# The limit of "unit-variance" ridge regression estimator when $\lambda\to\infty$

Consider ridge regression with an additional constraint requiring that $\hat{\mathbf y}$ has unit sum of squares (equivalently, unit variance); if needed, one can assume that $\mathbf y$ has unit sum of squares as well:

$$\hat{\boldsymbol\beta}_\lambda^* = \arg\min\Big\{\|\mathbf y - \mathbf X \boldsymbol \beta\|^2+\lambda\|\boldsymbol\beta\|^2\Big\} \:\:\text{s.t.}\:\: \|\mathbf X \boldsymbol\beta\|^2=1.$$

What is the limit of $\hat{\boldsymbol\beta}_\lambda^*$ when $\lambda\to\infty$?

Here are some statements that I believe are true:

1. When $\lambda=0$, there is a neat explicit solution: take OLS estimator $\hat{\boldsymbol\beta}_0=(\mathbf X^\top \mathbf X)^{-1}\mathbf X^\top \mathbf y$ and normalize it to satisfy the constraint (one can see this by adding a Lagrange multiplier and differentiating): $$\hat{\boldsymbol\beta}_0^* = \hat{\boldsymbol\beta}_0 \big/ \|\mathbf X\hat{\boldsymbol\beta}_0\|.$$

2. In general, the solution is $$\hat{\boldsymbol\beta}_\lambda^*=\big((1+\mu)\mathbf X^\top \mathbf X + \lambda \mathbf I\big)^{-1}\mathbf X^\top \mathbf y\:\:\text{with \mu needed to satisfy the constraint}.$$I don't see a closed form solution when $\lambda >0$. It seems that the solution is equivalent to the usual RR estimator with some $\lambda^*$ normalized to satisfy the constraint, but I don't see a closed formula for $\lambda^*$.

3. When $\lambda\to \infty$, the usual RR estimator $$\hat{\boldsymbol\beta}_\lambda=(\mathbf X^\top \mathbf X + \lambda \mathbf I)^{-1}\mathbf X^\top \mathbf y$$ obviously converges to zero, but its direction $\hat{\boldsymbol\beta}_\lambda \big/ \|\hat{\boldsymbol\beta}_\lambda\|$ converges to the direction of $\mathbf X^\top \mathbf y$, a.k.a. the first partial least squares (PLS) component.

Statements (2) and (3) together make me think that perhaps $\hat{\boldsymbol\beta}_\lambda^*$ also converges to the appropriately normalized $\mathbf X^\top \mathbf y$, but I am not sure if this is correct and I have not managed to convince myself either way.

#A geometrical interpretation

The estimator described in the question is the Lagrange multiplier equivalent of the following optimization problem:

$$\text{minimize f(\beta) subject to g(\beta) \leq t and h(\beta) = 1 }$$

\begin{align} f(\beta) &= \lVert y-X\beta \lVert^2 \\ g(\beta) &= \lVert \beta \lVert^2\\ h(\beta) &= \lVert X\beta \lVert^2 \end{align}

which can be viewed, geometrically, as finding the smallest ellipsoid $$f(\beta)=\text{RSS }$$ that touches the intersection of the sphere $$g(\beta) = t$$ and the ellipsoid $$h(\beta)=1$$

## Comparison to the standard ridge regression view

In terms of a geometrical view this changes the old view (for standard ridge regression) of the point where a spheroid (errors) and sphere ($$\|\beta\|^2=t$$) touch. Into a new view where we look for the point where the spheroid (errors) touches a curve (norm of beta constrained by $$\|X\beta\|^2=1$$). The one sphere (blue in the left image) changes into a lower dimension figure due to the intersection with the $$\|X\beta\|=1$$ constraint.

In the two dimensional case this is simple to view. When we tune the parameter $$t$$ then we change the relative length of the blue/red spheres or the relative sizes of $$f(\beta)$$ and $$g(\beta)$$ (In the theory of Lagrangian multipliers there is probably a neat way to formally and exactly describe that this means that for each $$t$$ as function of $$\lambda$$, or reversed, is a monotonous function. But I imagine that you can see intuitively that the sum of squared residuals only increases when we decrease $$||\beta||$$.)

