Strategic information transmission with sender’s approval

Abstract

We consider sender–receiver games in which the sender has finitely many types and the receiver makes a decision in a compact set. The new feature is that, after the cheap talk phase, the receiver makes a proposal to the sender, which the latter can reject in favor of an outside option. We focus on situations in which the sender’s approval is absolutely crucial to the receiver, namely, on equilibria in which the sender does not exit at the approval stage. A nonrevealing equilibrium without exit may not exist. Our main results are that if the sender has only two types or if the receiver’s preferences over decisions do not depend on the type of the sender, there exists a (perfect Bayesian Nash) partitional equilibrium without exit, in which the sender transmits information by means of a pure strategy. The previous existence results do not extend: we construct a counter-example (with three types for the sender and type-dependent utility functions) in which there is no equilibrium without exit, even if the sender can randomize over messages. We establish additional existence results for (possibly mediated) equilibria without exit in the three type case.

Appendix

1.1 Proof of Propositions 3 and 4

For the sake of completeness, we first explicitly recall the conditions to be satisfied by an equilibrium of $\Gamma (v_{0})$ whether they involve exit of some types on equilibrium path or not.

Let us fix a pair of strategies $\sigma :K\rightarrow \Delta (M)$ and $\tau :M\rightarrow X$. Player 1’s equilibrium conditions can be written as

$$\begin{aligned} U_{+}^{k}(\tau (m))\ge U_{+}^{k}(\tau (m^{\prime }))\text { }\forall k\in K \text {, }\forall m\in M\text { s.t. }\sigma (m\mid k)>0,\forall m^{\prime }\in M\text {.} \end{aligned}$$

(26)

Player 2’s equilibrium conditions can be written as

$$\begin{aligned} \sum \limits _{k}p_{m}^{k}(\sigma )W^{k}(v_{0},\tau (m))\ge \sum \limits _{k}p_{m}^{k}(\sigma )W^{k}(v_{0},x)\text { }\forall m\in M\text { s.t. }P_{\sigma }(m)>0\text {, }\forall x\in X\text {.} \end{aligned}$$

We deduce that the necessary and sufficient conditions for $(\sigma ,\tau )$ to be a NEE in $\Gamma (v_{0})$ are (10) for player 1 and, recalling (4),

$$\begin{aligned} \forall m\in M\text { s.t.}P_{\sigma }(m)>0\text {, }\tau (m)\in \left[ \arg \max _{x\in X}\sum \limits _{k}p_{m}^{k}(\sigma )W^{k}(v_{0},x)\right] \cap X( \text {supp }p_{m}(\sigma )) \end{aligned}$$

(27)

for player 2. These conditions imply constrained optimization, namely (11).

In the game $\Gamma$, we focus on NEE; the conditions for $(\sigma ,\tau )$ to be a NEE are thus just (10) and (11).¹²

Proof of Proposition 3

If $(\sigma ,\tau )$ satisfies (27) in $\Gamma (v_{0})$, the same holds in $\Gamma (z_{0})$ since for every $x\in X$,

$$\begin{aligned} \sum \limits _{k}p_{m}^{k}(\sigma )W^{k}(z_{0},x)\le \sum \limits _{k}p_{m}^{k}(\sigma )W^{k}(v_{0},x) \end{aligned}$$

and for every $x\in X($supp $p_{m}(\sigma ))$,

$$\begin{aligned} \sum \limits _{k}p_{m}^{k}(\sigma )W^{k}(z_{0},x)=\sum \limits _{k}p_{m}^{k}(\sigma )W^{k}(v_{0},x)=\sum \limits _{k}p_{m}^{k}(\sigma )V^{k}(x)\text {.} \end{aligned}$$

Furthermore, (11) must hold and player 1’s equilibrium conditions (10) are the same in $\Gamma (v_{0})$ and $\Gamma (z_{0})$ or $\Gamma$ as long as player 2’s strategy remains unchanged. $\square$

