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~

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)

