Assume that P(B) > 0. Prove that if P(A1|B) < P(A1) then P(Ai | B) > P(Ai) for some i = 2, … , k.

Suppose k events from a partition of the sample space Ω, i.e., they are disjoint and ∪ i=1 to k over Ai = Ω. Assume that P(B) > 0. Prove that if P(A1|B) < P(A1) then P(Ai | B) > P(Ai) for some i = 2, ... , k.

I used the law of total probability to prove P(A1|B) < P(A1) but I have no idea how to prove the that P(Ai | B) > P(Ai) for some i = 2, ... , k.

Thank you.

• The law of total probability cannot be used to prove that $P(A_1\mid B) < P(A_1)$ (as you are claiming you did). If you do believe that you have a valid proof, then consider that the same proof can be adapted to prove that $P(A_i\mid B) < P(A_i)$ for all choices of $i$, not just $i=1$, and so you will have disproved what you set out to do. – Dilip Sarwate Apr 19 '15 at 20:31
• That is why I am unable to understand how to prove that if for i=1 P(Ai | B) < P (Ai) then P(Ai | B) > P (Ai | B) for i=2, .. , k. Have you got any suggestions? – Octavian Apr 19 '15 at 20:41

Then $P(A_1\mid B)<P(A_1)$ and $P(A_i\mid B)\leq P(A_i)$ for $i=2,\dots,k$ leading to: $$1=P(\Omega\mid B)=\sum_{i=1}^kP(A_i\mid B)<\sum_{i=1}^kP(A_i)=P(\Omega)=1$$ A contradiction is found.