Timeline for in Euclidean space defined by multivariate normal distribution, what fraction of points falls within n-ball (centered at origin) tangent to point p?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 28, 2021 at 11:10 | comment | added | Dieter Kadelka | In the term $nr = 3$ $nr$ has the meaning number of rows = 3. Here $\Sigma$ is the covariance matrix with 3 rows and 3 columns. The dimension of the elements of $\Sigma$ is $(m/sec)^2$ if f.i. the $X_i$ is $m/sec$. Your assumption in the last remark is not correct. Actually $\sigma = \sqrt 5$. I think that this discussion is not of interest to mathematicians, although important for understandig your model. We may chat, if you like. | |
| Feb 28, 2021 at 5:41 | comment | added | aputnamist2 | I also notice that for the list (1,0.5,0.3,0.5,1,1.4,0.3,1.4,5) the last value is a 5. That represents 5 sigma so I would expect an answer like 0.9999997133 (fraction of points inside the n-ball tangent to that coordinate) or 0.0000002867 (fraction of points outside of the n-ball). The provided answer of P = 0.30 seems plainly wrong for what I'm asking. | |
| Feb 27, 2021 at 22:10 | comment | added | aputnamist2 |
thank you so much! I was able to run the code you shared. But I don't exactly understand what n = 3 #dimension of Sigma means. In the third line of your code you wrote Sigma = matrix(c(1,0.5,0.3,0.5,1,1.4,0.3,1.4,5),nr=3) which I would expect is a vector representing 9-dimensional space (with sigmas of 1,0.5,0.3,0.5,1,1.4,0.3,1.4,5). So what does nr = 3 mean there? Where does a dimensionality of 3 come in? I would expect the dimensionality to be 9 for this example! I hope you can explain in a way that I can understand.
|
|
| Feb 18, 2021 at 7:46 | comment | added | Dieter Kadelka | @aputnamist2: If you want to implement the algorithm do not use the unedited version. | |
| Feb 17, 2021 at 20:12 | history | edited | Dieter Kadelka | CC BY-SA 4.0 |
added 23 characters in body
|
| Feb 17, 2021 at 13:14 | history | edited | Dieter Kadelka | CC BY-SA 4.0 |
Added an example
|
| Feb 17, 2021 at 13:11 | comment | added | Dieter Kadelka |
The original answer with dcoga gave the density, not the distribution function pcoga.
|
|
| Feb 17, 2021 at 13:09 | history | edited | Dieter Kadelka | CC BY-SA 4.0 |
Added an example
|
| Feb 17, 2021 at 12:01 | comment | added | aputnamist2 | Thank you this seems to be what I want. For putting it into practice, at the end you mention the coga library and the dcoga_approx routine. Could you show me how I could use these to do the calculation you suggest, such as with an example? I am not skillful enough to put it into R (programming language) due to not being familiar enough with the techniques you mention and with the programming language. From what you've written it looks like the result will be what I want. | |
| Feb 17, 2021 at 0:53 | history | edited | Dieter Kadelka | CC BY-SA 4.0 |
added 197 characters in body
|
| Feb 16, 2021 at 22:59 | history | answered | Dieter Kadelka | CC BY-SA 4.0 |