The solution $$\beta_\lambda$$ for $$\lambda=0$$ is as you argued on a line between 0 and $$\beta_{LS}$$

The solution $$\beta_\lambda$$ for $$\lambda \to \infty$$ is (indeed as you commented) in the loadings of the first principal component. This is the point where $$\lVert \beta \rVert^2$$ is the smallest for $$\lVert \beta X \rVert^2 = 1$$. It is the point where the circle $$\lVert \beta \rVert^2=t$$ touches the ellipse $$|X\beta|=1$$ in a single point.

In this 2-d view the edges of the intersection of the sphere $$\lVert \beta \rVert^2 =t$$ and spheroid $$\lVert \beta X \rVert^2 = 1$$ are points. In multiple dimensions these will be curves

(I imagined first that these curves would be ellipses but they are more complicated. You could imagine the ellipsoid $$\lVert X \beta \rVert^2 = 1$$ being intersected by the ball $$\lVert \beta \rVert^2 \leq t$$ as some sort of ellipsoid frustum but with edges that are not a simple ellipses)

##Regarding the limit $$\lambda \to \infty$$

At first (previous edits) I wrote that there will be some limiting $$\lambda_{lim}$$ above which all the solutions are the same (and they reside in the point $$\beta^*_\infty$$). But this is not the case

Consider the optimization as a LARS algorithm or gradient descent. If for any point $$\beta$$ there is a direction in which we can change the $$\beta$$ such that the penalty term $$|\beta|^2$$ increases less than the SSR term $$|y-X\beta|^2$$ decreases then you are not in a minimum.

• In normal ridge regression you have a zero slope (in all directions) for $$|\beta|^2$$ in the point $$\beta=0$$. So for all finite $$\lambda$$ the solution can not be $$\beta = 0$$ (since an infinitesimal step can be made to reduce the sum of squared residuals without increasing the penalty).
• For LASSO this is not the same since: the penalty is $$\lvert \beta \rvert_1$$ (so it is not quadratic with zero slope). Because of that LASSO will have some limiting value $$\lambda_{lim}$$ above which all the solutions are zero because the penalty term (multiplied by $$\lambda$$) will increase more than the residual sum of squares decreases.
• For the constrained ridge you get the same as the regular ridge regression. If you change the $$\beta$$ starting from the $$\beta^*_\infty$$ then this change will be perpendicular to $$\beta$$ (the $$\beta^*_\infty$$ is perpendicular to the surface of the ellipse $$|X\beta|=1$$) and $$\beta$$ can be changed by an infinitesimal step without changing the penalty term but decreasing the sum of squared residuals. Thus for any finite $$\lambda$$ the point $$\beta^*_\infty$$ can not be the solution.

##Further notes regarding the limit $$\lambda \to \infty$$

The usual ridge regression limit for $$\lambda$$ to infinity corresponds to a different point in the constrained ridge regression. This 'old' limit corresponds to the point where $$\mu$$ is equal to -1. Then the derivative of the Lagrange function in the normalized problem

$$2 (1+\mu) X^{T}X \beta + 2 X^T y + 2 \lambda \beta$$ corresponds to a solution for the derivative of the Lagrange function in the standard problem

$$2 X^{T}X \beta^\prime + 2 X^T y + 2 \frac{\lambda}{(1+\mu)} \beta^\prime \qquad \text{with \beta^\prime = (1+\mu)\beta}$$

