Find the cumulative function and the density of $m_n$ and $M_n$, where $m_n=\min(X_1,X_2,…,X_n)$ and $M_n=\max(X_1,X_2,…,X_n)$. [closed]

Let be $X_1,X_2,...,X_n$ be uniformly distributed random variables i.i.d.

a) Find the cumulative function and the density of $m_n \text{ and } M_n$ , where $m_n=min(X_1,X_2,...,X_n)$ and $M_n=max(X_1,X_2,...,X_n)$.

b) Let $Z_n=n(1-M_n)$. Show that $Z_n \xrightarrow{d} Z$, where $Z$ is a random variable with cumulative function $F_Z(z)=1-e^{-z}$

closed as off-topic by Davide Giraudo, Stefan Mesken, Namaste, астон вілла олоф мэллбэрг, ShaileshNov 26 '16 at 3:29

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If $X_i\rightarrow f$ Then, 
a). The CMF are \begin{align} F_{M_n}(x)&=P(\{\omega:M_n(\omega)<x\}) \\ &=P(\{\omega:X_1(\omega)<x\}\cap\cdots\cap\{\omega:X_n(\omega)<x\}) \\ &=P(\{\omega:X_1(\omega)<x\})\cdots P(\{\omega:X_n(\omega)<x\}) \\ &=\prod_{k=1}^nF_{X_k}(x) \\ &=F_{X}^n(x)\tag{if $X_n$ is i.i.d} \end{align} \begin{align} F_{m_n}(x)&=P(\{\omega:m_n(\omega)<x\}) \\ &=P(\{\omega:X_1(\omega)<x\}\cup\cdots\cup\{\omega:X_n(\omega)<x\}) \\ &=1-P(\{\omega:X_1(\omega)<x\}^c\cap\cdots\cap\{\omega:X_n(\omega)<x\}^c) \\ &=1-P(\{\omega:X_1(\omega)>x\}\cap\cdots\cap\{\omega:X_n(\omega)>x\}) \\ &=1-\prod_{k=1}^nP(X_k>x) \\ &=1-\prod_{k=1}^n(1-F_{X_k}(x)) \\ &=1-(1-F_{X}(x))^n\tag{if $X_n$ is i.i.d} \end{align} The density are $$f_{M_n}(x)=F_{M_n}'(x)=nF_{X}^{n-1}(x)f_X(x)$$ $$f_{m_n}(x)=F_{m_n}'(x)=n(1-F_{X}(x))^{n-1}f_X(x)$$ Here $F_{X_n}(x)=x$ for $X_n$ is uniformly distributed random variables. So $$F_{M_n}(x)=x^n, \quad F_{m_n}(x)=1-(1-x)^n$$
b). $$F_{Z_n}(z)=P(n(1-M_n)<z)=P(M_n>1-\frac{z}{n})=1-\left(1-\frac{z}{n}\right)^n\to1-e^{-z}$$ Last step uses the facts: $$\lim_{x\to\infty}\left(1-\frac{1}{x}\right)^x=\frac1{e}\quad\text{and} \quad \lim_{n\to\infty}\left(1-\frac{z}{n}\right)^n=\lim_{n\to\infty}\left(\left(1-\frac{z}{n}\right)^{n/z}\right)^z=e^{-z}$$ Thus $Z_n\xrightarrow{d} Z$, where $Z$ is a random variable with cumulative function $F_Z(z)=1-e^{-z}$.