Why the statement: if $P(A|B) > P(A)$ then $P(B|A)>P(B)$ intuitively makes sense?

Question

I know how to prove the following statement but I can't understand why this makes sense intuitively:

If $P(A|B)>P(A),$ then $P(B|A)>P(B)$

I would appreciate it if someone can explain it intuitively.

Answer 1

See @MariaMazur's answer for the mathematical proof.

One intuition is that if $P(A|B)>P(A)$ then the events $A$ and $B$ have a positive correlation. So, we should also expect $P(B|A)>P(B)$.

In addition, the following work of art might provide more insight (or not).

Answer 2

Yes, if $P(A)\cdot P(B)\ne 0$ then (remember also that probability is non negative) $${P(A\cap B)\over P(B)}>P(A)\iff {P(A\cap B)\over P(A)}>P(B)$$

so $$ P(A|B)>P(B) \iff P(B|A)>P(B)$$

Answer 3

Note that if A and B are independent, we have $$P(A\cap B)= P(A)P(B)$$ Both your statements imply that $A$ and $B$ are dependent with positive correlation.

Answer 4

Interpret $P(A)$ and $P(B)$ as the probability that a random point in a sample space is in shape $A$ and $B$, respectively.

Because $P(A|B)>P(A)$, we know that $A$ proportionally takes up more of the area of $B$ compared to not $B$.

Now, to link this to the second statement, let us look at the area outside of $A$. Because this takes up the rest of what $A$ didn't take up, $P(\text{not B}|\text{not A})$ is more likely to occur than $P(\text{B}|\text{not A})$ proportionally.

$P(B)$ includes the cases "the point is in $A$" and "the point is not in $A$." The case "the point is not in $A$," even though the probability is weighted between the cases, will bring down $P(B)$ to lower than the probability of $P(B|A)$.

Therefore, $P(B|A)>P(B)$.

In the diagram below, $A$ and $B$ are the ellipses and the sample space is the square.