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Purely subjective extended Bayesian models with Knightian unambiguity

Abstract

This paper provides a model of belief representation in which ambiguity and unambiguity are endogenously distinguished in a purely subjective setting where objects of choices are, as usual, maps from states to consequences. Specifically, I first extend the maxmin expected utility theory and get a representation of beliefs such that the probabilistic beliefs over each ambiguous event are represented by a non-degenerate interval, while the ones over each unambiguous event are represented by a number. I then consider a class of the biseparable preferences. Two representation results are achieved and can be used to identify the unambiguity in the context of the biseparable preferences. Finally a subjective definition of ambiguity is suggested. It provides a choice theoretic foundation for the Knightian distinction between ambiguity and unambiguity.

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Fig. 1

Notes

  1. 1.

    This distinction goes by many names: risk versus uncertainty (Knight 1921); unambiguity versus ambiguity (Ellsberg 1961); precise versus vague probability (Savage 1972), and so forth.

  2. 2.

    See, for example Halevy (2007), Machina (2009) and Placido et al. (2011).

  3. 3.

    A good reference about biseparable preference is Luce (2000).

  4. 4.

    The general criticism of Epstein–Zhang’s definition is in Wakker (2008).

  5. 5.

    Kopylov (2007) discusses another possible structure, called mosaic, of collection of unambiguous events.

  6. 6.

    The comparison of different approach is discussed in Alon and Schmeidler (2014).

  7. 7.

    SEU representation result under our setting is introduced in Kobberling and Wakker (2003).

  8. 8.

    The intense discussion appears in Ghirardato and Marinacci (2001).

  9. 9.

    Note the above theorem will hold if we replace A5\(^*\) by binary comonotonic tradeoff consistency and the A9 by binary tradeoff consistency. This is because as Kobberling and Wakker (2003) show that with weak order, continuity and monotonicity assumptions, A5\(^*\) and A9 imply binary comonotonic tradeoff consistency and binary tradeoff consistency, respectively. We adopt traditional version of Bisymmetry to make the connection with Alon and Schmeidler (2014) and Ghirardato and Marinacci (2001) clear.

  10. 10.

    Notice that the midpoint operation \(\langle x;y\rangle \sim ^*\langle y;z\rangle \) defined in (4) is also valid here. The technique tool we provide here is consistent with those used in Ghirardato and Marinacci (2001) and Alon and Schmeidler (2014), which will make the connection transparent.

  11. 11.

    Another view about complementary additivity appears at Wakker (2004). He claims that many departures from ambiguity neutrality are induced by ambiguity insensitivity, not necessarily ambiguity aversion. He further concludes that complementary additivity is actually ambiguous insensitive.

References

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Acknowledgments

I am deeply indebted to David Schmeidler for inspiration, guidance, and support. I am also thankful for Yaron Azieli, Chew Soo Hong, and Ani Guerdjikova for helpful discussion. This research was supported by ANR and Labex.

Author information

Correspondence to Xiangyu Qu.

Appendix

Appendix

Proof of Section 3

We first state two facts without proofs.

Fact 1

Axioms 1–3 imply that for every \(f\in \mathcal {F}\), there exists \(x\in X\) such that \(f\sim \bar{x}\).

Fact 2

Axioms 1, 3, and 4 imply that there exist two consequences \(x,y\) such that \(x\succ y\).

Proof of Theorem 1: (i) implies (ii)

We first show that Axiom 7 (unambiguity independence) implies Certainty independence, which is introduced by Alon and Schmeidler (2014). Therefore, by Theorem 7 in them, the preference relation \(\succsim \) is a MEU preference. Then we show that all the probabilities have to agree on the set of unambiguous events.

Recall the Axiom of certainty independence:

Axiom 7 \({}^\prime \) Suppose that a pair of acts \(f,g\) and consequences \(x,a,w\) satisfy \(\langle x;a\rangle \sim ^*\langle a;w\rangle \), and \(\langle f(s);g(s)\rangle \sim ^*\langle g(s);w\rangle \) for all \(s\in S\). Then \(g\sim \bar{a}\Leftrightarrow f\sim \bar{x}\).

Lemma 1

Axiom 7 implies Axiom 7\(^{\prime }\).

