### Mortal Kolle

Discussion -Good evening all

I have encountered a counter-intuitive result while thinking about Bayesian networks and decided to ask the members of this group

Suppose A is the probability space of all possible events (with P(A)=1, of course)

Now suppose A is partitioned into A1 and A2 such that P(A1)=P(A2)=0.5 and let a be some event in A

According to Bayes' Rule,

P(A1|a) +P (A2|a) = P(A1)/P(a) *P(a|A1) + P(A2)/P(a)*P(a|A2) =

= 0.5/P(a) *(P(a|A1) + P(a|A2)) = 0.5 since P(A1)= P(A2) = 0.5

P(A1+A2)=1 and A1 and A2 are disjoint, yet P(A1+A2|a) ~= P(A1|a) + P(A2|a) = 0.5

But A1 and A2 are disjoint and P(A1) + P(A2) = 1 so a must be fully contained in the union of A1 and A2 since it is contained in the universal probability space A.

Likewise, P(a|A1) + P(a|A2) = (P(a)/0.5) * (P(A1|a) +P (A2|a))

My head is spinning from this. Is there a rationalization I don''t know about?

I have encountered a counter-intuitive result while thinking about Bayesian networks and decided to ask the members of this group

Suppose A is the probability space of all possible events (with P(A)=1, of course)

Now suppose A is partitioned into A1 and A2 such that P(A1)=P(A2)=0.5 and let a be some event in A

According to Bayes' Rule,

P(A1|a) +P (A2|a) = P(A1)/P(a) *P(a|A1) + P(A2)/P(a)*P(a|A2) =

= 0.5/P(a) *(P(a|A1) + P(a|A2)) = 0.5 since P(A1)= P(A2) = 0.5

P(A1+A2)=1 and A1 and A2 are disjoint, yet P(A1+A2|a) ~= P(A1|a) + P(A2|a) = 0.5

But A1 and A2 are disjoint and P(A1) + P(A2) = 1 so a must be fully contained in the union of A1 and A2 since it is contained in the universal probability space A.

Likewise, P(a|A1) + P(a|A2) = (P(a)/0.5) * (P(A1|a) +P (A2|a))

My head is spinning from this. Is there a rationalization I don''t know about?

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