What is the smallest $\lambda$ that gives a 0 component in lasso?

Define the lasso estimate $$\hat\beta^\lambda = \arg\min_{\beta \in \mathbb{R}^p} \frac{1}{2n} \|y - X \beta\|_2^2 + \lambda \|\beta\|_1,$$ where the $i^{th}$ row $x_i \in \mathbb{R}^p$ of the design matrix $X \in \mathbb{R}^{n \times p}$ is a vector of covariates for explaining the stochastic response $y_i$ (for $i=1, \dots n$).

We know that for $\lambda \geq \frac{1}{n} \|X^T y\|_\infty$, the lasso estimate $\hat\beta^\lambda = 0$. (See, for instance, Lasso and Ridge tuning parameter scope.) In other notation, this is expressing that $\lambda_\max = \frac{1}{n} \|X^T y\|_\infty$. Notice that $\lambda_\mathrm{max} = \sup_{\hat\beta^\lambda \ne 0} \lambda.$ We can see this visually with the following image displaying the lasso solution path:

Notice that on the far right hand side of the plot, all of the coefficients are zero. This happens at the point $\lambda_\mathrm{max}$ described above.

From this plot, we also notice that on the far left side, all of the coefficient are nonzero: what is the value of $\lambda$ at which any component of $\hat\beta^\lambda$ is initially zero? That is, what is $$\lambda_\textrm{min} = \min_{\exists j \, \mathrm{ s.t. } \, \hat\beta_j = 0} \lambda$$ equal to, as a function of $X$ and $y$? I'm interested in a closed form solution. In particular, I'm not interested in an algorithmic solution, such as, for instance, suggesting that LARS could find the knot through computation.

Despite my interests, it seems like $\lambda_\mathrm{min}$ may not be available in closed form, since, otherwise, lasso computational packages would likely take advantage of it when determining the tuning parameter depth during cross validation. In light of this, I'm interested in anything that can be theoretically shown about $\lambda_\mathrm{min}$ and (still) particularly interested in a closed form.

• This is stated and proven in the glmnet paper: web.stanford.edu/~hastie/Papers/glmnet.pdf – Matthew Drury Jul 6 '17 at 4:59
• @MatthewDrury Thanks for sharing this! However, this paper doesn't seem to share what you seem to suggest they do. In particular, notice that my $\lambda_\max$ is their $\lambda_\min$. – user795305 Jul 6 '17 at 12:06
• Are you sure we need [tuning-parameter] tag? – amoeba Jul 6 '17 at 12:47
• you a right, a closed form for the lasso solution does in general not exist (see stats.stackexchange.com/questions/174003/…). however, lars at least tells you whats going on and under which exact conditions/at which time you can add/delete a variable. i think something like this is the best you can get. – chRrr Jul 7 '17 at 8:14
• @chRrr I'm not sure that's completely fair to say: we know that $\hat\beta^\lambda = 0$ for $\lambda \geq \frac{1}{n} \|X^t y\|_\infty$. That is, in the extreme case of the solution being 0, we have a closed form. I'm asking if similar is true in the extreme case of the lasso estimate being dense (ie no zeros). Indeed, I'm not even interested in the exact entries of $\hat\beta_\lambda$---just whether they're zero or not. – user795305 Jul 7 '17 at 12:24

The lasso estimate described in the question is the lagrange multiplier equivalent of the following optimization problem:

$${\text{minimize } f(\beta) \text{ subject to } g(\beta) \leq t}$$

\begin{align} f(\beta) &= \frac{1}{2n} \vert\vert y-X\beta \vert\vert_2^2 \\ g(\beta) &= \vert\vert \beta \vert\vert_1 \end{align}

This optimizazion has a geometric representation of finding the point of contact between a multidimensional sphere and a polytope (spanned by the vectors of X). The surface of the polytope represents $g(\beta)$. The square of the radius of the sphere represents the function $f(\beta)$ and is minimized when the surfaces contact.

The images below provides a graphical explanation. The images made use of the following simple problem with vectors of length 3 (for simplicity in order to be able to make a drawing):

$$\begin{bmatrix} y_1 \\ y_2 \\ y_3\\ \end{bmatrix} = \begin{bmatrix} 1.4 \\ 1.84 \\ 0.32\\ \end{bmatrix} = \beta_1 \begin{bmatrix} 0.8 \\ 0.6 \\ 0\\ \end{bmatrix} +\beta_2 \begin{bmatrix} 0 \\ 0.6 \\ 0.8\\ \end{bmatrix} +\beta_3 \begin{bmatrix} 0.6 \\ 0.64 \\ -0.48\\ \end{bmatrix} + \begin{bmatrix} \epsilon_1 \\ \epsilon_2 \\ \epsilon_3\\ \end{bmatrix}$$ and we minimize $\epsilon_1^2+\epsilon_2^2+\epsilon_3^2$ with the constraint $abs(\beta_1)+abs(\beta_2)+abs(\beta_3) \leq t$