Proof

Suppose Axiom 7 holds. Let \(x=y\) and \(z=w\). Then first condition in Axiom 7 becomes \(\langle f(s);g(s)\rangle \sim ^*\langle g(s); w\rangle \) for all \(s\in S\), and the second condition becomes \(\langle x;a\rangle \sim ^*\langle a; w\rangle \), which are the assumption on the Axiom \(7'\). Then by (i) in Axiom 7, we have \(g\sim \bar{a}\Leftrightarrow f\sim \bar{x}\).\(\square \)

Proposition 1

(Proposition 12 from Alon and Schmeidler 2014) Suppose \(\succsim \) satisfies Axioms 1–5. Then the following is true:

  1. (1)

    there exists a continuous function \(u:X\rightarrow \mathbb {R}\) such that for all \(x,y\in X, x\succsim y\Leftrightarrow u(x)\ge u(y)\). Furthermore, \(u\) is unique up to a positive linear transformation.

  2. (2)

    \(\langle a; b\rangle \sim ^*\langle c;d\rangle \) implies \(u(a)-u(b)=u(c)-u(d)\).

  3. (3)

    there exists a unique continuous representation function \(V_c\) on \(\mathcal {F}\) such that \(J(\bar{x}=u(x)\) for all \(x\in X\).

The lemma below is an analog to Lemma 3.3 from Gilboa and Schmeidler (1989), and follows the Lemma 23 and other results from Alon and Schmeidler (2014). Denoted by \(B\) is the set of all real-valued functions on \(S\). For any \(\gamma \in \mathbb {R}, \bar{gamma}\in B\) is the constant function.

Lemma 2

There exists a functional \(I: B\rightarrow \mathbb {R}\) such that

  1. (i)

    For any \(f\in \mathcal {F}, I(u\circ f)=V_c(f)\).

  2. (ii)

    \(I\) is monotonic, i.e., for any \(a,b\in B, a\ge b\Leftarrow I(a)\ge I(b)\).

  3. (iii)

    \(I\) is superadditive and homogeneous of degree \(1\).

  4. (iv)

    \(I\) satisfies certainty independence: for any \(a\in B\) and constant function \(\bar{\gamma }, I(a+\bar{\gamma })=I(a)+\gamma \).

The next lemma will show that the functional \(I\) satisfies unambiguity independence: \(I\) is additive on an arbitrary act and an unambiguous act. Note that for any \(\beta ,\gamma \in \mathbb {R}\) and any \(E\in \Sigma , \beta 1_E+\gamma 1_{E^\mathrm{c}}\) is a binary function.

Lemma 3

\(I\) satisfies unambiguity independence: for any \(E\in \mathcal {U}\), any \(a\in B\), and any binary function \(\beta 1_E+\gamma 1_{E^\mathrm{c}}, I(a+\beta 1_E+\gamma 1_{E^\mathrm{c}})=I(a)+\beta I(1_E)+\gamma I(1_{E^\mathrm{c}})\).

Proof

Since \(I\) satisfies homogeneous of degree one and certainty independence, to show unambiguity independence it suffices to prove that \(I(a+1_E)=I(a)+I(1_E)\).

Let \(f\) be in \(\mathcal {F}\). Let \(z.w\in X\) be such that \(u\circ (zEw)=1_E\). According to Fact 1, there exists \(\xi \in X\) such that \(\bar{\xi }\sim zEw\). Therefore \(I(1_E)=u(\xi )\).

Let \(\eta ,x,y\in X\) be such that \(u(\eta )=\frac{1}{2}[u(x)+u(z)]\) and \(u(\eta )=\frac{1}{2}[u(y)+u(w)]\). Notice that such construction is always possible: we let \(x=w\) and \(y=z\), and by continuity of \(u\), there is a \(\eta \) such that \(u(\eta )=\frac{1}{2}[u(w)+u(z)\).

By Fact 1 again, there exists \(\psi \in X\) such that \(\bar{\psi }\sim xEy\). So \(I(u\circ (xEy))=u(\psi )\). Therefore by (ii) of Axiom 7, we have

$$\begin{aligned} u(\eta )=\frac{1}{2}[u(\xi )+u(\psi )] \end{aligned}$$
(8)

Now we construct an act \(g\) by,

$$\begin{aligned} u(g(s))=\left\{ \begin{array}{l@{\quad }l} \textstyle \frac{1}{2}[u(f(s))+u(z)], &{} \hbox {if } x\in E, \\ \textstyle \frac{1}{2}[u(f(s))+u(w)], &{} \hbox {if } x\in E^\mathrm{c}.\\ \end{array} \right. \end{aligned}$$

