Probability

How popular is Bayes' theorem among philosophers? As a physicist, it has had a profound effect on my thinking, and seems to reflect the way we intuitively deal with new evidence presented to us. As a reminder, Bayes' theorem states: Probability(A given B) = Probability(B given A)*Probability(A)/Probability(B) For example, if A is "A revolutionary new theory" and B is "Data from my experiment", then Bayes' theorem tells us that we have to take into account our initial (prior) belief in the theory P(A), given our background knowledge, before even looking at our data.

Comments

Bayes' Theorem is very popular among philosophers of science who work on the bearing of evidence on theory. As you say, it has some attractive features. In your formulation, "A" stands for the theory and "E" for the evidence. To keep this straight, I'm going to use "T" and "E" respectively.

If we take the P(T given E) to be the probability that theory T has after you observe evidence E and P(T) the probability the theory had before, then the difference between these is naturally taken to be the degree to which the evidence supports the theory, and Bayes theorem plausibly says that this will be greater the greater the probability of P(E given T) -- where this probability peaks at unity if T entails E -- and the smaller P(E), the probability of the evidence before you observed it. In other words, this take on Bayes theorem says that you get the strongest support from surprising evidence which would however have to be true if your theory is true. And that sounds right.

Of course Bayesianism has its critics. One point they often make is that the probabilities P(T) and P(E) have to be construed simply as actual degrees of belief, and this makes the whole mechanism too subjective. Another is that the Bayesian mechanism runs into trouble in cases where the scientist knows the evidence before she constructs the theory. In this case, Bayes theorem seems not to allow that this evidence supports the theory. The impact of the evidence will already be reflected in P(T), so there will be no gain in moving from P(T) to P(T given E).

Read another response by Peter Lipton

Read another response about Probability