Characteristic Function of Symmetric Difference
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Theorem
Let $A, B \subseteq S$.
Then:
- $\chi_{A \symdif B} = \chi_A + \chi_B - 2 \chi_{A \cap B}$
where:
- $\chi$ denotes characteristic function
- $\symdif$ denotes symmetric difference.
Proof
By definition of symmetric difference:
- $A \symdif B = \paren {A \cup B} \setminus \paren {A \cap B}$
Thus:
- $\chi_{A \symdif B} = \chi_{A \mathop \cup B} - \chi_{\paren {A \mathop \cup B} \mathop \cap \paren {A \mathop \cap B} }$
by Characteristic Function of Set Difference.
But by Intersection is Subset of Union and Intersection with Subset is Subset:
- $\paren {A \cup B} \cap \paren {A \cap B} = A \cap B$
Hence it follows that:
- $\chi_{A \symdif B} = \chi_{A \mathop \cup B} - \chi_{A \mathop \cap B}$
which by Characteristic Function of Union: Variant 2 becomes:
- $\chi_{A \symdif B} = \chi_A + \chi_B - 2 \chi_{A \mathop \cap B}$
as desired.
$\blacksquare$
Sources
- 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $\S 2$: Problem $6 \ \text{(i)}$