Skip to main content
3 votes
Accepted

Which is a better form of regularization: lasso (L1) or ridge (L2)?

The following graph shows the constraint region (green), along with contours for Residual sum of squares (red ellipse). These are iso-lines signifying that points on an ellipse have the same RSS. ...
s_bh's user avatar
  • 370
3 votes
Accepted

Why does L1 regularization yield sparse features?

In L1 regularization, the penalty term you compute for every parameter is a function of the absolute value of a given weight (times some regularization factor). Thus, irrespective of whether a weight ...
Daniel B.'s user avatar
  • 825
2 votes

Does L1/L2 Regularization help reach an optimum result faster?

I am not aware of any empirical results regarding this question. But in theory, adding a regularization term shall make the learning task actually even harder, since there is suddenly a second loss ...
Daniel B.'s user avatar
  • 825
2 votes
Accepted

What are the consequences when we multiply, instead of add, a penalty term?

Well let's consider one, Ridge regression. We have 2 terms: the regression loss $L^{pred} = \sum(f(x) - y)^2$, which we can see that it is a sum of squared values, thus $L^{pred} \ge 0$ the ...
Alberto's user avatar
  • 2,632
1 vote
Accepted

Would either $L_1$ or $L_2$ regularisation lower the MSE on the training and test data?

The answer is largely the same whether we consider $\ell_1$ or $\ell_2$ regularisation, so I will just speak generally about regularisation. Mean square error for training data Given some training ...
htl's user avatar
  • 1,010

Only top scored, non community-wiki answers of a minimum length are eligible