Is $2\bar x(1 - \bar x) - \sum_{i=1}^n 2 x_i (1-x_i) = 2 \sum_{i=1}^n (x_i - \bar x)^2$ true?

Question

In Nei 1973, right after equation (9), the author says:

[..] it can be shown that $H_T = 2\bar x(1 - \bar x)$ and $D_{st} = 2 \sigma_x^2$, where $\bar x$ and $\sigma_x^2$ are the mean and variance of the frequency of an allele among subpopulations, respectively.

I have a hard time to prove $D_{st} = 2 \sigma_x^2$. From equation (8), $D_{st} = H_T-H_S$, where $H_S = \sum_{i=1}^n 2 x_i (1-x_i)$

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So in short (from what I understand), the following should hold true

$$2\bar x(1 - \bar x) - \sum_{i=1}^n 2 x_i (1-x_i) = 2 \sum_{i=1}^n (x_i - \bar x)^2$$

, where $\bar x = \frac{\sum_{i=1}^n x_i}{n}$. $0<x_i<1 \forall i$ as the $x_i$ represent allele frequencies and $n$ is an strictly positive integer as it represents the number of subpopulations. Is this equation true?

Answer 1

Your relationship is not correct unless $\bar{x} = 0$ or $n=1$.

Try it with $n=3, x_i = i$. Here $\bar{x} =2$, $\sum x_i(1-x_i) = -8$, and $\sum(x_i-\bar{x})^2 = 2$. Your relation then reads $$ -4 - (-16) = 4 $$ In fact, the discrepancy is always $$2\left( \frac1{n}-1\right)\sum_{i=1}^nx_i$$