Which of the following best explains why the slope m that minimizes E(m)=∑i=1n(yi−mxi)2 must satisfy ∑i=1nxi(yi−mxi)=0?
a) Because squaring the errors makes the largest deviation dominate, the largest individual error must be zero at the minimum.
b) Minimizing forces the unweighted sum of residuals to vanish, so ∑i=1n(yi−mxi)=0 and hence ∑xi(yi−mxi)=0.
c) Setting the derivative of E with respect to m equal to zero makes the rate of change of total squared error vanish, so an extremum occurs when ∑i=1nxi(yi−mxi)=0.
d) At the minimum each residual is individually 0, implying their xi–weighted sum is also 0.