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In the original paper (section 2.2.), we can see that derivation of the $W$ statistic goes along these lines:

  1. Let let $m_i$ be the expected value of the $i$-th order statistic of a random sample of $n$ numbers from the standard normal distribution $N(0, 1)$ ($i=1,2,\ldots,n$) and let $V$ be the corresponding covariance matrix. Let $m^T=[m_1,m_2,\ldots,m_n]$.
  2. If the ordered sample $y^T=[y_1,y_2,\ldots,y_n]$ comes from $N(\mu,\sigma^2)$ and the ordered sample $[x_1,x_2,\ldots,x_n]$ from $N(0, 1)$, then $$y_i=m_i+\sigma x_i$$ and the best linear unbiased estimator for $\sigma$ is $$\hat{\sigma}=\frac{m^TV^{-1}y}{m^TV^{-1}m}$$
  3. In that case, it should hold that $\hat{\sigma}^2=\frac{1}{n-1}\sum_{i=1}^{n}(y_i-\overline{y})^2$, the usual estimator of variance.

Why wasn't the $W$ statistic defined simply as the corresponding ratio: $$\frac{\hat{\sigma}^2}{\frac{1}{n-1}\sum_{i=1}^{n}(y_i-\overline{y})^2}$$?

Instead, $S^2=\sum_{1}^{n}(y_i-\overline{y})^2$ is introduced as the estimator of $(n-1)\sigma^2$, and $\hat{\sigma}^2$ is multiplied by certain $R^2=m^TV^{-1}m$ and $C^{-1}=(m^TV^{-1}V^{-1}m)^{-1/2}$ to get $$b=\frac{R^2}{C}\hat{\sigma}^2=a^Ty\quad a^T=\frac{m^TV^{-1}}{(m^TV^{-1}V^{-1}m)^{1/2}}$$ so that $W$ could be defined as follows: $$W=\frac{b^2}{S^2}$$

So, my questions are:

  1. Why bother with all those manipulations to get $W$ in that particular form?
  2. What those multiplications and divisions mean? The authors say that $C^2$ is there to make sure that the linear coefficients are normalized, but what about multiplication with $R^2$ and why was the estimator of $(n-1)\sigma^2$ used? What it looks like to me is that $R^2=m^TV^{-1}m$ somehow corresponds to $n-1$ and that $n-1$ was deliberately removed so that the whole statistic doesn't depend on $n$.
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