Some penalty functions and approximations are well studied, such as the LASSO ($L_1$) and the Ridge ($L_2$) and how these compare in regression.

I've been reading about the Bridge penalty, which is the $\sum \|\beta_{j}\|^{\gamma}$ generalized penalty. Compare that to the LASSO, which has $\gamma = 1$, and the Ridge, with $\gamma = 2$, making them special cases.

Wenjiang [1] compared the Bridge penalty when $\gamma \geq 1$ to the LASSO, but I couldn't find a comparison to the Elastic Net regularization, a combination of the LASSO and the Ridge penalties, given as $\sum \lambda_{2} \|\beta\|^{2}+\lambda_{1}\|\beta\|_{1}$.

This is an interesting question because the Elastic Net and this specific Bridge have similar constraint forms. Compare these unit circles using the different metrics ($p$ is the power of the Minkowski distance):

Unit circles in for different powers of the Minkowski distance

$p = 1$ corresponds to the LASSO, $p = 2$ to the Ridge, and $p = 1.4$ to one possible Bridge. The Elastic Net was generated with equal weighting on $L_1$ and $L_2$ penalties. These figures are useful to identify sparsity, for example (which Bridge clearly lacks while Elastic Net preserves it from LASSO).

So how does the Bridge with $1<\gamma <2$ compares to Elastic Net regarding regularization (other than sparsity)? I have special interest in supervised learning, so perhaps a discussion about feature selection/weighting is pertinent. Geometric argumentation is welcome as well.

Perhaps, more important, is the Elastic Net always more desirable in this case?

[1] Fu, W. J. (1998). Penalized regressions: the bridge versus the lasso. Journal of computational and graphical statistics, 7(3), 397-416.

EDIT: There's this question How to decide which penalty measure to use ? any general guidelines or thumb rules out of textbook which superficially mentions LASSO, Ridge, Bridge and Elastic Net, but there are no attempts to compare them.

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    $\begingroup$ Only tangentially related, but if the $L_1$ norm penalty is the MAP estimate of a Bayesian regression with independent Laplace priors on coefficients, and the $L_2$ is the same for Gaussian priors, I wonder if the Bridge penalty is equivalent to a Subbotin prior... stats.stackexchange.com/questions/201038/… $\endgroup$
    – Sycorax
    Commented Jul 19, 2016 at 14:41
  • $\begingroup$ @RichardHardy There is no need to write lasso in all capitals, see my comment here. $\endgroup$
    – amoeba
    Commented Jul 19, 2016 at 17:44
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    $\begingroup$ Keep in mind that Bridge regression allows for $\gamma<1$ which gives a non-convex regression. These are nice in particular when trying to select groups of covariates, especially from sparse data. Or in general you could have pre-defined groups of covariates, which you would $L^2$ regularize so that no particular group is large, and then $L^1$ regularize single group coefficients to achieve sparsity. I.e. if you write $\beta=(a_1,\cdots,a_k)$, where $a_i=(\beta_{i_1},\beta_{i_2,},\cdots,\beta_{i_r})$ then you could do $\lambda_1 \|\beta\|^{\gamma_i}+\lambda_2\sum_i \|a_i\|^{\nu_i}$. $\endgroup$
    – Alex R.
    Commented Jul 19, 2016 at 18:19
  • $\begingroup$ @AlexR. I should actually make it clear I refer to $\gamma \geq 1$. I didn't know $\gamma < 1$ was also called Bridge. $\endgroup$
    – Firebug
    Commented Jul 19, 2016 at 18:38
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    $\begingroup$ @amoeba, OK, fine. I normally do not edit if the use of capitals is consistent throughout the post, but this time there were both "LASSO" and "lasso", so I just went for "LASSO" which was the first form in the post. I always think about the acronym, that is why I used all capitals; but as you say, simple "lasso" may be better. $\endgroup$ Commented Jul 20, 2016 at 6:12

1 Answer 1


How bridge regression and elastic net differ is a fascinating question, given their similar-looking penalties. Here's one possible approach. Suppose we solve the bridge regression problem. We can then ask how the elastic net solution would differ. Looking at the gradients of the two loss functions can tell us something about this.

Bridge regression

Say $X$ is a matrix containing values of the independent variable ($n$ points x $d$ dimensions), $y$ is a vector containing values of the dependent variable, and $w$ is the weight vector.

The loss function penalizes the $\ell_q$ norm of the weights, with magnitude $\lambda_b$:

$$ L_b(w) = \| y - Xw\|_2^2 + \lambda_b \|w\|_q^q $$

The gradient of the loss function is:

$$ \nabla_w L_b(w) = -2 X^T (y - Xw) + \lambda_b q |w|^{\circ(q-1)} \text{sgn}(w) $$

$v^{\circ c}$ denotes the Hadamard (i.e. element-wise) power, which gives a vector whose $i$th element is $v_i^c$. $\text{sgn}(w)$ is the sign function (applied to each element of $w$). The gradient may be undefined at zero for some values of $q$.

