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Given a vector $X \in R^n = \{x_1, x_2, ..., x_n\}$ drawn uniformly such that:

  • $x_i \in [0, 1]$ for all $i$; and
  • $\sum x_i = 1$,

how would you find the probability that any of the $x_i > y$, for some $y \in [0, 1]$?

Where I've gotten:

The problem should be symmetric w.r.t. the $x$s, so the result should be equivalent to $n $ x $P(x_0 > y)$.

We can consider a process of dividing up the total interval [0, 1] into chunks. First, you choose an order to assign values to the $x$s. For example, we could say $x_4 = 0.3, x_9 = 0.1...$

In this view, $P(x_0 > y)$ = $P(x_0 > y| x_0 \text{ is chosen first})P(x_0 \text{ is chosen first}) + P(x_0 > y| x_0 \text{ is chosen second})P(x_0 \text{ is chosen second})$...

This quickly becomes a nasty recursive problem, and I'm curious if there is already an elegant solution for it, or at least a well-documented one.

RedPanda
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  • Could you clarify what is the joint distribution for $\mathbf{X}$? – Lucas Prates Aug 09 '21 at 00:33
  • If $X$ is uniformly distributed on the sphere then I imagine a volume argument will work. –  Aug 09 '21 at 04:29
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    Take a look at this: https://stats.stackexchange.com/questions/162560/distribution-of-the-largest-fragment-of-a-broken-stick-spacings – soakley Aug 09 '21 at 18:54
  • What do you mean by $x_i \in U(0, 1)$ – kjetil b halvorsen Aug 10 '21 at 00:00
  • Ah, my intention with $x_i \in U(0, 1)$ is to say that, marginally, the elements of the vector have a uniform distribution. I think @LarsvanderLaan's suggestion of a uniform distribution on the sphere in $R^n$ is the correct way of stating this. – RedPanda Aug 11 '21 at 20:19
  • @soakley This is exactly what I was looking for! Is there a way I can give you credit for finding the correct answer? – RedPanda Aug 11 '21 at 20:27
  • Your expression for the $L^1$ norm is incorrect, so which do you intend: that the $x_i$ sum to unity (as written) or that their *absolute values* sum to unity (as implied by use of the norm)? Could you clarify that by "sphere" you mean the $L^1$ sphere and not the (usual) $L^2$ sphere? Previous comments and edits strongly suggest your statement of this problem is not the one you intend. – whuber Aug 11 '21 at 21:00
  • @whuber thanks for keeping me honest! I wasn't being careful with my writing. Fixing now. – RedPanda Aug 12 '21 at 21:29
  • There's still an ambiguity. Are you trying to say that $x$ is uniform on the unit simplex or that its distribution is conditional on the sum of *iid* uniform components lying on the simplex? The two distributions are not the same and the answer will be different, too. – whuber Aug 12 '21 at 21:39

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