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Let $X$ be, for simplicity, a finite set (with the discrete topology). Denote with $M(X)$ the set of probability measures on $X$ endowed with the weak topology. For $\mu\in M(X)$ and a (necessarily measurable) function $f:X\rightarrow[-1,1]$ denote with $E_{\mu}(f)$ the expected value of $f$.

For a (closed) set $A\subseteq M(X)$, define the lower expectation of $A$, denoted by $E(A)$, as the following functional of type $(X\rightarrow [-1,1])\rightarrow [-1,1]$:

$$E(A)(f)= \displaystyle \inf \{ E_{\mu}(f) \ | \ \mu \in A \}.$$

For a (closed) set $A\subseteq M(X)$ denote with $H(A)$ its convex hull of $A$, defined as expected.

It is easy to see that:

Proposition: For all $A,B\subseteq M(X)$, if $H(A)=H(B)$ then $E(A)=E(B)$.

Now my question is about the inverse direction of the previous statement.

QUESTION 1: Is it true that for $A,B\subseteq M(X)$, if $E(A)=E(B)$ then $H(A)=H(B)$?

QUESTION 2: What about restricting attention to functions $f$ of type $X\rightarrow [0,1]$?

Remark: note that, restricting even further to characteristic functions $f:X\rightarrow\{0,1\}$, the statement of QUESTION $1$ is not true anymore and the following is an example:

Example. Consider $X=\{a,b,c\}$, $\mu_{1}= \{ a\mapsto 0.3, b\mapsto 0.3, c\mapsto 0.4\}$, $\mu_{2}=\{ a\mapsto 0.4, b\mapsto 0.3, c\mapsto 0.3\}$ and $\mu_{3}=\{a\mapsto 0.5, b\mapsto 0.4, c\mapsto 0.1 \}$. Now consider $A=\{\mu_{1},\mu_{2}\}$ and $B=\{\mu_{1},\mu_{2},\mu_{3}\}$. Now $H(A)\neq H(B)$ because $\mu_{3}$ is not a convex combination of $\mu_{1}$ and $\mu_{2}$. Yet, for every set $Y\subseteq X$ (i.e., function $f:X\rightarrow\{0,1\}$, it holds that $E(A)(Y)=E(B)(Y)$.

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  • $\begingroup$ I know that I should probably find an answer to this question is the book "Statistical Reasoning with Imprecise Probabilities" of Peter Walley, 1991. Apparently he refers to my "lower expecations" as lower previsions. However I do not have access to this book unfortunately. $\endgroup$ – IamMeeoh Apr 23 '13 at 16:49
  • $\begingroup$ In finite case, if $H(A) \ne H(B)$ use hahn banach to seperate say $b$ and $H(A)$. The dual element is actually a function, and normalizing it to satisfy your conditions is no problem. This argument breaks down for non-finite sets. $\endgroup$ – mike Apr 23 '13 at 18:43
  • $\begingroup$ Hello mike, thank you very much for your answer. I guess you refer to the Hahn-Banach separation theorem (en.wikipedia.org/wiki/…). Would you mind expanding a bit your answer plese? I've difficulties following your argument. $\endgroup$ – IamMeeoh Apr 23 '13 at 19:47
  • $\begingroup$ finite case, measures and fctns both n-dimensional vectors, suppose there is a $b \in H(B), b \notin H(A)$ hahn banach, as in your ref, gives $X : \langle X,b \rangle < t <0 < \langle X,a \rangle $ for all $a \in H(A)$, so $E(B)(X) <0, E(A)(X) \ge 0$ $\endgroup$ – mike Apr 23 '13 at 20:16
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There is a one-to-one correspondence between convex sets of probability distributions and affinely superadditive lower expectations (or lower previsions in Walley's terminology). You should check http://sites.poli.usp.br/p/fabio.cozman/research/credalsetstutorial/introduction/node5.html

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  • $\begingroup$ Thanks for the answer and for the link. It seems to me that every lower expectation $E(A)$, generated as in my question by a set $A$ of probability measures, is affinely homogeneous and superadditive. Am I right? So the question becomes: Is it the case that every affinely homogeneous superadditive expectation (or prevision) $e$ arises as the lower expectation $E(A)$ of some set convex $A$ of probability measures? I.e., $\forall e. \exists A .\ conv(A)\ \wedge \ p = E(A)$? $\endgroup$ – IamMeeoh Apr 25 '13 at 6:15
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The first question has already been answered in the comments and by @Kenny: yes! Taking the convex closure of the set of probability measures has no influence on the lower expectation, so $E(A)=H(A)$ (and also $E(B)=H(B)$). Therefore, $E(A)=E(B)$ implies $H(A)=H(B)$ because the functionals $E$ and $H$ are identical.

As for the second question, the answer is yes too.

To see why, choose any variable $g\colon X\to\mathbb{R}$. Note that $g$ is bounded, and in particular has a minimum and a maximum, because $X$ was assumed to be finite. Let \begin{align} \alpha&:=\min g \\ \beta&:=\max g - \min g \\ f&:=\begin{cases} \frac{g-\alpha}{\beta} & \text{if $\beta>0$} \\ 0 & \text{if $\beta=0$} \end{cases} \end{align} Note that $f\colon X\to[0,1]$ and $g=\alpha+\beta f$ with $\beta\ge 0$. Also, $$E(A)(g)=E(A)(\alpha+\beta f)=\alpha +\beta E(A)(f)$$ In other words, $E(A)$ is completely determined by its restriction to variables of the form $f\colon X\to [0,1]$.

Now assume $E(A)(f)=E(B)(f)$ for all $f\colon X\to [0,1]$. Then for any $g\colon X\to\mathbb{R}$, by the previous part, we know that there are $f\colon X\to [0,1]$, $\alpha\in\mathbb{R}$ and $\beta\in\mathbb{R}_{\ge 0}$ (as constructed before) such that $g=\alpha+\beta f$, and $$E(A)(g)=\alpha +\beta E(A)(f)=\alpha +\beta E(B)(f)=E(B)(g)$$

Note that we never used countable additivity. So these arguments also extend to infinite $X$, bounded variables (using $\sup$ and $\inf$ instead of $\max$ and $\min$), and closed sets of finitely additive probability charges (under an appropriate topology).

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