Sum of Deviations from Mean

Theorem

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

Let $\overline x$ denote the arithmetic mean of $S$.

Then:

$\ds \sum_{i \mathop = 1}^n \paren {x_i - \overline x} = 0$

For brevity, let us write $\ds \sum$ for $\ds \sum_{i \mathop = 1}^n$.

Then:

$\ds \sum \paren {x_i - \overline x}$

$=$

$\ds x_1 - \overline x + x_2 - \overline x + \cdots + x_n - \overline x$

Definition of Summation

$\ds $

$=$

$\ds x_1 - \sum \frac {x_i} n + x_2 - \sum \frac {x_i} n + \cdots + x_n - \sum \frac {x_i} n$

$\ds $

$=$

$\ds \paren {x_1 + x_2 + \cdots + x_n} - n \paren {\sum \frac {x_i} n}$

$\ds $

$=$

$\ds \sum x_i - \sum x_i$

$\ds $

$=$

$\ds 0$

$\blacksquare$