The regulation of merchant fees in credit card markets

Abstract

This paper provides a theory on how to regulate the level of merchant fees in credit card markets. In particular, we discuss how to regulate the merchant fee in a closed payment system with heterogeneous merchants. We find that tourist test is not a valid approach. Our model suggests that the regulation should be based on costs of networks or banks and market elasticity. Moreover, we provide an alternative understanding on how to regulate the interchange fee in an open payment system. The initial public offerings of Visa and MasterCard do not merely change themselves from not-for-profit associations to for profit companies, but also change the mechanism in determining the fee level and structure borne by end users. These changes in industry urge an updated understanding of the optimality of interchange fees.

Appendix

See Table 2.

Table 2 The comparison with literature

1.1 Calculation for Table 1

In what follows, we calibrate the model assuming $b_{i}$ is drawn from uniform distribution between $\left[ 0,1\right]$. We also assume $1\le c\le 2$. $c\ge 1$ ensures that there is no negative solutions in the results. $c\le 2$ ensures that total cost is also no greater than the maximum sum of transaction benefits. The user surplus of consumers is $v_{B}\left( p_{B}\right) =E\left( b_{B}\mid b_{B}\ge p_{B}\right) -p_{B}$. With uniform distribution, it is equal to $\frac{1-p_{B}}{2}$. All the second order conditions for the following solutions are guaranteed by increasing hazard rates—a property that uniform distributions have. Corner solutions are also omitted. The profit, transaction volume and welfare with uniform distributions are given by

$$\begin{aligned} \pi= & {} \left( p_{B}+p_{S}-c\right) \left( 1-\overset{\wedge }{b_{B}}\right) \left( 1-\overset{\wedge }{b_{S}}\right) \\ T= & {} \left( 1-\overset{\wedge }{b_{B}}\right) \left( 1-\overset{\wedge }{b_{S}}\right) \\ W= & {} \frac{1}{2}\left( 1-\overset{\wedge }{b_{B}}\right) \left( 1-\overset{\wedge }{b_{S}}\right) \left( 2-2c+\overset{\wedge }{b_{B}}+\overset{\wedge }{b_{S}}\right) \end{aligned}$$

Therefore, the optimizing problem are given by

Monopoly (a)

$$\begin{aligned} \underset{p_{B}\,p_{S}}{max}\;\pi= & {} \left( p_{B}+p_{S}-c\right) \left( 1-\overset{\wedge }{b_{B}}\right) \left( 1-\overset{\wedge }{b_{S}}\right) \\ s.t.\;\;\overset{\wedge }{b_{B}}= & {} p_{B};\;\; \overset{\wedge }{b_{S}}=p_{S}-\frac{1-p_{B}}{2} \end{aligned}$$

Monopoly (b)

$$\begin{aligned} \underset{p_{B}\,p_{S}}{max}\;\pi= & {} \left( p_{B}+p_{S}-c\right) \left( 1-\overset{\wedge }{b_{B}}\right) \left( 1-\overset{\wedge }{b_{S}}\right) \\ s.t.\;\;\overset{\wedge }{b_{B}}= & {} p_{B};\;\;\overset{\wedge }{b_{S}}=p_{S} \end{aligned}$$

Association (a)

$$\begin{aligned} \underset{p_{B}\,p_{S}}{max}\;T= & {} \left( 1-\overset{\wedge }{b_{B}}\right) \left( 1-\overset{\wedge }{b_{S}}\right) \\ s.t.\;\;\overset{\wedge }{b_{B}}= & {} p_{B};\;\;\overset{\wedge }{b_{S}}=p_{S}- \frac{1-p_{B}}{2};\;\;p_{B}+p_{S}=c \end{aligned}$$

Association (b)

$$\begin{aligned} \underset{p_{B}\,p_{S}}{max}\;T= & {} \left( 1-\overset{\wedge }{b_{B}}\right) \left( 1-\overset{\wedge }{b_{S}}\right) \\ s.t.\;\;\overset{\wedge }{b_{B}}= & {} p_{B};\;\;\overset{\wedge }{b_{S}} =p_{S};\;\;p_{B}+p_{S}=c \end{aligned}$$

Regulator (a)

$$\begin{aligned} \underset{p_{B}\,p_{S}}{max}\;W= & {} \frac{1}{2}\left( 1-\overset{\wedge }{b_{B}}\right) \left( 1-\overset{\wedge }{b_{S}}\right) \left( 2-2c+\overset{\wedge }{b_{B}}+\overset{\wedge }{b_{S}}\right) \\ s.t.\;\;\overset{\wedge }{b_{B}}= & {} p_{B};\;\;\overset{\wedge }{b_{S}} =p_{S}-\frac{1-p_{B}}{2};\;\;p_{B}+p_{S}=c \end{aligned}$$

