I'm stuck in this exercise:
$(\Omega, \mathbb{P})$ is a discrete probability space and $A_1, …,A_n$ are $n \in \mathbb{N^*}$ events. There are
$\epsilon_1, …, \epsilon_n \in [0,1]$ with $\mathbb{P}(A_i) \ge 1- \epsilon_i$ for every $i = 1,…,n$
Show $$\mathbb{P}(\bigcap_{i=1}^{n}A_i) \ge 1-\sum_{i=1}^n\epsilon_i$$ with $$\mathbb{P(A_i) \ge 1-\epsilon_i}$$
Thoughts/Attempt:
Intuitively the left-hand side we have the intersection of some potentially non-disjoint events. Their correct probability is given by something similar to the sieve formular, it is basically about the Exclusion-Inclusion principle:
$$\mathbb{P}(\bigcap_{i=1}^{n}A_i) = \sum_{I:I \subseteq \{1,…,n\}, I \ne \emptyset}(-1)^{1+|I|}\cdot \mathbb{P}(A^I), A^I := \bigcup_{i \in I}A_i $$
So we get
$$\mathbb{P}(\bigcap_{i=1}^{n}A_i) = \sum_{I:I \subseteq \{1,…,n\}, I \ne \emptyset}(-1)^{1+|I|}\cdot \sum_{i \in I}(1-\epsilon_i)$$
I don't know what to make of this alternating series
Best Answer
$\mathbb{P}(\bigcap_{i=1}^{n}A_i) =1-\mathbb{P}(\bigcup_{i=1}^{n}A^{c}_i)\geq 1- \sum P(A_i^{c}) \geq 1- \sum \epsilon_i$.
[Note that $\mathbb{P}(\bigcup_{i=1}^{n}A^{c}_i) \leq \sum P(A_i^{c})$ and $P(A_i^{c})=1-P(A_i) \leq \epsilon_i$]