I'm looking for literature about negative ridge regression.

In short, it is a generalization of linear ridge regression using negative $\lambda$ in the estimator formula: $$\hat\beta = ( X^\top X + \lambda I)^{-1} X^\top y.$$ The positive case has a nice theory: as a loss function, as a constraint, as a Bayes prior... but I feel lost with the negative version with only the above formula. It happens to be useful for what I am doing but I fail to interpret it clearly.

Do you know any serious introductory text about negative ridge? How can it be interpreted?

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    $\begingroup$ I don't know of any introductory text that talks about it, but this source may be enlightening, especially the discussion at the bottom of page 18: jstor.org/stable/4616538?seq=1#page_scan_tab_contents $\endgroup$ Mar 1, 2018 at 17:20
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    $\begingroup$ In case that link dies in the future, the full citation is: Björkström, A. & Sundberg, R. "A generalized view on continuum regression". Scandinavian Journal of Statistics, 26:1 (1999): pp.17-30 $\endgroup$ Mar 1, 2018 at 17:21
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    $\begingroup$ Thanks a lot. This gives a clear interpretation of ridge via CR when $\lambda<-\lambda_1$.(largest eigenvalue of the covariance matrix). Still looking for an interpretation with $\lambda>-\lambda_1$... $\endgroup$ Mar 1, 2018 at 18:54
  • $\begingroup$ Note in this development of ridge regression from Tikhonov regularization that the Tikhonov regularization $\Gamma^{T} \Gamma$ becomes $\alpha^2 I$ for ridge regression. Subsequently, $\alpha^2$ is usually replaced by $\lambda$. The only way to make this negative is for $\alpha$ to be imaginary, i.e., a multiple of $i=\sqrt{-1}$. OK, Now what? Where do you want to go with it? $\endgroup$
    – Carl
    Apr 15, 2018 at 18:05
  • $\begingroup$ Negative ridge mentioned here: stats.stackexchange.com/questions/328630/… with some links $\endgroup$ Apr 15, 2018 at 18:54

1 Answer 1


Here is a geometric illustration of what is going on with negative ridge.

I will consider estimators of the form $$\hat{\boldsymbol\beta}_\lambda = (\mathbf X^\top \mathbf X + \lambda \mathbf I)^{-1}\mathbf X^\top\mathbf y$$ arising from the loss function $$\mathcal L_\lambda = \|\mathbf y - \mathbf X\boldsymbol\beta\|^2 + \lambda \|\boldsymbol\beta\|^2.$$ Here is a rather standard illustration of what happens in a two-dimensional case with $\lambda\in[0,\infty)$. Zero lambda corresponds to the OLS solution, infinite lambda shrinks the estimated beta to zero:

enter image description here

Now consider what happens when $\lambda\in(-\infty, -s^2_\max)$, where $s_\mathrm{max}$ is the largest singular value of $\mathbf X$. For very large negative lambdas, $\hat{\boldsymbol\beta}_\lambda$ is of course close to zero. When lambda approaches $-s^2_\max$, the term $(\mathbf X^\top \mathbf X + \lambda \mathbf I)$ gets one singular value approaching zero, meaning that the inverse has one singular value going to minus infinity. This singular value corresponds to the first principal component of $\mathbf X$, so in the limit one gets $\hat{\boldsymbol\beta}_\lambda$ pointing in the direction of PC1 but with absolute value growing to infinity.

What is really nice, is that one can draw it on the same figure in the same way: betas are given by points where circles touch the ellipses from the inside:

enter image description here

When $\lambda\in(-s^2_\mathrm{min},0]$, a similar logic applies, allowing to continue the ridge path on the other side of the OLS estimator. Now the circles touch the ellipses from the outside. In the limit, betas approach the PC2 direction (but it happens far outside this sketch):

enter image description here

The $(-s^2_\mathrm{max}, -s^2_\mathrm{min})$ range is something of an energy gap: estimators there do not live on the same curve.

UPDATE: In the comments @MartinL explains that for $\lambda<-s^2_\mathrm{max}$ the loss $\mathcal L_\lambda$ does not have a minimum but has a maximum. And this maximum is given by $\hat{\boldsymbol\beta}_\lambda$. This is why the same geometric construction with the circle/ellipse touching keeps working: we are still looking for zero-gradient points. When $-s^2_\mathrm{min}<\lambda\le 0$, the loss $\mathcal L_\lambda$ does have a minimum and it is given by $\hat{\boldsymbol\beta}_\lambda$, exactly as in the normal $\lambda>0$ case.

But when $-s^2_\mathrm{max}<\lambda<-s^2_\mathrm{min}$, the loss $\mathcal L_\lambda$ does not have either maximum or minimum; $\hat{\boldsymbol\beta}_\lambda$ would correspond to a saddle point. This explains the "energy gap".

The $\lambda\in(-\infty, -s^2_\max)$ naturally arises from a particular constrained ridge regression, see The limit of "unit-variance" ridge regression estimator when $\lambda\to\infty$. This is related to what is known in the chemometrics literature as "continuum regression", see my answer in the linked thread.

The $\lambda\in(-s^2_\mathrm{min},0]$ can be treated in exactly the same way as $\lambda>0$: the loss function stays the same and the ridge estimator provides its minimum.

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    $\begingroup$ Thank you for the interesting graphs. When $\lambda < -s_\text{max}^2$, the solution you have graphed is the global maximum of the cost function, not a global minimum. Similarly, when $-s_\text{max}^2 < \lambda < 0$, the point you have graphed should be a saddle point of the cost function. $\endgroup$
    – Martin L
    Apr 20, 2018 at 7:22
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    $\begingroup$ Consider only the quadratic terms in the cost function. They can be written as $$ \beta^T (X^T X + \lambda I) \beta.$$ Let $\lambda < - s_\text{max}^2$, then the matrix in parentheses has only negative eigenvalues. Let $- s_\text{max}^2 < \lambda < 0$, and the matrix has both positive and negative eigenvalues. These eigenvalues influence whether the point is a saddle point, minimum, or maximum of the cost function. $\endgroup$
    – Martin L
    Apr 20, 2018 at 11:19
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    $\begingroup$ That's very helpful, thanks a lot. I made an update to my answer. $\endgroup$
    – amoeba
    Apr 20, 2018 at 14:05
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    $\begingroup$ Thank you. In particular for realizing that the saddle point only holds when $-s_\text{max}^2 < \lambda < - s_\text{min}^2$. When $\lambda > -s_\text{min}^2$, the solution is indeed still a global minimum since then, $X^T X + \lambda I$ is positive definite. My earlier comment was thus partially incorrect. $\endgroup$
    – Martin L
    Apr 20, 2018 at 14:27

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