## How is the sum of differences squared related to common statistics [closed]

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 $\sum_{i, j\in{1,2,\ldots,n}} (x_i-x_j)^2 = \sum_{i, j\in{1,2,\ldots,n}} (x_i-\bar x)^2 + (x_j-\bar x)^2 + (x_i-\bar x)(x_j-\bar x)$ $=2 n \sum_i (x_i- \bar x)^2 + \sum_{i, j\in{1,2,\ldots,n}} (x_i-\bar x)(x_j-\bar x) =2 n \sum_i (x_i- \bar x)^2 .$

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I think it is zero. $$\sum_i\sum_j (x_i-\bar x)(x_j-\bar x)=\sum_i(x_i-\bar x)\sum_j (x_j-\bar x)$$ $$=\sum_i(x_i-\bar x)\big(\sum_j x_j-\sum_j\bar x\big)=\sum_i(x_i-\bar x)\big(n\bar x-n\bar x\big)=0$$ – BR Dec 14 2011 at 2:16
I don't see any research interest in this question. Voting to close, as per our faq. – Gerry Myerson Dec 14 2011 at 4:13