Timeline for Expected absolute value of the average of two points from the disc
Current License: CC BY-SA 4.0
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 24, 2021 at 18:52 | comment | added | esg | 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.) | |
| Jun 18, 2021 at 20:43 | comment | added | Moritz Firsching | Thanks, it's good to check it numerically in this way. I wonder what $0.367$ really is.. | |
| Jun 18, 2021 at 20:17 | comment | added | Michael Lugo | Sampling uniformly doesn't work! That's why I take the square root. See for example mathworld.wolfram.com/DiskPointPicking.html | |
| Jun 18, 2021 at 17:25 | history | answered | Michael Lugo | CC BY-SA 4.0 |