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{\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

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

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


\vspace{.25in}

{\bf Answer}

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

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

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

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 $h(z)=  ne^{-\frac{z}{2}} \cdot
\frac{1}{2n} =  \frac{1}{2}e^{-\frac{z}{2}},\ \ z>0$

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

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