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how can I prove that (if it is correct) ?

$\sum_{u,v \in P \times P} \frac{|u-v|^2}{N^2} = 2 \cdot Var(P)$

where $N$ is the number of elements of $P$. $P$ is a list of numbers.

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2 Answers 2

up vote 0 down vote accepted

Note that

$$ \begin{align}\sum_{u,v \in P \times P} |u-v|^2 &= \sum_{i=1}^N \sum_{j=1}^N (u_i-v_j)^2 \\ &= \sum_{i=1}^N\sum_{j=1}^N \{ (u_i- \bar u)-(v_j- \bar v)\}^2\\&= \sum_{i=1}^N\sum_{j=1}^N\{ (u_i-\bar u)^2+ (v_j-\bar v)^2 -2(u_i-\bar u)(v_j-\bar v)\} \\&= \sum_{i=1}^N\sum_{j=1}^N (u_i-\bar u)^2 + \sum_{i=1}^N\sum_{j=1}^N(v_j-\bar v)^2 -2\sum_{i=1}^N(u_i-\bar u)\sum_{j=1}^N(v_j-\bar v) \\ &= \sum_{j=1}^NNS^2+\sum_{i=1}^NNS^2=2N^2S^2 \end{align}$$

Hence we get $\sum_{u,v \in P \times P} \frac{|u-v|^2}{N^2} = 2.Var(P)$

Just note that since $u,v \in P \times P$ we have $\bar u=\bar v$. Again we know that $\sum_{i=1}^N(u_i-\bar u)=0=\sum_{j=1}^N(v_j-\bar v)$ and $S^2=Var(P)$.

This is a measure of mutual differences and this is known as Gini's Mean Difference.

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Just write $(u - v)^2 = \left( (u - \mu) - (v - \mu) \right)^2$ and go from there: $$ \begin{align*} \frac{1}{N^2} \sum_{\substack{u \in P \\ v \in P}} \left (u - \mu)^2 -2 (u - \mu) (v - \mu) + (v - \mu)^2 \right) &= 2 \sigma^2 - 2 \frac{1}{N^2} \left(\sum_{u \in P} (u - \mu) \right) \cdot \left(\sum_{v \in P} (v - \mu) \right) \\ &= 2 \sigma^2 \end{align*} $$

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