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Probabilities: frequencies viewed in perspective A. Billot , I. Gilboa , D. Samet , D. Schmeidler. Where do probabilities come from?. Are the answers we have any better?. The Bayesianist : by updating. The logician : by logical derivation (Carnap, Harsanyi).
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Probabilities:frequencies viewed in perspectiveA. Billot, I.Gilboa, D. Samet, D. Schmeidler
Where do probabilities come from? Are the answers we have any better? The Bayesianist: by updating. The logician: by logical derivation (Carnap, Harsanyi). The classical statitstician: by computing frequencies.
Evidence 1 2 . . . m case typesC memory I 5 12 . . . 3 A memory: I: C Z+I 0 The set of memories: I
Probability formation 3 Ω = {1,2,3,…,n} 1 2 case typesC memory I Probability formation function p : I Δ(Ω) 1 2 . . . m 5 12 . . . 3 States of the world . p(I) Δ(Ω)
The combination axiom Ω = {1,2,3,…,n} memory I + J case typesC memory I memory J 9 18 . . . 11 1 2 . . . m 5 12 . . . 3 4 6 . . . 8 + 3 . States of the world . p(I + J) . p(J) p(I) 1 2 Δ(Ω)
The combination axiom For everyI, JI, p(I + J) [p(I), p(J)].Ifp(I) p(J), then p(I + J) (p(I), p(J))
The main theorem (ii) For each case j, there exists a probability vector pj in Δ(Ω), not all of them on the same line, and positive weight sj, such that for each memory I j ≤ mI(j)sjpjp(I) = _________________j ≤ mI(j) sj The probabilities pjare uniquely determined, the weights sj - up to multiplication by the same scalar. The following two statements are equivalent: (i) The probability formation functionp satisfies the combination axiom, and its range does not lie on a line.
The probability formation function Denote by the vector of frequencies of cases in I. I ________ FI = j ≤ mI(j) Since p(I) depends on FI only, we can write… Dividing nominator and denominator by j ≤ mI(j), yiealds j ≤ mFI(j)sjpjp(I) = _________________j ≤ mFI(j) sj j ≤ mI(j)sjpjp(I) = _________________j ≤ mI(j) sj p(FI)
Frequencyof cases Probabilityof states Probability = Frequencyin perspective F1s1p1 +F2s2p2 +F3s3p3 For case 3 s2p2 s3p3 . 3 3 . . I For case 2 Δ(Ω) F = (F1, F2, F3) . p2 . . . 2 p3 2 . For case 1 . . s1p1 p1 p(F) = p(I) 1 1