Inclusion-Exclusion Principle

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Theorem

Let $\SS$ be an algebra of sets.

Let $A_1, A_2, \ldots, A_n$ be finite sets.

Let $f: \SS \to \R$ be an additive function.


Then:

\(\ds \map f {\bigcup_{i \mathop = 1}^n A_i}\) \(=\) \(\ds \sum_{i \mathop = 1}^n \map f {A_i}\)
\(\ds \) \(\) \(\, \ds - \, \) \(\ds \sum_{1 \mathop \le i \mathop < j \mathop \le n} \map f {A_i \cap A_j}\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \sum_{1 \mathop \le i \mathop < j \mathop < k \mathop \le n} \map f {A_i \cap A_j \cap A_k}\)
\(\ds \) \(\) \(\, \ds \cdots \, \) \(\ds \)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \paren {-1}^{n - 1} \map f {\bigcap_{i \mathop = 1}^n A_i}\)


Corollary

Let $\SS$ be an algebra of sets.

Let $A_1, A_2, \ldots, A_n$ be finite sets which are pairwise disjoint.

Let $f: \SS \to \R$ be an additive function.


Then:

$\ds \map f {\bigcup_{i \mathop = 1}^n A_i} = \sum_{i \mathop = 1}^n \map f {A_i}$


Proof

Proof by induction:

For all $n \in \N_{>0}$, let $\map P N$ be the proposition:

\(\ds \map f {\bigcup_{i \mathop = 1}^n A_i}\) \(=\) \(\ds \sum_{i \mathop = 1}^n \map f {A_i}\)
\(\ds \) \(\) \(\, \ds - \, \) \(\ds \sum_{1 \mathop \le i \mathop < j \mathop \le n} \map f {A_i \cap A_j}\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \sum_{1 \mathop \le i \mathop < j \mathop < k \mathop \le n} \map f {A_i \cap A_j \cap A_k}\)
\(\ds \) \(\) \(\, \ds \cdots \, \) \(\ds \)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \paren {-1}^{n - 1} \map f {\bigcap_{i \mathop = 1}^n A_i}\)


$\map P 1$ is true, as this just says $\map f {A_1} = \map f {A_1}$.


Basis for the Induction

$\map P 2$ is the case:

$\map f {A_1 \cup A_2} = \map f {A_1} + \map f {A_2} - \map f {A_1 \cap A_2}$

which is the result Additive Function is Strongly Additive.

This is our basis for the induction.


Induction Hypothesis

Now we need to show that, if $\map P r$ is true, where $r \ge 2$, then it logically follows that $\map P {r + 1}$ is true.


So this is our induction hypothesis:

\(\ds \map f {\bigcup_{i \mathop = 1}^r A_i}\) \(=\) \(\ds \sum_{i \mathop = 1}^r \map f {A_i}\)
\(\ds \) \(\) \(\, \ds - \, \) \(\ds \sum_{1 \mathop \le i \mathop < j \mathop \le r} \map f {A_i \cap A_j}\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \sum_{1 \mathop \le i \mathop < j \mathop < k \mathop \le r} \map f {A_i \cap A_j \cap A_k}\)
\(\ds \) \(\) \(\, \ds \cdots \, \) \(\ds \)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \paren {-1}^{r - 1} \map f {\bigcap_{i \mathop = 1}^r A_i}\)


Then we need to show:

\(\ds \map f {\bigcup_{i \mathop = 1}^{r + 1} A_i}\) \(=\) \(\ds \sum_{i \mathop = 1}^{r + 1} \map f {A_i}\)
\(\ds \) \(\) \(\, \ds - \, \) \(\ds \sum_{1 \mathop \le i \mathop < j \mathop \le r + 1} \map f {A_i \cap A_j}\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \sum_{1 \mathop \le i \mathop < j \mathop < k \mathop \le r + 1} \map f {A_i \cap A_j \cap A_k}\)
\(\ds \) \(\) \(\, \ds \cdots \, \) \(\ds \)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \paren {-1}^r \map f {\bigcap_{i \mathop = 1}^{r + 1} A_i}\)


Induction Step

This is our induction step:

\(\ds \map f {\bigcup_{i \mathop = 1}^{r + 1} A_i}\) \(=\) \(\ds \map f {\bigcup_{i \mathop = 1}^r A_i \cup A_{r + 1} }\)
\(\ds \) \(=\) \(\ds \map f {\bigcup_{i \mathop = 1}^r A_i} + \map f {A_{r + 1} } - \map f {\bigcup_{i \mathop = 1}^r A_i \cap A_{r + 1} }\) Basis for the Induction


Consider $\ds \map f {\bigcup_{i \mathop = 1}^r A_i \cap A_{r + 1} }$.

