In the scikit-learn library, Ridge class, there is a note that reads: "Regularization improves the conditioning of the problem and reduces the variance of the estimates."

Given the expression of beta for ridge regression:

$\beta = (X^T X + \lambda I_d)^{-1} X^T y$

How does the addition of $\lambda I_d$ to the moments matrix improve the condition number of the inverse? And how exactly does it reduce the variance of the estimates (mathematically)?


2 Answers 2


The additional term $\lambda I_d$ can be interpreted as adding some bias and it is the bias that reduces the variance.

  1. The addition of $\lambda I_d$ is equivalent to adding a penalty term to the least squares optimization problem $$\min_{\beta}: \overbrace{ (y-X\beta )^T(y-X\beta ) }^{\text{least squares}}+ \overbrace{\lambda (\beta^T\beta)}^{\text{additional penalty}}$$

  2. The penalty term is equivalent to constraining the parameters to a particular maximum size.

    $$\min_{\beta}: { (y-X\beta )^T(y-X\beta ) } \qquad \text{and} \qquad \beta^T\beta < c$$

  3. Constraining the parameters will reduce the variance of the parameters.

  • $\begingroup$ Is there a way to prove that the variance of the biased estimates is lower than that of the OLS? Can we quantify the bias/variance tradeoff? $\endgroup$
    – Paca
    Sep 18, 2022 at 15:59
  • $\begingroup$ @Paca possibly you might do this with an eigendecomposition. You can rewrite $$X^TX = Q\Lambda Q^T$$ where $Q$ is a matrix with eigenvectors and $\Lambda$ a matrix with eigenvalues on the diagonal. Then $$X^TX + \lambda I= Q\Lambda^\dagger Q^T$$ where $\Lambda^\dagger = \Lambda + \lambda I$ and for the inverse you get $$(X^TX + \lambda I)^{-1} = Q(\Lambda^\dagger)^{-1} Q^T$$ and so it is easy to see how the inverse changes due to the additional $\lambda I$ term. The terms on the diagonal matrix $(\Lambda^\dagger)^{-1}$ are smaller... $\endgroup$ Sep 18, 2022 at 19:06
  • $\begingroup$ ...I imagine that you can use that fact to show that the variance of the expression $$(X^TX + \lambda I)^{-1} X^TY = Q (\Lambda^\dagger)^{-1} Q^TX^TY$$ is smaller. $\endgroup$ Sep 18, 2022 at 19:08
  • 1
    $\begingroup$ But a smaller variance is not always better. It may be more difficult to show that it also means a smaller error (which is variance + bias). $\endgroup$ Sep 18, 2022 at 19:10

$\DeclareMathOperator{\rid}{\hat{\boldsymbol\beta}_\text{ridge}}\DeclareMathOperator{\ols}{\hat{\boldsymbol\beta}}\DeclareMathOperator{\bias}{\hat{\boldsymbol\beta^\ast}} \DeclareMathOperator{\tr} {trace}\DeclareMathOperator{\xx}{\mathbf X^\mathsf T\mathbf X}$

Condition number (cf. $\rm [I],$ chapter $4,$ p. $189$) $\kappa:\mathcal M\to \mathbb R^{\geq 0}$ is defined as

$$\kappa(\mathbf A) := \Vert \mathbf A \Vert \left\Vert \mathbf A^{-1}\right\Vert,\tag 1$$ where $\Vert \mathbf A\Vert := \displaystyle\max_{\Vert \mathbf x\Vert = 1} \Vert\mathbf{Ax} \Vert.$ Corresponding to $\Vert \mathbf x\Vert^2 =\sum_{i=1}^n |x_i|^2, $ the matrix norm subordinate to it can be shown (cf.$\rm [II],$ chapter $1,$ p. $60$) to be $\Vert \mathbf A \Vert= \sqrt{\lambda_1}, $ the largest eigenvalue of $\mathbf A^\mathsf T\mathbf A; $ if $\bf A$ is symmteric, then $$ \kappa(\mathbf A) =\frac{\lambda_\max(\mathbf A) }{\lambda_\min(\mathbf A)},\tag{1.I}$$

Eigenvalues of $\xx+ k\mathbf I$ are of the form $\lambda_i+ k, ~\lambda_i$ being an eigenvalue of $\xx;$ then from $\rm(1.I), $ $$\kappa(\mathbf A) =\frac{\lambda_\max + k }{\lambda_\min + k}.\tag 2$$ Since $k> 0,$ one is able to circumvent the scenario of dividing the numerator by a very small number $\lambda_\min, $ i.e. at least there would be $k:$ this would be alleviating from what would have been a very large condition number.

