The difference of $\sum_{i=1}^{n}X_{i}$ and $\sum_{i=1}^{n}X_{(i)}$

Question

Here is a exercise from *Mathematical Statistics. Jun Shao. Second edition. EX2.20

Let $X_1,..., X_n$ be $i.i.d.$ random variables having the exponential distribution $E(a,\theta)$, $a\in R$, and $\theta > 0$.

$$ f(x,a,\theta)=\frac{1}{\theta}e^{-(x-a)/\theta}I(x>a) $$

$$ F(x)=1-e^{-(x-a)/\theta} $$

Show that the smallest order statistic, $X_{(1)}$, has the exponential distribution $E(a,\theta/n)$ and that $2\sum\limits_{i=1}^{n}(X_{i}-X_{(1)})/\theta$ has the chi-square distribution $\chi^2_{2n-2}$.

What I know is:

The p.d.f. of $X_{(1)}$ is: $$ f(x)=\frac{n!}{(n-1)!}\left[\frac{1}{\theta}e^{-(x-a)/\theta}\right]\left[e^{-(x-a)/\theta}\right]^{n-1}=\frac{1}{\theta/n}\exp\left\{-\frac{x-a}{\theta/n}\right\}\sim E(a,\theta/n) $$

I don't know how to caculate the p.d.f. of $2\sum\limits_{i=1}^{n}(X_{i}-X_{(1)})/\theta$ ;

But I know how to caculate the p.d.f. of $2\sum\limits_{i=1}^{n}(X_{(i)}-X_{(1)})/\theta$ .

My question is that can we assume $\sum\limits_{i=1}^{n}X_{i}$ and $\sum\limits_{i=1}^{n}X_{(i)}$ are the same?

If we can not, can you give one example to explain? and how to solve the question above.

Answer 1

Hint: If the data is $X_1, X_2, \dotsc, X_n$ and its order statistics is $X_{(1)} \le X_{(2)} \le \dotsm \le X_{(n)}$, can you explain that the order stats is simply some permutation of the data $X_i$ in its original order, that the two collections are equal as sets, so the sum must be the same?

$1 +2+3$ is the same as $1+3+2$.