Timeline for Normalizing constant calculation of Strauss Process
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 9, 2023 at 14:23 | vote | accept | Fiodor1234 | ||
| Sep 9, 2023 at 12:47 | comment | added | Xi'an | And if instead by $p$ you mean the unnormalised version, the average converges to the integral of $p^2$ over the integral of $p$. | |
| Sep 9, 2023 at 12:36 | comment | added | Fiodor1234 | Yes my notation wasn't good, I understand what you mean now. | |
| Sep 9, 2023 at 12:33 | comment | added | Xi'an | As written it involves $p(\cdots)$ but $p$ is missing its normalising constant. | |
| Sep 9, 2023 at 12:28 | comment | added | Fiodor1234 | @Xi'an Why the summation is not possible? This would be the normalising constant for resampling as jbowman said. If I'm not mistaken | |
| Sep 9, 2023 at 12:22 | answer | added | Xi'an | timeline score: 2 | |
| Sep 9, 2023 at 12:14 | comment | added | Xi'an | The summation$$\sum_{j=1}^{M}p(\mu_{1,j}, \mu_{2,j},..., \mu_{K^{j},j},K^{k})$$is not possible since $p$ is only known up to a normalising constant. | |
| Sep 9, 2023 at 12:13 | comment | added | Xi'an | Is there a missing indicator in the exponent of $a$ ? The notation $\sum_{ij}|\mu_{i}-\mu_{j}|<d$ is ambiguous. | |
| Sep 8, 2023 at 18:36 | comment | added | jbowman | The normalizing constant you are calculating isn't the normalizing constant of the Strauss process, which is indeed intractable, but simply the normalizing constant for resampling from your sample. | |
| Sep 8, 2023 at 18:17 | history | edited | User1865345 | CC BY-SA 4.0 |
deleted 18 characters in body
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| Sep 8, 2023 at 16:54 | history | asked | Fiodor1234 | CC BY-SA 4.0 |