added together,  is maximized at this point. The order statistics PDF for $0\leq y-x\leq \beta$ is given by $$\begin{aligned}
f_{(k,N)}^{\mathrm{\mathrm{\mathcal{U},\mathcal{N}}}}&(y;\alpha,\beta,\sigma^2,x,k) =\nonumber\\ &N\begin{pmatrix}N-1\\k-1\end{pmatrix}\left(\frac{\alpha\exp(-\frac{(y-x)^2}{2\sigma^2})}{\sqrt{2\pi\sigma^2}}+\frac{1-\alpha}{\beta}\right)\nonumber\\
&\times\left(\frac{(1-\alpha)(y-x)}{\beta}+\frac{\alpha}{2}(1+\mathrm{Erf}\left[\frac{y-x}{\sqrt{2\sigma^2}}\right])\right)^{k-1}\nonumber\\
&\times\left(1+\frac{(\alpha-1)(y-x)}{\beta}-\frac{\alpha}{2}(1+\mathrm{Erf}\left[\frac{y-x}{\sqrt{2}\sigma}\right])\right)^{N-k},\end{aligned}$$ where $\mathrm{Erf}(\cdot) = \frac{2}{\sqrt{\pi}}\int_{0}^{\cdot}e^{-t^2}dt$ is the error function. In order to find the best order statistic estimator, we maximize the likelihood function $\ell(k\mid y=x,a,\beta,\sigma^2)$

$$\begin{aligned}
    \ell(k\mid y=x,\alpha,\beta,\sigma^2) &= N\begin{pmatrix}N-1\\k-1\end{pmatrix}2(2-\alpha)^{-k}(1-\frac{\alpha}{2})^N\alpha^{k-1}\nonumber\\&\times\left(\frac{1-\alpha}{\beta}+\frac{\alpha}{\sqrt{2\pi}\sigma}\right).
    \label{eq:best_order_likelihood_general}
    \end{aligned}$$

Noting that $\left(\frac{1-\alpha}{\beta}+\frac{\alpha}{\sqrt{2\pi}\sigma}\right)$ is always positive and independent of $k$, we extract it from the likelihood function. Simplifying  by means of manipulating the terms, we get

$$\begin{aligned}
        2(2-\alpha)^{-k}&=2^{1-k}(1-\frac{\alpha}{2})^{-k},\\
        \alpha^{k-1}&=(2\frac{\alpha}{2})^{k-1}=2^{k-1}(\frac{\alpha}{2})^{k-1}.
        \end{aligned}$$

the likelihood function to be maximized can be re-written as $$\begin{aligned}
    \ell(k\mid y=x,\alpha,,\sigma^2) \propto \begin{pmatrix}N-1\\k-1\end{pmatrix}(\frac{\alpha}{2})^{k-1}(1-\frac{\alpha}{2})^{N-k}.
    \label{eq:best_order_likelihood_modified}
    \end{aligned}$$

In order to find the maximum likelihood estimate $\hat{k} = \operatorname*{arg\,max}_k \ell(k\mid y-x=0)$, we note that  is a binomial distribution with probability of success $\frac{\alpha}{2}$. Hence, the maximum of the function is given at the mode of the distribution, $$\begin{aligned}
\hat{k}=\lfloor \frac{N\alpha}{2}\rfloor+1\quad \mathrm{or}\quad \lceil\frac{N\alpha}{2}\rceil.\end{aligned}$$ This gives the best order statistic estimator for the case when noise is a mixture of normal and uniform distribution as $$\begin{aligned}
\hat{x}^{\mathrm{\mathrm{\mathcal{U},\mathcal{N}}}}(y_{1:N},\alpha)= y^{(\hat{k})}_{1:N}.\end{aligned}$$

                 distribution                                          bias                                                       MSE
  ------------------------------------------ --------------------------------------------------------- ----------------------------------------------------------
            $\mathcal{U}[0,\beta]$                              $\frac{\beta}{N+1}$                                  $\frac{2\beta^2}{(N+1)(N+2)}$
        \[2mm\] $\mathrm{Exp}(\beta)$                            $\frac{\beta}{N}$                                       $\frac{2\beta^2}{N^2}$
      \[2mm\] $\mathrm{Rayleigh}(\beta)$                $\frac{\sqrt{\pi}\beta}{\sqrt{2N}}$                               $\frac{2\beta^2}{N}$
   \[2mm\] $\mathrm{Weibull}(\beta,\alpha)$   $\beta N^{-\frac{1}{\alpha}}\Gamma(1+\frac{1}{\alpha})$   $\beta^2N^{-\frac{2}{\alpha}}\Gamma(1+\frac{2}{\alpha})$
   \[2mm\] $\mathrm{Pareto}(\beta,\alpha)$               $\frac{N\alpha\beta}{N\alpha -1}$                        $\frac{N\alpha\beta^2}{N\alpha -2}$

  : Bias 