Show that varrho(A,B)=P(Atriangle B) satisfies all axioms of…

Show that \[ \varrho(A,B)=P(A\triangle B) \] satisfies all axioms of a metric space except the identity axiom ($\varrho(A,B)=0$ need not imply $A=B$), and that for arbitrary events $A,B,C$ \[ \varrho(A,B)+\varrho(B,C)\ge\varrho(A,C). \]

Step 1: The metric axioms

A function $d(x,y)$ is a metric if for all $x,y,z$:

1. $d(x,y)\ge0$ and $d(x,x)=0$;
2. $d(x,y)=0$ implies $x=y$ (identity);
3. $d(x,y)=d(y,x)$ (symmetry);
4. $d(x,y)+d(y,z)\ge d(x,z)$ (triangle inequality).

Step 2: Axioms 1 and 3 hold

• $\varrho(A,B)=P(\cdot)\ge0$, and $A\triangle A=\varnothing$, so $\varrho(A,A)=0$.
• $A\triangle B = A\overline{B}+\overline{A}B = B\triangle A$, so $\varrho(A,B)=\varrho(B,A)$.

Step 3: Axiom 2 fails in general

$\varrho(A,B)=0$ means only that $A$ and $B$ differ on an event of probability zero. Example: choose a point uniformly in $[0,1]$, $A=\{\tfrac12\}$ ("the point is exactly $\tfrac12$"), $B=\varnothing$. Then $A\triangle B=A$ and $\varrho(A,B)=P(A)=0$, but $A\ne B$. So $\varrho$ is a pseudometric; it becomes a metric if we identify events that differ by a null event.

Step 4: Axiom 4 — the triangle inequality

Claim: $A\triangle C\subseteq (A\triangle B)+(B\triangle C)$.

Take $\omega\in A\triangle C$; say $\omega\in A$, $\omega\notin C$ (the other case is symmetric). If $\omega\in B$, then $\omega\in B\triangle C$. If $\omega\notin B$, then $\omega\in A\triangle B$. Either way $\omega$ is in the right side.

Probability is monotone and $P(X+Y)\le P(X)+P(Y)$, so

\[ \varrho(A,C) = P(A\triangle C) \le P(A\triangle B)+P(B\triangle C) = \varrho(A,B)+\varrho(B,C). \quad\blacksquare \]

Step 5: The exact difference — eight atoms

Split $\Omega$ into the $8$ atoms $A^{x}B^{y}C^{z}$ ($x,y,z\in\{0,1\}$, $A^1=A$, $A^0=\overline{A}$). On an atom, $\omega\in A\triangle B$ exactly when $x\ne y$. Count how each atom contributes to $\varrho(A,B)+\varrho(B,C)$ and to $\varrho(A,C)$:

Only the atoms where $B$ disagrees with both $A$ and $C$ (while $A$ and $C$ agree) contribute to the difference, and they contribute twice:

\[ \boxed{\varrho(A,B)+\varrho(B,C)-\varrho(A,C) = 2\Big[P\big(A\overline{B}C\big)+P\big(\overline{A}B\overline{C}\big)\Big]\ \ge 0.} \]

This gives the triangle inequality again and shows exactly when it becomes an equality (see Problem 64).

Step 6: Example

Roll a die: $A=\{1,2,3\}$, $B=\{2,3,4\}$, $C=\{3,4,5\}$. Then $\varrho(A,B)=P(\{1,4\})=\tfrac26$, $\varrho(B,C)=P(\{2,5\})=\tfrac26$, $\varrho(A,C)=P(\{1,2,4,5\})=\tfrac46$. Here $\tfrac26+\tfrac26=\tfrac46$: equality, because $A\overline{B}C=\overline{A}B\overline{C}=\varnothing$.