Gap to independence Given a probability space $(\Omega, \mathcal {A}, P)$, what are the minimum and maximum of the quantity
$$
     P(A_1 \cap \cdots \cap A_n) -  P(A_1) \cdots P(A_n)  
$$
over $A_1, \ldots, A_n \in \mathcal {A}$, $n \geq 1$?
When $n = 2$, it is easily seen, from the Cauchy-Schwarz inequality (since
$$
    P(A_1  \cap A_2) -  P(A_1) P(A_2 ) = E ((1_{A_1} -P(A_1)) (1_{A_2} -P(A_2))) \, 
$$
and $ E ((1_{A_i} -P(A_i))^2) = P(A_i) - P(A_i)^2 \leq \frac 14$, $i=1,2$),
that $-\frac 14$ and $\frac 14$ are lower and upper bounds,
achieved on simple examples (on $[0,1]$ with $P$ the Lebesgue measure e.g.).
But now for arbitrary $n \geq 3$?
 A: The suggestions in the comment by usul are correct.
Indeed, let
\begin{equation}
    p:=P(B),\quad B:=\bigcap_1^n A_j,\quad p_j:=P(A_j). 
\end{equation}
We want to find the extreme values of
\begin{equation}
    d:=p-\prod_1^n p_j. 
\end{equation}
Clearly, $p_j\ge p$ for all $j$ and hence
\begin{equation}
    d\le p-p^n\le\max_{0\le p\le1}(p-p^n)=r-r^n,
\end{equation}
where $r:=1/n^{1/(n-1)}$. On the other hand, if $A_1=\dotsb=A_n$ and $p_j=r$ for all $j$, then $d=r-r^n$. So,
\begin{equation}
    \max d=r-r^n=\frac{n-1}{n^{n/(n-1)}}  
\end{equation}
(so that $\max d\to1$ as $n\to\infty$).
Next,
\begin{equation}
    B^c=\bigcup_1^n A_j^c,
\end{equation}
where $^c$ denotes the complement. So,
\begin{equation}
    1-p=P(B^c)\le\sum_1^n P(A_j^c)=n-\sum_1^n p_j,
\end{equation}
so that $\sum_1^n p_j\le n-(1-p)$ and hence, by the AM–GM inequality,
\begin{equation}
    d\ge p-\Bigl(\frac{n-(1-p)}n\Bigr)^n=p-\Bigl(1-\frac{1-p}n\Bigr)^n=:f(p). 
\end{equation}
Since $f(p)$ is increasing in $p\in[0,1]$, we have $f(p)\ge f(0)=-\bigl(1-\frac1n\bigr)^n$. So,
$d\ge-\bigl(1-\frac1n\bigr)^n$. On the other hand, if the sets $A_1^c,\dotsc,A_n^c$ form a partition of $\Omega$ and $P(A_j^c)=\frac1n$ for all $j$, then $p=0$ and $p_j=1-\frac1n$ for all $j$, whence $d=-\bigl(1-\frac1n\bigr)^n$.
So,
\begin{equation}
    \min d=-\Bigl(1-\frac1n\Bigr)^n 
\end{equation}
(so that $\min d\to-1/e$ as $n\to\infty$).
