Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have a question bout the proof of the independence of gap RVs. Given the independent exponentially distributed random variables $\xi_1$, $\xi_2$ ~ $\text{Exp}(\lambda)$, and a corresponding order statistic $(\xi_{(1)},\xi_{(2)})$ whereby $\xi_{(1)}<\xi_{(2)}$, (i.e. $\xi_1>\xi_2 \Rightarrow \xi_{(2)}=\xi_1 $ and $\xi_{(1)}=\xi_2$)

I would like to know the distribution of the gaps $\Delta_{(1)}=\xi_{(1)}-0=\xi_{(1)}$ and $\Delta_{(2)}=\xi_{(2)}-\xi_{(1)}$

Why can you, without loss of generality take $\lambda =1$?

My attempt is to show that for every $A,B\in \mathcal{B}$, i.e. every Borel subset, that the relation $P(\Delta_{(1)}\in A,\Delta_{(2)}\in B)=P(\Delta_{(1)}\in A)P(\Delta_{(2)}\in B)$ holds.

Equivalently, $P(\Delta_{(1)}>x,\Delta_{(2)}>y)=P(\Delta_{(1)}>x)P(\Delta_{(2)}>y)$ for all $x,y \in \mathbb{R}$. (Is the reasoning behind this $\cup_{x\in \mathcal{R}} [x,\infty)=\mathcal{B}$?)

So $P(\Delta_{(1)}>x,\Delta_{(2)}>y)=P(\xi_{(1)}>x,\xi_{(2)}-\xi_{(1)}>y)=P(\xi_{(1)}>x,\xi_{(2)}>x+y)$.

$\Rightarrow P(\xi_{(1)}>x,\xi_{(2)}>x+y)=P(\xi_1>x,\xi_2>x+y)$, why is this?

Obviously, since $\xi_1,\xi_2$ are independent $P(\xi_1>x,\xi_2>x+y)=\exp(-2x+y)$

So then, $P(\xi_{(1)}>x,\xi_{(2)}>x+y)=\exp(-2x+y)$ but why does this mean that $P(\xi_{(1)}>x=\exp(-2x)$ and $P(\xi_{(2)}>x+y)=\exp(-x-y)$?

share|improve this question
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.