In the book 'Pattern Recognition and Machine Learning' by Bishop (p.257 ff.) he considers a weight decay regularizer of the error function $$\hat E(w)=E(w)+\frac{\lambda}{2}w^tw$$

where $w$ is a weight vector corresponding to all of the weights in the network.

He then states that for a rescaling of weights $w\rightarrow w/a$ this regularizer is not invariant.

An invariant regularizer is given by

$$\frac{\lambda_1}{2}\sum_{w\in W_1} w^2+\frac{\lambda_2}{2}\sum_{w\in W_2} w^2$$

where we now only consider one hidden layer with its corresponding weights $W_1$ and the weights for the output layer $W_2$. In this example we only have two layers.

He writes that for a rescaling of the weights $w\rightarrow w/a$ and a simultaneous rescaling of $\lambda_1\rightarrow \sqrt{a}\lambda_1$ this regularizer is invariant. (I don't consider the second term, as it works in the same way, just look at the first sum)

However, if I do the transformation i get for the first sum $$\frac{\lambda_1\sqrt{a}}{2}\frac{1}{a^2}\sum_{w\in W_1}w^2$$

which is clearly not invariant. I cannot find my mistake.


1 Answer 1


Actually, you've got it correct. The actual transformations should be the following, instead of $\sqrt{\alpha}$ and $\frac{1}{\sqrt{c}}$, respectively: $$\lambda_1\rightarrow a^2\lambda_1, \ \ \ \lambda_2\rightarrow \frac{1}{c^2}\lambda_2$$ Otherwise, it yields another loss function, and the solutions change. See this (unofficial) errata.

  • $\begingroup$ It is strange that such a simple error is not even in his official errata... Are you 100% sure this is correct? $\endgroup$ Mar 30, 2019 at 11:36
  • 1
    $\begingroup$ I find it strange, too. But, the math is simple. $\endgroup$
    – gunes
    Mar 30, 2019 at 11:36
  • $\begingroup$ Do you by any chance know, if invariant regularizer have a special name? $\endgroup$ Mar 30, 2019 at 12:03

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