Interchange Fees and Innovation in Payment Systems

Abstract

We analyze the impact of interchange fees on consumers’ and merchants’ incentives to adopt an innovative payment instrument, in a setting with adoption externalities between consumers and merchants. We show that consumer adoption decreases with the interchange fee for high degrees of externality, and varies non-monotonically with it for low degrees of externality. The profit-maximizing interchange fee coincides with the social optimum when externalities are strong, whereas it is too high when they are weak. We also compare the issuers’ incentives to innovate when they cooperate and when they make their innovation decisions independently.

Appendices

Appendix

1.1 Appendix 1: Stage 3 (Adoption of the Payment Instrument)

1.1.1 Appendix 1.1: Proof of Lemma 1

We have $\partial \widetilde{n}_{S}/\partial p_{i}=-\left( \beta /t\right) \mu (m)\pi \left( m\right) <0$, as $\pi \left( m\right) >0$ and $\mu (m)>0$ . Similarly, we have $\partial \widetilde{n}_{S}$$/\partial m<0$. We also find that $\partial \widetilde{n}_{S}/\partial \theta _{i}=-\partial \widetilde{n}_{S}/\partial p_{i}>0$. $\square$

1.1.2 Appendix 1.2: Proof of Lemma 2

From (2), we have $\partial \widetilde{D} _{B}/\partial p_{i}=-(\beta /t)\mu (m)<0$. We also have $\partial \widetilde{ D}_{B}/\partial m=(2\alpha _{B}\beta /t)\partial \widetilde{n}_{S}/\partial m<0$, as $\partial \widetilde{n}_{S}$$/\partial m<0$. Finally, we find that $\partial \widetilde{D}_{B}/\partial \theta _{i}=(\beta /t)\mu (m)>0$. $\square$

1.1.3 Appendix 1.3: Proof of Lemma 3

From (6), we have

$$\begin{aligned} \frac{\partial \widetilde{D}_{B}^{i}}{\partial p_{i}}=-\frac{1}{2t}-\frac{ \beta }{t}+\frac{\beta \alpha _{B}}{t}\frac{\partial \widetilde{n}_{S}}{ \partial p_{i}}<0{,} \end{aligned}$$

since $\partial \widetilde{n}_{S}/\partial p_{i}<0$. Furthermore, we have $\partial \widetilde{D}_{B}^{i}/\partial \theta _{i}=-\partial \widetilde{D} _{B}^{i}/\partial p_{i}>0$. For $j\ne i$, we have

$$\begin{aligned} \frac{\partial \widetilde{D}_{B}^{i}}{\partial p_{j}}=\frac{1}{2t}+\frac{ \beta \alpha _{B}}{t}\frac{\partial \widetilde{n}_{S}}{\partial p_{j}}=\frac{ \mu (m)}{2t^{2}}\left[ t-2\beta (1+\beta )\alpha _{B}\pi \left( m\right) \right] {.} \end{aligned}$$

Since $t-2\beta \left( 1+\beta \right) (b_{S}-m)\alpha _{B}\ge t-2\beta \left( 1+\beta \right) b_{S}\alpha _{B}$, and since from Assumption 2 (iii), $t-2\beta \left( 1+\beta \right) b_{S}\alpha _{B}>0$, we have $t-2\beta \left( 1+\beta \right) (b_{S}-m)\alpha _{B}>0$. Since $\mu (m)>0$, we can conclude that $\partial \widetilde{D}_{B}^{i}/\partial p_{j}>0$. Finally, we have $\partial \widetilde{D}_{B}^{i}/\partial \theta _{j}=-\partial \widetilde{D}_{B}^{i}/\partial p_{j}<0$. $\square$

1.2 Appendix 2: Stage 2 (Prices)

1.2.1 Appendix 2.1: Second-Order Condition

The second-order condition holds if and only if

$$\begin{aligned} \left( 1+a\frac{\partial \widetilde{n}_{S}}{\partial p_{i}}\right) \frac{ \partial \widetilde{D}_{B}^{i}}{\partial p_{i}}\le 0{.} \end{aligned}$$

This inequality always holds as $\partial \widetilde{D}_{B}^{i}/\partial p_{i}\le 0$ (Lemma 3) and because we have necessarily $1+a\partial \widetilde{n}_{S}/\partial p_{i}\ge 0$ from Assumption 2 (ii). Indeed, if $t>2\beta b_{S}(b_{S}+\alpha _{B})$, for all $m\le b_{S}$ and $a\le b_{S}$, we have $t>\beta b_{S}(a+2\alpha _{B})$, which implies that $1+a\partial \widetilde{n} _{S}/\partial p_{i}\ge 0$. $\square$

