Looking at Average distance of the mean of n random complex numbers in a unit disc, I tried to figure out  what is the expected absolute value $|\frac{z_1 + z_2}{2}|$ of two numbers $z_1, z_2\in\mathbb{C}$ drawn uniformly from the unit disc.  We can look at the quadruple integral $$\int_0^1r_1\int_0^{2\pi}\int_0^1r_2\int_0^{2\pi}|\frac{1}{2}(r_1e^{\phi_1} + r_2e^{\phi_2})| d\phi_2 dr_2 d\phi_1dr_1 $$ This can be reduced to two integrals, so that we can actually calculate the values with high precision (see the following code)

from mpmath import mp, sqrt, ellipe, quad, pimp.dps = 15
def ellipe_integrand(r, s):
    return  sqrt(r*r + 2*r*s + s*s)*(ellipe(+pi, 4*r*s/(r*r + 2*r*s + s*s)))
def integrand_2(s):
    return quad(lambda r: 2*r*ellipe_integrand(r, s), [0, s, 1])mp.dps=150quad(lambda s: s*integrand_2(s), [0,1])
# after ~40 minutes on my machine we get:
mpc(real='1.42222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222182628758433670366371075596783355537463385287071932198285251', imag='4.76485167346043265351378514322413241842941120327748264250820934786695559445396622361703586368896605436500687259315655829464952759337107787438513621789267e-246')

The result of that integration looks suspiciously like $\frac{64}{45}$. To get the expected absolute volume we still have to divide by the area of the disc:$$ \frac{64}{45\pi} \approx 0.45270739368361339952038048248181862978690743677285389866003155$$ (I haven't tried running numerical simulation of simply drawing two points repeatedly, but I would hope that one could recover the first few digits here..)

Let's denote the expected absolute value of the mean of n points by $\operatorname{exp\_abs}(n)$. It is easy to see that $\operatorname{exp\_abs}(1) = \frac{2}{3}$.

Question 1 Is $ \operatorname{exp\_abs}(2) = \frac{64}{45\pi}$, like the numerical evidence would suggest? How to prove that?

Question 2 Is there some reason to expect that $\operatorname{exp\_abs}(n)\in\mathbb{Q}(\pi)$?

Question 3 What is the value of  $\operatorname{exp\_abs}(3)$? Here I would be interested in numerical evidence if the exact value is not easily derived. 

  • 2
    $\begingroup$ It would take some computation to prove, but it shouldn't be too surprising for exp_abs(n) $\in \mathbb{Q}(\pi)$. In particular, you can obtain the distribution of the averages of $n$ uniformly drawn points by repeatedly convolving the uniform measure with itself (and then scaling). This will introduce $\pi$ everywhere but it's conceivable that all the other constants are rational . $\endgroup$
    – Gabe K
    Jun 20, 2021 at 14:50
  • 1
    $\begingroup$ This phenomena also appears in another problem about drawing from a disk (or $n$-ball), which is the probability that the uniformly drawn points form the vertices of an acute triangle. See "Acute Triangles in the n-Ball" by Glen Hall for details on that. $\endgroup$
    – Gabe K
    Jun 20, 2021 at 14:52

3 Answers 3


The easiest way seems to be to take the integration variable to be $x= |\frac{z_1+z_2}{2}|$ and then integrate over the position of the point furthest from the origin:

enter image description here

This gives $$ \operatorname{exp\_abs}(2) = \frac{\int_0^1 \mathrm{d}x \, x^2 \int_x^1 \mathrm{d}y \sqrt{1-y^2} }{\int_0^1 \mathrm{d}x \, x \int_x^1 \mathrm{d}y \sqrt{1-y^2} } = \frac{2}{3}\frac{\int_0^1 \mathrm{d}y\, y^3\sqrt{1-y^2} }{\int_0^1 \mathrm{d}y \,y^2 \sqrt{1-y^2} } = \frac{64}{45\pi}.$$

  • $\begingroup$ Nice, thanks! I can immediately see the second and third equality, but didn't get the first equality right away. $\endgroup$ Jun 18, 2021 at 18:21
  • 1
    $\begingroup$ @MoritzFirsching, I have added an illustration that should make the first equality clear. $\endgroup$ Jun 20, 2021 at 12:09
  • $\begingroup$ Yes that illustration makes it clear! This answers Question 1, still trying to generalize that approach to $n=3$. $\endgroup$ Jun 22, 2021 at 6:47

(Too long for a comment) For the record: $$\mathrm{exp\_abs}(3)=\frac{4}{3 \pi^2}\,I_3=0.3671989447$$ where $$I_3=\int_{-1}^1\int_{-1}^1 \int_{-1}^1 |x+y+z|\sqrt{1-x^2}\sqrt{1-y^2}\sqrt{1-z^2}\,dx\,dy\,dz=2.718081241$$
(I couldn't express by known constants)


The exact value of $\mathrm{exp\_abs}(3)$ can be deduced from the following (amazing!) result: if $X,Y,Z$ are independent and identically uniformly distributed on $S^3$ (the unit sphere in $\mathbb{R}^4$), then \begin{align*} \mathbb{E}|X+Y+Z|=W_3(1,1) &=\frac{476}{525}A+\frac{52}{7\,\pi^2}\frac{1}{A}\\ \end{align*} with $A:=\frac{3}{16}\frac{2^{1/3}}{\pi^4}\Gamma(\frac{1}{3})^6$. ( See https://scholarship.claremont.edu/jhm/vol6/iss1/7/ (on page 100). Timothy Budd pointed to this paper in the related post.)

