Let $X_{1}, X_{2}, \ldots, X_{n}$ be a sample from distribution $F(x; \mu, \sigma)$ where $\mu$ and $\sigma$ are location and scale parameters respectively and let $X_{1:n}, X_{2:n}, \ldots, X_{n:n}$ be the corresponding order statistics of the sample and define $$Z_{r:n} = \frac{ X_{r:n} - \mu }{\sigma}$$ with $E(Z_{r:n}) = \alpha$, $\operatorname{Var}(Z_{r:n}) = V_{r,r}$ and $\operatorname{Cov}(Z_{r:n}, Z_{s:n}) = V_{r,s}$.
Now $X_{r:n} = \mu + \sigma Z_{r:n}$ we have
$E(X_{r:n}) = \mu + \sigma \alpha_{r}$ $\operatorname{Var}(X_{r:n}) = \sigma^{2}V_{r,r}$ and $\operatorname{Cov}(X_{r:n}, X_{s:n}) = \sigma^2 V_{r,s}$.
Now writing $z = [ Z_{1:n},Z_{2:n}, \ldots, Z_{n:n} ]'$ we have $x = \mu 1 + \sigma z$,
$E(z) = \alpha$, $\text{Cov} = \text{V}$ and $E(x) = \mu 1 + \sigma \alpha$ and $\operatorname{Cov}(x) = \sigma^2 V$.
where $1$ is $(n \times 1)$ vector of $1$’s, $\alpha$ is $(n \times 1)$ vector of $E(Z_{r:n})$ and $V$ is $(n \times n)$ matrix of variances and covariances of $Z_{r:n}$. Using the fact that least square estimate of parameters of the model $y = X \beta + \epsilon$ with $\operatorname{Cov}(\epsilon) = V$ are $\hat{\beta} = (X' V^{-1} X )^{-1} X' V^{-1} y $. The estimates of $\mu$ and $\sigma$ are
$E(x) = \mu 1 + \sigma \alpha$ then $E(x) = \begin{bmatrix} 1 & \alpha \end{bmatrix} \begin{bmatrix} \mu \\ \sigma \end{bmatrix} = X \beta$
After simplification
$ \hat{\mu} = \frac{1}{\delta} \{ -\alpha' V^{-1} (1\alpha' - \alpha 1') V^{-1} y \} $
and
$ \hat{\sigma} = \frac{1}{\delta} \{ 1' V^{-1} (1\alpha' - \alpha 1') V^{-1} y \}, $
where $\delta = (1' V^{-1}1) (\alpha' V^{-1} \alpha) - (1' V^{-1} \alpha)^2$
If the parent distribution is symmetrical, then we have $\alpha_{r} = -\alpha_{n+r-1}$ and hence we have $1' V^{-1} \alpha = 0$ and hence the ordered least square estimators reduces to $\hat{\mu} = \dfrac{ 1' V^{-1} y }{ 1' V^{-1} 1 }$ and $\hat{\sigma} = \dfrac{ \alpha' V^{-1} y }{ \alpha' V^{-1} \alpha }$.
- Kindly give analytical explanation in detail of this argument (If the parent distribution is symmetrical then we have $\alpha_r = -\alpha_{n+r-1}$ and hence we have ${ 1' V^{-1} \alpha = 0}$)
