A simple model of copyright levies: implications for harmonization

Abstract

Copyright levies are used as a way of compensating rightholders for the private use made of their protected works. This paper builds a simple model of copyright levies and investigates welfare implications of the harmonization of levy rates. The result is that, when the policy-maker places sufficient weight on the interests of collecting societies, harmonization could reduce social welfare. When countries are asymmetric, the country with a larger proportion of foreign consumption and more inefficient tax system loses more from harmonization. A calibration exercise using European data shows that harmonization would increase aggregate social welfare. However, in some countries, policy-makers are worse off although consumers are better off with harmonization. Especially, when larger countries have higher decision weights, the policy-makers are worse off overall, and hence would not agree to harmonize the levy rates.

Appendix

Proof of Proposition 1

It is straightforward to check that the columns of the Jacobian of the constraints are independent, and thus the constraint qualification condition holds at any solution candidate. Form a Lagrangian and derive the full set of first-order conditions as follows:

(18)

(i) κ ^∗=τ ^∗, τ ^∗=0, κ ^∗=0. Then, λ ₁+λ ₂=1+Γ′(0)−u′(0)<0. This is a contradiction.

(ii) κ ^∗=τ ^∗, τ ^∗>0, κ ^∗>0. Since λ ₃=0, λ ₁=δ−u′(0)<0. This is a contradiction.

(iii) κ ^∗<τ ^∗, τ ^∗>0, κ ^∗=0. Since λ ₁=λ ₂=0, u′(τ ^∗)=1+Γ′(τ ^∗) and λ ₃=u′(τ ^∗)−δ. Because λ ₃≥0, the condition holds if u′(τ ^∗)≥δ. Let $\bar{\delta}=u^{\prime }(\tau^{\ast })$. Then, $\bar{\delta}=1+\varGamma^{\prime }(\tau^{\ast })>1$ because τ ^∗>0. The next case shows that if $\bar{\delta}<\delta$, κ ^∗≠0 at the optimum.

(iv) κ ^∗<τ ^∗, τ ^∗>0, κ ^∗>0. Since λ ₁=λ ₂=λ ₃=0, u′(τ ^∗−κ ^∗)=1+Γ′(τ ^∗) and u′(τ ^∗−κ ^∗)=δ. Combining the two, Γ′(τ ^∗)=δ−1. For κ ^∗>0, it must be $\delta >u^{\prime }(\tau^{\ast })=1+\varGamma^{\prime }(\tau^{\ast })=\bar{\delta}>1$ because τ ^∗>0. There are no other solution candidates.

It is straightforward to check that Berge’s theorem of the maximum applies, and thus the solution (τ ^∗,κ ^∗) is upper hemicontinuous. Since u′′(l)<0 and Γ′′(τ)>0, the solution is single-valued, hence, continuous. □

Proof of Proposition 2

(i) Suppose $\delta \leq \bar{\delta}$. Then, τ ^∗ is determined by u′(τ ^∗)=1+Γ′(τ ^∗) and hence independent of δ. For δ>1, τ ^∗ is determined by Γ′(τ ^∗)=δ−1. By totally differentiating both sides, $\varGamma^{\prime \prime }(\tau^{\ast })\frac{\partial \tau^{\ast }}{\partial \delta }=1$. Since Γ′′>0, $\frac{\partial \tau^{\ast }}{\partial \delta }>0$. Also, u′(τ ^∗−κ ^∗)=δ holds true, where u′′(τ−κ)<0. By totaling differentiating both sides, $u^{\prime \prime }(\tau^{\ast }-\kappa^{\ast }) [ \frac{\partial \tau^{\ast }}{\partial \delta }-\frac{\partial \kappa^{\ast }}{\partial \delta } ] =1$. Thus, $\frac{\partial \kappa^{\ast }}{\partial \delta }=\frac{\partial \tau^{\ast }}{\partial \delta }-\frac{1}{u^{\prime \prime }}>0$.

(ii) Define $\bar{\delta}_{A}=u^{\prime }(\bar{\tau}_{A})=1+\varGamma_{A}^{\prime }(\bar{\tau}_{A})$ and $\bar{\delta}_{B}=u^{\prime }(\bar{\tau}_{B})=1+\varGamma_{B}^{\prime }(\bar{\tau}_{B})$. It can be readily shown that $\bar{\tau}_{A}<\bar{\tau}_{B}$ and $\bar{\delta}_{A}>\bar{\delta}_{B}$. Thus, if $\delta \leq \bar{\delta}_{B}$, $\tau_{A}^{\ast }=\bar{\tau}_{A}<\bar{\tau}_{B}=\tau_{B}^{\ast }$. If $\bar{\delta}_{B}<\delta \leq \bar{\delta}_{A}$, by 2.(i), $\tau_{A}^{\ast }=\bar{\tau}_{A}<\bar{\tau}_{B}<\tau_{B}^{\ast }$, and also $\kappa_{A}^{\ast }=0<\kappa_{B}^{\ast }$. If $\bar{\delta}_{A}<\delta$, since $1+\varGamma_{A}^{\prime }(\tau_{A}^{\ast })=\delta =1+\varGamma_{B}^{\prime }(\tau_{B}^{\ast })$, $\tau_{A}^{\ast }<\tau_{B}^{\ast }$; from $u^{\prime }(\tau_{A}^{\ast }-\kappa_{A}^{\ast })=\delta =u^{\prime }(\tau_{B}^{\ast }-\kappa_{B}^{\ast })$, $\kappa_{A}^{\ast }<\kappa_{B}^{\ast }$.

