So, I understand why simple linear or logistic regression will have infinite solutions in this case (good answers here and here). But while LASSO will only select n features, Elastic net does not have this limitation. This answer explains how regularization limits the potential solutions to a problem so that building a model can be possible. Is the same concept true of Elastic Net? If regularization limits the possible solutions, then how is the "final" solution chosen from that space?

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    $\begingroup$ Have a look at web.stanford.edu/~hastie/TALKS/enet_talk.pdf $\endgroup$ Commented May 21, 2019 at 19:33
  • $\begingroup$ @GabrielRomon I looked through this but perhaps the section I need went over my head... where in here do they address the problem of selecting more predictors than observations? $\endgroup$ Commented Jun 11, 2019 at 15:36
  • $\begingroup$ A lot is said on page 9. For the details you definitely want to check the original paper. If something in the paper is unclear, don't hesitate to ask. $\endgroup$ Commented Jun 11, 2019 at 16:17
  • $\begingroup$ With penalization the effective p is much lower than the apparent p. $\endgroup$ Commented Sep 30, 2023 at 15:44
  • $\begingroup$ Near duplicate: stats.stackexchange.com/questions/274225/…. $\endgroup$
    – whuber
    Commented Sep 30, 2023 at 16:44

1 Answer 1


One way to look at this is that (as long as $\lambda_2\neq 0$) the L2 penalty is equivalent to adding $p$ examples: $$ \Vert y - X \beta \Vert^2 + \lambda_2 \Vert \beta \Vert^2 = \Vert \tilde y - \tilde X \beta \Vert^2 $$ with $$ \tilde X = \begin{bmatrix}X\\ \sqrt{\lambda_2} I_{p\times p}\end{bmatrix} \quad \tilde y = \begin{bmatrix}y\\ 0_{p\times 1}\end{bmatrix}. $$ So, in general, $n\times p$ ridge regression is equivalent to $(n+p)\times p$ non-regularised regression. Similarly $n\times p$ elastic-net regression is equivalent to $(n+p)\times p$ Lasso regression.


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