\frac{1}{N(\sqrt{N}-1)}\sum_{k=1}^{N}y_k\nonumber\\&= \frac{1}{\sqrt N-1}(\sqrt{N}y^{(1)}_{1:N} - \bar{y}).\end{aligned}$$ Asymptotically, for large $N$, the sample mean and minimum order statistic are independent and the estimator MSE is given by $$\begin{aligned}
{\mathrm{MSE}}\left(\hat{x}^{\mathrm{Rayleigh}}(y_{1:N})\right)=\frac{(1+N)(4-\pi)\beta^2}{2N(\sqrt{N}-1)^2}.
\label{eq:minimum_orde_unknown_rayleigh}\end{aligned}$$

Weibull distribution {#subsec:weibull_distribution}
--------------------

Weibull distribution is a generalization of the Rayleigh, distribution that is parameterized by two parameters–scale parameter $\beta$ and shape parameter $\alpha>0$. In fact Weibull distribution is obtained by relaxing the assumption $\alpha=2$ in the Rayleigh distribution and its density function is given by

\[eq:weibull\] $$\begin{aligned}
    f^{\mathrm{Weibull}}(y_k;x,\beta,\alpha)=\left\{\begin{matrix}
    \frac{\alpha}{\beta}\left(\frac{y_k-x}{\beta}\right)^{\alpha-1}\exp(-(\frac{y_k-x}{\beta})^\alpha)&y_k> x, \\ 
    0&y_k\leq x. 
    \end{matrix}\right.
    \label{eq:weibull_pdf}
    \end{aligned}$$ and the CDF, for $y_k\geq x$ is given by $$\begin{aligned}
    F^{\mathrm{Weibull}}(y_k;x,\beta,\alpha)=1-\exp(-(\frac{y_k-x}{\beta})^\alpha).
    \label{eq:weibull_cdf}
    \end{aligned}$$ The BLUE estimator, in case of Weibull-distributed measurement noises is given by $$\begin{aligned}
\hat{x}_{\mathrm{BLUE}}^{\mathrm{Weibull}}(y_{1:N},\beta,\alpha) &= \frac{1}{N}\sum_{k=1}^{N}y_k - \beta\Gamma(1+\frac{1}{\alpha}),
&\quad 
{\mathrm{MSE}}(\hat{x}_{\mathrm{BLUE}}^{\mathrm{Weibull}}(y_{1:N},\beta,\alpha)) &= \frac{\beta^2}{N}\left[\Gamma(\frac{\alpha+2}{\alpha})-\left(\Gamma(\frac{\alpha+1}{\alpha})\right)^2\right].\end{aligned}$$

The marginal density of the $k$:th order statistic is given by $$\begin{aligned}
f^{\mathrm{Weibull}}_{(k,N)}(y;x,\beta,\alpha) = \frac{N\alpha}{\beta}\begin{pmatrix}N-1\\k-1\end{pmatrix}(\frac{y-x}{\beta})^{\alpha-1}\left(1-\exp^{-(\frac{y-x}{\beta})^\alpha}\right)^{k-1}\exp\left(-(N-k+1)(\frac{y-x}{\beta})^\alpha\right).
\label{eq:weibull_order}\end{aligned}$$ Hence, the first order statistic density in case of $e_k\sim\mathrm{Weibull}(\beta,\alpha)$, is another Weibull distribution, $$\begin{aligned}
f^{\mathrm{Weibull}}_{(1,N)}(y;x,\beta,\alpha) = f^{\mathrm{Weibull}}(y;x,\bar{\beta},\alpha),\end{aligned}$$ where $\bar{\beta}=\sqrt[-\alpha]{N}\beta$. This gives the MSE of the minimum order statistic estimator as $$\begin{aligned}
{\mathrm{MSE}}(\hat{x}_{\mathrm{min}}^{\mathrm{Weibull}}(y_{1:N})) = \beta^2N^{\frac{-2}{\alpha}}\Gamma(\frac{\alpha+2}{\alpha})\end{aligned}$$ Given $N$ independent observations, the joint density is given by $$\begin{aligned}
f^{\mathrm{Weibull}}(y_{1:N};x,\beta,\alpha)=(\frac{\alpha}{\beta})^N\prod_{k=1}^{N}\left(\frac{y_k-x}{\beta}\right)^{\alpha-1}\exp(-\sum_{k=1}^{N}(\frac{y_k-x}{\beta})^\alpha)\sigma(\min y_k - x)
\label{eq:weibull_pdf_joint}\end{aligned}$$ Since  cannot be factorized using Neyman-Fisher factorization, RBLS is not applicable. Additionally, 