Index: The Book of Statistical ProofsGeneral Theorems ▷ Probability theory ▷ Probability ▷ Probability under exclusivity

Theorem: Let $A$ and $B$ be two statements about random variables. Then, if $A$ and $B$ are mutually exclusive, their joint probability is zero:

$\label{eq:prob-exc} p(A,B) = 0 \; .$

Proof: If $A$ and $B$ are mutually exclusive, then the probability of their disjunction is the sum of the marginal probabilities:

$\label{eq:exc} p(A \vee B) = p(A) + p(B) \; .$

The law of marginal probability implies that

$\label{eq:prob-marg} \begin{split} p(A) &= p(A,B) + p(A,\overline{B}) \\ p(B) &= p(A,B) + p(\overline{A},B) \end{split}$

where $\overline{A}$ and $\overline{B}$ are the complements of $A$ and $B$. The probability of the disjunction $p(A \vee B)$ can also be expressed as the probability of a disjunction of three mutually exclusive statements

$\label{eq:prob-exc-s1} p(A \vee B) = p\left([A \wedge \overline{B}] \vee [\overline{A} \wedge B] \vee [A \wedge B] \right) \; ,$

such that the definition of exclusivity can be applied to give

$\label{eq:prob-exc-s2} \begin{split} p(A \vee B) &\overset{\eqref{eq:prob-exc-s1}}{=} p\left([A \wedge \overline{B}] \vee [\overline{A} \wedge B] \vee [A \wedge B] \right) \\ &\overset{\eqref{eq:exc}}{=} p(A,\overline{B}) + p(\overline{A},B) + p(A,B) \\ &= [p(A,\overline{B}) + p(A,B)] + [p(\overline{A},B) + p(A,B)] - p(A,B) \\ &\overset{\eqref{eq:prob-marg}}{=} p(A) + p(B) - p(A,B) \; . \end{split}$

Since $A$ and $B$ are mutually exclusive, we obtain:

$\label{eq:prob-exc-qed} \begin{split} p(A \vee B) &\overset{\eqref{eq:prob-exc-s2}}{=} p(A) + p(B) - p(A,B) \\ p(A \vee B) &\overset{\eqref{eq:exc}}{=} p(A \vee B) - p(A,B) \\ p(A,B) &= 0 \; . \end{split}$
Sources:

Metadata: ID: P242 | shortcut: prob-exc | author: JoramSoch | date: 2021-07-23, 17:19.