# If $E_P(X) \in$ span($E_{P_i}(X)$), $1 \leq i \leq n$, for all integrable $X$, then is $P$ a convex combination of the $P_i$?

Let $(\Omega, \mathfrak{F}, P)$ be a probability space and let $R = \{P_1,...,P_n \}$ be a finite set of probability measures on $(\Omega, \mathfrak{F})$, each of which is absolutely continuous with respect to $P$. I'm wondering if the following result holds.

Suppose that for all random variables $X$ which are integrable with respect to $P$ and $P' \in R$ we have that $E_P(X)$ lies in the interval spanned by $E_{P_i}(X)$, $1 \leq i \leq n$. Then $P$ is a convex combination of the members of $R$.

In this paper the result is shown for the case where $\Omega$ is finite. The proof relies on identifying random variables and probability measures with members of $\mathbb{R}^n$ and proceeds by appealing to some basic facts about convex sets. How might one approach this problem for general $(\Omega, \mathfrak{F})$?

Addendum. Here is what I've come up with. I would appreciate feedback.

We begin with a general measurable space $(\Omega, \mathfrak{F})$ and the space $V$ of all bounded random variables on $(\Omega, \mathfrak{F})$. The space $V$ is a (real) vector space with respect to pointwise addition and scalar multiplication, and if we equip $V$ with the $\sup$-norm, it becomes a Banach space (it is complete). We now identify probability measures $P$ on $(\Omega, \mathfrak{F})$ with continuous linear functionals in the dual space $V'$ via the embedding $P \mapsto \int (\cdot) dP$. Hence, for $X \in V$, we have $P(X) = E_P(X)$, where $E_P$ is expected value with respect to $P$. In order to consider sets of probability measures with certain topological properties, we endow $V'$ with its weak*-topology, which is the topology generated by the evaluation functionals $\lambda_X$ in the double dual $(V')'$, defined by $\lambda_X(\phi) = \phi(X)$, where $X \in V$ and $\phi \in V'$. That is, the weak*-topology is the weakest topology that makes the evaluation functionals continuous, and, moreover, a linear functional $\lambda$ on $V'$ is continuous if and only if it is an evaluation functional. Now, $V'$ is a locally convex topological vector space in the weak*-topology, and the set of probability measures on $(\Omega, \mathfrak{F})$ is weak*-compact in $V'$. It follows that any weak*-closed set of probability measures is weak*-compact. This much is just my attempt to summarize some facts that I learned by studying chapter 14 of Royden's Real Analysis, together with the thought that probability measures can be identified with continuous linear functionals in $V'$ (I also consulted Chapter 3 and Appendices D and E of Walley's Statistical Reasoning with Imprecise Probabilities). If what I have said so far is acceptable, then I believe that I've answered my question here.

Theorem Let $(\Omega, \mathfrak{F}, P)$ be an arbitrary probability space and $R$ an arbitrary set of probability measures on $(\Omega, \mathfrak{F})$ with $|R| \geq 2$. Then, the following two assertions are equivalent. (a) $E_P(X)$ lies strictly inside the interval spanned by $\{E_{P'}(X) \}_{P' \in R}$, i.e., $$\inf_{P' \in R} \{E_{P'}(X) \} < E_P(X) < \sup_{P' \in R} \{E_{P'}(X) \}.$$ (b) $P$ is in the interior of the convex weak*-closure of $R$, i.e. $P \in \text{int}(\text{co}R).$

Proof. We note that the convex weak*-closure $\text{co}R$ of $R$ is a convex, weak*-compact subset of $V'$ with non-empty interior, because $|R| \geq 2$. We begin by supposing that $P \notin \text{int}(\text{co}R)$. By a standard separating hyperplane result, there exists a continuous linear functional $\lambda$ on $V'$ and $\alpha \in \mathbb{R}$ such that $\lambda(P) \geq \alpha$ and $\lambda(P') \leq \alpha$ for all $P' \in \text{co}R$. But the continuous linear functionals on $V'$ consist of all and only the evaluation functionals. So for some $X \in V$, we have $\lambda = \lambda_X$, and hence $\lambda_X(P) \geq \alpha \geq \lambda_X(P')$ for all $P' \in \text{co}R$. But by the definition of the evaluation functional, this implies \begin{equation}\label{eqn: spanning violation} E_P(X)=P(X) \geq \alpha \geq P'(X) = E_{P'}(X) \end{equation} for all $P' \in \text{co}R$. Hence, $E_P(X) \geq \sup_{P' \in \text{co}R}\{E_{P'}(X) \} \geq \sup_{P' \in R}\{E_{P'}(X) \}$, and (a) is violated.

