# Extreme points of set of probability measures $\mathcal{P}= \{F: \int_{\mathbb{R}} |x|^k dF(x)=c \}$

I am interested in finding the extreme points of the following set of distributions \begin{align} \mathcal{P}= \left\{F: \int_{\mathbb{R}} |x|^k dF(x)=c \right\} \end{align} where $k,c>0$.

I know that this paper by Winkler is a standard reference on this question.

I was trying to extrac the answer from this paper and got that the set of extrem points is given by \begin{align} ex \mathcal{P}= \left\{F \in \mathcal{P} : F= (1-t) \delta_{x_1}+t \delta_{x_1}, t\in[0,1], x_1,x_2 \in \mathbb{R} \right\} \end{align}

However, not sure if this is correct since in this question, for a very similar set, it was pointed out that it must be combination of three mass points instead of two.

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 $|x|^k=c$.
What was pointed out is not the existence of three-point masses as extreme points, but that the proof (that only convex combinations of two point masses are extreme) was obvious for probability measures on a three point set instead of $\mathbb R$.
• A ok. Thanks. Can you explain where the condition $x +y \neq 0$ is coming from? Is this from the linear independence? – Boby Jul 24 '17 at 13:11
• Yes. Take $c=1$. Then $\delta_1$ and $\delta_{-1}$ are extreme, but not $(1-t)\delta_1+t\delta_{-1}$ since it is a convex combination of them. – Jean Duchon Jul 24 '17 at 13:38
• I have one more question. Do you know what are the extreme points of as set of probability measures with a bounded support? That is set to $\mathcal{P}=\{ F: |{\rm supp}(F)| \le A\}$ – Boby Aug 4 '17 at 13:43
• If you mean a support of at most $A$ points, the set of such probability measures is not convex. If you mean $\supp(F)\subset A$, the set is convex with $\delta_a,a\in A$ as extreme points. – Jean Duchon Aug 4 '17 at 14:23
• Sorry the was very bad notation. I mean distributions supported on the bounded interval of reals. All $F$ such that $F([a,b])=1$. I vaguely remember that it should be a set of singletons. – Boby Aug 4 '17 at 14:31