# Probability Measure on Finite Sample Space

## Theorem

Let $\Omega = \set {\omega_1, \omega_2, \ldots, \omega_n}$ be a finite set.

Let $\Sigma$ be a $\sigma$-algebra on $\Omega$.

Let $p_1, p_2, \ldots, p_n$ be non-negative real numbers such that:

- $p_1 + p_2 + \cdots + p_n = 1$

Let $Q: \Sigma \to \R$ be the mapping defined as:

- $\forall A \in \Sigma: \map Q A = \ds \sum_{i: \omega_i \in A} p_i$

Then $\struct {\Omega, \Sigma, Q}$ constitutes a probability space.

That is, $Q$ is a probability measure on $\struct {\Omega, \Sigma}$.

### Corollary

Let $\Omega = \set {\omega_1, \omega_2, \ldots, \omega_n}$ be a finite set.

Let $\Sigma$ be the power set of $\Omega$.

Let $p_1, p_2, \ldots, p_n$ be non-negative real numbers such that:

- $p_1 + p_2 + \cdots + p_n = 1$

Let $Q: \Sigma \to \R$ be the mapping defined as:

- $\forall A \in \Sigma: \map Q A = \ds \sum_{i: \omega_i \in A} p_i$

Then $\struct {\Omega, \Sigma, Q}$ constitutes a probability space.

That is, $Q$ is a probability measure on $\struct {\Omega, \Sigma}$.

## Proof

Recall the Kolmogorov axioms:

\((1)\) | $:$ | \(\ds \forall A \in \Sigma:\) | \(\ds 0 \) | \(\ds \le \) | \(\ds \map \Pr A \le 1 \) | The probability of an event occurring is a real number between $0$ and $1$ | |||

\((2)\) | $:$ | \(\ds \map \Pr \Omega \) | \(\ds = \) | \(\ds 1 \) | The probability of some elementary event occurring in the sample space is $1$ | ||||

\((3)\) | $:$ | \(\ds \map \Pr {\bigcup_{i \mathop \ge 1} A_i} \) | \(\ds = \) | \(\ds \sum_{i \mathop \ge 1} \map \Pr {A_i} \) | where $\set {A_1, A_2, \ldots}$ is a countable (possibly countably infinite) set of pairwise disjoint events | ||||

That is, the probability of any one of countably many pairwise disjoint events occurring | |||||||||

is the sum of the probabilities of the occurrence of each of the individual events |

First we determine that $\Pr$ as defined is actually a probability measure.

By definition, we have that $\map \Pr A$ is the sum of some subset of $\set {p_1, p_2, \ldots, p_n}$.

Thus $0 \le \map \Pr A \le 1$ and Axiom $(1)$ is fulfilled trivially by definition.

Let $A \in \Sigma$ be such that:

- $A = \set {\omega_{r_1}, \omega_{r_2}, \ldots, \omega_{r_k} }$

We have that:

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

From Simple Events are Mutually Exclusive, $\set {\set {\omega_{r_1} }, \set {\omega_{r_2} }, \ldots, \set {\omega_{r_k} } }$ constitutes a set of pairwise disjoint events.

Hence:

- $\map \Pr {\set {\omega_{r_1} } \cup \set {\omega_{r_2} } \cup \cdots \cup \set {\omega_{r_k} } } = \ds \sum_{i \mathop = 1}^k \map \Pr {\omega_{r_1} }$

and it is seen that axiom $(3)$ is fulfilled.

Then we have that:

\(\ds \map \Pr \Omega\) | \(=\) | \(\ds \map \Pr {\bigcup_{i \mathop = 1}^n \omega_i}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \ds \sum_{i \mathop = 1}^n \map \Pr {\omega_i}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \ds \sum_{i \mathop = 1}^n p_i\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds 1\) |

Hence axiom $(2)$ is satisfied.

$\blacksquare$

## Sources

- 1986: Geoffrey Grimmett and Dominic Welsh:
*Probability: An Introduction*... (previous) ... (next): $1$: Events and probabilities: $1.3$: Probabilities: Exercise $6$