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May 28

subgroup of the free group

What is the name of the subgroup of the free group on n letters that comprises words where each letter has a total of zero power? Thus $a^{2}b{a}^{-1}{b}^{-1}{a}^{-1}$ would be a member but $a^{2}b{a}^{-1}b{a}^{-1}$ would not (because the total power of $b$ is nonzero). Robinh (talk) 01:32, 28 May 2018 (UTC)

It would be the commutator subgroup of the original group. It's the kernel of the natural homomorphism from the free group to the free abelian group with the same generators. --RDBury (talk) 11:00, 28 May 2018 (UTC)

Thanks for this! How do I generate $a^{2}b{a}^{-1}{b}^{-1}{a}^{-1}$ from just commutators? Robinh (talk) 22:07, 28 May 2018 (UTC)

Got it! If $[x,y]=xy{x}^{-1}{y}^{-1}$ then it is $[a,b][ba{b}^{-1},a]$. Thank you very much indeed. It is not at all obvious how to generate any given element of my subgroup. Robinh (talk) 22:17, 28 May 2018 (UTC)

Mean squared prediction error

I asked this on the article’s talk page and at WikiProject Statistics, but got no response. Throughout the article Mean squared prediction error, shouldn’t every summation sign and every instance of ${\sigma }^2$ and ${\hat{\sigma }}^2$ be multiplied by 1/n? And how about in each of the two right-hand side terms of the first equation in the Estimation section? Loraof (talk) 16:30, 28 May 2018 (UTC)

I’ve gone ahead and changed it. Loraof (talk) 23:32, 28 May 2018 (UTC)