Discrete Uniform Distribution gives rise to Probability Measure

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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:

\((\text I)\)   $:$     \(\ds \forall A \in \Sigma:\)    \(\ds \map \Pr A \)   \(\ds \ge \)   \(\ds 0 \)      
\((\text {II})\)   $:$      \(\ds \map \Pr \Omega \)   \(\ds = \)   \(\ds 1 \)      
\((\text {III})\)   $:$     \(\ds \forall A \in \Sigma:\)    \(\ds \map \Pr A \)   \(\ds = \)   \(\ds \sum_{\bigcup \set e \mathop = A} \map \Pr {\set e} \)      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.

$\blacksquare$


Sources

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