Sequential Mergers and Delayed Monopolization in Triopoly

Abstract

Under triopoly and Cournot competition, we study an infinite horizon Markov perfect equilibrium merger game in which in each period one of the firms (“the Buyer”) selects a bid price and then the two sellers accept or reject this offer with some probability. The possibility of a “war of attrition” equilibrium in which the seller who outlasts the other is then able to sell in the following period at a greater price, is a distinct feature of the model. Delayed monopolization is all the more likely when the discount factor is small and the ratio duopoly/ triopoly profits is important. Two other equilibria are shown to be possible: an unmerged and an immediate monopolization equilibrium. Each equilibrium is shown to correspond to a different set of parameter values. The two special cases of linear and constant price elastic demand functions are fully characterized.

Appendix

Let us define $\rho =rB-\pi (3)\Leftrightarrow B=(b\rho +\pi (3))/r.$ Let us then substitute this value for B in $V_{A}(B,3)$ as defined by Eq. (7). We only need to consider values of $\rho \in [0,\pi (2)-\pi (3)]$.³⁷ One obtains a function $F(\rho )=\frac{N(\rho )}{D(\rho )}$ where

$$\begin{aligned} D(\rho )= & {} \rho ^{2}\beta -\left( \pi (2)-\pi (3)\right) ^{2},\\ N(\rho )= & {} -2\rho ^{3}\beta -2\rho (\beta (\pi (1)-2\pi (2))+\pi (2)-2\pi (3))(\pi (2)-\pi (3))-\nonumber \\{} & {} \quad \left( \pi (2)-\pi (3)\right) ^{2}\pi (3)+\nonumber \\{} & {} \quad \rho ^{2}\left( (-1+3\beta )\pi (1)-4\left( -1+\beta )\pi (2)-(3+2\beta \right) \pi (3)\right) .\nonumber \end{aligned}$$

(13)

This function has several useful properties:

(i) it has two asymptotes for $\rho =-\frac{1}{\sqrt{\beta }}(\pi [2]-\pi [3])$ and for $\rho =\frac{1}{\sqrt{\beta }}(\pi [2]-\pi [3])$ whenever $\beta >0.$ Note that

$$\begin{aligned} \left[ 0,\pi [2]-\pi [3]\right] \subset \left[ -\frac{1}{\sqrt{\beta }}(\pi [2]-\pi [3]),\frac{1}{\sqrt{\beta }}(\pi [2]-\pi [3])\right] , \end{aligned}$$

Of course, the parts of the above function at the left and right of the asymptotes are irrelevant. We intend to analyze the properties of the function between the asymptotes. For that purpose the properties of $F(\rho )$ outside this interval are informative.

(ii) $F(\rho )\rightarrow +\infty$ as $\mu \rightarrow -\infty$ and $F(\rho )\rightarrow -\infty$ as $\rho \rightarrow +\infty ;$

(iii) Differentiating (13) wrt $\rho ,$ we obtain $F^{\prime }(\rho )=\frac{2}{D(\rho )^{2}}\left[ G(\rho )\right]$ where

$$\begin{aligned}{} & {} G(\rho ) =-\rho ^{4}\beta ^{2}+\rho ^{2}\beta (\beta (\pi (1)-2\pi (2))+4\pi (2)-5\pi (3))+ \end{aligned}$$

(14)

$$\begin{aligned}{} & {} \rho (\pi (2)-\pi (3))^{2}((1-3\beta )\pi (1)+4(-1+\beta )\pi (2+3\pi (3))(1+\beta )+\nonumber \\{} & {} \quad (\beta ((\pi (1)-2\pi (2))+\pi (2)-2\pi (3))(\pi (2)-\pi (3))^{2}. \end{aligned}$$

(15)

The extrema of $F(\rho )$ are the solutions of $G(\rho )=0.$ There are at most four real solutions of this equation. The sum of the four roots, whether real or complex, equals zero.

Let us then define L $=N(-\frac{1}{\sqrt{\beta }}(\pi (2)-\pi (3)))$ and $R=N(\frac{1}{\sqrt{\beta }}(\pi (2)-\pi (3))$ and denote $\widetilde{\beta }=\left[ \frac{\pi (1)-4\pi (2)+3\pi (3)}{2\left( \pi (1)-2\pi (2)\right) }\right] ^{2}$.³⁸

When $\pi (1)-4\pi (2)+3\pi (3>0,$ it is easy to see that $R<0$ for all $\beta \in [|0,1)$ and that $L>0$ iff $\beta >\widetilde{\beta }$ (resp $L<0$ if. $\beta <\widetilde{\beta }$). There are accordingly two cases.³⁹

Case (a) occurs when $\beta >\widetilde{\beta }$ so that $L>0$ and $R<0$.

An example is pictured below.

Fig. 12

The function $F(\rho )$ when $\pi (1)=4,\pi (2)=1.5,\pi (3)=1,\beta =0.132$

In case (a) there is one (real) root $<L$, one (real) root $>R.$

and either zero or two real roots in (L, R).

Case (b) occurs when $\beta <\widetilde{\beta }$ so that L and R are both $<0$ An example is pictured below.

Fig. 13

The function $F(\rho )$ when $\pi (1)=4,\pi (2)=1.5,\pi (3)=1,\beta =0.132$

In the limit case, when $\beta =\widetilde{\beta }$, there is only one asymptote for $\rho =\frac{2(\pi (1)-2\pi (2))(\pi (2)-\pi (3)}{\pi (1)-4\pi (2)+3\pi (3}>0.$ For all relevant purposes, this limit case resembles case (b).

Fig. 14

The function $F(\rho )$ when $\pi (1)=4,\pi (2)=1.5,\pi (3)=1,\beta =\widetilde{\beta }=0.25$

When $\pi (1)-4\pi (2)+3\pi (3<0$ Figs. 12–16 below illustrate the different cases,⁴⁰ it is easy to see that $R<0$ for all $\beta <1$ and $L<0$ if $\beta <\widetilde{\beta }$ (resp. $L>0$ if $\beta >\widetilde{\beta }$).

Case (a) occurs (again) when $\beta >\widetilde{\beta }.$ so that $L>0$ and $R<0$. It follows that case (a) always obtains when $\beta >\widetilde{\beta }.$

Case (c) occurs when $\beta <\widetilde{\beta }$ so that R and L are both $<0.$ An example is pictured below.

Fig. 15

The function F($\rho$) in the linear demand case for $\beta =0.011$

In the limit case, when $\beta =\widetilde{\beta }$, there is only one asymptote for $\rho =\frac{2(\pi (1)-2\pi (2))(\pi (2)-\pi (3)}{\pi (1)-4\pi (2)+3\pi (3}<0.$ For all relevant purposes, this limit case looks like case (c).

Fig. 16

The Function F($\rho$) in the linear demand case when $\beta =\widetilde{\beta }=1/64$

Finally, in the special case when $\pi (1)-4\pi (2)+3\pi (3=0,$ we have $\widetilde{\beta }=0$ and so case (a) obtains for all $\beta >0.$

Note that the condition $\pi (1)-4\pi (2)+3\pi (3>0$ ( resp. $<0$) is the one that ensures that $F(\rho )$ is convex (resp. concave) in $\rho$ when $\beta =0.$

Whenever $\beta >\widetilde{\beta }$, we obtain case (a). The local extrema of $F(\rho )$ are the roots of a fourth order polynomial which has at most four real roots, so that there are at most four local extrema. There are obviously a local minimum in the interval $\ \left( -\infty ,-\frac{1}{\sqrt{\beta }}(\pi [2]-\pi [3])\right)$ and a local maximum in the interval $\left( \frac{1}{\sqrt{\beta }}(\pi [2]-\pi [3]),+\infty \right) .$ There are either no extremum or a local maximum and then a local minimum in the interval $\left( -\frac{1}{\sqrt{\beta }} (\pi [2]-\pi [3]),\frac{1}{\sqrt{\beta }}(\pi [2]-\pi [3])\right) .$ In case (b) $F(\rho )$ is convex in the interval $\left( -\frac{1}{\sqrt{\beta }}(\pi [2]-\pi [3]),\frac{1}{\sqrt{\beta }}(\pi [2]-\pi [3])\right)$ while in case (c) it is concave in this interval.

Proof of Lemma 2

(i) Suppose that Condition ($\lnot$B) holds then F’$(0)>0.$ Since under Condition (A)), $F(\pi [2]-\pi [3])\le F(0),$ then a war of attrition (delayed monopolization) is the only possible equilibrium. (ii) Suppose then that Condition (B) holds. so that $F^{\prime }(0)\le 0.$ For a war of attrition to be an equilibrium there should then be a minimum $\rho _{0}$ and a maximum $\rho _{1}>\rho _{0}$ of $F(\rho )$ in $(0,\pi [2]-\pi [3]).$ That already makes two roots. Now consider the three possible cases:

In case (a) there would necessarily be in addition one root (a maximum) in [L, 0],  one root (a minimum) in $\left( \rho _{1},R\right) ,$ another root (a minimum) $<L$ and one another (a maximum) $>R$. Therefore, there would be six roots, two more than the maximum number.

In case (b) there would necessarily be in addition to $\rho _{0}$ and $\rho _{1},$ one root (a minimum) in $(\rho _{1},R)$ and there is one root (a maximum) greater than R,  so that there would be four (real) roots greater than zero, which is impossible since the roots must sum to zero.

In case (c) notice first that $G^{\prime }(0)<0$ for all $\beta \in [0,\widetilde{\beta }].$ Indeed $G^{\prime }(0)=$ $(-1+b)^{2} (ab(1-3\beta )+4b(-1+\beta )+3(1+\beta )$ is linear in $\beta .$ For $\beta =0$ one obtains $G^{\prime }(0)=$ $(-1+b)^{2}(3-4b+ab)<0.$ For $\beta =\widetilde{\beta },$ one obtains $G^{\prime }(0)=(-1+b)^{2}\frac{(3-4b+ab)(ab-3)^{2}}{4b^{2}(a-2)^{2}b^{2}}<0.$ It follows that $G^{\prime }(0)<0$ for all $\beta \in [0,\widetilde{\beta }].$ Accordingly, if a war of attrition is an equilibrium there must be two strictly positive real values of $\rho \in (0,\pi [2]-\pi [3]),$ namely two inflection points, such that $G^{\prime \prime }(\rho )=0$ However, computing directly the two values of $\rho$ such that $G^{\prime \prime }(\rho )=0,$ we obtain

$$\begin{aligned} \rho ^{I}=\mp \frac{\sqrt{b-1}\sqrt{-5+4b-2b\beta +ab\beta }}{\sqrt{6\beta }}, \end{aligned}$$

so that either (a) there is no corresponding real value or (b) one of them is negative, a contradiction in both cases.

We can conclude that when Condition (B) holds, an unmerged equilibrium is the only possible equilibrium. $\blacksquare$

Proof of Lemma 3

(i) When Condition $(\hbox {C}_{1})$ holds we have $F^{\prime }(\pi [2]-\pi [3])<0$ so that, given that Condition ($\lnot A)$ means that $F(\pi [2]-\pi [3])>F(0),$ the only possible equilibrium is a war of attrition equilibrium.

(ii) When Condition (B) holds so that $F^{\prime }(0)\le 0,$ the proof that one cannot obtain a war of attrition equilibrium is exactly the same in as point (ii) of the proof of Lemma 2. Given Condition ($\lnot A),$ it can be concluded this time that immediate monopolization is the only possible equilibrium.