1.2.2 Appendix 2.2: Prices are Strategic Complements

Let $p_{i}^{BR}$ denote the best-response of issuer i. From the implicit function theorem, since $\partial ^{2}\Pi _{i}/\partial p_{i}^{2}<0$, $\partial p_{i}^{BR}/\partial p_{j}$ has the sign of $\partial ^{2}\Pi _{i}/\partial p_{i}\partial p_{j}$. From the FOC (7), we have

$$\begin{aligned} \frac{\partial ^{2}\Pi _{i}}{\partial p_{i}\partial p_{j}}=\left( 1+a\frac{ \partial \widetilde{n}_{S}}{\partial p_{i}}\right) \frac{\partial \widetilde{ D}_{B}^{i}}{\partial p_{j}}+a\frac{\partial \widetilde{n}_{S}}{\partial p_{j} }\frac{\partial \widetilde{D}_{B}^{i}}{\partial p_{i}}{.} \end{aligned}$$

(12)

From Lemma 1, we have $\partial \widetilde{n} _{S}/\partial p_{j}=\partial \widetilde{n}_{S}/\partial p_{i}\le 0$. From Lemma 3, we have $\partial \widetilde{D}_{B}^{i}/\partial p_{i}<0$ and $\partial \widetilde{D} _{B}^{i}/\partial p_{j}>0$. Finally, from the FOC, we have necessarily $(1+a\partial \widetilde{n}_{S}/\partial p_{i})\ge 0$. Hence, $\partial ^{2}\Pi _{i}/\partial p_{i}\partial p_{j}>0$, which proves that prices are strategic complements. $\square$

1.3 Appendix 3: Stage 1 (Interchange Fee)

1.3.1 Appendix 3.1: Proof of Proposition 1

First, we compute the derivative of the issuer’s price in the symmetric equilibrium at $a=0$. We have

$$\begin{aligned} \left. \frac{\partial p^{*}}{\partial a}\right| _{a=0}=-\frac{ t+2\beta (\theta +v_{B}-k_{I})}{t(t(4\beta +1)-2b_{S}\alpha _{B}\beta )^{2}} \mu (0)E(t), \end{aligned}$$

where $E\left( t\right) =\left( 2\alpha _{B}\beta \left( 1+\beta \right) +b_{S}\left( 2\beta +1\right) (3\beta +1)\right) t^{2}-2\alpha _{B}\beta \left( 1+\beta \right) (5b_{S}\beta +2\alpha _{B}\beta +2b_{S})b_{S}t+4b_{S}^{3}\alpha _{B}^{2}\beta ^{2}\left( 1+\beta \right) ^{2}$. Therefore, $\left. \partial p^{*}/\partial a\right| _{a=0}<0$ if and only if $E\left( t\right) >0$. As the coefficient of the $t^{2}$ factor in E is positive, this second-order polynomial is U-shaped, and it has two positive roots. The highest root is

$$\begin{aligned} t^{+}=\frac{b_{S}\alpha _{B}\beta \left( 1+\beta \right) \left( 5b_{S}\beta +2\alpha _{B}\beta +2b_{S}\right) +\sqrt{b_{S}^{2}\alpha _{B}^{2}\beta ^{4}\left( 1+\beta \right) ^{2}\left( b_{S}^{2}+12\alpha _{B}b_{S}+4\alpha _{B}^{2}\right) }}{2\alpha _{B}\beta \left( 1+\beta \right) +b_{S}\left( 2\beta +1\right) (3\beta +1)}{.} \end{aligned}$$

Since $\sqrt{b_{S}^{2}\alpha _{B}^{2}\beta ^{4}\left( 1+\beta \right) ^{2}\left( b_{S}^{2}+12\alpha _{B}b_{S}+4\alpha _{B}^{2}\right) }\le \sqrt{ b_{S}^{2}\alpha _{B}^{2}\beta ^{4}\left( 1+\beta \right) ^{2}\left( 9b_{S}^{2}+12\alpha _{B}b_{S}+4\alpha _{B}^{2}\right) }$ and $\sqrt{ b_{S}^{2}\alpha _{B}^{2}\beta ^{4}\left( 1+\beta \right) ^{2}\left( 9b_{S}^{2}+12\alpha _{B}b_{S}+4\alpha _{B}^{2}\right) }=b_{S}\alpha _{B}\beta ^{2}\left( 1+\beta \right) \left( 3b_{S}+2\alpha _{B}\right)$, we have

$$\begin{aligned} t^{+}<\frac{2b_{S}\alpha _{B}\beta \left( 1+\beta \right) \left( b_{S}+4b_{S}\beta +2\alpha _{B}\beta \right) }{2\alpha _{B}\beta \left( 1+\beta \right) +b_{S}\left( 2\beta +1\right) (3\beta +1)}\equiv \overline{t}^\pm \end{aligned}$$

We find that $\overline{t}^{+}<4\alpha _{B}\beta b_{S}$, and therefore under Assumption 2 (i), we always have $t^{+}<\overline{t}^{+}<t$. This implies that under Assumption 2 (i), we have $E(t)>0$. Since $v_{B}>k_{I}$ from Assumption 1, it follows that $\left. \partial p^{*}/\partial a\right| _{a=0}<0$.

Second, since $v_{B}>k_{I}$ from Assumption 1,

$$\begin{aligned} \left. \frac{\partial p^{*}}{\partial a}\right| _{a=b_{S}}=\frac{ \left[ t+2\beta \left( v_{B}+\theta -k_{I}\right) \right] \left[ b_{S}\left( 2\beta +1\right) (3\beta +1)-2\alpha _{B}\beta \left( 1+\beta \right) \right] }{t\left( 1+4\beta \right) ^{2}}>0 \end{aligned}$$

if and only if $b_{S}\left( 2\beta +1\right) (3\beta +1)>2\alpha _{B}\beta \left( 1+\beta \right)$. $\square$

1.3.2 Appendix 3.2: Proof of Proposition 2

For all parameter values, since $v_{B}>k_{I}$ from Assumption 1, consumer adoption decreases with a at $a=b_{S}$, as

$$\begin{aligned} \left. \frac{dD_{B}^{*}}{da}\right| _{a=b_{S}}=-\frac{2\beta \left[ t+\beta \left( 2v_{B}+\theta _{1}+\theta _{2}-2k_{I}\right) \right] \left[ 6\beta ^{2}\left( b_{S}+\alpha _{B}\right) +\left( 5b_{S}+4\alpha _{B}\right) \beta +\left( b_{S}+\alpha _{B}\right) \right] }{t^{2}(4\beta +1)^{2}}<0{.} \end{aligned}$$

We now compute $dD_{B}^{*}/da$ at $a=0$ for low and high values of $\alpha _{B}$. First, we have

$$\begin{aligned} \left. \frac{dD_{B}^{*}}{da}\right| _{a=0{, }\alpha _{B}=0}= \frac{2b_{S}\beta \left( 2\beta +1\right) \left( 3\beta +1\right) \left[ t+\beta \left( 2v_{B}+\theta _{1}+\theta _{2}-2k_{I}\right) \right] }{ t^{2}(4\beta +1)^{2}}>0{,} \end{aligned}$$

and hence, $D_{B}^{*}$ is increasing with a at $a=0$ for low values of $\alpha _{B}$. Second, we compute $dD_{B}^{*}/da$ at $a=0$ at the maximum value of $\alpha _{B}$ that is given by Assumption 2 (i). We have

$$\begin{aligned} \left. \frac{dD_{B}^{*}}{da}\right| _{a=0{, }\alpha _{B}=t/(4\beta b_{S})}=\frac{2\left( t+\left( 2v_{B}+\theta _{1}+\theta _{2}-2k_{I}\right) \right) \left[ 4(b_{S})^{2}\beta \left( 3\beta +1\right) (5\beta +1)-t(17\beta ^{2}+6\beta +1)\right] }{b_{S}t^{2}(7\beta +1)^{2}} {.} \end{aligned}$$

Since $v_{B}>k_{I}$ from Assumption 1, this expression has the sign of $4(b_{S})^{2}\beta \left( 3\beta +1\right) (5\beta +1)-t(17\beta ^{2}+6\beta +1)$, which is positive for high values of $b_{S}$, and negative otherwise. To sum up, if $\alpha _{B}$ is high and $b_{S}$ is low, $D_{B}^{*}$ is decreasing both at $a=0$ and at $a=b_{S}$. Otherwise, $D_{B}^{*}$ is increasing at $a=0$ and decreasing at $a=b_{S}$. $\square$

1.3.3 Appendix 3.3: Proof of Proposition 3

We have

$$\begin{aligned} \frac{dn_{S}^{*}}{da}=\frac{-G}{\left( 4(b_{S}-a)^{2}{\alpha } _{B}(a+{\alpha }_{B})\beta ^{2}(1+\beta )+t^{2}(1+4\beta )-2(b_{S}-a)t\beta (a+3a\beta +{\alpha }_{B}(2+5\beta ))\right) ^{2}} {,} \end{aligned}$$

and hence, $dn_{S}^{*}/da$ has the sign of $-G$. We find that $G=(t+(2\left( v_{B}-k_{I}\right) +\theta _{1}+\theta _{2})\beta )G_{1}$, and therefore, since $v_{B}>k_{I}$ from Assumption 1, G has the sign of $G_{1}$ . $G_{1}$ is a polynomial of degree 2 of $\alpha _{B}$ and it has an inverted bell curve, as the coefficient in $(\alpha _{B})^{2}$ is positive if $a<b_{S}$, since

$$\begin{aligned} G_{1}=4\left( b_{S}-a\right) ^{2}\beta ^{2}\left( \beta +1\right) \left[ t(3\beta +1)-2\left( b_{S}-a\right) ^{2}\beta \left( \beta +1\right) \right] (\alpha _{B})^{2}+\cdots {,} \end{aligned}$$

and $t(3\beta +1)>2\left( b_{S}-a\right) ^{2}\beta \left( \beta +1\right)$ from Assumption 2 (ii). We compute the discriminant of $G_{1}$ and show that it is strictly negative if t is sufficiently high, which proves that $dn_{S}^{*}/da>0$. The discriminant of $G_{1}$ is equal to $\Delta _{G_{1}}=-16\left( b_{S}-a\right) ^{2}t^{2}\beta ^{4}\left( \beta +1\right) \times g$, and hence, it has the opposite sign to that of g. We have $g=2\left( 4\beta +1\right) t^{2}-2\left( b_{S}-a\right) (2b_{S}\beta \left( 3\beta +1\right) -a\left( 2\beta +1\right) \left( 5\beta +1\right) )t-\left( b_{S}-a\right) ^{2}\beta \left( \beta +1\right) (4a(2\beta +1)(b_{S}-a)+b_{S}^{2}\beta )$. Note that $\left. g\right| _{t=0}<0$. However, the first two terms of g can rearranged so that

$$\begin{aligned} g=2t\left[ t\left( 4\beta +1\right) -\left( b_{S}-a\right) \left( 2b_{S}\beta \left( 3\beta +1\right) -a\left( 2\beta +1\right) \left( 5\beta +1\right) \right) \right] +g_{0}{,} \end{aligned}$$

where $g_{0}$ contains terms that are independent of t. Since the first term in the equation above is increasing in t, then g is positive for sufficiently high values of t. We can derive a sufficient condition for $g\ge 0$, using the fact that $\left( b_{S}-a\right) (2b_{S}\beta (3\beta +1)-a(2\beta +1)(5\beta +1))\le 2b_{S}^{2}\beta \left( 3\beta +1\right)$, and that $-g_{0}\le b_{S}^{2}\beta \left( \beta +1\right) \left[ \beta b_{S}^{2}+4b_{S}^{2}\left( 2\beta +1\right) \right] =b_{S}^{4}\beta \left( \beta +1\right) \left( 9\beta +4\right)$. For $g\ge 0$ to hold, it is then sufficient that

$$\begin{aligned} 2t\left[ t\left( 4\beta +1\right) -2\beta b_{S}^{2}\left( 3\beta +1\right) \right] >\beta b_{S}^{4}\left( \beta +1\right) \left( 9\beta +4\right) { .} \end{aligned}$$

A simpler sufficient condition can be obtained by using the fact that $t>2\beta b_{S}^{2}$ from Assumption 2 (ii), and replacing for $t=2\beta b_{S}^{2}$ in the term into brackets in the above expression. Rearranging the expression, we obtain

$$\begin{aligned} t>\frac{\left( \beta +1\right) \left( 9\beta +4\right) }{4\beta }b_{S}^{2} {.} \end{aligned}$$

If this (sufficient) condition holds, then $dn_{S}^{*}/da<0$ for all a . $\square$

1.3.4 Appendix 3.4: Proof of Proposition 4

First, we show that when $\alpha _{B}=0$, the derivative of the joint profits in the symmetric equilibrium is strictly positive at $a=0$. We have

$$\begin{aligned} \left. \frac{d\sum \Pi _{i}}{da}\right| _{\alpha _{B}=0,\, a=0}=\frac{\beta b_{S}\left( 2\beta +1\right) \left( 8\beta +3\right) \left( t+2\beta \left( v_{B}-k_{I}+\theta \right) \right) ^{2}}{t^{2}\left( 4\beta +1\right) ^{3}}>0 {.} \end{aligned}$$

Therefore, for low values of the degree of S2B externality (around 0), the profit-maximizing interchange fee is strictly positive. Second, we show that when $\alpha _{B}$ is high, the derivative of the joint profits in the symmetric equilibrium is strictly negative at $a=0$. From Assumption 2, $\overline{\alpha }_{B}=t/(4\beta b_{S})$ is a maximum for $\alpha _{B}$. We have

$$\begin{aligned} \left. \frac{d\sum \Pi _{i}}{da}\right| _{\alpha _{B}=\overline {\alpha } _{B},\,a=0}=\frac{8\left( t+2\beta \left( v_{B}-k_{I}+\theta \right) \right) ^{2}\left[ \beta b_{S}^{2}\left( 3\beta +1\right) \left( 13\beta +3\right) -t\left( 10\beta ^{2}+5\beta +1\right) \right] }{b_{S}t^{2}\left( 7\beta +1\right) ^{3}}{,} \end{aligned}$$

which has the sign of $\beta b_{S}^{2}\left( 3\beta +1\right) \left( 13\beta +3\right) -t\left( 10\beta ^{2}+5\beta +1\right)$. This expression decreases with t, and we find that it is negative for $t=4\beta b_{S}^{2}$ . Therefore, the derivative of $\Pi _{1}+\Pi _{2}$ is strictly negative at $a=0$, and assuming that joint profits are concave, it implies that $a^{\pi }=0$.