Proof of Proposition 4

Let $(\sigma ,\tau )$ be a NEE in $\Gamma$. By definition, constrained optimization (11) holds, so that in particular $\tau (m)\in X($supp $p_{m}(\sigma ))$ for every m such that $P_{\sigma }(m)>0$. Let us keep player 1’s strategy, $\sigma$, fixed. Player 2’s strategy $\tau$ remains a best reply to $\sigma$ in $\Gamma (v_{0})$, with $v_{0}\le \min _{k\in K}\min _{x\in X}V^{k}(x)$, provided that $v_{0}$ is such that optimality of no exit holds, namely, recalling (27),

$$\begin{aligned} \forall m\in M\text { s.t. }P_{\sigma }(m)>0,\sum \limits _{k}p_{m}^{k}(\sigma )V^{k}(\tau (m))\ge \max _{x\in X}\sum \limits _{k}p_{m}^{k}(\sigma )W^{k}(v_{0},x)\text {.} \end{aligned}$$

These can be viewed as finitely many inequalities over $v_{0}$, which have a solution in ${\mathbb {R}}$, since the right hand sights are well-defined, for every $v_{0}\le \min _{k\in K}\min _{x\in X}V^{k}(x)$, by Lemma 1.¹³ Hence there exists $v_{0}$ such that $(\sigma ,\tau )$ is a NEE of $\Gamma (v_{0})$ and from Proposition 3, in $\Gamma (z_{0})$, for every $z_{0}\le v_{0}$. $\square$

Application to a mechanism design problem

In Sect. 3, we have established that, under various reasonable assumptions, $\Gamma$ has a partitional NEE $(\sigma ,\tau )$, in which both $\sigma$ and $\tau$ are pure. In this case, we can easily compute the highest ex ante expected utility that player 2, interpreted here as the principal, can obtain at a partitional NEE of $\Gamma$.¹⁴ We will show below that there exists $v_{0}\in {\mathbb {R}}$ such that, for every $z_{0}\le v_{0}$, the highest ex ante expected utility player 2 can obtain at an arbitrary partitional equilibrium of $\Gamma (z_{0})$ (which can involve exit or not) is the same as in $\Gamma$. More precisely,

Corollary of Propositions 3 and 4

There exists $v_{0}\in {\mathbb {R}}$ such that, for every $z_{0}\le v_{0}$ , the best partitional NEE for player 2 in $\Gamma$ remains the best partitional equilibrium for player 2 in $\Gamma (z_{0})$. $\square$

Proof

Let $v_{NE}^{*}$ be the highest ex ante expected utility player 2 can obtain at a partitional NEE of $\Gamma$. This number is well-defined if $\Gamma$ has a partitional NEE. Let $(\sigma ^{*},\tau ^{*})$ achieve the expected utility $v_{NE}^{*}$ for player 2. Using Proposition 4, there exists $v_{0}$ sufficiently small such that for every $z_{0}\le v_{0}$, $(\sigma ^{*},\tau ^{*})$ is a NEE of $\Gamma (z_{0})$ with the same expected utility $v_{NE}^{*}$ for player 2. By Proposition 3, for every such $z_{0}$, there does not exist any equilibrium without exit giving a higher expected utility to player 2 (because such an equilibrium would still be a NEE of $\Gamma$, with the same expected utilities).

Let us consider the partitional equilibria $(\sigma ,\tau )$ of $\Gamma (v_{0})$ in which exit possibly occurs, i.e., in which the set

$$\begin{aligned} K_{E}=\left\{ k\in K:U^{k}(\tau \circ \sigma (k))<u_{0}^{k}\right\} \ne \emptyset \text {,} \end{aligned}$$

i.e., $p_{E}=_{def}\sum _{k\in K_{E}}p^{k}>0$. The highest expected utility player 2 can achieve at such an equilibrium is

$$\begin{aligned} p_{E}v_{0}+(1-p_{E}){\overline{v}} \end{aligned}$$

where

$$\begin{aligned} {\overline{v}}=\max _{k\in K}\max _{x\in X}V^{k}(x)\text {.} \end{aligned}$$

If $v_{0}$ is such that, for every $p_{E}$ that can arise given the prior p ,

$$\begin{aligned} p_{E}v_{0}+(1-p_{E}){\overline{v}}\le v_{NE}^{*}\text {,} \end{aligned}$$

$$\begin{aligned} \text {namely, }v_{0}\le \frac{1}{p_{E}}\left[ v_{NE}^{*}-(1-p_{E}) {\overline{v}}\right] \end{aligned}$$

(28)

then $(\sigma ^{*},\tau ^{*})$ will guarantee the highest possible equilibrium utility to player 2, in every game $\Gamma (z_{0})$ with $z_{0}\le v_{0}$.