(iii) Define $\bar{\delta}_{A}=u_{A}^{\prime }(\bar{\tau}_{A})=1+\varGamma^{\prime }(\bar{\tau}_{A})$ and $\bar{\delta}_{B}=u_{B}^{\prime }(\bar{\tau}_{B})=1+\varGamma^{\prime }(\bar{\tau}_{B})$. It can be readily shown that $\bar{\tau}_{A}>\bar{\tau}_{B}$ and $\bar{\delta}_{A}>\bar{\delta}_{B}$. Thus, if $\delta \leq \bar{\delta}_{B}$, $\tau_{A}^{\ast }=\bar{\tau}_{A}>\bar{\tau}_{B}=\tau_{B}^{\ast }$. If $\bar{\delta}_{B}<\delta <\bar{\delta}_{A}$, by 2.(i), $\tau_{A}^{\ast }=\bar{\tau}_{A}$ and since $1+\varGamma^{\prime }(\tau_{B}^{\ast })=\delta <\bar{\delta}_{A}=1+\varGamma^{\prime }(\bar{\tau}_{A})$, $\bar{\tau}_{A}>\tau_{B}^{\ast }$. Also, $\kappa_{A}^{\ast }=0<\kappa_{B}^{\ast }$. If $\bar{\delta}_{A}\leq \delta$, $1+\varGamma^{\prime }(\tau_{A}^{\ast })=\delta =1+\varGamma^{\prime }(\tau_{B}^{\ast })$, thus $\tau_{A}^{\ast }=\tau_{B}^{\ast }$; from $u_{A}^{\prime }(\tau_{A}^{\ast }-\kappa_{A}^{\ast })=\delta =u_{B}^{\prime }(\tau_{B}^{\ast }-\kappa_{B}^{\ast })$, $\kappa_{A}^{\ast }<\kappa_{B}^{\ast }$. □

Proof of Proposition 3

It is straightforward to show that $\tau_{i}^{\ast }$ is characterized by $(1-\gamma_{i})\delta =1+\varGamma^{\prime }(\tau_{i}^{\ast })$ in the noncooperative equilibrium and by δ=1+Γ′(τ ^h) in the cooperative equilibrium.

(i) Since γ ₁=γ ₂>0, it follows that $\tau_{1}^{\ast }=\tau_{2}^{\ast }=\tau^{\ast }<\tau^{h}$. Substituting into the first-order conditions, $u^{\prime }(\tau^{\ast }-\kappa_{1}^{\ast })=\delta =u^{\prime }(\tau^{\ast }-\kappa_{2}^{\ast })$. Since u′is a one-to-one function, $\kappa_{1}^{\ast }=\kappa_{2}^{\ast }=\kappa^{\ast }$. Then, $SW_{1}^{\ast }=SW_{2}^{\ast }=u(\tau^{\ast }-\kappa^{\ast })-\tau^{\ast }-\varGamma (\tau^{\ast })$. Similarly, we have $u^{\prime }(\tau^{h}-\kappa_{1}^{h})=\delta =u^{\prime }(\tau^{h}-\kappa_{1}^{h})$, so that $\kappa_{1}^{h}=\kappa_{2}^{h}=\kappa^{h}$. Therefore, τ ^∗−κ ^∗=τ ^h−κ ^h, and $SW_{1}^{h}=SW_{2}^{h}=u(\tau^{h}-\kappa^{h})-\tau^{h}-\varGamma (\tau^{h})$. ΔSW ₁=ΔSW ₂=τ ^∗+Γ(τ ^∗)−τ ^h−Γ(τ ^h)<0.

