Proving two different expressions of non-centrality parameters are equivalent 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.
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:
$${\color{red}2}\sum_{i=1}^{K}\xi_i(\mu_i - \bar{\mu})^2 = \sum_{i,j}\xi_i\xi_j(\mu_i - \mu_j)^2.$$
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:
$${\color{red}2}\sum_{i=1}^{K}\xi_i(\mu_i - \bar{\mu})^2 = \sum_{i,j}\xi_i\xi_j(\mu_i - \mu_j)^2.$$