Coarse correlated equilibria in linear duopoly games

Abstract

For duopoly models, we analyse the concept of coarse correlated equilibrium using simple symmetric devices that the players choose to commit to in equilibrium. In a linear duopoly game, we prove that Nash-centric devices, involving a sunspot structure, are simple symmetric coarse correlated equilibria. Any small unilateral perturbation from such a structure fails to be an equilibrium.

Appendix

We collect the proofs of some our results in this section.

Proof of Lemma 1

The profit of a firm is given by $\Pi ( \overline{q},\,\overline{q})=a\overline{q}-2b\overline{q}^{2}.$

For $k=2m+1,$ we have $\overline{q}=q_{1}+md.$ Substituting the value of $\overline{q},$ we get

$$\begin{aligned} \Pi (\overline{q},\,\overline{q})=aq_{1}-2bq_{1}^{2}-4bmdq_{1}+amd-2bm^{2}d^{2}. \end{aligned}$$

The expected payoff from any $k$-SSCD with equi-distant quantities and an anti-diagonal probability distribution is given by

$$\begin{aligned} E_{d}(\Pi )&= \sum \limits _{i=1}^{m}p_{i}\left[a\left(q_{i}+q_{2m+2-i}\right)-b\left(q_{i}^{2}+q_{2m+2-i}^{2}\right)-2bq_{i}q_{2m+2-i}\right]\\&+\left(1-2\sum \limits _{i=1}^{m}p_{i}\right)\left[aq_{m+1}-2bq_{m+1}^{2}\right]\\&= aq_{1}-2bq_{1}^{2}-4bmdq_{1}+amd-2bm^{2}d^{2}=\Pi (\overline{q},\,\overline{q }). \end{aligned}$$

Similarly for $k=2m,$ we have $\overline{q}=q_{1}+\frac{d(2m-1)}{2}.$ In this case,

$$\begin{aligned} \Pi (\overline{q},\,\overline{q})=aq_{1}-2bq_{1}^{2}-2bd(2m-1)q_{1}+\frac{ ad(2m-1)}{2}-\frac{bd^{2}(2m-1)^{2}}{2} \end{aligned}$$

and

$$\begin{aligned}&E_{d}(\Pi )=\sum \limits _{i=1}^{m}p_{i}\left[a\left(q_{i}+q_{2m+1-i}\right)-b\left(q_{i}^{2}+q_{2m+1-i}^{2}\right)\right.\\&\qquad \quad \left.-\,2bq_{i}q_{2m+1-i}\right] =aq_{1}-2bq_{1}^{2}-2bd(2m-1)q_{1}\\&\qquad \quad +\,\frac{ad(2m-1)}{2}-\frac{bd^{2}(2m-1)^{2} }{2}=\Pi (\overline{q},\,\overline{q}). \end{aligned}$$

$\square$

Proof of Lemma 2

For any $k,$ suppose a player deviates from the $k$-SSCD with equi-distant quantities and an anti-diagonal probability distribution to play an alternate strategy $q.$ We have

$$\begin{aligned} E_{d}(\Pi \mid q)=aq-bq^{2}-bq\sum \limits _{i=1}^{k}q_{i}\left(\sum \limits _{j=1}^{k}p_{ij}\right)=aq-bq^{2}-bq\overline{q}, \end{aligned}$$

which is maximised at $q=\frac{a-b\overline{q}}{ 2b},$ with $E_{d}(\Pi ^{*})=\mathrm{Max}E_{d}(\Pi \mid q)=\frac{a^{2}}{4b}-\frac{a }{2}\overline{q}+\frac{b}{4}\overline{q}^{2},$ for any $k,$ which proves the lemma. $\square$

Proof of Proposition 2

Given a distribution $P,$ consider a $k$-SSCD with equi-distant quantities $[P;\,q_{1};\,d].$ We will use the equilibrium condition $A_{P}(q_{c})\ge 0,$ in Proposition 1 for such a $k$ -SSCD. Substituting the values of $q_{i}=q_{1}+(i-1)d,\,1\le i\le k,$ and simplifying, the expression $A_{P}(q_{c})$ in Proposition 1 becomes

$$\begin{aligned}&\!\!\!\!\frac{3a}{2}\sum \limits _{i=1}^{k}\left[q_{1}+(i-1)d\right]\left(\sum \limits _{j=1}^{k}p_{ij}\right)\\&\!\!\!\!-\,b\left[q_{1}+\sum \limits _{i=1}^{k}(i-1)d\right]^{2}\left[\sum \limits _{j=1}^{k}p_{ij}+p_{ii}+\frac{1}{4}\left(\sum \limits _{j=1}^{k}p_{ij}\right)^{2}\right]\\&\!\!\!\!-\,b\sum \limits _{1\le i<j\le k}\left[q_{1}+(i-1)d\right]\left[q_{1}+(j-1)d\right]\left[2p_{ij}+\frac{1 }{2}\sum \limits _{s=1}^{k}p_{is}\sum \limits _{s=1}^{k}p_{js}\right]-\frac{a^{2}}{4b }\\&= \frac{3a}{2}q_{1}+\frac{3ad}{2}\left[\sum \limits _{i=1}^{k}( i-1) \sum \limits _{j=1}^{k}p_{ij}\right]-bq_{1}^{2}\left[2\sum \limits _{i=1}^{k}\sum \limits _{j=1}^{k}p_{ij}+\left(\frac{1}{2}\sum \limits _{i=1}^{k}\sum \limits _{j=1}^{k}p_{ij}\right)^{2}\right]\\&\!\!\!\!-\,bq_{1}d\left[4\sum \limits _{i=1}^{k}(i-1)\sum \limits _{j=1}^{k}p_{ij}+\frac{1}{2 }\sum \limits _{i=1}^{k}(i-1)\sum \limits _{j=1}^{k}p_{ij}\left(\sum \limits _{i=1}^{k}\sum \limits _{j=1}^{k}p_{ij}\right)\right]\\&\!\!\!\!-\,bd^{2}\!\left[\!\sum \limits _{i=1}^{k}(i-1) ^{2}\!\left\{ \!p_{ii}\!+\!\sum \limits _{j=1}^{k}p_{ij}+\frac{1}{4}\left(\sum \limits _{j=1}^{k}p_{ij}\!\right)^{2}\!\right\} {+}2\sum _{1\le i<j\le k}(i-1) (j-1)p_{ij}\!\right]\\&\!\!\!\!-\,\frac{bd^{2}}{2}\left[\sum \limits _{1\le i<j\le k}(i-1)(j-1)\sum \limits _{s=1}^{k}p_{is}\sum \limits _{s=1}^{k}p_{js}\right]-\frac{ a^{2}}{4b}\\&= \frac{3a}{2}q_{1}+\frac{3ad}{2}\left[\sum \limits _{i=1}^{k}(i-1) \sum \limits _{j=1}^{k}p_{ij}\right]-\frac{9b}{4}q_{1}^{2}-\frac{9bd}{2} q_{1}\left[\sum \limits _{i=1}^{k}(i-1)\sum \limits _{j=1}^{k}p_{ij}\right]\\&\!\!\!\!-\,bd^{2}\!\left[\!\sum \limits _{i=1}^{k}(i-1) ^{2}\!\left\{ \!p_{ii}+\sum \limits _{j=1}^{k}p_{ij}+\frac{1}{4}\!\left(\sum \limits _{j=1}^{k}p_{ij}\!\right)^{2}\!\right\} {+}2\sum _{1\le i<j\le k}(i-1) (j-1) p_{ij}\!\right]\\&\!\!\!\!-\,\frac{bd^{2}}{2}\left[\sum \limits _{1\le i<j\le k}(i-1)(j-1)\sum \limits _{s=1}^{k}p_{is}\sum \limits _{s=1}^{k}p_{js}\right]-\frac{ a^{2}}{4b}. \end{aligned}$$

To be an equilibrium, the above expression needs to be $\ge 0,$ for some values of $q_{1}$ and $d.$ For any $d>0,$ we can view this expression as a function of $q_{1}.$ Thus, a necessary and sufficient condition for the existence of such an equilibrium is that the maximum value of this function be $\ge 0.$ Using the first order condition, this is maximised at $\widehat{ q_{1}}=\frac{1}{3b}[a-3bd\{\sum \nolimits _{i=1}^{k}(i-1) \sum \nolimits _{j=1}^{k}p_{ij}\}].$

Using $\widehat{q_{1}},$ the maximum value of $A_{P}(q_{c})$ becomes

$$\begin{aligned}&bd^{2}\left[2\left\{ \sum \limits _{i=1}^{k}(i-1) \sum \limits _{j=1}^{k}p_{ij}\right\} ^{2}\right.\\&\quad \left.-\sum \limits _{i=1}^{k}(i-1) ^{2}\left(p_{ii}+\sum \limits _{j=1}^{k}p_{ij}\right)-2\sum _{1\le i<j\le k}(i-1)(j-1) p_{ij}\right]. \end{aligned}$$

