Average distance to the middle

The previous post looked at the average ℓp distance between points in the ℓp disk. This post looks at a related question, the average distance to the center. Unlike the previous post, we will look at dimension n greater than 2.

The paper [1] cited earlier ends with this statement:

It should be pointed out that J. S. Lew proved a more general result: in n-dimensional space, the average distance from a point in the unit ℓp ball to the center is n/(n + 1) for all p.

There is a paper by by J. S. Lew listed in the references with the same author and title as [2], but in place of the journal and page number it says “this Review, to appear.” So perhaps the authors of [1] talked to the authors of [2] and knew that they intended to prove the result quoted above. But [2] came out a year later, and did not include results for dimensions higher than n = 2, unless I’ve overlooked something.

I imagine the theorem above, if it’s true, is tedious to prove for general p. But we can show that it’s true for p = 2.

For n = 2, we use polar coordinates. The distance to the origin is simply r, the volume element is r dr dθ, and the area of the unit disk is π, and so the average distance to the origin is

\frac{1}{\pi}\int_0^{2\pi} \int_0^1 r^2\, dr\, d\theta = \frac{2}{3}

For n = 3, we use spherical coordinates. The distance to the origin is again r, the volume element is now

r^2 \, \sin \varphi \, dr \, d\varphi \, d\theta

and the volume of the unit ball is 4π/3, and so the average distance to the origin is

\frac{3}{4\pi}\int_0^{2\pi} \! \int_0^\pi \! \int_0^1 r^3 \sin\varphi \, dr\, d\varphi\, d\theta = \frac{3}{4}

Finally, for general n we use hyperspherical coordinates. In n dimensions, we have an r that ranges from 0 to 1 and a θ that ranges from 0 to 2π as before, and we have n-2 φ’s that run from 0 to π.

The volume element in hyperspherical coordinates is

r^{n-1} \sin^{n-2}(\varphi_1) \sin^{n-3}(\varphi_2) \cdots \sin(\varphi_{n-2}) \, dr \, d\varphi_1 \, \cdots d\varphi_{n-2} \, d\theta

We could find the volume of the n-sphere by integrating this, but we don’t have to. The integral for the average distance will have an r term with exponent n and the integral for the volume will have an r term with exponent n-1. All the other terms not involving r are the same in both integrals, so they cancel out when we take the ratio.

The average distance calculation reduces to

\frac{\int_0^1 r^n \, dr}{\int_0^1 r^{n-1} \, dr} = \frac{1/(n+1)}{1/n} = \frac{n}{n+1}

which proves the claim at the top of the post for p = 2.

Related posts

[1] C. K. Wong and Kai-Ching Chu. Distances in lp Disks. SIAM Review, Vol. 19, No. 2 (Apr., 1977), pp. 320-324.
[2] John S. Lew, James C. Frauenthal, and Nathan Keyfitz. On the average distances in a circular disc. SIAM Review, Vol. 20, No. 3 (Jul., 1978), pp. 584-592.

3 thoughts on “Average distance to the middle

  1. Austin Buchanan

    I used to read all of your posts via email. Now that you have switched, the emails only show a snippet of the post (one or two sentences), making me click a link to read the full post. This takes me out of my email client and discourages me from reading. Since the switch, I have found that I read 10% of the posts. Can you make it to where the email includes the full blog post?

  2. In fact, you can apply a similar argument for any p, or any norm for that matter. All you need is that the volume of a ball of radius r is proportional to r^n, which follows from the linearity property of a norm (that is, a ball of radius r is just a ball of radius 1 scaled by a factor of r, regardless of what that ball looks like an a particular norm).

  3. @Austin: I’m sorry but I don’t believe I have that option.

    @Nathan: That’s slick.

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