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".

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

1 Answer 1


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

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


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

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.

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

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.