Pictorially understanding Bonferroni's inequalities

Question

I understand in general what the Bonferroni's inequalities show (finding the upper and lower bounds in a finite probability of unions). Though, I don't understand why it works pictorially. Let's assume for this example n=3

First (which makes sense to me), the sum of the individual probabilities summed together will be greater than the union as you're "counting" the union between sets twice... $$P\left(\bigcup_{i=1}^n A_i\right) \le \sum_{i=1}^n P(A_i)$$

...so, to make up for this correction, you are subtracting the union of two elements. In doing so you are "overcorrecting" and thus subtracting the union of two sets and thus ignore where the intersection of all three events occurs:

$$P\left(\bigcup_{i=1}^n A_i\right) \ge \sum_{i=1}^n P(A_i)-\sum_{1 \le i < j \le n}^n P(A_i \cap A_j)$$

However, in the case of where n=3, does this not satisfy the inclusion-exclusion principle of probability because subtracting out the intersection of all three events should be sufficient to satisfy our original innequality:

$$P\left(\bigcup_{i=1}^n A_i\right) = \sum_{i=1}^n P(A_i) - \sum_{1 \le i < j \le n} P(A_i \cap A_j)+\sum_{1\le i < j < k \le n} P(A_i \cap A_j \cap A_k) $$

Thus, how does this show for the Bonferroni innequalities when n=3?

Answer 1

After letting the problem sit for a few hours, I think I may finally understand this. Bonferroni's inequalities are best thought of as an approximation to an unknown n (as long as n satisfies the notion of a finite set).

Thus as we approach n we know to alternatively flip the inequality sign as we are continuously over and under correcting.

If one would like to look at a finite n (as used in the original question, n=3) then simply stop at the n-1 sigma and notice whether we have over or under approximated (in this case we would have under approximated).

Referring to my original question, the statement:

$$P\left(\bigcup_{i=1}^n A_i\right) = \sum_{i=1}^n P(A_i) - \sum_{1 \le i < j \le n} P(A_i \cap A_j)+\sum_{1\le i < j < k \le n} P(A_i \cap A_j \cap A_k) $$

should be left as:

$$P\left(\bigcup_{i=1}^n A_i\right) \le \sum_{i=1}^n P(A_i) - \sum_{1 \le i < j \le n} P(A_i \cap A_j)+\sum_{1\le i < j < k \le n} P(A_i \cap A_j \cap A_k) $$

as it is technically correct and would serve as a lower bound estimation for when n+1 (or n=4).