1
$\begingroup$

I see the definition of expected calibration error being

$$\sum_{m=1}^{M}\frac{|B_m|}{n}|accuracy(B_m)-confidence(B_m)|$$

Where $B_m$ represents a outputs of the model that predicted class $m$ in a multi-classifcation problem. Then I read this nice article with an example of calculating it. So the binning is determined by what the model outputs as the max prob. So for example, if input 0 leads to a 90% prediction of "dog," that'll go into the "dog" bin. Once this is done the accuracy as in how many of them correctly estimated the class subtracted by the average confidence or predictive probability by the model is take. I'm a little bit confused about the binning part. What does binning the predictions in this way tell us exactly? Why not do the bins via the true classes? For example you take all the rows where the true class is $m$ then calculate the accuracy and confidence for class $m$?

$\endgroup$
0
$\begingroup$

The expected calibration error is (often) defined as:

$$\text{ECE}(f)=\mathbb{E}_{f(x)}\left[\left|f(X)-\mathbb{E}[Y \mid f(X)]\right|\right]$$

where $f(x) \in [0,1]$ is the NN output and $\mathbb{E}[Y \mid f(X)]=\mathbb{P}[Y=1 \mid f(X)]$ in the binary case. This is extended to multi-class problems by simply regarding every class as a binary classification objective and averaging over them (see https://arxiv.org/pdf/1909.10155.pdf).

Your formula is just an estimator of the expectation above:

$$\text{ECE}(f)=\mathbb{E}_{f(x)}\left[\left|f(X)-\mathbb{E}[Y \mid f(X)]\right|\right] \approx \sum_{m=1}^M \frac{|B_m|}{n}|conf(B_m)-acc(B_m)|$$

Because the $f(x)$ is continuous, we need to discretise it: Let us partition $[0,1]$ into $M$ intervals. Then we define $B_m$ to be the set data points that have predictions falling into the $m$th interval (see https://arxiv.org/abs/1706.04599 for more formal definition of $B_m$). So $m$ indicites the groupping based on confidence, not the predicted class itself.

Note: The $ECE$ is not a good metric as it has a lot of shortcomings. For instance, if $f(X) > 0.9$ for all data points, then the equal width binning of the $ECE$ is useless. Another issue is that the estimated $ECE$ tends to underestimate the true calibration error, which has been shown by multiple papers in different ways.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.