Timeline for Marginal distribution of uniform distribution over sphere
Current License: CC BY-SA 4.0
13 events
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| May 24, 2021 at 10:52 | comment | added | whuber♦ | This comes from observation $(8)$ and the dimensional scaling (volumes in $d-2$ space are proportional to the $d-2$ power of the size). That's why the factor $(1-r^2)^{(d-2)/2-1}$ appears in the density. | |
| May 24, 2021 at 10:47 | history | edited | whuber♦ | CC BY-SA 4.0 |
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| May 23, 2021 at 23:31 | comment | added | student_t | Thanks and lastly, is there any geometrical intuition for why the relative density size "flips"? For $d=3$, it's saying the density is higher towards the perimeter (actually at infinity on the perimeter $0^{-1/2}$!), but then for high dimensions $d$, it's pretty much concentrated at the center of the circle. | |
| May 23, 2021 at 18:38 | comment | added | whuber♦ | I agree that's odd--and it's a good check to make. A review of the analysis indicates it applies only to $d\ge 3.$ I have now noted that in the answer. | |
| May 23, 2021 at 18:37 | history | edited | whuber♦ | CC BY-SA 4.0 |
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| May 23, 2021 at 0:49 | comment | added | student_t | Something I'm not quite sure I understand is the case when $d = 2$ and the density is $\propto 0^{-1}$...which seems odd since things are well defined for the $d=2$ case. Any insights from others would be much appreciated! | |
| May 4, 2021 at 2:39 | comment | added | student_t | Also, could you please say a bit more about how $(Z_1, ..., Z_d)$ spherically symmetric implies this for $X$'s in (6)? Thanks! This is needed to argue that $\theta$ is distributed uniformly over $[0, 2\pi)$ in (7). I initially thought that once we know the joint of the $Z$'s then we could use change of variables to obtain the joint of $X$'s, but $Z$ to $X$ is not one to one... | |
| Apr 30, 2021 at 22:14 | comment | added | student_t | Ah ok, btw I was referring to the answer that says it's: $$f_k(z_1,\dots,z_k) =\frac{2^{(k-n)/2}\Gamma(n/2)}{\pi^{n/2}\Gamma((n-k)/2)}(1-s_1)^{(n-k)/2-1}$$ for $s_1=\sum_1^k z_j^2\in(0,1)$ for k = 2. | |
| Apr 30, 2021 at 21:00 | comment | added | whuber♦ | @student_t Unfortunately, the order of answers varies for each user, so "third answer" doesn't identify any of them. Nevertheless, if you detect any difference in the constant, just perform the kinds of checks I have done here: integrate the density (numerically if you have to) and also compare it to simulated data. | |
| Apr 30, 2021 at 20:10 | comment | added | student_t | Thanks for your fast reply! Sorry for the confusion and just to clarify, I was referring to the third answer to that question, and not the first, where there is an explicit form at the end of the answer. | |
| Apr 30, 2021 at 19:11 | comment | added | whuber♦ | @Student_t The Mathematics post doesn't explicitly give the constant, so there is no difference to note. The constant in my answer here is correct: look at the example and estimate the area under the curve. Clearly it's close to $1,$ as it should be. The fact that the curve agrees with the histogram shows the constant is correct. | |
| Apr 30, 2021 at 19:06 | comment | added | student_t | Hi, thank you for your detailed response! I actually realized that my question is a special case of mathoverflow.net/questions/359643/… With this said, interestingly, it looks like the constants differ in your and Iosif's answer (the key dependence on $(1 - x_1^2 - x_2^2)^{d / 2 - 2}$ is the same). And I was wondering if there might be a difference in the derivation that lead to this? | |
| Apr 21, 2021 at 15:27 | history | answered | whuber♦ | CC BY-SA 4.0 |