Regulator (b)

$$\begin{aligned} \underset{p_{B}\,p_{S}}{max}\;W= & {} \frac{1}{2}\left( 1-\overset{\wedge }{b_{B}}\right) \left( 1-\overset{\wedge }{b_{S}}\right) \left( 2-2c+\overset{\wedge }{b_{B}}+\overset{\wedge }{b_{S}}\right) \\ s.t.\;\;\overset{\wedge }{b_{B}}= & {} p_{B};\;\;\overset{\wedge }{b_{S}} =p_{S};\;\;p_{B}+p_{S}=c \end{aligned}$$

Social planner (a)

$$\begin{aligned} \underset{\overset{\wedge }{b_{B}},\,\overset{\wedge }{b_{S}}}{max}\;W= & {} \frac{1}{2} \left( 1-\overset{\wedge }{b_{B}}\right) \left( 1-\overset{\wedge }{b_{S}}\right) \left( 2-2c+\overset{\wedge }{b_{B}}+\overset{\wedge }{b_{S}}\right) \\ where\,\,p_{B}= & {} \overset{\wedge }{b_{B}};\;\;p_{S}= \overset{\wedge }{b_{S}}+\frac{1-p_{B}}{2} \end{aligned}$$

Social planner (b)

$$\begin{aligned} \underset{\overset{\wedge }{b_{B}},\,\overset{\wedge }{b_{S}}}{max}\;W= \frac{1}{2}\left( 1-\overset{\wedge }{b_{B}}\right) \left( 1-\overset{\wedge }{b_{S}}\right) \left( 2-2c+\overset{\wedge }{b_{B}}+\overset{\wedge }{b_{S}}\right) \end{aligned}$$

Solving all the problems yields results in Table 1.

1.2 Model for policy implication

1.2.1 Uniform distribution

The monopoly (b) solves

$$\begin{aligned} \underset{p_{B}\,p_{S}}{max}\;\pi= & {} \left( p_{B}+p_{S}-c\right) \left( 1-\overset{\wedge }{b_{B}}\right) \left( 1-\overset{\wedge }{b_{S}}\right) \\ s.t.\;\;\overset{\wedge }{b_{B}}= & {} p_{B};\;\;\overset{\wedge }{b_{S}} =p_{S}-\frac{1-p_{B}}{2};\;p_{B}+p_{S}=k \end{aligned}$$

Or

$$\begin{aligned} \underset{p_{B}}{max}\;\pi= & {} \frac{1}{2}\left( k-c\right) \left( 1-p_{B}\right) \left( 3-2k+p_{B}\right) \end{aligned}$$

Solving the problem yields $p_{B}=k-1, p_{S}=1, \overset{\wedge }{b_{B}}=k-1, \overset{\wedge }{b_{S}}=\frac{k}{2}, \pi =\frac{1}{2}\left( k-c\right) \left( 2-k\right) ^{2}, T=\left( 2-k\right) ^{2}$ and the welfare is

$$\begin{aligned} W=\frac{1}{4}\left( 2-k\right) ^{2}\left( 1-2c+\frac{3}{2}k\right) \end{aligned}$$

Solving$\frac{dW}{dk}=0$ yields $k=\frac{2+8c}{9}$ and $W\mid _{k=\frac{2+8c}{9}}=\frac{32}{243}\left( 2-c\right) ^{3}.$

1.2.2 Generalized Pareto distribution

Generally, the GPD is given by

$$\begin{aligned} c.d.f.\;\;F\left( b\right)= & {} 1-\left( 1-\frac{e\left( b-\mu \right) }{\delta }\right) ^{\frac{1}{e}}\\ p.d.f.\;\;f\left( b\right)= & {} \frac{1}{e}\left( -\frac{e}{\delta }\right) \left( 1-\frac{e\left( b-\mu \right) }{\delta }\right) ^{\frac{1}{e}-1} \end{aligned}$$

where $e>0, \mu<b<\frac{\delta }{e}$. The quasi-demand is given by

$$\begin{aligned} D\left( b\right) =1-F\left( b\right) =\left( 1-\frac{e\left( b-\mu \right) }{\delta }\right) ^{\frac{1}{e}} \end{aligned}$$

To calculate the average net consumer surplus, we have

$$\begin{aligned} v\left( \overset{\wedge }{b}\right) =E\left( b\mid b \ge \overset{\wedge }{b}\right) -\overset{\wedge }{b}=\frac{\int _{\overset{\wedge }{b}}^{\bar{b}}bdF\left( b\right) }{\int _{\overset{\wedge }{b}}^{\bar{b}}dF\left( b\right) }-\overset{\wedge }{b}=\frac{\int _{\overset{\wedge }{b}}^{\bar{b}}D\left( b\right) db}{D\left( \overset{\wedge }{b}\right) } \end{aligned}$$

