Timeline for Why are isotropic random vectors called isotropic if they aren't?
Current License: CC BY-SA 4.0
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| when toggle format | what | by | license | comment | |
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| Dec 2, 2020 at 16:08 | vote | accept | lamlame | ||
| Nov 23, 2020 at 0:38 | comment | added | lamlame | If the definition of isotropic random vectors implies the existence of such rotational invariances then this makes sense. I didn't know that. If X has components x_i, where P(x_1 = 1)=1 and P(x_i = 1)=P(x_i = -1) = 1/2 for i=2,...n then X satisfies E XX^T = I_n. The distribution has rotational symmetries, but all such rotations leave the 1st component fixed. This seems allowed with your definition of isotropy, but I'm not sure what norm $\mu$ to choose to make it isotropic for Carlo Beenakker's answer. | |
| Nov 22, 2020 at 23:43 | history | answered | R W | CC BY-SA 4.0 |