The images show:

• The red surface depicts the constraint, a polytope spanned by X.
• And the green surface depicts the minimalized surface, a sphere.
• The blue line shows the lasso path, the solutions that we find as we change $t$ or $\lambda$.
• The green vector shows the OLS solution $\hat{y}$ (which was chosen as $\beta_1=\beta_2=\beta_3=1$ or $\hat{y} = x_1 + x_2 + x_3$.
• The three black vectors are $x_1 = (0.8,0.6,0)$, $x_2 = (0,0.6,0.8)$ and $x_3 = (0.6,0.64,-0.48)$.

We show three images:

1. In the first image only a point of the polytope is touching the sphere. This image demonstrates very well why the lasso solution is not just a multiple of the OLS solution. The direction of the OLS solution adds stronger to the sum $\vert \beta \vert_1$. In this case only a single $\beta_i$ is non-zero.
2. In the second image a ridge of the polytope is touching the sphere (in higher dimensions we get higher dimensional analogues). In this case multiple $\beta_i$ are non-zero.
3. In the third image a facet tof the polytope is touching the sphere. In this case all the $\beta_i$ are nonzero.

The range of $t$ or $\lambda$ for which we have the first and third cases can be easily calculated due to their simple geometric representation.

Case 1: Only a single $\beta_i$ non-zero

The non-zero $\beta_i$ is the one for which the associated vector $x_i$ has the highest absolute value of the covariance with $\hat{y}$ (this is the point of the parrallelotope which closest to the OLS solution). We can calculate the Lagrange multiplier $\lambda_{max}$ below which we have at least a non-zero $\beta$ by taking the derivative with $\pm\beta_i$ (the sign depending on whether we increase the $\beta_i$ in negative or positive direction ):

$$\frac{\partial ( \frac{1}{2n} \vert \vert y - X\beta \vert \vert_2^2 - \lambda \vert \vert \beta \vert \vert_1 )}{\pm \partial \beta_i} = 0$$

$$\lambda_{max} = \frac{ \left( \frac{1}{2n}\frac{\partial ( \vert \vert y - X\beta \vert \vert_2^2}{\pm \partial \beta_i} \right) }{ \left( \frac{ \vert \vert \beta \vert \vert_1 )}{\pm \partial \beta_i}\right)} = \pm \frac{\partial ( \frac{1}{2n} \vert \vert y - X\beta \vert \vert_2^2}{\partial \beta_i} = \pm \frac{1}{n} x_i \cdot y$$

which equals the $\vert \vert X^Ty \vert \vert_\infty$ mentioned in the comments.

where we should notice that this is only true for the special case in which the tip of the polytope is touching the sphere (so this is not a general solution, although generalization is straightforward).

Case 3: All $\beta_i$ are non-zero.

In this case that a facet of the polytope is touching the sphere. Then the direction of change of the lasso path is normal to the surface of the particular facet.

The polytope has many facets, with positive and negative contributions of the $x_i$. In the case of the last lasso step, when the lasso solution is close to the ols solution, then the contributions of the $x_i$ must be defined by the sign of the OLS solution. The normal of the facet can be defined by taking the gradient of the function $\vert \vert \beta(r) \vert \vert_1$, the value of the sum of beta at the point $r$, which is:

$$n = - \nabla_r ( \vert \vert \beta(r) \vert \vert_1) = -\nabla_r ( \text{sign} (\hat{\beta}) \cdot (X^TX)^{-1}X^Tr ) = -\text{sign} (\hat{\beta}) \cdot (X^TX)^{-1}X^T$$

and the equivalent change of beta for this direction is:

$$\vec{\beta}_{last} = (X^TX)^{-1}X n = -(X^TX)^{-1}X^T [\text{sign} (\hat{\beta}) \cdot (X^TX)^{-1}X^T]$$

which after some algebraic tricks with shifting the transposes ($A^TB^T = [BA]^T$) and distribution of brackets becomes

$$\vec{\beta}_{last} = - (X^TX)^{-1} \text{sign} (\hat{\beta})$$

we normalize this direction:

$$\vec{\beta}_{last,normalized} = \frac{\vec{\beta}_{last}}{\sum \vec{\beta}_{last} \cdot sign(\hat{\beta})}$$

