probability axioms

The Kolmogorov 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. An alternative approach to formalising probability, favoured by some Bayesians, is given by Cox's theorem.

Axioms

The assumptions as to setting up the axioms can be summarised as follows: Let $(\Omega, F, P)$ be a measure space with $P(E)$ being 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:
:$P(E)\in\mathbb, P(E)\geq 0 \qquad \forall E \in F$

where $F$ is the event space. It follows 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
::$P\left(\bigcup_^\infty E_i\right) = \sum_^\infty P(E_i).$
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 remaining two axioms. Four of the immediate corollaries and their proofs are shown below:

Monotonicity

:$\quad\text\quad A\subseteq B\quad\text\quad P(A)\leq P(B).$

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

:$P(A)+P(B\setminus A)+\sum_^\infty P(E_i)=P(B).$

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 probability of the empty set''

Define $E_i := \varnothing$ for $i \in \N$, then these are disjoint, and $\bigcup_^\infty E_i = \varnothing = E_1$, hence by the third axiom $\sum_^\infty P(E_i) = P(E_1)$; subtracting $P(E_1)$ (which is finite by the first axiom) yields $\sum_^\infty P(E_i) = 0$. From this together with the first axiom follows $0 \leq P(E_2) \leq \sum_^\infty P(E_i) = 0$, thus $P(E_2) = P(\varnothing) = 0$.

The complement rule

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

''Proof of the complement rule''

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

$P(A \cup A^c)=P(A)+P(A^c)$ ''... (by axiom 3)''

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

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

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

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'' $P(E^c)\geq0$:

$1-P(E) \geq 0$

$\Rightarrow 1 \geq P(E)$

$\therefore 0\leq P(E)\leq 1$

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

: $P\left(A^\right) = P(\Omega\setminus A) = 1 - P(A)$

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.

We may define:

: $\Omega = \$
: $F = \$

Kolmogorov's axioms imply that:

: $P(\varnothing) = 0$
The probability of ''neither'' heads ''nor'' tails, is 0.

: $P(\^c) = 0$
The probability of ''either'' heads ''or'' tails, is 1.

: $P(\) + P(\) = 1$
The sum of the probability of heads and the probability of tails, is 1.

See also

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References

Further reading

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Formal definition
of probability in the Mizar system, and th
list of theorems
formally proved about it.

{{DEFAULTSORT:Probability Axioms
Probability theory
Mathematical axioms