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.
But what about the minimum?
This question is an extension of "Find range of possible values for probability of intersection given individual probabilities".

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.