• +1. Thanks a lot, this is super helpful! I will need some time to think it through. Apr 13 '18 at 19:36
• It's worth pointing out that the red and the black ellipsoids have the same shape: this is why the point where they touch lies on the line connecting their centers. Nice graphical proof of point #1 in my question. Apr 13 '18 at 20:00
• +5 (I started a bounty that I will happily award to your answer). I have also posted my own answer because I did some algebraic derivations and it was too much to add to the question. I am not convinced by your conclusion that the there will be some finite $\lambda_\text{lim}$ after which the solution will not change anymore and will be given by PC1. I don't see it algebraically, and I don't quite understand your argument for why it should exist. Let's try to figure it out. Apr 16 '18 at 14:11
• @amoeba, you were right about the finite $\lambda_{\lim}$ not existing. I argued too much intuitively and jumped to quickly from a particular condition for the regular ridge regression to the constrained ridge regression. Regular RR has a zero slope (in all directions) for $|\beta|^2$ in the point $\beta = 0$. I thought that (since $\beta^*_\infty \neq 0$) you do not get this with the constrained regression. However because $\beta$ is constrained to the ellipsoid $|X\beta| =1$ you can not 'move' $\beta$ in all directions. Apr 17 '18 at 10:57
• This makes sense. I think the issue is more or less clarified for me. Thanks a lot, this was really helpful. I will wait for several days with awarding the bounty and accepting an answer (perhaps this will attract some further attention to this thread). Apr 17 '18 at 11:21

This is an algebraic counterpart to @Martijn's beautiful geometric answer.

First of all, the limit of $$\hat{\boldsymbol\beta}_\lambda^* = \arg\min\Big\{\|\mathbf y - \mathbf X \boldsymbol \beta\|^2+\lambda\|\boldsymbol\beta\|^2\Big\} \:\:\text{s.t.}\:\: \|\mathbf X \boldsymbol\beta\|^2=1$$ when $\lambda\to\infty$ is very simple to obtain: in the limit, the first term in the loss function becomes negligible and can thus be disregarded. The optimization problem becomes $$\lim_{\lambda\to\infty}\hat{\boldsymbol\beta}_\lambda^* = \hat{\boldsymbol\beta}_\infty^* = \operatorname*{arg\,min}_{\|\mathbf X \boldsymbol\beta\|^2=1}\|\boldsymbol\beta\|^2 \sim \operatorname*{arg\,max}_{\| \boldsymbol\beta\|^2=1}\|\mathbf X\boldsymbol\beta\|^2,$$ which is the first principal component of $\mathbf X$ (appropriately scaled). This answers the question.

Now let us consider the solution for any value of $\lambda$ that I referred to in point #2 of my question. Adding to the loss function the Lagrange multiplier $\mu(\|\mathbf X\boldsymbol\beta\|^2-1)$ and differentiating, we obtain

$$\hat{\boldsymbol\beta}_\lambda^*=\big((1+\mu)\mathbf X^\top \mathbf X + \lambda \mathbf I\big)^{-1}\mathbf X^\top \mathbf y\:\:\text{with \mu needed to satisfy the constraint}.$$

How does this solution behave when $\lambda$ grows from zero to infinity?

• When $\lambda=0$, we obtain a scaled version of the OLS solution: $$\hat{\boldsymbol\beta}_0^* \sim \hat{\boldsymbol\beta}_0.$$

• For positive but small values of $\lambda$, the solution is a scaled version of some ridge estimator: $$\hat{\boldsymbol\beta}_\lambda^* \sim \hat{\boldsymbol\beta}_{\lambda^*}.$$

• When $\lambda=\|\mathbf X\mathbf X^\top \mathbf y\|$, the value of $(1+\mu)$ needed to satisfy the constraint is $0$. This means that the solution is a scaled version of the first PLS component (meaning that $\lambda^*$ of the corresponding ridge estimator is $\infty$): $$\hat{\boldsymbol\beta}_{\|\mathbf X\mathbf X^\top \mathbf y\|}^* \sim \mathbf X^\top \mathbf y.$$

• When $\lambda$ becomes larger than that, the necessary $(1+\mu)$ term becomes negative. From now on, the solution is a scaled version of a pseudo-ridge estimator with negative regularization parameter (negative ridge). In terms of directions, we are now past ridge regression with infinite lambda.

