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Jun 8, 2016 at 18:00 vote accept Banach space fan
Jun 8, 2016 at 17:42 comment added Banach space fan @davik, Got it, makes sense. Thanks!
Jun 8, 2016 at 17:38 comment added davik It makes no sense for z to be random, you are showing the matrix, which is just a regular matrix, is positive definite.
Jun 8, 2016 at 17:33 comment added Banach space fan @davik, So in this conext, does the condition of definiteness require all nonzero, non-random vectors $z$?
Jun 8, 2016 at 17:15 answer added Robert Israel timeline score: 4
Jun 8, 2016 at 17:12 comment added davik I'm fairly sure that 1) your z cannot depend on x since it is not a random vector and 2) your covariance can be singular but only if the distribution is trapped in a lower dimension, for example if the last entry of x is 0 with probability 1. But this is in a sense the only way for it to be singular
Jun 8, 2016 at 17:04 review First posts
Jun 8, 2016 at 17:10
Jun 8, 2016 at 17:01 history asked Banach space fan CC BY-SA 3.0