Sequential payments and optimal pricing in payment systems

Abstract

This paper studies the intraday payment behaviour between heterogeneous banks as well as optimal intraday pricing schemes. The paper shows the social optimality of payment sequencing, which allows a bank to delay payments until the bank receives payments from the counterparty. The payment sequencing allows a bank with high liquidity cost to ‘recycle’ payment inflow from another bank with lower liqudity cost, reducing the aggregate cost of funding of banks to settle all payments. But we also see that the banks have an incentive to delay payments more than the payment sequencing requires. This underscores the importance of social planner’s role reducing settlement delay, while leaving socially efficient payment sequencing. In this context, we compare two different pricing schemes, a standard throughput guideline and a time-varying intraday tariff, to discuss the optimal incentive mechanisms in payment systems for the ‘socially efficient sequential settlement’.

Appendices

Appendix 1: Proof of Proposition 2

The first-order conditions of the function 6, with respect to $d_{1}$ and $d_{2}$ are:

$$\begin{aligned} d_{1}=\frac{(-1)\alpha }{\mu }-\frac{\delta }{\mu }-\frac{\gamma }{\mu } \varepsilon _{1}-\frac{\eta }{\mu }\varepsilon _{2}-\frac{2\beta }{\mu } d_{2}+\frac{\lambda _{1}-\lambda _{3}}{\mu } \end{aligned}$$

That of $d_{2}$ is defined in the same way. Substituting these and assume $\lambda _{j}=0$ for all $j\in \{1,2,3,4\}$, we have:

$$\begin{aligned} d_{1}=\frac{(-1)\left( \alpha +\delta \right) }{\mu }-\frac{\gamma }{\mu } \varepsilon _{1}-\frac{\eta }{\mu }\varepsilon _{2}-\frac{2\beta }{\mu } \left\{ \frac{(-1)\alpha }{\mu }-\frac{\delta }{\mu }-\frac{\eta }{\mu } \varepsilon _{1}-\frac{\gamma }{\mu }\varepsilon _{2}-\frac{2\beta }{\mu } d_{1}\right\} \end{aligned}$$

Solving for $d_{1}$,

$$\begin{aligned} d_{1}=\frac{(-1)\left( \alpha +\delta \right) }{\mu +2\beta }+\frac{2\beta \gamma -\mu \eta }{\mu ^{2}-4\beta ^{2}}\varepsilon _{2}+\frac{2\beta \eta -\mu \gamma }{\mu ^{2}-4\beta ^{2}}\varepsilon _{1} \end{aligned}$$

From A3 and A6, we have:

$$\begin{aligned} d_{1}= & {} \frac{2\beta \gamma +\mu \gamma }{\mu ^{2}-4\beta ^{2}}\varepsilon _{2}+\frac{-2\beta \gamma -\mu \gamma }{\mu ^{2}-4\beta ^{2}}\varepsilon _{1}\\= & {} \frac{\gamma }{\mu -2\beta }\left( \varepsilon _{2}-\varepsilon _{1}\right) \end{aligned}$$

The first-best solution is, therefore,

$$\begin{aligned} d_{1}^{1st}=\max \left[ \frac{\gamma }{\mu -2\beta }\left( \varepsilon _{2}-\varepsilon _{1}\right) ,0\right] \end{aligned}$$

$\gamma \left( \mu -2\beta \right) ^{-1}<0$ thus $d_{1}^{1st}$ is a positive function of $\varepsilon _{1}$ and a negative function of $\varepsilon _{2}$. $\lambda _{1}$ and $\lambda _{2}$ can be non-zero if $d_{1}^{1st}=0$: the derivation of these is skipped. Since we have assumed a sufficiently large T, $\lambda _{3}=\lambda _{4}=0$ in any case. $\square$

Appendix 2: Proof of Proposition 4

We keep the assumptions made in Proposition 1, which ensures internal solutions, but we still need to formulate the Kuhn–Tucker conditions because we have the tariff function.

$$\begin{aligned}&\alpha +\mu d_{1}+\beta d_{2}+\gamma \varepsilon _{1}+\frac{\partial t}{ \partial d_{1}}-\lambda _{1}=0 \\&\quad \lambda _{1}\ge 0\text { and }\lambda _{1}d_{1}=0 \end{aligned}$$

The best response function $d_{1}^{BR}$ is then derived.

$$\begin{aligned} d_{1}^{BR}(E[d_{2}],\varepsilon _{1})=(-1)\frac{\alpha }{\mu }-\frac{\beta }{ \mu }\cdot E\left[ d_{2}(d_{1},\varepsilon _{2})\right] -\frac{\gamma }{\mu } \varepsilon _{1}-\frac{1}{\mu }\left( \frac{\partial t}{\partial d_{1}} -\lambda _{1}\right) \end{aligned}$$

Substitute this into the equation $E[d_{2}^{BR}(E[d_{1}],\varepsilon _{2})]$ , which is defined in a similar way, and we have:

$$\begin{aligned} E[d_{2}^{BR}(E[d_{1}],\varepsilon _{2})]=\frac{(-1)\alpha }{\mu +\beta }+ \frac{\beta }{\mu ^{2}-\beta ^{2}}\left( \frac{\partial t}{\partial d_{1}} -\lambda _{1}\right) -\frac{\mu }{\mu ^{2}-\beta ^{2}}\left( \frac{\partial t }{\partial d_{2}}-\lambda _{2}\right) \end{aligned}$$

$E[d_{1}^{BR}(E[d_{2}],\varepsilon _{1})]$ is defined in the same way. We will design the function t to eliminate the source of deadweight loss $(-1)\alpha \left( \mu +\beta \right) ^{-1}$, so we have the following simultaneous equations:

$$\begin{aligned} \frac{\alpha }{\mu +\beta }=\frac{\beta }{\mu ^{2}-\beta ^{2}}\left( \frac{ \partial t}{\partial d_{j}}-\lambda _{j}\right) -\frac{\mu }{\mu ^{2}-\beta ^{2}}\left( \frac{\partial t}{\partial d_{i}}-\lambda _{i}\right) \end{aligned}$$

for $i\ne j\in \{1,2\}$. By solving these equations, we have

$$\begin{aligned} \frac{\partial t}{\partial d_{1}}-\lambda _{1}=\frac{\partial t}{\partial d_{2}}-\lambda _{2}=(-1)\alpha \end{aligned}$$

Integrating these, with $\lambda _{i}\cdot d_{i}=0$, we have $t(d_{i})=t_{0}-\alpha d_{i}$. $\square$

Appendix 3: Proof of Proposition 5

The aggregate loss of the system under the tariff $t(d_{i})$ for given $\varepsilon _{1}$ and $\varepsilon _{2}$, $L(d_{1}^{**},d_{2}^{**},c,\varepsilon _{1},\varepsilon _{2})$, is defined as follows:

$$\begin{aligned}&L(d_{1}^{**},d_{2}^{**},c,\varepsilon _{1},\varepsilon _{2}) \\&\quad =\alpha \cdot \max \left[ -\frac{\gamma }{\mu }\varepsilon _{1},0\right] +\beta \cdot \max \left[ -\frac{\gamma }{\mu }\varepsilon _{1},0\right] \cdot \max \left[ -\frac{\gamma }{\mu }\varepsilon _{2},0\right] \\&\qquad +\,\gamma \cdot \max \left[ -\frac{\gamma }{\mu }\varepsilon _{1},0\right] \cdot \varepsilon _{1} +\,\delta \cdot \max \left[ -\frac{\gamma }{\mu }\varepsilon _{2},0\right] +\eta \cdot \max \left[ -\frac{\gamma }{\mu }\varepsilon _{2},0\right] \\&\qquad \cdot \varepsilon _{1}+\frac{1}{2}\mu \cdot \max \left[ -\frac{\gamma }{\mu } \varepsilon _{1},0\right] ^{2} +\,\alpha \cdot \max \left[ -\frac{\gamma }{\mu }\varepsilon _{2},0\right] +\beta \cdot \max \left[ -\frac{\gamma }{\mu }\varepsilon _{2},0\right] \\&\qquad \cdot \max \left[ -\frac{\gamma }{\mu }\varepsilon _{1},0\right] +\gamma \cdot \max \left[ -\frac{\gamma }{\mu }\varepsilon _{2},0\right] \cdot \varepsilon _{2} \\&\qquad +\,\delta \cdot \max \left[ -\frac{\gamma }{\mu }\varepsilon _{1},0\right] +\eta \cdot \max \left[ -\frac{\gamma }{\mu }\varepsilon _{1},0\right] \cdot \varepsilon _{2}+\frac{1}{2}\mu \cdot \max \left[ -\frac{\gamma }{\mu } \varepsilon _{2},0\right] ^{2} \end{aligned}$$

