On the coalitional stability of monopoly power in differentiated Bertrand and Cournot oligopolies

Abstract

In this article, we revisit the classic comparison between Bertrand and Cournot competition in the presence of a cartel of firms that faces outsiders acting individually. This competition setting enables to deal with both non-cooperative and cooperative oligopoly games. We concentrate on industries consisting of symmetrically differentiated products where firms operate at a constant and identical marginal cost. First, while the standard Bertrand–Cournot rankings still hold for Nash equilibrium prices, we show that the results may be altered for Nash equilibrium quantities and profits. Second, we define cooperative Bertrand and Cournot oligopoly games with transferable utility on the basis of their non-cooperative foundation. We establish that the core of a cooperative Cournot oligopoly game is strictly included in the core of a cooperative Bertrand oligopoly game when the number of firms is lower or equal to 25. Moreover, we focus on the aggregate-monotonic core, a subset of the core, that has the advantage to select point solutions satisfying both core selection and aggregate monotonicity properties. We succeed in comparing the aggregate-monotonic cores between Bertrand and Cournot competition regardless of the number of firms. Finally, we study a class of three-firm oligopolies with asymmetric costs in which the core inclusion property mentioned above still holds. We also provide numerical examples to illustrate the difficulty to generalize this result to an arbitrary number of firms because of negative equilibrium quantities.

Appendix

1.1 Geometrical properties of the reaction functions in price space

• Comparison of the y-intercepts of $R_I^B(p_j)$ and $\bar{R}_I^C(p_j)$ given by (9) and (13), respectively:

  $$\begin{aligned} \begin{aligned} \frac{(n+rs)nV}{2n^2(1+r)+nr^2s-r^2s^2}-\frac{nV}{2(n+r(n-s))}&=\frac{nVr^2s(n-s)}{2(n+r(n-s))(2n^2(1+r)+nr^2s-r^2s^2)}\\&>0\text{. } \end{aligned} \end{aligned}$$

• Comparison of the slopes of $R_I^B(p_j)$ and $\bar{R}_I^C(p_j)$ given by (9) and (13), respectively:

  $$\begin{aligned} \begin{aligned} \frac{(n+rs)r(n-s)}{2n^2(1+r)+nr^2s-r^2s^2}-\frac{r(n-s)}{2(n+r(n-s))}&=\frac{r^3(n-s)^2s}{2(n+r(n-s))(2n^2(1+r)+nr^2s-r^2s^2)}\\&>0\text{. } \end{aligned} \end{aligned}$$

• Comparison of the y-intercepts of $R_O^B(p_i)$ and $\bar{R}_O^C(p_i)$ given by (10) and (14), respectively:

  $$\begin{aligned} \begin{aligned} \frac{(n+r)nV}{n^2(2+r)+nr(s+1)+r^2s}&-\frac{nV}{2n+r(n+s-1)} \\&=\frac{nV(n-1)r^2}{(2n+r(n+s-1))(n^2(2+r)+nr(s+1)+r^2s)} >0\text{. } \end{aligned} \end{aligned}$$

• Comparison of the slopes of $R_O^B(p_i)$ and $\bar{R}_O^C(p_i)$ given by (10) and (14), respectively:

  $$\begin{aligned} \begin{aligned} \frac{(n+r)rs}{n^2(2+r)+nr(s+1)+r^2s}&-\frac{rs}{2n+r(n+s-1)} \\&=\frac{r^3(n-1)s}{(2n+r(n+s-1))(n^2(2+r)+nr(s+1)+r^2s)} >0\text{. } \end{aligned} \end{aligned}$$

1.2 Geometrical properties of the reaction functions in quantity space

• Comparison of the y-intercepts of $R_I^C(q_j)$ and $\bar{R}_I^B(q_j)$ given by (11) and (15), respectively:

  $$\begin{aligned} \begin{aligned} \frac{(n+r(n-s))(1+r)nV}{2n^2(1+r)+nr^2s-r^2s^2}-\frac{(1+r)nV}{2(n+rs)}&=\frac{nVr^2(1+r)(n-s)s}{2(n+rs)(2n^2(1+r)+nr^2s-r^2s^2)}\\&>0\text{. } \end{aligned} \end{aligned}$$

