I have a sequence of numbers $x_1, x_2, \ldots, x_n$. I have computed the sum of the differences squared as $\sum_{i, j\in{1,2,\ldots,n}} (x_i-x_j)^2$. Intuitively $\sum_{i, j\in{1,2,\ldots,n}} (x_i-x_j)^2$ is a measure of variance (taken informally). I am looking into making this statement more precise.
Answer: Thanks to a comment by BR, I have worked out the answer.
Clearly, we can rewrite $(x_i-x_j)$ as $ (x_i-\bar x + \bar x - x_j)$ which leads to $\sum_{i, j\in{1,2,\ldots,n}} (x_i-x_j)^2 = \sum_{i, j\in{1,2,\ldots,n}} (x_i-\bar x)^2 + (x_j-\bar x)^2 + (x_i-\bar x)(x_j-\bar x)$ $=2 n \sum_i (x_i- \bar x)^2 + \sum_{i, j\in{1,2,\ldots,n}} (x_i-\bar x)(x_j-\bar x) =2 n \sum_i (x_i- \bar x)^2 . $
