Consequentialism and dynamic consistency in updating ambiguous beliefs

Abstract

By proposing the notions of upper-constrained dynamic consistency and lower-constrained dynamic consistency that are weaker axioms than dynamic consistency, this paper axiomatizes the Dempster–Shafer updating rule and naive Bayes’ updating rule within the framework of Choquet expected utility. Based on the notion of conditional comonotonicity, this paper also provides an axiomatization of consequentialism under Choquet expected utility. Furthermore, based on the idea of the mean-preserving rule, this paper provides a unified approach for distinguishing capacity updating rules (the Dempster–Shafer updating rule, naive Bayes’ updating rule, and Fagin–Halpern updating rule) according to the degree of dynamic consistency.

Appendices

Appendix

Definition 9

An operator $I:{\mathbb {R}}^{\Omega } \rightarrow {\mathbb {R}}$ is comonotonic additive if $I(x+y)=I(x)+I(y)$ whenever x and y are comonotonic.

It is well known that Choquet integrals satisfy comonotonic additivity. Furthermore, Schmeidler (1986) shows that if an operator $I:{\mathbb {R}}^{\Omega }\rightarrow {\mathbb {R}}$ satisfies comonotonic additivity and monotonicity (i.e., $x\ge y$ on $\Omega$ implies $I(x)\ge I(y)$ for all $x,\,y\in {\mathbb {R}} ^{\Omega }$), then I is a Choquet integral. As mentioned in Schmeidler (1986), if an operator $I:{\mathbb {R}}^{\Omega }\rightarrow {\mathbb {R}}$ satisfies comonotonic additivity and monotonicity, then it satisfies positive homogeneity of degree 1 (that is, $I(tx)=tI(x)$ for all $t\ge 0$). Moreover, it is shown that if an operator $I:{\mathbb {R}} ^{\Omega }\rightarrow {\mathbb {R}}$ satisfies comonotonic additivity and positive homogeneity of degree 1, then I is a Choquet integral.

Definition 10

Choquet integrals with respect to a capacity v are said to be A-conditionally comonotonic additive if $\int _{\Omega }(x+y)\hbox {d}v=\int _{\Omega }x\hbox {d}v+\int _{\Omega }y\hbox {d}v$ whenever x and y are A-conditionally comonotonic.

For A-conditional comonotonicity, we can show the following proposition.

Proposition 2

Let $A \in {\mathcal {F}}$. Let v be a capacity such that $v(S\cap A)=v(S)$ for any $S \in {\mathcal {F}}$. Then, the Choquet integral of x with respect to such a capacity v, $\int _{\Omega } x \hbox {d}v$, is A-conditionally comonotonic additive. That is, $\int _{\Omega }(x+y)\hbox {d}v=\int _{\Omega }x\hbox {d}v+\int _{\Omega }y\hbox {d}v$ if x and y are A-conditionally comonotonic. Moreover, $\int _{\Omega }x\hbox {d}v=\int _{A}x|_{A}\hbox {d}v|_{A}$ where $x|_A$ and $v|_A$ are the restriction of a function x and the restriction of a capacity v on a set A, respectively.

Proof

Let $\langle \omega _{1},\omega _{2},\ldots ,\omega _{n} \rangle$ be a permutation of all the elements of $\Omega$ satisfying $x(\omega _{1})\ge x(\omega _{2})\ge \cdots \ge x(\omega _{n})$. Moreover, let $A=\{\omega _{i(1)},\omega _{i(2)},\ldots , \omega _{i(k)}\}$, where $\{i(1),i(2),\ldots ,i(k)\}\subseteq \{1,2,\ldots n\}$ and $i(1)<i(2)<\cdots <i(k)$ for $k=|A|$. Then,

