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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.

Arya McCarthy
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Inforz
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  • Original problem discussed at https://stats.stackexchange.com/q/272385/119261. – StubbornAtom Aug 25 '23 at 10:16

  • 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 $$2\sum\limits_{i=1}^{n}(X_{(i)}-X_{(1)})/\theta$$This is exactly the trick, use the memoryless property of the exponential distribution, well done.

     – Sextus Empiricus Aug 25 '23 at 11:05

  • But I know how to caculate the p.d.f. of $$2\sum\limits_{i=1}^{n}(X_{(i)}-X_{(1)})/\theta$$ Here you actually use the same principle as the one that you are asking for. The sum of ordered exponentially distributed variables $(X_{(i)}-X_{(1)})$ is similar to the sum of unordered exponentially distributed variables.

     – Sextus Empiricus Aug 25 '23 at 11:17
  • If you shuffle a deck of cards, the sum of the numbered cards is the same every time. It’s also the same if you sort the cards. – Sycorax Aug 25 '23 at 12:50

1 Answers1

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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 $1+3+2$.

kjetil b halvorsen
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  • When we observe the data, we can know the $\sum X_{i}$ is equal to the $\sum X_{(i)}$. But $X_{i}$ and $X_{(i)}$ are random variables, I am still a little confused. – Inforz Aug 25 '23 at 04:04
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    But whatever is the data, the same argument can be made. There is no possibility of an outcome where the result does not hold. – kjetil b halvorsen Aug 25 '23 at 04:05
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    I think I know what confused me. For $X_{i}$ and $X_{(i)}$, they have different distribution. But for $\sum X_{i}$ and $\sum X_{(i)}$, they have the same distribution. And $X_{i}$ are i.i.d. but $X_{(i)}$ are not independent. – Inforz Aug 25 '23 at 06:51