# Order statistics uniformly spaced?

Suppose you have $$n$$ i.i.d. random variables $$X_i$$ that take values in $$[0,1]$$, and have an absolutely continuous distribution. Let $$X_{(1)}\le X_{(2)}\le \dots \le X_{(n)}$$ be the random variables arranged in increasing order.

Is it true that the $$X_{(j)}$$ will be uniformly spaced ? That is, do we have something like $$X_{(j)} \approx C \cdot \frac{j}{n}$$ almost surely?

• Why would they be uniformly spaced? – David Jun 8 at 11:00
• Beta distributions satisfy your criteria but most are not uniform. – Nick Cox Jun 8 at 11:07
• I see. Thank you, I'm really novice at that, – Lonewolf Jun 8 at 13:19
• For a uniform distribution on $(0,1)$, expected values of order statistics are equally spaced. – BruceET Jun 8 at 17:20

Comment: Following @NickCox's Comment, here are histograms of large samples $$(n = 50,000)$$ from distributions $$\mathsf{Beta}(2,2)$$ and $$\mathsf{Beta}(1,3).$$ [You can read more about beta distributions on Wikipedia.]

In each panel of the figure, the beta density function is shown as a black curve, and the sample deciles are shown as vertical red lines. That is, about 5000 (sorted) observations lie between neighboring vertical lines.

par(mfrow=c(1,2))
x = rbeta(50000, 2, 2)
q = quantile(x, (0:10)/10 )
hist(x, prob=T, col="skyblue2", main="BETA(2,2)")

• BruceET I am not sure I am seeing the relationship between the OP's question and your answer. Shouldn't you be presenting the distribution of gaps between $X_{j}$ & $X_{j+1}$ for all $X_{j}$ from 1 to $n-1$? That is, what is the distribution of $X_{j+1}-X_{j}$? Caveat emptor: I am just now imbibing caffeine, so I may be an idiot. ;) – Alexis Jun 8 at 17:02
• Maybe should have given a reminder that sample deciles are closely related to order statistics. Here roughly, $X_{(1)}, X_{(5001)}, \dots, X_{(50,000)}.$ (Haven't even gotten to caffeine yet here.) – BruceET Jun 8 at 17:14
• (1) X <- sort(runif(n=1000)) (2) DX <- X[2:1000] - X[1:999] (3) hist(DX) was along the lines of what I was thinking. The expected value of any particular gap may be $C=\frac{1}{n}$, but I wonder if the gaps themselves are not i.i.d? Since if one gap is very large, that necessarily shrinks the average size of the remaining gaps. (I am learning here, thank you for entertaining my wondering. :) – Alexis Jun 8 at 17:21