Consider the quantity

$$h_n:=\frac{S_{n-2}}{V_{n-2}} \int_{r=0}^1 r^{n-2} \sqrt{1-r^2} dr$$

where $S_{n-2}$, $V_{n-2}$ is respectively the surface and volume of the hypersphere in $\mathbb R^{n-1}$.

Question: How can we understand $h_n$? In particular, is $h_n$ increasing? Does it tend to $1$ as $n$ goes to infinity?

PS: The formula i quote above is not the average height of unit hemisphere. By average height of unit hemisphere i mean the integral $\int_{0}^1 |x_1| d \mu$, where $\mu$ is the uniform measure on $S^{D-1}$ and $x=(x_1,\dots,x_D)$ is the coordinate representation of a point $x \in S^{D-1}$. This integral i have calculated to be $\frac{(D-1)!!}{(D-2)!!} \frac{2}{\pi}$ if $D$ is even, and $\frac{(D-1)!!}{(D-2)!!}$ if $D$ is odd.

  • 1
    $\begingroup$ I am getting different values for the average height. Have you checked your calculation? $\endgroup$ – Douglas Zare Mar 11 '15 at 14:26
  • $\begingroup$ @DouglasZare: I apologize for the confusion. Please see my edit in the question. Is this the same formula you were getting? $\endgroup$ – Manos Dec 21 '16 at 6:27

Your expression equals $$\frac{\sqrt{\pi } (n-1) \Gamma \left(\frac{n-1}{2}\right)}{4 \Gamma \left(\frac{n}{2}+1\right)},$$ which goes to zero as $n$ goes to infinity.

  • $\begingroup$ The ratio in front can be (essentially) found on the Wiki page on the $n$-sphere (en.wikipedia.org/wiki/N-sphere), The integral is standard, though you can use Mathematica $\endgroup$ – Igor Rivin Mar 11 '15 at 1:15
  • $\begingroup$ So the conclusion is that the average height of the unit hemisphere goes to zero. Interesting. $\endgroup$ – Manos Mar 11 '15 at 1:18
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    $\begingroup$ On a high-dimensional sphere most of the mass is concentrated around the equator, where the height is lowest. That's a heuristic explanation for why the average height goes to zero. $\endgroup$ – Martin M. W. Mar 11 '15 at 1:19
  • $\begingroup$ @MartinM.W.: Thanks for this explanation. $\endgroup$ – Manos Mar 11 '15 at 1:20

There is a simple argument that the average height goes to $0$: Consider a point uniformly distributed on the intersection of the sphere with the first orthant. This has the same average height equal to the average of every other coordinate. For any $\varepsilon \gt 0$, at most $\frac{1}{\varepsilon^2}$ of the coordinates can be at least $\epsilon$ or else the sum of the squares of the coordinates would be greater than $1$. By symmetry, the probability that each coordinate is greater than $\varepsilon$ is at most $\frac{1}{\varepsilon^{2}n}$. The average of each coordinate is at most $\varepsilon + \frac{1}{\epsilon^{2}n}$. If we choose $\epsilon = \frac{1}{\sqrt[3]{n}}$ then this upper bound for $h_n$ is $\frac{2}{\sqrt[3]n}$.

This bound is not sharp. A better heuristic to understand the average is that the distribution of each coordinate of a sphere of radius $\sqrt{n}$ approaches a standard normal distribution as $n\to \infty$ (much more is known), so each coordinate of a unit sphere has a distribution approximated by $N(0,\frac{1}{n})$. If $X \sim N(0,1), E[|X|] = \sqrt{\frac{2}{\pi}}$, so we can approximate the average by $\sqrt{\frac{2}{\pi n}}.$


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