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In reading section 2.2, page 14 of this book, I came across the term "singular distribution".

Apparently, a multivariate Gaussian distribution is singular if and only if it's covariance matrix is singular. One way (only way?) the covariance matrix can be singular is if one of the diagonal entries is zero. A Gaussian distribution approaches a Dirac delta as the variance goes to zero. Is the Dirac distribution an example of a singular distribution? I don't think so, since Wikipedia says the Lebesgue integral of the density function of a singular distribution is zero.

Please confirm.

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It turns out that there are two common definitions for singular distribution; see this article (Singular distribution - Springer Online Reference Works).

According to one definition, it is a probability distribution on $\mathbb{R}^n$ concentrated on a set of Lebesgue measure zero and giving probability zero to every one-point set. This is the case in the Wikipedia article you linked. Sometimes, a singular distribution is defined without the latter requirement; under this definition every discrete distribution is singular (with respect to Lebesgue measure). In any case, a singular distribution does not have a probability density function. (Distributions with probability density functions are called absolutely continuous.) As for the case of a multivariate Gaussian distribution, if the covariance matrix is singular, then the distribution is continuous singular (hence has no density, and gives probability zero to every one-point set).

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I find only the expression "this Gaussian is singular" on page 14 of your reference, but not the definition of "singular distribution".

But to answer your question:

The delta distribution is not a singular distribution, it is a discrete probability distribution. It does not have a Radon-Nikodym density with respect to the Lesbegue measure, because the Lesbegue measure of a single point is zero, and the delta distribution is concentrated on a single point.

Don't get confused if people write stuff like $$ \int_{\mathbb{R}} \delta_0(x) d x = 1 $$ This is not correct in the strict sense. Instead, the "density function" of the delta distribution concentrated on zero - which is not a density in the sense of Radon-Nikodym - would be $$ f(x) = 0 \; \text{for} \; x \neq 0 $$ and $$ f(0) = \infty $$ and therefore we would have $$ \int_{\mathbb{R}} f(x) d x = 0 $$

But: For a discrete probability distribution, it is possible to name an at most countable set of points such that each point can be assigned a finite probability, such that the probability of any set is equal to the sum of the probabilities of the points it does contain.

This is not possible for a singular probability distribution like the Cantor distribution. The Cantor distribution is not concentrated on a countable set of points. Therefore the terms "singular distribution" and "discrete probability distribution" are different, and the delta distribution is a discrete one, not a singular one.

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