The existence of \(g\) is guaranteed by the continuity of \(u\). Therefore, \(I(u\circ g)=I(\frac{1}{2}(u\circ f+u\circ (zEw)))\). So \(f\sim xEy\) and \(\bar{\psi }\sim xEy\) imply \(f\sim \bar{\psi }\). Hence \(I(u\circ f)=u(\psi )\). Axiom 7 implies \(g\sim \bar{\eta }\). Thus,

$$\begin{aligned} I\left( \frac{1}{2} u\circ f+\frac{1}{2}u\circ (zEw)\right) =\frac{1}{2}I(u\circ f+1_E)=\frac{1}{2}[u(\xi )+u(\psi )] \end{aligned}$$

That is, \(I(u\circ f+1_E)=I(u\circ f)+I(1_E)\).\(\square \)

Lemma 4

(Theorem 7 of Alon and Schmeidler 2014) \(\succsim \) on \(\mathcal {F}\) is represented by a MEU functional.

Lemma 5

For every \(E\in \mathcal {U}\) and every \(P,Q\in K\), we have \(P(E)=Q(E)\).

Proof

Let \(E\) be in \(\mathcal {U}\). By lemma 4, \(I(1_E)=\min \{\int 1_E\mathrm {d}P| P\in K\}=\widetilde{P}(E)\) and \(I(1_{E^\mathrm{c}})=\min \{\int 1_{E^\mathrm{c}}\mathrm {d}P|P\in K\}=\widehat{P}(E^\mathrm{c})\). So for every \(Q\in K\), we must have \(Q(E)\ge \widetilde{P}(E)\) and \(Q(E^\mathrm{c})\ge \widehat{P}(E^\mathrm{c})\). Hence \(\widetilde{P}(E)\le Q(E)\le \widehat{P}(E)\). Since \(I\) is unambiguity independent, \(I(1)=I(1_E)+I(1_{E^\mathrm{c}})\). That is \(\widetilde{P}(E)+\widehat{P}(E^\mathrm{c})=1\). So \(\widetilde{P}(E)=\widehat{P}(E)\). Thus, for any \(P,Q\in K\), we must have \(P(E)=Q(E)\) for all \(E\in \mathcal {U}\).\(\square \)

Proof of Theorem 1: (ii) implies (i)

We first define a functional \(I: B\rightarrow \mathbb {R}\) by \(I(b)=\min _{P\in K}\{\int _S bdP\}\). For any \(E\in \mathcal {U}\), and any \(P,Q\in K, P(E)=Q(E)\). We denote \(P^*(E)\) the probability of \(E\) on \(K\). Hence the preference relation is represented by: for any \(f\in \mathcal {F}, V_c(f)=I(u\circ f)\), and for any \(xEy\in \mathcal {F}^u, V_c(xEy)=u(x)P^*(E)+u(y)P^*(E^\mathrm{c})\).

By Theorem 7 of Alon and Schmeidler (2014), (ii) implies Axioms 1,2,3,4,5,6, and 8. For the rest, we only need to show that Axiom 7 (Unambiguity independence) holds. If \(E\in \mathcal {U}\) and \(b\in B\), then \(I(b+1_E)=I(b)+I(1_E)\). If \(\alpha >0\), then \(I(\alpha b)=\alpha I(b)\). Let \(E\in \mathcal {U}\), and \(f,g\in \mathcal {F}\), and \(x,y,z,w,a,b,c\in X\) satisfy the following conditions:

  1. (a)

    \(\langle f(s),g(s)\rangle \sim ^*\langle g(s),z\rangle \) for all \(s\in E\) and \(\langle f(s),g(s)\rangle \sim ^*\langle g(s),w\rangle \) for all \(s\in E^\mathrm{c}\);

  2. (b)

    \(\langle x,a\rangle \sim ^*\langle a,z\rangle \) and \(\langle y,a\rangle \sim ^*\langle a,w\rangle \)

According to the representation, condition (a) implies that

$$\begin{aligned} u(g(s))=\left\{ \begin{array}{l@{\quad }l} \textstyle \frac{1}{2}[u(f(s)+u(z)], &{} \hbox {if }s\in E; \\ \textstyle \frac{1}{2}[u(f(s)+u(w)], &{} \hbox {if }s\in E^\mathrm{c}. \\ \end{array} \right. \end{aligned}$$

And condition (b) implies that

$$\begin{aligned} u(a)=\frac{1}{2}[u(x)+u(z)]\quad \text { and }\quad u(a)=\frac{1}{2}[u(y)+u(w)] \end{aligned}$$

