1
$\begingroup$

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 $n \geq 3$?

$\endgroup$
1
  • 1
    $\begingroup$ 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 $-(1-1/n)^n$. $\endgroup$
    – usul
    Commented Feb 13, 2023 at 12:05

1 Answer 1

2
$\begingroup$

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 $n\to\infty$).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.