## Axioms - *Non-negativity:* Probabilities are always $\ge0$. This implies that no probability can be gt;1$, because the normalization axiom says that the sum of all [[Events]] $=1$. $ \mathbf P(A) \ge 0 \implies \mathbf P(A)\le1 $ - *Normalization:* The sum of the probabilities from the whole [[Sample Space]] $=1$. This implies that the probability of the empty set $\emptyset=0$ $ \mathbf P(\Omega)=1 \implies \mathbf P(\emptyset)=0 $ ^88c7a5 - *Additivity:* Probabilities of disjoint events can be added up to get the probability of the union. This can be extended from 2 to any $k$ sets. $ \begin{align} \mathbf P(A \cup B)&=\mathbf P(A)+\mathbf P(B) \\[8pt] \mathbf P(A \cup B \cup C)&= \mathbf P(A)+ \mathbf P(B) + \mathbf P(C)\\ \mathbf P(A_1 \cup A_2 \cup \dots \cup A_k)&= \sum_{i=1}^k \mathbf P(A_i) \end{align} $ ## Consequences of Axioms The event $A$ and its complement cover the whole sample space. $ A \, \cup \, A^C= \Omega $ - *Additivity* → we can sum the probabilities of disjoint events (which $A,A^C$ are). - *Normalization* → we know that their sum $=1$, since they cover the whole sample space. $ \mathbf P(A)+ \mathbf P(A^C)=1 \implies \mathbf P(A^C) = 1 - \mathbf P(A) $ We can apply the same logic to $\{\Omega, \Omega^C \}$, highlighting the fact that the complement of $\Omega$ is the empty set. $ \begin{align} \mathbf P(\Omega)+ \mathbf P(\Omega^C)=1 \\ 1+\mathbf P(\Omega^C)=1 \\ \mathbf P(\Omega^C)=0 \end{align} $ ## Countable Additivity Axiom This axiom extends the additivity axiom from finite to infinitely countable events (which can form a sequence). $ \mathbf P(A_1 \cup A_2 \cup \dots \cup A_\infty)= \mathbf P(A_1)+ \mathbf P(A_2)+\dots+ \mathbf P(A_\infty) $ A sequence can also go to infinity and can still remain countable. However, this is not true for a continuous range between $[0,1]$, which is not countable and thus cannot form a sequence.