$$\begin{aligned}&\int _{\Omega }x(\omega )\hbox {d}v \\&\quad =\sum _{1\le p\le n-1}(x(\omega _{p})-x(\omega _{p+1}))v(\{\omega _{1},\ldots ,\omega _{p}\})+x(\omega _{n})v(\Omega ) \\&\quad =\sum _{1\le p\le n-1}(x(\omega _{p})-x(\omega _{p+1}))v(\{\omega _{1},\ldots ,\omega _{p}\}\cap A)+x(\omega _{n})v(\Omega \cap A) \\&\quad =\sum _{1\le p<i(1)}(x(\omega _{p})-x(\omega _{p+1}))v(\emptyset ) \\&\qquad +\sum _{1\le q\le k-1}\left( \sum _{i(q)\le p<i(q+1)}(x(\omega _{p})-x(\omega _{p+1}))\right) v(\{\omega _{i(1)},\ldots ,\omega _{i(q)}\}) \\&\qquad +\left( \sum _{i(k)\le p\le n-1}(x(\omega _{p})-x(\omega _{p+1}))\right) v(A)+x(\omega _{n})v(A) \\&\quad =\sum _{1\le q\le k-1}(x(\omega _{i(q)})-x(\omega _{i(q+1)}))v(\{\omega _{i(1)},\ldots ,\omega _{i(q)}\})+x(\omega _{i(k)})v(A) \\&\quad =\int _{A}x|_A \hbox {d}v |_A. \end{aligned}$$

The last formula is the Choquet integral on A. Thus, the proposition is shown since the Choquet integral on A is A-conditionally comonotonic additive. $\square$

Proof of Proposition 1

Proof

Suppose that $f|_A=g|_A$. Then, by Proposition 2,

$$\begin{aligned} \int _{\Omega } u(f)\hbox {d}v_A =\int _A u(f|_A)\hbox {d}v_A |_A =\int _A u(g|_A)\hbox {d}v_A |_A=\int _{\Omega } u(g)\hbox {d}v_A. \end{aligned}$$

Next, we show the converse. Note that $1_S|_A=1_{S \cap A}|_A$. Thus, by consequentialism, $1_S \sim _A 1_{S \cap A}$. So, $v_A (S)=\int _{\Omega } 1_S d v_A=\int _{\Omega }1_{S \cap A} \hbox {d}v_A=v_A (S \cap A)$. $\square$

Proof of Theorem 2

Proof

(If part) We assume that there exist capacities v and $v_A$ on ${\mathcal {F}}$ and a non-constant affine real-valued function u on Y such that for all f and g in $L_{0}$, $f\succeq _{A}g\Leftrightarrow \;\int _{\Omega }u(f(\omega ))\hbox {d}v_{A}(\omega )\ge \int _{\Omega }u(g(\omega ))\hbox {d}v_{A}(\omega )$, where it holds that $v_{A}(S)=v_{A}(S\cap A)$ for all $S \in {\mathcal {F}}$.

Clearly, Axiom 5 holds since u is non-constant. Axiom 6 holds since u does not depend on a certain event A. Axioms 1, 3, and 4 hold by Theorem 1. Moreover, to show Axiom 7, note that $(y_{1},\{\omega \};y_{3},\Omega \backslash \{\omega \})=y_{1}1_{\{\omega \}}+y_{3}1_{\Omega \backslash \{\omega \}}$. Here, by $\omega \in A^{c}$, $1_{\{\omega \}}$ and $1_{\Omega \backslash \{\omega \}}$ are A-conditionally comonotonic. Therefore, by Proposition 2, $\int _{\Omega }u(y_{1},\{\omega \};y_{3},\Omega \backslash \{\omega \})\hbox {d}v_{A}=u(y_{1})v_{A}(\{\omega \})+u(y_{3})v_{A}(\Omega \backslash \{\omega \})$. Similarly, $\int _{\Omega }u(y_{2},\{\omega \};y_{3},\Omega \backslash \{\omega \})\hbox {d}v_{A}=u(y_{2})v_{A}(\{\omega \})+u(y_{3})v_{A}(\Omega \backslash \{\omega \})$. On the other hand, $v_{A}(\{\omega \})=v_{A}(\{\omega \}\cap A)=0$. Hence, Axiom 7 holds.

Finally, suppose that $f,\,g,\,h\in L_{0}$ are pairwise A-conditionally comonotonic and that $\int _{\Omega }u(f)\hbox {d}v_{A}\ge \int _{\Omega }u(g)\hbox {d}v_{A}$. Then,

$$\begin{aligned}&\int _{\Omega }u(\alpha f+(1-\alpha )h)\hbox {d}v_{A}-\int _{\Omega }u(\alpha g+(1-\alpha )h)\hbox {d}v_{A} \\&\quad =\alpha \int _{\Omega }u(f)\hbox {d}v_{A}+(1-\alpha )\int _{\Omega }u(h)\hbox {d}v_{A}-\alpha \int _{\Omega }u(g)\hbox {d}v_A-(1-\alpha )\int _{\Omega }u(h)\hbox {d}v_{A} \\&\quad =\alpha \left( \int _{\Omega }u(f)\hbox {d}v_{A}-\int _{\Omega }u(g)\hbox {d}v_{A} \right) \ge 0, \end{aligned}$$

from the A-conditionally comonotonic additivity of $v_{A}$ shown in Proposition 2. Thus, Axiom 2 holds. $\square$

Now, by showing Lemmas 4, 5, and 11–14, we prove the only if part of Theorem 2. In the following, if the proofs for the unconditional preference $\succeq$ are the same as those for the conditional preferences $\succeq _A$, we omit them.

Proof of Lemma 4

Proof

If $f,\,g\in L_{0}$ are comonotonic, then f and g are A-conditionally comonotonic. Thus, Axiom 2 implies Comonotonic Independence Axiom. By Theorem 1, Comonotonic Independence Axiom together with Axioms 1, 3, 4, 5 implies that there exist a capacity $v_{A}$ on ${\mathcal {F}}$ and an affine real-valued function $u_{A}$ on Y such that for all f and g in $L_{0}$,

$$\begin{aligned} f\succeq _{A}g\Leftrightarrow \;\int _{\Omega }u_{A}(f(\omega ))\hbox {d}v_{A}(\omega )\ge \int _{\Omega }u_{A}(g(\omega ))\hbox {d}v_{A}(\omega ). \end{aligned}$$

Moreover, by Axiom 6, for A and $\Omega$, we can show the existence of the common utility function u such that it represents $\succeq$ and $\succeq _A$ on $L_{c}$ since u is determined by the preference on $L_{c}$. Moreover, by Axiom 5, u is non-constant. $\square$

Proof of Lemma 5

Proof

We prove this lemma by mimicking the arguments of Schmeidler (1986,1989). Let $I(a)=\int _{\Omega }a(s)\hbox {d}v_{A}$. Let a, b$\in {\mathbb {R}} ^{\Omega }$ be A-conditionally comonotonic. By Axiom 2, for all pairwise A-conditionally comonotonic functions a, b, c, it holds that $I(a)\ge I(b)$ implies $I(\alpha a+(1-\alpha )c)\ge I(\alpha b+(1-\alpha )c)$. First, let us prove the following claim: If $x,y\in {\mathbb {R}} ^{\Omega }$ are A-conditionally comonotonic, then $I(\alpha x+(1-\alpha )y)=\alpha I(x)+(1-\alpha )I(y)$ for all $\alpha \in [0,1]$.

Indeed, for any $\varepsilon >0$, $(I(x)+\varepsilon )1_{\Omega }$ satisfies $I((I(x)+\varepsilon )1_{\Omega })>I(x)$ and $\ (I(y)+\varepsilon )1_{\Omega }$ satisfies $I((I(y)+\varepsilon )1_{\Omega })>I(y)$ since $I(\lambda 1_{\Omega })=\lambda$. Hence,

$$\begin{aligned}&\alpha I(x)+(1-\alpha )I(y)+\varepsilon \\&\quad = I(\alpha (I(x)+\varepsilon )1_{\Omega }+(1-\alpha )(I(y)+\varepsilon )1_{\Omega }) \\&\quad>I(\alpha x+(1-\alpha )(I(y)+\varepsilon )1_{\Omega }) \\&\quad >I(\alpha x+(1-\alpha )y). \end{aligned}$$

The first inequality holds since $\alpha (I(x)+\varepsilon )1_{\Omega }$, x, and $(1-\alpha )(I(y)+\varepsilon )1_{\Omega }$ are pairwise A-conditionally comonotonic. The second inequality holds since x, y, and $(1-\alpha )(I(y)+\varepsilon )1_{\Omega }$ are pairwise A-conditionally comonotonic.

Since $\varepsilon$ is any positive number, we obtain that

$$\begin{aligned} \alpha I(x)+(1-\alpha )I(y)\ge I(\alpha x+(1-\alpha )y). \end{aligned}$$

Furthermore, the converse inequality can be shown by using a similar argument for $\varepsilon <0$. Therefore, it is proved that

$$\begin{aligned} I(\alpha x+(1-\alpha )y)=\alpha I(x)+(1-\alpha )I(y). \end{aligned}$$

Then, our claim is proved.

