Can cross-holdings benefit consumers?

Abstract

Common wisdom suggests that cross-holdings can lead to significant output contraction, and thus hurt consumers. On the contrary, we demonstrate that cross-holdings may increase industry output and benefit consumers in an asymmetric Cournot oligopoly with the presence of a welfare-maximizing tax/subsidy policy. The government will strategically use the tax/subsidy policy to regulate the market outcomes in anticipation of the adverse effect of cross-holdings, which could raise industry output and benefit consumers in certain situations depending on the cost distributions and cross-holding structures.

Appendix

1.1 Proof of Lemma 1

Without cross-holdings, we have

$$\begin{aligned} \tau ^{1*}|_{\delta =0}=\frac{a (c m+d n) - c^2 m (n (N+2)+1)+2 c d m n (N+2)-d^2 n (m (N+2)+1)}{a (c m+d n)-(N+1) \left( c^2 m+ d^2n + mn (c-d)^2\right) }, \end{aligned}$$

where $N= m+n$. Both the denominator and numerator are linear in a. Therefore, straightforward calculations yields

$$\begin{aligned} \tau ^{1*}|_{\delta =0} {\left\{ \begin{array}{ll}<0, \text { when } T_2<a<T_1;\\ \in [0, 1), \text { when } a \le T_2, \end{array}\right. } \end{aligned}$$

where $T_1= \frac{(m+n+1) \left( c^2\,m+d^2n +mn (c-d)^2\right) }{c m+d n},$ and $T_2= \frac{c^2\,m +d^2 n + m n (c-d)^2 (m+n+2) }{c m+d n}.$ Notice that we have $T_1-T_2= cm + dn$.

Further, taking derivative with respect to $\delta$ over $\tau ^{1*}$ yields

$$\begin{aligned} \frac{\partial \tau ^{1*}}{\partial \delta }=-\frac{n (c m +d n (1-\delta )) \left( a d (c m +d n (1-\delta ))+(d (m+1)-c m) D_2 \right) }{\left( a (c m+d n (1-\delta ))-(1+ m + n (1-\delta ))D_1\right) ^2}, \end{aligned}$$

where $D_2= c d m (1+ m + 3 n (1-\delta ))- c^2\,m (2+ m + 2 n (1-\delta )) - d^2 n (m+1) (1-\delta ).$ It can be calculated that $D_2<0$ for any $c>d>0$, $m>0$, $n>0$ and $0<\delta <1/2$. Therefore, if $d/c<m/(1+m)$, we must have $\partial \tau ^{1*}/\partial \delta <0$. If $d/c>m/(1+m)$, we have $\partial \tau ^{1*}/\partial \delta <0$ when $a d (c m +d n (1-\delta ))+(d (m+1)-c m) D_2>0$ and $\partial \tau ^{1*}/\partial \delta >0$ when $a d (c m +d n (1-\delta ))+(d (m+1)-c m) D_2<0$.

1.2 Proof of Lemma 2

Without cross-holdings, we have

$$\begin{aligned} \tau ^{3*}|_{\delta =0} = \frac{m \left( c^2+(m+2) n (c-d)^2+n^2 (c-d)^2\right) + n d^2 -a (c m+d n)}{(m+n+1) \left( c^2 m+ d^2 n+ (c-d)^2 m n\right) -a (c m+d n)}. \end{aligned}$$

Both the denominator and numerator are linear in a. Therefore, we can simplify the condition of $\tau ^{3*}$ being tax or subsidy into intervals in terms of a. Straightforward calculations yields

$$\begin{aligned} \tau ^{3*}|_{\delta =0}{\left\{ \begin{array}{ll}<0 \text {, when } T_4<a<T_3;\\ \in [0, 1) \text { when } a \le T_4, \end{array}\right. } \end{aligned}$$

where $T_3=\frac{(m+n+1) \left( c^2\,m+ \left( (c-d)^2\,m +d^2\right) n\right) }{c m+d n}$, and $T_4=\frac{m \left( c^2+ (c-d)^2 (m+2) n + (c-d)^2 n^2\right) +d^2 n}{c m+d n}$. The distance between the two cutoffs is $T_3-T_4= cm+dn$.