The problem with $\ols$ is that, albeit being unbiased, the variance is large and the point estimate is unstable. One tradeoff could be that of allowing biased estimators, say $\bias,$ in that, as echoed in Sextus Empiricus' post, the variance of $\bias$ can be lowered such that the mean square error of $\bias$ is smaller than the variance of $\ols$ (cf.$\rm[III]$, chapter $9,$ p. $305$).

If $L_1^2:= \left(\ols-\boldsymbol\beta\right) ^\mathsf T\left(\ols-\boldsymbol\beta\right), $ then since $\mathbb E[\boldsymbol\varepsilon ]=\mathbf 0,~\mathbb E\left[\boldsymbol\varepsilon\boldsymbol\varepsilon^\mathsf T\right]=\sigma^2\mathbf I, $ \begin{align}\mathbb E\left[ L^2_1\right]&= \sigma^2\tr\left(\xx\right)^{-1}\\&= \sigma^2\sum_{i=1}^p\frac1{\lambda_i},\tag 3\end{align} $\lambda_i$ being the eigenvalues of $\xx.$

Now, $$\rid= \underbrace{\left[\xx + k\mathbf I\right]^{-1}}_{:=\mathbf W}\mathbf X^\mathsf T\mathbf y; \tag{4.I}$$ equivalently $$\rid =\underbrace{\left[\mathbf I +k\left(\xx\right)^{-1}\right]^{-1}}_{:=\mathbf Z}\ols.\tag{4.II}$$ As $\mathbf Z=\mathbf W\xx, $ $$\mathbf Z= \mathbf I-k\mathbf W. \tag 5$$

If $L^2(k):= \left(\rid-\boldsymbol\beta\right)^\mathsf T \left(\rid-\boldsymbol\beta\right), $ \begin{align}\mathbb E\left[L^2(k)\right]&= \mathbb E\left[\left(\ols-\boldsymbol\beta\right)^\mathsf T \mathbf Z^\mathsf T\mathbf Z\left(\ols-\boldsymbol\beta\right)\right]+ \left(\mathbf Z{\boldsymbol\beta}-\boldsymbol\beta\right)^\mathsf T \left(\mathbf Z{\boldsymbol\beta}-\boldsymbol\beta\right)\\ &= \mathbb E\left[{\boldsymbol\varepsilon}^\mathsf T\mathbf X\left(\xx\right)^{-1}\mathbf Z^\mathsf T\mathbf Z\left(\xx\right)^{-1}\mathbf X^\mathsf T\boldsymbol\varepsilon\right]+ \left(\mathbf Z{\boldsymbol\beta}-\boldsymbol\beta\right)^\mathsf T \left(\mathbf Z{\boldsymbol\beta}-\boldsymbol\beta\right)\\&= \sigma^2\tr\left[\left(\xx\right)^{-1}\mathbf Z^\mathsf T\mathbf Z\right]+ \boldsymbol\beta^\mathsf T\left(\mathbf Z-\mathbf I\right)^\mathsf T \left(\mathbf Z-\mathbf I\right) {\boldsymbol\beta}\\ &\overset{(5)}{=} \sigma^2\left[\tr\mathbf W-k\tr\mathbf W^2\right]+ k^2\boldsymbol\beta^\mathsf T\mathbf W^{2}\boldsymbol\beta\\ &= \underbrace{\sigma^2\sum_{i=1}^p \frac{\lambda_i}{(\lambda_i + k)^2}}_{\gamma_1(k) }+ \underbrace{k^2\sum_{i=1}^p \frac{\alpha_i^2}{(\lambda_i + k)^2}}_{\gamma_2(k)};\tag 6 \end{align} where $\boldsymbol\alpha = \mathbf P\boldsymbol\beta,~\mathbf P $ being the orthogonal matrix such that $\xx = \mathbf{ P\Lambda P}^\mathsf T, ~\mathbf\Lambda :=\operatorname{diag}(\lambda_i).$

