Timeline for Average distance of the mean of $n$ random complex numbers in a unit disc
Current License: CC BY-SA 4.0
14 events
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| Oct 1, 2022 at 5:50 | comment | added | D.W. | Cross-posted: mathoverflow.net/q/394883/37212, stats.stackexchange.com/q/510167/2921, math.stackexchange.com/q/4030410/14578. Please do not post the same question on multiple sites. | |
| Jun 26, 2021 at 8:28 | comment | added | Emil Jeřábek | $E|z|^2=\frac1{2n}$ follows by direct computation, using linearity of expectation and independence of the $z_i$’s: $E|z|^2=E\frac1{n^2}\bigl(\sum_iz_i\bigr)\bigl(\sum_i\overline z_i\bigr)=\frac1{n^2}\sum_{i,j}Ez_i\overline z_j=\frac1{n^2}\sum_iE|z_i|^2=\frac1{2n}$. Then $\Pr(|z|\ge t)=\Pr(|z|^2\ge t^2)\le\frac1{2nt^2}$ by Markov’s inequality, hence $E|z|=\int_0^\infty\Pr(|z|\ge t)\,dt\le\frac1{\sqrt{2n}}+\int_{1/\sqrt{2n}}^\infty\frac{dt}{2nt^2}=\frac2{\sqrt{2n}}=\sqrt{\frac2n}$. But note that the bound in esg’s answer, albeit asymptotic, is better: $E|z|=\sqrt{\frac\pi{8n}}(1+o(1))$. | |
| Jun 26, 2021 at 7:02 | comment | added | AgnostMystic | @EmilJeřábek ,Yes i am interested in the bounds as well.Could you kindly explain it a bit more how we go about finding these bounds? | |
| Jun 23, 2021 at 8:40 | history | edited | YCor | CC BY-SA 4.0 |
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| Jun 22, 2021 at 16:51 | answer | added | esg | timeline score: 5 | |
| Jun 22, 2021 at 16:37 | comment | added | Emil Jeřábek | I’m not sure if you are also interested in bounds, but it’s easy to compute $E|z|^2=\frac1{2n}$, which implies $E|z|<\sqrt{\frac2n}$ or so. | |
| Jun 22, 2021 at 14:59 | answer | added | Timothy Budd | timeline score: 2 | |
| Jun 21, 2021 at 10:18 | comment | added | AgnostMystic | @TimothyBudd could you kindly say how to start | |
| Jun 18, 2021 at 15:45 | comment | added | Timothy Budd | Along the same lines as Kluyver,"A local probability problem" (1906), one can derive that $|\sum_{i=1}^n z_i |$ has density $\rho(x) = \frac{1}{2\pi} \int_0^\infty dt\, t\,J_0(x\, t) ( 2J_1(t) / t)^n$. | |
| Jun 18, 2021 at 12:35 | comment | added | Moritz Firsching | I calculated numerically the value for $n=2$ and I have a conjecture as to what the exact value might be: mathoverflow.net/q/395659 | |
| Jun 10, 2021 at 6:51 | comment | added | Steve | For $n=1$ it is simply $\int_0^1 \sqrt{x} dx = 2/3$ because the direction does not matter and the distance from zero of each $z_i$ is distributed as the square root of a uniformly distributed random variable on $[0, 1]$. The case $n=2$ seems much harder. | |
| Jun 9, 2021 at 11:59 | comment | added | Moritz Firsching | The value for $n=1$ is $\frac{2}{3}$, which can be shown by solving a two-dimensional integral. Solving the 4-dimensional integral for $n=2$ should be withing reach, perhaps. | |
| Jun 9, 2021 at 8:54 | comment | added | Steve | This seems like a difficult version of an "expected distance of random walk from the origin" type of result, but I can't find a clearly related result. Usually the length of each step is fixed (here, it is random), and only the direction is random. | |
| Jun 9, 2021 at 5:06 | history | asked | AgnostMystic | CC BY-SA 4.0 |