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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?

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  • $\begingroup$ Singular measures are not absolutely continuous, that's all. For details, research the Lebesgue decomposition theorem. $\endgroup$
    – whuber
    Commented Oct 4, 2022 at 15:00
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    $\begingroup$ Thank you for the tidy answer. But, I am not yet sure then what is the relationship between singular probability and singular covariance? $\endgroup$
    – M.C. Park
    Commented Oct 4, 2022 at 16:02
  • $\begingroup$ There's a good answer in the duplicate thread at stats.stackexchange.com/a/91056/919. $\endgroup$
    – whuber
    Commented Oct 4, 2022 at 16:12
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    $\begingroup$ Thank you every time! $\endgroup$
    – M.C. Park
    Commented Oct 4, 2022 at 16:32
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    $\begingroup$ That reads like a prolix summary of the duplicate answer. $\endgroup$
    – whuber
    Commented Oct 5, 2022 at 15:04

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