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By Inclusion-Exclusion Principle, we know that if $A_1,\ldots,A_n$ are $n$ sets, then $$\left|\bigcup_{i=1}^nA_i\right|=\sum_{k=1}^n(-1)^{k+1}\left(\sum_{1\leq i_1<\cdots<i_k\leq n}|A_{i_1}\cap\cdots\cap A_{i_k}|\right).$$

I wonder, what if we cut the outer sum on the right to just $\sum_{k=1}^r$ for some $r\leq n$? Will we guarantee inequality in one way or the other?

For $r=1$, it is obvious by Boole's Inequality that the left-hand side is $\leq$ the right-hand side.

I think that the same should hold for $r$ odd, and the reverse inequality should hold for $r$ even. Any proof/counterexamples for that?

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Aren't these the Bonferroni inequalities, which are in the same Wikipedia article that you linked to in your question? This old question asked "How to prove Bonferroni inequalities"; Did's answer gives some hints:

A proof is there. The main idea is that this is the integrated version of analogous pointwise inequalities and that, for every $k$, $$ S_k=\mathbb E\left({T\choose k}\right),\qquad T=\sum_{i=1}^n\mathbf 1_{A_i}. $$ Hence the result follows from the stronger inequalities asserting that, for every positive integer $N$, $$ \sum_{i=0}^k(-1)^ia_i,\qquad a_i={N\choose i}, $$ is nonnegative when $k$ is even and nonpositive when $k$ is odd. In turn, this fact follows from the properties that the sequence $(a_i)_{0\leqslant i\leqslant N}$ is unimodal and that $\sum\limits_{i=0}^N(-1)^ia_i=0$.

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  • $\begingroup$ No reason to downvote; better to see whether OP can discuss relation of Bonferroni to the question at hand. $\endgroup$ – Gerry Myerson Oct 4 '13 at 13:04
  • $\begingroup$ I think it's just the same as the Bonferroni you linked to, just the count replaced by probability. How do you prove Bonferroni though? $\endgroup$ – PJ Miller Oct 4 '13 at 13:14
  • $\begingroup$ The link in my answer is resurrected (and I took the liberty to modify your answer accordingly). $\endgroup$ – Did Oct 4 '13 at 18:11

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