Collusive stability of cross-holding with cost asymmetry

Abstract

Recent empirical research demonstrates that cross-holding engenders significant anti-competitive effect even though it confers no decisive influence on the target firm. This research has obtained wide attention, which leads to both calls for and actual changes in antitrust policy. However, the effects of cross-holding on competition are not well established. This paper examines the collusive effect of cross-holding with asymmetry in cost functions across firms in an infinitely repeated Cournot duopoly game. The two firms have a different share of capital that affects marginal costs. We find that cross-holding facilitates collusion when a small firm (i.e., a firm with a small capital capacity) increases its passive ownership in a big firm (i.e., a firm with a large capital capacity), and the size of initial ownership is relatively small. Unexpectedly, collusion becomes more difficult to be sustained when a big firm increases ownership in a small firm, or when a small firm increases ownership in a big firm but the initial ownership is relatively large. Our result that cross-holding may either facilitate or hinder collusion, under certain circumstances, generates important implications for antitrust policy.

Appendix

1.1 Proof of Proposition 1

In the following, we show that ${\underline{k}}(\theta )<{\overline{k}}(\theta )$ holds for all $\theta <0.5$. Denote by $T(\theta , k)={(\pi _{1}^{M}/\pi _{1}^{N})}/{(\pi _{2}^{M}/\pi _{2}^{N})}$. Simple calculations yield that

$$\begin{aligned} T(\theta , k)=\frac{(1+k)^2 \left( 3-2 k\right) \left( \theta +(1-\theta ) k\right) }{3 \theta +k_1 \left( 4+\theta -\theta ^2+k \left( 4-9 \theta +2 \theta ^2-(7+(\theta -5) \theta ) k+2 k^2\right) \right) }. \end{aligned}$$

We then obtain that $\partial ^{2} T(\theta , k) /\partial k \partial \theta <1$, which implies that ${\underline{k}}(\theta )<{\overline{k}}(\theta )$ since both ${\underline{k}}(\theta )$ and ${\overline{k}} (\theta )$ decrease in $\theta$.

1.2 Proof of Lemma 1

Based on (13) and (14), it can be shown that $0< \widehat{\delta _1 }<1$ and $0< \widehat{\delta _2 }<1$ when ${\underline{k}}(\theta )<k<{\overline{k}}(\theta )$. Furthermore, straightforward calculations lead to

$$\begin{aligned} \frac{\partial \widehat{\delta _1 }}{\partial \theta }=\frac{3(1-\theta )(1-k)(1+2 k) \left( 3+(3-\theta )(1-k)k\right) \left( 3 (2-\theta )^2 (1+\theta )+k X\right) }{-k \left( -3 \theta (-4+\theta (2+\theta ))+k \left( 36+(-3+\theta ) \theta (4+\theta (7+\theta ))+k Y\right) \right) ^2}<0, \end{aligned}$$

where

$$\begin{aligned} X= & {} (2-\theta ) \left( 24-9 \theta +\theta ^3\right) +k \left( {15+\theta (-45+\theta (54+\theta (-23+3 \theta )))}\right. \\&\left. { +k \left( -75+\theta (132+\theta (-90+(28-3 \theta ) \theta ))+(-3+\theta )^3 (-1+\theta ) k\right) }\right) >0, \\ \end{aligned}$$

$Y=54-3 \theta (23+\theta (-7+(-3+\theta ) \theta ))-(-3+\theta )^2 (-1+\theta ) k \left( -3 (1+\theta )+(-1+\theta ) k\right)$.

Similarly, we obtain that ${\partial \widehat{\delta _2}}/{\partial \theta }>0$, where

$$\begin{aligned} \frac{\partial \widehat{\delta _{2}}}{\partial \theta }=\frac{18k\left( 3+(3-\theta )(1-k)k\right) \left( 3+k-2k^{2}\right) {}^{2}}{(1-k)\left( 3-3\theta -(3-\theta )k\right) ^{2}\left( k\left( 3(-3+\theta )+k\left( 18-4\theta -(3-\theta )k\right) \right) -18\right) ^{2}}. \end{aligned}$$

1.3 Proof of Lemma 2

The justification of existence of ${\tilde{k}}(\theta )$ in $\left( {\underline{k}}(\theta ),{\overline{k}}(\theta )\right)$ which solves $\widehat{\delta _{1}}=\widehat{\delta _{2}}$ is based on the following facts: (i) $0<\widehat{\delta _{1}}<1$ and $0<\widehat{\delta _{2}}<1$ when ${\underline{k}} (\theta )<k<{\overline{k}}(\theta )$; (ii) ${\partial \widehat{\delta _{1}}}/{ \partial \theta }<0$ and ${\partial \widehat{\delta _{2}}}/{\partial \theta } >0$ by Lemma 1; and (iii) $\widehat{\delta _{1}}=\widehat{\delta _{2}}$ when $\theta =0$ and ${k=0.5}$, or equivalently, ${\tilde{k}}(0)=0.5$. Furthermore, simple calculations yield that ${\partial \widehat{\delta _{1}}}/{ \partial k}<0$ and ${\partial \widehat{\delta _{2}}}/{\partial k}>0$ based on (12) and (13). Consequently, $\widehat{\delta _{1}}<\widehat{\delta _{2}}$ when $k>{\tilde{k}}(\theta )$, while $\widehat{\delta _{1}}>\widehat{\delta _{2}}$ otherwise. Implied by the two sets of partial derivatives: ${\partial \widehat{\delta _{i}}}/{\partial \theta }$ and ${\partial \widehat{\delta _{i}}}/{\partial k}$ for $i=1,2$, we can obtain that ${\tilde{k}}(\theta )$ decreases in $\theta$.