Arithmetic Mean is Weighted Mean with Equal Weights

Theorem

Let $S = \sequence {x_1, x_2, \ldots, x_n}$ be a set of real numbers.

Let $W$ be a weight function to be applied to the terms of $S$ such that:

$\forall x \in S: \map W x = c$

where $c$ is a real constant.

Let $\bar x$ denote the weighted mean of $S$ with respect to $W$.

Then:

$\bar x = A_n$

where $A_n$ denotes the arithmetic mean of $S$.

Proof

\(\ds \bar x\)

\(=\)

\(\ds \dfrac {\ds \sum_{i \mathop = 1}^n \map W {x_i} x_i} {\ds \sum_{i \mathop = 1}^n \map W {x_i} }\)

Definition of Weighted Mean

\(\ds \)

\(=\)

\(\ds \dfrac {\ds \sum_{i \mathop = 1}^n c x_i} {\ds \sum_{i \mathop = 1}^n c}\)

by hypothesis

\(\ds \)

\(=\)

\(\ds \dfrac {c \ds \sum_{i \mathop = 1}^n x_i} {c \ds \sum_{i \mathop = 1}^n 1}\)

Linear Combination of Indexed Summations

\(\ds \)

\(=\)

\(\ds \dfrac {c \ds \sum_{i \mathop = 1}^n x_i} {c \cdot n}\)

Summation of Unity over Elements

\(\ds \)

\(=\)

\(\ds \dfrac 1 n \ds \sum_{i \mathop = 1}^n x_i\)

dividing top and bottom by $c$

\(\ds \)

\(=\)

\(\ds A_n\)

Definition of Arithmetic Mean

$\blacksquare$