The multiple priors of the open-minded decision maker

Abstract

A multiple-prior decision maker is open-minded if he/she can describe, as subjective uncertainty, all convex sets of distributions over payoff relevant consequences. Theorem 1: open-mindedness is equivalent to the ability to subjectively describe both the uniform distribution on an interval and the set of all distributions on an interval. Theorem 2: sets of priors that fail either condition cannot describe a dense class of problems. The use of open-minded sets of priors to model decision makers allows the objective and the subjective approaches to uncertainty to inform each other. It also changes the implications of previously used axioms for multi-prior preferences because the axioms must apply to a larger set of problems.

Appendix: Proofs

We will repeatedly use two mathematical results, the Borel isomorphism theorem and the Blackwell–Dubins extension of Skorohod’s representation theorem.

• Two measurable spaces are measurably isomorphic if there is a bijection between them that is measurable in both directions. Let B be a measurable subset of a complete separable metric space (M, d) and $B^{\prime }$ a measurable subset of a complete separable metric space $(M^{\prime },d^{\prime })$. The Borel isomorphism theorem (Dellacherie and Meyer 1978, Theorem III.20) or (Dudley 2002, Theorem 13.1.1) says that for B and $B^{\prime }$ are measurably isomorphic if and only if they have the same cardinality.

• Blackwell and Dubins (1983) show that for any complete separable metric space (M, d) there exists a jointly measurable $\mathbf{b }_M: {\varDelta }(M) \times [0, 1] \rightarrow M$ such that

  • (1) for all $p \in {\varDelta }(M)$, the image measure of $\lambda$ under the function $\mathbf{b }_M(p,\cdot )$ is equal to p, that is, $\mathbf{b }_M(p,\lambda ) = p$, and

  • (2) if $\rho (p_n,p) \rightarrow 0$, then $\mathbf{b }_M(p_n,\cdot ) \rightarrow \mathbf{b }_M(p,\cdot )$ almost everywhere $\lambda$, that is, $\lambda (\{x \in [0, 1]: \mathbf{b }_M(p_n,x) \rightarrow \mathbf{b }_M(p,x)\}) = 1$.

This extends Skorohod (1956, Thm. 3.1.1), which worked for a single p or for a single sequence $p_n$ with $\rho (p_n,p) \rightarrow 0$.

We will also use the canonical set of priors, ${\varPi }^\circ$, defined as the closed convex hull of the set of uniform distributions, $\lambda _x$, on the line segments $\{x\} \times [0, 1] \subset [0, 1] \times [0, 1]$. Every element of ${\varPi }^\circ$ can be expressed as an integral of the $\lambda _x$ with respect to some countably additive probability $\eta$ on the set of $\lambda _x$, viz $\mu (E) = \int _{[0, 1]} \lambda _x(E)\,{\mathrm{d}}\eta (x)$.

Proof of Theorem 1

Fix $A \in {\mathcal {K}}_{{\varDelta }({\mathbb {X}})}$ and let $\psi :[0, 1] \leftrightarrow A$ be a measurable isomorphism. Define $f_A(s) = \mathbf{b }_{{\mathbb {X}}}(\varphi (f_{\varDelta }(s)),f_\lambda (s))$ where $\mathbf{b }_{{\mathbb {X}}}(\cdot ,\cdot )$ is the Blackwell–Dubins function. We verify that $f_A({\varPi }) = A$.

$f_A({\varPi }) \subset A$: For each $\mu \in {\varPi }$, $f_A(\mu ) \in A$ because $\psi ([0, 1]) = A$, A is a closed convex set, and $f_\lambda (\mu ) = \lambda$.

$A \subset f_A({\varPi })$: Fix $p \in A$ and let $x = \psi ^{-1}(p)$. By definition, there exists $\mu _x \in {\varPi }$ such that $f_{\varDelta }(\mu _x)$ is the point mass $\delta _x$. Since $f_\lambda (\mu _x) = \lambda$, $f_A(\mu _x) = p$. $\square$

The extension of Theorem 1 to the ability to represent closed convex sets of probabilities on complete separable metric (csm) spaces passes through the observation that every complete separable metric space that is homeomorphic to a $G_\delta$ is a compact metric space.

Proof of Corollary 1

Let ${\varPi }^{\prime }$ be a strongly zero-one set of non-atomic priors and let ${\varPi }$ be its closed convex hull. Recall that for a random variable X taking values in [0, 1] and having continuous cdf $F_X(\cdot )$, the random variable $Y = F_X(X)$ has the uniform distribution.

From the characterization of strongly zero-one sets in Dumav and Stinchcombe (2016)[Theorem 1 and §3.2], we obtain simultaneously two measurable isomorphisms $\varphi :{\varPi }^{\prime } \leftrightarrow [0, 1]$ and $g : {\varOmega }^{\prime } \leftrightarrow [0, 1]$ such that for all $r \in [0,1]$ and for all $\mu \in \varphi ^{-1}(r)$, $\mu (g^{-1}(r))=1$. Consider the mapping $\mu \mapsto F_\mu (t) := \mu (g^{-1}([0, t]))$ from ${\varPi }^{\prime }$ to the cdf of the distribution $g(\mu )$. Because $\mu$ is non-atomic and $g(\cdot )$ is a measurable isomorphism, the distribution $g(\mu )$ has a continuous cdf. The function $f_\lambda (\omega ^{\prime }) := F_{\varphi ^{-1}(g(\omega ^{\prime }))}(g(\omega ^{\prime }))$ is measurable, as it is a composite of two measurable functions, and for each $\mu \in {\varPi }^{\prime }$, $f_\lambda (\mu ) = \lambda$.

By the Borel isomorphism theorem, there exists a measurable bijection $\psi :{\varPi }\leftrightarrow [0, 1]$. Define $f_{\varDelta }(\omega ) = \psi (\varphi (\omega ))$. For each $\mu \in {\varPi }^{\prime }$, $f_{\varDelta }(\mu )$ is point mass on some $x \in [0, M]$. Since ${\varPi }$ is the closed convex hull of ${\varPi }^{\prime }$, $f_{\varDelta }({\varPi }) = {\varDelta }([0, 1])$. $\square$

Proof of Corollary 2

Let $A,A_n$ be a sequence in ${\mathcal {K}}_{{\varDelta }({\mathbb {X}})}$. We first show the existence of a sequence of measurable functions, $h,h_n:[0, 1] \rightarrow {\varDelta }({\mathbb {X}})$, with $h([0, 1]) = A$, $h_n([0, 1]) = A_n$ and $h_n(r) \rightarrow h(r)$ for each $x \in [0, 1]$, and then show how this yields the desired result. Let $\psi :[0, 1] \leftrightarrow [0, 1]^{{\mathbb {N}}}$ be a measurable isomorphism, and let $\{v_n: {n \in {\mathbb {N}}}\}$ be a countable sup-norm dense set of continuous functions mapping ${\mathbb {X}}$ to [0, 1]. The identification $p \leftrightarrow \{\int v_n\,{\mathrm{d}}p: {n \in {\mathbb {N}}}\}$ defines a linear homeomorphism between ${\varDelta }({\mathbb {X}})$ and a compact convex $K \subset [0, 1]^{{\mathbb {N}}}$.

Metrize the product topology on $[0,1]^{{\mathbb {N}}}$ with a strictly convex metric in the following way. For every points $\{x_n\}$ and $\{y_n\}$ of $[0,1]^{{{\mathbb {N}}}}$, let $\rho (x, y) := \sum _n \frac{1}{2^{n}}d(x_n - y_n)$ where d is any (strictly convex) metric on [0, 1] and bounded by 1, (for instance Euclidean distance on [0, 1]). By Engelking (1989, Theorem 4.2.2), $\rho$ metrizes the Cartesian product space $[0,1]^{{{\mathbb {N}}}}$ and it is strictly convex as a sum of strictly convex metrics. Now define the mapping n(A, v) as the nearest point in A to v. Because the metric is strictly convex, this mapping is jointly continuous. Finally, define $h(r) = n(A,\psi (r))$ and $h_n(r) = n(A_n,\psi (r))$.

Given the sequence of functions, $h, h_n$, define $f(s) = \mathbf{b }_{{\mathbb {X}}}(h(g(s)), f(s))$ and $f_n(s) = \mathbf{b }_{{\mathbb {X}}}(h_n(g(s)), f(s))$, where $\mathbf{b }_{{\mathbb {X}}}(\cdot ,\cdot )$ is the Blackwell–Dubins function. Because $[\rho (p_n,p) \rightarrow 0]$ implies that $\mathbf{b }_{{\mathbb {X}}}(p_n,\cdot )$ converges to $\mathbf{b }_{{\mathbb {X}}}(p,\cdot )$ almost everywhere $\lambda$ and $\rho _H(A_n,A) \rightarrow 0$, the result follows. $\square$

Proof of Lemma 2

Let A denote the set of probabilities p on [0, M] satisfying $p(\{x_0\}) \ge a$ for some $x_0 \in [0, M]$ and let $p_n$ be a sequence in A with $\rho (p_n,p) \rightarrow 0$. We show that $p \in A$. Let $x_n \in [0, M]$ satisfy $p_n(\{x_n\}) \ge {\varvec{a}}$. Any subsequence of $p_n$ has a further subsequence, still converging to p, for which $x_n \rightarrow x$ for some $x \in [0, M]$. By the definition of $\rho (\cdot ,\cdot )$, $p(\{x\}) \ge {\varvec{a}}$. If the closed set A had a non-empty interior, it would have to contain a probability having a density with respect to Lebesgue measure, contradicting the existence of an atom. $\square$

Proof of Theorem 2

Because the composition of measurable functions is measurable and ${\mathbb {F}}$ fails to be open-minded, no open-minded subset of ${\mathcal {K}}_{{\varDelta }({\mathbb {X}})}$ can belong to ${\mathcal {R}}({\mathbb {F}})$. It is therefore sufficient to show that the class of open-minded sets is dense in ${\mathcal {K}}_{{\varDelta }({\mathbb {X}})}$. To this end, pick arbitrary $A \in {\mathcal {K}}_{{\varDelta }({\mathbb {X}})}$ and $\epsilon > 0$. Let $A_f$ denote a finite set of extreme points for A such that $d(A,\text {co}(A_f)) < \epsilon /2$. For any $\delta > 0$ and $x \in {\mathbb {X}}$, let $B_\delta (x)$ denote the necessarily uncountable, open ball with radius $\delta > 0$ around $x \in {\mathbb {X}}$. By the Borel isomorphism theorem, there exists $\varphi _{x,\delta } : [0, 1] \times [0, 1] \leftrightarrow B_\delta (x)$ where $\varphi _{x,\delta }$ is a measurable bijection with measurable inverse. Let ${\varPi }_{x,\delta }$ denote the open-minded set $\varphi _{x,\delta }({\varPi }^\circ )$ where ${\varPi }^\circ$ is the canonical open-minded set of priors given above. Pick $\delta < \epsilon /2$ such that the points in $A_f$ are at least $2\delta$ from each other. Since the support sets are disjoint, the closed convex hull of the set $\cup _{x \in A_f} {\varPi }_{x,\delta }$ is open-minded and within $\epsilon$ of A. $\square$

The following argument closely parallels the proof for countably additive priors, but unlike the countably additive case, it does not extend to complete separable metric spaces.

Proof of Lemma 3

Fix $p \in {\varDelta }^{fa}({\mathbb {X}})$. Let ${\mathcal {E}}_n$ be a nested sequence of measurable partitions of ${\mathbb {X}}$ into elements having maximal diameter less than $1/2^n$. Construct a corresponding sequence of nested partitions of S, ${\mathcal {A}}_n$ having $\mu (A_n) = p(E_n)$ for each $E_n \in {\mathcal {E}}_n$. Pick $x_{k,n} \in E_{k,n} \in {\mathcal {E}}_n$ and define $f_n = \sum _k x_{k,n} 1_{E_{k,n}}$. For each s, $f_n(s)$ is a Cauchy sequence, hence converges. The function $f(s) := \lim _n f_n(s)$ is measurable, we must show that $f(\mu )$ is continuously equivalent to p. Let $v:{\mathbb {X}} \rightarrow {{\mathbb {R}}}$ be a continuous, hence bounded, function. Being the uniform limit of the $f_n$, f satisfies $\int _S v(f(s))\,{\mathrm{d}}\mu (s) = \int _{\mathbb {X}} v(x)\,{\mathrm{d}}p(x)$ for all continuous $v:{\mathbb {X}} \rightarrow {{\mathbb {R}}}$. $\square$

Proof of Theorem 3

Let $\{\mu _n: {n \in {\mathbb {N}}}\}$ be the countably infinite set of non-atomic probabilities supported on with disjoint supports, let $A \in {\mathcal {K}}_{{{\varDelta }^{fa}({\mathbb {X}})}}$, and let $\{p_n: {n \in {\mathbb {N}}}\}$ be a countable dense subset of A. From Lemma 3, there exists a measurable $f:S \rightarrow {\mathbb {X}}$ with $f(\mu _n) = p_n$. Taking weak$^*$ closure in ${\varDelta }^{fa}({\mathbb {X}})$ and using the continuity of the $\mu \mapsto f(\mu )$ mapping, $\rho _H(f({\varPi }),A) = 0$. For the second statement, let $A^{\prime }$ be the set of countably additive probabilities on ${\mathbb {X}}$ at distance 0 from A. Pick a measurable $f:S \rightarrow {\mathbb {X}}$ such that $f({\varPi }^{ca}) = A^{\prime }$. Again, taking weak$^*$ closure in ${\varDelta }^{fa}({\mathbb {X}})$ and using the continuity of the $\mu \mapsto f(\mu )$ mapping, $\rho _H(f({\varPi }),A) = 0$. $\square$

Proof of Proposition 1

Let $A = f(S)$ and $B = g(S)$. Define the non-degenerate interval $[a_f, b_f] = \{ \int v\,{\mathrm{d}}p: p \in A\}$ and $[a_g, b_g] = \{ \int v\,{\mathrm{d}}q: q \in B\}$. We stratify A and B as follows: For $u \in [a_f, b_f]$, define $A_u = \{p \in A: \int v\,{\mathrm{d}}p = u\}$; for $u \in [a_g, b_g]$, define $B_u = \{q \in B: \int v\,{\mathrm{d}}q = u\}$. It is easy to show that there exists a jointly measurable $m_f:[0, 1] \times [0, 1] \rightarrow A$ such that for all $x \in [0, 1]$, $m_f(x,\cdot )$ is a measurable isomorphism between [0, 1] and $A_{a_f + (b_f-a_f)x}$. In a similar fashion, there exists a jointly measurable $m_g:[0, 1] \times [0, 1] \rightarrow B$ such that $m_g(x,\cdot )$ is a measurable isomorphism between [0, 1] and $B_{a_g + (b_g-a_g)x}$. For later purposes, note that $x \mapsto a_f + (b_f-a_f)x$ and $x \mapsto a_g + (b_g-a_g)x$ are positive affine transformations of each other.

