Strategic information transmission despite conflict

Abstract

We analyze a standard Crawford and Sobel’s cheap talk game in a two dimensional framework, with uniform prior, quadratic preferences and a binary disclosure rule. Information might be credibly revealed by the Sender to the Receiver when players are able to strategically set aside their conflict. We exploit the few symmetries of the game parameters to derive multiple continua of equilibria, when varying the Sender’s bias over the entire euclidean space. In particular, credible information might be revealed whatever the bias. Then we show that the equilibria exhibited characterize the game’s full set of pure strategy equilibria.

Appendix: Proofs

1.1 Example 1

Given the Sender’s strategy ${\mathfrak {m}}_C$, we derive from (1) the induced actions

$$\begin{aligned} \varvec{a}_C(m_1)=\left( \frac{2}{3},\frac{1}{3}\right) \text {, and }\varvec{a}_C(m_2)=\left( \frac{1}{3},\frac{2}{3}\right) . \end{aligned}$$

For ${\mathfrak {m}}_A$ (resp. the two ${\mathfrak {m}}_H$), we obtain $\varvec{a}_A(m_1)=\left( \frac{2}{3},\frac{2}{3}\right)$ and $\varvec{a}_A(m_2)=\left( \frac{1}{3},\frac{1}{3}\right)$ (resp. $\varvec{a}_H(m_1)=\left( \frac{3}{4},\frac{1}{2}\right)$ and $\varvec{a}_H(m_2)=\left( \frac{1}{4},\frac{1}{2}\right)$, or $\varvec{a}_H(m_1)=\left( \frac{1}{2},\frac{3}{4}\right)$ and $\varvec{a}_H(m_2)=\left( \frac{1}{2},\frac{1}{4}\right)$).

Reciprocally, given the Receiver’s strategy $\varvec{a}:m\mapsto \varvec{a}(m)$, for instance $\varvec{a}(m_1)=\varvec{a}_C(m_1)=\left( \frac{2}{3},\frac{1}{3}\right)$, and $\varvec{a}(m_2)=\varvec{a}_C(m_2)=\left( \frac{1}{3},\frac{2}{3}\right)$ for the game associated to the bias $\varvec{b}=(b,b)$, $b\in {\mathbb {R}}$, from (2), we have ${\mathfrak {m}}(\varvec{\theta })=m_1$ if and only if (up to a null measure set)

$$\begin{aligned}&-\Vert \varvec{a}_C(m_1)-(\varvec{\theta }+\varvec{b})\Vert ^2\ge -\Vert \varvec{a}_C(m_2)-(\varvec{\theta }+\varvec{b})\Vert ^2\\&\quad \iff -\left( \frac{2}{3}-(\theta _1+b)\right) ^2-\left( \frac{1}{3}-(\theta _2+b)\right) ^2\\&\qquad \qquad \quad \ge -\left( \frac{1}{3}-(\theta _1+b)\right) ^2-\left( \frac{2}{3}-(\theta _2+b)\right) ^2\\&\quad \iff \theta _1\ge \theta _2 \end{aligned}$$

and similarly ${\mathfrak {m}}(\varvec{\theta })=m_2$ if and only if $\theta _1<\theta _2$. Therefore, strategies ${\mathfrak {m}}_C$ and $\varvec{a}_C$ are best response in equilibrium. Proofs for the profiles of strategies $({\mathfrak {m}}_{A},\varvec{a}_A)$ and $({\mathfrak {m}}_H,\varvec{a}_H)$ are similar.

1.2 Example 2

Consider the Sender’s strategy

$$\begin{aligned} {\mathfrak {m}}_{C(c)}(\varvec{\theta })={\left\{ \begin{array}{ll}m_1\text {, if }\theta _1\ge \theta _2+c,\\ m_2\text {, if }\theta _1<\theta _2+c,\end{array}\right. } \end{aligned}$$

for some $c\le 0$. Then from (1), we obtain, for $c>-1$,

$$\begin{aligned} \varvec{a}(m_1)={\mathbb {E}}[\varvec{\theta }|\theta _1\ge \theta _2+c]=\left( \frac{1}{3}\frac{c^3+3c^2+3c-2}{c^2+2c-1},\frac{1}{3}\frac{-c^3+3c-1}{c^2+2c-1}\right) , \end{aligned}$$

and

$$\begin{aligned} \varvec{a}(m_2)={\mathbb {E}}[\varvec{\theta }|\theta _1<\theta _2+c]=\left( \frac{c+1}{3},\frac{2-c}{3}\right) . \end{aligned}$$

Reciprocally, given the Receiver’s strategy $\varvec{a}_{C(c)}$ defined by the above actions, from (2), message $m_1$ is disclosed if and only if $-\Vert \varvec{a}_{C(c)}(m_1)-(\varvec{\theta }+\varvec{b})\Vert ^2\ge -\Vert \varvec{a}_{C(c)}(m_2)-(\varvec{\theta }+\varvec{b})\Vert ^2$, where $\varvec{b}=(b_1,b_2)\in {\mathbb {R}}^2$. Set $\varvec{b}=(b_1,b_1+\varDelta b)$ with $\varDelta b=b_2-b_1\in {\mathbb {R}}$. Then in equilibrium, we must have

$$\begin{aligned} \theta _1\ge \theta _2+\frac{1}{3}\frac{c(c+2)(2c-1)}{c^2+2c-1}+\varDelta b, \end{aligned}$$

so that c must solve $\varDelta b=\frac{1}{3}\frac{c^3+3c^2-c}{c^2+2c-1}$. This defines a strictly increasing function $\varDelta b\mapsto c(\varDelta b)$ from $\varDelta b\in (-\frac{1}{2},0]$ to $(-1,0]$. A similar bijection is obtained in case $c\ge 0$ and $\varDelta b\in [0,\frac{1}{2})$.

1.3 Example 3

It is straightforward to verify that each strategy of the profile of strategies $({\mathfrak {m}},\varvec{a})$ stated in the example is the best response of the other, i.e. satisfies (1) and (2).

1.4 Lemma 1

Let $({\mathfrak {m}},\varvec{a})$ be such that $({\mathfrak {m}},\varvec{a})\in {\mathcal {E}}(\varvec{b})$ for some $\varvec{b}\in {\mathbb {R}}^2$. Let $\varvec{b}'\in {\mathbb {R}}^2$. From $\Vert \varvec{x}\Vert ^2=\varvec{x}\cdot \varvec{x}$, given $\varvec{a}$, we obtain for instance ${\mathfrak {m}}'(\varvec{\theta })=m_1$iff

$$\begin{aligned}&-\Vert \varvec{a}(m_1)-(\varvec{\theta }+\varvec{b}')\Vert ^2\ge -\Vert \varvec{a}(m_2)-(\varvec{\theta }+\varvec{b}')\Vert ^2\\&\quad \iff -\Vert (\varvec{a}(m_1)-\varvec{b}-\varvec{\theta })-(\varvec{b}'-\varvec{b})\Vert ^2\ge -\Vert (\varvec{a}(m_2)-\varvec{b}-\varvec{\theta })-(\varvec{b}'-\varvec{b})\Vert ^2\\&\quad \iff -\Vert \varvec{a}(m_1)-(\varvec{\theta }+\varvec{b})\Vert ^2\ge -\Vert \varvec{a}(m_2)-(\varvec{\theta }+\varvec{b})\Vert ^2-2(\varvec{b}'-\varvec{b})\cdot (\varvec{a}(m_1)-\varvec{a}(m_2)). \end{aligned}$$

Therefore given $\varvec{a}$, the Sender of $\varGamma _{\varvec{b}'}$ uses the same disclosure strategy ${\mathfrak {m}}'$ as the Sender of $\varGamma _{\varvec{b}}$ who uses ${\mathfrak {m}}$ for any observed $\varvec{\theta }\in \varTheta$iff$(\varvec{b}'-\varvec{b})\cdot (\varvec{a}(m_1)-\varvec{a}(m_2))=0$. Reciprocally, strategies ${\mathfrak {m}}'$ and ${\mathfrak {m}}$ induce the same actions.

1.5 Lemma 2

Consider a line $(OT)$ with $O={\mathbb {E}}[\varTheta ]=(\frac{1}{2},\frac{1}{2})$ and $T\in \{0\}\times (\frac{1}{2},1)$ (see Figure 6 in the main text). We seek to show that for any such line, there exists a game $\varGamma _{\varvec{b}_T}$ such that $(OT)$ supports the two induced actions of the Receiver of $\varGamma _{\varvec{b}_T}$, for some profile $({\mathfrak {m}}_T,\varvec{a}_T)$ of equilibrium strategies.

Since for any profile $({\mathfrak {m}},\varvec{a})$ of equilibrium strategies, line $(\varvec{a}(m_1)\varvec{a}(m_2))$ is perpendicular to the Sender’s set of indifferent states ${\mathcal {L}}=\{\varvec{\theta },\, U_S(\varvec{\theta },\varvec{a}(m_1))=U_S(\varvec{\theta },\varvec{a}(m_2))\}$, a necessary equilibrium condition is the orthogonality of ${\mathcal {L}}$ and $(OT)$. Given ${\mathcal {L}}_T$ perpendicular to $(OT)$, let us denote $\varTheta _1({\mathcal {L}}_T)$ and $\varTheta _2({\mathcal {L}}_T)$ the two regions situated on the sides of ${\mathcal {L}}_T$, with for instance $O\in \varTheta _1({\mathcal {L}}_T)$.²⁰ Then we have to show that the players’ strategies ${\mathfrak {m}}_T$ and $\varvec{a}_T$ given by

$$\begin{aligned} {\mathfrak {m}}_T(\theta )={\left\{ \begin{array}{ll}m_1\text { if }\varvec{\theta }\in \varTheta _1({\mathcal {L}}_T),\\ m_2\text { if }\varvec{\theta }\in \varTheta _2({\mathcal {L}}_T),\end{array}\right. }\text { and }\varvec{a}_T(m)={\left\{ \begin{array}{ll}{\mathbb {E}}[\varTheta _1({\mathcal {L}}_T)]\text { if }m=m_1,\\ {\mathbb {E}}[\varTheta _2({\mathcal {L}}_T)]\text { if }m=m_2,\end{array}\right. } \end{aligned}$$

define a profile of equilibrium strategies which satisfies $\varvec{a}_T(m_i)\in (OT)$ for $i\in \{1,2\}$. Note that it is sufficient to show that $\varvec{a}_T(m_2)\in (OT)$, since we necessarily have $\varvec{a}_T(m_1)\in (O\varvec{a}_T(m_2))=(OT)$.

To do this, let us consider the orthogonal coordinate system ${\mathcal {R}}_T(O,x,y)$, with $(y^-Oy^+)$ supported by $(OT)$ (see Figure 11).

Fig. 11

${\mathcal {L}}_T(\ell )$ and subsequent $\varvec{a}_2(T,\ell )$

Given $\ell \in {\mathbb {R}}$, let ${\mathcal {L}}_T(\ell )$ be the line perpendicular to $(OT)$ and situated at a distance $\ell$ of O, with $O\notin \varTheta _2({\mathcal {L}}_T(\ell ))$, and let

$$\begin{aligned} \varvec{a}_2(T,\ell )={\mathbb {E}}[\varvec{\theta }|\varvec{\theta }\in \varTheta _2({\mathcal {L}}_T(\ell ))] \end{aligned}$$

denote the corresponding expectation. Let ${\overline{\ell }}$ be such that $(0,1)\in {\mathcal {L}}_T({\overline{\ell }})$. For any $T\in \{0\}\times (\frac{1}{2},1)$, if $\ell$ is sufficiently close to ${\overline{\ell }}$, then $\varTheta _2({\mathcal {L}}_T(\ell ))\subset \{(x,y),\, x>0\},$ and thus $x(\varvec{a}_2(T,\ell ))>0$ for any such $\ell$. Therefore, by continuity, if there exists ${\underline{\ell }}$ such that

$$\begin{aligned} x(\varvec{a}_2(T,{\underline{\ell }}))<0, \end{aligned}$$

(7)

then there is some $\ell ^*\in ({\underline{\ell }},{\overline{\ell }})$ such that $x(\varvec{a}_2(A,\ell ^*))=0$. Let us show that for any $T\in \{0\}\times (\frac{1}{2},1)$,

$$\begin{aligned} x(\varvec{a}_2(T,0))<0, \end{aligned}$$

(8)

i.e. ${\underline{\ell }}=0$ satisfies Condition (7).

Let us set $\varTheta ^+=\{\varvec{\theta }\in \varTheta ,\, x(\varvec{\theta })>0,y(\varvec{\theta })>0\}$ and $\varTheta ^-=\{\varvec{\theta }\in \varTheta ,\, x(\varvec{\theta })<0,y(\varvec{\theta })>0\}$, and $\varvec{a}_2^+={\mathbb {E}}[\varTheta ^+]$, $\varvec{a}_2^-={\mathbb {E}}[\varTheta ^-]$ (see Figure 12).

Fig. 12

$\varvec{a}_2(T,0)=\frac{\varvec{a}_2^++\varvec{a}_2^-}{2}$

Then from the law of iterated expectations, we have

$$\begin{aligned} (|\varTheta ^+|+|\varTheta ^-|)\varvec{a}_2(T,0)=|\varTheta ^+|\varvec{a}_2^++|\varTheta ^+|\varvec{a}_2^-, \end{aligned}$$

where $|\cdot |$ denotes the Lebesgue measure. Now note that rotation $\rho _{\frac{\pi }{2}}$ of angle $\frac{\pi }{2}$ and centre O transforms $\varTheta ^+$ to $\varTheta ^-$ point wise, so that for any $\varvec{\theta }\in \varTheta$, we have $(x(\varvec{\theta }),y(\varvec{\theta }))\in \varTheta ^+\,{\textit{iff}}\,(-y(\varvec{\theta }),x(\varvec{\theta }))\in \varTheta ^-$. Then on the one hand, we have $|\varTheta ^+|=|\varTheta ^-|$, which gives

$$\begin{aligned} \varvec{a}_2(T,0)=\frac{\varvec{a}_2^++\varvec{a}_2^-}{2}, \end{aligned}$$

and on the other hand, we have $\rho _{\frac{\pi }{2}}(\varvec{a}_2^+)=\varvec{a}_2^-$, which gives

$$\begin{aligned} {\left\{ \begin{array}{ll}x(\varvec{a}_2^+)=y(\varvec{a}_2^-),\\ y(\varvec{a}_2^+)=-x(\varvec{a}_2^-).\end{array}\right. } \end{aligned}$$

Consequently, Condition (8) is satisfied iff$\frac{x(\varvec{a}_2^-)+x(\varvec{a}_2^+)}{2}<0$, i.e.

$$\begin{aligned} x(\varvec{a}_2^+)<y(\varvec{a}_2^+). \end{aligned}$$

(9)

Now let $T'$ be the point such that $\rho _{\frac{\pi }{2}}(T')=T$ (see Figure 13), and let us partition the set $\varTheta ^+$ according to triangles $\varDelta =(OTT')$ and $\varDelta '=\varTheta ^+\setminus (OTT')$. Again, from the law of iterated expectations, $\varvec{a}_2^+$ is situated on line $(GG')$, where $G={\mathbb {E}}[\varvec{\theta }|\varvec{\theta }\in \varDelta ]$ and $G'={\mathbb {E}}[\varvec{\theta }|\varvec{\theta }\in \varDelta ']$ are the respective mass centres of triangles $\varDelta$ and $\varDelta '$.²¹

Fig. 13

$y(\varvec{a}_2^+)>x(\varvec{a}_2^+)$

Note that from $OT=OT'$, median (OI) of $(OTT')$ has equation $x=y$ in ${\mathcal {R}}(0,x,y)$. Therefore $x(G)=y(G)$ and Condition (9) holds iff

$$\begin{aligned} x(G')<y(G'). \end{aligned}$$

(10)

Finally, note that $G'$ is situated on the median line of $\varDelta '$ which passes through I and the point (0, 1), between I and (0, 1) which satisfy $y(I)=x(I)$ and $x((0,1))<y((0,1))$ respectively. Therefore $G'\in \{(x,y),\, y>x\}$ and Condition (10) holds.

Since (10) hold, so do Conditions (9), (8) and (7). Therefore we have $x(\varvec{a}_2(T,0))<0$, and we obtain the existence of $\ell ^*\in (0,{\overline{\ell }})$ such that $\varvec{a}_2(T,\ell ^*)\in (OT)$ as wanted.

