Consider the set of Borel-measurable probability measures over the interval $[0,1]$ with a given mean, say 1/2. To be precise, I'm talking about the following set $$M=\left(\mu\in \Delta([0,1]):\int x\mu(dx)=1/2\right)$$

This is a convex set, with convex combinations defined as $\mu=\alpha \eta +(1-\alpha)\xi$ when $\mu(E)=\alpha \eta(E)+(1-\alpha)\xi(E)$ for every Borel set $E\subset[0,1]$.

My question is: what is the set of extreme points of this convex set?

My conjecture is that it is the subset of probability measures with support consisting of at most two points. I'm fairly convinced that this is true, but so far I haven't been able to come up with an elegant argument to show this and I thought that perhaps this is a well-known result. It is not hard to show for the case of atomless measures or for the case of discrete measures, but I'd like to have the general case.

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    $\begingroup$ It is not too hard to check for the case when the support is of size 3. Then it is not hard to reduce the general case to this case, by partitioning the support into 3 non-null parts. $\endgroup$ Jan 12 '12 at 21:36

This question (where you prescribe a set of moments, on an arbitrary measure space) is completely answered in this very cool paper. (G. Winkler, Extremal points of moment sets).

  • $\begingroup$ Great reference, thank you so much! $\endgroup$ Jan 12 '12 at 23:05
  • $\begingroup$ I know this is an old post, but I hope I will get a response. I was checking the paper you suggested. I have a question about as set of distribution on a real with finite $k$-th finite moment that is $\mathcal{P}=\{ F: \int_{\mathbb{R}} |x|^k dF(x)=c \}$. Does this paper assert that the set of extreme poins is a convex combination of all two singletons, that is: $ ex \mathcal{P} = \{ F: F= (1-t) \delta_{x_1}+t \delta_{x_2}, t\in[0,1], x_1,x_2 \in \mathbb{R} \}$. Thanks for your help. $\endgroup$
    – Boby
    Jul 23 '17 at 20:28
  • $\begingroup$ What the paper asserts in this case is $ex\mathcal P=\{F:F=(1-t)\delta_{x}+t\delta_y, t\in[0,1], (1-t)|x|^k+t|y|^k=c, x+y\neq0\}$. This includes singletons $\delta_x$ with $|x|^k=c$. $\endgroup$ Jul 24 '17 at 8:22

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