# Probability axioms

The standard probability axioms are the foundations of probability theory introduced by Russian mathematician Andrey Kolmogorov in 1933. These axioms remain central and have direct contributions to mathematics, the physical sciences, and real-world probability cases.
There are several other (equivalent) approaches to formalising probability. Bayesians will often motivate the Kolmogorov axioms by invoking Cox's theorem or the Dutch book arguments instead.

## Kolmogorov axioms

The assumptions as to setting up the axioms can be summarised as follows: Let $(\Omega ,F,P)$ be a measure space such that $P(E)$ is the probability of some event $E$, and $P(\Omega )=1$. Then $(\Omega ,F,P)$ is a probability space, with sample space $\Omega$, event space $F$ and probability measure $P$.

### First axiom

The probability of an event is a non-negative real number:
    $P(E)\in \mathbb {R} ,P(E)\geq 0\qquad \forall E\in F$
where $F$ is the event space. It follows (when combined with the second axiom) that $P(E)$ is always finite, in contrast with more general measure theory. Theories which assign negative probability relax the first axiom.

### Second 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.
    $P(\Omega )=1$

### Third axiom

This is the assumption of σ-additivity:
    Any countable sequence of disjoint sets (synonymous with mutually exclusive events) $E_{1},E_{2},\ldots$ satisfies
       $P\left(\bigcup _{i=1}^{\infty }E_{i}\right)=\sum _{i=1}^{\infty }P(E_{i}).$
Some authors consider merely finitely additive probability spaces, in which case one just needs an algebra of sets, rather than a σ-algebra. Quasiprobability distributions in general relax the third axiom.

## Consequences

From the Kolmogorov axioms, one can deduce other useful rules for studying probabilities. The proofs of these rules are a very insightful procedure that illustrates the power of the third axiom, and its interaction with the prior two axioms. Four of the immediate corollaries and their proofs are shown below:

### Monotonicity

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

#### Proof of monotonicity

In order to verify the monotonicity property, we set $E_{1}=A$ and $E_{2}=B\setminus A$, where $A\subseteq B$ and $E_{i}=\varnothing$ for $i\geq 3$. From the properties of the empty set ($\varnothing$), it is easy to see that the sets $E_{i}$ are pairwise disjoint and $E_{1}\cup E_{2}\cup \cdots =B$. Hence, we obtain from the third axiom that
    $P(A)+P(B\setminus A)+\sum _{i=3}^{\infty }P(E_{i})=P(B).$
Since, by the first axiom, the left-hand side of this equation is a series of non-negative numbers, and since it converges to $P(B)$ which is finite, we obtain both $P(A)\leq P(B)$ and $P(\varnothing )=0$.

### The probability of the empty set

    $P(\varnothing )=0.$
In many cases, $\varnothing$ is not the only event with probability 0.

#### Proof of the probability of the empty set

$P(\varnothing \cup \varnothing )=P(\varnothing )$ since $\varnothing \cup \varnothing =\varnothing$,
$P(\varnothing )+P(\varnothing )=P(\varnothing )$ by applying the third axiom to the left-hand side (note $\varnothing$ is disjoint with itself), and so
$P(\varnothing )=0$ by subtracting $P(\varnothing )$ from each side of the equation.

### The complement rule

$P\left(A^{\complement }\right)=P(\Omega -A)=1-P(A)$

#### Proof of the complement rule

Given $A$ and $A^{\complement }$ are mutually exclusive and that $A\cup A^{\complement }=\Omega$:
$P(A\cup A^{\complement })=P(A)+P(A^{\complement })$ ... (by axiom 3)
and, $P(A\cup A^{\complement })=P(\Omega )=1$ ... (by axiom 2)
$\Rightarrow P(A)+P(A^{\complement })=1$
$\therefore P(A^{\complement })=1-P(A)$

### The numeric bound

It immediately follows from the monotonicity property that
    $0\leq P(E)\leq 1\qquad \forall E\in F.$

#### Proof of the numeric bound

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

## Further consequences

Another important property is:
    $P(A\cup B)=P(A)+P(B)-P(A\cap B).$
This is called the addition law of probability, or the sum rule. That is, the probability that an event in A or B will happen is the sum of the probability of an event in A and the probability of an event in B, minus the probability of an event that is in both A and B. The proof of this is as follows:
Firstly,
    $P(A\cup B)=P(A)+P(B\setminus A)$. (by Axiom 3)
So,
    $P(A\cup B)=P(A)+P(B\setminus (A\cap B))$ (by $B\setminus A=B\setminus (A\cap B)$).
Also,
    $P(B)=P(B\setminus (A\cap B))+P(A\cap B)$
and eliminating $P(B\setminus (A\cap B))$ from both equations gives us the desired result.
An extension of the addition law to any number of sets is the inclusion–exclusion principle.
Setting B to the complement A$^c$ of A in the addition law gives
    $P\left(A^{c}\right)=P(\Omega \setminus A)=1-P(A)$
That is, the probability that any event will not happen (or the event's complement) is 1 minus the probability that it will.

## Simple example: coin toss

Consider a single coin-toss, and assume that the coin will either land heads (H) or tails (T) (but not both). No assumption is made as to whether the coin is fair or as to whether or not any bias depends on how the coin is tossed.
We may define:
    $\Omega =\{H,T\}$
    $F=\{\varnothing ,\{H\},\{T\},\{H,T\}\}$
Kolmogorov's axioms imply that:
    $P(\varnothing )=0$
The probability of neither heads nor tails, is 0.
    $P(\{H,T\}^{c})=0$
The probability of either heads or tails, is 1.
    $P(\{H\})+P(\{T\})=1$
The sum of the probability of heads and the probability of tails, is 1.