Timeline for Why shouldn't the denominator of the covariance estimator be n-2 rather than n-1?
Current License: CC BY-SA 4.0
21 events
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| Jan 27, 2021 at 15:50 | answer | added | Karen Huynh | timeline score: 1 | |
| Sep 4, 2018 at 7:17 | history | edited | Ferdi | CC BY-SA 4.0 |
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| Dec 16, 2016 at 21:16 | history | post merged (destination) | |||
| Dec 16, 2016 at 6:35 | answer | added | Carl | timeline score: 3 | |
| Oct 28, 2016 at 21:31 | history | post merged (destination) | |||
| Nov 24, 2015 at 7:54 | answer | added | Uditg_ucla | timeline score: 9 | |
| Apr 17, 2015 at 21:36 | history | edited | whuber♦ |
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| Mar 19, 2015 at 21:39 | history | tweeted | twitter.com/#!/StackStats/status/578672285776773121 | ||
| Mar 19, 2015 at 21:21 | comment | added | amoeba | @whuber: I would perhaps say that it shows that what counts as "a parameter" depends on the situation. In this case variance is computed over $n$ observations and so each observation -- or the total mean -- can be seen as one parameter, even if it is a multivariate mean, as ttnphns said. However, in other cases when e.g. a test considers linear combinations of dimensions, each dimension of each observation becomes "a parameter". You are right that this is a tricky issue. | |
| Mar 19, 2015 at 18:54 | history | edited | whuber♦ | CC BY-SA 3.0 |
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| Mar 19, 2015 at 17:12 | comment | added | Henry | The denominator of your definition of sample variance is $n−1$ probably because this makes it an unbiased estimate of the population variance. The same is true of the sample covariance using the same denominator. But there are other definitions of sample statistics, with different denominators. | |
| Mar 19, 2015 at 16:41 | comment | added | ttnphns | @whuber, You are right at that. If it were only $n$ (independent observations) which matters we wouldn't spend more df in multivariate tests than in univariate ones. | |
| Mar 19, 2015 at 16:31 | comment | added | whuber♦ | @ttnphns That's not true: the bivariate mean is obviously two parameters because it requires two real numbers to express it. (Indeed it is a single vector parameter, but saying so only disguises the fact it has two components.) This shows up explicitly in the degrees of freedom for pooled-variance t-tests, for instance, where $2$ is subtracted, not $1$. What's interesting about this question is how it reveals just how vague, unrigorous, and potentially misleading is the common "explanation" that we subtract $1$ from $n$ because one parameter has been estimated. | |
| Mar 19, 2015 at 15:34 | answer | added | Silverfish | timeline score: 23 | |
| Mar 19, 2015 at 15:20 | comment | added | ttnphns | A bi/multivariate mean (expectation) is one, not 2 parameters. | |
| Mar 19, 2015 at 15:16 | answer | added | whuber♦ | timeline score: 41 | |
| Mar 19, 2015 at 14:19 | comment | added | whuber♦ | If you did that, you would have two conflicting definitions for the variance: one would be the first formula and the other would be the second formula applied with $Y=X$. | |
| Mar 19, 2015 at 14:06 | answer | added | statchrist | timeline score: 8 | |
| Mar 19, 2015 at 13:57 | answer | added | Elvis | timeline score: 10 | |
| Mar 19, 2015 at 13:21 | history | edited | MYaseen208 | CC BY-SA 3.0 |
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| Mar 19, 2015 at 13:13 | history | asked | MYaseen208 | CC BY-SA 3.0 |