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I came across a particular proof

And I have not been able to follow through these steps

Where it is mentioned that the expansion of the term $\displaystyle f \left({\bigcup_{i \mathop = 1}^r A_i}\right)$ is to be found

Which is $\displaystyle \left({-1}\right)^{s-1} \sum_{{I} \subseteq \left[{1 \,.\,.\,r}\right] \atop {\left|{I}\right| = s}} f \left({\bigcap_{i \mathop \in I} A_i}\right) - \left({-1}\right)^{s-2} \sum_{{J} \subseteq \left[{1 \,.\,.\,r}\right] \atop {\left|{J}\right| = s-1}} f \left({\bigcap_{i \mathop \in J} A_i \cap A_{r+1}}\right)$

where : $I$ ranges over all sets of $s$ elements out of $\left[{1 \,.\,.\, r}\right]$ : $J$ ranges over all sets of $s-1$ elements out of $\left[{1 \,.\,.\, r}\right]$ : $1 \le s \le r$

How is it obtained and how does this equals to this?

$\displaystyle \left({-1}\right)^{s-1} \sum_{{I'} \subseteq \left[{1 \,.\,.\,r+1}\right] \atop {\left|{I'}\right| = s}} f \left({\bigcap_{i \mathop \in I'} A_i}\right)$ where, $I'$ ranges over all sets of $s$ elements out of $\left[{1 \,.\,.\, r+1}\right]$ and $1 \le s \le r+1$.

Andrés E. Caicedo
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1 Answers1

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If your question was about the intuition into the inclusion-exclusion principle, here's a quick explanation. Let $f$ be a probability measure. In a probability framework, suppose $P(\bigcup_{i} A_i)$ is the chance of occurrence of at least one of events. Then, you add up the chance of occurrence of each of the events $\sum_i P(A_i)$. But, since the events are not disjoint, two events can occur simultaneously. $P(A_i \cap A_j)$ has been counted twice - once in $P(A_i)$ and again in $P(A_j)$. So, we subtract $\sum_i P(A_i \cap A_j)$ to ensure that these events are counted only once. However, three events could occur together. $P(A_i \cap A_j \cap A_k)$ was counted three times in the first sum $\sum_i P(A_i)$, and subtracted three times in the second sum $\sum_i P(A_i \cap A_j)$, so you must add it back once.

Hence, the alternating additions and subtractions in the inclusion-exclusion formula.

Quasar
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