Definition of Variance

If random variable $X$ has expected value (mean) $μ = E (X)$ , then the variance $V a r (X)$ of $X$ is given by:

V a r (X) = E [(X - μ)^{2}]

In practice we collect data and estimate the $V a r (X)$ as either

s_{n}^{2} = \frac{1}{n} \sum_{i = 1}^{n} {(x_{i} - \bar{x})}^{2} = (\frac{1}{n} \sum_{i = 1}^{n} x_{i}^{2}) - {\bar{x}}^{2}

(which makes clear the nature of the variance as a mean), or as

s^{2} = \frac{1}{n - 1} \sum_{i = 1}^{n} {(x_{i} - \bar{x})}^{2} = \frac{1}{n - 1} \sum_{i = 1}^{n} {(x_{i} - \frac{1}{n} \sum_{j = 1}^{n} x_{j})}^{2} = \frac{1}{n - 1} \sum_{i = 1}^{n} {(\frac{1}{n} \sum_{j = 1}^{n} (x_{i} - x_{j}))}^{2}

, (which provides a correction to the slight bias in the estimated value).

Let's play around with that last formula a little bit:

s^{2} = \frac{1}{n - 1} \sum_{i = 1}^{n} (\frac{1}{n} \sum_{j = 1}^{n} (x_{i} - x_{j})) (\frac{1}{n} \sum_{k = 1}^{n} (x_{i} - x_{k}))

Replacing the mean

\bar{x}

by the sum which defines it,

s^{2} = \frac{1}{n^{2} (n - 1)} \sum_{i = 1}^{n} \sum_{j = 1}^{n} \sum_{k = 1}^{n} (x_{i} - x_{j}) (x_{i} - x_{k}) .

Adding an appropriate form of zero (a favorite trick!),

s^{2} = \frac{1}{n^{2} (n - 1)} \sum_{i = 1}^{n} \sum_{j = 1}^{n} \sum_{k = 1}^{n} (x_{i} - x_{j}) (x_{i} - x_{j} + x_{j} - x_{k}),

which is

s^{2} = \frac{1}{n (n - 1)} \sum_{i = 1}^{n} \sum_{j = 1}^{n} (x_{i} - x_{j})^{2} - \frac{1}{n^{2} (n - 1)} \sum_{i = 1}^{n} \sum_{j = 1}^{n} \sum_{k = 1}^{n} (x_{j} - x_{i}) (x_{j} - x_{k}) .

Notice that the second sum in the previous expression as exactly

s^{2}

, so

s^{2} = \frac{1}{2 n (n - 1)} \sum_{i = 1}^{n} \sum_{j = 1}^{n} (x_{i} - x_{j})^{2},

or, finally,

s^{2} = \frac{1}{n (n - 1)} \sum_{i = 1}^{n} \sum_{j = i + 1}^{n} (x_{i} - x_{j})^{2},

This you can think of as

s^{2} = average distinct pair deviation squared

(which we knew!;)