1.3.5 Appendix 3.5: Proof of Corollary 1

(i) When $\alpha _{B}=0$, we find that $a^{S}=0$, $a^{\pi }=a^{B}=b_{S}/2$ and that $a^{V}\in \left[ 0,b_{S}/2\right)$. Therefore, $a^{S}\le a^{V}<a^{B}=a^{\pi }$. We begin by showing that $a^{S}=0$. We have

$$\begin{aligned} \frac{dn_{S}^{*}}{da}=-\frac{\left( 2\beta +1\right) \left( t+2\beta \left( v_{B}-k_{I}+\theta \right) \right) \left( \left( 4\beta +1\right) t-2\left( b_{S}-a\right) ^{2}\beta \left( 3\beta +1\right) \right) }{\left[ \left( 4\beta +1\right) t-2a\left( b_{S}-a\right) \beta \left( 3\beta +1\right) \right] ^{2}}{.} \end{aligned}$$

As the denominator is positive and $v_{B}>k_{I}$ under Assumption 1, $dn_{S}^{*}/da<0$ iff $\left( 4\beta +1\right) t-2\left( b_{S}-a\right) ^{2}\beta \left( 3\beta +1\right) >0$, which is always true under Assumption 2 (ii). Therefore, $dn_{S}^{*}/da<0$ for all $a\ge 0$, which implies that $a^{S}=0$. Second, we prove that $a^{B}=b_{S}/2$. We have

$$\begin{aligned} D_{B}^{*}=\frac{\left( 2\beta +1\right) \left( t+2\beta \left( v_{B}-k_{I}+\theta \right) \right) }{\left( 4\beta +1\right) t-2a\left( b_{S}-a\right) \beta \left( 3\beta +1\right) }{.} \end{aligned}$$

As the numerator is independent of a, the maximum of $D_{B}^{*}$ in a is given by the minimum of the denominator, at $a^{B}=b_{S}/2$. Third, we show that $a^{\pi }=b_{S}/2$. We find that

$$\begin{aligned} \frac{d\Pi }{da}=\frac{\left( b_{S}-2a\right) \beta \left( 2\beta +1\right) \left( t+2\beta \left( v_{B}-k_{I}+\theta \right) \right) ^{2}\left( \left( 8\beta +3\right) t-2a\left( b_{S}-a\right) \beta \left( 3\beta +1\right) \right) }{\left[ \left( 4\beta +1\right) t-2a\left( b_{S}-a\right) \beta \left( 3\beta +1\right) \right] ^{3}}{.} \end{aligned}$$

From Assumption 2 (ii), we have $\left( 8\beta +3\right) t-2a\left( b_{S}-a\right) \beta \left( 3\beta +1\right) >0$, and the denominator is positive. Therefore, $d\Pi /da=0$ iff $a=b_{S}/2$. Since $\left. d^{2}\Pi /da^{2}\right| _{a=b_{S}/2}<0$, we have $a^{\pi }=b_{S}/2$. Finally, since $a^{B}=b_{S}/2$ and $a^{S}=0$, then $a^{V}\in [0,b_{S}/2]$. Besides,

$$\begin{aligned} \left. \frac{d(n_{S}^{*}D_{B}^{*})}{da}\right| _{a=b_{S}/2}=- \frac{4\left( 2\beta +1\right) ^{2}\left( t+2\beta \left( v_{B}-k_{I}+\theta \right) \right) ^{2}}{\left[ 2\left( 4\beta +1\right) t-b_{S}^{2}\beta \left( 3\beta +1\right) \right] ^{2}}<0{.} \end{aligned}$$

Therefore, $a^{V}\in [0,b_{S}/2)$.

(ii) Under Assumption 3, and from Propositions 2 and 3, consumer and merchant adoption decrease with the interchange fee if $\alpha _{B}$ is high and $b_{S}$ is low. In this case, consumer and merchant adoption are maximized when the interchange fee is set to zero. As from Proposition 4, we have $a^{\pi }=0$, consumer and merchant adoption is maximized at the profit-maximizing interchange fee.