• When $\lambda\to\infty$, the term $\big((1+\mu)\mathbf X^\top \mathbf X + \lambda \mathbf I\big)^{-1}$ would go to zero (or diverge to infinity) unless $\mu = -\lambda/ s^2_\mathrm{max} + \alpha$ where $s_\mathrm{max}$ is the largest singular value of $\mathbf X=\mathbf{USV}^\top$. This will make $\hat{\boldsymbol\beta}_\lambda^*$ finite and proportionate to the first principal axis $\mathbf V_1$. We need to set $\mu = -\lambda/ s^2_\mathrm{max} + \mathbf U_1^\top \mathbf y -1$ to satisfy the constraint. Thus, we obtain that $$\hat{\boldsymbol\beta}_\infty^* \sim \mathbf V_1.$$

Overall, we see that this constrained minimization problem encompasses unit-variance versions of OLS, RR, PLS, and PCA on the following spectrum:

$$\boxed{\text{OLS} \to \text{RR} \to \text{PLS} \to \text{negative RR} \to \text{PCA}}$$

This seems to be equivalent to an obscure (?) chemometrics framework called "continuum regression" (see https://scholar.google.de/scholar?q="continuum+regression", in particular Stone & Brooks 1990, Sundberg 1993, Björkström & Sundberg 1999, etc.) which allows the same unification by maximizing an ad hoc criterion $$\mathcal T = \operatorname{corr}^2(\mathbf y, \mathbf X \boldsymbol\beta)\cdot \operatorname{Var}^\gamma(\mathbf X\boldsymbol\beta) \;\;\text{s.t.}\;\;\|\boldsymbol\beta\|=1.$$ This obviously yields scaled OLS when $\gamma=0$, PLS when $\gamma=1$, PCA when $\gamma\to\infty$, and can be shown to yield scaled RR for $0<\gamma<1$ and scaled negative RR for $1<\gamma<\infty$, see Sundberg 1993.

Despite having quite a bit of experience with RR/PLS/PCA/etc, I have to admit I have never heard about "continuum regression" before. I should also say that I dislike this term.

A schematic that I did based on the @Martijn's one: Update: Figure updated with the negative ridge path, huge thanks to @Martijn for suggesting how it should look. See my answer in Understanding negative ridge regression for more details.

• "Continuum regression" seems to be one of a surprisingly broad category of techniques aimed at unifying PLS and PCA within a common framework. I had never heard about it, incidentally, until researching negative ridge (I provide a link to the Bjorkstron & Sundberg, 1999, paper in the first comment of the negative ridge question you link to), though it seems to be rather widely discussed in the chemometric literature. There must be some historical reason why it has developed seemingly in isolation from other fields of statistics. (1/3) Apr 16 '18 at 14:46
• One paper you may want to read is de Jong et al. (2001). Their formulation of "canonical PLS" seems on a quick glance to be equivalent to yours, though I admit I haven't rigorously compared the math yet (they also provide a review of several other PLS-PCA generalizations in the same vein). But it may be insightful to see how they have explicated the problem. (2/3) Apr 16 '18 at 14:50
• In case that link dies, the full citation is: Sijmen de Jong, Barry M. Wise, N. Lawrence Ricker. "Canonical partial least squares and continuum power regression." Journal of Chemometrics, 2001; 15: 85-100. doi.org/10.1002/… (3/3) Apr 16 '18 at 15:02
• ah, ok, then $\lambda^*$ and $1+\mu^*$ go to $\pm$ infinity but their ratio remains $s_{max}^2$. In any case, the negative ridge regression path should be in the (negative) sector between the PLS and PCA vectors such that their projection onto the ellipse $|X\beta=1|$ is between the points PLS and PCA. (the norm going to infinity makes sense as the $\mu$ goes to infinity as well, so the path continues to the lower right, initially tangent to, negative, PLS and eventually to PCA) Apr 19 '18 at 14:55
• It would add to the visualization. I imagine the current three RR path points (where the circle and ellipsoid touch) continuing downwards to the right and eventually, at infinity, the circle $|\beta|^2=t_{\infty}$ and ellipsoid $|X (\beta - \hat\beta)|^2 =RSS$ should 'touch' in direction of the the spot where the circle $|\beta|^2=t_{pca}$ touches the ellipsoid $|X \beta|^2 =1$ Apr 19 '18 at 15:07