Talk:Least squares

I think the notation ${\displaystyle Y}${\displaystyle Y} here is a little confusing, because ${\displaystyle Y}${\displaystyle Y} makes it seem like it could be a matrix, when in fact it has to be a vector:

${\displaystyle L(D,{\vec {\beta }})=||X{\vec {\beta }}-Y||^{2}=(X{\vec {\beta }}-Y)^{T}(X{\vec {\beta }}-Y)=Y^{T}Y-Y^{T}X{\vec {\beta }}-{\vec {\beta }}^{T}X^{T}Y+{\vec {\beta }}^{T}X^{T}X{\vec {\beta }}}${\displaystyle L(D,{\vec {\beta }})=||X{\vec {\beta }}-Y||^{2}=(X{\vec {\beta }}-Y)^{T}(X{\vec {\beta }}-Y)=Y^{T}Y-Y^{T}X{\vec {\beta }}-{\vec {\beta }}^{T}X^{T}Y+{\vec {\beta }}^{T}X^{T}X{\vec {\beta }}}

I'd consider replacing ${\displaystyle Y}${\displaystyle Y} with ${\displaystyle {\vec {y}}}${\displaystyle {\vec {y}}}. — Preceding unsigned comment added by Fadmmatt (talk • contribs) 17:53, 26 July 2020 (UTC) Reply

Moreover, this equation seems incorrect about the beta$${\displaystyle =Y^{\mathsf {T}}Y-X^{\mathsf {T}}Y{\boldsymbol {\beta }}-X^{\mathsf {T}}Y{\boldsymbol {\beta }}+X^{\mathsf {T}}X{\boldsymbol {\beta }}^{2}}$${\displaystyle =Y^{\mathsf {T}}Y-X^{\mathsf {T}}Y{\boldsymbol {\beta }}-X^{\mathsf {T}}Y{\boldsymbol {\beta }}+X^{\mathsf {T}}X{\boldsymbol {\beta }}^{2}} 77.147.1.34 (talk) 22:31, 28 November 2024 (UTC) Reply

This is another entry presenting data in an overly-complex way. See this Investopedia article on the same topic for an easy-to-read summary of the same topic: https://www.investopedia.com/terms/l/least-squares.asp

Please consider revising this article for clarity and ease of comprehension. AliasSeaThinker (talk) 21:18, 12 February 2025 (UTC) Reply

The symbols S and L are each used in adjacent sections for the sum of residuals squared. I think it would be better to choose one or the other. In addition, D is explained in L(D,b), but not explicity defined, nor is it necessary. I advocate just using S and S(b). Chris2crawford (talk) 19:24, 8 August 2026 (UTC) Reply