statistics is equal to the dimension of the parameter, then the MVU estimator is given by $\hat{x}=g(T(y_{1:N}))$ for any function $g(\cdot)$ that satisfies $$\begin{aligned}
{\mathbb{E}}(g(T)) = x.\end{aligned}$$ Hence, the problem of MVU estimator turns into the problem of finding a complete sufficient statistic. The Neyman-Fisher theorem [@article:fisher_22; @article:AMS_halmos_49] gives the sufficient statistic $T(y_{1:N})$, if the PDF can be factorized as follows $$\begin{aligned}
f(y_{1:N};\Psi) =g(T(y_{1:N}),\Psi)h(y_{1:N}),\end{aligned}$$ where $\psi$ is the union of the noise hyper parameter(s) $\theta$ and $x$. The estimators in this work are derived in the order statistics framework.

Uniform Distribution {#sec:uniform_distribution}
====================

As the first scenario, consider the case in which the additive noise $e_k$ has a uniform distribution with a positive support, $p_e(\theta)=\mathcal{U}[0,\beta]$, $\beta>0$ and $\theta=\beta$. The BLUE is given by

$$\begin{aligned}
    \hat{x}_{\mathrm{BLUE}}^{\mathcal{U}}(y_{1:N},\beta) &= \frac{1}{N}\sum_{k=1}^N y_k- \frac{\beta}{2}.
    \label{eq:sample_mean_example}
    \end{aligned}$$

The MSE of BLUE for this case is given by $$\begin{aligned}
    {\mathrm{MSE}}\left(\hat{x}_{\mathrm{BLUE}}^{\mathcal{U}}(y_{1:N},\beta)\right) &= \frac{1}{N^2}\sum_{k=1}^{N}{\mathrm{Var}}\left(y_k-\frac{\beta}{2}\right)= \frac{\beta^2}{12N}.
    \label{eq:mse_blue_uniform}
    \end{aligned}$$

In order to find the MSE of the minimum order statistics estimator, $\hat{x}_{\mathrm{min}}^{\mathcal{U}}(y_{1:N}) $, we need to find the first two moments of the estimator. Let $\tilde{y}_k= \frac{1}{\beta}y_k$. Since $y_k\sim\mathcal{U}[x,x+b]$, then for any constant $\beta>0$, $\tilde{y_k}\sim\mathcal{U}[\frac{x}{\beta},\frac{x}{\beta}+1]$. Hence, $f(\tilde{y}_k)=1$ and $F(\tilde{y}_k)=\frac{1}{\beta}(y_k-x)$. From  we get,

$$\begin{aligned}
    f&_{(k,N)}^{\mathcal{U}[0,\beta]}(\tilde{y}) = N\begin{pmatrix}N-1\\k-1\end{pmatrix}\left(\frac{\tilde{y}-x}{\beta}\right)^{k-1}\left(\frac{\beta-(\tilde{y}-x)}{\beta}\right)^{N-k}\nonumber\\
    &= \frac{(N)!}{(k-1)!(N-k)!}\left(\frac{\tilde{y}-x}{\beta}\right)^{k-1}\left(\frac{\beta-(\tilde{y}-x)}{\beta}\right)^{N-k}.
    \end{aligned}$$

since $N\in\mathbb{N}^+$, 