Let ${\underline{k}}$ be the type with the smallest prior probability, namely, such that $p^{{\underline{k}}}=\min \left\{ p^{1},\cdots ,p^{K}\right\}$. The inequality (28) will hold at every $p_{E}$ that can arise given the prior p as soon as it holds at $p_{E}=p^{{\underline{k}}}$: we just have to require

$$\begin{aligned} v_{0}\le \frac{1}{p^{{\underline{k}}}}\left[ v_{NE}^{*}-(1-p^{\underline{k }}){\overline{v}}\right] \text {.} \end{aligned}$$

The previous result is quite intuitive: an upper bound on the receiver’s expected utility at an equilibrium of $\Gamma (v_{0})$ with exit is obtained when the receiver’s proposal is rejected by only the least likely type, while the best possible utility is achieved at all the other types. If $v_{0}$ is sufficiently low, the best equilibrium utility for the receiver in $\Gamma (v_{0})$ will be not be achieved at an equilibrium with exit, but rather at an equilibrium without exit, which is in turn is necessarily a NEE of $\Gamma$. $\square$

1.2 Proof of Proposition 9

For simplicity, in this section, we assume that $u_{0}^{k}=0$ for each k. This is w.l.o.g. since we can translate the payoffs of each type of the sender. We start with preliminaries.

1.2.1 Mappings and multi-valued mappings

Recalling definitions (1) and (9), let for each p in $\Delta (K)$:

$f(p)=\sup \{\sum _{k\in K}p^{k}V^{k}(x),x\in X(\mathrm {supp}\;p)\}\in {\mathbb {R}}\cup \{-\infty \}$ and

$\Phi (p)=\{(U^{k}(x))_{k\in K},x\in Y(p)\}\;\subset {\mathbb {R}}^{K}$.

The sets $Y(p)\subset X(\mathrm {supp}\;p)$ and $\Phi (p)$ are convex compact subsets of ${\mathbb {R}}$ and ${\mathbb {R}}^{K}$, respectively. If $X(\mathrm { supp}\;p)\ne \emptyset$, then $f(p)\in {\mathbb {R}}$, $Y(p)\ne \emptyset$ and $\Phi (p)\ne \emptyset$. For each $u\in \Phi (p)$, we have $u^{k}\ge 0$ for each $k\in \mathrm {supp}\;p$. At an equilibrium of $\Gamma$, if the belief of the receiver (after having received the message of the sender) is p, then he has to propose a decision in Y(p), inducing a vector payoff in $\Phi (p)$ for the different types of player 1.

We will use in the sequel the following three lemmas (Lemma 12 is a simple mean-value theorem for correspondences).

Lemma 10

The mapping f is u.s.c. and convex.

If $p_{n}\xrightarrow [n \rightarrow \infty ]{}p\in \Delta (K)$ with $\mathrm {supp} \;p_{n}=\mathrm {supp}\;p$ for each n, then $f(p_{n}) \xrightarrow [n \rightarrow \infty ]{}f(p)$.

Proof

Suppose $p_n\xrightarrow [n \rightarrow \infty ]{} p$. Then for n large enough, $\mathrm {supp}\; p_n\supset \mathrm {supp}\; p$ so $X( \mathrm {supp}\; p_n)\subset X( \mathrm {supp}\; p)$. It follows that $\limsup _n f(p_n)\le f(p)$. (whether $f(p)=-\infty$ or not)

If $\mathrm {supp}\; p_n= \mathrm {supp}\; p$ for each n, then $|f(p_n)-f(p)|\le \sup _{x \in X} \sum _{k \in K} |p_n^k-p^k| |V^k(x)|$, and $f(p_n)\xrightarrow [n \rightarrow \infty ]{} f(p)$.

