Consider events $A, B, C$ such that $P(A\mid B) > P(A)$ and $P(B\mid C) > P(B)$.

Question

Hey guys I have a problem that I'm having trouble solving. Here is the question:

Consider events $A, B, C$ such that $P(A\mid B) > P(A)$ and $P(B\mid C) > P(B)$. Does it follow that $P(A\mid C) > P(A)$? Either prove it to be so or provide a counterexample.

And here is what I have so far:

Partial solution

I would greatly appreciate it if you guys can give a hit or a suggestion to complete the problem.

Answer 1

Your partial solution must work for an appropriate choice of $p_i$'s. However, it's probably best to think about the problem this way.

The problem asks you to assume that knowing $C$ makes $B$ more likely, and knowing $B$ makes $A$ more likely. Then it asks if it follows that $C$ makes $A$ more likely.

There are at least a couple of approaches that could yield counterexamples.

First, it would be possible to construct an example in which $A$ never occurs when $C$ does.

Second, you could imagine an example in which $B$ is a low-probability event which is the intersection of much more likely events $A$ and $C$.

Answer 2

$A=\{1,2\}, B=\{2,3\}, C=\{3,4\}, \Omega=\{1,2,3,4,5\}$ with the uniform probability measure.

$\Pr(A)=\Pr(B)=\Pr(C)=\frac{2}{5}$

$\Pr(A\mid B)=\Pr(B\mid C)=\frac{1}{2}>\frac{2}{5}$

$\Pr(A\mid C)=0\not > \frac{2}{5}$