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Let $X_1$,$X_2$ be iid r.v, with Exp(a) distribution. Show that $X_{(2)}-X_{(1)}$ and $X_{(1)}$ are independent and find their distributions. ($X_{(1)}$ denotes the order statistic)

Their individual distributions can be easily found, where $X_{(2)}-X_{(1)} \in Exp(a)$ and $X_{(1)} \in Exp(a/2)$. I am struggling to show independence though, since I would think that this would require knowing the joint distribution of $X_{(2)}-X_{(1)}$ and $X_{(1)}$. Am I missing something obvious?

I am mostly interested in hints, not full solutions if possible!

Laplacinator
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    Can you compute the joint PDF of $(X_{(1)},X_{(2)})$? The rest follows by a change of variable. – Did Oct 02 '16 at 15:43
  • https://math.stackexchange.com/questions/2240822/let-u-operatornamemin-x-y-and-v-operatornamemax-x-y-show-that?noredirect=1&lq=1 – StubbornAtom Jul 14 '20 at 19:36

2 Answers2

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The joint probability density function is: $$\quad f_{[X_{(1)}, X_{(2)}-X_{(1)}]}(y,z) \\ = f_{[X_{(1)},X_{(2)}]}(y,z+y)~~ \\ = 2~f_X(y)~f_X(z+y) $$

If we have independence this will be equal to the product: $$f_{X_{(1)}}(y)~f_{[X_{(2)}-X_{(1)}]}(z)$$

Is it?

Graham Kemp
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HINT

Try working the problem a second time, replacing the exponential distributions with uniform distributions on $[0,1]$. (call these variables $Y$ instead of $X$). Stare at $$ P(Y_{(2)}-Y_{(1)} > \alpha | Y{(1)} = y) $$ and see if that expression contains a $y$.

Now with your exponential $X$s, stare at $$ P(X_{(2)}-X_{(1)} > \alpha | X{(1)} = x) $$

The special thing about the exponential distribution is how that expression depends on $x$.

Mark Fischler
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