Sum of Squared Deviations from Mean

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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:

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


Corollary 1

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


Corollary 2

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


Proof 1

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
\(\ds \) \(=\) \(\ds \sum \paren {x_i^2 - \overline x^2}\)

$\blacksquare$


Proof 2

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.

$\blacksquare$