1.3.6 Appendix 3.6: Proof of Proposition 5

The merchants’ surplus is

$$\begin{aligned} MS=\int _{0}^{n_{S}^{*}}(\pi (a)D_{B}^{*}-F_{S})dF_{S}=\frac{ (n_{S}^{*})^{2}}{2}{.} \end{aligned}$$

Define $u^{*}=v_{B}+\theta +\alpha _{B}n_{S}^{*}-p^{*}$. In the symmetric equilibrium, consumer surplus is

$$\begin{aligned} CS=2\int _{0}^{1/2}(u^{*}-tx)dx+2\beta \int _{0}^{u^{*}/t}(u^{*}-ty)dy=u^{*}-\frac{t}{4}+\beta \frac{(u^{*})^{2}}{t}{.} \end{aligned}$$

Finally, define total user surplus as $TUS=MS+CS$. Under the assumption that TUS is concave in a, if $\left. dTUS/da\right| _{a=a^{\pi }}\le 0$ and $a^{\pi }>0$, we have $a^{\pi }\ge a^{W}$.

(i) If $\alpha _{B}=0$, we have

$$\begin{aligned} \left. \frac{dTUS}{da}\right| _{\alpha _{B}=0}=-\frac{(2\beta +1)^{2}(2(v_{B}+\theta -c_{I})\beta +t)^{2}y(a)}{(t(4\beta +1)+2a(a-b_{S})\beta (3\beta +1))^{3}}, \end{aligned}$$

where

$$\begin{aligned} y(a)=\beta ((2a+b_{S})t+6(a-b_{S})^{3}\beta )+(at+2\beta (a-b_{S})^{3}). \end{aligned}$$

Since $t(4\beta +1)+2a(a-b_{S})\beta (3\beta +1)$ is minimal for $a=b_{S}/2$ and positive for $a=b_{S}/2$ (as by Assumption $t\ge 4\beta b_{S}^{2}$), $\left. dTUS/da\right| _{\alpha _{B}=0}$ has the sign of $-y(a)$. Since y is increasing in a, y is minimal for $a=0$. We have $y(0)=\beta b_{S}(t-b_{S}^{2}(6\beta +2))$. There are two cases. If $t\ge b_{S}^{2}(6\beta +2)$, we have $y(a)\ge 0$ for all $a\in \left[ 0,b_{S} \right]$. Therefore, we have $\left. dTUS/da\right| _{\alpha _{B}=0}\ge 0$ for all $a\in \left[ 0,b_{S}\right]$, which implies that $a^{W}\le a^{\pi }$. If $t\in \left[ 4\beta b_{S}^{2},b_{S}^{2}(6\beta +2) \right] ,$we denote by

$$\begin{aligned} \widetilde{a}=b_{S}-\sqrt{t(2\beta +1)/(6\beta (3\beta +1))} \end{aligned}$$

the level of interchange fee such that $y(\widetilde{a})=0$. We have $y(a)\le 0$ for $a\in \left[ 0,\widetilde{a}\right]$ and $y(a)\ge 0$ for $a\in \left[ \widetilde{a},b_{S}\right]$. Therefore, we have $\left. dTUS/da\right| _{\alpha _{B}=0}\ge 0$ for $a\in \left[ 0,\widetilde{a} \right]$ and $\left. dTUS/da\right| _{\alpha _{B}=0}\le 0$ for $a\in \left[ \widetilde{a},b_{S}\right]$. Therefore, if $a^{\pi }\le \widetilde{ a}$, we have $\left. dTUS/da\right| _{(\alpha _{B}=0,a=a^{\pi })}\ge 0$ . Provided that TUS is concave in a, this implies that $a^{\pi }\le a^{W}$. If $a^{\pi }\ge \widetilde{a}$, we have $a^{\pi }\ge a^{W}$.

(ii) From Assumption 2, $\overline{\alpha }_{B}=t/(4\beta b_{S})$ is a maximum for $\alpha _{B}$. In this case, we have $a^{\pi }=0$. We have

$$\begin{aligned} \left. \frac{dTUS}{da}\right| _{(a=0,\,\alpha _{B}=\overline{\alpha } _{B})}=\frac{2(3\beta +1)(2(v_{B}+\theta -c_{I})\beta +t)^{2}h(t)}{ b_{S}t^{3}\beta (7\beta +1)^{3}}{,} \end{aligned}$$

where $h(t)=-t^{2}(17\beta ^{2}+6\beta +1)-16b_{S}^{2}\beta ^{3}t+8b_{S}^{4}\beta ^{2}(3\beta +1)(5\beta +1)$. We have $h(4b_{S}\beta )\le 0$. Since h is decreasing with t, this implies that $h(t)\le 0$ for all $t\ge 4b_{S}\beta$. Therefore, we have

$$\begin{aligned} \left. \frac{dTUS}{da}\right| _{(a=0, \alpha _{B}=\overline{\alpha } _{B})}\le 0{.} \end{aligned}$$

Provided that TUS is decreasing with a, we can conclude that $a^{\pi }=a^{W}$ if the degree of S2B externality is high. It is not possible to prove analytically that TUS decreases with a. However, we ran simulations and checked that this was true for large sets of parameter values. We also checked that TUS is decreasing for all $\beta$, and that it is still decreasing if we increase $\alpha _{B}$ or $b_{S}$.