More generally, taking derivative with respect to $\delta$ over $\tau ^{3*}$ yields

$$\begin{aligned} \frac{\partial \tau ^{3*}}{\partial \delta }=\frac{-m (cm (1-\delta ) +d n) F }{\left( (1+n+m(1-\delta ))H_2(\delta )-a (c m(1-\delta ) +d n) \right) ^2}, \end{aligned}$$

where $F=a c (c m (1-\delta )+d n)-L$, and

$$\begin{aligned} L=(c (1+n )-d n) \left( c^2 m (m (1-\delta ) +2 (1+n)) \!-\! c d m (1+3 n+ \delta +m (1-\delta ))\!+\! d^2 (m+1) n \right) . \end{aligned}$$

The sign of $\partial \tau ^{3*}/\partial \delta$ depends solely on the sign of F. We have $\partial \tau ^{3*}/\partial \delta >0$ if $F<0$, and vice versa. Taking derivative of F with respect to $\delta$ yields

$$\begin{aligned} \frac{\partial F}{\partial \delta } = c m ((d+(c-d) m) (c+(c-d) n)-a c). \end{aligned}$$

When $a<{(d+(c-d) m) (c+(c-d) n)}/{c}$, $\partial F/\partial \delta >0$, which indicates F increases with $\delta$ in (0, 1/2). We find that $F|_{\delta =0}<0$ and $F|_{\delta =1/2}<0$. Hence, $F<0$ for any $\delta \in (0, 1/2)$, which implies $\partial \tau ^{3*}/\partial \delta >0$ in this case. When $a>{(d+(c-d) m) (c+(c-d) n)}/{c}$, $\partial F/\partial \delta <0$, which indicates F decreases with $\delta$ in (0, 1/2). We find that

$$\begin{aligned} F|_{\delta =0}{\left\{ \begin{array}{ll}<0 \text {, if }\frac{(d+(c-d) m) (c+(c-d) n)}{c}<a<G;\\>0 \text {, if } a>G, \end{array}\right. } \end{aligned}$$

where $G=\frac{c+(c-d) n}{c (c m+d n)}\left( c^2 m (m+2 n+2)-c d m (m+3 n+1)+d^2 n (m+1) \right)$; and

$$\begin{aligned} F|_{\delta =1/2}{\left\{ \begin{array}{ll}<0 \text {, if }\frac{(d+(c-d) m) (c+(c-d) n)}{c}<a<H;\\>0 \text { if } a>H, \end{array}\right. } \end{aligned}$$

where $H=\frac{c+(c-d) n}{c (c m+2 d n)}\left( c^2 m (m+4 n+4)-c d m (m+6 n+3)+2 d^2 n (m+1) \right)$. Since

$$\begin{aligned} H-G=\frac{2 m^2 (c-d) (c+c n-d n)^2}{(c m+d n) (c m+2 d n)}>0, \end{aligned}$$

given that $\partial F/\partial \delta < 0$, we have the following results:

• 1) when ${(d+(c-d) m) (c+(c-d) n)}/{c}<a<G$, we have $F|_{\delta =0}<0$ and $F|_{\delta =1/2}<0$. Then $F<0$ for any $\delta \in (0, 1/2)$. Thus, $\partial \tau ^{3*}/\partial \delta >0$;

• 2) when $G<a<H$, we have $F|_{\delta =0}>0$ and $F|_{\delta =1/2}<0$. Then F decreases from a positive number to a negative number as $\delta$ increases from 0 to 1/2. Thus, $\partial \tau ^{3*}/\partial \delta$ would correspondingly be negative and then positive.

• 3) when $a>H$, we have $F|_{\delta =0}>0$ and $F|_{\delta =1/2}>0$. Then $F>0$ for any $\delta \in (0, 1/2)$. Thus, $\partial \tau ^{3*}/\partial \delta <0$.