Therefore, by condition (a) we have:

$$\begin{aligned} V_c(g)&= I(u\circ g)\\&= I\left( \frac{1}{2}[u\circ f+u(z)]1_E+\frac{1}{2}[u\circ f+ u\circ (w)]1_{E^\mathrm{c}}\right. \\&= I\left( \frac{1}{2} u\circ f+\frac{1}{2}[u(z)1_E+u(w)1_{E^\mathrm{c}}]\right) \\&= \left. \frac{1}{2}I(u\circ f)+\frac{1}{2}[u(z)I(1_E)+u(w)I(1_{E^\mathrm{c}})]\right) \\&= \frac{1}{2}V_c(f)+\frac{1}{2}\left[ u(z)P^*(E)+u(w)P^*(E^\mathrm{c})\right] \end{aligned}$$

Also by condition (b) we have

$$\begin{aligned} U(a)=\frac{1}{2}\left[ u(x)P^*(E)+u(y)P^*(E^\mathrm{c})]+\frac{1}{2}[u(z)P^*(E)+u(w)P^*(E^\mathrm{c})\right] \end{aligned}$$

Hence \(g\sim \bar{a}\) iff

$$\begin{aligned} \frac{1}{2}V_c(f)+\frac{1}{2}\left[ u(z)P^*(E)+u(w)P^*(E^\mathrm{c})\right]&= \frac{1}{2}\left[ u(x)P^*(E)+u(y)P^*(E^\mathrm{c})\right] \\&\quad +\frac{1}{2}\left[ u(z)P^*(E)+u(w)P^*(E^\mathrm{c})\right] \end{aligned}$$

That is \(V_c(f)=u(x)P^*(E)+u(y)P^*(E^\mathrm{c})\). Therefore \(f\sim xEy\).

Now we assume that \(b\sim zEw\) and \(c\sim xEy\). Then we have

$$\begin{aligned} u(b)=u(z)P^*(E)+u(w)P(E^\mathrm{c})\quad \text { and }\quad u(c)=u(x)P^*(E)+u(y)P(E^*) \end{aligned}$$

By condition (b), \(\frac{1}{2}[u(b)+u(c)]=u(a)\). Hence \(\langle c;a\rangle \sim ^*\langle a;b\rangle \). Therefore, Axiom 7 (unambiguity independence) is satisfied.

Proof of Section 4

We begin by listing a theorem of Ghirardato and Marinacci (2001) a là Alon and Schmeidler (2014).

Theorem 5

Suppose that structure assumptions on \(S\) and \(X\) hold. The following statements are equivalent for \(\succsim \) on \(\mathcal {F}\):

  1. (i)

    \(\succsim \) satisfies Axioms 1,2,3,4\(^*\),4\(^{**}\), and \(5^*\).

  2. (ii)

    \(\succsim \) is a biseparable preference with representation as in (5).

Proof of Theorem 2: (i) \(\Rightarrow \) (ii)

By Theorem 5, \(\succsim \) is a biseparable preference. We only need to show for each \(E\in \mathcal {U}, \rho (E)+\rho (E^\mathrm{c})=1\). Let \(x,y,z\) be consequences such that \(x\succ y\succ z\). By Axiom 9 and Theorem 2,

$$\begin{aligned} u(\Psi _E(\Psi _E(x,z),\Psi _E(y,y)))=u(\Psi _E(\Psi _E(x,y),\Psi _E(z,y))) \end{aligned}$$
(9)

Note that \(\Psi _E(y,y)=y\). Since \(\succsim \) satisfies monotonicity, we have \(\Psi _E(x,y)\succ \Psi _E(z,y)\). Therefore (5) implies:

$$\begin{aligned}&u\big (\Psi _E(\Psi _E(x,y),\Psi _E(z,y))\big )\\&\quad = u(\Psi _E(x,y))\rho (E)+u(\Psi _E(z,y))(1-\rho (E)) \\&\quad = \big [u(x)\rho (E)+u(y)(1-\rho (E))\big ] \rho (E)+\big [u(z)(1-\rho (E^\mathrm{c}))+u(y)\rho (E^\mathrm{c})\big ] (1-\rho (E))\\&\quad = u(x)\rho (E)^2+u(y)\big (\rho (E)+\rho (E^\mathrm{c})-\rho (E)^2-\rho (E)\rho (E^\mathrm{c})\big )\\&\qquad +u(z)(1-\rho (E))(1-\rho (E^\mathrm{c})) \end{aligned}$$

Now consider the left of (9). There are two cases need to be discussed: \(\Psi _E(x,z)\succsim y\) and \(y\succsim \Psi _E(x,z)\).

Case 1 \(\Psi _E(x,y)\succsim y\).