The equilibrium condition thus requires the above to be $\ge 0.$ As both $b$ and $d$ are $>0,$ this leads to the statement in the proposition. $\square$

Proof of Proposition 3

Consider any Nash-centric device for the quadratic duopoly game. The expected payoff from following a Nash-centric device is given by

$$\begin{aligned} E_{NC}(\Pi )=(a-b^{\prime })q_{NE}-(1+a^{\prime }+\gamma )q_{NE}^{2}-c^{\prime }-2\delta ^{2}(2-\gamma )\sum \limits _{i=1}^{m}p_{i}(m+1-i)^{2}, \end{aligned}$$

for any odd $k,$ and

$$\begin{aligned} E_{NC}(\Pi ){=}(a-b^{\prime })q_{NE}-(1+a^{\prime }+\gamma )q_{NE}^{2}-c^{\prime }-2\delta ^{2}(2-\gamma )\sum \limits _{i=1}^{m}p_{i}(2m{+}1{-}2i)^{2}, \end{aligned}$$

for any even $k.$

Let $E_{NC}(\Pi \mid q)$ denote the expected payoff of the deviant from playing $q.$ For any $k,\,E_{NC}(\Pi \mid q)=aq-q^{2}-\gamma qq_{NE}-[a^{\prime }q^{2}+b^{\prime }q+c^{\prime }].$ As in the proof of Proposition 1, for the Nash-centric device to be a $k$-SSCCE, we must have, $E_{NC}(\Pi )\ge E_{NC}(\Pi \mid q),$ for all $q\in Q,$ which holds true if and only if $E_{NC}(\Pi )\ge \mathrm{Max}_{q\in Q}E_{NC}(\Pi \mid q).$ Note that $E_{NC}(\Pi \mid q)$ is maximised at $q^{*}=\frac{a-b^{\prime }-\gamma q_{NE}}{2(1+a^{\prime })},$ and hence, $\mathrm{Max}_{q\in Q}E_{NC}(\Pi \mid q)=\frac{(a-b^{\prime })^{2}}{4(1+a^{\prime })}+\frac{\gamma ^{2}q_{NE}^{2}}{ 4(1+a^{\prime })}-\frac{(a-b^{\prime })\gamma q_{NE}}{2(1+a^{\prime })}-c^{\prime }.$

Now, substituting the value of $q_{NE}$ in the above expressions, we have for $k=2m+1,\,E_{NC}(\Pi )-\mathrm{Max}_{q\in Q}E_{NC}(\Pi \mid q)=-2\delta ^{2}(2-\gamma )\sum \nolimits _{i=1}^{m}p_{i}(m+1-i)^{2}<0,$ and for $k{=}2m,\,E_{NC}(\Pi )-\mathrm{Max}_{q\in Q}E_{NC}(\Pi \mid q){=}-2\delta ^{2}(2-\gamma )\sum \nolimits _{i=1}^{m}p_{i}(2m+1-2i)^{2}{<}0.$

Hence, the Nash-centric device is not an equilibrium for this game. $\square$

Proof of Proposition 4

For $k=2m+1,$ from Proposition 2, the equilibrium condition for the off-diagonal-probability-perturbed Nash-centric device to be a $k$-SSCCE becomes

$$\begin{aligned}&2\left[p_{21}+\sum \limits _{i=1}^{m}(i-1)p_{i}+m\left(1-2p_{12}-2\sum \limits _{i=1}^{m}p_{i}\right)+\sum \limits _{i=1}^{m}(2m+1-i)p_{i}\right]^{2}\\&-\left[p_{21}+\sum \limits _{i=1}^{m}(i-1)^{2}p_{i}+m^{2}\left(1-2p_{12}-2\sum \limits _{i=1}^{m}p_{i}\right)+\sum \limits _{i=1}^{m}(2m+1-i)^{2}p_{i}\right]\\&-\left[m^{2}\left(1-2p_{12}-2\sum \limits _{i=1}^{m}p_{i}\right)+2\sum \limits _{i=1}^{m}(i-1)(2m+1-i)p_{i}\right]\ge 0. \end{aligned}$$

After simplification, the LHS of the above turns out to be $(2m-1)^{2}p_{21}(2p_{21}-1),$ which is always $<0,$ as $p_{21}<\frac{1}{2}.$

Similarly for $k=2m,$ the equilibrium condition turns out to be $4(m-1)^{2}p_{21}(2p_{21}-1),$ which is always $<0,$ as $p_{21}<\frac{1}{2}.$

