Timeline for Doubt about proof of positive semi-definite matrix implies covariance matrix
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Oct 23, 2019 at 15:52 | comment | added | whuber♦ | There is an approach to this that doesn't require independence: you can explicitly write down the characteristic function of a $p$-variate Normal distribution that has $\Sigma$ for its covariance matrix. Unless $\Sigma$ is diagonal, the components of that Normal distribution are not independent. | |
| Sep 5, 2019 at 21:10 | history | became hot network question | |||
| Sep 5, 2019 at 13:38 | comment | added | Fr1 | Here independence is not required. See the good answer below by jld | |
| Sep 5, 2019 at 13:37 | comment | added | Fr1 | Diagonal covariance matrix means that the variables are not linearly correlated. It does not mean they are independent (i.e. intuitively it does not mean that any function applied to such variables will preserve the 0 correlation). So independence implies a diagonal cov matrix, but the converse does NOT hold true. | |
| Sep 5, 2019 at 13:33 | answer | added | jld | timeline score: 3 | |
| Sep 5, 2019 at 13:24 | comment | added | Juan Corredor | Are you sure? Why? I know that's true if $X_i$'s are normal distributed but in general it does not need to be true. Take $X$ a Standard Normal r.v and $Y=X^2/\sqrt{2}$. Then $\text{Cov}(Z) = I$ where $Z=(X,Y)$. | |
| Sep 5, 2019 at 13:05 | history | asked | Juan Corredor | CC BY-SA 4.0 |