Variance/bias trade off regularisation penalty - why does it take this form?

In machine learning, if we estimate weights using a loss function

$$L(W) = ||Y-F_W(X)||^2$$

(where $$W$$ is a weight matrix) we may add a "regularisation penalty" to control for the "variance/bias trade-off" so that

$$L(W) = ||Y-F_W(X)||^2 + \lambda \phi(W).$$

Now, one of the penalties proposed frequently is of the form

$$\phi(W) = \sum_{i,j} |W_{ij}|^2$$

but I cannot find a good explanation of why such a penalty is used, including in the papers I have come across it.

Why does penalising larger values of $$W$$ affect the "bias/variance trade-off" of a model? Could someone give me an intuitive explanation for why we would prefer smaller values of $$W_{ij}$$ and why it's relevant to the bias/variance trade-off?

• $\phi(W) = \sum_{i,j} |W_{ij}|^2$ is convex and when added to a loss function that is convex, the resultant is convex. And, it is differentiable, so easy to work with mathematically. Mar 11, 2019 at 11:31
• That makes sense as far as finding a unique solution, but I don't understand how that relates to bias/variance of the resulting estimator Mar 11, 2019 at 11:34
• By adding a regularization term one is adding bias in the model and trying to avoid overfitting (variance). There is a high chance to fall into the trap of underfitting (bias) when one does that. Mar 11, 2019 at 11:46
• Have a look here: - stats.stackexchange.com/questions/141555/… Mar 11, 2019 at 11:48

2 Answers

why such a penalty is used

It is used to penalize the complexity of model (function), or equivalently to encourage the model to be as simple as possible (roughly has fewer weights involved). This way, over-fitting is avoided and generalization of model to unseen data is improved.

Why does penalizing larger values of W affect the "bias/variance trade-off" of a model?

By forcing the parameters to be close to zero, the variance of model which is a function of its parameters would decrease, since parameters are less free to change compared to when their value is not penalized. On the other hand, the true function may need a wide range of weights to be estimated correctly, i.e. more flexible/complex model. However, by this penalty, we are allowing less flexibility (less complexity) thus moving the model away from the best configuration of weights required to estimate the true function, i.e. more bias.

Why does penalising larger values of $$W$$ affect the "bias/variance trade-off" of a model? Could someone give me an intuitive explanation for why we would prefer smaller values of $$Wij$$ and why it's relevant to the bias/variance trade-off?

A penalty term is generally an hyper-parameter of the model -- a larger penalty term reduces the complexity of the resulting function.

A common example is weight decay used with neural networks and linear regression also known as ridge regression -- in that case the $$L_2$$ norm of the parameters are added to the loss function, which basically leads to smaller parameter magnitudes.

In terms of neural networks -- increasing the penalty term i.e. increasing the weight decay hyper-parameter corresponds to reducing the effective capacity by forcing the solution to be in a zero-centered hyper-sphere of smaller radius -- which may improve generalization.

Improvement in generalization can be seen as decrease in the expected error from the model -- which is composed of bias, variance and irreducible error. So, basically you trade an increase in bias for decrease in variance, hopefully more decrease on the variance side -- don't underfit!

For some intuition:

Bayesian interpretation: A regularization term can also be interpreted as an a-priori probability distribution on $$F$$ -- a set of functions from which one can be picked to minimize the expected generalization error of the chosen function-- in which case the weight decay is a scale parameter (inverse variance) of that distribution. And, by varying the scale you can make the distribution more concentrated or more dispersed.