3
$\begingroup$

I am currently trying to understand the MSE of ridge regression. First, I am calculating the MSE mathematically, but I found it quite vague. After reviewing some books and articles I understood that

$$ \begin{aligned} \text{MSE}(\hat{\beta_R}) &= E[||\hat{\beta}_R-{\beta}||^2] \\ \Rightarrow\hat{\beta_{R}}-\beta &= ((X^TX+\lambda)^{-1}X^TX-I)\beta+e \\ \Rightarrow||\hat{\beta}_R-{\beta}||^2 &= (\hat{\beta_R}-{\beta})^T(\hat{\beta_R}-{\beta}) \end{aligned} $$

After that I got stuck because of the norm and expectation calculation. I tried to solve it, but it becomes so complicated.

I have checked books like: "The Elements of Statistical Learning" and "An Introduction to Statistical Learning".

Can anyone please clarify MSE of ridge regression or guide me to a good source?

$\endgroup$
2
  • 2
    $\begingroup$ Look at page 12 here. $\endgroup$ Commented Jul 23, 2016 at 15:16
  • $\begingroup$ @Greenparker, Many thanks , actually I looked at this before, but it was not clear for me. $\endgroup$
    – jeza
    Commented Jul 23, 2016 at 15:20

1 Answer 1

5
$\begingroup$

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

The good source would be straight from horse's mouth. (cf. $\rm [I]$).

$$\rid= \underbrace{\left[\xx + k\mathbf I\right]^{-1}}_{:=\mathbf W}\mathbf X^\mathsf T\mathbf y; \tag{1.I}$$ equivalently $$\rid =\underbrace{\left[\mathbf I +k\left(\xx\right)^{-1}\right]^{-1}}_{:=\mathbf Z}\ols.\tag{1.II}$$ As $\mathbf Z=\mathbf W\xx, $ $$\mathbf Z= \mathbf I-k\mathbf W. \tag 2$$

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{(2)}{=} \sigma^2\left[\tr\mathbf W-k\tr\mathbf W^2\right]+ k^2\boldsymbol\beta^\mathsf T\mathbf W^{2}\boldsymbol\beta\\ &= {\sigma^2\sum_{i=1}^p \frac{\lambda_i}{(\lambda_i + k)^2}}+ {k^2\sum_{i=1}^p \frac{\alpha_i^2}{(\lambda_i + k)^2}};\tag 3 \end{align} where $\boldsymbol\alpha = \mathbf P\boldsymbol\beta,~\mathbf P $ being the orthogonal matrix such that $\xx = \mathbf{ P\Lambda P}^\mathsf T, ~\mathbf\Lambda :=\operatorname{diag}(\lambda_i).$


Reference:

$\rm [I]$ 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.

$\endgroup$
1
  • $\begingroup$ How do you know that Z=WXtX hold? and also, i cannot understand the formula (2) as well. Do I have to just write down the matrix calculation? $\endgroup$
    – user408967
    Commented Mar 8 at 11:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.