• Comparison of the absolute value of the slopes of $R_I^C(q_j)$ and $\bar{R}_I^B(q_j)$ given by (11) and (15), respectively:

  $$\begin{aligned} \begin{aligned} \frac{(n+r(n-s))r(n-s)}{2n^2(1+r)+nr^2s-r^2s^2}-\frac{r(n-s)}{2(n+rs)}&=\frac{r^3(n-s)^2s}{2(n+rs)(2n^2(1+r)+nr^2s-r^2s^2)}\\&>0\text{. } \end{aligned} \end{aligned}$$

• Comparison of the y-intercepts of $R_O^C(q_i)$ and $\bar{R}_O^B(q_i)$ given by (12) and (16), respectively:

  $$\begin{aligned} \begin{aligned}&\frac{(n(1+r)-r)(1+r)nV}{n^2(2+3r+r^2)-nr(1+r)(s+1)+r^2s} -\frac{nV(1+r)}{2n+r(n-s+1)} \\&\quad =\frac{nV(n-1)r^2(1+r)}{(2n+r(n-s+1))(n^2(2+3r+r^2)+nr(-rs-s-r-1)+r^2s)} >0\text{. } \end{aligned} \end{aligned}$$

• Comparison of the absolute value of the slopes of $R_O^C(q_i)$ and $\bar{R}_O^B(q_i)$ given by (12) and (16), respectively:

  $$\begin{aligned} \begin{aligned}&\frac{(n(1+r)-r)rs}{n^2(2+3r+r^2)-nr(1+r)(s+1)+r^2s} -\frac{rs}{2n+r(n-s+1)} \\&\quad =\frac{r^3(n-1)s}{(2n+r(n-s+1))(n^2(2+3r+r^2)+nr(-rs-s-r-1)+r^2s)} >0\text{. } \end{aligned} \end{aligned}$$

1.3 The asymptotic reaction functions in quantity space

By substituting s by n / k into (11), (12), (15), and (16), and taking the limit $n\rightarrow \infty$, the cartel asymptotic reaction functions are expressed as:

$$\begin{aligned} R_I^C(q_j)=\frac{(1+r)kV-r(k-1)q_j}{2(k+r)}\text{, } \end{aligned}$$

and

$$\begin{aligned} \bar{R}_I^B(q_j)=\frac{(r(k-1)+k)((1+r)kV-r(k-1)q_j)}{(k-1)r^2+2k^2r+2k^2}\text{. } \end{aligned}$$

The asymptotic reaction functions of any outsider are given by:

$$\begin{aligned} R_O^C(q_i)=\bar{R}_O^B(q_i)=\frac{(1+r)kV-rq_i}{2k+(k-1)r}\text{. } \end{aligned}$$

• Comparison of the y-intercepts of $R_I^C(q_j)$ and $\bar{R}_I^B(q_j)$ in the asymptotic case:

  $$\begin{aligned} \frac{(r(k-1)+k)(1+r)kV}{(k-1)r^2+2k^2r+2k^2}-\frac{(1+r)kV}{2(k+r)} =\frac{(k-1)kr^2(1+r)V}{2(k+r)((k-1)r^2+2k^2r+2k^2)} >0\text{. } \end{aligned}$$

• Comparison of the absolute value of the slopes of $R_I^C(q_j)$ and $\bar{R}_I^B(q_j)$ in the asymptotic case:

  $$\begin{aligned} \frac{(r(k-1)+k)(k-1)r}{(k-1)r^2+2k^2r+2k^2}-\frac{r(k-1)}{2(k+r)}=\frac{(k-1)^2r^3}{2(k+r)((k-1)r^2+2k^2r+2k^2)} >0\text{. } \end{aligned}$$

1.4 Proofs

In this subsection, the expressions (17)–(22) are all available in Wang and Zhao (2010) by letting all marginal costs be zero.