Observation $1.$ $\gamma_1(k) ,$ the variance, is decreasing function of $k$ and $\gamma^\prime_1(k) \to -\infty$ as $k\to +0, ~\lambda_\min\to 0.$

\begin{align}\lim_{k\to +0}\gamma^\prime_1(k)&= \lim_{k\to +0}-2\sigma^2\sum_{i=1}^p\frac{\lambda_i}{(\lambda_i + k)^3} \\ &= -2\sigma^2\sum_{i=1}^p\frac{ 1}{\lambda_i^2}.\tag 7\end{align}

Observation $2.$ $\gamma_2(k) ,$ the squared bias is increasing function of $k$ and $\gamma^\prime_2(k) \to 0$ as $k\to +0.$

\begin{align}\lim_{k\to +0}\gamma^\prime_2(k)&= \lim_{k\to +0}2k\sum_{i=1}^p\frac{\lambda_i\alpha_i^2}{(\lambda_i + k)^3} \\ &= 0.\tag 8\end{align}

These two observations indicate there are "admissible" values or $k$ for which $\operatorname{MSE}(\rid) $ is lesser than the variance of $\ols.$

Theorem 1. (cf. $\rm[IV]$) There exists $k > 0$ such that $\mathbb E\left[L_1^2(k)\right] < \mathbb E\left[L_1^2(0)\right].$

Proof. Note $\gamma_1(0)= \sigma^2\sum_{i=1}^p\frac1{\lambda_i}, ~\gamma_2(0) = 0.$

Now, \begin{align}\frac{\mathrm d}{\mathrm dk}\mathbb E\left[L_1^2(k)\right] &= -2\sigma^2\sum_{i=1}^p\frac{\lambda_i}{(\lambda_i + k)^3}+ 2k\sum_{i=1}^p\frac{\lambda_i\alpha_i^2}{(\lambda_i + k)^3}; \tag 9\end{align} based on both the observations, it suffices to show that there exists $k>0$ such that $\frac{\mathrm d}{\mathrm dk}\mathbb E\left[L_1^2(k)\right] <0.$ From $(9), $ the value of $k$ should be $$k < \frac{\sigma^2}{\alpha^2_\max}.\tag{10}$$ $\square$

It must be reiterated again for the sake of gravity, also as mentioned in Sextus Empiricus' comment, the agenda is not to solely decrease the variance but rather lower the mean square error of $\rid$ than the variance of the least square estimator $\ols.$ This is done by introducing a little bias and substantially reducing the variance for certain $k>0$ as shown above.


$[\rm I]$ Algebraic Eigenvalue Problem, J. H. Wilkinson, Oxford University Press, $1965.$

$[\rm II]$ Computational Methods of Linear Algebra, V. N. Faddeeva, Dover Publications, $1959.$

$\rm[III]$ Introduction to Linear Regression Analysis, Douglas C. Montgomery, Elizabeth A. Peck, G. Geoffrey Vining, John Wiley & Sons, $2012.$

$\rm [IV]$ Ridge Regression: Biased Estimation for Nonorthogonal Problems, Arthur E. Hoerl, Robert W. Kennard, Technometrics $42,$ no. $1~ (2000): ~80–86. $ https://doi.org/10.2307/1271436.

  • 1
    $\begingroup$ Nice post. This proves it very rigorously. I got stuck in my comments with a reasoning that used an eigendecomposition where the eigenvalues are reduced, but I could not convert it into a prove that the variance is reduced let alone the expectation value of the squared error... $\endgroup$ Sep 19, 2022 at 7:43
  • $\begingroup$ It took me a while to grab those old books in the library and typeset them. As for your posts, they do provide valid assertions. $\endgroup$ Sep 19, 2022 at 7:45
  • $\begingroup$ ... one problem with this question is that it is difficult to have the 'helicopter view' of what is happening in the bias variance trade-off. I find the case of estimating the mean of a population much simpler (Bias / variance tradeoff math). $\endgroup$ Sep 19, 2022 at 7:45
  • $\begingroup$ I do agree. Nevertheless, the paper by Hoerl and Kennard provided a lucid language to show the machinery of the bias variance trade in light of the ridge trace. $\endgroup$ Sep 19, 2022 at 7:50

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