By the fact that Intersection Distributes over Union, this can be written:

$\ds \map f {\bigcup_{i \mathop = 1}^r \paren {A_i \cap A_{r + 1} } }$

To this, we can apply the induction hypothesis:

\(\ds \map f {\bigcup_{i \mathop = 1}^r \paren {A_i \cap A_{r + 1} } }\) \(=\) \(\ds \sum_{i \mathop = 1}^r \map f {A_i \cap A_{r + 1} }\)
\(\ds \) \(\) \(\, \ds - \, \) \(\ds \sum_{1 \mathop \le i \mathop < j \mathop \le r} \map f {A_i \cap A_j \cap A_{r + 1} }\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \sum_{1 \mathop \le i \mathop < j \mathop < k \mathop \le r} \map f {A_i \cap A_j \cap A_k \cap A_{r + 1} }\)
\(\ds \) \(\) \(\, \ds \cdots \, \) \(\ds \)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \paren {-1}^{r - 1} \map f {\bigcap_{i \mathop = 1}^r A_i \cap A_{r + 1} }\)


At the same time, we have the expansion of the term $\ds \map f {\bigcup_{i \mathop = 1}^r A_i}$ to take into account.


So we can consider the general term of $s$ intersections in the expansion of $\ds \map f {\bigcup_{i \mathop = 1}^{r + 1} A_i}$:

$\ds \paren {-1}^{s - 1} \sum_{\substack {I \subseteq \closedint 1 r \\ \card I = s} } \map f {\bigcap_{i \mathop \in I} A_i} - \paren {-1}^{s - 2} \sum_{\substack {J \subseteq \closedint 1 r \\ \card J = s - 1} } \map f {\bigcap_{i \mathop \in J} A_i \cap A_{r + 1} }$

where:

$I$ ranges over all sets of $s$ elements out of $\closedint 1 r$
$J$ ranges over all sets of $s - 1$ elements out of $\closedint 1 r$
$1 \le s \le r$


Messy though it is, it can be seen that this expression is merely:

$\ds \paren {-1}^{s - 1} \sum_{\substack {I' \subseteq \closedint 1 {r + 1} \\ \card {I'} = s} } \map f {\bigcap_{i \mathop \in I'} A_i}$

where this time, $I'$ ranges over all sets of $s$ elements out of $\closedint 1 {r + 1}$ and $1 \le s \le r + 1$.


This is the required term in $s$ intersections in the expansion of $\ds \map f {\bigcup_{i \mathop = 1}^{r + 1} A_i}$.


Just to check, we can see the first term is:

$\ds \sum_{i \mathop = 1}^r \map f {A_i} + \map f {A_{r + 1} } = \sum_{i \mathop = 1}^{r + 1} \map f {A_i}$

As the expression $\ds \map f {\bigcup_{i \mathop = 1}^r A_i \cap A_{r + 1} }$ consists only of intersections of two or more elements of $\SS$, we see it does not contribute to this first term.


Finally, let us make sure of the last term.

This is:

$\ds -\paren {-1}^{r - 1} \map f {\bigcap_{i \mathop = 1}^r A_i \cap A_{r + 1} }$

which works out as:

$\ds \paren {-1}^r \map f {\bigcap_{i \mathop = 1}^{r + 1} A_i}$


We've done enough.


So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.


Therefore:

\(\ds \map f {\bigcup_{i \mathop = 1}^n A_i}\) \(=\) \(\ds \sum_{i \mathop = 1}^n \map f {A_i}\)
\(\ds \) \(\) \(\, \ds - \, \) \(\ds \sum_{1 \mathop \le i \mathop < j \mathop \le n} \map f {A_i \cap A_j}\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \sum_{1 \mathop \le i \mathop < j \mathop < k \mathop \le n} \map f {A_i \cap A_j \cap A_k}\)
\(\ds \) \(\) \(\, \ds \cdots \, \) \(\ds \)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \paren {-1}^{n - 1} \map f {\bigcap_{i \mathop = 1}^n A_i}\)

$\blacksquare$

Examples

$3$ Events in Event Space

Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.

Let $A, B, C \in \Sigma$.

Then:

\(\ds \map \Pr {A \cup B \cup C}\) \(=\) \(\ds \map \Pr A + \map \Pr B + \map \Pr C\)
\(\ds \) \(\) \(\, \ds - \, \) \(\ds \map \Pr {A \cap B} - \map \Pr {B \cap C} - \map \Pr {A \cap C}\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \map \Pr {A \cap B \cap C}\)


$3$ Events in Event Space: Example

Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.

Let $A, B, C \in \Sigma$ such that:.

Then:

\(\ds \map \Pr A\) \(=\) \(\ds \dfrac 5 {10}\)
\(\ds \map \Pr B\) \(=\) \(\ds \dfrac 7 {10}\)
\(\ds \map \Pr C\) \(=\) \(\ds \dfrac 6 {10}\)
\(\ds \map \Pr {A \cap B}\) \(=\) \(\ds \dfrac 3 {10}\)
\(\ds \map \Pr {B \cap C}\) \(=\) \(\ds \dfrac 4 {10}\)
\(\ds \map \Pr {A \cap C}\) \(=\) \(\ds \dfrac 2 {10}\)
\(\ds \map \Pr {A \cap B \cap C}\) \(=\) \(\ds \dfrac 1 {10}\)

The probability that exactly $2$ of the events $A$, $B$ and $C$ occur is $\dfrac 6 {10}$.


Context

This result is usually quoted in the context of combinatorics, where $f$ is the cardinality function.

It is also seen in the context of probability theory, in which $f$ is taken to be a probability measure.


Historical Note

The Inclusion-Exclusion Principle, in various forms, has been attributed to:


Sources

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