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58 votes
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Is Tikhonov regularization the same as Ridge Regression?

Tikhonov regularizarization is a larger set than ridge regression. Here is my attempt to spell out exactly how they differ. Suppose that for a known matrix $A$ and vector $b$, we wish to find a ...
Carl's user avatar
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50 votes
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Is regression with L1 regularization the same as Lasso, and with L2 regularization the same as ridge regression? And how to write "Lasso"?

Yes. Yes. LASSO is actually an acronym (least absolute shrinkage and selection operator), so it ought to be capitalized, but modern writing is the lexical equivalent of Mad Max. On the other hand, ...
Sycorax's user avatar
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47 votes
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The proof of shrinking coefficients using ridge regression through "spectral decomposition"

The question appears to ask for a demonstration that Ridge Regression shrinks coefficient estimates towards zero, using a spectral decomposition. The spectral decomposition can be understood as an ...
whuber's user avatar
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46 votes
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If only prediction is of interest, why use lasso over ridge?

You are right to ask this question. In general, when a proper accuracy scoring rule is used (e.g., mean squared prediction error), ridge regression will outperform lasso. Lasso spends some of the ...
Frank Harrell's user avatar
45 votes

Why do we only see $L_1$ and $L_2$ regularization but not other norms?

In addition to @whuber's comments (*). The book by Hastie et al Statistical learning with Sparsity discusses this. They also uses what is called the $L_0$ "norm" (quotation marks because this is not ...
kjetil b halvorsen's user avatar
38 votes
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Is ridge regression useless in high dimensions ($n \ll p$)? How can OLS fail to overfit?

A natural regularization happens because of the presence of many small components in the theoretical PCA of $x$. These small components are implicitly used to fit the noise using small coefficients. ...
Benoit Sanchez's user avatar
38 votes
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What are the implications of scaling the features to xgboost?

XGBoost is not sensitive to monotonic transformations of its features for the same reason that decision trees and random forests are not: the model only needs to pick "cut points" on features to split ...
Sycorax's user avatar
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36 votes

L1 regression estimates median whereas L2 regression estimates mean?

This explanation is a summation of muratoa and Yves's comments on D.W.'s answer. Though it is based on calculus, I found it straightforward and easy to understand. Assuming we have $y_1, y_2, ... y_k$...
chefwen's user avatar
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34 votes

Why is glmnet ridge regression giving me a different answer than manual calculation?

The difference you are observing is due to the additional division by the number of observations, N, that GLMNET uses in their objective function and implicit standardization of Y by its sample ...
skijunkie's user avatar
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34 votes
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Why Lasso or ElasticNet perform better than Ridge when the features are correlated

Suppose you have two highly correlated predictor variables $x,z$, and suppose both are centered and scaled (to mean zero, variance one). Then the ridge penalty on the parameter vector is $\beta_1^2 + ...
kjetil b halvorsen's user avatar
33 votes

Why L1 norm for sparse models

The Figure 3.11 from Elements of Statistical Learning by Hastie, Tibshirani, and Friedman is very illustrative: Explanations: The $\hat{\beta}$ is the unconstrained least squares estimate. The red ...
Zhanxiong's user avatar
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32 votes

Ridge, lasso and elastic net

To summarize, here are some salient differences between Lasso, Ridge and Elastic-net: Lasso does a sparse selection, while Ridge does not. When you have highly-correlated variables, Ridge regression ...
balaks's user avatar
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32 votes
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In regression, why not use regularization by default?

In short, regularization changes the distribution of the test statistic, rendering tests of hypothesis moot. In instances where we want to use regression to make inferences about interventions, we ...
Demetri Pananos's user avatar
27 votes
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Why regularization parameter called as lambda in theory and alpha in python?

No difference. It's just a symbol. Sometimes mathematics uses symbols by convention, but there's no rule or requirement that you must use a certain symbol for a concept. In this particular case, the ...
Sycorax's user avatar
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25 votes

Is Tikhonov regularization the same as Ridge Regression?

