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I have a uniform random variable $X \sim \mathcal{U}(a,b)$ and I take a sample of $N$ i.i.d. realization from it: $\{X_1, X_2, ..., X_N\}$. I'm interested in the statistics of the interval between every two successive realizations after they are sorted out in increasing order. More precisely, if the sorted realizations are $\{X'_1, X'_2, ..., X'_N\}$ (with $X'_i < X'_{i+1}$), I want to know what is the distribution of $X_{i+1} - X_i$ with respect to the one of $X.$

UPDATE: The question is updated for clarification. Thanks for the constructive comments.

subset
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  • The sequence $X_{i+1}-X_i$ is i.i.d. and you can compute the density by convolving the density of $X$ with the density of $-X$. – Math1000 Dec 21 '19 at 19:56
  • Rather than "$N$ samples", you should refer to a sample consisting of $N$ observations. If you intend them to be independent, I would mention that. And you should write $X_{i+1}-X_i$ rather than $x_{i+1}-x_i.$ Without attention to that last distinction, you can't even understand something like $\Pr(X_i\le x_i). \qquad$ – Michael Hardy Dec 21 '19 at 20:13
  • And now to clarify your question: If you get $X_1=0.57, ,, X_2 = 0.83,,, X_3 = 0.42,$ then are you asking about $0.83 - 0.57$ and $0.42-0.83,$ or did you mean to sort them into increasing order and consider $0.57-0.42$ and $0.83 - 0.57 \text{ ?}$ If the former, then there may be no point in considering any value of $N$ except $N=3.$ And if the latter, then you have a somewhat more complicated question. $\qquad$ – Michael Hardy Dec 21 '19 at 20:16
  • @Math1000 I forgot to mention that I'd like to know the distribution of the difference in the elements of sorted sequence. – subset Dec 22 '19 at 23:40

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For future reference, the sorted sample values are called the order statistics of the sample.

Consider $n+1$ points independently uniformly randomly chosen on a circle. Uniformly randomly choose one of them to break the circle into an interval, and the other $n$ as points on that interval. The distribution of these $n$ points is that of $n$ points independently uniformly randomly chosen on the interval. By the symmetry of the circle, all distances between the points are identically distributed.

The probability density of each distance $l$ is proportional to the volume that it leaves for the remaining $(n+1)-2=n-1$ points, which is $(L-l)^{n-1}$ (where $L=b-a$ is the length of the interval / the perimeter of the circle). Then normalization yields the density $\frac nL\left(1-\frac lL\right)^{n-1}$.

joriki
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