Decompostion of conditional probabilities

[torquerotates]

So If B=B1UB2U...UBn would Pr(A|B)=Pr(A|B1)Pr(B1)+Pr(A|B2)Pr(B2)+...+Pr(A|Bn)Pr(Bn)?

I haven't found a formula for this but it makes intuitive sense.

 

[chiro]

So If B=B1UB2U...UBn would Pr(A|B)=Pr(A|B1)Pr(B1)+Pr(A|B2)Pr(B2)+...+Pr(A|Bn)Pr(Bn)?

I haven't found a formula for this but it makes intuitive sense.

Hey torquerotates.

Before I give my take on the answer, do you assume that all the B1, B2, etc are disjoint? In other words is Bi AND Bj = NullSet for i != j?

 

[torquerotates]

Yes they are disjoint.

 

[chiro]

Yes they are disjoint.

For two events I get the following:

P(A|B1UB2) = P(A and (B1UB2))/P(B1UB2) = P((A and B1) U (A and B2))/P(B1UB2) = P(A and B1)/P(B1UB2) + P(A and B2)/P(B1UB2) - P(A and B1 and B2)/P(B1UB2)

= P(A and B1)/(P(B1) + P(B2)) + P(A and B1)/(P(B1) + P(B2)) - 0

= 1/P(B)[P(A and B1) + P(A and B2)]

= 1/P(B)[P(A and B1]P(B1)/P(B1) + P(A and B2)P(B2)/P(B2)]

= 1/P(B)[P(A|B1)P(B1) + P(A|B2)P(B2)] if we assume P(Bi) > 0 and is a valid probability.

If the above is right then your answer is wrong in general but right when B is the universal probability space.

Because your probability P(B) may not be one, it means that the above needs to be corrected using this information. Just for an example consider P(B1) = P(B2) = 1/4 and P(A|B1) = P(A|B2). But since 1/4[P(A|B1) + P(A|B2)] = 1/2P(A|B) for this example since P(B1) = P(B2) and since B1 and B2 are disjoint we have:

Then under these conditions we have:

1/P(B) x [P(A|B1)P(B1) + P(A|B2)P(B2)]
= 1/(1/2) x [P(A|B1)x1/4 + P(A|B1)x1/4]
=(1/4)/(1/2) x [P(A|B1) + P(A|B2)]
= 1/2 x [(1/2)/(1/4)]xP(A|B)
= 1/2 x 2 x P(A|B)
= P(A|B) for this example.

If we did not 'normalize' by P(B) we would not have gotten the right answer.

While I think the above is right, I would like others to check if they could just to make sure, but I'm only using a few identities which are basically distribution over sets and multiplication by 'x/x' terms for P(Bi)/P(Bi).

 

[Stephen Tashi]

it makes intuitive sense.

Did you do any examples?

Let
A = the roll of a fair die produces an even number
B = the face that comes up is a 1 or a 2
B1 = the face that comes up is a 1
B2 = the face that comes up is a 2