Yes, regularization penalizes models that are more complex than needed. But does it also penalize models that are simpler than needed?

  • 2
    $\begingroup$ Given we use an appropriate testing procedure to select our regularisation parameter strength, it should not penalise any models unnecessarily. (+1) $\endgroup$
    – usεr11852
    Mar 31, 2019 at 12:04

1 Answer 1


For regularization terms similar to $\left\|\theta\right\|_2^2$ in effect, no they don't, they only push toward simplicity, i.e. parameters closer to zero.

Error terms such as $\sum_i \left\|y_i - f_{\theta}(x_i)\right\|_2^2$ are responsible for fighting back toward complexity (penalizing over-simplification), since the simplest model, i.e. $\theta = 0$, leads to a high error.

We balance these two forces by using a regularization parameter ($\lambda$) in a summation like $$\frac{1}{N}\sum_{i=1}^{N} \left\|y_i - f_{\theta}(x_i)\right\|_2^2 + \lambda\left\|\theta\right\|_2^2,$$ where higher $\lambda$ forces the model toward more simplicity.

  • $\begingroup$ So, regularization like L2, L1 correspond to the first case, right? $\endgroup$
    – alienflow
    Mar 31, 2019 at 12:05
  • 1
    $\begingroup$ @alienflow yes they all force toward zero (most simple). $\endgroup$
    – Esmailian
    Mar 31, 2019 at 12:06

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