Incentives in experiments with objective lotteries

Abstract

Azrieli et al. (J Polit Econ, 2018) provide a characterization of incentive compatible payment mechanisms for experiments, assuming subjects’ preferences respect dominance but can have any possible subjective beliefs over random outcomes. If instead we assume subjects view probabilities as objective—for example, when dice or coins are used—then the set of incentive compatible mechanisms may grow. In this paper we show that it does, but the added mechanisms are not widely applicable. As in the subjective-beliefs framework, the only broadly-applicable incentive compatible mechanism (assuming all preferences that respect dominance are admissible) is to pay subjects for one randomly-selected decision.

Appendices

Appendix 1: Proof of Lemma 3

For all of the appendices, recall that $x\succeq y$ indicates that either $x\succ y$ or $x=y$.

Suppose that $E\in SI({\mathscr {D}})$, and that E is not a singleton. Let $\{x,y\}\subseteq E$ be arbitrary. Consider the following two linear orders, $\succeq$ and $\succeq '$, which are identical except in their ranking of x and y (which are adjacent): They rank all elements of $X\backslash E$ above all elements of E, and they rank x and y above all elements of $E\backslash \{x,y\}$. However, $x\succ y$ and $y\succ ' x$. It is clear that if there is no $D\in {\mathscr {D}}$ such that $\{x,y\}\subseteq D \subseteq E$, then for all $D\in {\mathscr {D}}$, we have ${\text {dom}}_{\succeq }D={\text {dom}}_{\succeq '}D$, yet ${\text {dom}}_{\succeq }E=x\ne y ={\text {dom}}_{\succeq '}E$, contradicting sure identification.

Conversely, suppose that for every pair $\{x,y\}\subseteq E$, there exists $D\in {\mathscr {D}}$ for which $\{x,y\}\subseteq D\subseteq E$. Suppose by means of contradiction that there exist $\succeq$ and $\succeq '$ for which for all $D\in {\mathscr {D}}$, ${\text {dom}}_{\succeq }D={\text {dom}}_{\succeq '}D$, but ${\text {dom}}_{\succeq }E\ne {\text {dom}}_{\succeq '}E$. Let $w={\text {dom}}_{\succeq }E$ and $z={\text {dom}}_{\succeq '}E$. There exists $D'\in {\mathscr {D}}$ for which $\{w,z\}\subseteq D'\subseteq E$. As a consequence, $w={\text {dom}}_{\succeq }D'$ and $z={\text {dom}}_{\succeq '}D'$, contradicting the fact that ${\text {dom}}_{\succeq }D={\text {dom}}_{\succeq '}D$ for all $D\in {\mathscr {D}}$.

Appendix 2: Proof of Theorem 2

2.1 Step 1: Restrictions on $\varphi$

Recall that $L(x,\succeq )$ and $U(x,\succeq )$ are the lower- and upper-contour sets of x according to $\succeq$, respectively. Let $r(x,\succeq ) = |U(x,\succeq )|$ be the rank of x in $\succeq$. Two elements x, y are adjacent in $\succeq$ if $|r(x,\succeq )-r(y,\succeq )|=1.$ A switch of x, y in an order $\succeq$ is the replacement of the order of x, y, where x, y are adjacent in $\succeq$. Denote the obtained order by $\succeq ^{xy}$.²⁷

Lemma 5

$\varphi$is incentive compatible with respect to${\mathscr {E}}^\mathrm {mon}$if and only if it has the following two properties:

• (1) For every$\succeq$and everyx, ywith$r(x,\succeq )=r(y,\succeq )-1$,

  • (a) $\varphi (\mu (\succeq ))(z) = \varphi (\mu (\succeq ^{xy}))(z)$for every$z\ne x,y$.

