Timeline for How to generate random points in the volume of a sphere with uniform nearest neighbour distances
Current License: CC BY-SA 3.0
20 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 5, 2014 at 15:22 | comment | added | user1449306 | Uh, this is a new question ... If you want to test for uniformity of your points, you can inspect the data visually by using histograms or qq-plots of each dimension. This usually says quite a lot. Or you apply tests like $\chi^2$ or others, but I'm not the right person to help you here ... | |
| Feb 5, 2014 at 15:08 | comment | added | Jolfaei | Correct. Now, we go back to the first problem. Can I apply the CSR test to any empirical data with a non-uniform distribution? How can I check the uniformity of spherical data? | |
| Feb 5, 2014 at 14:57 | comment | added | user1449306 | Well, you place your points independently of each other with a uniform probability over the ball. Hence you get a constant, position independent density and have met the Poisson-requirements of constant rate (=density) and independence of the events (points). | |
| Feb 5, 2014 at 14:49 | comment | added | Jolfaei | Clustering is a way of unsupervised learning. By clustering, I can make groups of close points and study whether there is some relationship between the points. In a uniform distribution, I cant learn much about points. | |
| Feb 5, 2014 at 14:37 | comment | added | Jolfaei | You are mentioning that for a uniform distribution of sample points $X$ in a 3D space, the probability that a sample area of a specific size will contain exactly $x$ points can be represented by the Poisson’s exponential function. My question is why? how did you infer the Poisson function? | |
| Feb 5, 2014 at 14:34 | comment | added | user1449306 | Concerning clustering: For example, you select some random points as initial clusters and start assigning the nearest neighbours to the clusters. At some point you end up with all points assigned to one cluster ... Actually, I don't understand why you're interested in clustering in the current context. | |
| Feb 5, 2014 at 14:27 | comment | added | user1449306 | Consider a small sphere $S$ inside your ball with uniformly distributed points. The probability to find exactly $k$ points in $S$ is $P(k) = \frac{\delta^k \exp(-\delta)}{k!}$ - the Poisson distribution with $\delta$ depending on volume and point density. | |
| Feb 5, 2014 at 14:14 | comment | added | Jolfaei | "why shouldn't it be possible to cluster uniformly distributed points?!", We normally use distance metrics to cluster close points. A uniform distribution of distances implies no pattern. | |
| Feb 5, 2014 at 14:10 | comment | added | Jolfaei | "if you have uniformly distributed points, the number of points in a certain area V is Poisson distributed." can you explain more? | |
| Feb 5, 2014 at 13:58 | comment | added | user1449306 | Look, if you have uniformly distributed points, the number of points in a certain area $V$ is Poisson distributed. You're mixing up different things. And clustering depends on the set of rules you apply, why shouldn't it be possible to cluster uniformly distributed points?! | |
| Feb 5, 2014 at 13:54 | comment | added | Jolfaei | In addition to that, assuming homogeneous distribution of Poisson points, the probability distribution function is $f(r)$=$\frac{\rho}{ k}(4 \pi r^2)$ $\exp (-\frac{\rho}{k}$ $(\frac{4}{3} \pi r^3))$, where $\frac{\rho}{ k}$ is the point intensity | |
| Feb 5, 2014 at 13:40 | comment | added | Jolfaei | Also, I disagree with your second point. How can you cluster populated points, if they have uniform distribution? | |
| Feb 5, 2014 at 13:38 | comment | added | Jolfaei | That formula is for 2D points with Poisson distribution (Refer to Clark-Evans testing procedure). It cant be applied for any other distributions. To test CSR, one need to firstly fit the empirical data to an empirical distribution and then check CSR with respect that. Otherwise, your test results would be wrong. | |
| Feb 5, 2014 at 13:33 | comment | added | user1449306 | Just read the article. The poisson distribution describes the amount of points per region, not the point distribution. | |
| Feb 5, 2014 at 13:26 | comment | added | user1449306 | I didn'd calculate the distribution, I just adapted the formula of the wikipedia article to your situation. $\lambda$ is connected with the point density and the number of dimensions of your problem. | |
| Feb 5, 2014 at 13:25 | comment | added | Jolfaei | I think this answer may not be correct. In Complete spatial randomness, the empirical data is assumed to have has Poisson distribution, and randomness is checked with the Poisson theoretical distribution. Also, calculation are done in a 2D space not 3D. This is not my case. | |
| Feb 5, 2014 at 13:24 | history | edited | user1449306 | CC BY-SA 3.0 |
added 51 characters in body
|
| Feb 5, 2014 at 13:20 | comment | added | Jolfaei | How did you calculate the probability distribution function? what is $Lambda$? | |
| Feb 5, 2014 at 13:15 | history | edited | user1449306 | CC BY-SA 3.0 |
added 3 characters in body
|
| Feb 5, 2014 at 13:07 | history | answered | user1449306 | CC BY-SA 3.0 |