The space of probability measures and its intersection with hyperplanes in the space of measures

Question

• Let $X$ be some uncountable standard Borel space (e.g., the real line).
• Let $D$ be the set of Borel probability measures on $X$.
• Let $M$ be the set of signed Borel measures on $X$
• Now let $p_1,...,p_N$ be a finite sequence of linearly independent probability measures in $D$.
• Let $A$ be the set of all possible linear (including but not limited to convex) combinations of $p_1,...,p_N$. $A$ will be some subset of $M$.

Can it be shown that the intersection of $A$ with $D$ is an $(N-1)$-dimensional simplex?

In other words, do there exist linearly independent $q_1, ... , q_N \in D$ such that $A\cap D = co(q_1,...,q_N)$?

The question is easily resolved when $X$ is finite. I am interested in a generalization to the case when $X$ is uncountable.

I mentioned hyperplanes in the title because $A \cap D$ should be a subset of all linear combinations of the form $\sum_{k=1}^{N} \alpha_k p_k$ where $\alpha_k\in\mathbb{R}$ and $\sum_{k=1}^N \alpha_k=1$. Obviously, these are not the same as convex combinations, which would require that $\alpha_k \geq 0$ as well. Anyhow, the set of $(\alpha_1,...,\alpha_N)$ for which $\sum_{k=1}^{N} \alpha_k p_k \in D$ fall on the hyperplane defined by the equation $\sum_{k=1}^N \alpha_k=1$.

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Case: $N=2$

Let $p_1, p_2 \in X$ and $p_1 \neq p_2$.

The definition of $A$ is given as follows:

$A=\lbrace \alpha_1 p_1 + \alpha_2 p_2 \in D | \alpha_1,\alpha_2\in\mathbb{R}\rbrace$

If $q \in D$, then $q(X)=1$. Therefore, if $\alpha_1 p_1+\alpha_2 p_2 \in D$, then $\alpha_1 p_1(X) +\alpha_2 p_2(X)=1$. Since $p_1(X)=1=p_2(X)$, we can rewrite $A$ as:

$A=\lbrace \alpha p_1 + (1-\alpha) p_2 \in D | \alpha\in\mathbb{R}\rbrace = \lbrace p_1 + (1-\alpha)(p_2 - p_1) \in D | \alpha\in\mathbb{R}\rbrace$

Let $B=\lbrace \alpha \in \mathbb{R} | p_1 + (1-\alpha)(p_2-p_1) \in D\rbrace$.

That is, $\alpha \in B \iff p_1 + (1-\alpha)(p_2-p_1) \in D$

Claim: $B$ is convex. Proof: Let $\alpha,\beta\in B$ and $\gamma\in[0,1]$. Then $p_1 + (1-(\gamma\alpha+(1-\gamma)\beta))(p_2-p_1)$ $= \gamma(p_1 + (1-\alpha)(p_2-p_1)) + (1-\gamma)(p_1 + (1-\beta)(p_2-p_1))\in D$ since it is a convex combination of probability measures.

Claim: $B$ has an upper bound. Proof: Suppose not. Since $p_1 \neq p_2$, there exists a Borel set $E$ such that $p_1(E) \neq p_2(E)$. Then $\lim_{\alpha\to\infty} p_1(E) + (1-\alpha)(p_2(E)-p_1(E)) = \infty$ if $p_1(E) > p_2(E)$ and $=-\infty$ if $p_1(E) < p_2(E)$.

Claim: The least upper bound of $B$ is in $B$. Proof: Suppose not. Let $\bar{b}$ be the least upper bound of $B$. It is straightforward that $\bar{b} > 1$. $q=p_1 + (1-\bar{b})(p_2-p_1)$ is clearly a probability measure if $q(E) \geq 0$ for all Borel subsets $E$ of $X$. Therefore there exists $E$ such that $q(E) < 0$. Since $p_1(E) \geq 0$, we have $(1-\bar{b})(p_2(E)-p_1(E)) < -p_1(E)$. However, then there exists $b < \bar{b}$ such that $(1-b)(p_2(E)-p_1(E)) < -p_1(E)$. This contradicts $\bar{b}$ being a least upper bound of $B$ since $B$ is convex.

Claim: $B$ has a greatest lower bound, which is in $B$. Proof: Similar to above.

Claim: $B$ is an interval. Proof: It is a convex, bounded set on the real line that contains its LUB and GLB.

Claim: There exist $q_1, q_2 \in D$ such that $A = co(q_1,q_2)$. Proof: Let $B=[a,b]$. Let $q_1=p_1 + (1-a)(p_2-p_1)$ and $q_2=p_1 + (1-b)(p_2-p_1)$.

Answer 1

Something is wrong in the statement of the question. If $X$ has four points, then a measure on $X$ is a list of four non-negative numbers that sum to 1. But if you take the plane subtended by $(1,0,1,0)$, $(1,0,0,1)$, and $(0,1,1,0)$, then it also contains $(0,1,0,1)$, and these four measures are the four extremal ones in this plane. So the assertion, as stated, does not hold in the finite case after all. The space of all measures is a simplex when $X$ is finite; all I'm then saying is that you can intersect a plane with a tetrahedron to get a square. The geometry also shows you that the counterexample is stable with respect to small variations of the numbers.

The only condition that comes to mind to make it true is to say that the measures have disjoint support. Then it's easy to show that they subtend a simplex of measures in general.

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In the comments, the poster asks whether the intersection is always a polytope. However, Anton Petrunin's example disproves that possibility. If you take the three measures to be proportional to $1$, $1+(\cos x)$, and $1+(\sin x)$ on $[0,2\pi]$, then the set of non-negative linear combinations is indeed a round circle.

In fact, I suspect that you can create any finite-dimensional convex body whatsoever by such a construction.

Answer 2

Here is a counterexample.

Take $X=\mathbb S^1=\mathbb R/2{\cdot}\pi{\cdot}\mathbb Z$ and $$p_i=\tfrac1{2{\cdot}\pi}{\cdot}[1+\cos(x-\alpha_i)]$$ for $\alpha_1=0$, $\alpha_2=\pi/2$.

In this case $$A=\{\,\tfrac1{2{\cdot}\pi}{\cdot}[1+\lambda{\cdot}\cos(x-\alpha)]\mid \alpha\in \mathbb S^1, \ \lambda\ge 0\,\}$$ and $$A\cap D=\{\,\tfrac1{2{\cdot}\pi}{\cdot}[1+\lambda{\cdot}\cos(x-\alpha)]\mid \alpha\in\mathbb S,0\le\lambda\le1\,\}$$ The set of extremal point $A\cap D$ is $$\{\,\tfrac1{2{\cdot}\pi}{\cdot}[1+\cos(x-\alpha)]\mid \alpha\in\mathbb S\,\}$$ and it is NOT finite...