Why does the natural extension of a lower prevision dominate it?

Question

On p. 48 in Troffaes and de Cooman's Lower Previsions there's a claim that "It is clear that" the natural extension $\underline{E}_\underline{P}$ of lower prevision $\underline{P}$ dominates (42) it, i.e. that $\underline{E}_\underline{P}(f) \geq \underline{P}(f)$ for all bounded functions ("gambles") in the domain of $\underline{P}$.

For the last week or two, I've been trying to understand why, without success. Since I think that familiarity with lower previsions is uncommon, but familiarity with ideas needed to answer my question are, I'll give (too many? too few?) details (with page numbers in parentheses).

A lower prevision $\underline{P}$ (38) is a function $\sup\{\mu \in \mathbb{R}: f-\mu \in {\cal D}\}$ from a set $\cal D$ that includes all bounded functions that are nowhere negative, includes no functions that are negative somewhere, and is closed under finite addition and multiplication by non-negative reals (30f)).

A lower prevision that avoids sure loss is (among other, equivalent, things, 42) one where the set ${\cal A}_{\underline{P}} = \{f-\mu : f \in \mbox{domain of } \underline{P} \mbox{ and } \mu<\underline{P}\}$ (42) is such that $\sum_{k=1}^n \lambda_k f_k$ is nowhere negative, for any $f_k \in {\cal A}_{\underline{P}}$ and any nonnegative reals $\lambda_k$.

The natural extension $\underline{E}_{\underline{P}}$ (47) of a lower prevision $\underline{P}$ that avoids sure loss, is a function from any bounded function to a real number:

$$\underline{E}_{\underline{P}} = \sup\{ \alpha \in \mathbb{R} : f - \alpha \geq \sum_{k=1}^n \lambda_k [f_k - \underline{P}(f_k)], n \in \mathbb{N}, f_k \in \mbox{domain of } \underline{P}, \lambda_k \in \mathbb{R}, \lambda_k \geq 0 \}$$

I see that $f-\alpha$ dominates $f-\underline{P}(f)$ for any $f$ in the domain of $\underline{P}$ (letting $n=1$ and $\lambda_1 = 1$). But why should the $\sup\alpha$ be greater than or equal to $\underline{P}(f)$? Off the top of my head (and also after much pondering and exploration), if anything, I'd think the inequality should go the other way (since both $\alpha$ and $\underline{P}(f)$ are subtracted in the definition of $\underline{E}_{\underline{P}}$. Sorry for all of the detail--I'm not sure that all of it is relevant. I think I'm missing something obvious, or that there's some basic fact about suprema that I don't understand.

[Note: Most of the tags I've attached to this question are not quite appropriate, but they're the closest tags we have at the moment. I don't have sufficient privilege on this site to create a new tag, but I've posted an "answer" in Tag management 2015 in meta.math.SE proposing a new tag "imprecise-probability", which I believe would be the correct tag.]

Answer 1

More detailed answer can be found in Walley's book "Statistical Reasoning with Imprecise probabilities". Although I know that it does not use the concept of lower prevision as maximal buying price of a "gamble" (i.e., random variable), I often find easier to explain that in terms of dominated probabilities of $\underline{P}$ over a finite set.

If you take a finite set $\mathcal{X}=\{x_1,\ldots,x_n\}$ and a (possibly finite) set $\mathcal{D}$ of real-valued functions over $\mathcal{X}$, then the value $\underline{P}(f)$ of a lower prevision on $\mathcal{D}$ can be interpreted as a lower bound of the expectation of $f$, that is you can read it as $$\underline{P}(f) \leq \sum_{x_i \in \mathcal{X}} f(x_i)p(x_i)$$ where $p(x_i)$ is a probability mass over $\mathcal{X}$. A lower prevision avoids sure loss iff the set $$\mathcal{M}(\underline{P})=\{p : \underline{P}(f) \leq \sum_{x_i \in \mathcal{X}} f(x_i)p(x_i) \forall f \in \mathcal{D} \}$$ is non-empty, that is if there is at least a probability whose expectations dominate the lower bounds $\underline{P}(f)$ for all functions of $\mathcal{D}$. Provided this is the case, the natural extension of any function $g$ (and in particular functions within $\mathcal{D}$ then amounts to compute $$\underline{\mathbb{E}}(g)=\min_{p \in \mathcal{M}(\underline{P})} \sum_{x_i \in \mathcal{X}} g(x_i)p(x_i).$$ That is, you search the minimal expectation value among all probabilities dominated by $\underline{P}$. Since $\underline{P}(f)$ may not be a tight lower bound of $\sum_{x_i \in \mathcal{X}} f(x_i)p(x_i)$, then it follows that $\underline{\mathbb{E}}(f) \geq \underline{P}(f)$. You can see that by considering $\underline{P}(f)$ as bounds over a linear constraint, and $\underline{\mathbb{E}}(f)$ as the minimal value of some objective functions delimited by such constraints. It may be the case that the constraint given by $\underline{P}(f)$ is not saturated in your solution. Actually, the way it is formalised in your question and in Troffaes/De Cooman is the dual version of this linear constraint problem.