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~
Best Answer
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: