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Let $X_1$ and $X_2$ be independent, $\text{Exp}(a)$-distributed random variables. Show that $X_{(1)}$ and $X_{(2)}−X_{(1)}$ are independent, and determine their distributions.

Although it looks like a duplicate(Old question) I have a different question. I'm more concerned about the bounds of integration.

I found the joint to be: $$f_{U, V}(u,v)=2a^2e^{-a(2u+v)}, \quad \text{ with } \quad U=X_{(1)} , V= X_{(2)} - X_{(1)} $$

Hypothetically speaking, I think that the bounds are found by writing: $X_{(1)}=U$ then I know automatically that $0\le U\le\infty$.

But for $V$, I think I do a similar thing with $X_{(2)}=V+U$, $ 0\le V+U \le \infty$ $ \implies -V\le U \le \infty $

Sorry about the fundamentality of this question, I feel like because of that I'm getting:

$X_{(2)} - X_{(1)}\sim \text{Exp}(a)$ which is true, but for $f_U(u)=2ae^{-au}$ instead of $f_U(u)=\frac{a}{2}e^{-\frac{a}{2}u}$

Questions:

How to find the bounds of integration? Is my way of doing it correct? Furthermore, how do you work with this stuff without having to do the transformations with the jacobians?

Mahamad A. Kanouté
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    Similar question with $a=1$: https://math.stackexchange.com/questions/2240822/let-u-operatornamemin-x-y-and-v-operatornamemax-x-y-show-that?noredirect=1&lq=1. – StubbornAtom Feb 22 '19 at 09:54
  • I checked it but couldn't find anything to answer my question – Mahamad A. Kanouté Feb 22 '19 at 09:59
  • It's not clear what you are asking. In any case, $X_2-X_1$ can't be an exponential, since it doesn't have always positive sign. You might want to write $X_1=aY_1$, $X_2=aY_2$ with $Y_i$ being $Exp(1)$ and do the calculations there first, just to avoid the annoying parameter $a$ and clear your ideas – Lucio Feb 22 '19 at 10:02
  • ah I see sorry. In the question the difference between $X_i$ and $X_{(i)}$ was never defined, I didn't realize the they were different objects. – Lucio Feb 22 '19 at 10:11
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    @MahamadA.Kanouté Regarding "how do you work with this stuff without having to do the transformations with the jacobians?", my linked post does have the answers. – StubbornAtom Feb 22 '19 at 10:11
  • @StubbornAtom I did get it now. But more importantly, I always fail when taking the marginal of such joints. Hence my question. Sorry if it wasn't clear enough – Mahamad A. Kanouté Feb 22 '19 at 10:15
  • The information $-V\le U \le \infty$ is useless because both $U$ and $V$ are non-negative by definition. Please elaborate on how you arrived at $f_U(u)=\frac{a}{2}e^{-\frac{a}{2}u}$ for people to help pointing out where you've gone wrong. – Lee David Chung Lin Feb 22 '19 at 13:59

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