# Range of values for the probability of the intersection of a finite number of events

If we have $$P(A_i)$$ for a finite number of events $$A_i$$ (with $$i=1,\dots,n$$), then what is the range of values that their intersection $$P(\cap_{i=1}^n A_i)$$ can take?

I think the maximum is $$\min_i P(A_i)$$. Given the event $$A_j$$ with $$j=\arg\min_i P(A_i)$$, there is no way to increase the probability from $$P(A_j)=\min_i P(A_i)$$ by intersecting $$A_j$$ with other events.

This question is an extension of "Find range of possible values for probability of intersection given individual probabilities".

• Hint: consider the maximum probability of the union of the complements of the $A_i.$
– whuber
Jun 29, 2021 at 15:29
• @whuber, let me try. $P(\cup A^c_i)\leq \sum_i P(A^c_i)$. Whenever the latter is $>1$ it is not very useful, though. Jun 29, 2021 at 16:10
• Let's look at it even more simply, then: consider the case $n=2.$ What condition on the two probabilities would guarantee the intersection of the two events must have a nonzero probability? Now generalize.
– whuber
Jun 29, 2021 at 16:21
• The range tag is not suitable for this question.
– whuber
Dec 4, 2021 at 14:41

First, probabilities are monotonic. This means the probability of any subset of an event cannot exceed the probability of the event itself.

Let $$\mathcal S$$ be an at most countable collection of events in a common probability space. (This generalizes the question, which concerns only finite collections.) Then, since $$\cap \mathcal S \subset E$$ for all $$E\in \mathcal S$$ (that is part of the definition of set intersection), monotonicity implies

$$\Pr(\cap \mathcal S) \le \Pr(E)$$

for all $$E\in S.$$ That is equivalent (by definition of the minimum, which I use interchangeably with "infimum" here) to

$$\Pr(\cap \mathcal S) \le \min_{E\in\mathcal S}(\Pr(E)).$$

Monotonicity also implies that for any collection $$\mathcal T$$ of events,

$$\Pr(\cup T) \le \sum_{F\in\mathcal T} \Pr(F).$$

Apply this to the set of complements of the original events, $$\mathcal T = \{\Omega\setminus E\mid E\in\mathcal S\},$$ allowing for DeMorgan's Law $$\Omega\setminus \cup\mathcal T = \cap\mathcal S.$$ From two applications of the complement rule we obtain

\begin{aligned} \Pr(\cap S) &= \Pr(\Omega\setminus \cup T) = 1 - \Pr(\cup T) \\ &\ge 1 - \sum_{F\in\mathcal T}\Pr(F) \\ &= 1 - \sum_{E\in \mathcal S}\Pr(\Omega\setminus E) \\ &= 1 - \sum_{E\in\mathcal S}(1-\Pr(E)). \end{aligned}

Thus,

$$1 - \sum_{E\in\mathcal S}(1-\Pr(E))\, \le\, \Pr(\cap \mathcal S)\, \le\, \min_{E\in\mathcal S}(\Pr(E)).$$

Often the left hand side is negative: in such cases, we may replace it with $$0$$ (because probabilities are non-negative). Obviously the lower bound can be positive only if $$\mathcal S$$ has nonempty intersection.

Let's show these are the tightest possible general bounds. It suffices to provide examples. Let $$\Omega$$ be the probability space on the set $$\{0,1,2\}$$ with the totally discrete sigma algebra and uniform probability measure. Let $$\mathcal S = \{\{0,1,2\},\,\{2\}\}.$$ Here, $$\cap \mathcal S = \{2\}$$ has probability $$1/3$$ and $$1-\sum_{E\in\mathcal S}(1-\Pr(E)) = 1 - \left(1-\Pr(\{0,1,2\}\, + \, 1 - \Pr(\{2\})\right) = 1 - (0 + 2/3) = 1/3.$$ The lower bound is attained.

Moreover, since $$1/3$$ is the smallest of the probabilities of the members of $$\mathcal S,$$ the upper bound is also attained, QED.