The LASSO problem works by minimizing

$$\min_\beta (\frac{1}{2}\left\rVert y-X\beta\right\rVert^2_2+\lambda\left\rVert\beta\right\rVert_1)$$

Here in this webpage I found that the minimal value of the $\lambda$ parameter that makes $0$ all the $\beta$ coefficients in the model is computed as $\lambda=\left\lVert X^ty\right\rVert_\infty$, so any value of $\lambda$ greater than this will send all the $\beta$ coefficients to zero.

On the other side, if our variables are grouped into $m$ groups of size $p_l$ the group LASSO works by solving:

$$\min_{\vec{\beta}}\left\{\frac{1}{2} \left\lVert\vec{y}-\sum_{l=1}^mX^{(l)}\vec{\beta^{(l)}}\right\rVert_2^2 +\lambda\sum_{l=1}^m\sqrt{p_l}\left\lVert\vec{\beta^{(l)}}\right\rVert_2\right\},$$

Is there a way to find the minimal value for the $\lambda$ parameter in the case of the group LASSO? Consider for instance the gglasso package and the BostonHousing dataset. I will consider a random grouping structure for the data:

library(mlbench, gglasso)
x <- data.matrix(BostonHousing[,-14])
y <- BostonHousing[,14]
index <- c(1,1,2,2,2,2,2,2,2,3,3,3,3)
fit = gglasso(x, y, loss="ls", nlambda=5, intercept=TRUE, group = index)
> 389.1944
> crim      zn   indus    chas     nox      rm     age     dis     rad     tax ptratio       b   lstat 
   0       0       0       0       0       0       0       0       0       0       0       0       0 

I would really appreciate any insight, article or book that shows how to find this minimal lambda, and that demonstrates why is it computed that way, not just for the group LASSO but also for the LASSO.

  • $\begingroup$ I gather you mean $\left\lVert\vec{\beta^{(l)}}\right\rVert_1$ and not $\left\lVert\vec{\beta^{(l)}}\right\rVert_2$ in the latter sum, right? $\endgroup$ Commented Apr 14, 2018 at 10:15
  • 2
    $\begingroup$ No, I meant using norm2. If the group $l$ in question has size 1 (this means, is a group of just one element, norm2 is actually norm1, but if the size of the group is bigger than 1, then norm2 is used so that whole group is included or excluded from the model. It is the usual definition of the group LASSO penalization. $\endgroup$ Commented Apr 14, 2018 at 10:21
  • $\begingroup$ Sorry, you are right. The L2 norm is used in the group Lasso. $\endgroup$ Commented Apr 14, 2018 at 10:29

1 Answer 1


If a change of $\beta$ in any direction will not decrease the cost/objective function then you have found yourself in, at least, a local minimum. The calculations below will show for which λ the solution/point $\beta = 0$ stops to be a minimum.

Consider the effect of 'a change of $\beta^{(l)}$ by an infinitesimal distance $\partial l$' on the 'change of 1) the error/residual term and 2) the penalty term'.

$$\underbrace{ \frac{1}{2} \left\lVert\vec{y}-\sum_{l=1}^mX^{(l)}\vec{\beta^{(l)}}\right\rVert_2^2}_{\text{RSS term}} + \underbrace { \lambda\sum_{l=1}^m\sqrt{p_l}\left\lVert\vec{\beta^{(l)}}\right\rVert_2}_{\text{ penalty term}}$$

  • The penalty term will change by: $$\partial \left( \lambda \sqrt{p_l} \lVert\vec{\beta^{(l)}}\rVert_2 \right) = \left( \lambda \sqrt{p_l} \right) \partial l$$ (independent on the direction of change)
  • The error term will change by: $$\partial \frac{1}{2} RSS = \left( \lvert {X^{(l)}}^Ty \rvert \right) \partial l $$ where, if all beta are zero, then this term ${X^{(l)}}^Ty$ is the gradient of the RSS term. This gradient is the direction in which the directional derivative, change/minimization, will be greatest, and it's value is $\lvert {X^{(l)}}^Ty \rvert$ .

So per group the relative change of error term and penalty term (which needs to be greater than 1 or otherwise the overall cost term does not decrease) will be:

$$ \frac{ \lvert {X^{(l)}}^Ty \rvert } { \lambda \sqrt{p_l} } > 1$$


$$ \lambda_{min} = \underset{l}{max} \left(\frac{ \lvert {X^{(l)}}^Ty \rvert } { \sqrt{p_l} } \right) $$

which becomes $\lambda=\left\lVert X^ty\right\rVert_\infty$ if all group sizes $p_l$ are equal to one.

Note that, initially, a change of $\beta^{(l)}$ in two groups together is not beneficial. You could always improve the solution by shifting more weight to the group with a higher ratio for $\frac{ \lvert {X^{(l)}}^Ty \rvert } { \lambda \sqrt{p_l} }$.

This is most easily/intuitively seen in a geometrical viewpoint. The shape of iso-surface for beta is a polytope which makes contact with the iso-surface for the error which is a ellipsoid. They will initially contact in a point of the polytope (see for instance the graphical views in this answer or this answer ).

Instead of $\lambda_{min}$ it might be better to speak about the suppremum (the smallest upper bound).

We have for all $\lambda$ with non-zero $\beta$ that $\lambda < \lambda_{min}$

$$ \lambda < \underset{l}{max} \left(\frac{ \lvert {X^{(l)}}^Ty \rvert } { \sqrt{p_l} } \right) $$

  • $\begingroup$ Thank you for your answer, although I am afraid I dont understand all of what you are saying. When you say "a change of β(l) by an infinitesimal distance ∂l", do you mean taking the derivatives of the objective function over beta vector? I also dont understand how do you get that error term change. And my final question: The lambda should be a real number, but the lambda value you indicate is a vector, since t(X)*y has dimension l*1, am I right? $\endgroup$ Commented Apr 15, 2018 at 16:09
  • $\begingroup$ @ÁlvaroMéndezCivieta, I changed the vector thingy. Regarding the meaning of "a change of β(l) by an infinitesimal distance ∂l" you could see any introduction to least angle regression. The two links at the end of the question might help. (But to be quick, maybe the following comment may help: if a change of beta in any direction will not decrease the objective function then you have found yourself in, at least, a local minimum. The calculations above show for which $\lambda$ the point 0 stops to be a minimum.) $\endgroup$ Commented Apr 15, 2018 at 18:17
  • $\begingroup$ @LiangStats the inequality is, I believe, right. See my edit. $\endgroup$ Commented Apr 9, 2020 at 7:01

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