# 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 $$n \geq 3$$?

• Largest should be if $A_1 = ... = A_n$, when the expression is $q - q^n$, maximized at $q=(1/n)^{1/(n-1)}$. For smallest I'm not sure, but you can get $n$ events of probability $1-1/n$ each with empty intersection, making the expression $-(1-1/n)^n$.
– usul
Commented Feb 13, 2023 at 12:05

Indeed, let $$$$p:=P(B),\quad B:=\bigcap_1^n A_j,\quad p_j:=P(A_j).$$$$ We want to find the extreme values of $$$$d:=p-\prod_1^n p_j.$$$$
Clearly, $$p_j\ge p$$ for all $$j$$ and hence $$$$d\le p-p^n\le\max_{0\le p\le1}(p-p^n)=r-r^n,$$$$ where $$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$$. So, $$$$\max d=r-r^n=\frac{n-1}{n^{n/(n-1)}}$$$$ (so that $$\max d\to1$$ as $$n\to\infty$$).
Next, $$$$B^c=\bigcup_1^n A_j^c,$$$$ where $$^c$$ denotes the complement. So, $$$$1-p=P(B^c)\le\sum_1^n P(A_j^c)=n-\sum_1^n p_j,$$$$ so that $$\sum_1^n p_j\le n-(1-p)$$ and hence, by the AM–GM inequality, $$$$d\ge p-\Bigl(\frac{n-(1-p)}n\Bigr)^n=p-\Bigl(1-\frac{1-p}n\Bigr)^n=:f(p).$$$$ Since $$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$$. So, $$$$\min d=-\Bigl(1-\frac1n\Bigr)^n$$$$ (so that $$\min d\to-1/e$$ as $$n\to\infty$$).