Plugging into the GPD functions, we have

$$\begin{aligned} v\left( \overset{\wedge }{b}\right) =\frac{\int _{\overset{\wedge }{b}}^{\bar{b}}D \left( b\right) db}{D\left( \overset{\wedge }{b}\right) }= \frac{\int _{\overset{\wedge }{b}}^{\bar{b}} \left( 1-\frac{e\left( b-\mu \right) }{\delta }\right) ^{\frac{1}{e}}db}{D\left( \overset{\wedge }{b}\right) }= \frac{\delta -e\left( \overset{\wedge }{b}-\mu \right) }{1+e} \end{aligned}$$

For simplicity, let $\mu =0, \frac{\delta }{e}=1$. We have

$$\begin{aligned} c.d.f.\;\;F\left( b\right)= & {} 1-\left( 1-b\right) ^{\frac{1}{e}}\\ p.d.f.\;\;f\left( b\right)= & {} \frac{1}{e}\left( 1-b\right) ^{\frac{1}{e}-1}\\ D\left( \overset{\wedge }{b}\right)= & {} 1-\left( 1-\overset{\wedge }{b}\right) ^{\frac{1}{e}}\\ v\left( \overset{\wedge }{b}\right)= & {} \frac{e\left( 1-\overset{\wedge }{b}\right) }{1+e} \end{aligned}$$

Therefore, the regulation problem becomes

$$\begin{aligned} \underset{p_{B}\,p_{S}}{max}\;\pi= & {} \left( p_{B}+p_{S}-c\right) \left( 1-\overset{\wedge }{b_{B}}\right) ^{\frac{1}{e}} \left( 1-\overset{\wedge }{b_{S}}\right) ^{\frac{1}{e}}\\ s.t.\;\;\overset{\wedge }{b_{B}}= & {} p_{B};\;\;\overset{\wedge }{b_{S}}=p_{S}- \frac{e\left( 1-p_{B}\right) }{1+e};\;p_{B}+p_{S}=k \end{aligned}$$

Or

$$\begin{aligned} \underset{p_{B}}{max}\;\pi =\frac{1}{2}\left( k-c\right) \left( 1-\left( k-p_{S}\right) \right) ^{\frac{1}{e}} \left( 1-p_{S}+\frac{e}{1+e}\left( 1-\left( k-p_{S}\right) \right) \right) ^{\frac{1}{e}} \end{aligned}$$

Solving the problem yields $p_{B}=\frac{k-2e+ek}{2}, p_{S}=\frac{2e+k-ek}{2}, \overset{\wedge }{b_{B}}=\frac{k-2e+ek}{2}, \overset{\wedge }{b_{S}}=\frac{k}{2}$ and the welfare is

$$\begin{aligned} W= & {} \int _{\overset{\wedge }{b_{B}}}^{1}\int _{\overset{\wedge }{b_{S}}}^{1} \frac{1}{e^{2}}\left( 1-b_{B}\right) ^{\frac{1}{e}-1} \left( 1-b_{S}\right) ^{\frac{1}{e}-1}\left( b_{B}+b_{S}-c\right) db_{S}db_{B}\\ W= & {} \frac{1}{1+e}\left( 1-\overset{\wedge }{b_{B}}\right) ^{\frac{1}{e}} \left( 1-\overset{\wedge }{b_{S}}\right) ^{\frac{1}{e}} \left( 2e-c\left( 1+e\right) +\overset{\wedge }{b_{B}}+\overset{\wedge }{b_{S}}\right) \end{aligned}$$

Plugging into $\overset{\wedge }{b_{B}}=\frac{k-2e+ek}{2}, \overset{\wedge }{b_{S}}=\frac{k}{2}$, we have

$$\begin{aligned} W=\frac{2^{\frac{-\left( 2+e\right) }{e}}}{1+e} \left( 1+e\right) ^{\frac{1}{e}}\left( -2+k\right) ^{\frac{1}{e}} \left( 2k-2c\left( 1+e\right) +e\left( 2+k\right) \right) \end{aligned}$$

Solving$\frac{dW}{dk}=0$ yields $k=\frac{2\left( 2c+2ce+e^{2}\right) }{\left( 2+e\right) ^{2}}$. By plugging k into $p_{B}=\frac{k-2e+ek}{2}, p_{S}=\frac{2e+k-ek}{2}$, we finally have the customer fee and merchant fee given as follows:

$$\begin{aligned} p_{B}= & {} \frac{2c\left( 1+e\right) ^{2}-e\left( 4+3e\right) }{\left( 2+e\right) ^{2}}\\ p_{S}= & {} \frac{e\left( 4+5e\right) -2c\left( -1+e^{2}\right) }{\left( 2+e\right) ^{2}} \end{aligned}$$

Notably, all calculation rules out the corner solutions and the Mathematica code is given by the supplement material.