Sum of Squared Deviations from Mean/Proof 2

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:

$\ds \sum_{i \mathop = 1}^n \paren {x_i - \overline x}^2 = \sum_{i \mathop = 1}^n \paren { {x_i}^2 - {\overline x}^2}$

Proof

In this context, $x_1, x_2, \ldots, x_n$ are instances of a discrete random variable.

Hence the result Variance as Expectation of Square minus Square of Expectation can be applied:

$\var X = \expect {X^2} - \paren {\expect X}^2$

which means the same as this but in the language of probability theory.

$\blacksquare$