Definition D1
When P(B) > 0, the probability Q under the information that B has occurred is defined as Q(A) = P(A ∩ B) / P(B), written P(A|B).
Proof.
(1)If B has occurred, no outcome outside B has occurred. The new whole space is B, not S.convention
(2)Therefore what remains of an event A is only A ∩ B, and Q(A) is determined by P(A∩B).(1)
(3)Setting Q(A) = P(A∩B) gives Q(S) = P(S∩B) = P(B), which in general is not 1.(2), A7, E
(4)A probability must equal 1 on the whole space. So the scale is reset to P(B).(3), A2
(5)Define Q(A) = P(A∩B) / P(B). This definition applies only to B with P(B) > 0.(4), A4
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Lemma L1
If P(B) = 0, then for every event A, P(A ∩ B) = 0
Proof.
(1)B = (A∩B) ∪ (B∖A), and the two events are disjoint.A11
(2)P(B) = P(A∩B) + P(B∖A)(1), A3
(3)P(A∩B) ≥ 0 and P(B∖A) ≥ 0A1
(4)Suppose P(B) = 0.assumption
(5)Both terms are non-negative and their sum is 0, so P(A∩B) = 0(2), (3), (4), A12
(6)Then both numerator and denominator of D1 become 0, and by Proof 02 the value is not uniquely determined. This is why the proviso P(B) > 0 is there.(5), D1, A4
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Theorem T1
If P(B) > 0, then Q satisfies all three of Kolmogorov's axioms.
Proof.
(1)Suppose P(B) > 0.assumption
(2)P(A∩B) ≥ 0A1
(3)Q(A) = P(A∩B)/P(B) ≥ 0. This is P1.(1), (2), D1, A5
(4)S ∩ B = BA7
(5)Q(S) = P(S∩B)/P(B) = P(B)/P(B) = 1. This is P2.(1), (4), D1, A10, E
(6)Suppose X ∩ Y = ∅.assumption
(7)(X ∪ Y) ∩ B = (X∩B) ∪ (Y∩B)A8
(8)X∩B and Y∩B are disjoint.(6), A9
(9)P((X∪Y)∩B) = P(X∩B) + P(Y∩B)(7), (8), A3
(10)Q(X ∪ Y) = [P(X∩B) + P(Y∩B)] / P(B)(9), D1
(11)= P(X∩B)/P(B) + P(Y∩B)/P(B) = Q(X) + Q(Y). This is P3.(10), A6, D1
(12)Q satisfies all three axioms.(3), (5), (11)
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Corollary C1
Every proposition proved from the three axioms alone holds for P(A|B) as well.
Proof.
(1)Q satisfies the three axioms.T1
(2)Therefore any proposition resting only on the three axioms applies to Q.(1)
(3)By Proof 161, 0 ≤ P(A|B) ≤ 1(2)
(4)By Proof 162, P(Aᶜ|B) = 1 − P(A|B)(2)
(5)Therefore calling P(A∩B)/P(B) a probability is not a name but a credential.(1), D1
(6)The P(B) in the denominator is not a convention. It is the scale that makes Q(S) = 1, and without that scale the credential in (5) does not stand.(5), D1, A2
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