If $p=\lambda p_{1}+(1-\lambda )p_{2}$ with $\lambda \in (0,1)$, then $\mathrm {supp}\;p=\mathrm {supp}\;p_{1}\cup \;\mathrm {supp}\;p_{2}$ and $X( \mathrm {supp}\;p)=X(\mathrm {supp}\;p_{1})\cap X(\mathrm {supp}\;p_{2})$. If $f(p)=-\infty$ then $\lambda f(p_{1})+(1-\lambda )f(p_{2})\ge f(p)$. Consider x in $X(\mathrm {supp}\;p)$, we have $f(p_{1})\ge \sum _{k\in K}p_{1}^{k}V^{k}(x)$ and $f(p_{2})\ge \sum _{k\in K}p_{2}^{k}V^{k}(x)$, so $\lambda f(p_{1})+(1-\lambda )f(p_{2})\ge \sum _{k\in K}p^{k}V^{k}(x)$, and taking the supremum for x in $X(\mathrm {supp}\;p)$ we get $\lambda f(p_{1})+(1-\lambda )f(p_{2})\ge f(p)$. Hence f is convex. $\square$

Lemma 11

Consider a converging sequence $p_{n}\xrightarrow [n \rightarrow \infty ]{}p\in \Delta (K)$.

• (a) Assume $\limsup _{n}f(p_{n})=f(p)$. Then if $u_{n} \xrightarrow [n \rightarrow \infty ]{}u\in {\mathbb {R}}^{K}$, with $u_{n}\in \Phi (p_{n})$ for each n, we have $u\in \Phi (p)$,

• (b) Otherwise $\limsup _{n}f(p_{n})<f(p)$. Then there exists $n_{0}$ such that for each $u\in \Phi (p)$ and $n\ge n_{0}$, one can find $k\in \mathrm {supp} \;p_{n}\backslash \{\mathrm {supp}\;p\}$ such that $u^{k}<0$.

Proof

• (a) Without loss of generality we assume that $f(p_{n})\xrightarrow [n \rightarrow \infty ]{}f(p)$. Write $u_{n}=(U^{k}(x_{n}))_{k\in K}$ with $x_{n}$ in $Y(p_{n})$ for each n. By taking a converging subsequence we can assume that $x_{n}$ converges to some x in X. Since $p_{n}\xrightarrow [n \rightarrow \infty ]{}p$, for n large enough $\mathrm {supp} \;p_{n}\supset \mathrm {supp}\;p$ so $x\in X(\mathrm {supp}\;p)$. And $\sum _{k\in K}p_{n}^{k}V^{k}(x_{n})=f(p_{n})\xrightarrow [n \rightarrow \infty ]{}f(p)$ , so $\sum _{k\in K}p^{k}V^{k}(x)=f(p)$. Then x belongs to Y(p), and $u\in \Phi (p)$.

• (b) Assume that $\limsup _{n}f(p_{n})<f(p)$. We first claim that for n large enough, $Y(p)\cap X(\mathrm {supp}\;(p_{n}))=\emptyset$. Otherwise, we can find x in $Y(p)\cap X(\mathrm {supp}\;p_{n}))$ for infinitely many n’s, we have $f(p)=\sum _{k}p^{k}V^{k}(x)$ and $f(p_{n})\ge \sum _{k}p_{n}^{k}V^{k}(x)$ for infinitely many n’s, so $\limsup _{n}f(p_{n})\ge f(p)$ which is a contradiction. We have shown that there exists $n_{0}$ such that for $n\ge n_{0}$, $Y(p)\cap X(\mathrm {supp} \;(p_{n}))=\emptyset$. If $x\in Y(p)$ and $n\ge n_{0}$, then $x\notin X( \mathrm {supp}\;p_{n})$. So if $u\in \Phi (p)$ and $n\ge n_{0}$, there exists $k\in \mathrm {supp}\;p_{n}\backslash \{\mathrm {supp}\;p\}$ such that $u^{k}<0$. $\square$

Lemma 12

Let $F:[0,1]\rightrightarrows {\mathbb {R}}$ be a correspondence with non empty convex values and compact graph. If $F(0)\subset \{x\in {\mathbb {R}},x<0\}$ and $F(1)\subset \{x\in {\mathbb {R}},x>0\}$, there exists t in (0, 1) such that $0\in F(t)$.