Elastic net

The loss function is:

$$ L_e(w) = \|y - Xw\|_2^2 + \lambda_1 \|w\|_1 + \lambda_2 \|w\|_2^2 $$

This penalizes the $\ell_1$ norm of the weights with magnitude $\lambda_1$ and the $\ell_2$ norm with magnitude $\lambda_2$. The elastic net paper calls minimizing this loss function the 'naive elastic net' because it doubly shrinks the weights. They describe an improved procedure where the weights are later rescaled to compensate for the double shrinkage, but I'm just going to analyze the naive version. That's a caveat to keep in mind.

The gradient of the loss function is:

$$ \nabla_w L_e(w) = -2 X^T (y - Xw) + \lambda_1 \text{sgn}(w) + 2 \lambda_2 w $$

The gradient is undefined at zero when $\lambda_1 > 0$ because the absolute value in the $\ell_1$ penalty isn't differentiable there.


Say we select weights $w^*$ that solve the bridge regression problem. This means the the bridge regression gradient is zero at this point:

$$ \nabla_w L_b(w^*) = -2 X^T (y - Xw^*) + \lambda_b q |w^*|^{\circ (q-1)} \text{sgn}(w^*) = \vec{0} $$


$$ 2 X^T (y - Xw^*) = \lambda_b q |w^*|^{\circ (q-1)} \text{sgn}(w^*) $$

We can substitute this into the elastic net gradient, to get an expression for the elastic net gradient at $w^*$. Fortunately, it no longer depends directly on the data:

$$ \nabla_w L_e(w^*) = \lambda_1 \text{sgn}(w^*) + 2 \lambda_2 w^* -\lambda_b q |w^*|^{\circ (q-1)} \text{sgn}(w^*) $$

Looking at the elastic net gradient at $w^*$ tells us: Given that bridge regression has converged to weights $w^*$, how would the elastic net want to change these weights?

It gives us the local direction and magnitude of the desired change, because the gradient points in the direction of steepest ascent and the loss function will decrease as we move in the direction opposite to the gradient. The gradient might not point directly toward the elastic net solution. But, because the elastic net loss function is convex, the local direction/magnitude gives some information about how the elastic net solution will differ from the bridge regression solution.

Case 1: Sanity check

($\lambda_b = 0, \lambda_1 = 0, \lambda_2 = 1$). Bridge regression in this case is equivalent to ordinary least squares (OLS), because the penalty magnitude is zero. The elastic net is equivalent ridge regression, because only the $\ell_2$ norm is penalized. The following plots show different bridge regression solutions and how the elastic net gradient behaves for each.

enter image description here

Left plot: Elastic net gradient vs. bridge regression weight along each dimension

The x axis represents one component of a set of weights $w^*$ selected by bridge regression. The y axis represents the corresponding component of the elastic net gradient, evaluated at $w^*$. Note that the weights are multidimensional, but we're just looking at the weights/gradient along a single dimension.

Right plot: Elastic net changes to bridge regression weights (2d)

Each point represents a set of 2d weights $w^*$ selected by bridge regression. For each choice of $w^*$, a vector is plotted pointing in the direction opposite the elastic net gradient, with magnitude proportional to that of the gradient. That is, the plotted vectors show how the elastic net wants to change the bridge regression solution.

These plots show that, compared to bridge regression (OLS in this case), elastic net (ridge regression in this case) wants to shrink weights toward zero. The desired amount of shrinkage increases with the magnitude of the weights. If the weights are zero, the solutions are the same. The interpretation is that we want to move in the direction opposite to the gradient to reduce the loss function. For example, say bridge regression converged to a positive value for one of the weights. The elastic net gradient is positive at this point, so elastic net wants to decrease this weight. If using gradient descent, we'd take steps proportional in size to the gradient (of course, we can't technically use gradient descent to solve the elastic net because of the non-differentiability at zero, but subgradient descent would give numerically similar results).

Case 2: Matching bridge & elastic net

($q = 1.4, \lambda_b = 1, \lambda_1 = 0.629, \lambda_2 = 0.355$). I chose the bridge penalty parameters to match the example from the question. I chose the elastic net parameters to give the best matching elastic net penalty. Here, best-matching means, given a particular distribution of weights, we find the elastic net penalty parameters that minimize the expected squared difference between the bridge and elastic net penalties:

$$ \min_{\lambda_1, \lambda_2} \enspace E \left [ ( \lambda_1 \|w\|_1 + \lambda_2 \|w\|_2^2 - \lambda_b \|w\|_q^q )^2 \right ] $$

Here, I considered weights with all entries drawn i.i.d. from the uniform distribution on $[-2, 2]$ (i.e. within a hypercube centered at the origin). The best-matching elastic net parameters were similar for 2 to 1000 dimensions. Although they don't appear to be sensitive to the dimensionality, the best-matching parameters do depend on the scale of the distribution.

