# A Question on the Inclusion-Exclusion Principle in Probability Theory

Let $$n \geq 3$$ be a fixed integer and consider an arbitrary set of the events $$\{A_i, 1 \leq i \leq n \}$$ in the same probability space. Consider the following two sums of probabilities: \begin{align} S_1 ^{(n)} &= \sum_{1 \leq i \leq n} \mathbb{P}(A_i); \\ S_2 ^{(n)} &= \sum_{1 \leq i_1 < i_2 \leq n} \mathbb{P}(A_{i_1} \cap A_{i_2}). \end{align} The question is whether it holds that $$S_1 ^{(n)} \geq S_2 ^{(n)}$$ and $$2 S_1 ^{(n)} \geq 3 S_2 ^{(n)}$$?

Both inequalities hold for $$n = 3$$. But I could not have a proof based on induction according to this base case. Anyone has a idea? Thanks very much.

• Neither inequality can hold in general as the case when all the events $A_i$ are equal, shows. – Did Dec 25 '18 at 10:38
• @Did Thanks a lot. In case that the events are not identical nor disjoint, which would be of more interest, do the inequalities hold in general? – Richie Dec 25 '18 at 11:13
• The case of identical events serves to show that some specific condition must be added. Of course nonequal events shall also disprove the conjecture: assume for example that $P(A_i)=a$ and $P(A_i\cap A_j)=b$ for every $i$ and $j$ and consider the limit when $n\to\infty$. Once again, these trivial cases only help to show the inequality cannot hold in general. (Not sure I understand your mention of disjointness since disjoint events automatically fit the conjecture because then $S^{(n)}_2=0$.) – Did Dec 25 '18 at 12:13