# Squared Loss for Multilabel Classification

I have a classification problem with $K$ labels. I represent the correct label $y$ of an observation $x$ as a vector $y$ in $R^K$, with entries $y_{k'} = \delta_{kk'}$ if $x$ belongs to class $k$.

Given an observation $x$, I predict its label with a vector $f(x) \in R^K$, where the components $f_k(x)$ satisfy $f_k(x) \in (0,1)$ and $\sum_k f_k(x) = 1$. A larger value of some $f_k(x)$ means that $x$ is more likely to belong to class $k$.

We want to learn the best function $f$ by choosing an appropriate loss function $\ell$. I know that a common choice is the cross-entropy: $\ell(x,y) = -\sum_k y_k \log f_k(x)$. Is squared loss $\ell(x,y) = \frac{1}{2}||y - f(x)||_2^2$ ever used? If so, does it tend to produce classifiers with noticeably distinct performance profiles, compared to cross-entropy?

A comment on a related question warns against the use of squared loss.

• What model do you use? Logistic regression? – Łukasz Grad Mar 11 '17 at 9:08
• It's a question about a general classifier. It could be linear or nonlinear. – John Kleve Mar 11 '17 at 22:51

This scheme is called the Brier loss. It is a proper scoring rule, and hence only the optimal classifier is correct, etc. It corresponds, of course, to the $L_2$ distance between the predictive label distribution and the true label distribution (which is a point mass).
Deep learning types these days strongly prefer the cross-entropy loss, which corresponds to the KL divergence $KL( y \| \hat y)$. This will penalize giving very low probabilities to the correct class very harshly, perhaps encouraging a flattening out of predicted probabilities relative to the Brier loss.
Consider a $K$-way classification problem, where your estimate of the probability of the $i$th class is $\hat p_i$. Let $y$ be the correct label for a given instance $x$, and $B = (\hat p_y(x) - 1)^2$ the Brier loss. Then $$\nabla B = 2 (\hat p_y(x) - 1) \nabla \hat p_y(x),$$ whereas if $C(x, y, w) = - \log \hat p_y(x)$ is the cross-entropy loss, then $$\nabla C = - \frac{1}{\hat p_y(x)} \nabla \hat p_y(x).$$ Plotting these:
Another interesting property: suppose that there are three categories, with the first one being correct. Cross-entropy would value the predictions $(.8, .2, 0)$ and $(.8, .1, .1)$ equally, whereas Brier loss would prefer the second one. I don't know if that's of huge practical importance, but only caring about the true category seems like a reasonable criterion to me, and that leads to cross-entropy being the only proper scoring rule.
• @JohnKleve Yeah, basically. I edited in a little math showing what I think they mean, which more or less just reinforces what I said before about KL versus $L_2$ distances between probability distributions. – Dougal Mar 13 '17 at 22:17