Let's consider Ridge with only one predictor (extreme and simple case). I would like to proof that $V(B_r)=\sigma^2/(1+\lambda)$, so its variance it less than OLS variance, that is $V(B_{OLS})=\sigma^2$. I don't want the proof in the matrix form, but in it that mathematical way.

Thank you in advance to everyone!!!

  • 1
    $\begingroup$ Could you elaborate on what "not ... matrix but ... mathematical" might mean? BTW, a matrix proof appears at stats.stackexchange.com/questions/11009. It's a straightforward matter of (slight) changes in notation to make it "non-matrix," although the benefits of such a translation are dubious. $\endgroup$ – whuber Dec 1 '18 at 23:38

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