How can I calculate $E(N_n)$ and $V(N_n)$ with $N_n = \min(X_1, X_2,\cdots, X_n)$? Consider ($X_1, X_2,\cdots, X_n$) a random sample of a population of $X$ as a Weibull distribution of parameters $(0, \delta, 2), \delta\in\mathbb R^+$, (in short , $X \sim W (0, \delta, 2)$ where $X$ has distribution function $F(x) = 1 - e^{-\left(\frac x\delta\right)^2}$.
Consider $$\overline{X}_n = \frac1n\sum\limits_{i = 1}^n X_i\text{ and }N_n = \min(X_1, X_2, \cdots, X_n)$$

I already calculate $E(X)$ and $V(X)$ but now I'm trying to calculate $E(N_n)$ and $V(N_n)$. How can I calculate $E(N)$ and $V(N)$? I have no idea...

The solution is 


  $$E(N_n) = \frac{\sqrt\pi}2\frac\delta{\sqrt n}\text{ and }V(N_n) = \left(1 - \frac\pi4\right)\left(\frac\delta{\sqrt n}\right)^2 = \left(1 - \frac\pi4\right)\frac{\delta^2}n.$$


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For any $t&gt;0$, we have
$$
\{\min(X_1,\ldots,X_n)&gt;t\} = \bigcap_{i=1}^n\{X_i&gt;t\},
$$
and so 
$$
\mathbb P(N_n &gt; t) = \mathbb P\left(\bigcap_{i=1}^n\{X_i&gt;t\}\right).
$$
Since the $X_i$ are independent, 
$$\mathbb P\left(\bigcap_{i=1}^n\{X_i&gt;t\}\right) = \prod_{i=1}^n \mathbb P(X_i&gt;t).
$$
Since the $X_i$ all have $W(0,\delta,2)$ distribution,
$$
\prod_{i=1}^n \mathbb P(X_i&gt;t) = \mathbb P(X_1&gt;t)^n = (1-F(t))^n = e^{-\left(\frac x\delta\right)^{2n}}.
$$
It follows that
$$
\mathbb E[N_n] = \int_0^\infty (1-F_{N_n}(t))\ \mathsf dt = \int_0^\infty e^{-\left(\frac x\delta\right)^{2n}}\ \mathsf dt = \delta\cdot \Gamma \left(1+\frac{1}{2 n}\right),
$$
and
$$
\mathbb E[N_n^2] = \int_0^\infty 2t(1-F_{N_n}(t))\ \mathsf dt = \int_0^\infty 2te^{-\left(\frac x\delta\right)^{2n}}\ \mathsf dt = \delta^2 \Gamma \left(1+\frac{1}{n}\right).
$$
Therefore
\begin{align}
\operatorname{Var}(N_n) &amp;= \mathbb E[N_n^2] - \mathbb E[N_n]^2\\
&amp;= \delta^2 \Gamma \left(1+\frac{1}{n}\right) - \left(\delta\cdot \Gamma \left(1+\frac{1}{2 n}\right)\right)^2\\
&amp;= \delta ^2 \left(\Gamma \left(1+\frac{1}{n}\right)-\Gamma \left(1+\frac{1}{2 n}\right)^2\right)
\end{align}
For any $t&gt;0$, we have
$$
\{\min(X_1,\ldots,X_n)&gt;t\} = \bigcap_{i=1}^n\{X_i&gt;t\},
$$
and so 
$$
\mathbb P(N_n &gt; t) = \mathbb P\left(\bigcap_{i=1}^n\{X_i&gt;t\}\right).
$$
Since the $X_i$ are independent, 
$$\mathbb P\left(\bigcap_{i=1}^n\{X_i&gt;t\}\right) = \prod_{i=1}^n \mathbb P(X_i&gt;t).
$$
Since the $X_i$ all have $W(0,\delta,2)$ distribution,
$$
\prod_{i=1}^n \mathbb P(X_i&gt;t) = \mathbb P(X_1&gt;t)^n = (1-F(t))^n = e^{-\left(\frac x\delta\right)^{2n}}.
$$
It follows that
$$
\mathbb E[N_n] = \int_0^\infty (1-F_{N_n}(t))\ \mathsf dt = \int_0^\infty e^{-\left(\frac x\delta\right)^{2n}}\ \mathsf dt = \delta\cdot \Gamma \left(1+\frac{1}{2 n}\right),
$$
and
$$
\mathbb E[N_n^2] = \int_0^\infty 2t(1-F_{N_n}(t))\ \mathsf dt = \int_0^\infty 2te^{-\left(\frac x\delta\right)^{2n}}\ \mathsf dt = \delta^2 \Gamma \left(1+\frac{1}{n}\right).
$$
Therefore
\begin{align}
\operatorname{Var}(N_n) &amp;= \mathbb E[N_n^2] - \mathbb E[N_n]^2\\
&amp;= \delta^2 \Gamma \left(1+\frac{1}{n}\right) - \left(\delta\cdot \Gamma \left(1+\frac{1}{2 n}\right)\right)^2\\
&amp;= \delta ^2 \left(\Gamma \left(1+\frac{1}{n}\right)-\Gamma \left(1+\frac{1}{2 n}\right)^2\right)
\end{align}