Short Note on Ridge Regression

Author

Bas Machielsen

Published

November 23, 2022

Introduction

I am doing a couple of assignments involving penalized estimators such as Ridge regression, and I wanted to do a short derivation of its asymptotic covariance. In comparison to the existing resources which I could find, some details are left out, which I wanted to recapitulate more clearly. I’ll also contrast the variance of the Ridge estimator to the variance of the OLS estimator, illustrating a fact that also comes to the surface in many other resources, namely that the variance of the Ridge estimator is smaller than that of the OLS estimator.

Setting

I assume non-stochastic regressors \(X\), and a model with \(Y = X\beta + \epsilon\), \(\epsilon \sim \mathcal{N}(0, \sigma^2 I)\) with \(\mathbb{E}[X^T \epsilon] = 0\).

The ridge estimator can be expressed as:

\[ \hat{\beta}_{R} = (X^T X + \lambda I)^{-1} X^T y \]

It is easy to show that the Ridge estimator is biased for \(\lambda \neq 0\) by evaluating the expected value.

Consistency of the Ridge Estimator

Doing so also allows us to express $_{R} $ as:

\[ \hat{\beta}_R = (X^T X + \lambda I)^{-1} X^T X \beta \]

Taking the $ $ of this expression and applying Slutsky’s theorem then gives:

\[ \text{plim} \left((X^T X + \lambda I)^{-1} X^T X \beta \right) = \text{plim} (\frac{1}{n} X^T X + \frac{1}{n} \lambda I)^{-1} \cdot \text{plim} (\frac{1}{n} X^T X) \beta \]

After realizing that

\[ \text{plim} ( \frac{1}{n} \lambda I) = 0 \]

the above expression simplifies to \(\beta\), thus showing consistency.

Asymptotic Variance of the Ridge Estimator

The asymptotic variance variance of the Ridge regression around its \(\text{plim}\) can be obtained by rewriting the estimator in the following form:

\[ \hat{\beta}_R = (X^T X + \lambda I)^{-1} X^T y \]

\[ = (X^TX + \lambda I)^{-1} X^T (X\beta + \epsilon) \]

\[ = (X^TX + \lambda I)^{-1} X^T \beta + (X^T X + \lambda I)^{-1} X^T \epsilon \]

Which by the CLT converges to its \(\text{plim}\). The variance is then determined by its second part, since the first part is stochastic.

  • First, then, according to the (a) CLT, \(X^T \epsilon\) converges to its \(\text{plim}\) , which is zero by assumption, with a variance being equal to \(\sigma^2 X^T X\). Then, by the product limit normal rule (Cameron & Trivedi, 2005, Theorem A.17), the variance of \(\hat{\beta}_R\) is then equal to:

\[ \text{Var} (\hat{\beta}_R) = \sigma^2 (X^T X + \lambda I)^{-1} X^T X (X^T X + \lambda I)^{-1} \]

which can also be expressed as:

\[ \text{Var} (\hat{\beta}_R) = \sigma^2 (X^T X + \lambda I)^{-1} X^T X (X^T X)^{-1} X^T X (X^T X + \lambda I)^{-1} \]

Comparison of Variance with OLS Estimator

Now, I show the positive semidefiniteness of the matrix Var ${OLS} $ - Var $ {R}$:

  • Taking the previous expression, and defining \(W = X^T X (X^T X + \lambda I )^{-1}\), we can rewrite \(\text{Var}(\hat{\beta}_{R})\) in a simple form:

\[ \text{Var} ( \hat{\beta}_{R} ) = \sigma^2 W^T (X^T X)^{-1} W \]

The difference between \(\text{Var}(\beta_{OLS})\) - \(\text{Var}(\beta_{R})\) is then:

\[ \text{Var} (\hat{\beta_{OLS}}) - \text{Var} (\hat{\beta}_{R}) = \sigma^2 (X^T X)^{-1} - \sigma^2 W^T (X^T X)^{-1} W \newline = \sigma^2 \left( (X^T X)^{-1} - W^T (X^T X)^{-1} W \right) \]

It remains to show that

\[ \left( (X^T X)^{-1} - W^T (X^T X)^{-1} W \right) \]

is a positive semi-definite matrix.

Since \(X^T X\) is p.s.d., \((X^T X)^{-1}\) is also p.s.d. (A short proof is comparing the eigenvalues of a matrix and its inverse). Also, if you add \(\lambda I\) to a matrix, its eigenvalues increase with \(\lambda\). Hence with $> 0 $, we increase the already positive eigenvalues, and \(W\) is also p.s.d.

After some derivation, we can show that the difference in variances is equal to the following quadratic form:

\[ \sigma^2 (X^T X + \lambda I)^{-1} \left[ 2 \lambda I + \lambda^2 (X^T X)^{-1} \right] (X^T X + \lambda I)^{-1} \]

Since, by the preceding discussion, all matrices here are p.s.d., the final variance is positive semi-definite and the variance of the OLS estimator is larger than the variance of the Ridge estimator.

Conclusion

In this post, I have set out some properties of the Ridge estimator, arguably the easiest to understand shrinkage estimator. I have focused on some standard theoretical results, and try to explain this in a way that works for me. Thank you for reading!