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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 $a^{2}ba^{-1}b^{-1}a^{-1}$ would be a member but $a^{2}ba^{-1}ba^{-1}$ would not (because the total power of $b$ is nonzero). Robinh (talk) 01:32, 28 May 2018 (UTC)[reply]

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)[reply]

Thanks for this! How do I generate

a^{2}ba^{-1}b^{-1}a^{-1}

from just commutators? Robinh (talk) 22:07, 28 May 2018 (UTC)[reply]

Got it! If

\left[x,y\right]=xyx^{-1}y^{-1}

then it is

\left[a,b\right]\left[bab^{-1},a\right]

. 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)[reply]

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)[reply]

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