Information acquisition interactions in two-player quadratic games

Abstract

This paper considers two-player quadratic games to examine the relation between strategic interactions in actions and in information decisions. We analyze the role of external effects and of the relative intensities with which the players’ actions interact with the uncertain payoff-relevant parameter. We show that, under some conditions on the quadratic preferences, information choices become substitutes when actions are sufficiently complementary. When attention is restricted to beauty contest games, our results contrast qualitatively with the case studied by Hellwig and Veldkamp (Review of Economic Studies, 76(1)223–251, 2009), where the set of players is a continuum.

Appendix

Proof of Lemma 1

To ease notation, let us denote by $E_{i}[\cdot ]=E[\cdot |s_{i},\,\omega ]$ player $i$’s expectation operator conditioned on observing signal realization $s_{i}$ and on the information profile $\omega .$ From the specification of preferences in (1) and from Assumption 1(i), it follows that, for each information profile $\omega \in [0,\,1]^{2},$ player $i$’s optimal action strategy is given by:

$$\begin{aligned} \alpha ^{*}_{i}\left( s_{i},\,\omega \right) =\rho E_{i}[\tilde{\theta }] +\lambda E_{i}\left[ {\alpha }_{j}\left( \tilde{s}_{j},\,\omega \right) \right] . \end{aligned}$$

By iterating recursively, we obtain

$$\begin{aligned} \alpha ^{*}_{i}\left( s_{i},\,\omega \right) =\rho \sum ^{\infty }_{k=1}\lambda ^{k-1} E_{i}E_{j}E_{i}\cdots E_{p(k)}[\tilde{\theta }], \end{aligned}$$

(9)

where $E_{i}E_{j}E_{i}\cdots E_{p(k)}[\tilde{\theta }]$ denotes the $(k+1)$-order iterated expectations of $\tilde{\theta },$ and the player index $p(k)$ equals $i$ if $k$ even and equals $j$ if $k$ is odd.

In order to obtain a closed expression for optimal action strategies, we need an operator that allows us to keep track of the discounted nested expectations of the players. To do this, we make use of the knowledge index introduced by Calvó-Armengol and de Martí (2007) in their work on communication in networks. Given the assumed information structure, this knowledge index is specified for our two-player game in the form of the two by two matrix

$$\begin{aligned} \Omega := \begin{pmatrix} 0 &{} \omega _{1} \omega _{2} \\ \omega _{1} \omega _{2} &{} 0 \end{pmatrix}. \end{aligned}$$

Substitution of the posterior expectation $E_{i}[\tilde{\theta }]=E[\tilde{\theta }|s_{i},\,\omega ]$ given in (4) into the expression for player $i$’s optimal action strategy in (9) yields

$$\begin{aligned} \alpha ^{*}_{i}\left( s_{i},\,\omega \right)&= \rho \left( \lambda ^{0}\left[ \mu +\omega _{i}\left( s_{i}-\mu \right) \right] + \lambda ^{1}\left[ \mu +\omega _{j}\omega _{i}\left( s_{i}-\mu \right) \right] \right. \\&+\left. \lambda ^{2}\left[ \mu +\omega _{i}\omega _{j}\omega _{i}\left( s_{i}-\mu \right) \right] +\cdots \right) . \end{aligned}$$

Then, considering $\omega =(\omega _{1},\,\omega _{2})$ as a vector, the expression above for player $i$’s optimal action strategy can be rewritten as

$$\begin{aligned} \alpha ^{*}_{i}\left( s_{i},\,\omega \right)&= \frac{\rho \mu }{1-\lambda }+\rho \omega \cdot \sum ^{\infty }_{k=0} \lambda ^{k}\Omega ^{k} \cdot e_{i}\left( s_{i}-\mu \right) \\&= \frac{\rho \mu }{1-\lambda }+\rho \omega \cdot (I-\lambda \Omega )^{-1} \cdot e_{i}\left( s_{i}-\mu \right) , \end{aligned}$$

where $I$ denotes the two by two identity matrix and $e_{i}$ is the $i$th vector of the canonical basis of $\mathbb R ^{2}.$ To verify that the infinite sum $\sum ^{\infty }_{k=0} \lambda ^{k}\Omega ^{k}=(I-\lambda \Omega )^{-1}$ above is well-defined, we apply a linear algebra result shown by Debreu and Herstein (1953). According to this result, the convergence of $\sum ^{\infty }_{k=0} \lambda ^{k}\Omega ^{k}$ is guaranteed if $\left| \lambda \right|$ is strictly less than the inverse of the largest eigenvalue of $\Omega .$ It can be easily verified that the largest eigenvalue of $\Omega$ is $z=+\sqrt{\omega _{1}\omega _{2}} \in [0,\,1]$ so that $1/z \ge 1$ for all $\omega _{1},\,\omega _{2} \in [0,\,1].$ Since we are assuming that $\lambda \in (-1,\,1),$ it follows that $\left| \lambda \right| <1/z,$ as required.

