Suppose we have independently distributed $X_i \sim \text{Uniform}(0,a_i)$ where the $a_i>0$ are fixed numbers. I want to obtain the probability that $X_j=X_{n-i+1,n}$ where $X_{n-i+1,n}$ is the $n-i+1$-th order statistic of the sample $X_1,...,X_n$.

Obviously, $\{X_i\}_{i=1}^n\overset{d}{=}\{a_i U_i\}_{i=1}^{n}$ where $U_i$ are i.i.d standard uniform random variables. Is there a smarter way to calculate this probability other than integrating (and summing) over sets of the form $\{a_iU_i> a_jU_j>... \}$?


2 Answers 2


Denote $X_{(1)}$ as any first observation from the original sample, $X_{(2)}$ as any second observation etc. We can calculate the probability as follows: \begin{align}P(X_j&=X_{n-i+1,n})\\ &=\mathbb{E}(P(X_{(1)}\leq X_j,...,X_{(n-i)} \leq X_j, X_{(n-i+1)}>X_j,...,X_{(n-1)}>X_j \text{ for all possible combinations() }|X_j))\\ &=\int_{0}^{a_j} \frac{1}{a_j} \sum_{\text{any combination}}^{n-1 \choose n-i}\prod_{i=1}^{n-i} \frac{x}{a_i} \prod_{i=1}^{i-1} \frac{1-x}{a_i}dx\\ &=\left(\sum_{\text{any combination}}^{n-1 \choose n-i}\prod_{i=1}^{n-i} \frac{1}{a_i} \prod_{i=1}^{i-1} \frac{1}{a_i} \right)\int_{0}^{a_j} \frac{1}{a_j}x^{n-i}(1-x)^{i-1} dx \\ &=\left({n-1 \choose n-i}\prod_{i=1}^{n} \frac{1}{a_i} \right)\int_{0}^{a_j} x^{n-i}(1-x)^{i-1} dx \end{align}


I believe the $i-th$ order statistic of a uniform sample follows a $Beta(i, n-i+1)$ distribution.

Proof is likely integrating those sets you mention once so to show what the density ought to be.

  • 1
    $\begingroup$ You are answering a different question: your assertion holds only for identically distributed uniform$(0,1)$ variables. $\endgroup$
    – whuber
    Commented Nov 30, 2020 at 18:53
  • $\begingroup$ Indeed, you are right. Woops $\endgroup$
    – Three Diag
    Commented Dec 1, 2020 at 3:34

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