| Kolmogorov's Axioms of Probability
- (1) $\mathsf{P(\Omega)=1}$
- (2) $\mathsf{P(A)\geq 0}$, if $\mathsf{A \in \Omega}$
- (3) $\mathsf{A_{1}, A_{2}, \cdots \Rightarrow P(\cup_{k=1}^{\infty} A_{k})= \sum_{k=1}^{\infty}P(A_{k})}$
Example) Show that $\mathsf{P(\varnothing)=0}$
Let A be an event, then $A=A\cup \varnothing \cup \varnothing \cdots$, as A and $\varnothing$ are disjointed!
$\rightarrow P(A)=P(A \cup \varnothing \cup \varnothing \cdots) = P(A)+ P(\varnothing)+ \cdots$ (by Axiom 3) $\therefore P(\varnothing)=0$
| DeMorgan's Laws
- $(\mathsf{\bigcup_{i} A_{i})^c =\bigcap_{i} A_{i}^c}$, $\mathsf {(\bigcap_{i} A_{i})^c=\bigcup_{i} A_{i}^c}$
Proof
LHS $\rightarrow$ RHS
$\mathsf{(\bigcup_{i} A_{i})^c}$ occurs $\rightarrow \mathsf {(\bigcup_{i} A_{i})}$ does NOT occur $\rightarrow$ None of the $\mathsf {A_{i}}$ occurs $\rightarrow$ All $\mathsf {A_{i}^c}$ occur
$\mathsf {\therefore \bigcap_{i} A_{i}^c}$ occurs!
RHS $\rightarrow$ LHS
$\mathsf{\bigcap_{i} A_{i}}$ occurs $\rightarrow$ All $\mathsf{A_{1}^c, A_{2}^c...}$ occur $\rightarrow$ None of the A's occurs $\rightarrow$ $\mathsf{\bigcup A_{i}}$ does NOT occur
$\mathsf {\therefore (\bigcup A_{i})^c}$ occurs
No comments:
Post a Comment