  • (b) $\varphi (\mu (\succeq ))(x) > \varphi (\mu (\succeq ^{xy}))(x)$and$\varphi (\mu (\succeq ))(y) < \varphi (\mu (\succeq ^{xy}))(y)$whenever$\mu (\succeq )\ne \mu (\succeq ^{xy})$.

• (2) $\varphi (m)\in \Phi _{NR}$whenever$m\in M_{NR}$.

Proof

Assume $\varphi$ is incentive compatible and fix some $\succeq$ and some x, y with $r(x,\succeq )=r(y,\succeq )-1$. If $\mu (\succeq )=\mu (\succeq ^{xy})$ then the conditions are trivially true. Now assume that they differ. Let $z\ne x,y$ be some other element of X. Assume first that $r(z,\succeq )< r(x,\succeq )$, so z is ranked above x (and y) according to $\succeq$. Incentive compatibility implies that $\varphi (\mu (\succeq ))(U(z,\succeq )) \ge \varphi (\mu (\succeq ^{xy}))(U(z,\succeq ))$ and that $\varphi (\mu (\succeq ^{xy}))(U(z,\succeq ^{xy})) \ge \varphi (\mu (\succeq ))(U(z,\succeq ^{xy}))$. But since $U(z,\succeq ) = U(z,\succeq ^{xy})$ we get that they are equal, that is $\varphi (\mu (\succeq ))(U(z,\succeq )) = \varphi (\mu (\succeq ^{xy}))(U(z,\succeq ))$. The same argument applies to any z ranked above x (according to $\succeq$), which proves that $\varphi (\mu (\succeq ))(z) = \varphi (\mu (\succeq ^{xy}))(z)$ for any such z. A similar argument proves the assertion for elements z ranked below y. It follows that we must have $\varphi (\mu (\succeq ))(x) > \varphi (\mu (\succeq ^{xy}))(x)$ (and therefore $\varphi (\mu (\succeq ))(y) < \varphi (\mu (\succeq ^{xy}))(y)$) in order for $\varphi (\mu (\succeq )) \sqsupset \varphi (\mu (\succeq ^{xy}))$ to hold. This concludes the proof of property 1.

As for property 2, whenever $m\in M_{NR}$ incentive compatibility implies that $\varphi (\mu (\succeq )) \sqsupset \varphi (m)$ for every $\succeq$. First, this implies that $\varphi (m) \ne \varphi (\mu (\succeq ))$ for every $\succeq$. Second, assume that $\varphi (m)$ is not in $\Phi _{NR}$. Then by the separation theorem there is a vector $u\in {\mathbb {R}}^X$ such that $\sum _x u(x) \varphi (m)(x) > \sum _x u(x) \varphi (\mu (\succeq ))(x)$ for every $\succeq$. By boundedness of the set $\Phi _{NR}$, we can choose u such that $u(x)\ne u(y)$ whenever $x\ne y$. Let $\succeq _u$ be the order over X defined by u (formally, $u(x)>u(y)$ implies $x\succ y$). Then an expected utility maximizer with utilities $u(\cdot )$ prefers to report the non-rationalizable choices m over her true choices $\mu (\succeq _u)$. But this means $\varphi (\succeq _u)$ does not first-order stochastically dominate $\varphi (m)$ according to $\succeq _u$, a contradiction.

Conversely, assume that properties 1 and 2 are satisfied. Fix some $\succeq$ and consider some rationalizable deviation $m\in M_R$, $m\ne \mu (\succeq )$. Let $\succeq '\in \mu ^{-1}(m)$. Consider a minimal sequence of switches that starts at $\succeq$ and ends at $\succeq '$. This means that x and y are switched somewhere a long the path if and only if $x\succ y$ but $y\succ ' x$. Then property 1 implies that after any switch along the way we get a lottery that is dominated (relative to $\succeq$) by the previous one. This shows that $\varphi (\mu (\succeq ))$ dominates $\varphi (\mu (\succeq '))=\varphi (m)$. Finally, if $m\in M_{NR}$ then by property 2 and the above argument $\varphi (m)$ is a convex combination of lotteries that are dominated (relative to $\succeq$) by $\varphi (\mu (\succeq ))$, so it is dominated as well. This proves the lemma. $\square$

2.2 Step 2: Capacity representation

A capacity is a set function $v:2^X\rightarrow {\mathbb {R}}$ such that $v(\emptyset )=0$. A capacity v is normalized if $v(X)=1$ and monotone if $A\subseteq B$ implies $v(A)\le v(B)$.