Appendix 4: Quality Investments

1.1 Appendix 4.1: Second-Order Conditions

No-cooperation We have $d^{2}\Pi _{i}/d\theta _{i}^{2}=d^{2}\widetilde{\Pi }_{i}/d\theta _{i}^{2}-\rho$. Since $d^{2}\widetilde{\Pi }_{i}/d\theta _{i}^{2}$ is independent of $\theta _{i}$, the second-order condition is satisfied if $\rho$ is high enough.

No-cooperation We have $d^{2}(\Pi _{1}+\Pi _{2})/d\theta _{i}^{2}=d^{2}(\widetilde{\Pi } _{1}+\widetilde{\Pi }_{2})/d\theta _{i}^{2}-\rho$. Since $d^{2}(\widetilde{ \Pi }_{1}+\widetilde{\Pi }_{2})/d\theta _{i}^{2}$ is independent of $\theta _{i}$, $d^{2}(\Pi _{1}+\Pi _{2})/d\theta _{i}^{2}<0$ if $\rho$ is high enough. Furthermore, the determinant of the Hessian matrix is positive if

$$\begin{aligned} \frac{4\left( \beta +1\right) ^{2}\left[ \left( 2\beta +1\right) t-2\beta \left( \beta +1\right) \alpha _{B}\left( b_{S}-a\right) \right] \left[ t-\left( b_{S}-a\right) \left( a+2\alpha _{B}\right) \beta \right] }{t\left[ t\left( 4\beta +3\right) -2\beta \left( \beta +1\right) \left( b_{S}-a\right) \left( a+3\alpha _{B}\right) \right] ^{2}}>0{.} \end{aligned}$$

The denominator is positive. The numerator is also positive as $\left( 2\beta +1\right) t-2\beta (\beta +1)\alpha _{B}(b_{S}-a)>0$ from Assumption A2 (i), and $t-\left( b_{S}-a\right) \left( a+2\alpha _{B}\right) \beta >0$ from Assumption A2 (ii). Therefore, the SOC holds.

1.2 Appendix 4.2: Proof of Proposition 6

No-cooperation We begin by proving that equilibrium quality investments are decreasing with the interchange fee at $a=b_{S}$. We have

$$\begin{aligned} \left. \frac{d(\theta ^{NC})^{*}}{da}\right| _{a=b_{S}}=\frac{N_{1}}{ \left[ \rho t(4\beta +1)^{2}(4\beta +3)-2\beta (2\beta +1)\left( 8\beta ^{2}+8\beta +1\right) \right] ^{2}}{.} \end{aligned}$$

As the denominator is strictly positive from the SOC at $a=b_{S}$, $\left. d(\theta ^{NC})^{*}/da\right| _{a=b_{S}}$ has the sign of $N_{1}=-\rho \beta \left( 4\beta +1\right) [t+2\beta (v_{B}-k_{I})]N_{1}^{\prime }$, where $N_{1}^{\prime }$ is a sum of positive terms. Hence, from Assumption 1, $\left. d(\theta ^{NC})^{*}/da\right| _{a=b_{S}}<0$.

Now, we study the variations of quality investments with respect to the interchange fee at $a=0$. We have $\left. d(\theta ^{NC})^{*}/da\right| _{a=0}=N_{2}/D_{2}$. Since $D_{2}$ is a square, it is strictly positive, and therefore, $\left. d(\theta ^{NC})^{*}/da\right| _{a=0}$ has the same sign as $N_{2}$. When $\alpha _{B}=0$, we have

$$\begin{aligned} N_{2}=\rho t^{6}\beta (2\beta +1)(4\beta +1)b_{S}\left[ t+2\beta (v_{B}-k_{I})\right] \left[ 192\beta ^{4}+400\beta ^{3}+280\beta ^{2}+72\beta +5\right] >0{.} \end{aligned}$$

At the other extreme, when $\alpha _{B}\rightarrow \infty$, $N_{2}$ goes to $-\infty$, as $t-2\beta (v_{B}-k_{I})>0$.

