# Joint distribution of order statistics

Let $X_{(1)} ≤ \cdots ≤ X_{(n)}$ be order statistics corresponding to an $\operatorname{Exp}(λ)$ i.i.d. sample. Let $U_1 = X_{(1)}$ and $U_k = X_{(k)}-X_{(k-1)}$ ($k = 2, \ldots , n$) be the differences between the consecutive ones. How can we find the joint and the individual distributions of $U_1, U_2, \ldots , U_n$. Thanks.

So far I tried as follows: joint probability density function $f$ given by

\begin{align} f(u_1, u_2, \ldots, u_n) & = n! f(u_1) \, f(u_2) \, \cdots \, f(u_n), \quad u_1 \lt u_2 \lt \cdots \lt u_n \\[10pt] & =n!\lambda e^{-\lambda u_1}\lambda e^{-\lambda u_2} \cdots \lambda e^{-\lambda u_n} \\[10pt] & =n!\lambda e^{-\lambda x_1}\lambda e^{-\lambda (x_2-x_1)} \cdots \lambda e^{-\lambda (x_n-(x_{(n-1)})}, \quad x_1 \lt x_2 \lt \cdots \lt x_n \\[10pt] & =n!\lambda e^{-\lambda x_n} \end{align}

But I get stuck hereafter.

• Have a look at (math.kth.se/matstat/gru/sf2955/exponorderstats.pdf) – Jean Marie Oct 29 '16 at 20:51
• @MichaelHardy It's because of $U_k \sim Exp(\lambda)$ since $X_k \sim Exp(\lambda)$ already. Am I wrong? – Userabc Oct 29 '16 at 21:34
• If I'm not mistaken, $U_1 \sim \operatorname{Exp}(n\lambda)$ and $U_2 \sim \operatorname{Exp}((n-1)\lambda)$, and so on, and they're independent (whereas, with distributions other than the exponential, they would not be independent). $\qquad$ – Michael Hardy Oct 29 '16 at 21:57
• The support should not be $0\le u_1\le u_2 \le \cdots \le u_n$, because these are not order statistics, but differences between adjacent order statistics. There's no reason when $U_2$ should always be greater than $U_1. \qquad$ – Michael Hardy Oct 29 '16 at 21:59

Given memorylessness of the exponential distribution, you will have independently $$U_j \sim \operatorname{Exp}((n+1-j)\lambda)$$ (much as Michael Hardy said in comments) so each with marginal density on positive $u_j$ values $$f_j(u_j) = (n+1-j) \lambda \,e^{-(n+1-j) \lambda u_j}$$
which is going to make the joint density \begin{align} f(u_1, u_2, \ldots, u_n) & = f_1(u_1) \, f_2(u_2) \, \cdots \, f_n(u_n) \\ & = n \lambda e^{-n \lambda u_1}\, (n-1) \lambda e^{-(n-1) \lambda u_2} \cdots \lambda e^{- \lambda u_n} \\ & =n! \,\lambda^n\, e^{-\lambda (nu_1+(n-1)u_2+\cdots + u_n)} \end{align}
for $u_1, u_2, \ldots, u_n$ all positive (or at least non-negative)