Media bias, slant regulation, and the public-interest media

Abstract

This study addresses the role of government in reducing media bias that arises from the demand side. Introducing a public-interest media outlet reduces the equilibrium slants that would otherwise exist under laissez-faire. Subsidy for the truthful report and price regulation are designed to effectively remedy media bias. The socially optimal subsidy policy can reduce both slant and media prices.

Appendix

Proof of Proposition 1

By backward induction, in the final stage, solving the first-order conditions yields equilibrium prices

$$\begin{aligned} p_1^*=\frac{\phi ^2\left( 6b_2(z_2-z_1)+z_2^2-z_1^2\right) }{3(\chi +\phi )}, \quad p_2^*=\frac{\phi ^2\left( 6b_2(z_2-z_1)+z_1^2-z_2^2\right) }{3(\chi +\phi )}. \end{aligned}$$

(11)

Plugging $p_1^*$, $p_2^*$ into $\pi _1$ and $\pi _2$ and then solving $\partial \pi _1/\partial z_1=0$ and $\partial \pi _2/\partial z_2=0$ for $z_1$ and $z_2$ yield

$$\begin{aligned} z_1^*=\frac{-3b_2}{2}, \quad z_2^*=\frac{3b_2}{2}, \quad \text{ E }_d(s_1^*)=\frac{-3\phi b_2}{2(\chi +\phi )}, \quad \text{ E }_d(s_2^*)=\frac{3\phi b_2}{2(\chi +\phi )}, \end{aligned}$$

(12)

where $\text{ E }_d(\cdot )$ is the expected slant.

Proof of Proposition 2

Solving $u_b^1=u_b^3$ yields

$$\begin{aligned} \hat{b}_1=\frac{\phi ^2 z_1^2+(p_1-p_3)(\chi +\phi )-v_d\phi ^2}{2\phi ^2 z_1}. \end{aligned}$$

Solving $u_b^2=u_b^3$ yields

$$\begin{aligned} \hat{b}_2=\frac{\phi ^2 z_2^2+(p_2-p_3)(\chi +\phi )-v_d\phi ^2}{2\phi ^2 z_2}. \end{aligned}$$

The profit functions are as follows.

$$\begin{aligned} \pi _1&=\frac{p_1(\hat{b}_1-b_1)}{b_2-b_1}=\frac{p_1(\phi ^2z_1^2+(p_1-p_3)(\chi +\phi )-v_d\phi ^2+2b2\phi ^2z_1)}{4b_2\phi ^2z_1},\\ \pi _2&=\frac{p_2(b_2-\hat{b}_2)}{b_2-b_1}=\frac{p_2(2b_2\phi ^2z_2-\phi ^2z_2^2+(p_3-p_2)(\chi +\phi )+v_d\phi ^2)}{4b_2\phi ^2z_1},\\ \pi _3&=\frac{p_3(\hat{b}_2-\hat{b}_1)}{b_2-b_1}=\frac{p_3\left( \frac{\phi ^2z_2^2+(p_2-p_3)(\chi +\phi )-v_d\phi ^2}{2\phi ^2z_2} -\frac{\phi ^2z_1^2+(p_1-p_3)(\chi +\phi )-v_d\phi ^2}{2\phi ^2z_1}\right) }{2b_2}. \end{aligned}$$

Solving $\partial \pi _1/\partial p_1=0$, $\partial \pi _2/ \partial p_2=0$, and $\partial \pi _3/\partial p_3=0$ yields

$$\begin{aligned} p_1^{**}&=-\frac{(-z_1z_2^2-2v_dz_1-2z_2z_1^2+2v_dz_2-10z_2z_1b_2+3z_1^3+6z_1^2b_2)\phi ^2}{6(\chi +\phi )(z_1-z_2)},\\ p_2^{**}&=\frac{(-2z_1z_2^2+2v_dz_1-z_2z_1^2-2v_dz_2+10z_2z_1b_2-6z_2^2b_2+3z_2^3)\phi ^2}{6(\chi +\phi )(z_1-z_2)}, \\ p_3^{**}&=\frac{\phi ^2(z_1z_2^2-v_dz_1-z_2z_1^2+v_dz_2+4z_2z_1b_2)}{3(\chi +\phi )(z_1-z_2)}. \end{aligned}$$