Then, by computing the inverse of the matrix $(I-\lambda \Omega ),$ we obtain

$$\begin{aligned} \alpha ^{*}_{i}\left( s_{i},\,\omega \right) =\frac{\rho \mu }{1-\lambda }+ \frac{\rho \omega _{i}(1+\lambda \omega _{j})}{1-\lambda ^{2}\omega _{i}\omega _{j}}\left( s_{i}-\mu \right) , \end{aligned}$$

as stated.

Finally, it remains to show that the action strategy profile $\alpha ^{*}=(\alpha ^{*}_{1},\,\alpha ^{*}_{2})$ that we have obtained is unique. To do this, we follow the approach used by Calvó-Armengol and de Martí (2009) to prove uniqueness of Nash equilibrium for a one-stage, beauty contest game with an exogenously given information choice (Theorem 1). This approach makes use of two ingredients: (a) some available results of existence of potential games⁹ with team payoffs for quadratic games, and (b) a central result of team theory due to Radner (1962).

We begin by specifying the two-player team payoff function

$$\begin{aligned} \nu \left( a_{i},\,a_{j},\,\theta \right) :=-(1-\lambda )\left[ \left( a_{i}-\theta \right) ^{2}+\left( a_{j}-\theta \right) ^{2}\right] -\lambda \left( a_{i}-a_{j}\right) ^{2}. \end{aligned}$$

Now, consider a given information profile $\omega$ and a given signal realization $s_{i}.$ Because our payoff function $v$ is assumed to be common for both players, the cross-derivatives of $v$ with respect to the players’ actions are the same for both of them. Then, since $v$ is quadratic, it follows from Lemma 6 in Ui (2009) that the payoff $v$ admits $\nu$ as a potential. This result implies that, for a fixed pair $(s_{i},\,\omega ),$ a player’s decision problem in actions for our game can be solved using the team payoff function $\nu .$ Given this, we can then rely on a key result due to Radner (1962) to show that optimal actions are unique in the team game with payoffs $\nu .$ It follows from Theorem 4 in Radner (1962) that optimal actions are unique for the game with payoffs $\nu$ if the matrix $Q:=({\partial ^{2} \nu (a,\,\theta )}/{\partial a_{i}\partial a_{j}})$ is negative definite. For the above specification of $\nu ,$ we obtain

$$\begin{aligned} Q=2\begin{pmatrix} -1 &{} \lambda \\ \lambda &{} -1 \end{pmatrix}, \end{aligned}$$

a matrix whose eigenvalues are $z_{1,2}=-1 \pm \sqrt{1-(1-\lambda ^{2})}<0$ for each $\lambda \in (-1,\,1).$ Therefore, the matrix $Q$ above is definite negative and optimal actions $\alpha ^{*}_{i}(s_{i},\,\omega )$ are unique for the given information profile $\omega$ and the given signal realization $s_{i}.$ $\square$

1.1 Derivation of a closed expression for the value of information

Using the specification of preferences given in (1) together with the expression of player $i$’s ex-ante expected payoff in (6), we obtain

$$\begin{aligned} V_{i}&= E\left[ v\left( \alpha ^{*}_{i}\left( \tilde{s}_{i},\,\omega \right) ,\,\alpha ^{*}_{j}\left( \tilde{s}_{j},\,\omega \right) ,\,\tilde{\theta }\right) \right] -K\left( \omega _{i}\right) \\&= -E\left[ \alpha ^{*}_{i}\left( \tilde{s}_{i},\,\omega \right) ^{2}\right] -\pi E\left[ \alpha ^{*}_{j}\left( \tilde{s}_{j},\,\omega \right) ^{2}\right] -\delta E\left[ \tilde{\theta }^{2}\right] \\&+2\lambda E\left[ \alpha ^{*}_{i}\left( \tilde{s}_{i},\,\omega \right) \alpha ^{*}_{j}\left( \tilde{s}_{j},\,\omega \right) \right] \\&+2\rho E\left[ \alpha ^{*}_{i}\left( \tilde{s}_{i},\,\omega \right) \tilde{\theta }\right] +2 \phi E\left[ \alpha ^{*}_{j}\left( \tilde{s}_{j},\,\omega \right) \tilde{\theta }\right] -K\left( \omega _{i}\right) . \end{aligned}$$

