Average angle of vectors

Question

Given the following set ($n$ dimensional vectors with the length $1$ where each component is positive):

$ S=\{x\in\mathbb{R}_{\geq0}^n: \|x\|=1\} $

What is the average / expected angle between two of these vectors?

For the $1$-dimensional case it is trivial, but for the $2$-dimensional case it already seems to be hard to get this value.

Maybe someone can help me?

Thank you very much.

Answer 1

Edit: Following the answer by @Guangliang and a remark of Christian Blatter showing that I had made an error (see bottom remark), here is a completely modified answer.

In $\mathbb{R^2}$, taking two arbitrary unit vectors $(\cos(a),\sin(a))$ and $(\cos(b),\sin(b))$ in the first quadrant (i.e., $(a,b)$ in the domain $D:=[0,\pi/2] \times [0,\pi/2]$) with dot product

$$\tag{1}\cos(a)\cos(b) + \sin(a)\sin(b)=\cos(a-b).$$

a) the average value of $\gamma=cos(a-b)$ is $$\hat{\gamma}=\dfrac{8}{\pi^2}$$

b) (the question asked by the OP) the average value of their angle $c=a-b$ is

$$\hat{c}=\dfrac{\pi}{6}.$$

Explanation for a) :

The volume under the surface with equation $z=\cos(a-b)$ for $(a,b) \in D$ is

$$I=\int_{a=0}^{\pi/2}\int_{b=0}^{\pi/2}\cos(a-b)dadb=\int_{a=0}^{\pi/2}\int_{b=0}^{\pi/2}(\cos(a)\cos(b) + \sin(a)\sin(b))dadb=$$

$$=\int_{a=0}^{\pi/2}\cos(a)da\int_{b=0}^{\pi/2}\cos(b)db+\int_{a=0}^{\pi/2}\sin(a)da\int_{b=0}^{\pi/2}\sin(b)db=1+1=2.$$

As the area of domain $D$ is $A=\pi^2/4$, we have, by definition of an average value:

$$\hat{c}=\dfrac{I}{A}=\dfrac{2}{\pi^2/4}=\dfrac{8}{\pi^2}.$$

Explanation for b): Instead of working with the cosine of the angle, we work directly with the angle. It means that, instead of the previous integral, taking $(1)$ into account, we take:

$$\tag{2}I=\int_0^{\pi/2}\int_0^{\pi/2}\arccos(\cos(a-b))dadb=\int_0^{\pi/2}\int_0^{\pi/2}|a-b|dadb$$

(because $\forall x, \ \ \arccos(\cos(x))=|x|$).

$$I=\int_{b=0}^{\pi/2}\int_{a=b}^{\pi/2}|a-b|dadb=\int_{b=0}^{\pi/2}\int_{a=b}^{\pi/2}(a-b) dadb + \int_{a=0}^{\pi/2}\int_{b=a}^{\pi/2}(b-a) dadb=$$

$$=\frac{\pi^3}{48}+\frac{\pi^3}{48}=\frac{\pi^3}{24}$$

For the same reason as in part a), the average angle is

$$\hat{c}=\dfrac{I}{A}=\dfrac{\pi^3/24}{\pi^2/4}=\dfrac{\pi}{6}.$$

Remark: the error I had made in the first version of the text is to assume that $\hat{c}=\arccos(\gamma)$ which, as Guangliang and Christian Blatter objected, would mean that $E(\arccos(X))=\arccos(E(X))$ which isn't true.

Answer 2

In two dimensional case, the problem becomes computing the mean of $|x-y|$ where $x$ and $y$ are independently drawn from a uniform random distribution in $[0, \frac{\pi}{2}]$, which is $$\left(\frac{2}{\pi}\right)^2\int_0^{\pi/2}\int_0^{\pi/2}|x-y|\,dx\,dy = \frac{\pi}{6}$$.