2 ms·
Out of curiosity, is there a reason why you are using ELO proper, rather than one of the ELO variants that doesn't make assumptions about the distribution of re
by Alex3917 1y ago
Out of curiosity, is there a reason why you are using ELO proper, rather than one of the ELO variants that doesn't make assumptions about the distribution of results? E.g.:
https://github.com/pfmonville/whole_history_rating https://github.com/pfmonville/whole_history_rating
- npip99 1y agoHey! We actually did a lot of research into ELO consistency, i.e. to check whether or not the NxN pairwise matrix followed the ELO model. It was a long road that's probably grounds for an entirely separate blog post, but the TLDR is that we observe that: For each document, there is a secret hidden score "s" which is the "fundamental relevance according to the LLM". Then, when we sample (q, d1, d2) from the LLM, the LLM follows the statistical property that: - The "fundamental hidden preference" is `pref = s_{d1} - s_{d2}`, usually ranging between -4 and 4. - The LLM will sample a normal distribution around the `pref` with stddev ~0.2, which is some "inner noise" that the LLM experiences before coming to a judgement. - The preference will pass through the sigmoid to get a sampled_score \in [0, 1]. - There is an additional 2% noise. i.e., 0.98 * sampled_score + 0.02 * random.random() When we use Maximum Likelihood Estimation to find the most likely predicted "hidden scores" \hat{s} associated with each document, then we go ahead and sample pairwise matrices according to `0.98 * sigmoid( \hat{s}_1 - \hat{s}_2 + N(0, 0.02) ) + Uniform(0.02)`, then we get a pairwise matrix with virtually identical statistical properties to the observed pairwise matrices.
- slybot 1y agoMore confused, 1) 0.02 * random.random() != N(0, 0.02) 2) The LLM will sample a normal distribution, this only depends on your c parameter, the absolute scale doesn't matter neither in Bradley-Terry nor in Elo. So saying +-4 and claiming LLM reasoning in Standard normal is ridiculous. 3) > then we get a pairwise matrix with virtually identical statistical properties to the observed pairwise matrices. >>> then did you asked yourselves if I have "statistically identical" pair-wise matrix and observed pairwise matrix, the. why you even bother myself? You can simply use observed pairwise matrix...