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 $n \geq 3$?



Gap to independence
The suggestions in the [comment](https://mathoverflow.net/questions/440758/gap-to-independence#comment1137007_440758) 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 $n\to\infty$).


