# A theoretical explanation why ridge is superior to lasso in non-sparse models

There are several posts about the comparison of lasso vs. ridge. However I didn't find an explanation to my question. My question is why ridge is generating lower prediction errors in cases where the true model is non sparse. I know that lasso is designed to handle sparse models. But Lasso doesn't have to shrink coefficents exactly to zero.

Can someone can give me some intuition why ridge is performing better than lasso in situations, where the coefficients have similar size?

• Sparsity refers rather to the situation in which only few of the predictors considered are actually relevant to predict $Y$, not so much to the size of the conditions mentioned in the last sentence (although these are also important for Lasso's capability to find those "true" variables). – Christoph Hanck Aug 2 '18 at 12:18
• Could you extend your comment with some theory. Its still unclear, why Lasso is inferior in situations where the model is non-sparse as lasso doenst have to shrink coefficients exactly to zero – Leo96 Aug 2 '18 at 13:00