# Order statistics of independent but non-identical uniform distribution

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 $$\{a_iU_i> a_jU_j>... \}$$?

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.
• You are answering a different question: your assertion holds only for identically distributed uniform$(0,1)$ variables. – whuber Nov 30 '20 at 18:53