Then, substitution of the players’ optimal action strategies obtained in Lemma 1 (specified in (7)) into the expression above gives

$$\begin{aligned} V_{i}&= \xi -m^{2}_{i}E\left[ \left( \tilde{s}_{i}-\mu \right) ^{2}\right] -\pi m^{2}_{j} E\left[ \left( \tilde{s}_{j}-\mu \right) ^{2}\right] \\&+2\lambda m_{i}m_{j} E\left[ \left( \tilde{s}_{i}-\mu \right) \left( \tilde{s}_{j}-\mu \right) \right] +2\rho m_{i}E\left[ \left( \tilde{s}_{i}-\mu \right) \tilde{\theta }\right] \\&+2 \phi m_{j}E\left[ \left( \tilde{s}_{j}-\mu \right) \tilde{\theta }\right] -K\left( \omega _{i}\right) , \end{aligned}$$

where $\xi =2(\rho +\phi )\mu c +(2\lambda -1-\pi )c^{2}-\delta (\sigma ^{2}_{\theta }+\mu ^{2})$ is a term which does not depend on the players’ information choices. Now, using the assumed information structure, the expression above can be rewritten as

$$\begin{aligned} V_{i}\!=\!\xi \!-\!m^{2}_{i}\left[ \sigma ^{2}_{\theta }/\omega _{i}\right] \! -\!\pi m^{2}_{j}\left[ \sigma ^{2}_{\theta }/\omega _{j}\right] +2\lambda m_{i}m_{j}\sigma ^{2}_{\theta }+2\rho m_{i}\sigma ^{2}_{\theta }+2\phi m_{j}\sigma ^{2}_{\theta }\!-\!K\left( \omega _{i}\right) . \end{aligned}$$

By developing the expressions of $m_{i}$ and $m_{j}$ given in Lemma 1, we obtain, after rearranging terms,

$$\begin{aligned} V_{i}&= \xi +\frac{\sigma ^{2}_{\theta } \rho ^{2}}{(1-\lambda ^{2}\omega _{i}\omega _{j})^{2}} \big [-\omega _{i}\left( 1+\lambda \omega _{j}\right) ^{2}-\pi \omega _{j}\left( 1+\lambda \omega _{i}\right) ^{2} \\&+2\lambda \omega _{i} \omega _{j}\left( 1+\lambda \omega _{j}\right) \left( 1+\lambda \omega _{i}\right) \\&+2\omega _{i}\left( 1\!+\!\lambda \omega _{j}\right) \left( 1\!-\!\lambda ^{2}\omega _{i}\omega _{j}\right) \!+\!2\left( \frac{\phi }{\rho } \right) \omega _{j}\left( 1\!+\!\lambda \omega _{i}\right) \left( 1\!-\!\lambda ^{2}\omega _{i}\omega _{j}\right) \big ]\!-\!K\left( \omega _{i}\right) . \end{aligned}$$

The above expression gives us player $i$’s ex-ante expected payoff (when both players choose their optimal action strategies) as a function of the information profile $\omega .$ This is our expression of interest to analyze player $i$’s incentives to acquire information when player $j$ changes (locally) the amount of information that she acquires. This approach requires that we compute the second order derivative $\partial ^{2} V_{i}/\partial \omega _{i}\partial \omega _{j}$ from the expression above. Then, it can be verified that

$$\begin{aligned} \frac{\partial ^{2} V_{i}}{\partial \omega _{i}\partial \omega _{j}}&= \lambda \left[ \frac{\sqrt{2} \sigma _{\theta }\rho }{(1-\lambda ^{2}\omega _{i}\omega _{j})^{2}} \right] ^{2} \Bigg [\left[ 1-\pi + \left( \frac{\phi }{\rho } \right) \right] +[2-\pi ]\lambda \omega _{i} \nonumber \\&+ \left[ 1-2\pi + 2\left( \frac{\phi }{\rho } \right) \right] \lambda \omega _{j} + 4[1-\pi ]\lambda ^{2}\omega _{i}\omega _{j}+[1-2\pi ]\lambda ^{3}\omega ^{2}_{i}\omega _{j} \nonumber \\&+\left[ 2-\pi - 2\left( \frac{\phi }{\rho } \right) \right] \lambda ^{3}\omega _{i}\omega ^{2}_{j}+ \left[ 1-\pi -\left( \frac{\phi }{\rho } \right) \right] \lambda ^{4}\omega ^{2}_{i}\omega ^{2}_{j} \Bigg ]. \end{aligned}$$

