Suppose I have the following events $A_0, A_1, ... A_N$. These events are not independent. I want to know the probability of the event $A_0 \cap A_1 \cap ... \cap A_N$ occurring. However, I don't have the information I need to do this. All I know is the probability of each individual event (i.e. $P(A_0)$, $P(A_1)$, etc). Instead, I would like to create a lower bound on the probability.
From here I found the following inequality
$$P(B \cap C) \geq max(0, P(B) + P(C) - 1)$$
Using this inequality I can write (I believe, correct me if I am wrong)
$$P(A_0 \cap A_1 \cap ... \cap A_N) \geq max(0, P(A_0) + P(A_1 \cap ... \cap A_N) - 1)$$
Observing that
$$P(A_1 \cap ... \cap A_N) \leq min(P(A_1), ... , P(A_N))$$
I then can observe that
$$P(A_0 \cap A_1 \cap ... \cap A_N) \geq max(0, P(A_0) + min(P(A_1), ... , P(A_N)) - 1)$$
WLOG, if I were to sort $A_0, A_1, ... A_N$ such that $P(A_0)$ has the highest probability and $P(A_N)$ the lowest I would then have a "decent" bound given the information I have available (I say decent since most of the time there will be strong dependencies between events, i.e. event $A_0$ can only occur if $A_1$ has occurred. However, this statement is not true in general.).
My question(s) are the following.
(1) Is my reasoning correct?
(2) Is there a better way to do what I want?
Sorry, that doesn't work. You made an algebraic error when substituting, specifically, an error in the direction of the $\ge$ sign.
First you can drop the $\max(0, *)$ part - your LHS is a probability which is always non-negative anyway. After dropping that, what you have is similar to $X \ge Y - 1$ and $Y \le Z$, but you cannot combine them to get $X \ge Z - 1$. If you have $Y \ge Z$ then you can combine them, but you only have $Y \le Z$ which is the wrong direction.
Instead, you can apply the first inequality recursively to get:
$$P(A_0 \cap ... \cap A_N) \ge \sum^N_{j=0} P(A_j) - N$$
Without more info, this inequality is tight. E.g. imagine you have $N+1$ balls labelled $j \in \{0, 1, ..., N\}$ and one is picked at random, and $A_j = $ the event that ball $j$ is NOT picked. So $P(A_j) = {N \over N+1} $ and both sides $= 0$.
Follow-up details on the recursion:
$$P(A_0 \cap ...\cap A_N) \ge (P(A_0) - 1) + P(A_1 \cap ... \cap A_N) \ge (P(A_0) -1) + (P(A_1) - 1) + P(A_2 \cap ... \cap A_N) \ge ... \ge \sum^{N-1}_{j=0} (P(A_j) - 1) + P(A_N) = ( \sum^N_{j=0} P(A_j) ) - N$$