Proof of $\sum_{u \in P} \sum_{v \in P} m_u \cdot m_v |u-v|^2 = 2 \cdot Var(P)$

Question

How can we prove that, with $m_u$ being the mass of point $u$ (with a sum of 1) and $Var(P)$ being the weighted variance of points of $P$ considering these masses $m_u$ :

$\sum_{u \in P} \sum_{v \in P} m_u \cdot m_v |u-v|^2 = 2 \cdot Var(P)$

This question follows this one adding weigths.

Answer 1

\begin{align*} \sum_{v \in V}\sum_{v' \in V} (m_v \cdot m_{v'}) \|v - v'\|^2 & = \sum_{v \in V} \left(m_v \sum_{v' \in V} m_{v'} \|v - v'\|^2\right) \\ & = \sum_{v \in V} m_v I(V,v) \\ & = \sum_{v \in V} m_v \left(N \cdot \|v-g\|^2 + I(V)\right) \\ & = N \cdot \sum_{v \in V} m_v \|v-g\|^2 + \sum_{v \in V} m_v I(V) \\ & = N \cdot I(V) + N \cdot I(V) \\ & = 2N \cdot I(V) \\ & = 2 \cdot Var(V) \end{align*}