Can one know if $P(A \cap B) > P(A)P(B) $ or $P(A \cap B) < P(A)P(B) $? I understand the following definition:
$P(A \cap B) = P(A)P(B) $ if $A$ and $B$ are independent and
$P(A \cap B) = P(B|A)P(A)$ and $P(B|A)P(A)\neq P(A)P(B)$ if $A$ and $B$ are not independent.
My question is:

For the case of dependence, does $P(A \cap B) > P(A)P(B) $ or $P(A \cap B) < P(A)P(B) $ or is it not possible to know without any values ?

 A: There exists a rough estimate about $P(A\cap B)$ and $P(A)P(B)$ whether events $A$ and $B$ are dependence or not, which is 
$$ \left| P(A\cap B) - P(A)P(B) \right| \leq \frac{1}{4} $$
The proof of this inequality can be proved by discussing the relationship of events $A$ and $B$.
A: I guess this might help:
Assuming $\mathsf{P}(A)\neq 0,1$ and $\mathsf{P}(B)\neq 0,1$.
Consider the random variable $1_A$, where $1_A(\omega)=1$ if $\omega\in A$ and $1$ if $\omega\notin A$.
Now $\mathsf{Cov}(1_A,1_B)=\mathsf{E}(1_A\cdot 1_B)-\mathsf{E}(1_A)E(1_B)=\mathsf{E}(1_{A\cap B})-\mathsf{E}(1_A)\mathsf{E}(1_B)$
Now $\mathsf{E}(1_A)=1\cdot \mathsf{P}(A)+0\cdot \mathsf{P}(A^c)=\mathsf{P}(A)$
Hence $\mathsf{Cov}(1_A,1_B)=\mathsf{P}(A\cap B)-\mathsf{P}(A)\mathsf{P}(B)$
So $\mathsf{P}(A\cap B)>\mathsf{P}(A)\mathsf{P}(B)$ implies $\mathsf{Cov}(1_A,1_B)>0$ and $\mathsf{P}(A\cap B)<\mathsf{P}(A)\mathsf{P}(B)$ implies $\mathsf{Cov}(1_A,1_B)<0$.
So $\mathsf{Cov}(1_A,1_B)>0$ implies if $A$ occurs then it is more likely to $B$ also occurs or if $B$ occurs then it is more likely to $A$ also occurs.
Similarly, $\mathsf{Cov}(1_A,1_B)<0$ implies if $A$ occurs then it is less likely to $B$ also occurs or if $B$ occurs then it is less likely to $A$ also occurs.
A: Since $P(A\cap B)=P(B\mid A)P(A)$, the equation $P(A\cap B)>P(A)P(B)$ is equivalent to $$P(B\mid A)>P(B)$$ (assume that $P(A)\neq 0$ to avoid trivial cases). This equation, says that: given that $A$ occured, $B$ has become more probable. For example, let $A$ be the event "there are clouds in the sky" and $B$ the event "it rains". Assume that you know from past data the probability $P(B)$ (perhaps you are a meteorologist). Now, knowing that $A$ applies (you open your window and you see many clouds outside), then certainly this increases the probability of $B$. 
But it can also be the other way: let "A" be the event "there are no clouds in the sky" (you open your window and it is so sunny that you close it again). Then, certainly $P(B\mid A)<P(B)$.
