For A posteriori, why the probability is devided by total distance?

Question

For A posteriori, why the probability is devided by total distance?

jayahm opened this issue 4 years ago · 4 comments

Hi,

As per code for A posteiori,

        competences_masked = np.ma.sum(masked_preprocessed,
                                       axis=1) / np.ma.sum(masked_dist, axis=1)

Why the probability of correct classification is divided by the distance only?

I saw on the paper, it is divided the total probability multiple by the distance.

Answer 1 · 2020-10-15T05:10:16.000Z

Hello,

It is done in order to weigh the influence of each data point x_j in the region of competence in such a way that the closer ones have more influence in the competence level estimation.

Answer 2 · 2020-10-18T04:32:56.000Z

I see. But, I saw in your paper, the denominator is the product of probability and distance

In your code above, it is only distance.

Or, I miss some information somewhere?

Answer 3 · 2020-10-18T05:24:03.000Z

This confusion is just due to the notation used in the original paper one by Giacinto et al (and re-used by the review paper), which is quite confusing. The denominator is supposed to just filter out the distance of examples that belongs to the reference class (wl). Hence, we used a boolean mask for that in the code. I suggest you check the explanation of this technique in the book Combining Pattern Classifiers by Ludmila Kuncheva which also presents some examples of how the competence estimates should behave. The notation there since it is much clearer in my opinion and. The steps and examples shown in the book are exactly the ones used for the implementation of this technique in the library. Em dom., 18 de out. de 2020 às 00:33, jayahm <notifications@github.com> escreveu:

…

I see. But, I saw in your paper, the denominator is the product of probability and distance In your code above, it is only distance. Or, I miss some information somewhere? — You are receiving this because you commented. Reply to this email directly, view it on GitHub <#230 (comment)>, or unsubscribe <https://github.com/notifications/unsubscribe-auth/AD6SFZBPX5ZYVPRT7VSMEBLSLJVYJANCNFSM4SJ4IZTQ> .

Answer 4 · 2020-10-19T11:38:02.000Z

Yes, I had double-checked with the original paper before asking you.

I see.

Okay, thank you for the clarification.