Definition 11

A capacity v satisfies switch positivity if for every $x,y\in X$ and $A\subseteq X\setminus \{x,y\}$ the following holds: If $T(x,y,A)\ne \emptyset$ then $v(A\cup \{x,y\})+v(A) > v(A\cup \{x\})+v(A\cup \{y\})$; otherwise, $v(A\cup \{x,y\})+v(A) = v(A\cup \{x\})+v(A\cup \{y\})$.

If v satisfies switch positivity then it is supermodular, meaning $v(A\cup B)+v(A\cap B) \ge v(A) +v(B)$ for every $A,B\subseteq X$.

Lemma 6

If a mechanism$\varphi$is incentive compatible with respect to${\mathscr {E}}^\mathrm {mon}$then there exists a normalized and monotone capacityvthat satisfies switch positivity such that$\varphi (m)(x) = v(L(x,m)) - v(L(x,m)\setminus \{x\})$for every$m\in M_R$ and every $x\in X$.

Proof

Given $A\subseteq X$, consider some order $\succeq$ which ranks A at the bottom. Define $v(A) = \varphi (\mu (\succeq ))(A):= \sum _{x\in A} \varphi (\mu (\succeq ))(x)$. Notice first that, under incentive compatibility, v is well-defined in the sense that it does not depend on the particular order $\succeq$ used. Indeed, this follows from property (1a) in Lemma 5. It is also clear that v is normalized and monotone.

We now claim that v satisfies switch positivity. To see this, take any x, y and $A\subseteq X\setminus \{x,y\}$. Consider some order $\succeq$ with $L(x,\succeq )=A\cup \{x\}$ and $L(y,\succeq )=A\cup \{x,y\}$. We have $v(A\cup \{x\}) = \varphi (\mu (\succeq ))(A\cup \{x\})$ and $v(A\cup \{y\}) = \varphi (\mu (\succeq ^{xy}))(A\cup \{y\})$, so

$$\begin{aligned}v(A\cup \{x\}) + v(A\cup \{y\}) &= \varphi (\mu (\succeq ))(A\cup \{x\}) + \varphi (\mu (\succeq ^{xy}))(A\cup \{y\}) \\ &=v(A) + \varphi (\mu (\succeq ))(A) + \varphi (\mu (\succeq ))(x) + \varphi (\mu (\succeq ^{xy}))(y).\end{aligned}$$

Now, if $T(x,y,A)\ne \emptyset$ then $\mu (\succeq )\ne \mu (\succeq ^{xy})$ (see Remark 4.3), so by property (1b) of Lemma 5 we have $\varphi (\mu (\succeq ^{xy}))(y) < \varphi (\mu (\succeq ))(y)$. Thus,

$$\begin{aligned}v(A\cup \{x\}) + v(A\cup \{y\})&< v(A) + \varphi (\mu (\succeq ))(A) + \varphi (\mu (\succeq ))(x) + \varphi (\mu (\succeq ))(y)\\ &= v(A) + v(A\cup \{x,y\}),\end{aligned}$$

as required. On the other hand, if $T(x,y,A)=\emptyset$ then $\mu (\succeq )=\mu (\succeq ^{xy})$, so we get

$$\begin{aligned} v(A\cup \{x\}) + v(A\cup \{y\}) &= v(A) + \varphi (\mu (\succeq ))(A) + \varphi (\mu (\succeq ))(x) + \varphi (\mu (\succeq ))(y)\\ &= v(A) + v(A\cup \{x,y\}). \end{aligned}$$