Because ${\varPi }$ is open-minded, there exists $h_\lambda :S \rightarrow [0, 1]$ with $h_\lambda (\mu ) = \lambda$ for each $\mu \in {\varPi }$, and there exists $h_D:S \rightarrow [0, 1] \times [0, 1]$ such that $h_D({\varPi }) = {\varDelta }([0, 1] \times [0, 1])$. Define

$$\begin{aligned} f^{\prime }(s) = \mathbf{b }_{{\mathbb {X}}}(m_f(h_D(s)),h_\lambda (s)) \quad \text{ and } \quad g^{\prime }(s) = \mathbf{b }_{{\mathbb {X}}}(m_g(h_D(s)),h_\lambda (s)). \end{aligned}$$

For each $\mu \in {\varPi }$, $f^{\prime }(\mu )$ is a convex combination of probabilities in A and $g^{\prime }(\mu )$ is a convex combination of probabilities in B. Since A and B are convex, $f^{\prime }({\varPi }) \subset A$ and $g^{\prime }({\varPi }) \subset B$. For any $(x,y) \in [0, 1] \times [0, 1]$, let $\mu _{(x,y)} \in {\varPi }$ put mass 1 on $h_D^{-1}(x,y)$ (because $\delta _{(x,y)} \in {\varDelta }([0, 1] \times [0, 1]$, there exists such a $\mu _{(x,y)}$). Each $p \in A$ and $q \in B$ is of the form $f^{\prime }(\mu _{(x,y)})$ and $g^{\prime }(\mu _{(x,y)})$, respectively, so $A \subset f^{\prime }({\varPi })$ and $B \subset g^{\prime }({\varPi })$. Finally, $v_{f^{\prime }}(\cdot )$ is a positive affine transformation of $v_{g^{\prime }}(\cdot )$ because $x \mapsto a_f + (b_f-a_f)x$ and $x \mapsto a_g + (b_g-a_g)x$ are positive affine transformations of each other. $\square$

Proof of Proposition 2

Let $[a, b] = \{\int v \circ f\,{\mathrm{d}}\mu : \mu \in {\varPi }\} = \{\int v \circ g\,{\mathrm{d}}\mu : \mu \in {\varPi }\}$. The only increasing affine transformation of [a, b] with itself is the identity. Apply Proposition 1. $\square$

Proof of Lemma 1

F contains f(S) so that ${\varDelta }(F)$ contains $f({\varDelta }(S))$. The extreme points in ${\varDelta }(F)$ are the point masses $\delta _x$, $x \in F$. Because both ${\varDelta }(F)$ and $f({\varDelta }(S))$ are convex, it is sufficient to show that for every $\epsilon > 0$, there exists a probability in $f({\varDelta }(S))$ within $\rho$-distance of $\delta _x$. Since $x \in F$, there exists a sequence $s_n$ in S with $f(s_n) \rightarrow x$. Because singletons are measurable in S, ${\varDelta }(S)$ contains $\delta _{s_n}$, point mass on each $s_n$. Finally, $\rho (f(\delta _{s_n}),\delta _x) = \rho (\delta _{x_n},\delta _x) \rightarrow 0$. The second assertion is immediate. $\square$

Proof of Proposition 3

Let ${\varPi }$ be the canonical set of priors ${\varPi }^\circ \subset {\varDelta }([0, 1] \times [0, 1])$ given above, and let $Q \in {\varDelta }({\varPi })$ be the uniform distribution on $\{\lambda _x: x \in [0, 1]\}$. Fix a ${\mathfrak {p}} \in {\varDelta }({\varDelta }({\mathbb {X}}))$, and let $\varphi :[0, 1] \rightarrow {\varDelta }({\mathbb {X}})$ have the property that $\varphi (\lambda ) = {\mathfrak {p}}$. Define $f:[0, 1] \times [0, 1] \rightarrow {\mathbb {X}}$ by $f(x,y) = \mathbf{b }_{{\mathbb {X}}}(\varphi (x),y)$ where, again, $\mathbf{b }_{{\mathbb {X}}}:[0, 1] \times {\varDelta }({\mathbb {X}})\rightarrow {\mathbb {X}}$ is the Blackwell–Dubins function. When $\lambda _x$ is picked, $f(\lambda _x) = \varphi (x)$. Since Q picks the x according to $\lambda$, ${\mathfrak {p}}$ is the distribution of $f(\mu )$ when $\mu$ is distributed according to Q. $\square$