Penalty surface

Here's a contour plot of the total penalty imposed by bridge regression ($q=1.4, \lambda_b=100$) and best-matching elastic net ($\lambda_1 = 0.629, \lambda_2 = 0.355$) as a function of the weights (for the 2d case):

enter image description here

Gradient behavior

enter image description here

We can see the following:

  • Let $w^*_j$ be the chosen bridge regression weight along dimension $j$.
  • If $|w^*_j|< 0.25$, elastic net wants to shrink the weight toward zero.
  • If $|w^*_j| \approx 0.25$, the bridge regression and elastic net solutions are the same. But, elastic net wants to move away if the weight differs even slightly.
  • If $0.25 < |w^*_j| < 1.31$, elastic net wants to grow the weight.
  • If $|w^*_j| \approx 1.31$, the bridge regression and elastic net solutions are the same. Elastic net wants to move toward this point from nearby weights.
  • If $|w^*_j| > 1.31$, elastic net wants to shrink the weight.

The results are qualitatively similar if we change the the value of $q$ and/or $\lambda_b$ and find the corresponding best $\lambda_1, \lambda_2$. The points where the bridge and elastic net solutions coincide change slightly, but the behavior of the gradients are otherwise similar.

Case 3: Mismatched bridge & elastic net

$(q=1.8, \lambda_b=1, \lambda_1=0.765, \lambda_2 = 0.225)$. In this regime, bridge regression behaves similar to ridge regression. I found the best-matching $\lambda_1, \lambda_2$, but then swapped them so that the elastic net behaves more like lasso ($\ell_1$ penalty greater than $\ell_2$ penalty).

enter image description here

Relative to bridge regression, elastic net wants to shrink small weights toward zero and increase larger weights. There's a single set of weights in each quadrant where the bridge regression and elastic net solutions coincide, but elastic net wants to move away from this point if the weights differ even slightly.

$(q=1.2, \lambda_b=1, \lambda_1=173, \lambda_2 = 0.816)$. In this regime, the bridge penalty is more similar to an $\ell_1$ penalty (although bridge regression may not produce sparse solutions with $q > 1$, as mentioned in the elastic net paper). I found the best-matching $\lambda_1, \lambda_2$, but then swapped them so that the elastic net behaves more like ridge regression ($\ell_2$ penalty greater than $\ell_1$ penalty).

enter image description here

Relative to bridge regression, elastic net wants to grow small weights and shrink larger weights. There's a point in each quadrant where the bridge regression and elastic net solutions coincide, and elastic net wants to move toward these weights from neighboring points.

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    $\begingroup$ (+1) Great answer, thanks for the effort! Could you address one last thing: "is the Elastic Net always more desirable?". No need to be lengthy; $\endgroup$
    – Firebug
    Commented Jul 28, 2016 at 11:46
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    $\begingroup$ Bridge regression and elastic net are equivalent to MAP estimation with different kinds of priors on the weights. From this perspective, it seems like the better choice would be the prior that better matches the data-generating process, and that neither method could be better in all cases. $\endgroup$
    – user20160
    Commented Jul 29, 2016 at 5:07
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    $\begingroup$ +6, very nice answer. Regarding your above comment: what prior yields bridge regression? I know that Gaussian prior corresponds to ridge and Laplace prior to lasso. Can one somehow combine these priors to get something that corresponds to elastic net? $\endgroup$
    – amoeba
    Commented Jul 31, 2016 at 22:07
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    $\begingroup$ @amoeba The question wasn't directed to me, I know, but as GeneralAbrial said in the question, bridge probably corresponds to a Subbotin prior. Elastic net, as expected, is between Gaussian and Laplacian priors. See Li, Q., & Lin, N. (2010). The Bayesian elastic net. Bayesian Analysis, 5(1), 151-170. and Zou, H., & Hastie, T. (2005). Regularization and variable selection via the elastic net. Journal of the Royal Statistical Society: Series B (Statistical Methodology), 67(2), 301-320. for a brief comparison between elastic net and bridge regression. $\endgroup$
    – Firebug
    Commented Aug 1, 2016 at 19:22
  • 2
    $\begingroup$ @amoeba thanks for the bounty and drawing attention to this post, likewise for the other post about PCA vs. nonlinear dimensionality reduction. It's admirable that you use your rep to promote others' questions/answers, and it makes me glad if this post is at least of some small value to people. Others, thanks too for the kind words. $\endgroup$
    – user20160
    Commented Aug 3, 2016 at 2:22

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