Plugging $p_1^{**}$, $p_2^{**}$, $p_3^{**}$ into $\pi _1$ and $\pi _2$ yields

$$\begin{aligned} \pi _1&=-\frac{\phi ^2(-z_1z_2^2-2v_dz_1-2z_2z_1^2+2v_dz_2-10z_2z_1b_2+3z_1^3+6z_1^2b_2)^2}{144z_2b_2(\chi +\phi )(z_1-z_2)^2},\\ \pi _2&=\frac{\phi ^2(-2z_1z_2^2+2v_dz_1-z_2z_1^2-2v_dz_2+10z_2z_1b_2-6z_2^2b_2+3z_2^3)^2}{144z_2b_2(\chi +\phi )(z_1-z_2)^2}. \end{aligned}$$

When $v_d<\frac{9}{16}b_2^2$, solving $\partial \pi _1/\partial z_1=0$ and $\partial \pi _2/\partial z_2=0$ as $z_2=-z_1$ for symmetry yields four solutions: $z_1^{**1}=-\frac{3b_2}{8}-\frac{\sqrt{9b_2^2-16v_d}}{8}$, $z_1^{**2}=-\frac{3b_2}{8}+\frac{\sqrt{9b_2^2-16v_d}}{8}$, $z_1^{**3}=-2b_2+\sqrt{4b_2^2+v_d}$, $z_1^{**4}=-2b_2-\sqrt{4b_2^2+v_d}$.

We first show that $z_1^{**3}$ and $z_1^{**4}$ are not valid as follows. Since $z_1^{**3}$ is positive, it is obviously not valid. After some calculations, we find that the second-order condition of $z_1^{**4}$ is not satisfied, since

$$\begin{aligned}&\frac{\partial ^2\pi _1}{\partial z_1^2}|_{z_1=z_1^{**4}}\\&\quad =\frac{\left( 429b_2^2v_d+1352b_2^4+b_2(130v_d+676b_2^2)\sqrt{4b_2^2+v_d}+25v_d^2\right) \phi ^2}{72\left( \chi +\phi \right) b_2\left( 2b_2+\sqrt{4b_2^2+v_d}\right) ^3}>0. \end{aligned}$$

For $z_1^{**2}$, the second-order condition

$$\begin{aligned}&\frac{\partial ^2\pi _1}{\partial z_1^2}|_{z_1=z_1^{**2}}\\&\quad =-\frac{\left( 4016b_2^2v_d-3627b_2^4-b_2(264v_d-1209b_2^2)\sqrt{9b_2^2-16v_d}+2720v_d^2\right) \phi ^2}{72\left( \chi +\phi \right) b_2\left( -3b_2+\sqrt{9b_2^2-16v_d}\right) ^3}<0. \end{aligned}$$

which implies $v_d<\frac{155}{289}b_2^2$, and $p_3=-\frac{\left( -33b_2^2+11b_2\sqrt{9b_2^2-16v_d}+40v_d\right) \phi ^2}{96\left( \chi +\phi \right) }>0$,

implies $v_d>\frac{13}{50}b_2^2$, and

$$\begin{aligned} \hat{b}_1=-\frac{-33b_2^2+11b_2\sqrt{9b_2^2-16v_d}+40v_d}{24\left( -3b_2+\sqrt{9b_2^2-16v_d}\right) }<0, \end{aligned}$$

