Leave-one-out Cross Validation for Ridge Regression

$\newcommand{\bs}[1]{\boldsymbol{#1}}$ $\newcommand{\al}{\bs{\alpha}} \newcommand{\bphi}{\bs{\phi}} \newcommand{\x}{\bs{x}} \newcommand{\y}{\bs{y}}$

Given a dataset ${\x_i, y_i}_{i=1}^n \subset \mathcal{X} \times \mathbb{R}$ the goal of ridge regression is to learn a linear (in parameter) function $\widehat{f}(x) = \al^\top \bphi(\x)$, such that the squared-loss:

$\frac{1}{n} \sum_{i=1}^n ( \al^\top \bphi(\x_i) - y_i )^2 + \lambda \al^\top \al $

is minimized. Here $\lambda \geq 0$ is a regularization parameter and $\bphi(\cdot): \mathcal{X} \mapsto \mathbb{R}^D$ is a fixed basis function. The regularization parameter and parameter(s) in the basis function are often chosen by cross validation where data is partitioned into $K$ non-overlapping sets of which $K-1$ sets are used to learn $\widehat{f}$, and the last set is used for testing. This is repeated so that each set becomes a test set exactly once. The cross validation error $E_{CV}$ is the average error over the $K$ test sets. Finally, the parameter combination which minimizes $E_{CV}$ is selected. This parameter selection process is known as $K$-fold cross validation. In the special case where $K=n$, this process is also called leave-one-out cross validation (LOOCV) (so called because only one example is left out for testing).

In $K$-fold cross validation, it is necessary to learn the model $K$ times with the $K$ dataset partitions. In LOOCV, it turns out that there is a way to avoid learning the model $n$ times. Surprisingly, for each parameter combination the LOOCV error $E_{LOOCV}$ can be computed with the same computational complexity as learning a model only once.

In ridge regression, $E_{LOOCV} = \frac{1}{n} \| H \widetilde{H}^{-1} \y \|^2$ where \ $\begin{align*} H &= I - \Phi^\top L \\ \Phi &= \left( \bphi(\x_1) | \cdots | \bphi(\x_n) \right) \\ L &= \left( \Phi \Phi^\top + \lambda I \right)^{-1} \Phi \\ \end{align*}$ \ and $\widetilde{H}$ is a matrix with the same diagonal as $H$ and all zeros off-diagonal. $I$ is the identity matrix.

Proof By definition the LOOCV error is

$E_{LOOCV} = \frac{1}{n} \sum_{j=1}^n (\bphi_j^\top \widehat{\al}_j - y_j )^2 $,

where we let $\bphi_j := \bphi(\x_j)$ and $\widehat{\al}_j$ be the solution learned using the original dataset with the $j^{th}$ example excluded. The plan here is to express $\widehat{\al}_j$ in terms of $\widehat{\al}$ which is the solution when all examples are used, and is given by the normal equation:

$\widehat{\al} = \left( \Phi \Phi^\top + \lambda I \right)^{-1} \Phi\y = L\y $

Using the above equation but excluding the $j^{th}$ example, we have

where $Q = (\Phi \Phi^\top + \lambda I)$, $\beta_j = \bphi_j^\top Q^{-1} \bphi_j $, and we used Sherman-Morrison fomula to go from the first to the second line by viewing $-\bphi_j \bphi_j^\top$ as a rank-one update. Plugging $\widehat{\al}j$ in to $E{LOOCV}$ above yields

as desired. Q.E.D.