Sum of Squared Deviations from Mean/Corollary 2
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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:
- $\ds \sum_{i \mathop = 1}^n \paren {x_i - \overline x}^2 = \sum_{i \mathop = 1}^n x_i^2 - \frac 1 n \paren {\sum_{i \mathop = 1}^n x_i}^2$
Proof
For brevity, let us write $\ds \sum$ for $\ds \sum_{i \mathop = 1}^n$.
Then:
\(\ds \sum \paren {x_i - \overline x}^2\) | \(=\) | \(\ds \sum x_i^2 - n \overline x^2\) | Sum of Squared Deviations from Mean:Corollary $1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum x_i^2 - n \paren {\frac 1 n \sum x_i}^2\) | Definition of Arithmetic Mean | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum x_i^2 - n \paren {\frac 1 {n^2} } \paren {\sum x_i}^2\) | Exponent Combination Laws | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum x_i^2 - \frac 1 n \paren {\sum x_i}^2\) |
$\blacksquare$