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Should they be close to the middle (origin) or close its surface?

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    $\begingroup$ Could you rephrase your question as "a hypersphere is the set of points at a constant distance from a given point" [Wikipedia]. "In mathematics, a ball is the space bounded by a sphere." [Wikipedia] $\endgroup$
    – Xi'an
    Commented Feb 10, 2018 at 12:46
  • $\begingroup$ Closely related: stats.stackexchange.com/questions/99171/… $\endgroup$
    – Sycorax
    Commented Feb 10, 2018 at 17:27

1 Answer 1

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As pointed out by @Xi'an, the OP's question is actually about a uniform distribution on the $n$-dimensional ball of radius $r$, the set of points at distance no more than $r$ from the center of the ball, and not about a uniform distribution on the $n$-dimensional hypersphere which is the surface of the ball (the set of points at distance exactly $r$ from the center). Note that it is being assumed that the joint density of the $n$ random variables has constant value $V^{-1}$ where $V$ is the volume of the ball. This is not the same as assuming that the distance of the random point is uniformly distributed on $[0,r]$ (or $[0,r)$ for those who do not want to include the surface of the hypersphere).

Almost the entire volume of a $n$-dimensional ball lies close to the surface. This is because $V$ is proportional to the $n$-th power of the radius of the ball, and $r^n$ is a very rapidly increasing function. Even in $3$-space, $\frac 78 = 1 - \left(\frac 12\right)^3$th of the volume lies closer to the surface than to the origin, and this fraction gets closer and closer to $1$ as $n$ increases. Turning the calculation around, for a fixed proportion $\alpha$, say $\alpha=0.95$, $100\alpha\%$ of the volume lies in a shell of inner radius $\sqrt[n]{\alpha}r$ and outer radius $r$ and so $1-\sqrt[n]{\alpha}$, the relative thickness of the shell, decreases towards $0$ with increasing $n$ for any choice of $\alpha \in (0,1)$.

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    $\begingroup$ Do you mean ball? A hypersphere is the set of points at a constant distance from a given point. $\endgroup$
    – Xi'an
    Commented Feb 10, 2018 at 12:46
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    $\begingroup$ +1 Dilip Sarwate the OP also explicitly asked about the origin of the hypersphere; while your answer addresses a relationship to the surface, based on volume, how does your answer address uniformity with respect to distance from (i.e. how "close to") the origin? (It may do so implicitly, but maybe your answer could also make that explicit). $\endgroup$
    – Alexis
    Commented Feb 10, 2018 at 18:06
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    $\begingroup$ Some alternative, more explicit, viewpoint that could help to understand the question/answer would be expressing the relationship for the density function of the points lying within radii $r-\frac{1}{2} dr$ and $r+\frac{1}{2}dr$ $$f(r) = \frac{n}{R^n} r^{n-1} $$ with $n$ and $R$ the dimension and size of the ball. So we see the density is larger for larger values of $r$ and with larger $n$ this difference becomes more dramatic. $\endgroup$ Commented Feb 11, 2018 at 12:09

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