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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.
But what about the minimum?

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

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  • $\begingroup$ Hint: consider the maximum probability of the union of the complements of the $A_i.$ $\endgroup$
    – whuber
    Jun 29, 2021 at 15:29
  • $\begingroup$ @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. $\endgroup$ Jun 29, 2021 at 16:10
  • $\begingroup$ 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. $\endgroup$
    – whuber
    Jun 29, 2021 at 16:21
  • $\begingroup$ The range tag is not suitable for this question. $\endgroup$
    – whuber
    Dec 4, 2021 at 14:41

1 Answer 1

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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.

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  • $\begingroup$ The notation adopted in that answer is that of Halmos. $\endgroup$
    – whuber
    Dec 4, 2021 at 15:44
  • $\begingroup$ Beautiful. Thank you! $\endgroup$ Dec 4, 2021 at 17:05

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