To find the $\lambda_{min}$ below which all coefficients are non-zero. We only have to calculate back from the OLS solution back to the point where one of the coefficients is zero,

$$d = min \left( \frac{\hat{\beta}}{\vec{\beta}_{last,normalized}} \right)\qquad \text{with the condition that } \frac{\hat{\beta}}{\vec{\beta}_{last,normalized}} >0$$

,and at this point we evaluate the derivative (as before when we calculate $\lambda_{max}$). We use that for a quadratic function we have $q'(x) = 2 q(1) x$:

$$\lambda_{min} = \frac{d}{n} \vert \vert X \vec{\beta}_{last,normalized} \vert \vert_2^2$$

Images

a point of the polytope is touching the sphere, a single $\beta_i$ is non-zero:

a ridge (or differen in multiple dimensions) of the polytope is touching the sphere, many $\beta_i$ are non-zero:

a facet of the polytope is touching the sphere, all $\beta_i$ are non-zero:

Code example:

library(lars)
data(diabetes)
y <- diabetes$y - mean(diabetes$y)
x <- diabetes$x # models lmc <- coef(lm(y~0+x)) modl <- lars(diabetes$x, diabetes$y, type="lasso") # matrix equation d_x <- matrix(rep(x[,1],9),length(x[,1])) %*% diag(sign(lmc[-c(1)]/lmc[1])) x_c = x[,-1]-d_x y_c = -x[,1] # solving equation cof <- coefficients(lm(y_c~0+x_c)) cof <- c(1-sum(cof*sign(lmc[-c(1)]/lmc[1])),cof) # alternatively the last direction of change in coefficients is found by: solve(t(x) %*% x) %*% sign(lmc) # solution by lars package cof_m <-(coefficients(modl)[13,]-coefficients(modl)[12,]) # last step dist <- x %*% (cof/sum(cof*sign(lmc[]))) #dist_m <- x %*% (cof_m/sum(cof_m*sign(lmc[]))) #for comparison # calculate back to zero shrinking_set <- which(-lmc[]/cof>0) #only the positive values step_last <- min((-lmc/cof)[shrinking_set]) d_err_d_beta <- step_last*sum(dist^2) # compare modl[4] #all computed lambda d_err_d_beta # lambda last change max(t(x) %*% y) # lambda first change enter code here  note: those last three lines are the most important > modl[4] # all computed lambda by algorithm$lambda
[1] 949.435260 889.315991 452.900969 316.074053 130.130851  88.782430  68.965221  19.981255   5.477473   5.089179
[11]   2.182250   1.310435

> d_err_d_beta       # lambda last change by calculating only last step
xhdl
1.310435
> max(t(x) %*% y)    # lambda first change by max(x^T y)
[1] 949.4353

• Thanks for including the edits! So far in my reading, I'm stuck just past the "case 1" subsection. The result for $\lambda_\max$ derived there is wrong since it doesn't include an absolute value or a maximum. We know further that there's a mistake since in the derivation, there's a sign mistake, a place where differentiability is wrongly assumed, an "arbitrary choice" of $i$ to differentiate with respect to, and an incorrectly evaluated derivative. To be frank, there isn't one "$=$" sign that's valid. – user795305 Oct 22 '17 at 20:26
• I have corrected it with a plus minus sign. The change of the beta can be possitive or negative. Regarding the maximum and "arbitrary choice"... "for which the associated vector $x_i$ has the highest covariance with $\hat{y}$" – Martijn Weterings Oct 22 '17 at 20:55
• Thanks for the update! However, there's still problems. For instance, $\frac{\partial}{\partial \beta_i} \|y - X \beta\|_2^2$ is evaluated incorrectly. – user795305 Oct 22 '17 at 21:00
• If $\beta=0$ then $\frac{\partial}{\partial\beta_i} \vert \vert y - X\beta \vert \vert_2^2$ $$= \frac{\partial \vert \vert y - X\beta \vert \vert_2}{\partial\beta_i} 2 \vert \vert y - X\beta \vert \vert_2$$ $$= \frac{\partial \vert \vert y - s x_i \vert \vert_2}{\partial s} 2 \vert \vert y - X\beta \vert \vert_2$$ $$= 2 cor(x_i,y) \vert \vert x_i \vert \vert_2 \vert \vert y \vert \vert_2$$ $$= 2 x_i \cdot y$$ this correlation enters the equation because,if s=0 then only the change of $s x_i$ tangent to $y$ is changing the length of the vector $y - s x_i$ – Martijn Weterings Oct 22 '17 at 21:18
• Ah, okay, so there's a limit involved in your argument! (You're using both $\beta = 0$ and that a coefficient is nonzero.) Further, the second equality in the line with $\lambda_\max$ is misleading since the sign could change due to the differentiation of the absolute value. – user795305 Oct 22 '17 at 22:00