- 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 $q_2=p_1 + (1-b)(p_2-p_1)$.