2
$\begingroup$

Let $\mathcal{P}$ be a set of Borel probability measures on $[0, 1]$ equipped with the weak$^{*}$-topology as described here: What is the weak*-topology on a set of probability measures?. Consider for $\rho\in [0, 1]$, $$\mathcal{P}(\rho)=\left\{\nu\in \mathcal{P}:\int_{0}^{1}{td\nu(t)}=\rho\right\}.$$ I can show that this set is convex and compact for the weak$^{*}$-topology. I was wondering what the extreme points of $\mathcal{P}(m)$ are and found that the extreme points are of the form $$\nu=p\delta_{x}+q\delta_{y}$$ with $p, q\geq 0$, $x, y\in [0, 1], p+q=1, px+qy=\rho$ where $\delta_{x}$ denotes the Dirac measure at $x$. Can someone explain to me why they are of this form?

$\endgroup$
2
  • $\begingroup$ Shouldn't $px + py=\rho$? $\endgroup$
    – aduh
    Apr 27 at 10:45
  • $\begingroup$ You are correct. Thank you! $\endgroup$
    – Shaq155
    Apr 27 at 10:52
1
$\begingroup$

This might not satisfy you, but this is covered by a result of Dubins, cited in my answer to a previous MSE question. You have a compact convex set $\mathcal P$ of probability measures, its set of extreme points is the set of point masses. Your $\mathcal P(\rho)$ is the intersection of $\mathcal P$ and a codimension-1 affine space, namely the measures $\nu$ cut out by the single equation $\int_0^1 td\nu(t)=\rho$. (This equation is linear in $\nu$.) The Dubins result then tells us that the extreme points of $\mathcal P(\rho)$ are linear combinations of $1+1=2$ extreme points of $\mathcal P$, that is, of pairs of point masses.

I wrote "might not satisfy you" because this post only tells you that the fact you ask about has been noticd before but does not explain why the result is true. Maybe you can examine your proof and see how it squares with Dubins's?

$\endgroup$
6
  • $\begingroup$ Thank your for your answer! Could you elaborate why the condition for $\mathcal{P}$ gives a "linear space of co-dimension 1"? $\endgroup$
    – Shaq155
    Apr 27 at 11:25
  • $\begingroup$ I wrote "linear" and should have written "affine", as the defining equation is inhomogeneous. I've fixed the text; sorry about the mistake. $\endgroup$ Apr 27 at 11:50
  • $\begingroup$ No problem. Thank you! $\endgroup$
    – Shaq155
    Apr 27 at 11:53
  • $\begingroup$ Can you give an explanation why this affine space has codimension 1? $\endgroup$
    – Shaq155
    Apr 27 at 11:56
  • $\begingroup$ Not without knowing what you know already. Informally, as en.wikipedia.org/wiki/Codimension says: "The subspaces can be defined by the vanishing of a certain number of linear functionals, which if we take to be linearly independent, their number is the codimension." In your case, one equation, so codimension 1. $\endgroup$ Apr 27 at 12:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.