Assume that $P( A\mid B) < P( A)$ & $P( B\mid C) < P( B)$ . Is it true that $P(A\mid C)<P(A)$?

Question

If $P( A\mid B) < P( A)$ & $P( B\mid C) < P( B)$ , then is it true that $P(A\mid C)< P(A)$, where $P( A\mid B)$ is the conditional property of $A$ given $B$?

I tried the following:

• $P(A ∩ B) < P(A)P(B)$
• $P(A ∩ C) < P(A)P(C)$
• $P(C ∩ B) < P(C)P(B)$

by using $P(A\mid B)=P(A ∩ B)/P(B)$. But it led me nowhere.

Answer 1

The answer is "No". Try $A=(0,\frac12)$, $B=(\frac12,1)$, $C=(0,\frac12)$, in the probability space $\Omega=[0,1]$ endowed with the Borel sigma-algebra and the Lebesgue probability measure.

To see that both situations are possible, try $A=(0,\frac47)$, $B=(\frac17,\frac67)$, $C=(\frac37,1)$, in the same probability space.