(iii) What we already proved is pictured in Fig. 17. The blue area corresponds to immediate monopolization and the green area corresponds to delayed monopolization (war of attrition). The white area in between where conditions ($\lnot A$), ($\lnot B)$ and ($\lnot C_{1}$) hold is now under study.

Fig. 17

The areas corresponding to the different conditions when $\beta =0.5$

Fig. 18

Case $\beta =0.5$

When conditions ($\lnot B$) and ($\lnot C_{1})$ both hold we have $F^{\prime }(0)>0$ and $F^{\prime }(\pi [2]-\pi [3])>0$ (see the corresponding area in Fig. 17). Then there are either zero or two real roots in the interval $\left( 0,C\right) .$ In the latter case, the first is a maximum and a second is a minimum. In the former case, the only equilibrium is the immediate monopolization equilibrium. When there are two real roots $\rho _{1}$ and $\ \rho _{2},$ we need to compare $F(\rho _{1})$ with $F(\pi [2]-\pi [3]).$ This can only be done numerically⁴¹ in the space $\left( b,a\right)$ for specific values of $\beta .$This yields in each case the locus $a(b,\beta )$ along which $F(\rho _{1})$ $=F(\pi [2]-2\pi [3])$. This is pictured in red below for $\beta =0.5.$ Below the locus, i.e., for all $a\in (\frac{3b-1}{b},a(b,\beta )]$, $\ F(\rho _{1})\ge F(\pi [2]-\pi [3]),$ a war of attrition equilibrium obtains.⁴² Above we obtain an immediate monopolization equilibrium. Indeed since $\rho _{1}<b-1$

$$\begin{aligned} \frac{\partial \left[ F(\rho _{1})-F(\pi [2]-2\pi [3])\right] }{\partial a}=-\frac{(b-1-\rho _{1})(b-1+\rho _{1}(1-2\beta ))--}{(b-1)^{2} -\beta \rho _{1}^{2}}<0. \end{aligned}$$

This is illustrated in Fig. 18. $\blacksquare$

Proof of Proposition 4

$\widehat{\rho }$ is strictly positive iff $\left( \pi [1]-\pi [2]-2\pi [3]\right) >0$ (Condition ($\lnot \hbox {E}$)). It is strictly smaller than $\pi [2]-\pi [3]$ iff $\pi [1]-4\pi [2]+\pi [3]<0$ (Condition $\hbox {C}_{0}$). Iff condition (E) holds, then the global maximum occurs at $\rho =0,$ corresponding to an unmerged equilibrium. Iff condition ($\lnot C_{0}$) holds, the global maximum occurs at $\rho =\pi [2]-\pi [3],$ corresponding to an immediate monopolization equilibrium. Iff Conditions ($\hbox {C}_{0}$) and (E) hold the global maximum is an interior one at $\rho =\widehat{\rho }.$ $\blacksquare$

Proof of Proposition 5

In the case when $B\in \left[ 1/16r,1/9r\right] ,$ use Eq. (7) in which we set $\pi (1)=1/4,$ $\pi (2)=1/9$ and $\pi (3)=1/16.$ Denote $\rho =rB-\pi [3].$ Then the FOC wrt B obtains as

$$\begin{aligned} -&490+39284\beta -8586\beta ^{2}+214990848\rho ^{3}\beta ^{2}-859963392\rho ^{4}\beta ^{2}\nonumber \\&+ \rho ^{2}\left( 5515776\beta -18994176\beta ^{2}\right) +\rho \left( -14112-929376\beta -694656\beta ^{2}\right) =0, \end{aligned}$$

(16)

This equation has one (and only one) solution in the interval $\left[ 1/16,1/9\right]$ iff $\beta \ge 1/2$ which corresponds to a global maximum as is easily checked using Mathematica ManipulatePlot function. $\blacksquare$