A subset $S$ of a real vector space is convex if it is closed under finite mixtures: for any $\lambda_1,\ldots,\lambda_n>0$ such that $\lambda_1+\cdots+\lambda_n=1$ and any $x_1,\ldots,x_n\in S$, $\lambda_1x_1+\cdots+\lambda_nx_n\in S$. In a normed space, we can generalize and talk about countable mixtures, defined in the obvious way. I suspect that with still more structure, we can define even more general notions of mixture (using some kind of integration, perhaps). But how does it go? Are there examples of sets closed under countable mixtures but not arbitrary mixtures?

If it helps, what I am most interested in is mixtures of probability measures and mixtures of finitely-additive probability measures.


We can define a mixture of probability measure as follows. Assume everything required to make what follows make sense. Let $\{P_\mu (\cdot): \mu \in \mathcal M\}$ be a family of distributions and $Q(\cdot)$ a probability distribution on $\mathcal M$. Then we can mix over $\mu$ by doing $P(A) = \int P_\mu (A) \ Q(d\mu)$.

For a concrete example, we can write many symmetric distributions as a "scale mixture of normals" by considering an $N(0, \sigma^2)$ distribution and putting a suitable distribution on $\sigma^2$, including the Laplace and $t_\nu$ distributions (see e.g. Andrews and Mallows 1974). This sort of trick is very often used in Gibbs sampling schemes since portraying things as mixtures of normal lets us exploit its conjugacy properties. In particular, putting an inverse gamma distribution on $\sigma^2$ should result in some kind of $t$ distribution.

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    $\begingroup$ These mixtures of probability measures are also essential for de Finetti's Theorem, that shows that one can decompose an exchangeable process as a mixture of iid processes. $\endgroup$ – Michael Greinecker May 7 '12 at 15:56
  • $\begingroup$ Thanks. I don't know why I thought it was more subtle than this. Is it right that all that you need for this construction to work is that $Q$ measures enough sets to make each function $\mu\mapsto P_\mu(A)$ measurable? Also, is it right that if $Q$ is merely finitely additive the construction will still work but the mixture might be merely finitely additive even if all the measures being mixed are countably additive? $\endgroup$ – gmr May 7 '12 at 19:52
  • $\begingroup$ I think $\mu \to P_\mu(A)$ being measurable for each measurable $A$ should be enough. Sadly I know next to nothing about the theory of finite measures, though heuristically it looks to me like you need $Q$ to be countably additive for the obvious proof of $P$ being countably additive to go through. $\endgroup$ – guy May 7 '12 at 20:34

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