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Comment by etienneb on Looking for proof of conditional probability · 2013-10-25T00:58:08.205Z · score: 0 (0 votes) · LW · GW

Why is P(A∩B)/P(B) called conditional probability? Or, let's turn it the other way round (which is your question), why would conditional probability be given by P(A∩B)/P(B)? I think I was able to develop a proof, see below. Of course, double-checking by others would be required.

First, I would define conditional probability as the “Probability of A knowing that B occurs”, which is meaningful and I guess everybody would agree on (see also wikipedia).

Starting from there, “Probability of A knowing that B occurs” means the probability of A in a restricted space where B is known to occur. This restricted space is simply B. Now, in this restricted space (forget about the larger space Ω for a minute), we want to know the probability of A. Well, the occurrence of A in B is given by A∩B. In case of equiprobability of elementary events, the probability of A∩B in B is card(A∩B)/card(B). Now, this can be tweaked as card(A∩B)/card(Ω)×card(Ω)/card(B)=P(A∩B)/P(B), where P is the probability in the original space Ω. The latter result being Kolmogorov’s definition of conditional probability!

Note that we can also think in terms of areas, writing that the probability of A∩B in B is the area of A∩B divided by the area of B: area(A∩B)/area(B), which can be tweaked the same way, and works without requiring equiprobability (and the probability space could also be non-discrete).