关于概率子集族交事件的概率比不等式成立条件的技术问询

Hey there, let's dig into this probability inequality question—great query by the way! First, let's recap the setup to make sure we're aligned:

We have subsets ( A_i \subset B_i ) for all ( i = 1, \dots, n ), and for every pair of distinct indices ( i \neq j ), the inequality
$$\frac{P(A_{i}\cap A_{j})}{P(B_{i}\cap B_{j})} \leq \frac{P(A_i)}{P(B_i)}$$
holds true. We want to figure out when the following stronger inequality is satisfied for all ( i = 1, \dots, n ):
$$\frac{P(\bigcap\limits_{i=1}^{{n}A_i)}{P(\bigcap\limits_{i=1}}{n}B_i)}\leq \frac{P(A_i)}{P(B_i)}$$

Let's start with small cases to build intuition #

• For ( n = 2 ): This is exactly the pairwise condition given in the problem, so the statement is automatically true. No extra hoops to jump through here.
• For ( n \geq 3 ): The pairwise condition isn't always sufficient on its own—we need some additional structure on the events to guarantee the n-way intersection inequality holds.

Key conditions that ensure the inequality holds #

Here are the most common and practical scenarios where the desired inequality is satisfied:

1. Conditional independence relative to the intersection of other ( B_k )
   Suppose that for each ( i ), ( A_i ) is conditionally independent of ( \bigcap_{k \neq i} A_k ) given ( \bigcap_{k \neq i} B_k ). We can recursively expand the n-way intersection probability using this independence, and since each step respects the pairwise ratio bound, the final ratio will stay bounded by ( \frac{P(A_i)}{P(B_i)} ) for all ( i ).

2. Positive association of restricted events
   If the events ( {A_i \mid B_i} ) (meaning ( A_i ) considered only within the space of ( B_i )) are positively associated—knowing one ( A_j ) occurs (within ( B_j )) doesn't decrease the likelihood another ( A_k ) occurs (within ( B_k ))—then the intersection probability ratio won't exceed the individual ratios. This is a standard framework in combinatorial probability for these types of inequalities.

3. Recursive monotonicity of intersection ratios
   If for every ( k ) from 2 to ( n ), the ratio ( \frac{P(\bigcap_{m=1}^k A_m)}{P(\bigcap_{m=1}^k B_m)} \leq \frac{P(\bigcap_{m=1}^{k-1} A_m)}{P(\bigcap_{m=1}^{k-1} B_m)} ), we can use induction to extend this to the full n-way intersection. This recursive "shrinking" of the ratio directly implies the desired inequality for all ( i ).

A counterexample to show when the condition fails #

To see why we need extra conditions for ( n \geq 3 ), let's build a concrete case where pairwise conditions hold but the n-way inequality breaks:

• Let ( B_1 = {1,2,3,4} ), ( B_2 = {1,3,4,5} ), ( B_3 = {1,2,4,5} ) (all have probability 1, so we're working over the full space ( {1,2,3,4,5} )).
• Let ( A_1 = {1,2} ), ( A_2 = {1,3} ), ( A_3 = {1,4} ), with probabilities assigned as:
  • ( P(1) = 0.08 ), ( P(2) = 0.42 ), ( P(3) = 0.42 ), ( P(4) = 0.07 ), ( P(5) = 0.01 )
• Check pairwise ratios: For ( i=1,j=2 ), ( \frac{P(A_1 \cap A_2)}{P(B_1 \cap B_2)} = \frac{0.08}{0.08+0.42+0.07} ≈ 0.14 \leq 0.5 = \frac{P(A_1)}{P(B_1)} )—this holds for all pairs.
• Now compute the n-way ratio: ( \frac{P(\bigcap A_i)}{P(\bigcap B_i)} = \frac{0.08}{0.08+0.07} ≈ 0.53 > 0.5 = \frac{P(A_i)}{P(B_i)} )

This violates the desired inequality, proving that pairwise conditions alone aren't enough for ( n \geq 3 ).

Quick summary #

The inequality holds for all ( i ) if either:

• ( n = 2 ) (it's exactly the given pairwise condition), or
• For ( n \geq 3 ), we have additional structure like conditional independence, positive association of restricted events, or recursive monotonicity of intersection ratios.

备注：内容来源于stack exchange，提问作者Vishal Tripathy