Given the probability space $(\Omega,\mathfrak{A},P)$ and sequence of events $A_1,A_2…A_n\in \mathfrak{A}$ how do I go about proving the inequality:
$$\max \bigg\{0, \sum_{j=1}^{n}P(A_j) -n+1 \bigg\}\leq P\bigg( \bigcap_{j=1}^{n} A_j\bigg)\leq \min_{1 \leq j \leq n} P(A_j).$$
The inequality on the right-hand side makes some sense to me. If we consider $A_j$ a decreasing sequence the intersection of all the events is bound to be less than or equal to the smallest $A_j$ and so the inequality of the probability would follow from that but I'm not sure if such an assumption is too strong. We're not given whether the sequence is decreasing so I'm not sure how to proceed.
As for the left hand side I'm just not sure where to begin.
Best Answer
First inequality
It's immediate that $0 \leq P \left( \bigcap_{j=1}^n A_j \right)$, so we just need to show the inequality for the second argument in the $\max$. To that end, use De Morgan's law to get $$\begin{align*} P\left( \bigcap_{j=1}^n A_j \right) = P \left[\left( \bigcup_{j=1}^n A_j^c \right)^c\right] = 1 - P \left( \bigcup_{j=1}^n A_j^c \right) \geq 1 - \underbrace{\sum_{j=1}^n P(A_j^c)}_{\sum (1- P(A_j))} = \sum_{j=1}^n P(A_j) - n + 1 \end{align*}$$ where we used the union bound for the inquality.
Second inequality
Note that we have $\bigcap_{j=1}^n A_j \subseteq A_k$ for every $k = 1, \ldots , n$. Thus, by monotonicity, $$P \left( \bigcap_{j=1}^n A_j \right) \leq P(A_k) \qquad \text{for all } k$$ so that in particular: $$P \left( \bigcap_{j=1}^n A_j \right) \leq \min_{1 \leq k\leq n}P(A_k) $$