# Cutting the tail of inclusion-exclusion principle

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?

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 $\sum\limits_{i=0}^N(-1)^ia_i=0$.