In conclusion, we have

$$\begin{aligned} \frac{\partial \tau ^{3*}}{\partial \delta } {\left\{ \begin{array}{ll}>0, \text { when } a<G;\\<0, \text { when } G<a<\frac{L}{c (c m (1-\delta )+d n)};\\>0, \text { when } \frac{L}{c (c m (1-\delta )+d n)}<a<H;\\ <0, \text { when } a>H. \end{array}\right. } \end{aligned}$$

1.3 The expression for joint profit in Case 3

$\Pi ^{3*}= n \pi _i^{3*}+m \pi _j^{3*}$, straightforward calculations lead to

$$\begin{aligned} \Pi ^{3*}= \frac{\left( \begin{array}{c} a^2 \left( c m (c-d \delta )+d^2 n\right) (c m (1-\delta )+d n)^3+ a H_2(\delta ) \cdot \\ (c m (1-\delta )+d n)^2\left( c d m (\delta (2+m (1-\delta ) ) + n (4-\delta ) )\right. \\ \left. -c^2 m (2+ 2 n + m \delta (1-\delta ) ) - d^2 n ( 2+ m(2-\delta ) )\right) - H_2(\delta )^2\cdot \\ (c m (1-\delta ) +d n) \left( m \left( c m \delta ^2 (c-d)+ \delta (d+d m -c m) (c- c n +d n) \right. \right. \\ \left. \left. + (2 c d n-d^2 n) (m+n+2) -c^2 (n (m+n+2)+1) \right) -d^2 n\right) \end{array} \right) }{(c m (1-\delta ) + d n)^3 \left( (1+ n+ m (1-\delta )) H_2(\delta ) -a (c m (1-\delta ) + d n) \right) }. \end{aligned}$$

1.4 Detailed calculations for the case of unit tax

All settings remain consistent with the basic model, except that we have replaced the ad valorem tax $\tau$ with a unit tax t. Moreover, this tax may function as a tax when t is positive and as a subsidy when t is negative.

Case 1: Cross-holdings between efficient firms

In the second stage, each firm simultaneously chooses the optimal quantity to achieve profit maximization. The profit for each efficient firm is

$$\begin{aligned} \pi _i=(1-\delta )(a-Q-t-d) q_i+\frac{\delta }{n-1}\sum _{k \le n,k\ne i}(a-Q-t-d) q_k, \end{aligned}$$

while the profit for each inefficient firm is $\pi _j=(a-Q-t-c) q_j.$ Solving the first-order conditions leads to the equilibrium outputs as

$$\begin{aligned} q_i= \frac{(1-\delta ) (a+c m-d (m+1)-t)}{1+n+m-\delta n},\quad q_j=\frac{a+c ((\delta -1) n-1)-d \delta n+d n-t}{1+n+m-\delta n}. \end{aligned}$$

All the second-order conditions in this section are satisfied.

In the first stage, the government chooses the unit tax t to maximize social welfare

$$\begin{aligned} SW=(P-t)Q-d Q_i-c Q_j+Q^2/2+t Q. \end{aligned}$$

By solving the first-order condition, we obtain

$$\begin{aligned} t^{1*}=\frac{c m-a (m-\delta n+n)+d (1-\delta ) n}{(n+m-\delta n)^2}. \end{aligned}$$

Substituting it into the outputs, we get the equilibrium industry output as

$$\begin{aligned} Q^{1*}=a-d-\frac{m (c-d)}{m+n-\delta n}. \end{aligned}$$

Differentiating it with respect to $\delta$ leads to

$$\begin{aligned} \frac{\partial Q^{1*}}{\partial \delta }=-\frac{m n (c-d)}{(n+m-\delta n)^2}<0. \end{aligned}$$

