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Consider two events, A and B, on the same sample space. Prove that either $P(A|B) \leq P(A) \leq P(A|B^{c})$ or $P(A|B) \geq P(A) \geq P(A|B^{c})$.

Looking at the question, I believe I need to use the definition of conditional probability. But even then, I'm not exactly sure where to go with it. How should I start this?

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  • $\begingroup$ Is this homework? If so add te self study tag. $\endgroup$
    – Michael R. Chernick
    Commented Jan 12, 2017 at 7:24
  • $\begingroup$ What do E and F have to do with A and B? $\endgroup$
    – Michael R. Chernick
    Commented Jan 12, 2017 at 7:24
  • $\begingroup$ @MichaelChernick Sorry, didn't mean to put E and F––I meant A and B. $\endgroup$
    – Bing
    Commented Jan 12, 2017 at 7:30
  • $\begingroup$ I find it easier to reason about this circumstance by negating the assertion: is it really possible that the two conditional probabilities, $P(A\mid B)$ and $P(A\mid B^c)$, could both be strictly greater or strictly less than the probability of $A$ itself? $\endgroup$
    – whuber ♦
    Commented Jan 12, 2017 at 15:14

1 Answer 1

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So we have $ P(A)=P(A|B) P(B) +P(A|B^c)P(B^c)$

and $P(B)=P(B|A)P(A)+P(B|A^c) P(A^c)$ Note that whatever the sets A and B are either

(a) $ 1>= P(A)>= P(B)>=0$

or

(b) $ 1>=P(B)>=P(A)>=0$

Assuming (a) gives one of the inequalities and (b) gives the other.

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  • $\begingroup$ See if you can solve with the hints I have given you. $\endgroup$
    – Michael R. Chernick
    Commented Jan 12, 2017 at 7:46
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    $\begingroup$ Michael, the hints are more confusing than helping:$$P(A)=P(A|B) P(B) +P(A|B^c)P(B^c)$$is all that is needed to prove the result, while (a) and (b) do not separate between the two orderings. $\endgroup$
    – Xi'an
    Commented Jan 12, 2017 at 8:41

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