Relative Frequency is Probability Measure

Theorem
The relative frequency model is a probability measure.

Proof
We check all the Kolmogorov axioms in turn:

First Axiom
Let $n$ be the number of times a certain event $\omega$ is observed to happen.

Let $n'$ be the number of times $\omega$ is observed not to happen.

By Law of Excluded Middle and Principle of Non-Contradiction, $\omega$ either happened or did not, and not both.

Therefore $n + n'$ is the total number of observations.

It is supposed that at least one observation is actually made, so that $n + n' \ne 0$.

The relative frequency model says that the probability of $\omega$ occurring can be defined as:


 * $\Pr \left({\omega}\right) = \dfrac n {n + n'}$

the numerator and denominator of which are positive integers such that $n + n' \ge n$.

If $\omega$ is observed never to happen, then $n = 0$ and $\Pr \left({\omega}\right) = \dfrac 0 {0 + n'} = 0$.

If $\omega$ is observed to always happen, then $n' = 0$ and $\Pr \left({\omega}\right) = \dfrac n {n + 0} = 1$.

Otherwise, if $n, n' \ne 0$, from Mediant is Between:
 * $0 = \dfrac 0 {0 + n'} < \dfrac n {n + n'} < \dfrac n {n + 0} = 1$

Thus $\Pr$ is bounded:


 * $0 \le \Pr \left({\cdot}\right) \le 1$

Second Axiom
By hypothesis:


 * $\Pr \left({\Omega}\right) = \dfrac {n + n'}{n + n'} = 1$

Third Axiom
This is a proof by induction.

Basis for the Induction
The case $j = 2$ is verified as follows:

Let $A$ and $B$ be two pairwise disjoint events.

Let $p$ and $q$ be the number of times $A$ and $B$ have been observed, respectively.

Let $n$ be the total number of trials observed.

By the definition of pairwise disjoint, $A$ and $B$ never happened at the same time.

By the same reasoning as the proof for the first axiom, in all $n$ observations:


 * $A$ happened $p$ times
 * $B$ happened $q$ times
 * $A \lor B$ happened $p + q$ times.

By hypothesis:

This is the basis for the induction.

Induction Hypothesis
Let $A_1, A_2, \ldots, A_j$ be $j$ pairwise disjoint events.

By the definition of the relative frequency model, $j$ is finite.

Assume:
 * $\displaystyle \Pr \left({\bigcup_{i \mathop = 1}^j A_i}\right) = \Pr \left({A_1}\right) + \Pr \left({A_2}\right) + \cdots + \Pr\left({A_j}\right)$

This is our induction hypothesis.

Induction Step
This is our induction step:

Let $A_1, A_2, \ldots, A_j, A_{j+1}$ be $j+1$ pairwise disjoint events.

Define $C = A_1 \lor A_2 \lor A_3 \lor \cdots \lor A_j$.

Then $C$ and $A_{j+1}$ are also pairwise disjoint.

By the base case:


 * $\Pr \left({C \cup A_{j+1}}\right) = \Pr \left({C}\right) + \Pr \left({A_{j+1}} \right)$

By the definition of $C$, this equation is logically equivalent to:


 * $\displaystyle \Pr \left({\bigcup_{i \mathop = 1}^{j+1} A_i} \right) = \sum_{i \mathop = 1}^{j+1} \Pr \left({A_i} \right)$

By the definition of the relative frequency model, $j + 1$ is finite.

The result follows by the Principle of Mathematical Induction.