(10)

In addition, for the particular case of beauty contest payoffs ($\pi =\lambda ^{2},\,\delta =(1-\lambda )^{2}, \rho =(1-\lambda ),$ and $\phi =-\lambda (1-\lambda )$), the expression above yields

$$\begin{aligned} \frac{\partial ^{2}V_{i}}{\partial \omega _{i}\partial \omega _{j}}\bigg |_{\mathrm{bc}}&= \lambda \left[ \frac{\sqrt{2}\sigma _{\theta }(1-\lambda )}{(1-\lambda ^{2}\omega _{i}\omega _{j})^{2}} \right] ^{2} \Bigg [\left[ 1-\lambda -\lambda ^{2}\right] +\left[ 2-\lambda ^{2}\right] \lambda \omega _{i} \nonumber \\&+ \left[ 1-2\lambda -2\lambda ^{2}\right] \lambda \omega _{j} +4\left[ 1-\lambda ^{2}\right] \lambda ^{2}\omega _{i}\omega _{j}+\left[ 1-2\lambda ^{2}\right] \lambda ^{3}\omega ^{2}_{i}\omega _{j} \nonumber \\&+\left[ 2+2\lambda -\lambda ^{2}\right] \lambda ^{3}\omega _{i}\omega ^{2}_{j}+ \left[ 1+\lambda -\lambda ^{2}\right] \lambda ^{4}\omega ^{2}_{i}\omega ^{2}_{j} \Bigg ]. \end{aligned}$$

(11)

With the expressions for the second order derivatives of the ex-ante expected payoffs given by (10) and (11) at hand, we proceed to the proofs of the propositions.

Proof of Proposition 1

Notice that the expression in (10) for $\partial ^{2}V_{i}/\partial \omega _{i}\partial \omega _{j}$ can be rewritten as ${\partial ^{2} V_{i}}/{\partial \omega _{i}\partial \omega _{j}}= h(\lambda ,\,\rho ; \,\omega )f_{i}(\lambda ,\,\pi ,\,(\phi /\rho );\,\omega ),$ where

$$\begin{aligned} h(\lambda ,\,\rho ;\,\omega ):= \lambda \left[ \frac{\sqrt{2}\sigma _{\theta } \rho }{(1-\lambda ^{2}\omega _{i}\omega _{j})^{2}}\right] ^{2}, \end{aligned}$$

(12)

and

$$\begin{aligned} f_{i}(\lambda ,\,\pi ,\,(\phi /\rho );\,\omega )&:= [1-\pi + (\phi /\rho )] +[2-\pi ]\lambda \omega _{i}+[1-2\pi +2(\phi /\rho )]\lambda \omega _{j} \nonumber \\&+4[1-\pi ]\lambda ^{2}\omega _{i}\omega _{j}+[1-2\pi ]\lambda ^{3}\omega ^{2}_{i}\omega _{j} \nonumber \\&+[2-\pi - 2(\phi /\rho )]\lambda ^{3}\omega _{i}\omega ^{2}_{j}+[1-\pi -(\phi /\rho )] \lambda ^{4}\omega ^{2}_{i}\omega ^{2}_{j}.\nonumber \\ \end{aligned}$$

(13)

Note first that the sign of $\partial ^{2}V_{i}/\partial \omega _{i}\partial \omega _{j}$ coincides with the sign of $h$ when actions are complements, $\lambda \in (0,\,1),$ and the sign of $\partial ^{2}V_{i}/\partial \omega _{i}\partial \omega _{j}$ is different from the sign of $h$ when actions are substitutes, $\lambda \in (-1,\,0).$ In addition, note that the sign of the function $h$ is completely determined by the sign of $\lambda .$ Therefore, if $f_{i}>0,$ then information decisions feature the same type of strategic interactions as actions. Thus, we proceed by analyzing the sign of the function $f_{i}$ when $\lambda$ is close to zero.

• (i) The result follows directly from the specification of the function $\partial ^{2}V_{i}/\partial \omega _{i}\partial \omega _{j}$ in (10).

• (ii) Take a given $\omega \in (0,\,1)^{2}$ and suppose that the condition $\phi /\rho >-1+\pi$ holds. Then, we have

  $$\begin{aligned} f_{i}(0,\,\pi ,\,(\phi /\rho );\,\omega )=1-\pi +(\phi /\rho )>0. \end{aligned}$$

The result follows because $f_{i}(\lambda ,\,\pi ,\,(\phi /\rho );\,\omega )$ is a continuous function in $\lambda \in (-1,\,1),$ and (a) $h(\lambda ,\,\rho ;\,\omega )>0$ for each $\lambda \in (0,\,1)$ whereas (b) $h(\lambda ,\,\rho ;\,\omega )<0$ for each $\lambda \in (-1,\,0).$ $\square$

Proof of Proposition 2

As in the proof of Proposition 1, we use the fact that ${\partial ^{2} V_{i}}/{\partial \omega _{i}\partial \omega _{j}}= h(\lambda ,\,\rho ;\,\omega )f_{i}(\lambda ,\,\pi ,\,(\phi /\rho );\,\omega ),$ where the functions $h$ and $f_{i}$ are specified in (12) and (13), respectively. Note that the sign of the function $h$ is determined by the sign of $\lambda .$ Therefore, for $\lambda >0,$ it follows that information decisions are substitutes if $f_{i}<0.$ Analogously, for $\lambda <0,$ it follows that information decisions are complements if $f_{i}<0$ too. Thus, we proceed by studying the sign of the function $f_{i}$ when $\lambda$ approaches 1, (i) below, and when $\lambda$ approaches -1, (ii) below.

• (i) Take a given $\omega \in (0,\,1)^{2}.$ We have

  $$\begin{aligned} \lim _{\lambda \rightarrow 1}f_{i}(\lambda ,\,\pi ,\,(\phi /\rho );\,\omega )&= \left[ 1+2\omega _{i}+\omega _{j}+4\omega _{i}\omega _{j}+\omega ^{2}_{i}\omega _{j}+2\omega _{i}\omega ^{2}_{j}+\omega ^{2}_{i}\omega ^{2}_{j}\right] \\&+\left[ 1+2\omega _{j}-2\omega _{i}\omega ^{2}_{j}-\omega ^{2}_{i}\omega ^{2}_{j}\right] (\phi /\rho )\\&-\left[ 1\!+\!\omega _{i}+2\omega _{j}\!+\!4\omega _{i}\omega _{j}\!+\!2\omega ^{2}_{i}\omega _{j}\!+\!\omega _{i}\omega ^{2}_{j}\!+\!\omega ^{2}_{i}\omega ^{2}_{j}\right] \pi . \end{aligned}$$

  From the expression above, it follows that $\lim _{\lambda \rightarrow 1}f_{i}(\lambda ,\,\pi ,\,(\phi /\rho );\,\omega )<0$ if and only if ${\phi }/{\rho }<-\kappa _{c}(\omega )+\tau _{c}(\omega )\pi ,$ where $\kappa _{c}(\omega )$ and $\tau _{c}(\omega )$ are specified as

  $$\begin{aligned} \kappa _{c}(\omega )&:= \frac{1+2\omega _{i}+\omega _{j}+\omega _{i}\omega _{j}[4+\omega _{i}+2\omega _{j}+\omega _{i}\omega _{j}]}{[1-\omega ^{2}_{i}\omega ^{2}_{j}]+2\omega _{j}[1-\omega _{i}]},\\ \tau _{c}(\omega )&:= \frac{1+\omega _{i}+2\omega _{j}+\omega _{i}\omega _{j}[4+2\omega _{i}+\omega _{j}+\omega _{i}\omega _{j}]}{[1-\omega ^{2}_{i}\omega ^{2}_{j}]+2\omega _{j}[1-\omega _{i}]}. \end{aligned}$$

  Inspection of both expressions above indicates that $\kappa _{c}(\omega ),\,\tau _{c}(\omega )>0$ for each $\omega \in (0,\,1)^{2}.$ It remains to show that the subset of $X=[0,\,1] \times [-1,\,1]$ specified by the inequality ${\phi }/{\rho }<-\kappa _{c}(\omega )+\tau _{c}(\omega )\,\pi$ is nonempty. To do this, it suffices to verify that $\tau _{c}(\omega )-\kappa _{c}(\omega )>-1.$ Note that

  $$\begin{aligned} \tau _{c}(\omega )-\kappa _{c}(\omega )&= \frac{(\omega _{j}-\omega _{i})(1-\omega _{i}\omega _{j})}{(1-\omega _{i}\omega _{j})(1+\omega _{i}\omega _{j})+2\omega _{j}(1-\omega _{i})} >-1\\&\Leftrightarrow \left( 1-\omega _{i}\omega _{j}\right) \left[ 1+\omega _{j}+\omega _{i}\left( \omega _{j}-1\right) \right] +2\omega _{j}\left( 1-\omega _{i}\right) >0, \end{aligned}$$

  an inequality which holds for each $\omega \in (0,\,1)^{2}.$ Then, the result follows because $f_{i}(\lambda ,\,\pi ,\,(\phi /\rho );\,\omega )$ is a continuous function in $\lambda \in (0,\,1)$ and $h(\lambda ,\,\rho ;\,\omega )>0$ for each $\lambda \in (0,\,1).$

• (ii) Take a given $\omega \in (0,\,1)^{2}.$ We have

  $$\begin{aligned} \lim _{\lambda \rightarrow -1}f_{i}(\lambda ,\,\pi ,\,(\phi /\rho );\,\omega )&= \left[ 1-2\omega _{i}-\omega _{j}+4\omega _{i}\omega _{j}-\omega ^{2}_{i}\omega _{j} -2\omega _{i}\omega ^{2}_{j}+\omega ^{2}_{i}\omega ^{2}_{j}\right] \\&+\left[ 1-2\omega _{j}+2\omega _{i}\omega ^{2}_{j}-\omega ^{2}_{i}\omega ^{2}_{j}\right] (\phi /\rho )\\&-\left[ 1-\omega _{i}-2\omega _{j}+4\omega _{i}\omega _{j}-2\omega ^{2}_{i}\omega _{j}-\omega _{i}\omega ^{2}_{j}+\omega ^{2}_{i}\omega ^{2}_{j}\right] \pi . \end{aligned}$$

  From the expression above, it follows that $\lim _{\lambda \rightarrow -1}f_{i}(\lambda ,\,\pi ,\,(\phi /\rho );\,\omega )<0$ if and only if ${\phi }/{\rho }<-\kappa _{s}(\omega )+\tau _{s}(\omega )\pi ,$ where $\kappa _{s}(\omega )$ and $\tau _{s}(\omega )$ are specified as

  $$\begin{aligned} \kappa _{s}(\omega )&:= \frac{1-2\omega _{i}-\omega _{j}+\omega _{i}\omega _{j}[4-\omega _{i}-2\omega _{j}+\omega _{i}\omega _{j}]}{[1-\omega _{i}\omega _{j}][1+\omega _{j}(\omega _{i}-2)]},\\ \tau _{s}(\omega )&:= \frac{1-\omega _{i}-2\omega _{j}+\omega _{i}\omega _{j}[4-2\omega _{i}-\omega _{j}+\omega _{i}\omega _{j}]}{[1-\omega _{i}\omega _{j}][1+\omega _{j}(\omega _{i}-2)]}. \end{aligned}$$

  Notice that the sign of both $\kappa _{s}(\omega )$ and $\tau _{s}(\omega )$ depends on the value of $\omega \in (0,\,1)^{2}.$ Nevertheless, observe that

  $$\begin{aligned} \lim _{\omega _{j} \rightarrow 0} \kappa _{s}(\omega )&= 1-2\omega _{i} \in (-1,\,1)\quad \forall \omega _{i} \in (0,\,1), \\ \lim _{\omega _{j} \rightarrow 0} \tau _{s}(\omega )&= 1-\omega _{i} \in (0,\,1)\quad \forall \omega _{i} \in (0,\,1). \end{aligned}$$

  It trivially follows that $\lim _{\omega _{j} \rightarrow 0}[\tau _{s}(\omega )-\kappa _{s}(\omega )]=\omega _{i} \in (0,\,1).$ Therefore, for values of $\omega _{j}$ sufficiently close to 0, we have $\tau _{s}(\omega ),\,\kappa _{s}(\omega )>0,$ and $\tau _{s}(\omega )-\kappa _{s}(\omega ) \in (0,\,1).$ Then, the subset of $X$ specified by ${\phi }/{\rho }<-\kappa _{s}(\omega )+\tau _{s}(\omega )\pi$ is nonempty.

On the other hand, notice that

$$\begin{aligned} \lim _{\omega _{j} \rightarrow 1} \kappa _{s}(\omega )&= 0, \\ \lim _{\omega _{j} \rightarrow 1} \tau _{s}(\omega )&= \frac{(1-\omega _{i})^{2}}{(1-\omega _{i})^{2}}=1. \end{aligned}$$

Therefore, for values of $\omega _{j}$ sufficiently close to 1, it follows that $\lim _{\lambda \rightarrow -1}f_{i}(\lambda ,\pi ,\,(\phi /\rho );\,\omega )<0$ if and only if $\phi /\rho <\pi .$

The result follows because $f_{i}(\lambda ,\,\pi ,\,(\phi /\rho );\,\omega )$ is a continuous function in both $\lambda \in (-1,\,0)$ and in $\omega \in (0,\,1)^{2},$ and $h(\lambda ,\,\rho ;\,\omega )<0$ for each $\lambda \in (-1,\,0).$ $\square$

Proof of Proposition 3

The expression in (11) for $\partial ^{2} V_{i}/\partial \omega _{i}\partial \omega _{j}|_{\mathrm{bc}}$ can be rewritten as ${\partial ^{2} V_{i}}/{\partial \omega _{i}\partial \omega _{j}}|_{\mathrm{bc}}= q(\lambda ;\,\omega )g_{i}(\lambda ;\,\omega ),$ where

$$\begin{aligned} q(\lambda ;\,\omega ):=\lambda \left[ \frac{\sqrt{2} \sigma _{\theta }(1-\lambda )}{(1-\lambda ^{2}\omega _{i}\omega _{j})^{2}} \right] ^{2},\end{aligned}$$

and

$$\begin{aligned} g_{i}(\lambda ;\,\omega )&:= \left( 1-\lambda -\lambda ^{2}\right) +\left( 2-\lambda ^{2}\right) \lambda \omega _{i}+\left( 1-2\lambda -2\lambda ^{2}\right) \lambda \omega _{j}+4\left( 1-\lambda ^{2}\right) \lambda ^{2}\omega _{i}\omega _{j}\\&+ \left( 1-2\lambda ^{2}\right) \lambda ^{3}\omega ^{2}_{i}\omega _{j} +\left( 2+2\lambda -\lambda ^{2}\right) \lambda ^{3}\omega _{i}\omega ^{2}_{j}+\left( 1+\lambda - \lambda ^{2}\right) \lambda ^{4}\omega ^{2}_{i}\omega ^{2}_{j}. \end{aligned}$$

Note that the sign of $\lambda$ determines the sign of the function $q.$ We then turn to analyze the relation between $\lambda \in (-1,\,1)$ and the sign of $g_{i}.$

• (i) The result follows directly from the specification of the function $\partial ^{2}V_{i}/\partial \omega _{i}\partial \omega _{j}|_{\mathrm{bc}}$ in (11).

• (ii) Take a given $\omega \in (0,\,1)^{2}.$ We have $g_{i}(0;\,\omega )=1>0.$ Then, the result follows because $g_{i}(\lambda ;\,\omega )$ is a continuous function in $\lambda \in (-1,\,1),$ and (a) $q(\lambda ;\,\omega )>0$ for each $\lambda \in (0,\,1)$ whereas (b) $q(\lambda ;\,\omega )<0$ for each $\lambda \in (-1,\,0).$

• (iii) Take a given $\omega \in (0,\,1)^{2}.$ We have

  $$\begin{aligned} \lim _{\lambda \rightarrow 1}g_{i}(\lambda ;\,\omega )=-1+\omega _{i}\left[ 1+\omega _{i}\omega _{j}\left( \omega _{j}-1\right) \right] +3\omega _{j}\left( \omega _{i}\omega _{j}-1\right) , \end{aligned}$$

  an expression which is strictly negative for each $\omega \in (0,\,1)^{2}.$ Then, the result follows because $g_{i}(\lambda ;\,\omega )$ is a continuous function in $\lambda \in (0,\,1)$ and $q(\lambda ;\,\omega )>0$ for each $\lambda \in (0,\,1).$ $\square$