<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
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{\bf Question}

Let $X_1,\ldots,X_n$ be a random sample of size $n$ from the
distribution having pdf

$$f(x|\theta)=\ds\frac{1}{\theta}\exp\left(-\ds\frac{x}{\theta}\right),\
\ \theta&gt;0.$$

{\noindent}Let $Y_1=$min$(X_1,\ldots,X_n)$ and
$Y_n=$max$(X_1,\ldots,X_n)$.  Find the probability density
functions of $Y_1$ and $Y_n$.  Find $E(Y_1)$.  Find $P(Y_n \leq
4)$ when $n=5$.



\vspace{.25in}

{\bf Answer}

We have $f(x|\theta)=\ds\frac{1}{\theta}e^{-\frac{x}{\theta}},\ \
0&lt;x&lt;\infty$

Therefore $F(x)=\ds\int_0^x \ds\frac{1}{\theta}
e^{-\frac{u}{\theta}}\,du = 1-e^{-\frac{x}{\theta}},\ \
0&lt;x&lt;\infty$

[using: $\ds\int e^{mu} \,du = \frac{e^{mu}}{m} + const$]

pdf of $Y_1$ is

\begin{eqnarray*}
g(y_1) &amp; = &amp; n\{1-F(y_1)\}^{n-1}f(y_1),\ \ 0&lt;y_1&lt;\infty\\ &amp; = &amp;
n\left\{1-e^{-\frac{y_1}{\theta}}\right\}^{n-1}\ds\frac{1}{\theta}e^{-\frac{y_1}{\theta}},\
\ 0&lt;y_1&lt;\infty\\ &amp; = &amp;
\ds\frac{n}{\theta}e^{-\frac{(n-1)y_1}{\theta}} \cdot
e^{-\frac{y_1}{\theta}},\ \ 0&lt;y_1&lt;\infty\\ &amp; = &amp;
\ds\frac{n}{\theta}e^{-\frac{n}{\theta}y_1},\ \ 0&lt;y_1&lt;\infty
\end{eqnarray*}

We see that $Y_1 \sim$ Exponential with $\beta=\frac{\theta}{n}$

[Table of common distributions]

Therefore $E(Y_1)=\beta=\frac{\theta}{n}$

pdf of $Y_n$ is
\begin{eqnarray*} h(y_n) &amp; = &amp; n\{F(y_n\}^{n-1}f(y_n)\\ &amp; = &amp;
n\left\{1-e^{-\frac{y_n}{\theta}}\right\}\ds\frac{1}{\theta}
e^{-\frac{y_n}{\theta}}\\&amp; = &amp;
\ds\frac{n}{\theta}e^{-\frac{y_n}{\theta}}\left(1-e^{-\frac{y_n}{\theta}}\right)^{n-1},\
\ 0&lt;y_n&lt;\infty \end{eqnarray*}

$P(Y_n \leq y_n)=\{F(y_n)\}^n$

Therefore $P(Y_n \leq 4)= \{F(4)\}^n =
\left(1-e^{-\frac{4}{\theta}}\right)^n$ for any $n$

Therefore Answer $=\left(1-e^{-\frac{4}{\theta}}\right)^5$ when
$n=5$


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