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Homework Statement
Consider the exponential probability density function with location parameter ##\theta## :
$$f(x|\theta) = e^{-(x-\theta)}I_{(\theta,\infty)}(x)$$
Let ##X_{(1)}, X_{(2)},...,X_{(n)}## denote the order statistics.
Let ##Y_i = X_{(n)} - X_{(i)}##.
Show that ##X_{(1)}## is independent of ##(Y_1,Y_2,...,Y_{n-1})##
Homework Equations
The Attempt at a Solution
First, I consider the distribution of all order statistics for this distribution.
$$
\begin{align*}
f(x_1, x_2,..., x_n) &= n!e^{-\sum_{i=1}^{n}(x_i-\theta)}\\
&= n!\exp\left(n\theta - \sum_{i=1}^{n}x_i\right)\\\\
\end{align*}
$$
Next, I coerce the desired joint probability distribution function to resemble the known jpdf of order statistics.
$$
\begin{align*}
f(x_1, y_1,...,y_{n-1}) &= f(x_1, x_n-x_1,...,x_n-x_{n-1})\\
&= n!\exp(n\theta -x_1-x_2-...-x_n)\\
&= n!\exp(n\theta-x_1+(y_2-x_n)+...+(y_{n-1}-x_n)-(y_1+x_1))\\
&= n!\exp(n\theta-x_1+(y_2-y_1-x_1)+...+(y_{n-1}-y_1-x_1)-(y_1+x_1))\\
&= n!\exp(n\theta-nx_1-(n-1)y_1+y_2+...+y_{n-1})\\
&= n!\exp(n\theta-nx_1)\exp(-(n-1)y_1)\prod_{i=2}^{n-1}\exp(y_i)\\
&= g(x_1)h(y_1,...,y_{n-1})
\end{align*}
$$
The factorization into ##g## and ##h## prove independence.'
Okay, so I am really not sure if this type of argument holds... or if I need to pursue a more rigorous approach?