Sum of Squared Deviations from Mean/Corollary 1
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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:
- $\ds \sum_{i \mathop = 1}^n \paren {x_i - \overline x}^2 = \sum_{i \mathop = 1}^n x_i^2 - n \overline x^2$
Proof
\(\ds \sum_{i \mathop = 1}^n \paren {x_i - \overline x}^2\) | \(=\) | \(\ds \sum_{i \mathop = 1}^n \paren {x_i^2 - \overline x^2}\) | Sum of Squared Deviations from Mean | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{i \mathop = 1}^n x_i^2 - \sum_{i \mathop = 1}^n \overline x^2\) | Summation is Linear | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{i \mathop = 1}^n x_i^2 - n \overline x^2\) | Sum of Identical Terms |
$\blacksquare$