Finally, we need to show that $\varphi (m)(x) = v(L(x,m)) - v(L(x,m)\setminus \{x\})$ whenever $m\in M_R$. Fix $m\in M_R$ and some $x\in X$. We claim that there is $\succeq \in \mu ^{-1}(m)$ such that $L(x,m)=L(x,\succeq )$. Since $L(x,m)\subseteq L(x,\succeq )$ for all $\succeq \in \mu ^{-1}(m)$, it is sufficient to show that the reverse inclusion holds for some $\succeq \in \mu ^{-1}(m)$. To see this, start with an arbitrary $\succeq \in \mu ^{-1}(m)$ and consider the set $L(x,\succeq )\setminus L(x,m)$. If this set is empty we are done. Otherwise, take the highest ranked element (according to $\succeq$) in this set, say y. Then for any z ranked between y and x (including $z=x$) it cannot be that zR(m)y, so by a sequence of switches we can put y above x without changing the resulting choices. By repeating this procedure for every element in $L(x,\succeq )\setminus L(x,m)$ we get the desired order, say $\succeq '$. For this order we have

$$\begin{aligned}v(L(x,m)) - v(L(x,m)\setminus \{x\}) &= v(L(x,\succeq ')) - v(L(x,\succeq ')\setminus \{x\})\\ &= \varphi (m)(L(x,\succeq ')) - \varphi (m)(L(x,\succeq ')\setminus \{x\})\\ &= \varphi (m)(x), \end{aligned}$$

as needed. $\square$

2.3 Step 3: Weighting vector representation

Lemma 7

Given$\varphi$, if there exists a normalized and monotone capacityvthat satisfies switch positivity such that$\varphi (m)(x) = v(L(x,m)) - v(L(x,m)\setminus \{x\})$for every$m\in M_R$and every$x\in X$, then$\varphi$is a weighted set-selection mechanism that satisfies switch positivity.

Proof

Let v be a normalized and monotone capacity that satisfies switch positivity and represents $\varphi$ as in the assertion of the lemma. As is well known (see Gilboa and Schmeidler 1995, e.g.), any capacity can be uniquely represented as a linear combination of the ‘unanimity capacities’. That is, there is a unique vector $\{\lambda (E)\}_{E\subseteq X, E\ne \emptyset }$ such that $v(A) = \sum _{E\subseteq A} \lambda (E)$ for every $A\subseteq X$.

We first show that if B is not SI then $\lambda (B)=0$. If B is not SI then by Lemma 3 there are $x,y\in B$ such that for no $1\le i\le k$ it holds that $\{x,y\} \subseteq D_i \subseteq B$. This in turn implies that $T(x,y,B\setminus \{x,y\})$ is empty (see Remark 4.3). Since v satisfies switch positivity we get that

$$\sum _{\{E :\, \{x,y\} \subseteq E \subseteq B\}} \lambda (E) = v(B) - v(B\setminus \{x\}) - v(B\setminus \{y\}) + v(B\setminus \{x,y\}) =0.$$

But if $\{x,y\} \subseteq D_i \subseteq B$ for no i then for every set E in the sum on the left it is also true that $\{x,y\} \subseteq D_i \subseteq E$ for no i, which implies that every such E is not SI. By induction on the size of B we can therefore prove that $\lambda (B)=0$.

We next check that the vector $\lambda$ satisfies switch positivity. Take any $x,y\in X$ and $A\subseteq X\setminus \{x,y\}$. By the last paragraph,

$$\sum _{ \{E\in T(x,y,A)\}} \lambda (E) = v(A\cup \{x,y\}) - v(A\cup \{x\}) - v(A\cup \{y\}) + v(A).$$

If $T(x,y,A)\ne \emptyset$ then since v satisfies switch positivity we have that $v(A\cup \{x,y\}) - v(A\cup \{x\}) - v(A\cup \{y\}) + v(A)>0$. Thus, $\lambda$ satisfies switch positivity.