Case 2: Cross-holdings between inefficient firms

By analogy, we can easily obtain the equilibrium outcomes when inefficient firms hold passive ownership in each other. A simple way to do this is to switch n and m with each other, and switch d and c with each other in case 1. Thus, we obtain

$$\begin{aligned} t^{2*}=\frac{d n-a (m-\delta m+n)+c (1-\delta ) m}{(m+n-\delta m)^2},\quad Q^{2*}=a-c+\frac{n (c-d)}{m+n-\delta m}. \end{aligned}$$

Simple calculations lead to

$$\begin{aligned} \frac{\partial Q^{2*}}{\partial \delta }=\frac{m n (c-d)}{(m+n-\delta m)^2}>0. \end{aligned}$$

Case 3: Efficient firms hold shares in inefficient firms

The profit for each efficient acquiring firm is

$$\begin{aligned} \pi _i=(a-Q-t-d) q_i+\frac{\delta }{n}\sum _{j=1}^{m}{(a-Q-t-c) q_j}, \end{aligned}$$

while that for each inefficient acquired firm is $\pi _j=(1-\delta )(a-Q-t-c) q_j.$ Standard calculations by solving the first-order conditions yields

$$\begin{aligned} q_i=\frac{n (a+c m-d (m+1)-t)-\delta m (a-c-t)}{n (1+m+n-\delta m)},\quad q_j=\frac{a-c (n+1)+d n-t}{1+m+n-\delta m}. \end{aligned}$$

In the first stage, the government chooses a optimal unit tax t to maximize social welfare, which yields

$$\begin{aligned} t^{3*}=\frac{m \left( c (m+n) \delta +c-d (m+n+1) \delta -\delta ^2 m (c-d)\right) +d n-a (m-\delta m+n)}{(m+n-\delta m)^2}. \end{aligned}$$

The equilibrium industry output is

$$\begin{aligned} Q^{3*}=a-d-\frac{m (c-d)}{m+n-\delta m}. \end{aligned}$$

Differentiating $Q^{3*}$ with respect to $\delta$ leads to

$$\begin{aligned} \frac{\partial Q^{3*}}{\partial \delta }=-\frac{m^2 (c-d)}{(m+n-\delta m)^2}<0. \end{aligned}$$

Case 4: Inefficient firms hold shares in efficient firms

We can easily obtain the equilibrium outcomes in case 4 by switching n and m with each other, and switching d and c with each other in case 3. Thus, we obtain

$$\begin{aligned} t^{4*}= & {} \frac{n \left( d (m+n) \delta +d-c (m+n+1) \delta +\delta ^2 n (c-d)\right) +c m-a (m-\delta n+n)}{(m+n-\delta n)^2},\\ Q^{4*}= & {} a-c+\frac{n (c-d)}{m-\delta n+n}. \end{aligned}$$

Simple calculations lead to

$$\begin{aligned} \frac{\partial Q^{4*}}{\partial \delta }=\frac{n^2 (c-d)}{(m-\delta n+n)^2}>0. \end{aligned}$$

As in the basic model, numerous examples can be found to substantiate that, under certain circumstances, the firms have the incentive to increase cross-holdings in these four cases.

1.5 Simulations for the case of price competition

We conduct extensive numerical simulations and present the results in Tables 1, 2, 3, and 4. The Mathematica codes to solve each case and generate the tables are available upon request. In each table, we use $a=100, c=10, d=6$ and $\gamma =0.5$. We assign different values to m and d and increase $\theta$ from 0.00 to 0.45. In each case, we consider four values for m, which are listed in the first column, and four values for d, which are given in the second row at the top. The equilibrium industry output is provided in the last four columns. The "/" in the tables implies that these parameter values fail to satisfy the assumptions in our model.

As we see in these tables, an increase in the degree of cross-holdings has adverse effects on $Q^*$ and thus hurts consumers in cases 1 and 3, while improving $Q^*$ and thus benefiting consumers in cases 2 and 4.

Table 1 $Q^*$ in Case 1

Table 2 $Q^*$ in Case 2

Table 3 $Q^*$ in Case 3

Table 4 $Q^*$ in Case 4