Next, let us use this claim twice. First, let $\alpha =1/2$, $x=2a$, and $y=0$. Then, $I(a)=(1/2)I(2a)$. Similarly, let $\alpha =1/2$, $x=0$, and $y=2b$. Then, $I(b)=(1/2)I(2b)$. Second, let $\alpha =1/2$, $x=2a$, and $y=2b$. Then,

$$\begin{aligned} I(a+b)=\frac{1}{2}I(2a)+\frac{1}{2}I(2b)=I(a)+I(b). \end{aligned}$$

$\square$

Lemma 11

Let $v_A$ be the capacity in Lemma 4. Then, $v_{A}(S)=v_{A}(S\cap A)+v_{A}(S\cap A^{c})$ for all $S\in {\mathcal {F}}$.

Proof

Note that $1_{S\cap A}$ and $1_{S\cap A^{c}}$ are A-conditionally comonotonic. By Lemma 5, it holds that $\int _{\Omega }1_{S\cap A}+1_{S\cap A^{c}}\hbox {d}v_{A}=\int _{\Omega }1_{S\cap A}\hbox {d}v_{A}+\int _{\Omega }1_{S\cap A^{c}}\hbox {d}v_{A}$. This equation implies that $v_{A}(S)=v_{A}(S\cap A)+v_{A}(S\cap A^{c})$. $\square$

Lemma 12

Let $v_A$ be the capacity in Lemma 4. Then, $v_A(S)=\sum _{\omega \in S} v_A(\{\omega \})$ for every S with $S \subseteq A^c$.

Proof

Let $T_1 \subseteq A^c$ and $T_2 \subseteq A^c$ satisfy $T_1 \cap T_2 = \emptyset$. Then, $1_{T_1}$ and $1_{T_2}$ are A-conditionally comonotonic. Hence, by Lemma 5, $\int (1_{T_1} +1_{T_2})\hbox {d}v_A=\int 1_{T_1}\hbox {d}v_A+\int 1_{T_2}\hbox {d}v_A$, which implies $v_A(T_1 \cup T_2)=v_A(T_1)+v_A(T_2)$. Therefore, for every S with $S \subseteq A^c$, it holds that $v_A(S)=\sum _{\omega \in S} v_A(\{\omega \})$ since $\Omega$ is finite. $\square$

Lemma 13

Let $v_A$ be the capacity in Lemma 4. Then, $v_{A}(S)=0$ for every $S \subseteq A^c$.

Proof

Take an arbitrary y with $y^{*}\succ _{A}y\succ _{A}y_{*}$. Indeed, such $y^*$, y, and $y_*$ exist by A3, A5, and A6. And let $f=(y^{*},\{\omega \};y_{*},\Omega \backslash \{\omega \})$, $g=(y,\{\omega \};y_{*},\Omega \backslash \{\omega \})$. By A7, it holds that $f\sim _{A}g$. Hence, $\int _{\Omega }u(f(\omega ))\hbox {d}v_{A}(\omega )=\int _{\Omega }u(g(\omega ))\hbox {d}v_{A}(\omega )$, which leads that $(u(y^{*})-u(y_{*}))v_{A}(\{\omega \})+u(y_{*})=(u(y)-u(y_{*}))v_{A}(\{\omega \})+u(y_{*})$. Therefore, $v_{A}(\{\omega \})=0$ since $u(y^{*})-u(y)>0$. This result together with Lemma 12 proves the claim. $\square$

The following lemma is shown from Lemmas 11, 12, and 13, immediately.

Lemma 14

Let $v_A$ be the capacity in Lemma 4. Then, $v_{A}(S)=v_{A}(S\cap A)$ for all $S\in {\mathcal {F}}$.

Proof of Lemma 6

Proof

Let $f, g \in L_0$ and $A \in {\mathcal {F}} \backslash \{\Omega ,\emptyset \}$ with $v(A^c) \ne 1$. For ease of notation, let $\Omega _1 =\Omega ^{UC}(f,g;A)$. Note that for all $\omega \in \Omega _1$ and for all $\omega ' \in \Omega _1 ^c$, $f(\omega ) \succeq f(\omega ')$ and $g(\omega ) \succeq g(\omega ')$. Let $\langle \omega _1 ^f, \omega _2 ^f, \ldots , \omega _n ^f \rangle$ be a permutation of $\Omega$ satisfying $u(f(\omega _1 ^f)) \ge u(f(\omega _2 ^f)) \ge \cdots \ge u(f(\omega _n ^f))$. Similarly, let $\langle \omega _1 ^g, \omega _2 ^g, \ldots , \omega _n ^g \rangle$ be a permutation of $\Omega$ satisfying $u(g(\omega _1 ^g)) \ge u(g(\omega _2 ^g)) \ge \cdots \ge u(g(\omega _n ^g))$. Let $\Omega _1=\{\omega _1 ^f, \ldots , \omega _k ^f \}=\{\omega _1 ^g, \ldots , \omega _k ^g \}$. Then, by Axiom 8, it is possible to set $\omega _i ^f=\omega _i ^g$ for all $i \in \{1,\ldots ,k\}$.

