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I am doing my final thesis in the field of Deepfakes and their detection. The final outcome is to have a binary classifier which could predict which video was updated and which was not. In other words, if video is fake, output number close to 1 and if video is real, output number close to 0. One of the reasons I started this project is because Facebook released a massive dataset called DFDC and they announced in media that the best solution got only 82% average precision on validation dataset and 65% average precision on the testing dataset. Source: https://ai.facebook.com/datasets/dfdc/

However, they also released a research paper describing the results: https://arxiv.org/pdf/2006.07397.pdf This is the part where I am confused. Research paper does not mention anything about 82% and 65%. Contradictory, in their research paper's figure you can see that Average Precision of the best score was around 0.98 (on the public testing dataset)

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However, they also include the log loss scores and I managed to understand how Facebook calculates the values of 82% and 65% on public dataset and private dataset. What they do is described in the answer here: https://stats.stackexchange.com/questions/276067/whats-considered-a-good-log-loss

which basically if the log-loss is 0.42, they take average precision as:

$$ AP = \frac{1}{e^{0.42}} $$

I was wondering if it is acceptable to use the above method because when I evaluate my score with scikit library and use: average_precision_score, I get much higher scores if to compare with the method which Facebook used as in https://stats.stackexchange.com/questions/276067/whats-considered-a-good-log-loss

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  • $\begingroup$ Wrong link maybe for the AP-LL connection? At least, I don't see such a claim there. $\endgroup$ – Ben Reiniger Feb 5 at 0:52
  • $\begingroup$ @BenReinigerI think it is correct link. In the answer of the link it says that L(p) = 0.5. This means that probability to the right class is 1/e^0.5 = 0.607. In the facebook announcement they say that average precision was 82.5% on public and 65.2% on private, but I can see log loss of the best public and private scores here: kaggle.com/c/deepfake-detection-challenge/leaderboard which are 0.192 and 0.428. So 1/e^0.192=82.5% and 1/e^0.428=65.2%, but what I am unsure about is if this metric is actually average precision. $\endgroup$ – MichiganMagician Feb 11 at 10:07

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