The difference of two order statistics of exponential distribution If we have a random sample $X_1,X_2, \ldots, X_n \stackrel{\text{iid}}\sim f(x\mid\theta)=e^{-(x-\theta)}I(x >\theta)$. We want to prove 
$$2\sum X_i-2n X_{(1)} \sim \chi^2_{n-2}$$
where $X_{(1)}$ is the smallest order statistic.
I tried: 
$$2\sum X_i-2n X_{(1)} =2\left[\sum X_i-n X_{(1)}\right]=2\left[\sum X_{(i)}-n X_{(1)}\right]=2\left[\sum \left(X_{(i)}- X_{(1)}\right)\right]$$ 
And I was trying to find the distribution of 
$$X_{(i)}- X_{(1)}$$
And I searched that
$$X_{(i)}-X_{(i-1)} \sim \operatorname{Exp}\left(\frac{1}{n+1-i}\right) \text{ if } X_i \stackrel{\text{iid}}\sim \operatorname{Exp}(1)$$
Any ideas? Thank you~
 A: Comment: @spaceisdarkgreen, I simulated the case $n = 10,\, \theta = 0$
to see if this seems to work at all and to confirm your correction of the degrees of freedom.
The red curve is for $\mathsf{Chisq}(n-2)$ and the green for
$\mathsf{Chisq}(2n-2).$ Of course, this doesn't prove anything, but (to me
anyhow) it offers hope your argument might be made rigorous. 

R code in case it is of any use:
m = 10^5;  n = 10
x = rexp(m*n);  MAT = matrix(x, nrow=m)
t = rowSums(MAT);  v = apply(MAT, 1, min)
y = 2*t - 2*n*v
hist(y, prob=T, br= 25, col="skyblue2", ylim=c(0,.12))
 curve(dchisq(x, n-2), 0, 50, lwd=2, col="red", add=T)
 curve(dchisq(x, 2*n-2), lwd=2, col="darkgreen", add=T)

A: Let $X_1,X_2,\ldots,X_n$ be a random sample from exponential distribution with mean $1.$ 
Then joint probability density of order statistics $X_{(1)},X_{(2)},\ldots,X_{(n)}$ is 
$$f_{X_{(1)},X_{(2)},\ldots,X_{(n)}}(x_1,x_2,\ldots,x_n)= n! e^{-\sum_{i=1}^{n}x_i}, 0\leq x_1\leq x_2\leq \cdots \leq x_n \leq \infty$$
Let us consider transformation 
$$Y_1=nX_{(1)}, Y_2=(n-1)(X_{(2)}-X_{(1)}), Y_3=(n-2)(X_{(3)}-X_{(2)}),\ldots,Y_n= X_{(n)}-X_{(n-1)}$$
$$\Rightarrow X_{(1)}=\frac{Y_1}{n}, X_{(2)}=\frac{Y_1}{n}+\frac{Y_2}{n-1},\ldots, X_{(n)}=\frac{Y_1}{n}+\frac{Y_2}{n-1}+\frac{Y_3}{n-2}+\cdots+Y_n$$
Jacobian of above transformation is $\frac{1}{n!}$.
So joint probability density function of  $Y_1,Y_2,\ldots,Y_n$  is given by
$f_{Y_1,Y_2,\ldots,Y_n}(y_1,y_2,\ldots,y_n)= e^{-\sum_{i=1}^n y_i}; 0\leq y_1,y_2,\ldots,y_n\leq \infty $.
This follows, using factorization theorem, $Y_1,Y_2,Y_3,\ldots,Y_n$ are identically and independently distributed as exponential variate with mean $1.$
$\Rightarrow Y_i=(n-i+1)(X_{(i)}-X_{(i-1)}) \stackrel{\text{iid}}{\sim} \operatorname{exp}(1)$; $i=2,3,\ldots,n$.
Hence $\sum_{i=2}^{n} Y_i= \sum_{i=1}^n(X_i-X_{(1)})$ is sum of $(n-1)$ independent $exp(1)$ variates, so $\sum_{i=1}^n(X_i-X_{(1)})\sim \operatorname{gamma}(n-1)$.
Ref: "Order Statistics & Inference" by Balakrishnan & Cohen.
https://www.amazon.com/Order-Statistics-Inference-Estimation-Methods/dp/149330738X
A: First try to show that $$X_i - X_{(1)} \sim \begin{cases} \delta_0 & \text{if } X_i = X_{(1)}, \\ \chi^2_2 & \text{otherwise}. \end{cases}$$ Then think about how to show that $X_i-X_{(1)},\, X_j-X_{(1)}$ are independent. Then use standard properties of the chi-square distribution.