(ii) In the noncooperative equilibrium, since γ ₁>γ ₂>0, it follows that $\tau_{1}^{\ast }<\tau_{2}^{\ast }<\tau^{h}$. From the first-order conditions, $u^{\prime }((1-\gamma_{1})\tau_{1}^{\ast }-\kappa_{1}^{\ast }+\gamma_{2}\tau_{2}^{\ast })=\delta =u^{\prime }((1-\gamma_{2})\tau_{2}^{\ast }-\kappa_{2}^{\ast }+\gamma_{1}\tau_{1}^{\ast })$. Since u′is a one-to-one function, $(1-\gamma_{1})\tau_{1}^{\ast }-\kappa_{1}^{\ast }+\gamma_{2}\tau_{2}^{\ast }=(1-\gamma_{2})\tau_{2}^{\ast }-\kappa_{2}^{\ast }+\gamma_{1}\tau_{1}^{\ast }$. Then, $SW_{1}^{\ast }-SW_{2}^{\ast }=\tau_{2}^{\ast }+\varGamma (\tau_{2}^{\ast })-\tau_{1}^{\ast }-\varGamma (\tau_{1}^{\ast })>0$. In a similar fashion, it can be shown that $SW_{1}^{h}-SW_{2}^{h}=0$. Finally, since $u^{\prime }((1-\gamma_{2})\tau^{h}-\kappa_{2}^{h}+\gamma_{1}\tau^{h})=\delta$, $SW_{2}^{h}-SW_{2}^{\ast }=\tau_{2}^{\ast }+\varGamma (\tau_{2}^{\ast })-\tau^{h}-\varGamma (\tau^{h})<0$, it follows that ΔSW ₁<ΔSW ₂<0. □

Proof of Proposition 4

(i) It is straightforward to show that $\tau_{i}^{\ast }$ is characterized by $(1-\gamma )\delta =1+\varGamma_{i}^{\prime }(\tau_{i}^{\ast })$ in the noncooperative equilibrium. Since, $\varGamma_{1}^{\prime }(\tau )>\varGamma_{2}^{\prime }(\tau )$, $\tau_{1}^{\ast }<\tau_{2}^{\ast }$. It also satisfies $u^{\prime }((1-\gamma )\tau_{1}^{\ast }-\kappa_{1}^{\ast }+\gamma \tau_{2}^{\ast })=\delta =u^{\prime }((1-\gamma )\tau_{2}^{\ast }-\kappa_{2}^{\ast }+\gamma \tau_{1}^{\ast })$, where u′ is an injection. Thus, $SW_{1}^{\ast }-SW_{2}^{\ast }=\tau_{2}^{\ast }+\varGamma (\tau_{2}^{\ast })-\tau_{1}^{\ast }-\varGamma (\tau_{1}^{\ast })>0$. On the other hand, in the cooperative case, τ ^h is determined by $\delta =1+\varGamma_{1}^{\prime }(\tau^{h})/2+\varGamma_{2}^{\prime }(\tau^{h})/2$. Since $\varGamma_{1}^{\prime }(\tau )>\varGamma_{2}^{\prime }(\tau )$, it can be shown that $\tau^{h}>\tau_{1}^{\ast }$. It still holds that $u^{\prime }((1-\gamma )\tau^{h}-\kappa_{1}^{h}+\gamma \tau^{h})=\delta$. Then, $SW_{1}^{h}-SW_{1}^{\ast }=\tau_{1}^{\ast }+\varGamma (\tau_{1}^{\ast })-\tau^{h}-\varGamma (\tau^{h})<0$ because $\tau_{1}^{\ast }<\tau^{h}$. $\Delta SW_{2}-\Delta SW_{1}=\varGamma_{1}(\tau^{h})-\varGamma_{2}(\tau^{h})+SW_{1}^{h}-SW_{2}^{\ast }>0$, where the last inequality holds because Γ ₁(τ ^h)>Γ ₂(τ ^h).

(ii) It is straightforward to show that $\tau_{i}^{\ast }$ is characterized by $(1-\gamma )\delta =1+\varGamma^{\prime }(\tau_{i}^{\ast })$ in the noncooperative equilibrium, and τ ^h is characterized by δ=1+Γ′(τ ^h) in the cooperative case. Hence, $\tau_{1}^{\ast }=\tau_{2}^{\ast }=\tau^{\ast }<\tau^{h}$. Using the first-order conditions, $u_{1}^{\prime }(\tau^{\ast }-\kappa_{1}^{\ast })=\delta =u_{1}^{\prime }(\tau^{h}-\kappa_{1}^{h})$ and $u_{2}^{\prime }(\tau^{\ast }-\kappa_{2}^{\ast })=\delta =u_{2}^{\prime }(\tau^{h}-\kappa_{2}^{h})$, where u′ is an injection. Then, $SW_{1}^{h}-SW_{1}^{\ast }=\tau_{1}^{\ast }+\varGamma (\tau_{1}^{\ast })-\tau^{h}-\varGamma (\tau^{h})<0$. Also, $\Delta SW_{1}-\Delta SW_{2}=u_{1}(\tau^{h}-\kappa_{1}^{h})-u_{1}(\tau^{\ast }-\kappa_{1}^{\ast })-u_{2}(\tau^{h}-\kappa_{2}^{h})+u_{2}(\tau^{\ast }-\kappa_{2}^{\ast })=0$. □