Timeline for Connection between sum of normally distributed random variables and mixture of normal distributions
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 30, 2012 at 11:03 | vote | accept | vonjd | ||
| Jul 30, 2012 at 5:17 | answer | added | Michael R. Chernick | timeline score: 5 | |
| Jul 30, 2012 at 4:48 | comment | added | Douglas Zare | You don't have to go to higher moments to see that a not all mixtures are normal. Mixtures of normal distributions don't have to be unimodal, although they can be. In addition, since distinct normal distributions have distinct tail asymptotics, you can read off the components from the tails. | |
| Jul 29, 2012 at 21:56 | comment | added | cardinal | If you have suspicions that something might not be true, it's good to look for simple (counter)examples. Consider $X_1$ and $X_2$ independent Bernoulli random variables. A mixture of these is still Bernoulli (why?), but the sum of them could, in general, take the value $2$ (in addition to $0$ or $1$). So, clearly they can't be the same thing. :) | |
| Jul 29, 2012 at 17:15 | answer | added | assumednormal | timeline score: 19 | |
| Jul 29, 2012 at 17:05 | comment | added | assumednormal | I'll write it up now. | |
| Jul 29, 2012 at 16:49 | comment | added | Dilip Sarwate | @Max, that's an answer! | |
| Jul 29, 2012 at 16:48 | comment | added | Dilip Sarwate | Two independent random variables that are normally distributed have a jointly normal distribution. No "not necessarily jointly" but instead ""so that they are also jointly normal". | |
| Jul 29, 2012 at 16:43 | comment | added | assumednormal | A mixture of two normals is not the sum of two normal random variables. It's a random variable whose PDF and CDF are weighted sums of the individuals PDFs and CDFs. | |
| Jul 29, 2012 at 16:31 | history | asked | vonjd | CC BY-SA 3.0 |