Timeline for Minimization Proof of Conditioning on Gaussian is Gaussian
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Feb 18, 2019 at 18:03 | comment | added | Jochen Wengenroth | That $E(X|Y)$ is Gaussian for a Gaussian $n+m$-dimensional vector $(X,Y)$ follows from the fact that uncorrelated components of $(X,Y)$ are independent -- no need to calculate densities. And I don't believe that sums of Gaussian distributed random variables are Gaussian (you need that the joint distribution is Gaussian). | |
| Feb 18, 2019 at 13:57 | answer | added | JPR | timeline score: 1 | |
| Feb 18, 2019 at 11:39 | comment | added | ABIM | Updated, the trail of thought.... hopefully the proof sketch is clearer :) | |
| Feb 18, 2019 at 11:38 | history | edited | ABIM | CC BY-SA 4.0 |
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| Feb 18, 2019 at 11:31 | history | edited | ABIM | CC BY-SA 4.0 |
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| Feb 18, 2019 at 10:49 | history | edited | ABIM | CC BY-SA 4.0 |
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| Feb 18, 2019 at 10:24 | history | asked | ABIM | CC BY-SA 4.0 |