Since \(\succsim \) is a biseparable preference, by (5) we have:

$$\begin{aligned} u\big (\Psi _E(\Psi _E(x,z),y)\big )&= u(\Psi _E(x,z))\rho (E)+u(y)(1-\rho (E)) \nonumber \\&= u(x)\rho (E)^2+u(y)(1-\rho (E))+u(z)\rho (E)(1-\rho (E)) \end{aligned}$$
(10)

Because of (9), we get:

$$\begin{aligned}&u(x)\rho (E)^2+u(y)(1-\rho (E))+u(z)\rho (E)(1-\rho (E))\\&\quad = u(x)\rho (E)^2+u(y)\big (\rho (E)+\rho (E^\mathrm{c})\\&\qquad -\rho (E)^2-\rho (E)\rho (E^\mathrm{c})\big )+u(z)(1-\rho (E))(1-\rho (E^\mathrm{c})), \end{aligned}$$

which yields \(\big (u(y)-u(z)\big )\big (1-\rho (E)\big )\big (1-\rho (E)-\rho (E^\mathrm{c})\big )=0\). Since \(u(y)>u(z)\) and \(E\) is essential, we must have \(1-\rho (E)-\rho (E^\mathrm{c})=0\). Hence \(\rho (E)+\rho (E^\mathrm{c})=1\).

Case 2 \(y\succsim \Psi (xEy)\).

Similarly, by (5), we have:

$$\begin{aligned}&u\big (\Psi _E(\Psi _E(x,z),y)\big )= u(\Psi _E(x,z))(1-\rho (E^\mathrm{c}))+u(y)\rho (E^\mathrm{c}) \\&\quad =u(x)\rho (E)(1-\rho (E^\mathrm{c}))+u(y)\rho (E^\mathrm{c})+u(z)\rho (E)(1-\rho (E))(1-\rho (E^\mathrm{c})), \end{aligned}$$

Again, because of (9), we get:

$$\begin{aligned}&u(x)\rho (E)^2+u(y)(1-\rho (E))+u(z)\rho (E)(1-\rho (E))\\&\quad =u(x)\rho (E)(1-\rho (E^\mathrm{c}))+u(y)\rho (E^\mathrm{c})+u(z)\rho (E)(1-\rho (E))(1-\rho (E^\mathrm{c})), \end{aligned}$$

which yields \(\big (u(x)-u(y)\big )\big (1-\rho (E)-\rho (E^\mathrm{c})\big )\). Since \(u(x)>u(y)\) and \(E\) is essential, we must have \(1-\rho (E)-\rho (E^\mathrm{c})=0\). Hence \(\rho (E)+\rho (E^\mathrm{c})=1\).

Proof of Theorem 2: (ii) \(\Rightarrow \) (i)

Because of Theorem 5, we only need to show that Axiom 9 holds. Let \(E\) be in \(\mathcal {U}\) and \(x,x',y,y'\) be in \(X\). Since \(E\) is unambiguous event, we know that \(\rho (E)+\rho (E^\mathrm{c})=1\). So

$$\begin{aligned} u\big (\Psi _E(\Psi _E(x,x'),\Psi _E(y,y'))\big )&= u(\Psi _E(x,x'))\rho (E))+u(\Psi _E(y,y'))\rho (E^\mathrm{c}) \\&= u(x)\rho (E)^2+u(x')\rho (E)\rho (E^\mathrm{c})\\&\quad +\,u(y)\rho (E)\rho (E^\mathrm{c})+u(y')\rho (E^\mathrm{c})^2 \end{aligned}$$

and

$$\begin{aligned} u\big (\Psi _E(\Psi _E(x,y),\Psi _E(x',y'))\big )&= u(\Psi _E(x,y))\rho (E))+u(\Psi _E(x',y'))\rho (E^\mathrm{c}) \\&= u(x)\rho (E)^2+u(x')\rho (E)\rho (E^\mathrm{c})\\&\quad +\,u(y)\rho (E)\rho (E^\mathrm{c})+u(y')\rho (E^\mathrm{c})^2, \end{aligned}$$

Therefore, \(u\big (\Psi _E(\Psi _E(x,x'),\Psi _E(y,y'))\big )=u\big (\Psi _E(\Psi _E(x,y),\Psi _E(x',y'))\big )\). Hence \(\Psi _E(\Psi _E(x,x'),\Psi _E(y,y'))\sim \Psi _E(\Psi _E(x,y),\Psi _E(x',y'))\) and Axiom 9 holds.

