Timeline for Uniform sampling on a Riemannian manifold via tangent space and exponential map
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Sep 9, 2019 at 21:35 | history | suggested | Călin | CC BY-SA 4.0 |
Fix the link for rejection sampling
|
| Sep 9, 2019 at 19:29 | review | Suggested edits | |||
| S Sep 9, 2019 at 21:35 | |||||
| Sep 9, 2019 at 19:12 | vote | accept | Călin | ||
| Sep 9, 2019 at 19:12 | comment | added | Călin | You're right. Thanks! | |
| Sep 9, 2019 at 19:04 | comment | added | Iosif Pinelis | @Călin : To use this method, you will indeed have to have an expression for $h$, which I think can be computed by standard means in each particular case. The main difficulty here is to invert the map $\exp_x$, and you seem to be assuming, in the description of the 2nd step of your procedure, that this map is accessible. Then the inversion can be done at least numerically, in principle to any precision, I think. I also think something like this is unavoidable in order to reach your goal. | |
| Sep 9, 2019 at 18:33 | comment | added | Călin | Thanks for the answer, Iosif! But how can I use rejection sampling given that $h(\cdot)$ is an implicit density? I don't have an expression for it, right? | |
| Sep 9, 2019 at 15:39 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
added 16 characters in body
|
| Sep 9, 2019 at 15:27 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
added 11 characters in body
|
| Sep 9, 2019 at 15:20 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
added 331 characters in body
|
| Sep 9, 2019 at 15:11 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |