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 counterexample (with three types for the sender and typedependent 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.
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Notes
 1.
Except for the approval stage, our model behaves as a standard cheap talk game, in which Perfect Bayesian equilibrium is not restrictive.
 2.
We do not make any assumption beyond the fact that there are finitely many types; in particular, types can be “multidimensional,” with \(K\subset {\mathbb {R}}^{n_{1}}\), for some \(n_{1}\ge 1\).
 3.
As a typical example, \(X\subset {\mathbb {R}}^{n_{2}}\), for some \(n_{2}\ge 1\); for instance, player 2 has a finite set of actions A and \(X=\Delta (A)\) corresponds to the set of mixed strategies of player 2. In the latter case, observe that, according to the timing of \(\Gamma (v_{0})\), the proposal to player 1 is \(x\in X\), as opposed to \(a\in A\) selected according to x.
 4.
 5.
Pure strategies of player 2 (with respect to the set X) will suffice for our existence results. As indicated above, X can be a set of lotteries.
 6.
The observation that the principal’s ex ante expected utility is maximized when all types of the agent participate is also made in Bester and Strausz (2001), footnote 8.
 7.
 8.
The reasoning below holds, and is much simpler, if player 2’s proposal has to be in A.
 9.
The same conclusion obviously applies if player 2 is restricted to A at the outset, but is much stronger if the decision set is \(\Delta (A)\).
 10.
As noted in Sect. 2.2, for Nash equilibrium, player 1’s incentive compatibility condition (12), with \(U_{+}^{k}(x)\) in the right hand sight, is equivalent to (10), which only uses \( U^{k}(x)\). By contrast, for mediated equilibrium, (22) implies the condition obtained by replacing \(U_{+}^{k}(x)\) by \(U^{k}(x)\) in the right hand sight of (22) but the reverse is not true.
 11.
Condition (iii) in Sorin (1983) is an individual rationality condition for player 1 which has no exact counterpart in the above formulation of our equilibrium conditions. Indeed, once the final approval stage is captured in the utility functions \(U_{+}^{k}(.)\), \(k\in K\), player 1 has no action explicitly influencing the payoff.
 12.
 13.
The right hand sight of the inequalities can be rewritten as
$$\begin{aligned} \max _{L\varsubsetneq \text {supp}(p_{m}(\sigma ))}\max _{x\in X(L)}\left\{ \sum \limits _{k\in L}p_{m}^{k}(\sigma )V^{k}(x)+v_{0}\sum \limits _{k\in \text { supp}(p_{m}(\sigma ))\diagdown L}p_{m}^{k}(\sigma )\right\} \text {.} \end{aligned}$$  14.
The number of pure equilibrium payoffs is finite, in the same way as the number of partitions of K. Hence as soon as there is a pure equilibrium in \(\Gamma \), there is an equilibrium achieving the highest expected payoff for the receiver.
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This research started during the winter 2015–2016, while the first author was visiting Humboldt University, Berlin. We did not know of Shimizu (2013, 2017) at the time. These papers were pointed out to the first author by Daniel Krähmer on the occasion of a talk in Bonn in April 2018. Then we discovered Matthews (1989), among the references of Shimizu (2013, 2017).
We thank Anna Bogomolnaia, Ron Holzman, Ulrich Horst, Vincent Iehlé, Fr édéric Koessler, Daniel Krähmer, Ehud Lehrer, Ronny Razin, Antoine Salomon, Roland Strausz and Bertrand Villeneuve for stimulating conversations on the topic of this research. We also thank the participants of the workshops “Competition and Incentives” (Humboldt University, Berlin, June 9–10, 2016), “Game theory: Economics and Mathematics (GEM)” (South University of Denmark at Odense, October 5–6, 2018), “Signaling in Markets, Auctions and Games: a Multidisciplinary Approach” (Paris 2, May 22–23, 2019), “Frontiers in Design” (University College London (UCL), June 14–15, 2019), “Information Design and Splitting Games” (Centre d’Economie de la Sorbonne, Paris, June 17–19, 2019), as well as the audiences of seminars in Besançon, Bonn, Lancaster, London (LSE), ParisDauphine and Rome (LUISS). Jérô me Renault gratefully acknowledges funding from ANR3IA Artificial and Natural Intelligence Toulouse Institute, grant ANR17EUR0010 (Investissements d’Avenir program) and ANR MaSDOL.
Appendix
Appendix
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
Player 2’s equilibrium conditions can be written as
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),
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).^{Footnote 12}
Proof of Proposition 3
If \((\sigma ,\tau )\) satisfies (27) in \(\Gamma (v_{0})\), the same holds in \(\Gamma (z_{0})\) since for every \(x\in X\),
and for every \(x\in X(\)supp \(p_{m}(\sigma ))\),
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),
These can be viewed as finitely many inequalities over \(v_{0}\), which have a solution in \({\mathbb {R}}\), since the right hand sights are welldefined, for every \(v_{0}\le \min _{k\in K}\min _{x\in X}V^{k}(x)\), by Lemma 1.^{Footnote 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 \(\Gamma \).^{Footnote 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 welldefined 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
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
where
If \(v_{0}\) is such that, for every \(p_{E}\) that can arise given the prior p ,
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
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 \(\Gamma \). \(\square \)
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.
Mappings and multivalued 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 meanvalue 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 \( f(p_n)f(p)\le \sup _{x \in X} \sum _{k \in K} p_n^kp^k V^k(x)\), 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. \(\square \)
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:
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:
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 \(q_t=t \delta _2+(1t)p_{3}\) 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]:
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 \(e\ge 0\):
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. \(\square \)
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Forges, F., Renault, J. Strategic information transmission with sender’s approval. Int J Game Theory (2021). https://doi.org/10.1007/s00182021007571
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Keywords
 Approval
 Cheap talk
 Sender–receiver game
 Participation constraints
JEL Classification
 C72
 D82