0
$\begingroup$

First of all i denote $\{1,2,3,...,m\}$ by $[m]$

Let there be a collection of sets $\alpha=\{A_{1},A_{2},...,A_{m}\}$ such $\bigcup_{i\in[m]}A_{i}\subseteq [n]$ Consider any function $f:\mathcal{P}([n])\rightarrow \mathbb{C}$

It is well known that $$f(\bigcup_{i\in[m]}A_{i})=\sum_{i=1}^{m}(-1)^{i-1}\sum_{I\subseteq[m],|I|=i}f(\bigcap_{l\in I}A_{l})$$

That formula is called a inclusion-exclusion principle

But what if we associate to $\alpha$ a matrix $\beta=(x_{A_{i}}(j))_{i\in[m],j\in[n]}$

Where $x_{A_{i}}(j)=\begin{cases} 1 &\text{when } j\in A_{i}\\0 &\text{otherwise } \end{cases}$

How to express inclusion-exclusion principle using operations on matrix $\beta$?

By operation i mean for example adding, multiplying rows, columns, taking determinant etc.

In particular i am intrested in case when $f$ is probability measure P.

Here are the things i done in this case: Consider binomiall and independent distribution over all the numbers from the set $[n]$. Probability that we pick an element $i$ is equal to $p$.

I noticed that $f(A\cap B)=P(A\cap B)=p^{|A\cup B|}$ (where $A$ stands both for set and an event to include set , same with $B$)

But how to use a matrix there?

$\endgroup$
2
  • $\begingroup$ Row $i$ is the $\left\{0,1\right\}$-vector corresponding to the set $A_i$. Thus, multiplying rows $i$ and $j$ entrywise yields the $\left\{0,1\right\}$-vector corresponding to the set $A_i \cap A_j$, and similarly for larger products. But multiplying rows of a matrix entrywise is not a standard matrix operation, so I'm not sure how much you gain from this point of view. $\endgroup$ Apr 19, 2020 at 12:53
  • $\begingroup$ Also, the inclusion-exclusion principle doesn't hold for any arbitrary $f$. You need $f$ to be modular (i.e., obtainable by summing values at single elements). $\endgroup$ Apr 19, 2020 at 12:55

1 Answer 1

0
$\begingroup$

As darij grinberg noted, the inclusion-exclusion principle holds only if $f$ is additive over $[n]$ or, equivalently, is a (signed) measure over $[n]$, and then $$f(A)=\int x_A\,df$$ for any $A\subseteq[n]$.

Now, to get the inclusion-exclusion principle from properties of indicators $x_A$ of sets $A$, write \begin{align}x_{\bigcup_1^m A_j}&=1-\prod_1^m(1-x_{A_j}) \\ &=\sum_{i=1}^m(-1)^{i-1}\sum_{J\in\binom{[n]}i}\prod_{j\in J} x_{A_j} \\ &=\sum_{i=1}^m(-1)^{i-1}\sum_{J\in\binom{[n]}i}x_{\bigcap_1^m A_j}, \end{align} where $\binom{[n]}i$ is the set of all sets $J$ such that $J\subseteq[n]$ and $|J|=i$.
Integrating now with respect to $df$, we get the inclusion-exclusion principle: \begin{align}f(\bigcup_1^m A_j)&= \sum_{i=1}^m(-1)^{i-1}\sum_{J\in\binom{[n]}j}f(\bigcap_1^m A_j). \end{align}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.