Timeline for Linear combo of normals is normal; how about other distributions?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| May 31, 2018 at 6:12 | comment | added | Ben | That is the same question - you have $\alpha^\text{T} Z = \sum \alpha_i Z_i$ so the vector condition is just asking about linear combinations, which is the subject of the present answer. | |
| May 18, 2018 at 0:05 | comment | added | Jasha | I'd been thinking of the statement that $Z$ is multivariate normal if and only if $\alpha^T Z$ is univariate normal for all $\alpha\in\mathbb R^n$. I wonder if this property holds for other stable distributions (e.g. is it true that $X$ is multivariate cauchy iff every linear combination $\alpha^T X$ is univariate cauchy)? I will post another question about it. | |
| May 17, 2018 at 23:58 | vote | accept | Jasha | ||
| May 17, 2018 at 23:58 | comment | added | Jasha | Oh, I see. I'm not sure where I got that notion. Thanks again. | |
| May 17, 2018 at 23:57 | comment | added | Ben | Statement (3) is false for all distributional families. This can easily be seen by considering the case where $X_1 = \cdots = X_n$, in which case the variables have identical marginal distributions, their linear combination is a constant multiple of $X_i$, but they are not independent. | |
| May 17, 2018 at 23:57 | comment | added | Jasha | Great, thanks! How about the reverse implication about stability implying independence? Does (3) hold for the distributions in the Lévy alpha-stable family? | |
| May 17, 2018 at 10:32 | history | answered | Ben | CC BY-SA 4.0 |