implies $v_d>\frac{11}{25}b_2^2$. From these three conditions, we find that this equilibrium slant is valid only when $v_d\in \left( \frac{11}{25}b_2^2,\frac{155}{289}b_2^2\right)$. Finally, by the same calculation process, we can find that $z_1^{**1}$ is valid when $v_d\le \frac{9}{16}b_2^2$. This is because $\partial ^2\pi _1^*/\partial z_1^2<0$, $p_3(z_1=z_1^{**1})>0$, and $\hat{b}_1(z_1=z_1^{**1}<0)$. The second part of the proof is shown from $z_1^{**}=-\frac{3b_2}{8}-\frac{\sqrt{9b_2^2-16v_d}}{8}<0$, and $z_2^{**}=\frac{3b_2}{8}+\frac{\sqrt{9b_2^2-16v_d}}{8}>0$, and compare $z_1^*$, $z_1^{**1}$ and $z_1^{**2}$, we have $z_1^{**1}-z_1^*=\frac{9b_2}{8}-\frac{\sqrt{9b_2^2-16v_d}}{8}>0$, $z_1^{**2}-z_1^*=\frac{9b_2}{8}+\frac{\sqrt{9b_2^2-16v_d}}{8}>0$. These results imply that when a public-interest media outlet is introduced, the equilibrium slants will be smaller than in the duopoly case.

When $v_d>\frac{9}{16}b_2^2$, we have $\partial \pi _1/\partial z_1>0$, $\partial \pi _2/\partial z_2<0$ when $z_1$ and $z_2$ are close to $0$. Thus, the unique corner solution is $z_1^{**}=z_2^{**}=0$. Therefore, $p_1^{**}=p_2^{**}=p_3^{**}=0$ à la Bertrand. However, each one which deviates from its location will earn a positive profit. Therefore, there does not exist a symmetric equilibrium slant when $v_d>\frac{9}{16}b_2^2$.

Proof of Corollary 1

The average bias with a public-interest media outlet for the first equilibrium slant is

$$\begin{aligned} \text{ ARB }^{**}&=\int \limits _b\text{ E }_d\left[ (n-d)^2\right] \\&=\left[ (\hat{b}\!-\!b_1)\left( s_1^{**}(d_1)^2\right) \!+\!(b_2-\hat{b})\left( s_2^{**}(d_2)^2\right) \!+\!(\hat{b}_2\!-\!\hat{b}_1)(s_3)^2\right] \bigg /(b_2-b_1)\\&=\frac{b_2^3\phi ^2\left( 39+13\sqrt{9-16\frac{v_d}{b_2^2}}+40\frac{v_d}{b_2^2}\right) \left( 3+\sqrt{9-16\frac{v_d}{b_2^2}}+8\frac{v_d}{b_2^2}\right) }{128\left( 3+\sqrt{9-16\frac{v_d}{b_2^2}}\right) \left( \chi +\phi \right) ^2}<\text{ ARB }^*, \end{aligned}$$

since $s_3=0$ (the public-interest media), $z_2^{**}<z_2^*$, and $\hat{b}_2>\hat{b}_1$. Similar results apply to the second equilibrium slant.

Proof of Proposition 3

For the first equilibrium slant, plugging $z_1^{**}$ and $z_2^{**}$ into $p_1$, $p_2$, $p_3$ yields

$$\begin{aligned} p_1^{**}&=p_2^{**}=\frac{\phi ^2\left( 39b_2^2+40v_d+13b_2\sqrt{9b_2^2-16v_d}\right) }{96(\chi +\phi )},\\ p_3^{**}&=\frac{\left( 33b_2^2-40v_d+11b_2\sqrt{9b_2^2-16v_d}\right) \phi ^2}{96(\chi +\phi )}, \end{aligned}$$

Comparing the above equilibrium prices with the duopoly case in Proposition 1 yields the following result.

$$\begin{aligned} p_3^{**}-p_1^{**}&=-\frac{\left( 3b_2^2+40v_d+b_2\sqrt{9b_2^2-16v_d}\right) \,\phi ^2}{48(\chi +\phi )} <0,\end{aligned}$$

(13)

$$\begin{aligned} p_1^{**}-p_1^*&=\frac{\left( -537b_2^2+40v_d+13b_2\sqrt{9b_2^2-16v_d}\right) \,\phi ^2}{96(\chi +\phi )} <0. \end{aligned}$$

(14)

