Proving two different expressions of non-centrality parameters are equivalent

Question

I am stuck in proving $$\sum_{i=1}^{K}\xi_i(\mu_i - \bar{\mu})^2 = \sum_{i,j}\xi_i\xi_j(\mu_i - \mu_j)^2,$$ where $\bar{\mu} = \sum_{i=1}^{K}\xi_i\mu_i$ and $\sum_{i=1}^{K}\xi_i = 1$.

I am not sure whether it is true or not.

Answer 1

Let $P = \sum_{i=1}^K \xi_i \mu_i^2.$

Notice that:

$$\sum_{i=1}^K \sum_{j=1}^K \xi_i\xi_j(\mu_i - \mu_j)^2 = \sum_{i=1}^K \xi_i\sum_{j=1}^K \xi_j(\mu_i^2 + \mu_j^2 - 2\mu_i\mu_j) = \\ = \sum_{i=1}^K \xi_i\left(\sum_{j=1}^K \xi_j\mu_i^2 + \sum_{j=1}^K \xi_j\mu_j^2 - \sum_{j=1}^K \xi_j2\mu_i\mu_j\right)=\\ = \sum_{i=1}^K \xi_i\left(\mu_i^2 + \sum_{j=1}^K \xi_j\mu_j^2 - 2\mu_i \sum_{j=1}^K \xi_j\mu_j\right)=\\ = \sum_{i=1}^K \xi_i\left(\mu_i^2 + P - 2\mu_i \overline{\mu}\right)=\\ = \sum_{i=1}^K \xi_i\mu_i^2 + \sum_{i=1}^K \xi_iP - 2\sum_{i=1}^K \xi_i\mu_i \overline{\mu}=\\ = P + P - 2\overline{\mu}^2 = 2(P - \overline{\mu}^2). $$

On the other hand:

$$\sum_{i=1}^{K}\xi_i(\mu_i - \bar{\mu})^2 = \sum_{i=1}^{K}\xi_i \mu_i^2 + \sum_{i=1}^{K}\xi_i \bar{\mu}^2 - 2\sum_{i=1}^{K}\xi_i \mu_i \bar{\mu} = \\ = P + \bar{\mu}^2 - 2\bar{\mu}^2 = P - \bar{\mu}^2.$$

Therefore:

$${\color{red}2}\sum_{i=1}^{K}\xi_i(\mu_i - \bar{\mu})^2 = \sum_{i,j}\xi_i\xi_j(\mu_i - \mu_j)^2.$$