by $$\begin{aligned}
f^{\mathrm{Exp}}_{(k,N)}(y;x,\beta) = \frac{N}{\beta}\begin{pmatrix}N-1\\k-1\end{pmatrix}\left(1-\exp(-\frac{(y-x)}{\beta})\right)^{k-1}\exp\left(-\frac{(N-k+1)(y-x)}{\beta}\right).
\label{eq:exponential_order}\end{aligned}$$ The first order statistic density is then given by letting $k=1$ in  that results in another exponential distribution, $$\begin{aligned}
f^{\mathrm{Exp}}_{(1,N)}(y;x,\beta) = f^{\mathrm{Exp}}(y;x,\bar{\beta}),\end{aligned}$$ where $\bar{\beta}=\frac{\beta}{N}$. Hence, the MSE of the minimum order statistics estimator is given by $$\begin{aligned}
{\mathrm{MSE}}\left(\hat{x}_{\mathrm{min}}^{\mathrm{Exp}}(y_{1:N})\right) = \frac{2\beta^2}{N^2}.\end{aligned}$$ In order to find the MVU estimator, we re-write the PDF as $$\begin{aligned}
f(y_{1:N};x,\beta) &= \frac{1}{\beta^N}\exp\left[-\frac{1}{\beta}\sum_{k=1}^{N}y_k\right]\exp\left[-\frac{N}{\beta}x\right]\times\sigma(y^{(1)}_{1:N} - x).
\label{eq:exponential_pdf_2}\end{aligned}$$

### Known hyper parameter $\beta$ {#known-hyper-parameter-beta}

In case of the known hyper parameter $\beta$, the Neyman-Fisher factorization of PDF  gives

$$\begin{aligned}
    T(y_{1:N}) &= y^{(1)}_{1:N}\\
    h(y_{1:N}) &=   \frac{1}{\beta^N}\exp\left[-\frac{1}{\beta}\sum_{k=1}^{N}y_k\right].
    \label{eq:exponential_known_ss}
    \end{aligned}$$

The MVU estimator can then be obtained from a transformation of the minimum order statistic that makes it an unbiased estimator. Finally, in case of exponential noise with known hyper parameter of the distribution, the MVU estimator and its MSE are given by

$$\begin{aligned}
    \hat{x}_{MVU}^{\mathrm{Exp}}(y_{1:N},\beta) &= y^{(1)}_{1:N} - \frac{\beta}{N}
    \\
    {\mathrm{MSE}}\left(\hat{x}_{MVU}^{\mathrm{Exp}}(y_{1:N},\beta)\right) &= \frac{\beta^2}{N^2}.
    \end{aligned}$$

### Unknown hyper parameter $\beta$ {#unknown-hyper-parameter-beta}

If the hyper parameter $\beta$ is unknown, the factorization gives

$$\begin{aligned}
    T(y_{1:N}) = \begin{bmatrix}
    y^{(1)}_{1:N}\\ \sum_{k=1}^{N} y_k
    \end{bmatrix} = \begin{bmatrix}
    T_1(y_{1:N}) \\ T_2(y_{1:N})
    \end{bmatrix}.
    \label{eq:uniform_unknown_ss}
    \end{aligned}$$

Noting that sum of exponential random variables results in a Gamma distribution, we have $T_2(y_{1:N})\sim\Gamma(N,\beta)$. Hence, $$\begin{aligned}
    {\mathbb{E}}(T(y_{1:N})) = \begin{bmatrix}
    