Conversely, suppose that $P \in \text{int}(\text{co}R)$. Then there exists a collection $\{P_i \}_{i=1}^n$, $n \geq 2$, of probability measures in $R$ such that $P = \sum_{i=1}^n \beta_i P_i$, with $\beta_i > 0$ and $\sum_{i=1}^n \beta_i = 1$ (I'm not sure I can justify this step; suggestions here are would be appreciated). For an arbitrary random variable $X \in V$, we have \begin{align*} E_P(X) &= \int X dP = \int X d(\sum_{i=1}^n \beta_i P_i) \\ &= \sum_{i=1}^n \beta_i \int X dP_i = \sum_{i=1}^n \beta_i E_{P_i}(X). \end{align*} We note that this calculation relies on the fact that the $P_i$ are finite measures. We now see that $E_P(X)$ is a strict convex combination of the $E_{P_i}(X)$, so (a) is satisfied.

• I think this can be shown by a separation argument (Hahn-Banach): Suppose $P$ does not belong to the convex hull of $R$. Then, there is a random variable $X$, such that $E_P(X) < \gamma \le E_{P'}(X)$ for some $\gamma \in \mathbb R$ and all $P' \in R$. – gerw Jan 23 '17 at 22:07
• @gerw I think so too, I'm just not too confident about the details. If you feel like spelling things out more carefully in an answer, that would be great. – grndl Jan 23 '17 at 22:11
• @gerw For example, to make the argument rigorous, should I identify probability measures with linear functional on the space of random variables? – grndl Jan 24 '17 at 0:20
• For the Hahn-Banach argument, you need a space $Y$ in which all your probability measures live. Then, you get the separation vector in the dual or predual space (if it exists). I am not sure about a good choice of $Y$. In the case where $\Omega$ is a locally compact metric space, you can choose $Y = \mathcal{M}(\Omega) = C_0(\Omega)^*$. Otherwise, I am not sure. – gerw Jan 24 '17 at 12:17
• @gerw Thanks for your comments. I've asked a more general, related question here. Hopefully someone will provide some details for us. – grndl Jan 24 '17 at 16:18

I believe that you may reduce the first problem to the finite dimensional case in the following way:

Let $\Delta_n =\{\lambda\in {\Bbb R}^n : \sum_{i=1}^n \lambda_i=1, \lambda_1,...,\lambda_n\geq 0\}$

For $A\in \mathfrak{F}$ define $Z(A)=\{ \lambda \in \Delta_n^n: P(A)=\sum_{i=1}^n \lambda_i P_i(A)\}$

Then by hypothesis each $Z(A)$ is a non-empty, compact subset of $\Delta_n$. If $Z(\mathfrak{F})=\bigcap_{A\in \mathfrak{F}} Z(A)$ is non-empty then any $\lambda\in Z(\mathfrak{F})$ allows to write $P(A)=\sum_i \lambda_i P_i(A)$ on measurable sets which suffices.

So suppose instead that $Z(\mathfrak{F})=\emptyset$ then by the finite intersection principle for compact sets there must be a finite number of sets $A_1,...,A_n$ for which $Z(A_1)\cap ...\cap Z(A_n)=\emptyset$. Let ${\cal B}=\sigma(A_1,...,A_n)$ be the (finite) $\sigma$ algebra generated by this finite collection and consider the restriction of $P$ and the $P_i$'s to ${\cal B}$. We are then dealing with probability measures on a finite space which will satisfy the same hypotheses but $Z({\cal B})=\emptyset$ which apparently is in contradiction with the paper you cite.