Hence, the equilibrium condition is violated for any $k.$ $\square$

Proof of Proposition 5

For $k=2m+1,$ from Proposition 2, the equilibrium condition for the diagonal-probability-perturbed Nash-centric device to be a $k$-SSCCE becomes

$$\begin{aligned}&2\left[\sum \limits _{i=1}^{m}(i-1)p_{i}+m\left(1-p_{11}-2\sum \limits _{i=1}^{m}p_{i}\right)+\sum \limits _{i=1}^{m}(2m+1-i)p_{i}\right]^{2}\\&-\left[\sum \limits _{i=1}^{m}(i-1)^{2}p_{i}+m^{2}\left(1-p_{11}-2\sum \limits _{i=1}^{m}p_{i}\right)+\sum \limits _{i=1}^{m}(2m+1-i)^{2}p_{i}\right]\\&-\left[m^{2}\left(1-p_{11}-2\sum \limits _{i=1}^{m}p_{i}\right)+\left\{ 2\sum \limits _{i=1}^{m}(i-1)(2m+1-i)p_{i}\right\} \right]\ge 0. \end{aligned}$$

Simplifying, the LHS of the above turns out to be $-2m^{2}p_{11}(1-p_{11}),$ which is always $<0.$

Similarly for $k=2m,$ the equilibrium condition turns out to be $(2m-1)^{2}\left(\sum \nolimits _{i=1}^{m}p_{i}\right)\left[2\left(\sum \nolimits _{i=1}^{m}p_{i}\right)-1\right],$ which is $\ge 0$ if and only if $\sum \nolimits _{i=1}^{m}p_{i}\ge \frac{1}{2},$ which is not possible as $p_{11}>0.$

Hence, the equilibrium condition is violated for any $k.$ $\square$

Proof of Proposition 6

From Proposition 1, substituting the values of $q_{1}$ and other $q_{i}$ for $i\ne 1,$ (for any $k$), the expression $A_{P}(q_{c})$ becomes

$$\begin{aligned} \frac{3a}{2}q_{NE}-\frac{9b\varepsilon p_{1}}{2}q_{NE}-\frac{9b}{4} q_{NE}^{2}+\frac{3ap_{1}\varepsilon }{2}-b\varepsilon ^{2}\left(p_{1}+\frac{ p_{1}{}^{2}}{4}\right)-\frac{a^{2}}{4b}. \end{aligned}$$

For the quantity-perturbed Nash-centric device to be a $k$-SSCCE, we need the above expression to be $\ge 0.$ Now substituting $q_{NE}=\frac{a}{3b},$ the expression becomes $-b\varepsilon ^{2}(p_{1}+\frac{p_{1}{}^{2}}{4}),$ which is always $<0.$ Hence, the equilibrium condition is violated. $\square$

Proof of Proposition 7

Following Proposition 1, for the composite device (for any $k$) the expression $A_{P}(q_{c})$ becomes

$$\begin{aligned} \frac{3a\varepsilon }{2}q_{\varepsilon }&+ \frac{3a(1-\varepsilon )}{2} q_{NE} -b\left(2\varepsilon +\frac{\varepsilon ^{2}}{4}\right)q_{\varepsilon }^{2}\\&-\frac{ b\varepsilon (1-\varepsilon )}{2}q_{\varepsilon }q_{NE}-b\left[2(1-\varepsilon )+\left( \frac{1-\varepsilon }{2}\right)^{2}\right]q_{NE}^{2}-\frac{a^{2}}{4b}. \end{aligned}$$

Substituting $q_{NE}=\frac{a}{3b},$ and rearranging, we get, $A_{P}(q_{c})=q_{\varepsilon }[\frac{a\varepsilon (8+\varepsilon )}{6} ]-q_{\varepsilon }^{2}[\frac{b\varepsilon (8+\varepsilon )}{4}]-\frac{ a^{2}\varepsilon (8+\varepsilon )}{36b},$ which can be viewed as a (quadratic) function in $q_{\varepsilon }.$ This function is maximised at $q_{\varepsilon }=\frac{a}{3b}$ (the Nash equilibrium quantity) and the maximised value of the function is $0.$ Therefore, for any $q_{\varepsilon }>0,$ other than the Nash Equilibrium quantity, the value of $A_{P}(q_{c})$ is $<0.$

From Proposition  1, for the composite device to be a $(k+1)$-SSCCE, we need the above $A_{P}(q_{c})$ to be $\ge 0.$ However, from above, the value of the above function is $<0,$ for any $q_{\varepsilon }>0,$ other than the Nash Equilibrium quantity. Hence, the equilibrium condition is violated. $\square$