Sum of Squared Deviations from Mean/Proof 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:
- $\ds \sum_{i \mathop = 1}^n \paren {x_i - \overline x}^2 = \sum_{i \mathop = 1}^n \paren { {x_i}^2 - {\overline x}^2}$
Proof
For brevity, let us write $\ds \sum$ for $\ds \sum_{i \mathop = 1}^n$.
Then:
\(\ds \sum \paren {x_i - \overline x}^2\) | \(=\) | \(\ds \sum \paren {x_i - \overline x} \paren {x_i - \overline x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum x_i \paren {x_i - \overline x} - \overline x \sum \paren {x_i - \overline x}\) | Summation is Linear | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum x_i \paren {x_i - \overline x} - 0\) | Sum of Deviations from Mean | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum x_i \paren {x_i - \overline x} + 0\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum x_i \paren {x_i - \overline x} + \overline x \sum \paren {x_i - \overline x}\) | Sum of Deviations from Mean | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum \paren {x_i + \overline x} \paren {x_i - \overline x}\) | Summation is Linear | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum \paren {x_i^2 - \overline x^2}\) |
$\blacksquare$