Timeline for What structure is needed to define a Gaussian distribution on a given space?
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13 events
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| Jul 12, 2023 at 12:39 | comment | added | Learning math | @MichaelHardy Just found the question and your comment - what'd go wrong if we replace the inverse of the covariance matrix by it's pseudoinverse in the definition of the PDF? | |
| Mar 28, 2011 at 18:51 | comment | added | Tom LaGatta | A Gaussian measure with zero variance is just a point-mass measure supported on the mean. Consequently, if you expand the notion of "density function" to include the Dirac $\delta$-function, then there's no problem with defining a finite-dimensional Gaussian measure by way of its generalized density function. | |
| Mar 28, 2011 at 15:32 | answer | added | Tom LaGatta | timeline score: 5 | |
| Mar 28, 2011 at 15:26 | comment | added | Simon Lyons | Perhaps "in most textbooks" was a little strong. Some books do start with a non-negative definite matrix $\Sigma$ and then specify the density function. I expect this construction is much more common in applied maths. I've checked two textbooks I had to hand - one engineering text and one on machine learning. They both construct the mutlivariate normal in this way. | |
| Mar 28, 2011 at 14:08 | answer | added | Mark Meckes | timeline score: 6 | |
| Mar 28, 2011 at 13:51 | answer | added | Syang Chen | timeline score: 11 | |
| Mar 27, 2011 at 22:58 | comment | added | Michael Hardy | PS: I don't mean to suggest that defining them via the density can't or shouldn't be done. It would have to be a density with respect to a measure on a subspace of dimension in some cases lower than the dimension of the ambient space. Somehow it seems as if that could get messy, but I've never thought it through. | |
| Mar 27, 2011 at 22:51 | comment | added | Michael Hardy | One also see this: If $A$ is an $m\times n$ matrix, $b$ is an $m\times 1$ column vector, and $\mathbb{Z} = (Z_1,\dots,Z_n)^T$ has independent normally distributed components each with a $N(0,1)$ distribution, then $AZ$ has an $m$-dimensional normal distribution. In that case the expected value is $b$ and the variance is the $m\times m$ matrix $AA^T$ (which of course in some cases is singular). | |
| Mar 27, 2011 at 22:40 | comment | added | Michael Hardy | "In most textbooks, the normal distribution is defined on $\mathbb{R}^n$ by specifying its probability density function." Really? Specifically which books? The most usual definition in my experience is this: A random variable with values in $\mathbb{R}^n$ is normal iff its dot product with every constant (i.e. non-random) vector has a 1-dimensional normal distribution. How do the books you have in mind deal with the case where the variance $E((X-\mu)(X-\mu)^T)$ is singular? The vector of residuals (as opposed to errors) in linear regression is such a case. | |
| Mar 27, 2011 at 22:13 | comment | added | Simon Lyons | Yes, that's more or less equivalent to constructing a Brownian motion on the space. You need some generalisation of the Laplace operator. | |
| Mar 27, 2011 at 22:07 | comment | added | Suvrit | how about regarding a / the solution of the associated heat equation on that space / manifold? would that correspond to a generalized gaussian? | |
| Mar 27, 2011 at 19:08 | answer | added | jzadeh | timeline score: 1 | |
| Mar 27, 2011 at 18:41 | history | asked | Simon Lyons | CC BY-SA 2.5 |