Proof of Theorem 3: (i) \(\Rightarrow \) (ii)

Suppose that \(\succsim \) satisfies condition (i) in Theorem 3. Note that Axiom 10 implies Axiom 9. We are left to show that for all \(f\in \mathcal {F}, V(f)=\int _S u(f(s))\mathrm {d}P\) where \(P\) is a probability measure on \(\mathcal {U}\). Let \(P:\mathcal {U}\rightarrow [0,1]\) be defined by \(P(E)=\rho (E)\) for all \(E\in \mathcal {U}\). We first show that \(P\) is a probability measure on \(\mathcal {U}\).

$$\begin{aligned} \Psi _G(x,x,y)&\sim \Psi _G(\Psi _A(z,z'),\Psi _A(w,w'),\Psi _A(w,z'))\nonumber \\&\sim \Psi _G(z,w,w)A\Psi _G(z',w',z') \end{aligned}$$
(11)

Since \(u(x)=\rho (A)u(z)+(1-\rho (A))u(z')\), and \(u(x)=\rho (A)u(w)+(1-\rho (A))u(w')\) and \(u(y)=\rho (A)u(w)+(1-\rho (A))u(z')\), we have

$$\begin{aligned}&u(x)P(E\cup F)+ u(y)(1-P(E\cup F))=[u(z)P(E)+u(w)(1-P(E))] \rho (A)\\&\qquad +\,[u(z')(1-P(F))+u(w')P(F)](1-\rho (A))\\&\quad =[(u(z)-u(w))P(E)+u(w)]\rho (A)+[(u(w')-u(z'))P(F) +u(z')](1-\rho (A)) \end{aligned}$$

According to definition of \(u(y)\), we can simplify the above further as

$$\begin{aligned}{}[u(x)-u(y)]P(E\cup F)=[u(z)-u(w)]P(E)\rho (A) +[u(w')-u(z')]P(F)(1-\rho (A)) \end{aligned}$$

Since \(x\succ y\) and \(u(x)-u(y)=\rho (A)[u(z)-u(w)]=(1-\rho (A))[u(w')-u(z')]\), we can conclude that \(P(E\cup F)=P(E)+P(F)\). Hence \(P\) is a probability measure on \(\mathcal {U}\).

Let \(G=\{E_1,\ldots ,E_n\}\) be a ground partition and \(f_n=x_1E_1\ldots x_nE_n\). For \(n=2, f_2\in \mathcal {F^U}\) and Theorem 2 gives the desired result. For using the mathematical induction, we thus assume the conclusion is true for all \(n-1\) ground partition. Then it suffices to show that for \(n\) ground partition

$$\begin{aligned} V(f_n)=\sum ^n_{i=1}u(x_i)P(E_i) \end{aligned}$$

where \(P:\mathcal {U}\rightarrow [0,1]\) is such that \(P(E)=\rho (E)\) for all \(E\in \mathcal {U}\).

If \(x_i\sim x_j\) for some \(1\le i<j\le n\), then the desired result easily follows from \(E_i\cup E_j\in \mathcal {U}\) by \(\lambda \)-system assumption of \(\mathcal {U}\), and the hypothesis of the induction. Let \(G'=\{E_1,\ldots ,E_{n-2},E_{n-1}\cup E_n\}\) and \(G''=\{E_1\cup E_2, E_3,\ldots ,E_n\}\). So \(G'\) and \(G''\) are \(n-1\) ground partition since \(\mathcal {U}\) is a \(\lambda \)-system. Without loss of generality, we assume that \(x_1\succ \ldots x_n\).

Below we have three cases to discuss:

Case 1 There is some \(x,y\in X\) and \(E\in \Sigma \) such that \(xEx_2\sim x_1\) and \(x_{n-1}Ey\sim x_n\).

It follows from Axiom 10 and monotonicity that

$$\begin{aligned} f_n&\sim \Psi _G(x_1,\ldots ,x_n)\\&\sim \Psi _G(\Psi _E(x,x_2),\Psi _E(x_2,x_2),\ldots ,\Psi _E(x_{n-1},x{n-1}),\Psi _E(x_{n-1},u)\\&\sim \Psi _G(x,x_2,\ldots ,x_{n-1},x_{n-1})E\Psi _G(x_2,x_2,\ldots ,x_{n-1},y)\\&\sim \Psi _{G'}(x,x_2,\ldots ,x_{n-1})E\Psi _{G''}(x_2,\ldots ,x_{n-1},y) \end{aligned}$$

