Probability axioms

The probability of an event is a non-negative real number:

${\displaystyle P(E)\in \mathbb {R} ,P(E)\geq 0\qquad \forall E\in F}$

where ${\displaystyle F}$ is the event space. It follows that ${\displaystyle P(E)}$ is always finite, in contrast with more general measure theory. Theories which assign negative probability relax the first axiom.

This is the assumption of unit measure: that the probability that at least one of the elementary events in the entire sample space will occur is 1

${\displaystyle P(\Omega )=1.}$

This is the assumption of σ-additivity:

Any countable sequence of disjoint sets (synonymous with mutually exclusive events) ${\displaystyle E_{1},E_{2},\ldots }$ satisfies

${\displaystyle P\left(\bigcup _{i=1}^{\infty }E_{i}\right)=\sum _{i=1}^{\infty }P(E_{i}).}$

Some authors consider merely finitely additive probability spaces, in which case one just needs an algebra of sets, rather than a σ-algebra.[4] Quasiprobability distributions in general relax the third axiom.

${\displaystyle \quad {\text{if}}\quad A\subseteq B\quad {\text{then}}\quad P(A)\leq P(B).}$

If A is a subset of, or equal to B, then the probability of A is less than, or equal to the probability of B.

In order to verify the monotonicity property, we set ${\displaystyle E_{1}=A}$ and ${\displaystyle E_{2}=B\setminus A}$, where ${\displaystyle A\subseteq B}$ and ${\displaystyle E_{i}=\varnothing }$ for ${\displaystyle i\geq 3}$. It is easy to see that the sets ${\displaystyle E_{i}}$ are pairwise disjoint and ${\displaystyle E_{1}\cup E_{2}\cup \cdots =B}$. Hence, we obtain from the third axiom that

${\displaystyle P(A)+P(B\setminus A)+\sum _{i=3}^{\infty }P(E_{i})=P(B).}$

Since, by the first axiom, the left-hand side of this equation is a series of non-negative numbers, and since it converges to ${\displaystyle P(B)}$ which is finite, we obtain both ${\displaystyle P(A)\leq P(B)}$ and ${\displaystyle P(\varnothing )=0}$.

${\displaystyle P(\varnothing )=0.}$

In some cases, ${\displaystyle \varnothing }$ is not the only event with probability 0.

As shown in the previous proof, ${\displaystyle P(\varnothing )=0}$. However, this statement is seen by contradiction: if ${\displaystyle P(\varnothing )=a}$ then the left hand side ${\displaystyle [P(A)+P(B\setminus A)+\sum _{i=3}^{\infty }P(E_{i})]}$ is not less than infinity; ${\displaystyle \sum _{i=3}^{\infty }P(E_{i})=\sum _{i=3}^{\infty }P(\varnothing )=\sum _{i=3}^{\infty }a={\begin{cases}0&{\text{if }}a=0,\\\infty &{\text{if }}a>0.\end{cases}}}$

If ${\displaystyle a>0}$ then we obtain a contradiction, because the sum does not exceed ${\displaystyle P(B)}$ which is finite. Thus, ${\displaystyle a=0}$. We have shown as a byproduct of the proof of monotonicity that ${\displaystyle P(\varnothing )=0}$.

${\displaystyle P\left(A^{c}\right)=P(\Omega \setminus A)=1-P(A)}$

Given ${\displaystyle A}$ and ${\displaystyle A^{c}}$are mutually exclusive and that ${\displaystyle A\cup A^{c}=\Omega }$:

${\displaystyle P(A\cup A^{c})=P(A)+P(A^{c})}$ ... (by axiom 3)

and, ${\displaystyle P(A\cup A^{c})=P(\Omega )=1}$ ... (by axiom 2)

${\displaystyle \Rightarrow P(A)+P(A^{c})=1}$

${\displaystyle \therefore P(A^{c})=1-P(A)}$

It immediately follows from the monotonicity property that

${\displaystyle 0\leq P(E)\leq 1\qquad \forall E\in F.}$

Given the complement rule ${\displaystyle P(E^{c})=1-P(E)}$ and axiom 1 ${\displaystyle P(E^{c})\geq 0}$:

${\displaystyle 1-P(E)\geq 0}$

${\displaystyle \Rightarrow 1\geq P(E)}$

${\displaystyle \therefore 0\leq P(E)\leq 1}$