Timeline for Are linear combinations (i.e. "sums") of gaussian distributions also gaussian?
Current License: CC BY-SA 4.0
9 events
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| Oct 15, 2021 at 20:31 | comment | added | Abhinav Gupta | @stats555 (1) No, the linear combinations of Gaussian densities are not necessarily Gaussian. (2) Linear combinations of JOINTLY Gaussian RVs is necessarily Gaussian. The conditions 'jointly' is important (As Chris Huang has pointed out). I will edit my answer to include this condition. | |
| Oct 15, 2021 at 18:41 | history | edited | Abhinav Gupta | CC BY-SA 4.0 |
added 5 characters in body
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| Oct 15, 2021 at 18:25 | history | edited | Abhinav Gupta | CC BY-SA 4.0 |
A proof for a more general case added
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| Oct 15, 2021 at 17:31 | comment | added | stats_noob | @ AbhinavGupta : Thank you so much for your answer! So just to clarify: linear combinations of gaussian densities WILL NOT NECESSAIRLY have a gaussian distribution ... BUT linear combinations of gaussian random variables WILL NECESSAIRLY have a gaussian distribution? Thank you so much! | |
| Oct 15, 2021 at 17:29 | vote | accept | stats_noob | ||
| Oct 15, 2021 at 17:21 | comment | added | Chris Haug | It's true if they are jointly Gaussian. You can have distributions which have Gaussian marginals but are not jointly Gaussian. For example, my answer here: stats.stackexchange.com/questions/238226 . There, $X+Y$ has an atom at zero and so cannot be Gaussian, for example. | |
| Oct 15, 2021 at 16:03 | comment | added | Abhinav Gupta | @ChrisHaug I will add the proof for a more general case. But what are the conditions on join distribution? Can you please elaborate? | |
| Oct 15, 2021 at 12:08 | comment | added | Chris Haug | The first statement is not always true, it depends on their joint distribution. The proof that follows is for a very special case (independence). | |
| Oct 15, 2021 at 4:12 | history | answered | Abhinav Gupta | CC BY-SA 4.0 |