Proof of Proposition 3.2:

In quantity space, the intersections of reaction functions $R_I^C(q_j)$ and $R_O^C(q_i)$, and $\bar{R}_I^B(q_j)$ and $\bar{R}_O^B(q_i)$ given by (11), (12), (15) and (16), respectively, provide Nash equilibrium quantities produced by each cartel member in Bertrand and Cournot competition, respectively:

$$\begin{aligned} D_i(p_s^*,{\tilde{p}}_s) =\frac{V(2n(1+r)-r)(r(n-s)+n)}{4n^2+r(n(6n-2(s+1)) +r(n-s)(2n+s-2))} \end{aligned}$$

(17)

and

$$\begin{aligned} q_s^* =\frac{nV(2n+r)(1+r)}{4n^2+2rn(1+n)+rs((n+2)r+2n)-r^2s^2}\text{. } \end{aligned}$$

Calculating the difference between these two quantities leads to:

$$\begin{aligned} D_i(p_s^*,{\tilde{p}}_j)-q_s^* =\frac{Vr^2(n-s)p(r)}{AB}\text{, } \end{aligned}$$

where $A>0$ and $B>0$ denote the denominators of $D_i(p_s^*,{\tilde{p}}_s)$ and $q_s^*$ respectively, and p(r) is defined as:

$$\begin{aligned} \begin{aligned} p(r) =&r^2(2n((n-s)(s-1)+1)+s(s-2))\\&+\,r2n(s(2n-s)+n(s-1)+1)\\&+\,4n^2s\\&>0, \end{aligned} \end{aligned}$$

which concludes the proof. $\square$

Fig. 4

Asymptotic Nash equilibrium quantities

A similar argument to the one in the proof of Proposition 3.2 permits to determine Nash equilibrium quantities produced by each outsider in Bertrand competition:

$$\begin{aligned} D_j(p_s^*,{\tilde{p}}_s)=\frac{V(2n(1+r)-rs)(r(n-1)+n)}{4n^2+r(n(6n-2(s+1))+r(n-s)(2n+s-2))}. \end{aligned}$$

(18)

Proof of Proposition 3.3:

Consider the asymptotic case of an industry containing an infinite number of firms $n\rightarrow \infty$ with a cartel of significant size $s\rightarrow \infty /k=\infty$ as illustrated in Fig. 4. As a consequence, the asymptotic reaction functions of any cartel member $\bar{R}_I^B(q_j)$ and $R_I^C(q_j)$ differ for any $k>1$.¹⁰ By contrast, since each outsider acts as a price-taking profit maximizer, its asymptotic reaction functions $\bar{R}_O^B(q_i)$ and $R_O^C(q_i)$ are equal (see footnote 10). In quantity space, decompose the quantity change from B to C into two stages. First, it follows from Proposition 3.2 that each cartel member reduces its quantity from $D_i(p_s^*,{\tilde{p}}_s)$ to $q_s^*$ to achieve A which raises price for all outsiders. Second, in response, each atomic outsider must increase its own quantity from $D_j(p_s^*,{\tilde{p}}_s)$ to ${\tilde{q}}_s$. $\square$

Proof of Proposition 3.4:

In quantity space, decompose the quantity change from B to C into two stages as illustrated in Figs. 2 and 3. First, each cartel member reduces its quantity from $D_i(p_s^*,p_j^*)$ to $q_s^*$ to achieve A. By gross substitutability, this raises price for all outsiders, and hence benefits them. Second, facing $q_s^*$, each outsider reduces its quantity (Fig. 2) or raises it (Fig. 3) to ${\tilde{q}}_s$ to achieve C which is profit-maximizing.

When each outsider reduces its quantity from Bertrand to Cournot competition, a similar argument permits to conclude that each cartel member earns larger profits by decomposing the quantity change from B to C via D (Fig. 2). $\square$

Proof of Proposition 3.5:

In price space, the intersection of reaction functions $R_I^B(p_j)$ and $R_O^B(p_i)$ given by (9) and (10), respectively, provides Nash equilibrium prices charged by each cartel member:

$$\begin{aligned} p_s^*=\frac{\displaystyle V\big (2n(1+r)-r\big )n}{\displaystyle 2\big (2n+r(n+s-1)\big )\big (n+r(n-s)\big )-r^2s(n-s)}\text{, } \end{aligned}$$

and by each outsider:

$$\begin{aligned} {\tilde{p}}_s=\frac{\displaystyle V\big (2n(1+r)-r s\big )n}{\displaystyle 2\big (2n+r(n+s-1)\big )\big (n+r(n-s)\big )-r^2s(n-s)}\text{. } \end{aligned}$$

The profit of each cartel member at Bertrand–Nash equilibrium $(p_s^*,{\tilde{p}}_s)$ is expressed as:

$$\begin{aligned} \pi _i^B(p_s^*,{\tilde{p}}_s)=\frac{V^2(2n(1+r)-r)^2n(n+r(n-s))}{\big (4n^2+6n^2r-2nrs +2n^2r^2-r^2s^2-2nr-2nr^2+2r^2s-nr^2s\big )^2}. \end{aligned}$$

(19)

In quantity space, the intersection of reaction functions $R_I^C(q_j)$ and $R_O^C(q_i)$ given by (11) and (12), respectively, provides Nash equilibrium quantities produced by each cartel member:

$$\begin{aligned} q_s^*=\frac{\displaystyle V(2n+r)(1+r)n}{\displaystyle 4n^2+2rn(1+n)+rs((n+2)r+2n)-r^2s^2} \end{aligned}$$

(20)

and by each outsider:

$$\begin{aligned} {\tilde{q}}_s=\frac{\displaystyle V(2n+rs)(1+r)n}{\displaystyle 4n^2+2rn(1+n)+rs((n+2)r+2n)-r^2s^2}. \end{aligned}$$

(21)

The profit of each cartel member at Cournot–Nash equilibrium $(q_s^*,{\tilde{q}}_s)$ is expressed as:

$$\begin{aligned} \pi _i^C(q_s^*,{\tilde{q}}_s)=\frac{V^2((2n+r)^2n(n+rs)(1+r))}{(4n^2+2rn(1+n)+rs((n+2)r+2n)-r^2s^2)^2}. \end{aligned}$$

(22)

Calculating the difference between the profits given by (19) and (22) leads to:

$$\begin{aligned} \pi _i^C(q_s^*,{\tilde{q}}_s)-\pi _i^B(p_s^*,{\tilde{p}}_s) =\frac{\displaystyle nr^3(2+r)(n-s)V^2p(r)}{\displaystyle A^2B^2}\text{, } \end{aligned}$$

where $A^2$ and $B^2$ denote the denominators of $\pi _i^C(q_s^*,{\tilde{q}}_s)$ and $\pi _i^B(p_s^*,{\tilde{p}}_s)$ respectively, and p(r) is defined as:

$$\begin{aligned} \begin{aligned}p(r) =&r^4(s(n-s)(2n+s-2)^2) \\&+\,r^34n^2(n^2(-s^2+4s+1)+n(2s^3-3s^2-3s-2)-s^4-s^3+4s^2-s+1)\\&+\,r^24n^2(4n^3+n^2(-s^2+4s-3)+n(2s^3-3s^2-3s-2)-s^4-s^3+4s^2-s+1)\\&+\,r(32n^5-32n^4)+\,16n^5-16n^4{.} \end{aligned} \end{aligned}$$

It remains to study p(r). Note that for any n and any $s< n$, it holds that $p(r)>0$ when r is sufficiently small or sufficiently large. Two cases can occur:

• for any $r>0$, $p(r)>0$, i.e., p(r) has no positive root.

• there exists a root $r_1>0$ which implies that $p(r)<0$ at the neighborhood of $r_1$.

A simple algorithm permitting to compute roots of p(r) shows that no positive root appears for any $n\le 25$ and any $s< n$.¹¹$\square$

Proof of Proposition 3.6:

By substituting s by n / k into (19) and (22), the difference between these two profits becomes:

$$\begin{aligned} \pi _i^C(q_s^*,{\tilde{q}}_s)-\pi _i^B(p_s^*,{\tilde{p}}_s)=\frac{\displaystyle (k-1)k^3r^3(r+2)V^2p(n)}{\displaystyle A^2B^2}\text{, } \end{aligned}$$

where $(An/k^2)^2$ and $(Bn/k^2)^2$ denote the denominators of $\pi _i^C(q_s^*,{\tilde{q}}_s)$, and $\pi _i^B (p_s^*,{\tilde{p}}_s)$, respectively, and p(n) is defined as:

$$\begin{aligned} \begin{aligned}p(n) =&-n^4(4(r^3+r^2)(k^2-2k+1)) \\&+n^3k(r^3(16k^2-12k-4)+r^2(16k^3+16k^2-12k-4)+16k^3(2r+1))\\&+n^2(r^4(4k^3-3k-1)+r^3k^2(4k^2-12k+16)\\&+r^2k^2(-12k^2-12k+16) -16k^4(2r+1))\\&+nk(r^4(-8k^2+4k+4)+r^3k^2(-8k-4)+r^2k^3(-8k-4))\\&+4k^2r^2(r^2(k-1)+k^2(1+r))\text{. } \end{aligned} \end{aligned}$$

Thus, there exists ${\bar{n}}>0$, such that for any $n>{\bar{n}}$, it holds that $p(n)<0$ which concludes the proof. $\square$

Proof of Lemma 4.2:

First, for simplicity, we assume that the size s of cartel S is a real number in the interval $[1,n-1]$. Then, differentiating $\pi _i^C(q_s^*,{\tilde{q}}_s)$ with respect to s leads to:

$$\begin{aligned} \frac{d}{ds}\pi _i^C(q_s^*,{\tilde{q}}_s)=\frac{V^2nr^2(1+r)(2n+r)^2(3rs^2-(r(n+2)-2n)s-2n)}{(4n^2+2rn(1+n)+rs((n+2)r+2n)-r^2s^2)^3}\text{. } \end{aligned}$$

We aim to study the polynomial function of degree 2, $p:[1,n-1] \longrightarrow {\mathbb {R}}$, defined as:

$$\begin{aligned} p(s)=3rs^2-(r(n+2)-2n)s-2n\text{. } \end{aligned}$$

The discriminant of p(s) is given by:

$$\begin{aligned} \Delta = (n+2)^2r^2+(16n-4n^2)r+4n^2\text{, } \end{aligned}$$

and is positive for any $n\ge 3$ and any $r>0$. Hence, p(s) has two distinct real roots:

$$\begin{aligned} s_{1,2}= \frac{r(n+2)-2n \pm \sqrt{(n+2)^2r^2+(16n-4n^2)r+4n^2}}{6r}\text{. } \end{aligned}$$

We want to prove that $s_1<0$ and $1<s_2<n-1$. We proceed in three steps.

First, we distinguish two cases. If $r(n+2)-2n \ge 0$, then:

$$\begin{aligned} \begin{aligned} s_1&= \frac{\sqrt{(r(n+2)-2n)^2}-\sqrt{(n+2)^2r^2+(16n-4n^2)r+4n^2}}{6r}\\&= \frac{\sqrt{(n+2)^2r^2+(-8n-4n^2)r+4n^2}-\sqrt{(n+2)^2r^2 +(16n-4n^2)r+4n^2}}{6r}\\&<0 \text{. } \end{aligned} \end{aligned}$$

Otherwise, if $r(n+2)-2n < 0$, then we can easily verify that $s_1<0$.

Second, it holds that:

$$\begin{aligned} s_2-1= \frac{r(n-4)-2n + \sqrt{(n+2)^2r^2+(16n-4n^2)r+4n^2}}{6r}\text{. } \end{aligned}$$

We distinguish two cases. If $r(n-4)-2n \le 0$, then:

$$\begin{aligned} \begin{aligned} s_2-1&= \frac{-\sqrt{(2n-r(n-4))^2} +\sqrt{(n+2)^2r^2+(16n-4n^2)r+4n^2}}{6r}\\&= \frac{-\sqrt{(n-4)^2r^2+(16n-4n^2)r+4n^2}+\sqrt{(n+2)^2r^2 +(16n-4n^2)r+4n^2}}{6r}\\&>0 \text{. } \end{aligned} \end{aligned}$$

Otherwise, if $r(n-4)-2n > 0$, then we can easily verify that $s_2-1>0$.

Third, it holds that:

$$\begin{aligned} n-1-s_2= \frac{r(5n-8)+2n- \sqrt{(n+2)^2r^2+(16n-4n^2)r+4n^2}}{6r}\text{. } \end{aligned}$$

We distinguish two cases. If $n=3$, then:

$$\begin{aligned} \begin{aligned} 2-s_2&= \frac{7r+6-\sqrt{25r^2+12r+36}}{6r}\\&= \frac{\sqrt{49r^2+84r+36}-\sqrt{25r^2+12r+36}}{6r}\\&>0 \text{. } \end{aligned} \end{aligned}$$

Otherwise, if $n \ge 4$, then:

$$\begin{aligned} \begin{aligned} n-1-s_2&\ge \frac{r(5n-8)+2n - \sqrt{(n+2)^2r^2+4n^2}}{6r}\\&> \frac{r(5n-8)+2n - \sqrt{(n+2)^2r^2}-\sqrt{4n^2}}{6r}\\&=\frac{4n-10}{6}\\&>0 \text{. } \end{aligned} \end{aligned}$$

It follows from the above three steps that p(s) is strictly decreasing in the interval $[1,s_2]$ and strictly increasing in the interval $[s_2,n-1]$. This implies that $\pi _i^C(q_s^*,{\tilde{q}}_s)$ attains its maximum either at $s=1$ or at $s=n-1$ which proves the first part of Lemma 4.2.

It remains to compare the two following profits of each cartel member derived from (22):

$$\begin{aligned} \pi _i^C(q_1^*,{\tilde{q}}_1)=\frac{V^2n(1+r)(n+r)}{(2n+r(n+1))^2}\text{, } \end{aligned}$$

and

$$\begin{aligned} \pi _i^C(q_{n-1}^*,{\tilde{q}}_{n-1})=\frac{V^2n(1+r)(2n+r)^2(n+r(n-1))}{(4n^2(1+r)+3r^2(n-1))^2}\text{. } \end{aligned}$$

Calculating the difference between these two individual profits leads to:

$$\begin{aligned} \begin{aligned}&\pi _i^C(q_{n-1}^*,{\tilde{q}}_{n-1})- \pi _i^C(q_1^*,{\tilde{q}}_1) =V^2n(n-2)r^2(1+r) \\&\quad \times \frac{(n^2-6n+5)r^3+(4n^3-16n^2+8n)r^2 +(4n^4-8n^3-4n^2)r+4n^4-8n^3}{(2n+r(1+n))^2(3r^2(n-1)+4n^2(1+r))^2} \text{. }\end{aligned} \end{aligned}$$

We can verify that $\pi _i^C(q_{n-1}^*,{\tilde{q}}_{n-1})- \pi _i^C(q_1^*,{\tilde{q}}_1)$ is positive for any $n\ge 5$.¹² We conclude that for any $s\in \lbrace {1,\ldots ,n-2}\rbrace$, $\pi _i^C(q_{n-1}^*,{\tilde{q}}_{n-1})>\pi _i^C(q_s^*,{\tilde{q}}_s)$. $\square$

Proof of Proposition 4.3:

It follows from (7) that a Cournot oligopoly TU-game $(N,v^C )\in G$ is balanced, and so has a non-empty core, if and only if $v^C(N)\ge v_R^C(N)$ where $(N,v_R^C )$ is the root game of $(N,v^C )$. Furthermore, we deduce from Lemma 4.2 that the worth of the grand coalition $v_R^C(N)$ is obtained either at the balanced collection $\lbrace {\lbrace {k}\rbrace :k\in N}\rbrace$ or at the balanced collection $\lbrace {N\backslash {\lbrace {k}\rbrace } : k\in N}\rbrace$. Hence, it holds that:

$$\begin{aligned} v_R^C(N)=n\max \left\{ \pi _i^C(q_{1}^*,{\tilde{q}}_{1}),\pi _i^C(q_{n-1}^*,{\tilde{q}}_{n-1}).\right\} \end{aligned}$$

(23)

