Bayes theorem pAB pBA pA pB In general

Bayes' theorem p(A|B) = p(B|A) p(A) / p(B) In general p(A|B) (usually read 'probability of A given B’) = the probability of finding observation A, given that some piece of evidence B is present p(A) = the probability of the outcome occurring, without knowledge of the new evidence p(B) = the probability of the evidence arising, without regard for the outcome p(B|A) = the probability of the evidence turning up, given that the outcome obtains

Video Also mentions base rate fallacy *try not to laugh with hypothesitis

An example Ø There is a race between 2 horses: Fleetfoot and Dogmeat Ø Fleetfoot and Dogmeat have raced against each other on 12 previous occasions: Dogmeat won 5 and Fleetfoot won 7 Ø Therefore, the estimated probability of Dogmeat winning the next race is 5/12 = 0. 417 = 41. 7% Fleetfoot, on the other hand, appears to be a better bet at 7/12 = 58. 3% Ø BUT 3/5 of Dogmeat's previous wins, it had rained heavily However, it had rained only once on the days that he lost Ø On the day of the race in question, it is raining Ø Which horse should I bet on? Ø 4 possible situations: Raining Not Raining Dogmeat wins 3 2 Dogmeat loses 1 6 Source: http: //www. kevinboone. net/bayes. html

So, Bayes’ theorem in our example… p(A|B) = p(B|A) p(A) / p(B) ‘A’ = the outcome in which Dogmeat wins ‘B’ = the piece of evidence that it is raining p(A|B) = the probability of Dogmeat winning given that it is raining = what we want to find out p(B|A) = the probability that it is raining, given that Dogmeat wins ü p(B|A) = 3/5 = 0. 6 (since it was raining 3/5 times that Dogmeat won) ü p(A) = 5/2 = 0. 417 (because Dogmeat won on 5/12 occasions) ü p(B) = 4/12 = 0. 333 (since we know it rained on 4 days in total) Ø Therefore, p(A|B) = p(B|A)*p(A) / p. B) = 0. 6 * 0. 417 / 0. 333 = 0. 75

Patten et al. (2017)

Friston et al. (2014)

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