# Discrete Uniform Distribution gives rise to Probability Measure

## Theorem

Let $\EE$ be an experiment.

Let the probability space $\struct {\Omega, \Sigma, \Pr}$ be defined as:

$\Omega = \set {\omega_1, \omega_2, \ldots, \omega_n}$
$\Sigma = \powerset \Omega$
$\forall A \in \Sigma: \map \Pr A = \dfrac 1 n \card A$

where:

$\powerset \Omega$ denotes the power set of $\Omega$
$\card A$ denotes the cardinality of $A$.

Then $\Pr$ is a probability measure on $\struct {\Omega, \Sigma}$.

## Proof

From Power Set of Sample Space is Event Space we have that $\Sigma$ is an event space.

$\Box$

We check the axioms defining a probability measure:

 $(\text I)$ $:$ $\displaystyle \forall A \in \Sigma:$ $\displaystyle \map \Pr A$ $\displaystyle \ge$ $\displaystyle 0$ $(\text {II})$ $:$ $\displaystyle \map \Pr \Omega$ $\displaystyle =$ $\displaystyle 1$ $(\text {III})$ $:$ $\displaystyle \forall A \in \Sigma:$ $\displaystyle \map \Pr A$ $\displaystyle =$ $\displaystyle \sum_{\bigcup \set e \mathop = A} \map \Pr {\set e}$ where $e$ denotes the elementary events of $\EE$

Axiom $\text I$ is seen to be satisfied by the observation that the cardinality of a set is never negative.

Hence $\map \Pr A \ge 0$.

$\Box$

Then we have:

 $\ds \map \Pr \Omega$ $=$ $\ds \dfrac 1 n \card \Omega$ $\ds$ $=$ $\ds \dfrac 1 n \times n$ Definition of $\Omega$: it has been defined as having $n$ elements $\ds$ $=$ $\ds 1$

Axiom $\text {II}$ is thus seen to be satisfied.

$\Box$

Let $A = \set {\omega_{r_1}, \omega_{r_2}, \ldots, \omega_{r_k} }$ where $k = \card A$.

Then by Union of Set of Singletons:

$A = \set {\omega_{r_1} } \cup \set {\omega_{r_2} } \cup \cdots \cup \set {\omega_{r_k} }$

Hence:

 $\ds \map \Pr A$ $=$ $\ds \dfrac 1 n \card A$ $\ds$ $=$ $\ds \dfrac 1 n \card {\set {\omega_{r_1}, \omega_{r_2}, \ldots, \omega_{r_k} } }$ $\ds$ $=$ $\ds \dfrac 1 n \card {\set {\omega_{r_1} } \cup \set {\omega_{r_2} } \cup \cdots \cup \set {\omega_{r_k} } }$ $\ds$ $=$ $\ds \dfrac 1 n \paren {\underbrace {1 + 1 + \cdots + 1}_{\text {k times} } }$ $\ds$ $=$ $\ds \underbrace {\dfrac 1 n + \dfrac 1 n + \cdots + \dfrac 1 n}_{\text {k times} }$ $\ds$ $=$ $\ds \map \Pr {\set {\omega_{r_1} } } + \map \Pr {\set {\omega_{r_2} } } + \cdots + \map \Pr {\set {\omega_{r_k} } }$ $\ds$ $=$ $\ds \sum_{\bigcup \set e \mathop = A} \map \Pr {\set e}$

Hence Axiom $\text {III}$ is thus seen to be satisfied.

$\Box$

All axioms are seen to be satisfied.

Hence the result.

$\blacksquare$