The last thing to check is that $\lambda$ in fact represents the weighted set-selection mechanism $\varphi$ as in Definition 7. This follows from (for $m\in M_R$)

$$\varphi (m)(x)= v(L(x,m)) - v(L(x,m)\setminus \{x\}) = \sum _{\left\{ E:\, x\in E \subseteq L(x,m) \right\} } \lambda (E) = \sum _{\left\{ E:\, {\text {dom}}_m(E)=x \right\} } \lambda (E).$$

$\square$

2.4 Step 4: Weighted set-selection mechanisms are incentive compatible

Lemma 8

If$\varphi$is a weighted set-selection mechanism that satisfies switch positivity and satisfies$\varphi (m)\in \Phi _{NR}$whenever$m\in M_{NR}$then$\varphi$is incentive compatible with respect to${\mathscr {E}}^\mathrm {mon}$.

Proof

To check that $\varphi$ is incentive compatible we use Lemma 5. It follows immediately from the definition of a weighted set-selection mechanism that in a switch of adjacent two elements x, y the probability of any other element z being selected is not affected. Further, we have just showed above that the probability of x goes strictly up after a switch that increases the rank of x and changes the truthful message. This proves property 1 of Lemma 5. Property 2 of Lemma 5 is satisfied by assumption. $\square$

Appendix 3: Proof of Proposition 3

It will be convenient to think of any lottery f as a function $f:{\mathbb {R}}\rightarrow {\mathbb {R}}$, with f(x) being the probability assigned to x by f. For instance, the lottery l in the proposition is identified with the function $l(0)=l(3)=1/2$ and $l(x)=0$ otherwise. Addition of lotteries and multiplication of lotteries by scalars are performed pointwise (yielding functions which are not necessarily lotteries). The expected utility of a lottery f for a subject with utility function u (defined on dollar amounts) is denoted by $u\cdot f$.

The proof is broken into two claims:

Claim 1

If$\varphi$is incentive compatible then there is$a\in (0,1/2]$such that$\varphi (\delta _1,\delta _2)- \varphi (l,\delta _2) = -a \delta _0 + 2a \delta _1 -a \delta _3$.

Proof

Consider some strictly increasing utility function u such that $1/2u(0)+1/2u(3)=u(1)$. Then an expected utility maximizer (a special case of RDU preferences) with utility function u is indifferent between l and $\delta _1$, so both announcements $(\delta _1,\delta _2)$ and $(l,\delta _2)$ are truthful. By incentive compatibility this decision maker must also be indifferent between $\varphi (\delta _1,\delta _2)$ and $\varphi (l,\delta _2)$, that is $u \cdot \varphi (\delta _1,\delta _2) = u \cdot \varphi (l,\delta _2)$.

First, it cannot be true that there is $x\in {\mathbb {R}}\setminus \{0,1,3\}$ such that $\varphi (\delta _1,\delta _2)(x) \ne \varphi (l,\delta _2)(x)$. Indeed, let u be as above, and change the values of u in a small neighborhood of x (small enough that it does not affect any other point in the support of the two lotteries) to get a new increasing function ${\tilde{u}}$ that still satisfies $1/2{\tilde{u}}(0)+1/2{\tilde{u}}(3)={\tilde{u}}(1)$. Then it cannot be true that both equalities $u \cdot \varphi (\delta _1,\delta _2) = u \cdot \varphi (l,\delta _2)$ and ${\tilde{u}} \cdot \varphi (\delta _1,\delta _2) = {\tilde{u}} \cdot \varphi (l,\delta _2)$ hold simultaneously, a contradiction.