Carl has given a thorough answer that nicely explains the mathematical differences between Tikhonov regularization vs. ridge regression. Inspired by the historical discussion here, I thought it might ...
GeoMatt22's user avatar
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25 votes

Is ridge regression useless in high dimensions ($n \ll p$)? How can OLS fail to overfit?

Thanks everybody for the great ongoing discussion. The crux of the matter seems to be that minimum-norm OLS is effectively performing shrinkage that is similar to the ridge regression. This seems to ...
amoeba's user avatar
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24 votes
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Bridge penalty vs. Elastic Net regularization

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 ...
user20160's user avatar
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24 votes
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LASSO and ridge from the Bayesian perspective: what about the tuning parameter?

Penalized regression estimators such as LASSO and ridge are said to correspond to Bayesian estimators with certain priors. Yes, that is correct. Whenever we have an optimisation problem involving ...
Ben's user avatar
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23 votes

Is multicollinearity really a problem?

It's a problem for causal inference - or rather, it indicates difficulties in causal inference - but it's not a particular problem for prediction/forecasting (unless it's so extreme that it prevents ...
DHW's user avatar
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23 votes
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The proof of equivalent formulas of ridge regression

The classic Ridge Regression (Tikhonov Regularization) is given by: $$ \arg \min_{x} \frac{1}{2} {\left\| x - y \right\|}_{2}^{2} + \lambda {\left\| x \right\|}_{2}^{2} $$ The claim above is that ...
Royi's user avatar
  • 1,151
22 votes

Why will ridge regression not shrink some coefficients to zero like lasso?

Ridge Regression $L_{2}=(y-x\beta)^2+\lambda\sum_{i=1}^{p}\beta_{i}^2$ Will solve this equation only for one β for now and latter you can generalize this: So, $(y-xβ)^2+λβ^2$ this is our equation ...
Chetan Patil's user avatar
21 votes
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Proving Ridge Regression is strictly convex

"you can prove a function is strictly convex if the 2nd derivative is strictly greater than 0" That's in one dimension. A multivariate twice-differentiable function is convex iff the 2nd ...
Thomas Lumley's user avatar
19 votes

When to use Ridge regression and Lasso regression. What can be achieved while using these techniques rather than the linear regression model

In short, ridge regression and lasso are regression techniques optimized for prediction, rather than inference. Normal regression gives you unbiased regression coefficients (maximum likelihood ...
mzunhammer's user avatar
  • 1,188
19 votes

Why do we only see $L_1$ and $L_2$ regularization but not other norms?

I think the answer to the question depends a lot on how you define "better." If I'm interpreting right, you want to know why it is that these norms appear so frequently as compared to other options. ...
RedPanda's user avatar
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19 votes
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The limit of "unit-variance" ridge regression estimator when $\lambda\to\infty$

#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 ...
Sextus Empiricus's user avatar
19 votes
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Why lasso for feature selection?

First, be careful in specifying what you mean by "the most important features" in a dataset. See this page for different perspectives on this issue. For example, features that are deemed "unimportant" ...
EdM's user avatar
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18 votes

Is ridge regression useless in high dimensions ($n \ll p$)? How can OLS fail to overfit?

Here is an artificial situation where this occurs. Suppose each predictor variable is a copy of the target variable with a large amount of gaussian noise applied. The best possible model is an average ...
Jonny Lomond's user avatar
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17 votes
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Why are confidence intervals and p-values not reported as default for penalized regression coefficients

Little late to the party, but in case anyone stumbles across this question in the future. . . . Best answer: have a look at section 6 of the vignette for the penalized R package ("L1 and L2 ...
eac2222's user avatar
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17 votes

Understanding negative ridge regression

Here is a geometric illustration of what is going on with negative ridge. I will consider estimators of the form $$\hat{\boldsymbol\beta}_\lambda = (\mathbf X^\top \mathbf X + \lambda \mathbf I)^{-1}\...
amoeba's user avatar
  • 105k
16 votes

If only prediction is of interest, why use lasso over ridge?

I think the specific setup of the example you reference is key to understanding why lasso outperforms ridge: only 2 of 45 predictors are actually relevant. This borders on a pathological case: lasso,...
mbrig's user avatar
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