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I know that if $U$ is a uniform r.v. in $(0,1)$, then $U^a\sim Beta(1/a,1)$ with $a>0$.

On the other hand, if $U_{(1)}\leq \cdots\leq U_{(n)}$ are the uniform order statistics, then, with $U_{(0)}=0$ and $U_{(n+1)}=1$, we have $$(U_{(i+1)}-U_{(i)})_{i=0\ldots n}\sim Dir(1,\ldots,1)$$

My question is: Can we say something like $$(U_{(i+1)}^a-U_{(i)}^a)_{i=0\ldots n}\sim Dir(1/a,\ldots,1/a,1) ?$$

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  • $\begingroup$ I have edited your question to introduce more subscript brackets and other minor adjustments. You might want to check it is what you want $\endgroup$
    – Henry
    Commented Jul 12, 2023 at 9:08
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    $\begingroup$ $Dir(1/a,\ldots,1/a)$ is certainly not correct as that would make $\mathbb E (U_{(i+1)}^a-U_{(i)}^a) = \frac1{n+1}$ when these expectations will vary with both $a$ and $i$ $\endgroup$
    – Henry
    Commented Jul 12, 2023 at 9:21
  • $\begingroup$ $Dir(1/a,\ldots,1/a,1)$ will also be wrong $\endgroup$
    – Henry
    Commented Jul 12, 2023 at 15:36
  • $\begingroup$ Is there some particular reason ? $\endgroup$
    – Pierre
    Commented Jul 12, 2023 at 15:52
  • $\begingroup$ If $n$ and $a$ are both large, then the expectations will strictly increase with $i$ while they do not for that Dirichlet distribution $\endgroup$
    – Henry
    Commented Jul 12, 2023 at 16:02

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