Geometric mean of the odds = mean of the evidences.
Suppose you have probabilities in odds form; 1: 2^a and 1:2^b, corresponding to a and b bits, respectively. Then the geometric mean of the odds is 1: sqrt(2^a * 2^b) = 1 : 2^((a+b)/2), corresponding to ((a+b)/2) bits; the midpoint in the evidences.
For some more background as to why bits are the natural unit of probability, see for example this arbital article, or search Probability Theory, the Logic of Science. Bits are additive: you can just add or substract bits as you encounter new evidence, and this is a pretty big “wink wink, nod, nod, nudge, nudge” as to why they’d be the natural unit.
In any case, if person A has seen a bits of evidence, of which a’ are unique, and person B has seen b bits of evidence, of which b’ are unique, and they have both seen s’ bits of shared evidence, then you’d want to add them, to end up at a’+b’+s’, or a + b −2s’. So maybe in practice (a+b)/2 = s’ + (a’+b’)/2 ~ a’+b’+s’, when a’ + b’ small (or overestimated, which imho seems to often be the case; people overestimate the importance of their own private information; there is also some literature on this).
This corresponds to the intuition that if someone is at 5%, and someone else is at 3% for totally unrelated reasons, the aggregate should be lower than that. And this would be a justification for Tetlock’s extremizing.
Anyways, in practice, you might estimate s’ as the historical base rate (to which you and your forecasters have access), and take a’ b’ as the deviation from that.
I have a sense that that log-odds are an underappreciated tool, and this makes me excited to experiment with them more—the “shared and distinct bits of evidence” framework also seems very natural.
On the other hand, if the Goddess of Bayesian evidence likes log odds so much, why did she make expected utility linear on probability? (I am genuinely confused about this)
Geometric mean of the odds = mean of the evidences.
Suppose you have probabilities in odds form; 1: 2^a and 1:2^b, corresponding to a and b bits, respectively. Then the geometric mean of the odds is 1: sqrt(2^a * 2^b) = 1 : 2^((a+b)/2), corresponding to ((a+b)/2) bits; the midpoint in the evidences.
For some more background as to why bits are the natural unit of probability, see for example this arbital article, or search Probability Theory, the Logic of Science. Bits are additive: you can just add or substract bits as you encounter new evidence, and this is a pretty big “wink wink, nod, nod, nudge, nudge” as to why they’d be the natural unit.
In any case, if person A has seen a bits of evidence, of which a’ are unique, and person B has seen b bits of evidence, of which b’ are unique, and they have both seen s’ bits of shared evidence, then you’d want to add them, to end up at a’+b’+s’, or a + b −2s’. So maybe in practice (a+b)/2 = s’ + (a’+b’)/2 ~ a’+b’+s’, when a’ + b’ small (or overestimated, which imho seems to often be the case; people overestimate the importance of their own private information; there is also some literature on this).
This corresponds to the intuition that if someone is at 5%, and someone else is at 3% for totally unrelated reasons, the aggregate should be lower than that. And this would be a justification for Tetlock’s extremizing.
Anyways, in practice, you might estimate s’ as the historical base rate (to which you and your forecasters have access), and take a’ b’ as the deviation from that.
Thank you for pointing this out!
I have a sense that that log-odds are an underappreciated tool, and this makes me excited to experiment with them more—the “shared and distinct bits of evidence” framework also seems very natural.
On the other hand, if the Goddess of Bayesian evidence likes log odds so much, why did she make expected utility linear on probability? (I am genuinely confused about this)