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Suppose that X is a multivariate normal vector with the covariance matrix $\Omega$.

As we know, if $\Omega$ is singular, the density of X is not defined in the usual way (because the denominator is zero).

Measure theoretically, that a random vector does not have the density means that the distribution(probability measure) is not absolutely continuous with respect to the proper Lebesgue measure.

Here, I am not sure how "$\Omega$ is singular" is connected to "the probability measure is not absolutely continuous".

How are they linked?

M.C. Park
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  • Singular measures are not absolutely continuous, that's all. For details, research the Lebesgue decomposition theorem. – whuber Oct 04 '22 at 15:00
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    Thank you for the tidy answer. But, I am not yet sure then what is the relationship between singular probability and singular covariance? – M.C. Park Oct 04 '22 at 16:02
  • There's a good answer in the duplicate thread at https://stats.stackexchange.com/a/91056/919. – whuber Oct 04 '22 at 16:12
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    Thank you every time! – M.C. Park Oct 04 '22 at 16:32
  • @whuber How about this explanation: In the univariate normal case, consider the situation of variance zero (singular variance). Then, the normal distribution is a degenerate distribution with the probability one at the mean. Here, the one-dimensional Lebesgue measure of the mean is zero, but the distribution measure of that is one. This directly violate the abosolute continuity of the distribution measure. In my humble opinion, this is an example of the link between singular variance and absolute continuity. – M.C. Park Oct 05 '22 at 14:36
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    That reads like a prolix summary of the duplicate answer. – whuber Oct 05 '22 at 15:04
  • @whuber Thank you for the comment! – M.C. Park Oct 05 '22 at 16:54

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