[dododo]

i wonder if you are thinking of another monoid property of a gaussian?

suppose we have:

   x ~ N(0,1)
   y|x ~ N(x, 1)

then we have:

   y ~ N(0, 2)

i.e., gaussians are closed under marginalization.

however, i believe gaussians are not the only distribution with this property either: i think this property corresponds to the stable family of distributions: https://en.wikipedia.org/wiki/Stable_distributions

[mturmon]

Stable distributions are something else, not related to marginals or conditioning. They come up when studying laws of averages.

Gaussian distributions belong to the class of stable distributions, though, because of another of their properties: independent Gaussians, when added, are again Gaussian.

[dododo]

the particular property of the stable distribution, i was thinking of, is "closure under convolution" which is the above marginalisation (i believe?).

infinite divisibility is (yet another) property of gaussians!

[mturmon]

Nope. Closure under convolution is the same as closure under summation of the associated random variable, which is the defining property of stable distributions. This is explained in the first paragraphs of the wikipedia page you linked to ;-)

Closure under marginalization is something else.

It so happens that the functional form of the gaussian satisfies both, but the two properties are not at all the same.

  P1: X, Y gaussian => Z = X+Y gaussian
  P2: X, Y gaussian => X | Y gaussian

[noelwelsh]

Yes, that's what I was thinking of! Thanks for the link to stable distributions -- hadn't heard of them before.