Timeline for Extreme points of set of probability measures $\mathcal{P}= \{F: \int_{\mathbb{R}} |x|^k dF(x)=c \}$
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 16, 2018 at 17:42 | comment | added | Boby | I know that this is an old question, but I have one more quick question. The above result also holds for $\mathbb{R}^n$ where $n>1$, right? | |
| Aug 7, 2017 at 22:19 | comment | added | Boby | Please see the question I asked here: mathoverflow.net/questions/278196/… | |
| Aug 4, 2017 at 14:31 | comment | added | Boby | 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. | |
| Aug 4, 2017 at 14:23 | comment | added | Jean Duchon | 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. | |
| Aug 4, 2017 at 13:43 | comment | added | Boby | 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 $\mathcal{P}=\{ F: |{\rm supp}(F)| \le A\}$ | |
| Jul 25, 2017 at 22:07 | vote | accept | Boby | ||
| Jul 24, 2017 at 13:38 | comment | added | Jean Duchon | 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. | |
| Jul 24, 2017 at 13:11 | comment | added | Boby | A ok. Thanks. Can you explain where the condition $x +y \neq 0$ is coming from? Is this from the linear independence? | |
| Jul 24, 2017 at 8:32 | history | answered | Jean Duchon | CC BY-SA 3.0 |