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I am reading a paper that claims (without proof)

$$P(A \mid B) \leq P(A)$$ if and only if $$P(A \mid \overline{B}) \geq P(A)$$

for any two events $A$ and $B$.

This seems reasonable, but I can't seem to prove it directly from the definition of conditional probability. Perhaps there is some identity involving these terms that I'm forgetting?

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1 Answer 1

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Hint: $P(A)=P(B)P(A|B) + P(\bar B)P(A|\bar B)$.

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  • $\begingroup$ I apologize for being dense. If I assume $P(A \mid B) \leq P(A)$, I derive only that $P(A \mid B) \leq P(A \mid \overline{B})$ using your hint. $\endgroup$
    – user72986
    Commented Apr 17, 2013 at 3:45
  • $\begingroup$ In general, if $0\leq t\leq 1$ then $tX+(1-t)Y$ is between $X$ and $Y$. Let $t=P(B)$. Then $1-t=P(\bar B)$. $\endgroup$
    – Thomas Andrews
    Commented Apr 17, 2013 at 3:57
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    $\begingroup$ Essentially, what the above equation says is that $P(A)$ is a "weighted average" of $P(A|B)$ and $P(A|\bar B)$, and hence must be between these two values. $\endgroup$
    – Thomas Andrews
    Commented Apr 17, 2013 at 3:59
  • $\begingroup$ That's a very nice way to think of it. Thank you. $\endgroup$
    – user72986
    Commented Apr 17, 2013 at 4:13

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