Next, we define the game $\varGamma _{\varvec{b}_T}$. We have to find some $\varvec{b}_T\in {\mathbb {R}}^2$ such that strategies ${\mathfrak {m}}_T$ and $\varvec{a}_T$ given by

$$\begin{aligned} {\mathfrak {m}}_T(\varvec{\theta })={\left\{ \begin{array}{ll}m_1\text { if }\varvec{\theta }\in \varTheta _1({\mathcal {L}}_T(\ell ^*)),\\ m_2\text { if }\varvec{\theta }\in \varTheta _2({\mathcal {L}}_T(\ell ^*)),\end{array}\right. }\text { and }\varvec{a}_T(m)={\left\{ \begin{array}{ll}\varvec{a}_1(T,\ell ^*)\text { if }m=m_1,\\ \varvec{a}_2(T,\ell ^*)\text { if }m=m_2,\end{array}\right. } \end{aligned}$$

are equilibrium strategies of $\varGamma _{\varvec{b}_T}$. A sufficient condition is that ${\mathcal {L}}_T(\ell ^*)$ is the perpendicular bisector of $\varvec{a}_T(m_1)-\varvec{b}_T$ and $\varvec{a}_T(m_2)-\varvec{b}_T$. Let us set $\varvec{b}_T=\frac{\varvec{a}_T(m_1)+\varvec{a}_T(m_2)}{2}-\varvec{\theta }_T$, where $\varvec{\theta }_T=(OT)\cap {\mathcal {L}}_T(\ell ^*)$, so that $\varvec{\theta }_T$ is the mid-point of $\varvec{a}_T(m_1)-\varvec{b}_T$ and $\varvec{a}_T(m_2)-\varvec{b}_T$. Since ${\mathcal {L}}_T(\ell ^*)\perp (OT)$, ${\mathcal {L}}_T(\ell ^*)$ has the desired property.

1.6 Lemma 3

We show that strategies $({\mathfrak {m}}_T,\varvec{a}_T)$ extend continuously to one of the symmetric deviated strategies. By construction, when $T$ spans continuously the open segment $\{0\}\times (\frac{1}{2},1)$, line ${\mathcal {L}}_T$, and therefore²² sets ${\mathfrak {m}}_T^{-1}(m_i)$, $i\in \{1,2\}$, actions $\varvec{a}_T(m_1)$ and $\varvec{a}_T(m_2)$, and bias

$$\begin{aligned} T\mapsto \varvec{b}_T=\frac{\varvec{a}_T(m_1)+\varvec{a}_T(m_2)}{2}+\varvec{\theta }_T\in (OT) \end{aligned}$$

(11)

can all be chosen to move continuously.

When $T\rightarrow T_1=(0,1)$, we have: (a) actions tend to be onto the diagonal line $(OT_1)$, and (b) Sender’s set of indifferent states ${\mathcal {L}}_T$ tends to intersect the diagonal line orthogonally at some $\varvec{\theta }_{T_1}=(\theta _{T_1},1-\theta _{T_1})$, for some $0\le \theta _{T_1}\le \frac{1}{2}$.

If $\theta _{T_1}>0$, this necessarily identifies a (1, 1)-symmetric (deviated if $\theta _{T_1}<\frac{1}{2}$) comparative profile of strategies $({\mathfrak {m}}_{C(c)},\varvec{a}_{C(c)})$, $c\in (0,1)$, since, as shown in Example 2, when c spans (0, 1), the support of the induced actions and the sets of indifferent states associated with profiles $({\mathfrak {m}}_{C(c)},\varvec{a}_{C(c)})$ span all such orthogonal lines.

If $\theta _{T_1}=0$, this identifies a babbling equilibrium. Let us show that this case does not occur.

Let us set $T(0,\frac{1}{2}+t)$, where $t>0$ is sufficiently close to $T_1$, and let us denote (z(t), 1), with $0\le z(t)\le \frac{1}{2}$, the coordinates of the intersection of ${\mathcal {L}}_T$ and the upper border of $\varTheta$ (see Figure 14).

Fig. 14

$\varTheta _2({\mathcal {L}}_T)$ as $T\rightarrow T_1$

Suppose $\theta _{T_1}=0$. This implies that if $T$ is sufficiently close to $T_1$, then line ${\mathcal {L}}_T$ intersects the diagonal line $(OT_1)$ at some $\varvec{\theta }_T=(\theta _T,1-\theta _T)$ with $\lim \limits _{T\rightarrow T_1}\theta _T=\theta _{T_1}=0$, so that $\lim \limits _{t\rightarrow \frac{1}{2}}z(t)=0$.

Let us show that $\lim \limits _{t\rightarrow \frac{1}{2}}z(t)>0$, so that $\theta _{T_1}=0$ does not hold.

From ${\mathcal {L}}_T\perp (OT)$, it is easy to derive that the vertex of $\varTheta _2({\mathcal {L}}_T)={\mathfrak {m}}_T^{-1}(m_2)$ are given by (0, 1), (z(t), 1) and $(0,1-\frac{z(t)}{2t})$, as depicted in Figure 14. The coordinates of the mass centre $\varvec{a}(m_2)$ of $\varTheta _2({\mathcal {L}}_T)$ are given by $\varvec{a}(m_2)=\left( \frac{1}{3}z(t),1-\frac{1}{3}\frac{z(t)}{2t}\right)$. From the equilibrium condition $\varvec{a}(m_2)\in (OT)$, with $(OT)=\{(\theta _1,\theta _2)\in \varTheta ,\, \theta _2=-2t\theta _1+\frac{1}{2}+t\}$, we obtain, for any t sufficiently close to $\frac{1}{2}$,

$$\begin{aligned} 1-\frac{1}{3}\frac{z(t)}{2t}=-2t\frac{1}{3}z(t)+\frac{1}{2}+t. \end{aligned}$$

This gives $z(t)=\frac{t-\frac{1}{2}}{\frac{-1}{6t}+\frac{2}{3}t}=\frac{6t}{2(2t+1)}$, and in particular, $\lim \limits _{t\rightarrow \frac{1}{2}}z(t)=\frac{3}{4}>0$, as expected.

Let us now look at the other limit of $({\mathfrak {m}}_T,\varvec{a}_T)$, i.e. when $T\rightarrow T_0=(0,\frac{1}{2})$. The arguments are similar, but $\varTheta _2({\mathcal {L}}_T)$ has a more complex shape.

We have: (a) actions $\varvec{a}_T(m_1)$ and $\varvec{a}_T(m_2)$ tend to be supported on the horizontal line$(OT_0)$, and (b) line ${\mathcal {L}}_T$ of the Sender’s indifferent states tends to intersect the horizontal line orthogonally at some $\varvec{\theta }_{T_0}=(\theta _{T_0},\frac{1}{2})$, for some $0\le \theta _{T_0}\le \frac{1}{2}$.

Again, if $\theta _{T_0}>0$, this necessarily identifies a (0, 1)-symmetric (deviated if $\theta _{T_0}<\frac{1}{2}$) half-babbling profile of strategies $({\mathfrak {m}}_{H(c)},\varvec{a}_{H(c)})$, $c\in (0,1)$, whereas if $\theta _{T_0}=0$, this identifies a babbling equilibrium. We show that this cannot occur.

Suppose $\theta _{T_0}=0$. This implies that if $T$ is sufficiently close to $T_0$, the set of indifferent states ${\mathcal {L}}_T$ associated with strategies $({\mathfrak {m}}_T,\varvec{a}_T)$ intersects the horizontal line $(OT_0)$ at $\varvec{\theta }_T=(\theta _T,\frac{1}{2})$, with $\lim \limits _{T\rightarrow T_1}\theta _T=\theta _{T_1}=0$. Since ${\mathcal {L}}_T$ is orthogonal to $(OT)$, and since $(OT)$ tends to the horizontal line $(OT_0)$, we have that ${\mathcal {L}}_T$ tends to the vertical line supported by $(0,0)-(0,1)$.

Fig. 15

Partition of $\varTheta _2({\mathcal {L}}_T)$ and subsequent expectations

Let us consider $T(0,\frac{1}{2}+t)$, with $t>0$, sufficiently close to $T_0$, and let us parameterize the set $\varTheta _2({\mathcal {L}}_T)={\mathfrak {m}}_T^{-1}(m_2)$ through a partition consisting of a triangle with (0, 0) as vertex and hypotenuse parallel to ${\mathcal {L}}_T$, and a parallelogram of width z(t) for some $z(t)\in [0,\frac{1}{2})$, as depicted in Figure 15. Equality $\theta _{T_0}=0$ implies $\lim \limits _{t\rightarrow 0}z(t)=0$ and thus we seek to show $\lim \limits _{t\rightarrow 0}z(t)>0$ in order to obtain a contradiction.

From the law of iterated expectations, $\varvec{a}(m_2)$ might be written $\varvec{a}_T(m_2)=(1-\lambda (z(t),t))\varvec{a}^H_T+\lambda (z(t),t)\varvec{a}^L_T$, with $\lambda (z(t),t)=\frac{V(z(t))}{V(z(t))+t}\in [0,1]$, where V(z(t)) and t are the respective Lebesgue measure of the parallelogram and the triangle. The coordinates of their respective mass centre are given by $\varvec{a}^L_T=\left( \frac{2t+z(t)}{2},\frac{1}{2}\right)$ and $\varvec{a}^H_T=\left( \frac{2}{3}t,\frac{2}{3}\right)$ respectively. From $\varvec{a}_T(m_2)\in (OT)$, with $(OT)=\{(\theta _1,\theta _2)\in \varTheta ,\, \theta _2=-2t\theta _1+\frac{1}{2}+t\}$, we obtain, for any t sufficiently close to 0,

$$\begin{aligned} (1-\lambda (z(t),t))\frac{2}{3}+\lambda (z(t),t)\frac{1}{2}=-2t\left( (1-\lambda (z(t),t))\frac{2}{3}t+\lambda (z(t),t)\frac{2t+z(t)}{2}\right) +\frac{1}{2}+t.\nonumber \\ \end{aligned}$$

(12)

Let us set $\lambda =\lim \limits _{t\rightarrow 0}\lambda (z(t),t)\in [0,1]$. Then when $t\rightarrow 0$, Eq. (12) gives $(1-\lambda )\frac{2}{3}+\lambda \frac{1}{2}=\frac{1}{2}$, so that

$$\begin{aligned} \lambda =\lim \limits _{t\rightarrow 0}\lambda (z(t),t)=1. \end{aligned}$$

(13)

Now for $t>0$, Eq. (12) is

$$\begin{aligned} z(t)=\frac{-1}{6\lambda (z(t),t)}\frac{1-\lambda (z(t),t)}{t}+\frac{1}{\lambda (z(t),t)}-\frac{4}{3}t\frac{(1-\lambda (z(t),t))}{\lambda (z(t),t)}-2t, \end{aligned}$$

(14)

and note that

$$\begin{aligned} \frac{1-\lambda (z(t),t)}{t}=\frac{1-\frac{V(z(t))}{V(z(t))+t}}{t}=\frac{t}{t(V(z(t))+t)}=\frac{1}{V(z(t))+t}. \end{aligned}$$

If $\lim \limits _{t\rightarrow 0}z(t)=0$, then we have $\lim \limits _{t\rightarrow 0}V(z(t))=0$, and thus

$$\begin{aligned} \lim \limits _{t\rightarrow 0}\frac{1-\lambda (z(t),t)}{t}=\infty . \end{aligned}$$

Now (13) and (14) give the impossible $0=\infty$. Therefore $\lim \limits _{t\rightarrow 0}z(t)>0$ as expected.

1.7 Proposition 1

We show that the strategies $({\mathfrak {m}}_T,\varvec{a}_T)$ exhibited in Lemma 2 allows any bias $\varvec{b}\in {\mathbb {R}}^2$ to be associated with an influential equilibrium.

According to Lemma 3, maps

$$\begin{aligned} T\mapsto \varvec{b}_T=\frac{\varvec{a}_T(m_1)+\varvec{a}_T(m_2)}{2}+\varvec{\theta }_T, \end{aligned}$$

and $T\mapsto \varvec{a}_T(m_i)$, $i\in \{1,2\}$, can be extended continuously to the closed segment $\{0\}\times [\frac{1}{2},1]$, with non babbling strategies $({\mathfrak {m}}_T,\varvec{a}_T)$ for each $T$. According to the four symmetries of $\varTheta$, and according to Lemma 4, these maps extend continuously to the full border of $\varTheta$. Note that this needs a suitable choice of message assignment across the Ts. W.l.o.g., we assume $O={\mathbb {E}}[\varTheta ]\in {\mathfrak {m}}_{T}^{-1}(m_1)$ for all T.

Now let us consider the continuous maps

$$\begin{aligned} b_{\varTheta }:T\mapsto \Vert \varvec{b}_T\Vert , \end{aligned}$$

and, given $\varvec{b}\in {\mathbb {R}}^2$,

$$\begin{aligned} \pi _{\varvec{b}}:T\mapsto \left\| \varvec{b}\cdot \frac{\varvec{a}_T(m_1)-\varvec{a}_T(m_2)}{\Vert \varvec{a}_T(m_1)-\varvec{a}_T(m_2)\Vert }\right\| . \end{aligned}$$

Each map is defined on the full border of $\varTheta$. Map $\pi _{\varvec{b}}$ is the normalized norm of the projection of $\varvec{b}$ onto $(OT)$. Note that

$$\begin{aligned} b_{\varTheta }(T)=\pi _{\varvec{b}}(T)\,{ \textit{iff}}\,\Vert \varvec{b}_T\Vert \Vert \varvec{a}_T(m_1)-\varvec{a}_T(m_2)\Vert =\Vert \varvec{b}\cdot (\varvec{a}_T(m_1)-\varvec{a}_T(m_2))\Vert , \end{aligned}$$

and that $\Vert \varvec{b}_T\Vert \Vert \varvec{a}_T(m_1)-\varvec{a}_T(m_2)\Vert =\varvec{b}_T\cdot (\varvec{a}_T(m_1)-\varvec{a}_T(m_2))$ according to our choice $O\in {\mathfrak {m}}^{-1}_T(m_1)$ (the alternative choice would have given $\Vert \varvec{b}_T\Vert \Vert \varvec{a}_T(m_1)-\varvec{a}_T(m_2)\Vert =-\varvec{b}_T\cdot (\varvec{a}_T(m_1)-\varvec{a}_T(m_2))$). Then we obtain:

$$\begin{aligned} b_{\varTheta }(T)=\pi _{\varvec{b}}(T)\,{ \textit{iff}}\,{\left\{ \begin{array}{ll}\varvec{b}_T\cdot (\varvec{a}_T(m_1)-\varvec{a}_T(m_2))=\varvec{b}\cdot (\varvec{a}_T(m_1)-\varvec{a}_T(m_2)),\text { or }\\ \varvec{b}_T\cdot (\varvec{a}_T(m_1)-\varvec{a}_T(m_2))=-\varvec{b}\cdot (\varvec{a}_T(m_1)-\varvec{a}_T(m_2)).\end{array}\right. } \end{aligned}$$

Now consider the central symmetry $\rho$ of $\varTheta$. Strategies $\rho ({\mathfrak {m}}_T)$ and $\rho (\varvec{a}_T)$ are given by strategies ${\mathfrak {m}}_{\rho (T)}$ and $\varvec{a}_{\rho (T)}$, associated with bias $\varvec{b}_{\rho (T)}=\rho (\varvec{b}_{T})=-\varvec{b}_{T}$, and we have $\varvec{a}_{T}(m_1)=\varvec{a}_{\rho (T)}(m_2)$, $\varvec{a}_{T}(m_2)=\varvec{a}_{\rho (T)}(m_1)$. Then the second equation above might be written

$$\begin{aligned} \varvec{b}_{\rho (T)}\cdot (\varvec{a}_{\rho (T)}(m_1)-\varvec{a}_{\rho (T)}(m_2))=\varvec{b}\cdot (\varvec{a}_{\rho (T)}(m_1)-\varvec{a}_{\rho (T)}(m_2)). \end{aligned}$$

From Lemma 1, we obtain:

$$\begin{aligned} b_{\varTheta }(T)=\pi _{\varvec{b}}(T)\,{ \textit{iff}}\,[({\mathfrak {m}}_T,\varvec{a}_T)\in {\mathcal {E}}(\varvec{b}),\text { or }({\mathfrak {m}}_{\rho (T)},\varvec{a}_{\rho (T)})\in {\mathcal {E}}(\varvec{b})]. \end{aligned}$$

This means: if $\varvec{b}$ projects as $\varvec{b}_T$ or $-\varvec{b}_T$ onto $(OT)$, then respectively $({\mathfrak {m}}_T,\varvec{a}_T)$ or $({\mathfrak {m}}_{\rho (T)},\varvec{a}_{\rho (T)})$ are profiles of equilibrium strategies of $\varGamma _{\varvec{b}}$. In particular, we have:

$$\begin{aligned} \text {if }b_{\varTheta }(T)=\pi _{\varvec{b}}(T) \text { for some }T,\text { then } {\mathcal {E}}(\varvec{b})\ne \varnothing . \end{aligned}$$

(15)

Next, we apply Bolzano’s Theorem to show that there is always such a T, unless $\varvec{b}$ is small. Let $[{\underline{b}}_\varTheta ,{\overline{b}}_\varTheta ]\subset {\mathbb {R}}^+$ denotes the set of values spanned by $b_{\varTheta }$ (continuous) when $T$ spans the border of $\varTheta$ (compact). We have ${\underline{b}}_\varTheta \ge 0$, and according to Example 3, ${\overline{b}}_\varTheta >0$. Concerning $\pi _{\varvec{b}}$, since $(OT)$ spans every directions when $T$ spans the border of $\varTheta$, $\pi _{\varvec{b}}(T)$ spans $[0,\Vert \varvec{b}\Vert ]$ ($\pi _{\varvec{b}}$ has been normalized).

Note that if $T$ is such that $(OT)\perp \varvec{b}$, then $\pi _{\varvec{b}}(T)=0$, and thus we have

$$\begin{aligned} b_{\varTheta }(T)-\pi _{\varvec{b}}(T)=b_{\varTheta }(T)-0\ge 0 \end{aligned}$$

for any such $T$. Therefore, by continuity and Bolzano’s Theorem, if there exists $T$ such that

$$\begin{aligned} b_{\varTheta }(T)-\pi _{\varvec{b}}(T)\le 0 \end{aligned}$$

(16)

then there exists $T$ onto the border of $\varTheta$ such that $b_{\varTheta }(T)-\pi _{\varvec{b}}(T)=0$, which implies ${\mathcal {E}}(\varvec{b})\ne \varnothing$ according to (15).

First, let us consider the case of a large bias. If $\Vert \varvec{b}\Vert \ge {\overline{b}}_{\varTheta }$, choose $T$ such that $(OT)$ and $\varvec{b}$ have the same direction, so that $\pi _{\varvec{b}}(T)=\Vert \varvec{b}\Vert$. Then we have

$$\begin{aligned} b_{\varTheta }(T)-\pi _{\varvec{b}}(T)\le {\overline{b}}_{\varTheta }-\Vert \varvec{b}\Vert \le 0, \end{aligned}$$

and (16) holds for such a T. Thus ${\mathcal {E}}(\varvec{b})\ne \varnothing$ for all biases $\varvec{b}$ with $\Vert \varvec{b}\Vert \ge {\overline{b}}_\varTheta$.

Next, consider the case of a small bias. Suppose that (16) does not hold when $T$ spans the full border of $\varTheta$. In particular, it does not hold if $T_1=(0,1)$ or $T_2=(0,0)$, so that for $i\in \{1,2\}$, $b_\varTheta (T_i)-\pi _{\varvec{b}}(T_i)>0$, i.e.

$$\begin{aligned} \pi _{\varvec{b}}(T_i)<b_\varTheta (T_i). \end{aligned}$$

(17)

Note that $T_1$ and $T_2$ are such that $({\mathfrak {m}}_{T_1},\varvec{a}_{T_1})=({\mathfrak {m}}_{C(c)},\varvec{a}_{C(c)})$ for some $c\in (-1,0)$, and $({\mathfrak {m}}_{T_2},\varvec{a}_{T_2})=({\mathfrak {m}}_{A(c'},\varvec{a}_{A(c')})=\rho _{\frac{\pi }{2}}({\mathfrak {m}}_{C(c')},\varvec{a}_{C(c')})$ for some $c'\in (-1,0)$.²³ In particular, according to Example 2, we have $\varvec{b}_{T_i}=(b_1,b_2)$ for some $(b_1,b_2)\in {\mathbb {R}}^2$ such that $|b_2-b_1|<\frac{1}{2}$. Furthermore, since $\varvec{b}_{T_1}$ is supported on $(OT_1)$, we have $b_2=-b_1$. Hence $|b_2-b_1|=2|b_1|$ so that $|b_1|<\frac{1}{4}$. Then we obtain:

$$\begin{aligned} b_\varTheta (T_1)=\Vert \varvec{b}_{T_1}\Vert =\sqrt{b_1^2+b_2^2}=|b_1|\sqrt{2}<\frac{\sqrt{2}}{4}. \end{aligned}$$

Similarly, we have

$$\begin{aligned} b_\varTheta (T_2)=\Vert \varvec{b}_{T_2}\Vert <\frac{\sqrt{2}}{4}. \end{aligned}$$

Consequently, from (17), $\varvec{b}$ projects onto $(OT_1)$ and $(OT_2)$ as vectors the norm of them is less than $\frac{\sqrt{2}}{4}$. Since $(OT_1)$ and $(OT_2)$ are orthogonal, we can deduce

$$\begin{aligned} \Vert \varvec{b}\Vert =\sqrt{\pi _{\varvec{b}}(T_1)^2+\pi _{\varvec{b}}(T_2)^2}<\sqrt{b_\varTheta (T_1)^2+b_\varTheta (T_2)^2}<\sqrt{\frac{1}{4}}=\frac{1}{2}. \end{aligned}$$

Then, setting $\varvec{b}=(b'_1,b'_2)\in {\mathbb {R}}^2$, it is necessary that $|b'_1|<\frac{1}{2}$ and $|b'_2|<\frac{1}{2}$, so that $|b'_2-b'_1|\le |b'_2|-|b'_1|<\frac{1}{2}$. Again, according to Example 2, there is some $c''$ such that $({\mathfrak {m}}_{C(c'')},\varvec{a}_{C(c'')})$ is an equilibrium strategy of $\varGamma _{\varvec{b}}$, and thus ${\mathcal {E}}(\varvec{b})\ne \varnothing$.

1.8 Lemma 5

We show that given $T(0,\frac{1}{2}+t)\in \{0\}\times (\frac{1}{2},1)$, for some $t\in (0,\frac{1}{2})$, there is a unique line ${\mathcal {L}}_T$ such that (i) ${\mathcal {L}}_T$ is orthogonal to (OT), (ii) ${\mathcal {L}}_T$ partitions $\varTheta$ through $\varTheta =\varTheta _1\cup \varTheta _2$, with $|\varTheta _i|>0$, $i\in \{1,2\}$ and w.l.o.g. $O\in \varTheta _1$, and (iii) ${\mathbb {E}}[\varTheta _i]\in (OT)$, $i\in \{1,2\}$. Notice that the existence of ${\mathcal {L}}_T$ has already been established and it is sufficient to show that if conditions (i), (ii) and (iii) are satisfied for some ${\mathcal {L}}$, then ${\mathcal {L}}$ is uniquely determined.

Let us parameterize any line ${\mathcal {L}}$ satisfying these conditions through

$$\begin{aligned} {\mathcal {L}}=\{\varvec{\theta }\in \varTheta ,\,\theta _1=2e\theta _2+c-e\}, \end{aligned}$$

so that it passes through the points $(c,\frac{1}{2})$, with $c\le \frac{1}{2}$, and through $(c+e,1)$ and $(c-e,0)$ for some $e>0$ (see Figure 16).

Fig. 16

Parametrization of ${\mathcal {L}}_T$

Notice that Condition (i) gives $((c+e,1)-(c,\frac{1}{2}))\cdot ((\frac{1}{2},\frac{1}{2})-(0,\frac{1}{2}+t))=0$, i.e. $e=t$, and in particular,

$$\begin{aligned} e\in \left( 0,\frac{1}{2}\right) . \end{aligned}$$

(18)

Condition (ii) implies

$$\begin{aligned} c+e<1,\text { and }c-e<\frac{1}{2}, \end{aligned}$$

(19)

since otherwise, we would have $O\notin \varTheta _1$. To derive Condition (iii) with respect to c and e, let us compute the expectations with respect to each side of ${\mathcal {L}}_T$ and with respect to parameters c and e. We compute

$$\begin{aligned} E_{11}(c,e)&=\frac{\iint _{\theta _1>2e\theta _2+c-e}\theta _1\mathop {}\!\mathrm {d}\varvec{\theta }}{\iint _{\theta _1>2e\theta _2+c-e}\mathop {}\!\mathrm {d}\varvec{\theta }}=\frac{3c^2+e^2-3}{6(c-1)},&E_{12}(c,e)&=\frac{\iint _{\theta _1<2e\theta _2+c-e}\theta _1\mathop {}\!\mathrm {d}\varvec{\theta }}{\iint _{\theta _1<2e\theta _2+c-e}\mathop {}\!\mathrm {d}\varvec{\theta }}=\frac{3c^2+e^2}{6c},\\ E_{21}(c,e)&=\frac{\iint _{\theta _1>2e\theta _2+c-e}\theta _2\mathop {}\!\mathrm {d}\varvec{\theta }}{\iint _{\theta _1> 2e\theta _2+c-e}\mathop {}\!\mathrm {d}\varvec{\theta }}=\frac{3c+e-3}{6(c-1)},&E_{22}(c,e)&=\frac{\iint _{\theta _1<2e\theta _2+c-e}\theta _2\mathop {}\!\mathrm {d}\varvec{\theta }}{\iint _{\theta _1<2e\theta _2+c-e}\mathop {}\!\mathrm {d}\varvec{\theta }}=\frac{3c+e}{6c}. \end{aligned}$$

Now we have that ${\mathcal {L}}$ is orthogonal to $(\varvec{E}_1\varvec{E}_2)$, with $\varvec{E}_1(E_{11}(c,e),E_{21}(c,e))$ and $\varvec{E}_2(E_{12}(c,e),E_{22}(c,e))$, iff$(E_{11}-E_{12},E_{21}-E_{22})\cdot (e,\frac{1}{2})=0$, which gives

$$\begin{aligned} e^2=3c-3c^2-\frac{1}{2}. \end{aligned}$$

(20)

According to (18), $e^2$ spans $(0,\frac{1}{4})$ and therefore (20) has two solutions $c_1=\frac{3-\sqrt{3}\sqrt{1-4e^2}}{6}$ and $c_2=\frac{3+\sqrt{3}\sqrt{1-4e^2}}{6}$. Now in particular, (19) gives $c_2+e<1$, which implies $e<\frac{1}{4}$. It also gives $c_2-e<\frac{1}{2}$, which implies $e>\frac{1}{4}$. Hence, for any $e\in (0,\frac{1}{2})$, $c_2$ does not satisfy (19), and $c_1$ is the only solution to (20). Hence, there is a unique ${\mathcal {L}}$ that satisfies conditions (i), (ii) and (iii).