Proof

The sets $C_{+}=\{t\in [0,1],F(t)\cap {\mathbb {R}}_{+}\ne \emptyset \}$ and $C_{-}=\{t\in [0,1],F(t)\cap {\mathbb {R}}_{-}\ne \emptyset \}$ are closed because F is u.s.c. Since F has non empty values, $C_{+}$ and $C_{-}$ are non empty, and $C_{+}\cup C_{-}=[0,1]$. By connexity of [0, 1], one can find t in both sets, that is such that F(t) intersects both ${\mathbb {R}}_{+}$ and ${\mathbb {R}}_{-}$. Since F(t) is convex, it contains 0. $\square$

1.2.2 Existence of an equilibrium

For $k=2,3$, define $\delta _{k}$ as the Dirac measure on the state k, and $p_{-k}$ as the conditional probability on K knowing the state is not k:

$$\begin{aligned}&\delta _{2}=(0,1,0),\;\;\delta _{3}=(0,0,1),\;\;p_{-2}=\left( \frac{p_{1}}{ p_{1}+p_{3}},0,\frac{p_{3}}{p_{1}+p_{3}}\right) \;\; \\&\qquad \mathrm {and}\;\;p_{-3}=\left( \frac{ p_{1}}{p_{1}+p_{2}},\frac{p_{2}}{p_{1}+p_{2}},0\right) . \end{aligned}$$

Choose $u_{2}$ in $\Phi (\delta _{2})$, $u_{3}$ in $\Phi (\delta _{3})$, $u_{1,2}$ in $\Phi (p_{-3})$ and $u_{1,3}$ in $\Phi (p_{-2})$. These are vectors in ${\mathbb {R}}^{3}$, and to simply notations we write:

$$\begin{aligned} u_{2}=\left( \begin{array}{c} a \\ + \\ - \\ \end{array} \right) ,u_{3}=\left( \begin{array}{c} b \\ - \\ + \\ \end{array} \right) ,u_{1,2}=\left( \begin{array}{c} c\ge 0 \\ + \\ - \\ \end{array} \right) ,u_{1,3}=\left( \begin{array}{c} d\ge 0 \\ - \\ + \\ \end{array} \right) . \end{aligned}$$

with $a=u_{2}^{1}$, $b=u_{3}^{1}$, $c=u_{1,2}^{1}$ and $d=u_{1,3}^{1}$. Here $+$ means $\ge 0$, and − means $<0$. We have $u_{2}^{2}\ge 0$, $u_{3}^{3}\ge 0$, $c\ge 0$, $u_{1,2}^{2}\ge 0$, $d\ge 0$ and $u_{1,3}^{3}\ge 0$ since for each p and $u\in \Phi (p)$, we have $u^{k}\ge 0$ for each $k\in \mathrm {supp}\;p$. The subset $\{2,3\}$ is not in ${\mathcal {T}}$, this gives $u_{2}^{3}<0$, $u_{3}^{2}<0$, $u_{1,2}^{3}<0$ and $u_{1,3}^{2}<0$.

Suppose $a\le d$. Then a simple equilibrium exists. Player 1 uses the partition $\{\{2\}, \{1,3\}\}$ to communicate: he sends the message $m=2$ if the state is 2, and the message $m=\{1,3\}$ if the state is 1 or 3. Player 2 proposes $x_2$ in $Y(\delta _2)$ such that $u_2=(U^k(x_2))_{k \in K}$ after receiving $m=2$, and proposes $x_{1,3}$ in $Y(p_{-2})$ such that $u_{1,3}=(U^k(x_{1,3}))_{k \in K}$ after receiving $m=\{1,3\}$. By definition of $Y(\delta _2)$ and $Y(p_{-2})$, player 2 is in best reply. And no type of player 1 has an incentive to deviate, so we have an equilibrium where player 1 plays pure. If we suppose $b\le c$, we have a similar equilibrium where player 1 uses the partition $\{\{3\}, \{1,2\}\}$.

From now on, we assume that $a>d\ge 0$ and $b>c\ge 0$. Then $a\ge 0$. Consider any sequence $(p_{n})$ converging to the Dirac measure on state 2 such that $\mathrm {supp}\;p_{n}=\{1,2\}$ for each n. By Lemma 11 part (b), we must have $\limsup _{n}f(p_{n})\ge f(p)$, and since f is u.s.c., $\limsup _{n}f(p_{n})=f(p)$. This being true for any such sequence, $f(p_{n}) \xrightarrow [n \rightarrow \infty ]{}f(p)$. That is, the restriction of f to the set $\{p,\mathrm {supp}\;p\subset \{1,2\}\}$ is continuous at $\delta _{2}$. And by Lemma 11 part (a), the restriction of $\Phi$ to the segment $[p_{-3},\delta _{2}]$ has a closed graph. Similarly we have $b\ge 0$, and we can prove that the restriction of f to the set $\{p,\mathrm {supp} \;p\subset \{1,3\}\}$ is continuous at $\delta _{3}$, and the restriction of $\Phi$ to the segment $[p_{-2},\delta _{3}]$ has a closed graph.

