3, 2013 at 22:18 | comment | added | Gerry Myerson | Dividing by $n-1$, not by $n$? | |
| Nov 3, 2013 at 22:14 | history | asked | Joseph O'Rourke | CC BY-SA 3.0 |