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Suppose we have $n$ points on the circle. This will divide the circle into $n$ pieces: $x_1, x_2, \dots, x_n$. By symmetry we know that each $x_i$ has the same distribution.

How to find this distribution? Somehow for me it is equivalent to evaluate the marginal distribution for $X_i$, where each $X_i$ is uniformly distributed in $[0,1]$, and conditioned on $X_1 + \dots + X_n = 1$

I need an answer with good calculus calculation, as I am somewhat confused. Just for declaration, this is not a homework problem, but just a natural extension to these problems behind:

Probability that n points on a circle are in one semicircle

Random points on a circle

I find out that there is no clear solution with regard to $n$ points cases. And I really curious about the distribution of one variable among them.

EggTart
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  • Hi! To avoid down-votes and close-votes, please provide us some context for this question, such as: (a) Is this homework? (b) If so, what course are you taking? (c) What specific topic are you covering at the moment? (d) What do you know that you think might be connected? (e) If you're stuck, what are you stuck on? For example, do you know what to apply, but don't know how to apply it, or do you not know what to apply? Please [edit] these facts into your original post, not as responses to this comment, as comments may be deleted without warning. – Brian Tung Oct 14 '24 at 18:51
  • Fixing the first point by symmetry, won't this be the same as the distribution of the $n$ segments resulting from choosing $n-1$ points indepently uniformly from $[0,1]$? That must be a known problem. – Greg Martin Oct 14 '24 at 19:38
  • Note that the duplicate shifts the definition of $n$ by $1$. That is, they imagine a selection of $n-1$ points on the circle. – lulu Oct 14 '24 at 19:40
  • It should be a known problem. I just kind of stucked. I just don't understand about the rigorous way of doing this using conditional probability. – EggTart Oct 14 '24 at 19:49

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