Thus, there are $a,b,c \in {\mathbb {R}}$ such that $\varphi (\delta _1,\delta _2)- \varphi (l,\delta _2) = a \delta _0 + b \delta _1 +c \delta _3$. Consider a utility function u with $u(0)=1, u(1)=2$ and $u(3)=3$. By the same argument as above we get that $u \cdot \varphi (\delta _1,\delta _2) = u\cdot \varphi (l,\delta _2)$, which implies $a+2b+3c=0$. By considering another function u with $u(0)=1, u(1)=3$ and $u(3)=5$ gives $a+3b+5c=0$. Solving these two equations gives $b=-2a$ and $c=a$.

To conclude, we showed that $\varphi (\delta _1,\delta _2)- \varphi (l,\delta _2) = -a \delta _0 + 2a \delta _1 -a \delta _3$ for some $a\in {\mathbb {R}}$. The fact that $a>0$ follows by looking at an expected utility maximizer with the utility function $u(x)=x$ (who strictly prefers the lottery l over $\delta _1$). The fact that $a\le 1/2$ is obvious. $\square$

The proof of the above claim only considered EU preferences. The following claim makes use of RDU preferences which do not satisfy independence.

Claim 2

If$\varphi$is incentive compatible then$\varphi (\delta _1,\delta _2)=\delta _1$and$\varphi (l,\delta _2)=l$.

Proof

Let u be the function $u(x)=0$ for $x<1$, $u(1)=1$, and $u(x)=3$ for $x>1$. Consider an RDU decision maker with utility function u and a probability weighting function q satisfying $q(0)=0, q(1/2)=1/2, q(1)=1$. Any such decision maker prefers the lottery l over $\delta _1$ and prefers $\delta _2$ over l. Thus, for incentive compatibility, any such decision maker must prefer $\varphi (l,\delta _2)$ over $\varphi (\delta _1,\delta _2)$, that is $U_q(\varphi (l,\delta _2)) > U_q(\varphi (\delta _1,\delta _2))$.

By the definition of u we have that

$$U_q(h) = 1[q(H(1))-q(H(1-))] + 3[1-q(H(1))]$$

for every lottery h with cdf H. Denote by F the cdf of the lottery $\varphi (l,\delta _2)$ and by G the cdf of $\varphi (\delta _1,\delta _2)$. Then incentive compatibility requires that

$$U_q(\varphi (l,\delta _2)) - U_q(\varphi (\delta _1,\delta _2)) = 2[q(G(1))-q(F(1))] - [q(F(1-))-q(G(1-))]>0$$

(2)

for every q as above.

By the previous claim, there is $0<a\le 1/2$ such that $G(1)-F(1)= F(1-)-G(1-)=a$. We claim now that incentive compatibility can only hold if $a=1/2$., i.e. if $G(1)=1, F(1)=F(1-)=1/2$ and $G(1-)=0$. Indeed, in any other case it is possible to choose q (strictly increasing, $q(0)=0$, $q(1)=1$, $q(1/2)=1/2$) such that $q(G(1))-q(F(1))$ is much smaller than $q(F(1-))-q(G(1-))$, which violates (2). The precise construction of such q depends on which of the intervals [0, 1 / 2] or [1 / 2, 1] each one of these four numbers belongs to, but it is straightforward in all cases. For instance, assume that $G(1-)\in [0,1/2]$ and $F(1-), F(1), G(1)$ are all in (1 / 2, 1]. Then we can find q such that $q(F(1-)), q(F(1))$ and q(G(1)) are all close to 1, while $q(G(1-))$ is at most 1/2. Other cases are treated similarly. $\square$

To conclude the proof, notice that the same arguments can be applied to the choices in the second decision problem. That is, incentive compatibility requires that $\varphi (l,l)=l$ and $\varphi (l,\delta _2)=\delta _2$. Since we cannot have $\varphi (l,\delta _2)=l$ and $\varphi (l,\delta _2)=\delta _2$ at the same time, we conclude that an incentive compatible mechanism does not exist.