In the case of a classification problem where a cost matrix is used to maximize the model performance, it is common to do a rebalance technique.

Let's say for example that I have the following costs for the two classes.

C(a,a) = 0, C(b,b) = 0, C(a,b) = 2, C(b,a) = 1.

Then, with a Rebalancing technique, I would need examples of class b twice as the examples of class a.

But, what should by rebalancing strategy will be when there is a cost for (a,a) or (b,b)?

For instance,

C(a,a) = 0, C(b,b) = 2, C(a,b) = -2, C(b,a) = -10

How should I handle those cases?

  • $\begingroup$ Is C(a,b) the cost of "the model says a and the actual is b" or "the model says b and the actual is a"? $\endgroup$ – Juan Esteban de la Calle Apr 25 '19 at 15:04
  • $\begingroup$ If C(I,j) is the cost, then I is the prediction and j the actual class. In your case, the first is true. $\endgroup$ – Tasos Apr 25 '19 at 15:08

Is not very common to find cost functions where there is a cost associated with a correct answer C(b,b) (in your example).

But supposing there is, I think the solution to the classification could be trivial: I could say "All my predictions are 'b'" and in that way, I could have -10 as a cost many times and thus giving me a negative cost (depending on balance, of course).

I did not know the technique for applying cost you mention (rebalance accordingly), but for me, it would be more natural if the objective function changes to have this into account.

The following article talks about a possibility for addressing this (Instead of rebalancing, we should measure the cost sensitive matrix). And with XGBoost!

As far as I know, the cost function of a XGBoost classification can be personalized.

Cost Sensitive Classification

XGBoost change loss function - A related question in DS

| improve this answer | |
  • $\begingroup$ You can assume that C(a,b) and C(b,a) are costs but C(b,b) is a bonus. So, it's a positive value. $\endgroup$ – Tasos Apr 26 '19 at 10:03
  • $\begingroup$ Oh! So just ignore the first part. Read only the suggestion $\endgroup$ – Juan Esteban de la Calle Apr 26 '19 at 11:55

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