TITLE: Show that, $\lvert\mathbb P(A\cap B)-\mathbb P(A)\mathbb P(B)\rvert \le \dfrac{1}{4}$
QUESTION [7 upvotes]: If $(\Omega,\mathcal A,\mathbb P)$ is a probability space, show that for $A,B\in\mathcal A$
$\lvert\mathbb P(A\cap B)-\mathbb P(A)\mathbb P(B)\rvert \le \dfrac{1}{4}$

If $A, B$ are independent then the inequality is correct (LHS is 0). If not how can we express this dependency ?
Does it attain its maximum, if $\mathbb P(B)$ is 1 ?

REPLY [4 votes]: Start with:
$x=P\left[A\cap B\right]$, $y=P\left[A^{c}\cap B^{c}\right]$, $a=P\left[A\cap B^{c}\right]$
and $b=P\left[A^{c}\cap B\right]$ where $a,b,x,y\geq0$ and $a+b+x+y=1$.
Then: $$\left|P\left(A\cap B\right)-P\left(A\right)P\left(B\right)\right|=\left|x-\left(a+x\right)\left(b+x\right)\right|=\left|xy-ab\right|\leq\max\{xy,ab\}$$
Here $xy\leq\frac{1}{4}$ as a consequence of $x,y\geq 0$ and $x+y\leq 1$. Under these conditions $xy$ achieves its maximum if $x=y=\frac{1}{2}$.
Likewise $ab\leq\frac{1}{4}$ and we are ready.
Its maximum is attained if $P(A)=\frac{1}{2}$ and $A\in\{B,B^c\}$.