Machine Learning (Theory)


Loss Function Semantics

Tags: Machine Learning,Problems jl@ 9:00 am

Some loss functions have a meaning, which can be understood in a manner independent of the loss function itself.

  1. Optimizing squared loss lsq(y,y’)=(y-y’)2 means predicting the (conditional) mean of y.
  2. Optimizing absolute value loss lav(y,y’)=|y-y’| means predicting the (conditional) median of y. Variants can handle other quantiles. 0/1 loss for classification is a special case.
  3. Optimizing log loss llog(y,y’)=log (1/Prz~y’(z=y)) means minimizing the description length of y.

The semantics (= meaning) of the loss are made explicit by a theorem in each case. For squared loss, we can prove a theorem of the form:
For all distributions D over Y, if

y’ = arg miny’ Ey ~ D lsq (y,y’)
y’ = Ey~D y

Similar theorems hold for the other examples above, and they can all be extended to predictors of y’ for distributions D over a context X and a value Y.

There are 3 points to this post.

  1. Everyone doing general machine learning should be aware of the laundry list above. They form a handy toolkit which can match many of the problems naturally encountered.
  2. People also try to optimize a variety of other loss functions. Some of these are (effectively) a special case of the above. For example, “hinge loss” is absolute value loss when the hinge point is at the upper range. Some of the other losses do not have any known semantics. In this case, discovering a semantics could be quite valuable.
  3. The natural direction when thinking about how to solve a problem is to start with the semantics you want and then derive a loss. I don’t know of any general way to do this other than simply applying the laundry list above. As one example, what is a loss function for estimating the mean of a random variable y over the 5th to 95th quantile? (How do we do squared error regression which is insensitive to outliers?) Which semantics are satisfiable with a loss?


The Missing Bound

Sham Kakade points out that we are missing a bound.

Suppose we have m samples x drawn IID from some distribution D. Through the magic of exponential moment method we know that:

  1. If the range of x is bounded by an interval of size I, a Chernoff/Hoeffding style bound gives us a bound on the deviations like O(I/m0.5) (at least in crude form). A proof is on page 9 here.
  2. If the range of x is bounded, and the variance (or a bound on the variance) is known, then Bennett’s bound can give tighter results (*). This can be a huge improvment when the true variance small.

What’s missing here is a bound that depends on the observed variance rather than a bound on the variance. This means that many people attempt to use Bennett’s bound (incorrectly) by plugging the observed variance in as the true variance, invalidating the bound application. Most of the time, they get away with it, but this is a dangerous move when doing machine learning. In machine learning, we are typically trying to find a predictor with 0 expected loss. An observed loss of 0 (i.e. 0 training error) implies an observed variance of 0. Plugging this into Bennett’s bound, you can construct a wildly overconfident bound on the expected loss.

One safe way to apply Bennett’s bound is to use McDiarmid’s inequality to bound the true variance given an observed variance, and then plug this bound on the true variance into Bennett’s bound (making sure to share the confidence parameter between both applications) on the mean. This is a clumsy and relatively inelegant method.

There should exist a better bound. If we let the observed mean of a sample S be u(S) and the observed variance be v(S), there should exist a bound which requires only a bounded range (like Chernoff), yet which is almost as tight as the Bennett bound. It should have the form:

PrS ~ Dm ( Ex~D x <= f(u(S), v(S) ,d)) >= 1 – d

For machine learning, a bound of this form may help design learning algorithms which learn by directly optimizing bounds. However, there are many other applications both within and beyond machine learning.

(*) Incidentally, sometimes people try to apply the Bennett inequality when they only know the range of the random variable by computing the worst case variance within that range. This is never as good as a proper application of the Chernoff/Hoeffding bound.


Conditional Tournaments for Multiclass to Binary

This problem has been cracked (but not quite completely solved) by Alina, Pradeep, and I. The problem is essentially finding a better way to reduce multiclass classification to binary classification. The solution is to use a carefully crafted tournament, the simplest version of which is a single elimination tournament where the “players” are the different classes. An example of the structure is here:

For the single elimination tournament, we can prove that:
For all multiclass problems D, for all learned binary classifiers c, the regret of an induced multiclass classifier is bounded by the regret of the binary classifier times log2 k. Restated:

regmulticlass(D,Filter_tree_test(c)) <= regbinary (Filter_tree_train(D),c)


  1. Filter_tree_train(D) is the induced binary classification problem
  2. Filter_tree_test(c) is the induced multiclass classifier.
  3. regmulticlass is the multiclass regret (= difference between error rate and minimum possible error rate)
  4. regbinary is the binary regret

This result has a slight dependence on k which we suspect is removable. The current conjecture is that this dependence can be removed by using higher order tournaments such as double elimination, triple elimination, up to log2 k-elimination.

The key insight which makes the result possible is conditionally defining the prediction problems at interior nodes. In essence, we use the learned classifiers from the first level of the tree to filter the distribution over examples reaching the second level of the tree. This process repeats, until the root node is reached. Further details, including a more precise description and some experimental results are in the draft paper.

Powered by WordPress