If $X=(X_1,X_2,X_3,X_4)$ is uniform on $S^3$, and $U=(U_1,U_2)$ is uniform on $D^2,$ (the unit disk in $\mathbb{R}^2$), the distributions of $X_1$ resp. $U_1$ coincide, and the same will then hold for the first coordinate of i.i.d. sums.

Now, for any rotationally symmetric random vector $V=(V_1,V_2,...,V_k)$ with finite expectation of $|V|:=\sqrt{V_1^2+V_2^2+\ldots +V_k^2}$ it holds that $$\mathbb{E}(|V|)=\frac{\Gamma(\tfrac{1}{2})\Gamma(\tfrac{k+1}{2})}{\Gamma(\tfrac{k}{2})}\mathbb{E}(|V_1|)$$

Thus if $U,V,W$ are uniform on $D^2$, and $X,Y,Z$ are uniform on $S^3$ we have \begin{align*}\mathbb{E}(|U+V+W|)=\frac{\pi}{2} \mathbb{E}(|(U+V+W)_1|\\ \mathbb{E}(|X+Y+Z|)=\frac{3\pi}{4} \mathbb{E}(|(X+Y+Z)_1|\end{align*} and therefore \begin{align*} \mathbb{E}(|U+V+W|)=\frac{2}{3}\,\mathbb{E}(|X+Y+Z|)=\frac{2}{3}W_3(1,1) \end{align*} and $$\mathrm{exp\_abs}(3)=\frac{2}{9}\,W_3(1,1)\;\;.$$ (And $I_3=\frac{\pi^2}{6}\, W_3(1,1)$.)

  • $\begingroup$ That's cool! Is there a fast way of calculating more digits here? I tried a bit, but only found $2.71808124147..$. I guess one could only integrate over $[0,1]$ for the first integral and then multiply by $2$, because of symmetry. When only integrating over positive orthant $[0, 1]^3$, one can solve it symbolically to get $\frac{pi^2}{16}$, for orthants with mixed signs it seems to be more challenging to solve it symbolically. $\endgroup$ Jun 24, 2021 at 7:38
  • $\begingroup$ @Moritz Firsching: I have now found the exact value of $\operatorname{exp\_abs}(3)$ via an amazing result of Borwein and co-workers, please see above. $\endgroup$
    – esg
    Jun 25, 2021 at 18:49
  • $\begingroup$ Wow! That's just great. $\endgroup$ Jun 25, 2021 at 19:28

In R, I can simulate a million pairs of points in the unit disc:

n = 10^6
k = 2
theta = matrix(runif(k*n, 0, 2*pi), nrow = n)
r = sqrt(matrix(runif(k*n, 0, 1), nrow = n))
x = cos(theta)*r; y = sin(theta)*r;
d = sqrt(rowSums(x)^2+rowSums(y)^2)/k

and this returns for me 0.4526405. Here we choose a point in a disc uniformly by choosing its angle (uniform on $[0, 2\pi]$) and its distance from the origin (square root of uniform on $[0,1]$).

I similarly get $exp\_abs(1) \approx 0.666$, which is clearly $2/3$. I get $exp\_abs(3) \approx 0.367$ and $exp\_abs(4) \approx 0.316$ - on a million draws I wouldn't trust more decimal places than that.

  • 3
    $\begingroup$ Sampling uniformly doesn't work! That's why I take the square root. See for example mathworld.wolfram.com/DiskPointPicking.html $\endgroup$ Jun 18, 2021 at 20:17
  • $\begingroup$ Thanks, it's good to check it numerically in this way. I wonder what $0.367$ really is.. $\endgroup$ Jun 18, 2021 at 20:43
  • $\begingroup$ The best is a closed form expression. It looks like $I_3=\frac{\pi^2}{6}\, W_3(1,1)$, where $W_3(1,1)=\frac{476}{525}A+\frac{52}{7\,\pi^2}\frac{1}{A}$ with $A=\frac{3}{16}\frac{2^{1/3}}{\pi^4}\Gamma(\frac{1}{3})^6$ is the expected radial distance $\mathbb{E}|X+Y+Z|$ for $X,Y,Z$ uniform on $S^3$, as given in scholarship.claremont.edu/jhm/vol6/iss1/7 (on page 100). (Timothy Budd pointed to this paper in the related MO post.) $\endgroup$
    – esg
    Jun 24, 2021 at 18:52

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