# Where are most points in a uniformly distributed high-dimensional ball?

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

• 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] Commented Feb 10, 2018 at 12:46
• Closely related: stats.stackexchange.com/questions/99171/…
– Sycorax
Commented Feb 10, 2018 at 17:27

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 $\alpha \in (0,1)$.
• 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. Commented Feb 11, 2018 at 12:09