So \(f_n\sim \Psi _E(\Psi _{G'}(x,x_2,\ldots ,x_{n-1}),\Psi _{G''}(x_2,\ldots ,x_{n-1},y))\). Since \(xEx-2\sim x_1\) and \(x_{n-1}Ey\sim x_n\), Theorem 2 implies that

$$\begin{aligned}&u(x_1)=u(x)\rho (E)+u(x_2)(1-\rho (E))\quad \text { and }\quad u(x_n)=u(x_{n-1})\rho (E)\\&\qquad +\,u(y)(1-\rho (E)) \end{aligned}$$

Thus by the definition of \(u(f_n)\) and the hypothesis of induction, we get

$$\begin{aligned} V(f_n)&= u(\Psi _E(\Psi _{G'}(x,x_2,\ldots ,x_{n-1}),\Psi _{G''}(x_2,\ldots ,x_{n-1},y)) \\&= \rho (E)u(\Psi _{G'}(x,x_2,\ldots ,x_{n-1}))+(1-\rho (E))u(\Psi _{G''}(x_2,\ldots ,x_{n-1},y))\\&= \rho (E)\left( \displaystyle \sum _{i=2}^{n-1}u(x_i)P(E_i)+u(x)P(E_1)+u(x_{n-1})P(E_{n-1})\right) \\&+(1-\rho (E))\left( \displaystyle \sum _{i=2}^{n-1}u(x_i)P(E_i)+u(x_2)P(E_i)+u(y)P(E_n)\right) \\&= \displaystyle \sum _{i=2}^{n-1}u(x_i)P(E_i)+P(E_1)(\rho (E)u(x)+(1-\rho (E))u(x_2))\\&\quad +\,P(E_n)(\rho (E)u(x_{n-1})+(1-\rho (E))u(y))\\&= \displaystyle \sum _{i=1}^n u(x_i)P(E_i) \end{aligned}$$

Case 2 Either there is some \(x\in X\) and non-universal \(E\in \Sigma \) such that \(xEx_2\sim x_1\) or there is some \(y\in X\) and non-null \(F\in \Sigma \) such that \(x_{n-1}Fy\sim x_n\).

A similar proof as in Case 1 applies when \(xEx_2\sim x_1\) for some \(x\in X\) and non-universal \(E\in \Sigma \). Now suppose there exist \(y\) and non-null \(F\) satisfying \(x_{n-1}Fy\sim x_n\). First notice that such \(F\) is neither non-null nor universal. Then let \(z\sim x_1Fx_2\) for some \(z\in X\). So \(x_1\succ z\succ x_2\). Thus by Axiom 10,

$$\begin{aligned} zE_1x_2E_2\ldots x_nE_n&\sim \Psi _F(x_1,x_2)E_1\Psi _F(x_2,x_2)E_2\ldots \Psi _F(x_n,x_n)E_n \\&\sim \Psi _G(x_1,\ldots ,x_n)E\Psi _G(x_2,x_2,\ldots ,x_n)\\&\sim \Psi _G(x_1,x_2,\ldots ,x_n)F\Psi _{G''}(x_2,x_3,\ldots ,x_n) \end{aligned}$$

Since \(f_n\sim \Psi _G(x_1,\ldots ,x_n)\), Theorem 2 implies

$$\begin{aligned} u(\Psi _G(z,x_2,\ldots ,x_n))=\rho (F)u(\Psi _G(x_1,\ldots ,x_n))+(1-\rho (F))u(\Psi _{G''}(x_2,\ldots ,x_n)) \end{aligned}$$

Note that by case 1,

$$\begin{aligned} u(\Psi _G(z,x_2,\ldots ,x_n))=P(E_1)u(z)+\displaystyle \sum ^n_{i=2}u(x_i)P(E_i) \end{aligned}$$

Also by Theorem 2, \(u(z)=\rho (F)u(x_1)+(1-\rho (F))u(x_2))\). Therefore, we rearrange the above equation and derive:

$$\begin{aligned}&\rho (F)V(f_n)\\&\quad = P(E_1)u(z)+\displaystyle \sum ^n_{i=2}u(x_i)P(E_i) -(1-\rho (F))\left\{ P(E_1)u(x_2)+\sum ^n_{i=2}u(x_i)P(E_i)\right\} \\&\quad = \rho (F)\displaystyle \sum ^n_{i=1}u(x_i)P(E_i) \end{aligned}$$

Therefore we obtain the desired result.

Case 3 There is no \(x,y\in X\), non-universal \(E\in \Sigma \), and non-null \(F\in \Sigma \) such that \(xEx_2\sim x_1\) or \(x_{n-1}Fy\sim x_n\).