Cooperation We begin by proving that equilibrium quality investments are decreasing with the interchange fee at $a=b_{S}$. We have

$$\begin{aligned} \left. \frac{d(\theta ^{C})^{*}}{da}\right| _{a=b_{S}}=\frac{N_{3}}{ \left[ \rho t(4\beta +1)^{2}-4\beta ^{2}(2\beta +1)\right] ^{2}}{.} \end{aligned}$$

As the denominator is strictly positive, $\left. d(\theta ^{C})^{*}/da\right| _{a=b_{S}}$ has the sign of $N_{3}=-2\rho \beta ^{2}\left( 4\beta +1\right) [t+2\beta (v_{B}-k_{I})]N_{3}^{\prime }$, where $N_{3}^{\prime }$ is a sum of positive terms. Therefore, from Assumption 1, $\left. d(\theta ^{C})^{*}/da\right| _{a=b_{S}}<0$.

Now, we study the variations of the equilibrium quality investment with respect to the interchange fee at $a=0$. We have $\left. d(\theta ^{C})^{*}/da\right| _{a=0}=N_{4}/D_{4}$. Since $D_{4}$ is a square, it is strictly positive, and hence, $\left. d(\theta ^{C})^{*}/da\right| _{a=0}$ has the same sign as $N_{4}$. When $\alpha _{B}=0$, we have

$$\begin{aligned} N_{4}=2b_{S}\rho t^{4}\beta ^{2}(2\beta +1)(4\beta +1)(8\beta +3)\left[ t+2\beta (v_{B}-k_{I})\right] >0{,} \end{aligned}$$

whereas when $\alpha _{B}\rightarrow \infty$, $N_{4}$ goes to $-\infty$, as $t-2\beta (v_{B}-k_{I})>0$.

1.3 Appendix 4.3: Proof of Proposition 7

Let $\epsilon \left( a\right) \equiv (\theta ^{NC})^{*}-(\theta ^{C})^{*}$. We start by showing that $\epsilon \left( b_{S}\right) >0$. We have

$$\begin{aligned} \epsilon \left( b_{S}\right) =\frac{kt(2\beta +1)^{2}(4\beta +1)^{2}(2\beta \left( v_{B}-k_{I}\right) +t)}{\left( kt(4\beta +1)^{2}-4\beta ^{2}(2\beta +1)\right) \left( kt(4\beta +1)^{2}(4\beta +3)-2\beta (2\beta +1)\left( 8\beta ^{2}+8\beta +1\right) \right) }{.} \end{aligned}$$

In the no-cooperation scenario, if $a=b_{S}$, the SOC holds iff $kt(4\beta +1)^{2}(4\beta +3)^{2}>(2\beta +1)\left( 8\beta ^{2}+8\beta +1\right) ^{2}$. We find that if this is true, the denominator of $\epsilon \left( b_{S}\right)$ is strictly positive, hence $\epsilon \left( b_{S}\right) >0$.

Now, we prove that $\epsilon \left( 0\right) \ge 0$ if $\alpha _{B}$ is small or if it is high. First, we find that $\left. \epsilon \left( 0\right) \right| _{\alpha _{B}=0}=\epsilon \left( b_{S}\right) >0$. Second, from Assumption 2, $\alpha _{B}$ is bounded from above by $\overline{\alpha } _{B}=\min \left\{ t/(4\beta b_{S}),t/(2\beta (1+\beta )b_{S})\right\}$. Note that $\overline{\alpha }_{B}=t/(4\beta b_{S})$ if $\beta <1$, and $\overline{\alpha }_{B}=t/(2\beta (1+\beta )b_{S}$ otherwise. To begin with, assume that $\beta <1$. We have

$$\begin{aligned} \left. \epsilon \left( 0\right) \right| _{\alpha _{B}=\overline{\alpha } _{B}}=\frac{2kt(1-\beta )(3\beta +1)^{2}(7\beta +1)^{2}(2\beta \left( v_{B}-k_{I}\right) +t)}{\left( kt(7\beta +1)^{2}-16\beta ^{2}(3\beta +1)\right) \left( kt(7\beta +1)^{2}(5\beta +3)-4\beta (3\beta +1)\left( 17\beta ^{2}+14\beta +1\right) \right) }{.} \end{aligned}$$

The SOC in the no-cooperation scenario at $a=0$ and $\alpha _{B}=t/(4\beta b_{S})$ holds iff $kt(7\beta +1)^{2}(5\beta +3)^{2}>(3\beta +1)\left( 17\beta ^{2}+14\beta +1\right) ^{2}$. If this is true, then the denominator of $\left. \epsilon \left( 0\right) \right| _{\alpha _{B}=\overline{ \alpha }_{B}}$ is strictly positive, hence $\left. \epsilon \left( 0\right) \right| _{\alpha _{B}=\overline{\alpha }_{B}}>0$. Finally, we find that when $\beta >1$, $\left. \epsilon \left( 0\right) \right| _{\alpha _{B}= \overline{\alpha }_{B}}=0$. Therefore, $\left. \epsilon \left( 0\right) \right| _{\alpha _{B}=\overline{\alpha }_{B}}\ge 0$ for all $\beta$.