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Bayes' Theorem

Let A and B_j be sets. Conditional probability requires that

P(A intersection B_j)=P(A)P(B_j|A),
(1)

where intersection denotes intersection ("and"), and also that

P(A intersection B_j)=P(B_j intersection A)=P(B_j)P(A|B_j).
(2)

Therefore,

P(B_j|A)=(P(B_j)P(A|B_j))/(P(A)).
(3)

Now, let

S= union _(i=1)^NA_i,
(4)

so A_i is an event in S and A_i intersection A_j=emptyset for i!=j, then

A=A intersection S=A intersection ( union _(i=1)^NA_i)= union _(i=1)^N(A intersection A_i)
(5)
P(A)=P( union _(i=1)^N(A intersection A_i))=sum_(i=1)^NP(A intersection A_i).
(6)

But this can be written

P(A)=sum_(i=1)^NP(A_i)P(A|A_i),
(7)

so

P(A_i|A)=(P(A_i)P(A|A_i))/(sum_(j=1)^NP(A_j)P(A|A_j))
(8)

(Papoulis 1984, pp. 38-39).

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