Wikipedia:Reference desk/Archives/Mathematics/2018 May 28

= 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 $$a^2ba^{-1}b^{-1}a^{-1}$$ would be a member but $$a^2ba^{-1}ba^{-1}$$ 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  $$a^2ba^{-1}b^{-1}a^{-1}$$ from just commutators? Robinh (talk) 22:07, 28 May 2018 (UTC)
 * 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)

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)