Now let \(z\sim x_1Ex_2\) for some \(E\in \Sigma \) which is neither null nor universal. Then \(x_1\succ z\succ x_2\). Applying the result in Case 2, we can get the desired result.

Proof of Theorem 3: (ii) \(\Rightarrow \) (i)

Suppose Condition (ii) holds. Then \(\succsim \) satisfies Axiom 1–5\(^*\) is straightforward. The detailed proofs are referred to Alon and Schmeidler (2014). We are only left to show that Axiom 10 holds. Let \(\Omega =\{A_1,\ldots ,A_n\}\) be a ground partition. Then \(x_1A_1\ldots x_nA_n\) and \(y_1A_1\ldots x_nA_n\) have SEU representation where the probability is consistent with capacity on \(\Omega \). So \(V(x_1A_1\ldots x_nA_n)=\sum _iu(x_i)\rho (A_i)\) and \(V(y_1A_1\ldots y_nA_n)=\sum _iu(y_i)\rho (A_i)\). Therefore,

$$\begin{aligned}&V(\Psi _\Omega (x_1,\ldots ,x_n)E\Psi _\Omega (y_1,\ldots ,y_n))\\&\quad =u(\Psi _\Omega (x_1,\ldots ,x_n))\rho (E)+ u(\Psi _\Omega (y_1,\ldots ,y_n)(1-\rho (E))\\&\quad =\sum _i u(x_i)\rho (A_i)\rho (E)+\sum _iu(y_i)\rho (A_i)(1-\rho (E))\\&\quad = \sum _i V(x_iEy_i)\rho (A_i)\\&\quad = V(\Psi _E(x_1,y_1)A_1\ldots \Psi _E(x_n,y_n)A_n) \end{aligned}$$

Therefore, Axiom 10 holds.

Proof of Corollary 1

Condition (ii) implies Condition (i) is similar to the proof of Theorem 3. We only prove that Condition (i) implies (ii). Since MEU preference is also a biseparable preference, capacity \(\rho \), defined by \(\rho (E)=\min _{P\in K}P(E)\) for all \(E\in \mathcal {U}\), is complementarily additive on each \(E\in \mathcal {U}\). That is, \(\min _{P\in K}P(E)+\min _{P\in K}P(E^\mathrm{c})=1\). Hence for all \(E\in \mathcal {U}\) and all \(P,Q\in K, P(E)=Q(E)\).

Proof of Corollary 2

We already know that \(\succsim \) satisfies 1,2,3,\(4^*,4^{**}\), and 5.1 iff \(\succsim \) has a CEU representation. Since CEU preference is also a biseparable preference, by Theorem 3, it has a SEU representation on \(\mathcal {F^U}\).

Proof of Section 5

Proof of Theorem 4

Let the consequences \(x,x',y,y',z,z'\) satify the condition of Definition 3. Suppose that \(xAz'\sim zAy'\) and \(zAx'\sim yAz'\). Since preference is biseparable, by representation we have

$$\begin{aligned} u(x)\rho (A)+u(z')(1-\rho (A))&= u(z)(1-\rho (A^\mathrm{c}))+u(y')\rho (A^\mathrm{c})\\ u(z)(1-\rho (A^\mathrm{c}))+u(x')\rho (A^\mathrm{c})&= u(y)\rho (A)+u(z')(1-\rho (A)) \end{aligned}$$

Adding them together yields

$$\begin{aligned} u(x)\rho (A)+u(x')\rho (A^\mathrm{c})=u(y)\rho (A)+u(y')\rho (A^\mathrm{c}) \end{aligned}$$
(12)

Since event \(A\) is unambiguous, by definition \(xAx'\sim yAy'\), which implies

$$\begin{aligned} u(x)\rho (A)+u(x')(1-\rho (A))=u(y)\rho (A)+u(y')(1-\rho (A)) \end{aligned}$$

Hence, \([u(x')-u(y')(1-\rho (A)-\rho (A^\mathrm{c})=0\). Since \(x'\succ y'\) by assumption, we must have \(1-\rho (A)-\rho (A^\mathrm{c})=0\). Thus \(\rho \) is complementarily additive on event \(A\).

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Qu, X. Purely subjective extended Bayesian models with Knightian unambiguity. Theory Decis 79, 547–571 (2015). https://doi.org/10.1007/s11238-015-9489-9

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Keywords

  • Knightian distinction
  • Maxmin expected utility
  • Biseparable preference
  • Unambiguous event

JEL Classification

  • D80
  • D81