確率の演算

確率測度の演算

\(\forall A,B \in \mathcal{F}\)に対して,以下が成立する.

\[ \begin{align} & P(A^c) = 1 - P(A) \tag{P.1} \\ & A \subset B \to P(A) \leq P(B) \tag{P.2} \\ & P(A \cup B) = P(A) + P(B) - P(A \cap B) \tag{P.3} \end{align} \]

証明は以下の通り,

• (P.1) \[ \begin{align} &({\rm O}.2) \to A \cup A^c = \Omega \Rightarrow \\ & ({\rm M}.2) \to P(A \cup A^c) = P(\Omega) = 1 \Rightarrow \\ & ({\rm M}.3) \to P(A)+P(A^c) = 1 \Rightarrow \\ & P(A^c) = 1-P(A) \end{align} \]

• (P.2) \[ \begin{align} & A \subset B \to B = A \cup (B \cap A^c) \land \emptyset = A \cap (B \cap A^c) \Rightarrow \\ & P(B) = P(A \cup (B \cap A^c)) = P(A) + P(B \cap A^c) \Rightarrow\\ & ({\rm M}.1) \to P(A) \leq P(B) \end{align} \]

• (P.3) \[ \begin{align} P(A \cup B) &= P(((A \cup B) \cap (A \cap B)^c) \cup (A \cap B)) \\ &= P(((A \cup B) \cap (A \cap B)^c))+P(A \cap B) \\ &= P((A \cap (A \cap B)^c) \cup (B \cap (A \cap B)^c))+P(A \cap B) \\ &= P(A \cap (A \cap B)^c) + P(B \cap (A \cap B)^c)+P(A \cap B) \\ &= P((A^c \cup (A \cap B))^c) + P((B^c \cup (A \cap B))^c)+P(A \cap B) \\ &= 2-P(A^c \cup (A \cap B)) - P(B^c \cup (A \cap B))+P(A \cap B) \\ &= 2-P(A^c) - P(A \cap B) - P(B^c) -P(A \cap B)+P(A \cap B) \\ &= 1 - P(A^c) + 1 - P(B^c) - P(A \cap B) \\ &= P(A) + P(B) - P(A \cap B) \\ \end{align} \]