From (14), the numerator can be organized as $b_2^2\left[ -537+40\frac{v_d}{b_2^2}+13\sqrt{9-16\frac{v_d}{b_2^2}}\right]$. Since $v_d/b_2^2$ must be in the range of $[0,9/16]$, the above function should be negative and thus $p_1^{**}<p_1^*$. Moreover, (13) is obviously negative. For the second solution, similarly, we have

$$\begin{aligned} p_1^{**}&=p_2^{**}=\frac{\phi ^2\left( 39b_2^2+40v_d-13b_2\sqrt{9b_2^2-16v_d}\right) }{96(\chi +\phi )},\\ p_3^{**}&=\frac{\left( 33b_2^2-40v_d-11b_2\sqrt{9b_2^2-16v_d}\right) \phi ^2}{96(\chi +\phi )},\\ p_3^{**}-p_1^{**}&=-\frac{\left( 3b_2^2+40v_d-b_2\sqrt{9b_2^2-16v_d}\right) \,\phi ^2}{48(\chi +\phi )}<0,\\ p_1^{**}-p_1^*&=\frac{\left( -537b_2^2+40v_d-13b_2\sqrt{9b_2^2-16v_d}\right) \,\phi ^2}{96(\chi +\phi )}<0. \end{aligned}$$

The last inequality remains true since $[-537+40v_d/b_2^2-13\sqrt{9-16v_d/b_2^2}]>0$ for all $v_d/b_2^2\in [0,9/16]$. Thus, $p_3^{**}<p_1^{**}<p_1^*$.

Proof of Proposition 4

Solving $\partial \text{ SW }/\partial z_1=0$ and $\partial \text{ SW }/\partial z_2=0$ simultaneously yields the first best solution $(z_1^o,z_2^o)=(-\frac{1}{2}b_2,\frac{1}{2}b_2)$, and $(s_1^o,s_2^o)=\left( \frac{\phi }{\chi +\phi }(\frac{-b_1}{2}+d),\frac{\phi }{\chi +\phi }(\frac{b_2}{2}-d)\right)$ is shown from $s_i=\frac{\phi }{\chi +\phi }(z_i-d)$, $i=1, 2$.

Proof of Corollary 2

It is because $z_1^*$ and $z_2^*$ are $\frac{-3b_2}{2}$ and $\frac{3b_2}{2}$, and $z_1^0=\frac{-b_2}{2}$, $z_2^0=\frac{b_2}{2}$.

Proof of Proposition 5

Since $\partial \overline{\pi _1}/\partial z_1=\frac{\overline{p}}{4b_2}>0$, and $\partial \overline{\pi _2}/\partial z_2=-\frac{\overline{p}}{4b_2}<0$, both media outlets tend to locate at the center $(z_1^*=0,z_2^*=0)$ without incentive to deviate.

Proof of Proposition 6

Solving the first-order conditions in the first stage yields the following proposition.

$$\begin{aligned} z_1^*(t)=\frac{-3(\chi +\phi )b_2}{3t+2b_2(\chi +\phi )}, \quad z_2^*(t)=\frac{3(\chi +\phi )b_2}{3t+2b_2(\chi +\phi )}, \end{aligned}$$

(15)

and plugging $z_1^*$, $z_2^*$ in (15) into $p_1^*$ and $p_2^*$ in (11) yields the proposition. Moreover,

$$\begin{aligned} \frac{\partial z_2^*}{\partial t}&=-\frac{\partial z_1^*}{\partial t}=-\frac{9(\chi +\phi )b_2^2}{\left( 3t+2b_2(\chi +\phi )\right) ^2}<0, \\ \frac{\partial p_1^*}{\partial t}&= \frac{\partial p_2^*}{\partial t}=\frac{-36\phi ^2b_2^3}{(3t+2\chi b_2+2\phi b_2)^2}<0. \end{aligned}$$

Proof of Proposition 7

Solving $z_1^*(t)=z_1^o$ and $z_2^*(t)=z_2^o$ simultaneously yields the optimal slant tax rate $t^o=\frac{2}{3}\left( \chi +\phi \right) .$ Plugging $t^o$ and $z_1^o$ and $z_2^o$ into (11) yields $p_1^0=p_2^0=\frac{2\phi ^2b_2^2}{\chi +\phi }$ which are obviously lower than the prices ($p_1=p_2=\frac{6\phi ^2b_2^2}{\chi +\phi }$) in Proposition 1.