This is scikit GradientBoosting's binomial deviance loss function,

   def __call__(self, y, pred, sample_weight=None):
        """Compute the deviance (= 2 * negative log-likelihood). """
        # logaddexp(0, v) == log(1.0 + exp(v))
        pred = pred.ravel()
        if sample_weight is None:
            return -2.0 * np.mean((y * pred) - np.logaddexp(0.0, pred))
            return (-2.0 / sample_weight.sum() *
                    np.sum(sample_weight * ((y * pred) - np.logaddexp(0.0, pred))))

This loss functions is not similar between class with 0 and class with 1. Can anyone explain how this is considered OK.

For example, with no sample weigth, the loss function for class 1 is

-2(pred - log(1 + exp(pred))

vs for class 0


The plot for these two are not similar in terms of cost. Can anyone help me understand.


There are two observations needed to understand this implementation.

The first is that pred is not a probability, it is a log odds.

The second is a standard algebraic manipulation of the binomial deviance that goes like this. Let $P$ be the log odds, what sklearn calls pred. Then the definition of the binomial deviance of an observation is (up to a factor of $-2$)

$$y \log(p) + (1-y) \log(1 - p) = \log(1 - p) + y \log \left( \frac{p}{1-p} \right)$$

Now observe that $p = \frac{e^{P}}{1 + e^{P}}$ and $1-p = \frac{1}{1 + e^{P}}$ (a quick check is to sum them in your head, you'll get $1$). So

$$\log(1-p) = \log \left( \frac{1}{1 + e^{P}} \right) = - \log(1 + e^{P}) $$


$$ \log \left( \frac{p}{1-p} \right) = \log ( e^{P} ) = P $$

So altogether, the binomial deviance equals

$$y P - \log( 1 + e^{P} )$$

Which is the equation sklearn is using.

  • $\begingroup$ Thanks you. If i replace pred with log odds, the loss function is uniform for both the classes. $\endgroup$ – Kumaran Jun 21 '15 at 6:11
  • $\begingroup$ This same question came up for me recently. I was looking at gradientboostedmodels.googlecode.com/git/gbm/inst/doc/gbm.pdf page 10 where the gradient of the deviance is listed. But it seems like the gradient they show is for the log-lik not the negative log-lik. Is this correct - it seems to match your explanation here? $\endgroup$ – B_Miner Mar 29 '16 at 17:48
  • 1
    $\begingroup$ @B_Miner the link is broken $\endgroup$ – GeneX Jun 30 '18 at 11:43
  • $\begingroup$ Thanks a lot @Matthew Drury $\endgroup$ – Catbuilts Apr 2 '19 at 7:47

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