Cooperative oligopoly games with boundedly rational firms

Abstract

We analyze cooperative Cournot games with boundedly rational firms. Due to cognitive constraints, the members of a coalition cannot accurately predict the coalitional structure of the non-members. Thus, they compute their value using simple heuristics. In particular, they assign various non-equilibrium probability distributions over the outsiders’ set of partitions. We construct the characteristic function of a coalition in such an environment and we analyze the core of the corresponding games. We show that the core is non-empty provided the number of firms in the market is sufficiently large. Moreover, we show that if two distributions over the set of partitions are related via first-order dominance, then the core of the game under the dominated distribution is a subset of the core under the dominant distribution.

Appendix

Lemma A1

Assume the demand function is $Q=1-p^b$. Then the characteristic function is given by (13).

Proof

The profit function of coalition $i$ under a partition with $j$ members is

$$\begin{aligned} \pi _i^j=\left( 1-q_s-\sum \limits _{r=1, r\ne i}^j q_r^j-q_i^j\right) ^{\frac{1}{b}}q_i^j, \quad i=1,2,...,j \end{aligned}$$

(23)

Note that

$$\begin{aligned} \frac{\partial {\pi _i^j}}{\partial {q_i^j}}=0\Leftrightarrow b\left( 1-q_s-q_i^j-\sum \limits _{r=1, r\ne i}^j q_r^j\right) =q_i^j \end{aligned}$$

(24)

By symmetry, all $j$ outside coalitions produce the same. So let $q_r^j=q_i^j$, for all $r$. Therefore by (24), $b(1-q_s-jq_i^j)=q_i^j$ and hence

$$\begin{aligned} \tilde{q}_i^j=\frac{b(1-q_s)}{bj+1} \end{aligned}$$

(25)

The objective function of the deviant coalition $S$ is

$$\begin{aligned} \pi _f(S)= \sum \limits _{j=1}^{n-s}f_{n,s}(j) \left( 1-q_s-\sum \limits _{i=1}^j q_i^j\right) ^{\frac{1}{b}}q_s \end{aligned}$$

(26)

Note that ${\displaystyle {\frac{\partial {\pi _f(S)}}{\partial {q_s}}=0}}$ if

$$\begin{aligned} \sum \limits _{j=1}^{n-s}f_{n,s}(j)\left( 1-q_s-\sum _{i=1}^jq_i^j\right) ^{\frac{1}{b}}= \frac{1}{b}\sum \limits _{j=1}^{n-s}f_{n,s}(j)\left( 1-q_s- \sum _{i=1}^jq_i^j\right) ^{\frac{1}{b}-1}q_s \end{aligned}$$

(27)

Using (25), (27) becomes

$$\begin{aligned} \sum \limits _{j=1}^{n-s}f_{n,s}(j)\left( \frac{1-q_s}{1+bj}\right) ^{\frac{1}{b}}= \frac{1}{b}\sum \limits _{j=1}^{n-s}f_{n,s}(j)\left( \frac{1-q_s}{1+bj}\right) ^{\frac{1}{b}-1}q_s \end{aligned}$$

and hence

$$\begin{aligned} (1-q_s)^{\frac{1}{b}}\sum \limits _{j=1}^{n-s}f_{n,s}(j)\left( \frac{1}{1+bj}\right) ^{\frac{1}{b}}= \frac{1}{b}(1-q_s)^{\frac{1}{b}-1}\sum \limits _{j=1}^{n-s}f_{n,s}(j)\left( \frac{1}{1+bj}\right) ^{\frac{1}{b}-1}q_s \end{aligned}$$

Define ${\displaystyle {\psi _j=\frac{1}{bj+1}.}}$ Then rearranging the above relation gives

$$\begin{aligned} q_s\left( \sum \limits _{j=1}^{n-s}f_{n,s}(j)\psi _j^{\frac{1}{b}} +\frac{1}{b}\sum \limits _{j=1}^{n-s}f_{n,s}(j)\psi _j^{\frac{1}{b}-1}\right) = \sum \limits _{j=1}^{n-s}f_{n,s}(j)\psi _j^{\frac{1}{b}} \end{aligned}$$

(28)

Notice that

$$\begin{aligned}&\sum \limits _{j=1}^{n-s}f_{n,s}(j)\psi _j^{\frac{1}{b}} +\frac{1}{b}\sum \limits _{j=1}^{n-s}f_{n,s}(j)\psi _j^{\frac{1}{b}-1}= \sum \limits _{j=1}^{n-s}f_{n,s}(j)\psi _j^{\frac{1}{b}}\left( 1+\frac{1}{b\psi _j}\right) \nonumber \\&\qquad =\sum \limits _{j=1}^{n-s}f_{n,s}(j)\psi _j^{\frac{1}{b}}(1+(bj+1)/b)= \sum \limits _{j=1}^{n-s}f_{n,s}(j)\psi _j^{\frac{1}{b}}(1+j+1/b) \end{aligned}$$

(29)

Using (28) and (29) we get

$$\begin{aligned} q_s(f)=\frac{\sum \limits _{j=1}^{n-s}f_{n,s}(j)\psi _j^{\frac{1}{b}}}{\sum \limits _{j=1}^{n-s}f_{n,s}(j)\psi _j^{\frac{1}{b}}(1+j+1/b)} \end{aligned}$$

(30)

Using (30), (25) becomes

$$\begin{aligned} q_i^j(f)=b\psi _j\frac{\sum \limits _{j=1}^{n-s}\psi _j^ {\frac{1}{b}}(j+1/b)}{\sum \limits _{j=1}^{n-s}f_{n,s}(j)\psi _j^{\frac{1}{b}}(1+j+1/b)} \end{aligned}$$

(31)

Plugging (30) and (31) in (26) gives us (13). $\square$

Lemma A2

Assume the inverse demand $p(Q)$ is weakly concave. Then $Q_{j}^{-s}(w)< Q_j^{-s}(z)$, $j=1,2,...,n-s$.

Proof

Fix $Q_j^{-s}$. For notational convenience, define $F(q_s)\equiv \sum \limits _{j=1}^{n-s}w_{n,s}(j)\pi _s(q_s,Q_j^{-s})$ and $H(q_s)\equiv \sum \limits _{j=1}^{n-s}z_{n,s}(j)\pi _s(q_s,Q_j^{-s})$. Then $\tilde{q}_s(w)$ and $\tilde{q}_s(z)$ satisfy respectively the first-order conditions ${\displaystyle {\frac{\partial {F(q_s)}}{\partial {q_s}}=0}}$ and ${\displaystyle {\frac{\partial {H(q_s)}}{\partial {q_s}}=0}}$ or equivalently

$$\begin{aligned} \sum \limits _{j=1}^{n-s}w_{n,s}(j)p'\left( q_s+Q_{j}^{-s}\right) q_s+ \sum \limits _{j=1}^{n-s}w_{n,s}(j)p\left( q_s+Q_{j}^{-s}\right) -c=0 \end{aligned}$$

(32)

and

$$\begin{aligned} \sum \limits _{j=1}^{n-s}z_{n,s}(j)p'\left( q_s+Q_{j}^{-s}\right) q_s+ \sum \limits _{j=1}^{n-s}z_{n,s}(j)p\left( q_s+Q_{j}^{-s}\right) -c=0 \end{aligned}$$

(33)

The function $F(q_s)$ is strictly concave in $q_s$ (by assumptions A1-A3). Hence $\tilde{q}_s(w)>\tilde{q}_s(z)$ if and only if ${\displaystyle {\frac{\partial {F(\tilde{q}_s(z))}}{\partial {q_s}}>0}}$. By (32) we have that

$$\begin{aligned} \frac{\partial {F(\tilde{q}_s(z))}}{\partial {q_s}}\!>\!0\!\Leftrightarrow \! \sum \limits _{j=1}^{n-s}w_{n,s}(j)p'\left( \tilde{q}_s(z)\!+\!Q_{j}^{-s}\right) \tilde{q}_s(z)+\sum \limits _{j=1}^{n-s}w_{n,s}(j)p\left( \tilde{q}_s(z)\!+\!Q_{j}^{-s}\right) \!-\!c\!>\!0\nonumber \\ \end{aligned}$$

(34)

Solving for $\tilde{q}_s(z)$ by (33) and plugging in (34) we have that

$$\begin{aligned} \frac{\partial {F(\tilde{q}_s(z))}}{\partial {q_s}}>0&\Leftrightarrow \frac{\sum \limits _{j=1}^{n-s}w_{n,s}(j)p'\left( \tilde{q}_s(z)+Q_{j}^{-s}\right) }{-\sum \limits _{j=1}^{n-s}z_{n,s}(j)p'\left( \tilde{q}_s(z)+Q_{j}^{-s}\right) } \left( \sum \limits _{j=1}^{n-s}z_{n,s}(j)p\left( \tilde{q}_s(z)+Q_{j}^{-s}\right) -c\right) \nonumber \\&\,\,+\,\,\sum \limits _{j=1}^{n-s}w_{n,s}(j)p\left( \tilde{q}_s(z)+Q_{j}^{-s}\right) -c>0 \end{aligned}$$

(35)

We now use the concavity assumption and claim that

$$\begin{aligned} \frac{\sum \limits _{j=1}^{n-s}w_{n,s}(j)p'\left( \tilde{q}_s(z)+Q_{j}^{-s}\right) }{- \sum \limits _{j=1}^{n-s}z_{n,s}(j)p'\left( \tilde{q}_s(z)+Q_{j}^{-s}\right) }\ge -1 \end{aligned}$$

(36)

To show the above we can equivalently show

$$\begin{aligned} \sum \limits _{j=1}^{n-s}w_{n,s}(j)p'\left( \tilde{q}_s(z)+Q_{j}^{-s}\right) - \sum \limits _{j=1}^{n-s}z_{n,s}(j)p'\left( \tilde{q}_s(z)+Q_{j}^{-s}\right) \ge 0 \end{aligned}$$

(37)

We have

$$\begin{aligned}&\sum \limits _{j=1}^{n-s}w_{n,s}(j)p'\left( \tilde{q}_s(z)+Q_{j}^{-s}\right) - \sum \limits _{j=1}^{n-s}z_{n,s}(j)p'\left( \tilde{q}_s(z)+Q_{j}^{-s}\right) \\&\quad =\left( w_{n,s}(1)-z_{n,s}(1)\right) p'\left( \tilde{q}_s(z)+Q_1^{-s}\right) +\sum \limits _{j=2}^{n-s}\left( w_{n,s}(j)- z_{n,s}(j)\right) p'\left( \tilde{q}_s(z)+Q_{j}^{-s}\right) \\&\quad \ge \left( w_{n,s}(1)-z_{n,s}(1)\right) p'\left( \tilde{q}_s(z)+Q_2^{-s}\right) +\sum \limits _{j=2}^{n-s}\left( w_{n,s}(j)- z_{n,s}(j)\right) p'\left( \tilde{q}_s(z)+Q_{j}^{-s}\right) \end{aligned}$$

where the inequality holds because $w_{n,s}(1)-z_{n,s}(1)\ge 0$ and because the weak concavity of price implies that $p'\left( \tilde{q}_s(z)+Q_{1}^{-s}\right) \ge p'\left( \tilde{q}_s(z)+Q_{2}^{-s}\right)$ (recall that $Q_1^{-s}<Q_{2}^{-s}).$ If we continue the process of iterating $j$, we end up with (37). Since the latter condition holds, we have that

$$\begin{aligned}&\frac{\sum \limits _{j=1}^{n-s}w_{n,s}(j)p'\left( \tilde{q}_s(z)+Q_{j}^{-s}\right) }{-\sum \limits _{j=1}^{n-s}z_{n,s}(j)p'\left( \tilde{q}_s(z)+Q_{j}^{-s}\right) } \left( \sum \limits _{j=1}^{n-s}z_{n,s}(j)p\left( \tilde{q}_s(z)+Q_{j}^{-s}\right) -c\right) \nonumber \\&\qquad +\,\,\sum \limits _{j=1}^{n-s}w_{n,s}(j)p\left( \tilde{q}_s(z)+Q_{j}^{-s}\right) -c\nonumber \\&\quad \ge -\left( \sum \limits _{j=1}^{n-s}z_{n,s}(j)p\left( \tilde{q}_s(z)+Q_{j}^{-s}\right) -c\right) + \sum \limits _{j=1}^{n-s}w_{n,s}(j)p\left( \tilde{q}_s(z)+Q_{j}^{-s}\right) -c\nonumber \\&\quad =\sum \limits _{j=1}^{n-s}w_{n,s}(j)p\left( \tilde{q}_s(z)+Q_{j}^{-s}\right) -\sum \limits _{j=1}^{n-s}z_{n,s}(j)p\left( \tilde{q}_s(z)+Q_{j}^{-s}\right) \end{aligned}$$

(38)

But expression (38) can be written as

$$\begin{aligned}&\big (w_{n,s}(1)-z_{n,s}(1)\big )p\left( \tilde{q}_s(z)+Q_1^{-s}\right) + \sum \limits _{j=2}^{n-s}\big (w_{n,s}(j)-z_{n,s}(j)\big )p\left( \tilde{q}_s(z)+Q_{j}^{-s}\right) \\&\quad >\big (w_{n,s}(1)-z_{n,s}(1)\big )p(\tilde{q}_s(z)+Q_2^{-s})+ \sum \limits _{j=2}^{n-s}\big (w_{n,s}(j)-z_{n,s}(j)\big )p\left( \tilde{q}_s(z)+Q_{j}^{-s}\right) \end{aligned}$$

where the last inequality holds because $Q_1^{-s}<Q_2^{-s}$. Continuing the iterations on $j$, we end up with

$$\begin{aligned} \sum \limits _{j=1}^{n-s}w_{n,s}(j)p\left( \tilde{q}_s(z)+Q_{j}^{-s}\right) - \sum \limits _{j=1}^{n-s}z_{n,s}(j)p\left( \tilde{q}_s(z)+Q_{j}^{-s}\right) >0 \end{aligned}$$

(39)

Combining (35), (38) and (39) we conclude that ${\displaystyle {\frac{\partial {F(\tilde{q}_s(z))}}{\partial {q_s}}>0}}$ and hence $\tilde{q}_s(w)>\tilde{q}_s(z).$ But then $Q_{j}^{-s}(w)<Q_j^{-s}(z)$, since $Q_j^{-s}(w)$ and $Q_j^{-s}(z)$ emerge from $\tilde{Q}_j^{-s}$ for $q_s=\tilde{q}_s(w)$ and $q_s=\tilde{q}_s(z)$ respectively and commodities in a Cournot market are substitutes. $\square$