The initial probability p is on the segment $[\delta _2,p_{-2}]$, and also on the segment $[\delta _3,p_{-3}]$. For $t\in [0,1]$, define $q_t=t \delta _2+(1-t)p_{-3}$ and $q^{\prime }_t$ in $[p_{-2}, \delta _3]$ such that p belongs to the segment $[q_t,q^{\prime }_t]$. $q^{\prime }_t$ is uniquely defined for each t, $q^{\prime }_0=\delta _3$ and $q^{\prime }_1=p_{-2}$. We are going to construct an equilibrium with posteriors $q_t$ and $q^{\prime }_t$ for some appropriate t. We need player 1 of type 1 to be indifferent between splitting to $q_t$ and $q^{\prime }_t$.

Define the correspondence $F:[0,1]\rightrightarrows {\mathbb {R}}$, with for each t in [0, 1]:

$$\begin{aligned} F(t)=\{u^{1}-v^{1},u\in \Phi (q_{t}),v\in \Phi (q_{t}^{\prime })\}. \end{aligned}$$

F clearly has non empty convex compact values. We have seen that the restrictions of $\Phi$ to the segments $[p_{-3},\delta _{2}]$ and $[p_{-2},\delta _{3}]$ have closed graphs, moreover $q_{t}$ and $q_{t}^{\prime }$ are continuous in t, hence F has a closed graph. $F(0)=\{u^{1}-v^{1},u\in \Phi (p_{-3}),v\in \Phi (\delta _{3})\}$. If $F(0)\cap {\mathbb {R}}_{+}\ne \emptyset$, there exists a pure equilibrium where player 1 uses the partition $\{\{3\},\{1,2\}\}$, so we assume that F(0) is a subset of $\{x\in {\mathbb {R}},x<0\}$. Similarly, we assume that $F(1)=\{u^{1}-v^{1},u\in \Phi (\delta _{2}),v\in \Phi (p_{-2})\}$ is a subset of $\{x\in {\mathbb {R}},x>0\}$ (otherwise there exists an equilibrium where player 1 uses the partition $\{\{2\},\{1,3\}\}$). Then by Lemma 12 we can find $t^{*}$ in [0, 1] such that $0\in F(t^{*})$.

We can now conclude the proof. We can find x in $Y(q_{t^{*}})$, y in $Y(q_{t^{*}}^{\prime })$, $u=(U^{k}(x))_{k}\in \Phi (q_{t^{*}})$ and $u^{\prime }=(U^{k}(y))_{k}\in \Phi (q_{t^{*}}^{\prime })$ such that for some $e\ge 0$:

$$\begin{aligned} u=\left( \begin{array}{c} e \\ + \\ - \\ \end{array} \right) ,u^{\prime }=\left( \begin{array}{c} e \\ - \\ + \\ \end{array} \right) \end{aligned}$$

We have an equilibrium as follows. Player 1 sends a message so as to induce the posteriors $q_{t^{*}}$ and $q_{t^{*}}^{\prime }$ (type 2 sends the message 2, type 3 sends the message 3, and type 1 randomizes between the messages 2 and 3 so that the posteriors are $q_{t^{*}}$ after $m=2$ and $q_{t^{*}}^{\prime }$ after $m=3$). Player 2 then proposes x at $q_{t^{*}}$, and y at $q_{t^{*}}^{\prime }$. Player 2 is in best reply by construction. Type 1 of player 1 is indifferent. If type 2 of player 1 deviates and sends $q_{t^{*}}^{\prime }$, player 2 will propose y and type 2 will reject it, having the reserve payoff of 0, which is not better than the payoff without deviating. Similarly, player 1 of type 3 has no profitable deviation, and we have an equilibrium. $\square$