1
$\begingroup$

Let $X$ be a random variable with the distribution $F$ (cdf).

What are the extreme points of the sets of the form: \begin{align} P_1&=\left\{ F: \int |x|^k dF\le c \right\},\\ P_2&=\left\{ F: |X| \le d \right\},\\ P_3&=\left\{ F: \int |x|^k dF\le c, \, |X| \le d \right\}.\\ \end{align} In this question it was shown for the set $P_1$, the set of extrem points are all two mass disributions. What about $P_2$ and $P_3$?

It would also be nice if some one can provide a good reference where the subject of finding extreme points of a set of distributions can be found. I am familiar with this reference. However, was thinking maybe there is a more modern work or survey on this.

Edit 1; Here is the definition of an extrem point:

An extreme point of a convex set, $A$, is a point $x \in A$, with the property that if $x = ty + (1 − t)z$ with $y,z \in A $and $t \in [0, 1]$, then $y = x$ and/or $z = x$.

Note that all of the sets above are convex.

Improve this question
$\endgroup$
3
  • $\begingroup$ @MattF. I think we don't need a metric to define extreme points. It's convexity issue. Do you agree? $\endgroup$
    – Boby
    Commented Aug 9, 2017 at 12:36
  • $\begingroup$ Does F represent the pdf or the cdf? $\endgroup$
    – user44143
    Commented Aug 9, 2017 at 13:59
  • $\begingroup$ @Matt F. $F$ represents the cdf. $\endgroup$
    – Boby
    Commented Aug 9, 2017 at 15:56

1 Answer 1

Reset to default
0
+50
$\begingroup$

You already know (from the previous question on MathOverflow, and in particular this paper) that the extreme points of $P_1$ form a subset of the set that consists of: Dirac measures $\delta_x$ with $|x|^k \leqslant c$, and two-point distributions $F = (1-t) \delta_x + t \delta_y$ with $(1-t) |x|^k + t |y|^k \leqslant c$, $t \in (0, 1)$, $|x|^k < c < |y|^k$. Dirac measures are always extremal. It is easy to see that if $(1-t) |x|^k + t |y|^k < c$, then the corresponding distribution $F = (1-t) \delta_x + t \delta_y$ is not extremal: it is a convex linear combination of $(1-t-\varepsilon) \delta_x + (t+\varepsilon) \delta_y$ and $\delta_x$ for sufficiently small $\varepsilon > 0$. On the other hand, if $(1-t) |x|^k + t |y|^k = c$, then the distribution $F$ is extremal: if it could be written as a convex combination of two distributions $F_1, F_2$, both $F_1, F_2$ would be supported in $\{x, y\}$, and thus their $k$-th moments are necessarily equal to $c$. Thus necessarily $F_1 = F_2 = F$, that is, $F$ is extremal. This characterises the set of extremal points of $P_1$.

Dirac measures $\delta_x$ with $|x| \le d$ obviously form the set of extreme points of $P_2$.

Finally, $P_3 = P_1 \cap P_2$, and if $F \in P_3$ is written as a convex combination of $F_1$ and $F_2$, then necessarily $F_1$ and $F_2$ belong to $P_2$. Therefore, $F \in P_3$ is extremal for $P_3$ if and only if it is extremal for $P_1$.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Do you have a good reference on this subject? Also, never understood by why $\delta_x$ are extreme points in the case when $|x| \le d$. Could you explain this point a little? $\endgroup$
    – Boby
    Commented Aug 15, 2017 at 0:13
  • $\begingroup$ No, I do not have any reference. Dirac measures are the extreme points of the entire set of probability measures. Indeed, if $\delta_x=(1-t)\mu_1+t\mu_2$, then both $\mu_1$ and $\mu_2$ are necessarily supported in $\{x\}$, and so they are equal to $\delta_x$. Conversely, if $\mu$ is a measure with support $A$ that contains at least two points, then $A$ can be written as $A=A_1\cup A_2$ with $t:=\mu(A_2)\in(0,1)$ and so $\mu=(1-t)\mu_1+t\mu_2$ with $\mu_j(E)=\mu(E\cap A_j)/\mu(A_j)$. Thus, $\mu$ is not extremal. This argument is valid also when the support of $\mu$ is restricted to a given set. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Aug 15, 2017 at 21:00
  • $\begingroup$ In the last part, do you mean that $F \in \mathbb{P}_3$ is extremal for $P_3$ is and only if it is extremal for $P_2$? Or do you mean $P_1$? $\endgroup$
    – Boby
    Commented Aug 16, 2017 at 13:06
  • $\begingroup$ I meant $P_1$; it need not be extremal for $P_2$. If $F$ is extremal for $P_3$ and $F=(1-t)F_1+t F_2$, $F_1, F_2 \in P_1$, $t\in(0,1)$, then in fact $F_1, F_2 \in P_3$ (because the support of $F_1$ and $F_2$ is contained in the support of $F$) and therefore $F_1=F_2=F$. Hence, $F$ is extremal for $P_1$. Converse is always true: an extremal point $F$ of a set is also extremal for any subset that contains $F$. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Aug 16, 2017 at 16:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .