Difference of statistics of order of a exponential distribution

Question

Let $X_1, ..., X_n$ be a random sample of a exponential distribution with mean $1$. Is there an easy way to show the following:

$$(n - i + 1)(X_{(i)} - X_{(i-1)}) \stackrel{idd}{\sim} \mathrm{Exp}(1), \quad i = 2, \ldots, n \text{ ?}$$

I resolved using the joint distribution of $X_{(i-1)}, X_{(i)}$ and findind the moment generating function to $X_{(i)} - X_{(i-1)}$ but the calculation is extensive. I could not prove independence.

Answer 1

Let $X_1,X_2,...X_n$ be a random sample from exponential distribution with mean 1. Then joint probability density of order statistics $X_{(1)},X_{(2)},...X_{(n)}$ is $$f_{X_{(1)},X_{(2)},...X_{(n)}}(x_1,x_2,...,x_n)= n! e^{-\sum_{i=1}^{n}x_i}, 0\leq x_1\leq x_2\leq ...\leq x_n \leq \infty$$ Let us consider transformation

$Y_1=nX_{(1)}, Y_2=(n-1)(X_{(2)}-X_{(1)}), Y_3=(n-2)(X_{(3)}-X_{(2)}),....,Y_n= X_{(n)}-X_{(n-1)}$

$\Rightarrow X_{(1)}=\frac{Y_1}{n}, X_{(2)}=\frac{Y_1}{n}+\frac{Y_2}{n-1},...., X_{(n)}=\frac{Y_1}{n}+\frac{Y_2}{n-1}+\frac{Y_3}{n-2}+...+Y_n$

Jacobian of above transformation is $\frac{1}{n!}$.

So joint probability density function of $Y_1,Y_2,...,Y_n$ is given by

$f_{Y_1,Y_2,...,Y_n}(y_1,y_2,...,y_n)= e^{-\sum_{i=1}^{n}y_i}; 0\leq y_1,y_2,...,y_n\leq \infty $.

This follows, using factorization theorem, $Y_1,Y_2,Y_3,...,Y_n$ are identically and independently distributed as exponential variate with mean 1.

$\Rightarrow Y_i=(n-i+1)(X_{(i)}-X_{(i-1)})\stackrel{iid}{\sim} exp(1)$; $i=2,3,...,n$.