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.

Say we have $U_1 \dots U_n$ i.i.d. random variables uniform on $[0,1]$ and $Y_1 \dots Y_{n+1}$ i.i.d. random variables distributed as $Y_i \sim Exp(1)$. I know that the joint distribution of the order statistics $(U_{(1)}, \dots, U_{(n)})$ is equal in distribution to $(\frac{Y_1}{\sum_i^{n+1} Y_i}, \frac{Y_1+Y_2}{\sum_i^{n+1} Y_i}, \dots, \frac{Y_1+\dots+Y_n}{\sum_i^{n+1} Y_i})$. I can prove the result using a transformation of variables, Jacobians, etc., but this is rather tedious. Is there a more elegant way of deriving this statement? Maybe something with poisson processes?

share|improve this question
add comment

1 Answer

You are right that you can think of this problem in terms of Poisson processes. The first arrival time after $t = 0$ as well as the inter-arrival times in a Poisson process of rate $1$ are independent $\text{Exp}(1)$ random variables and so $Y_1$, $Y_1 + Y_2$, $\ldots$, $Y_1 + Y_2 + \cdots + Y_{n+1}$ can be taken to be the times of the first, second, $\ldots$, $(n+1)$-th arrivals after $t = 0$ in the process. The random variables $\frac{Y_1}{\sum_i^{n+1} Y_i}, \frac{Y_1+Y_2}{\sum_i^{n+1} Y_i}, \dots, \frac{Y_1+\dots+Y_n}{\sum_i^{n+1} Y_i}$ that you are looking at are the first $n$ arrival times "normalized" to a unit interval.

For $0 < t_1 < t_2 < \dots < t_n < 1$, the conditional probability that there is one arrival in each interval $(t_i, t_i + \Delta t_i)$ and none in the remaining time of total length $(1 - \sum_i \Delta t_i)$ given that there are $n$ arrivals in $(0, 1)$ is approximately $$ \begin{align*} \frac{\exp(-(1 - \sum_i \Delta t_i)) \prod_{i=1}^n \exp(-\Delta t_i)\Delta t_i/1!}{\exp(-1)\frac{1^n}{n!}} &= n! \Delta t_1\Delta t_2 \cdots \Delta t_n\\ &= f_{U_{(1)}, \dots, U_{(n)}}(t_1, t_2, \ldots , t_n)\Delta t_1\Delta t_2 \cdots \Delta t_n \end{align*} $$

share|improve this answer
add comment

Your Answer

 
discard

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.