Proving Conditional ProbabilitiesDate: 01/24/2001 at 20:09:10 From: Mike Walker Subject: Conditional Probability Dr. Math, I was hoping that you could give me a hand with a short proof. (Note: ^c means complement.) If P(B|A) > P(B), show that P(B^c|A) < P(B^c). I got a hint from my professor as well. He suggested adding P(B^c|A) to both sides of the given inequality. Any help with this proof would be greatly appreciated. Thanks in advance. MiKe Date: 01/24/2001 at 20:19:52 From: Doctor Schwa Subject: Re: Conditional Probability Hi Mike, First let's do it intuitively, then see if we can make it rigorous. The first inequality says "If A happens, B is more likely to happen than if it were to happen on its own." The second inequality says "If A happens, not-B is less likely to happen than if it were on its own." It sure seems like a tautology that if the first one is true, the second one must be too. Now down to definitions. P(B|A) = P(A and B) / P(A), so the first inequality can be rewritten P(A and B) > P(A) P(B), while the second inequality can be rewritten P(A and B^c) < P(A) P(B^c). Since it's not immediately obvious to me how to turn one into the other, I'll try your hint: add P(B^c | A) to both sides. Now we have P(B^c | A) + P(B | A) > P(B) + P(B^c | A). Rewriting the conditionals in terms of the definition, P(B^c and A)/P(A) + P(B and A)/P(A) > P(B) + P(B^c and A)/P(A). Multiplying by P(A), P(B^c and A) + P(B and A) > P(A) P(B) + P(B^c and A). Now the left side is just the probability that A happens somehow. It can happen with B, or without ... P(A) > P(A) P(B) + P(B^c and A) That is, P(A) - P(A) P(B) > P(B^c and A) P(A) (1 - P(B)) > P(B^c and A) P(A) P(B^c) > P(B^c and A) Wait a minute, isn't that the second inequality I was trying to prove? I didn't mean to solve the whole problem for you... I was just messing around, using your professor's hint and the definitions of conditional probability and complement ... I guess that's a good general strategy for proving things, taking a look at the definitions of the key terms in the proof. Personally I like the intuitive proof better. I'm left more convinced by it than by the half-page of algebra. - Doctor Schwa, The Math Forum http://mathforum.org/dr.math/ |
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