Regarding the more general problem, let $E$ denote the vectorspace of signed measures on $(\Omega, \mathfrak{F})$ with the variation norm and $\Lambda=\overline{B}_1(L^\infty(\Omega, \mathfrak{F}))$, i.e. the set of measurable maps: $X:\Omega \rightarrow [-1,1]$. The variation norm of $\mu\in E$ is then $\|\mu\|= \sup_{\Lambda\in \Lambda} |\mu(X)|$ showing that $\Lambda$ separates points in $E$. The set $\Lambda$ gives rise to a weak topology on $E$. Given $\mu\in E$, $X_1,...,X_m\in \Lambda$, $\epsilon>0$ we set $$\{ \nu \in E: |\nu(X_i)-\mu(X_i)| < \epsilon, i=1,...,m\}$$ Then the sets of this type forms a base (neighborhood of $\mu$) for the $\Lambda$-topology on $E$. In general it is weaker than the weak topology generated by the dual $E'$ since $\Lambda$ may be viewed as a subset of $E'$.

Now let $M\subset E$ be the subset of probability measures and consider a subset $R\subset M$ and $\mu\in M$. We set $S=co(R)$, the closed convex hull of $R$, closed for the $\Lambda$ topology.

If $\mu\notin S$ then by definition there is an open neighborhood $B$ of $\mu$ that does not intersect $S$. Thus, $|\nu(X_i)-\mu(X_i)|<\epsilon$, $i=1,...,m$. But by a finite dimension argument (see e.g. S Lang: Real and Functional Analysis, Thm 1.2, p. 85) there are real constants $c_1,...,c_m$ such that $X=c_1X_1+ \ldots c_m X_m$ separates $\mu$ from $S$, in other words, there is $a\in {\Bbb R}$ s.t.

$$\mu(X) > a \geq \nu(X)$$ for every $\nu\in S$. In your case, the condition you impose on $P$ tells us that the above can not happen, so $P\in co(R)$. (This follows pretty much the same line of arguments you give).

Now, the implications of this are not so obvious. In general you can not hope for $P$ to be a finite linear combination of elements in $R$. If e.g. you take $P_1,P_2,...$ to be a sequence in $M$ with disjoint support, then $P = \sum_k 2^{-k} P_k$ is in the closed convex hull but is not finitely generated. Worse, if $R$ is uncountable, you can not in general hope to write $P$ as a countable convex combination of elements in $R$.

Choquet's Theorem states that if $S$ is a metrizable compact convex subset of a locally compact Hausdorff topol vector space and $P\in S$ then there is a measure supported on the extremal points of $S$ (which here should be a subset of $R$ it seems) which represents $P$ (so you may write $P$ as an 'integral' over $R$). Now, what you need to metrize is the $\Lambda$ topology and this is not possible in our general setup. (It would be if $\Omega$ was compact Hausdorff and we looked at Borel measures.) I think that in the general setup you can not obtain further information on how to decompose $P$ with respect to $R$.

• Thanks for this. The paper I cite does something similar in the Appendix, but I prefer your argument. Ultimately, I'm interested in letting $R$ be an arbitrary set of probabilities, in which case your method doesn't seem to work (please correct me if I'm wrong). Do you think the argument I give in my Addendum is promising in that case? – grndl Feb 2 '17 at 19:01
• The topological interior of $co(R)$ may be empty (could e.g. be just a segment) but perhaps you mean relative interior? I wouldn't be too optimistic about this. On the other hand sounds plausible that $P$ lies in the closure of the convex hull. But the separation argument may be a bit more tricky than you describe. I'll have to think more deeply about this. – H. H. Rugh Feb 2 '17 at 20:01
• Ah you're right, of course. Thank you. – grndl Feb 2 '17 at 20:53
• I think that there is a problem when you claim that $M$ (the set of probability measures) is weak-* compact in $V'$, at least I don't see a reasonable justification. The closed unit ball in $V'$ is weak-* compact but is quite nasty in general and I don't see a reasonable way of showing that $M$ is weak-* closed in $V'$ (without some further structure on $\Omega$ like being locally compact). This has an impact of the sequel as $co(R)$ may contain other things than probability measures. Would be nice to have a counter example at hand but if it exists it probably needs the AoC – H. H. Rugh Feb 3 '17 at 11:37
• It could be finitely generated but it need not. For ex, take $\Omega={\Bbb N}$, $R$ to be the Dirac masses and $P=\sum c_i \delta_i$ with every $c_i>0$ and $\sum_i c_i=1$. (but I still haven't figured out a reasonable answer to the question whether or not $P$ may satisfy your condition but still be an extremal point). – H. H. Rugh Feb 13 '17 at 21:02