Note that

$$\begin{aligned} \int _{\Omega } u(f) \hbox {d}v_A ^{DS}= & {} \sum _{i=1}^n (u(f(\omega _i ^f))-u(f(\omega _{i+1} ^f)))v_A^{DS}(\{\omega _1^ f, \ldots , \omega _i ^f \}) \\ \text {and}\,\,\int _{\Omega } u(g) \hbox {d}v_A ^{DS}= & {} \sum _{i=1}^n (u(g(\omega _i ^g))-u(g(\omega _{i+1} ^g)))v_A^{DS}(\{\omega _1^g, \ldots , \omega _i ^g \}), \end{aligned}$$

where $u(f(\omega _{n+1} ^f)) = u(g(\omega _{n+1} ^g))=0$. Then, it follows that

$$\begin{aligned}&f \succeq _A g \Leftrightarrow \int _{\Omega } u(f) \hbox {d}v_A ^{DS} \ge \int _{\Omega } u(g) \hbox {d}v_A ^{DS} \\&\quad \Leftrightarrow \sum _{i=1}^n (u(f(\omega _i ^f))-u(f(\omega _{i+1} ^f)))(v(\{\omega _1^ f, \ldots , \omega _i ^f \} \cup A^c)-v(A^c)) \\&\quad \ge \sum _{i=1}^n (u(g(\omega _i ^g))-u(g(\omega _{i+1} ^g)))(v(\{\omega _1^g, \ldots , \omega _i ^g \} \cup A^c)-v(A^c)) \\&\quad \Leftrightarrow \sum _{i=1}^n (u(f(\omega _i ^f))-u(f(\omega _{i+1} ^f)))v(\{\omega _1^ f, \ldots , \omega _i ^f \} \cup A^c)- u(f(\omega _1^f))v(A^c) \\&\quad \ge \sum _{i=1}^n (u(g(\omega _i ^g))-u(g(\omega _{i+1} ^g)))v(\{\omega _1^g, \ldots , \omega _i ^g \} \cup A^c)-u(g(\omega _1 ^g))v(A^c), \end{aligned}$$

where the second equivalence holds by telescoping sums. Furthermore, by Axiom 8, the summation from $i=1$ to $k-1$ of the first term on the left-hand side of the inequality is equal to that on the right-hand side of the inequality and $u(f(\omega _1^f))v(A^c)=u(g(\omega _1^g))v(A^c)$. Moreover, by $A^c \subseteq \{\omega _1^f,\ldots ,\omega _k^f\}=\{\omega _1^g,\ldots ,\omega _k^g\}$, it follows that

$$\begin{aligned}&f \succeq _A g \\&\quad \Leftrightarrow \sum _{i=k}^n (u(f(\omega _i ^f))-u(f(\omega _{i+1} ^f)))v(\{\omega _1^ f, \ldots , \omega _i ^f \}) \\&\quad \ge \sum _{i=k}^n (u(g(\omega _i ^g))-u(g(\omega _{i+1} ^g)))v(\{\omega _1^g, \ldots , \omega _i ^g \}). \end{aligned}$$

Note that by Axiom 8,

$$\begin{aligned}&\sum _{i=1}^{k-1} (u(f(\omega _i ^f))-u(f(\omega _{i+1} ^f)))v(\{\omega _1^ f, \ldots , \omega _i ^f \})\\&\quad =\sum _{i=1}^{k-1} (u(g(\omega _i ^g))-u(g(\omega _{i+1} ^g)))v(\{\omega _1^g, \ldots , \omega _i ^g \}). \end{aligned}$$

By adding this to both sides of the above inequality, it follows that

$$\begin{aligned} f \succeq _A g \Leftrightarrow \int _{\Omega } u(f) \hbox {d}v \ge \int _{\Omega } u(g) \hbox {d}v \Leftrightarrow f \succeq g. \end{aligned}$$

$\square$