Oscar_Cunningham comments on Looking for proof of conditional probability - Less Wrong

-1 Post author: DanielLC 28 July 2011 02:24AM

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Comment author: Oscar_Cunningham 28 July 2011 08:21:05PM *  2 points [-]

Short answer: The Kolmogorov axioms are just mathematical. They have nothing inherently to do with the real world. P(A|B)=P(A∩B)/P(B) is the definition of P(A|B). There is a compelling argument that the beliefs of a rational agent should obey the Kolmogorov axioms, with P(A|B) corresponding to the degree of belief in A when B is known.

Long answer: I have a sequence of posts coming up.

Comment author: Wei_Dai 28 July 2011 09:06:23PM 0 points [-]

There is a compelling argument that the beliefs of a rational agent should obey the Kolmogorov axioms, with P(A|B) corresponding to the degree of belief in A when B is known.

Are you thinking of this one, or something else?

Comment author: Oscar_Cunningham 29 July 2011 07:10:25AM 1 point [-]

I was thinking of the Dutch book argument others have mentioned. But I think you may have misunderstood my point.The original poster has summed up what I wanted to say better than I could:

If I have a set of axioms, and I derive theorems from them, then anything that these axioms are true about, all the theorems are also true about. For example, suppose we took Euclid's first four postulates and derived a bunch of theorems from them. These postulates are true if you use them to describe figures on a plane, so the theorems are also true about those figures. This also works if it's on a sphere. It's not that a "point" means a spot on a plane, or two opposite spots on a sphere, it's just that the reasoning for abstract points applies to physical models.

Statistics isn't just those axioms. You might be able to find something else that those axioms apply to. If you do, every statistical theorem will also apply. It still wouldn't be statistics. Statistics is a specific application. P(A|B) represents something in this application. P(A|B) always equals P(A∩B)/P(B). We can find this out the same way we figured out that P(∅) always equals zero. It's just that the latter is more obvious than the former, and we may be able to derive the former from something else equally obvious.

I agree with the first paragraph but the second seems confused. We want to show that P(A|B) defined as P(A∩B)/P(B) tells us how much weight to assign A given B. DanielLC seems to be looking for an a priori mathematical proof of this, but this is futile. We're trying to show that there is a correspondence between the laws of probability and something in the real world (the optimal beliefs of agents) , so we have to mention properties of the real world in our arguments.