Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 45 / 2 / a / Solution

For an $n\times p$ design matrix, distinguish the coefficient vector from the fitted-response vector. The ridge regression objective and its solution areThe gradient is $2(X^{T}X+\lambda I_p)\beta -2X^{T}Y$. For $\lambda >0$, $X^{T}X+\lambda I_p$ is a positive-definite matrix, so this is the unique minimizer even if $X$ is rank deficient. The expression with an additional leading $X$ is instead the $n$-vector of fitted values,This distinction corrects the dimensional mismatch in the printed coefficient-estimator notation.

The quadratic penalty discourages large coefficients, exchanging some bias for lower variance. If $X=UD{V}^T$ is a reduced singular value decomposition, with positive singular values $d_k$, thenThe principal-component shrinkage by ridge regression therefore shrinks directions with small $d_k^2$ most strongly. These are precisely the poorly determined directions created by multicollinearity, where ordinary least-squares coefficients can have large variance or fail to be uniquely determined. A positive penalty stabilizes their inversion and sets coefficients in the null space to zero. It does not make the predictors independent or provide a universal risk improvement for every signal. Because the penalty depends on coefficient units, predictors are ordinarily standardized before choosing $\lambda$ by cross-validation. An unpenalized intercept in ridge regression is handled separately by centering; the displayed objective either has no intercept or applies the penalty to every component included in $\beta$.