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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 $F_Z(z)=1-e^{-z}$

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

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If $X_i\rightarrow f$ Then, $$ $$

\begin{eqnarray} F_{M_n}(x)&=& \mathbb{P}\{\omega:M_n(\omega)= \max(X_i(\omega))\leq x\}\\ &=& \mathbb{P}\{\omega:\forall ~i~X_i(\omega))\leq x\} \\&\overset{i.i.d}{=}&\prod \mathbb{P}\{\omega:X_i(\omega)\leq x\} \\& = & f(x)^n \end{eqnarray} On the other hand, \begin{eqnarray} F_{m_n}(x)&=& \mathbb{P}\{\omega:m_n(\omega)= \min(X_i(\omega))\leq x\}\\ &=& 1-\mathbb{P}\{\omega:m_n(\omega)= \min(X_i(\omega))\geq x\}\\ &=&1- \mathbb{P}\{\omega:\forall ~i~X_i(\omega))\geq x\} \\&\overset{i.i.d}{=}&1-\prod \mathbb{P}\{\omega:X_i(\omega)\geq x\} \\& = & 1-(1-f(x))^n. \end{eqnarray} The last question easily follows since \begin{eqnarray} F_{Z_n}(x)&=& \mathbb{P}\{\omega:Z_n(\omega)= n(1-M_n)(\omega)\leq x\}\\ &=& 1-F_{M_n}(1-\frac{x}{n}) = 1-[f(1-\frac{x}{n})]^n \end{eqnarray}

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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 $F_Z(z)=1-e^{-z}$.

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