<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}
\begin{document}
\parindent=0pt

{\bf Question}

Let $Y_1=$min$(X_1,\ldots,X_n)$ where $X_1,\ldots,X_n$ is a random
sample of size $n$ from the distribution with pdf

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

{\noindent}Show that $2n(Y_1 - \theta)$ has the $\chi^2$
distribution with 2 degrees of freedom.


\vspace{.25in}

{\bf Answer}

We have $f(x|\theta)=\ds e^{-(x-\theta)},\ \ x&gt;\theta$

Therefore \begin{eqnarray*} F(x) &amp; = &amp; \ds\int_{\theta}^x
e^{-(u-\theta)}\,du\\ &amp; = &amp; \ds\int_0^{x-\theta} e^{-z} \,dz,\ \
z={u-\theta}\\ &amp; = &amp; 1-e^{-(x-\theta)},\ \ x&gt;\theta
\end{eqnarray*}
\begin{eqnarray*} {\rm pdf\ of}\ Y_1 &amp; = &amp; g(y_1)\\ &amp; = &amp;
n\{1-F(y_1)\}^{n-1}f(y_1)\\ &amp; = &amp;
n\left\{e^{-(y_1-\theta)}\right\}^{n-1}  e^{-(y_1-\theta)},\ \
y_1&gt;\theta\\ &amp; = &amp; ne^{-n(y_1-\theta)},\ \ y_1&gt;\theta
\end{eqnarray*}

Let $z=2n(Y_1-\theta) \Rightarrow Y_1=\frac{Z}{2n}+\theta,\ \ Z&gt;0$

Therefore $\ds\frac{dy_1}{dz}=\ds\frac{1}{2n}$.  Therefore
$\left|\ds\frac{dy_1}{dz}\right|=\ds\frac{1}{2n}$

Therefore pdf of $Z$ is $h(z)=  ne^{-\frac{z}{2}} \cdot
\frac{1}{2n} =  \frac{1}{2}e^{-\frac{z}{2}},\ \ z&gt;0$

This is the pdf of $\chi^2$ with 2 degrees of freedom.

\end{document}
</pre></body></html>