It remains to show that $\pi _i^C(q_{n}^*,{\tilde{q}}_{n})\ge \max \lbrace {\pi _i^C(q_{1}^*,{\tilde{q}}_{1}),\pi _i^C(q_{n-1}^*,{\tilde{q}}_{n-1}})\rbrace$. Using the expression of Nash equilibrium profit of any cartel member in Cournot competition given by (22), one gets:

$$\begin{aligned} \begin{aligned} \pi _i^C(q_{n}^*,{\tilde{q}}_{n})-\pi _i^C(q_{1}^*,{\tilde{q}}_{1})&= \frac{(Vr(n-1))^2}{4(r(1+n)+2n)^2} \\&>0\text{, } \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \pi _i^C(q_{n}^*,{\tilde{q}}_{n})-\pi _i^C(q_{n-1}^*,{\tilde{q}}_{n-1})&= \frac{V^2r^2((5n^2-14n+9)r^2+(8n^3-16n^2+4n)r+8n^3-12n^2)}{4(3r^2(n-1)+4n^2(1+r))^2} \\&>0\text{. } \end{aligned} \end{aligned}$$

Hence, we conclude that $v^C(N)\ge v_R^C(N)$ which is equivalent to $C(N,v^C )\ne \emptyset$. $\square$

Proof of Theorem 4.5:

First, we determine the aggregate-monotonic cores of $(N,v^B)\in G$ and $(N,v^C)\in G$ respectively. It follows from Theorem 4.4 that the worth of the grand coalition $v_R^B(N)$ is obtained at the balanced collection $\lbrace {N\backslash {\lbrace {k}\rbrace } : k\in N}\rbrace$. Hence, it holds that:

$$\begin{aligned} v_R^B(N)=n\pi _i^B(p_{n-1}^*,{\tilde{p}}_{n-1}), \end{aligned}$$

(24)

where $(N,v_R^B)$ is the root game of $(N,v^B)$. Moreover, since oligopoly TU-games $(N,v^B)$ and $(N,v^C)$ are symmetric, the cores of their associated root games $C(N,v_R^B)$ and $C(N,v_R^C)$, respectively, are singletons.¹³ We deduce from (23) and (24) that:

$$\begin{aligned} C(N,v_R^B)=\left\{ (\pi _i^B(p_{n-1}^*,{\tilde{p}}_{n-1}))_{i=1}^n\right\} \text{, } \end{aligned}$$

and

$$\begin{aligned} C(N,v_R^C)=\Big \{( \max \Big \{ \pi _i^C(q_{1}^*,{\tilde{q}}_{1}),\pi _i^C(q_{n-1}^*,{\tilde{q}}_{n-1})\Big \})_{i=1}^n\Big \}. \end{aligned}$$

It follows from (8) that:

$$\begin{aligned} AC(N,v^B)=\lbrace {(\pi _i^B(p_{n-1}^*,{\tilde{p}}_{n-1}))_{i=1}^n + (v^B(N)-n\pi _i^B(p_{n-1}^*,{\tilde{p}}_{n-1}))\cdot \Delta _n }\rbrace \text{, } \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} AC(N,v^C)&=\lbrace {(\max \lbrace {\pi _i^C(q_{1}^*,{\tilde{q}}_{1}),\pi _i^C(q_{n-1}^*, {\tilde{q}}_{n-1}})\rbrace )_{i=1}^n} \\&{+(v^C(N)-n\max \lbrace {\pi _i^C(q_{1}^*, {\tilde{q}}_{1}),\pi _i^C(q_{n-1}^*,{\tilde{q}}_{n-1})}\rbrace ) \cdot \Delta _n }\rbrace \text{, } \end{aligned} \end{aligned}$$

where $\Delta _n$ denotes the unit-simplex.

Second, since $v^B(N)=v^C(N)$ it is sufficient to show that $\pi _i^C(q_{n-1}^*,{\tilde{q}}_{n-1}) > \pi _i^B(p_{n-1}^*,{\tilde{p}}_{n-1})$ to establish that $AC(N,v^C)\subset AC(N,v^B)$. Using the expressions of Nash equilibrium profits of any cartel member in Bertrand and Cournot competition given by (19) and (22), one gets:

$$\begin{aligned} \begin{aligned} \pi _i^C(q_{n-1}^*,{\tilde{q}}_{n-1})-\pi _i^B(p_{n-1}^*,{\tilde{p}}_{n-1})&=\frac{V^2nr^3(n-1)(2+r)}{(3r^2(n-1)+4n^2(1+r))^2}\\&>0\text{, } \end{aligned} \end{aligned}$$

which concludes the proof. $\square$