If $\varepsilon _{1}>0$ and $\varepsilon _{2}>0$, we have:

$$\begin{aligned} L(d_{1}^{**},d_{2}^{**},c,\varepsilon _{1},\varepsilon _{2})= & {} \alpha \frac{-\gamma }{\mu }\varepsilon _{1}+\beta \frac{-\gamma }{\mu } \varepsilon _{1}\frac{-\gamma }{\mu }\varepsilon _{2}+\gamma \frac{-\gamma }{ \mu }\varepsilon _{1}\varepsilon _{1} \\&+\,\delta \frac{-\gamma }{\mu }\varepsilon _{2}+\eta \frac{-\gamma }{\mu } \varepsilon _{2}\varepsilon _{1}+\frac{1}{2}\mu \left( \frac{-\gamma }{\mu } \varepsilon _{1}\right) ^{2} \\&+\,\alpha \frac{-\gamma }{\mu }\varepsilon _{2}+\beta \frac{-\gamma }{\mu } \varepsilon _{2}\frac{-\gamma }{\mu }\varepsilon _{1}+\gamma \frac{-\gamma }{ \mu }\varepsilon _{2}\varepsilon _{2} \\&+\,\delta \frac{-\gamma }{\mu }\varepsilon _{1}+\eta \frac{-\gamma }{\mu } \varepsilon _{1}\varepsilon _{2}+\frac{1}{2}\mu \left( \frac{-\gamma }{\mu } \varepsilon _{2}\right) ^{2} \\= & {} 2\frac{\beta \gamma ^{2}}{\mu ^{2}}\varepsilon _{1}\varepsilon _{2}-2 \frac{\eta \gamma }{\mu }\varepsilon _{1}\varepsilon _{2}-\frac{1}{2}\frac{ \gamma ^{2}}{\mu }\varepsilon _{1}^{2}-\frac{1}{2}\frac{\gamma ^{2}}{\mu } \varepsilon _{2}^{2} \\&-\,\left( \alpha +\delta \right) \frac{\gamma }{\mu }\varepsilon _{1}-\left( \alpha +\delta \right) \frac{\gamma }{\mu }\varepsilon _{2} \end{aligned}$$

From A3 ($\alpha +\delta =0$) and A6 ($\gamma +\eta =0$), we have:

$$\begin{aligned} L(d_{1}^{**},d_{2}^{**},c,\varepsilon _{1},\varepsilon _{2})= \frac{\gamma ^{2}}{\mu }\left\{ 2\frac{\beta +\mu }{\mu }\varepsilon _{1}\varepsilon _{2}-\frac{1}{2}\varepsilon _{1}^{2}-\frac{1}{2}\varepsilon _{2}^{2}\right\} \end{aligned}$$

(8)

Notice that $L(d_{1}^{**},d_{2}^{**},c,\varepsilon _{1},\varepsilon _{2})$ can be positive when $\left( \beta +\mu \right) /\mu >\frac{1}{2}$ and $\varepsilon _{1}\simeq \varepsilon _{2}>0$. If $\varepsilon _{1}>0$ and $\varepsilon _{2}\le 0$, $L(d_{1}^{**},d_{2}^{**},c,\varepsilon _{1},\varepsilon _{2})$ with A3 and A6 is:

$$\begin{aligned} L(d_{1}^{**},d_{2}^{**},c,\varepsilon _{1},\varepsilon _{2})= & {} \alpha \frac{-\gamma }{\mu }\varepsilon _{1}+\gamma \frac{-\gamma }{\mu } \varepsilon _{1}\varepsilon _{1}+\frac{1}{2}\mu \left( \frac{-\gamma }{\mu } \varepsilon _{1}\right) ^{2} \\&+\,\delta \frac{-\gamma }{\mu }\varepsilon _{1}+\eta \frac{-\gamma }{\mu } \varepsilon _{1}\varepsilon _{2} \\= & {} \frac{\gamma ^{2}}{\mu }\varepsilon _{1}\varepsilon _{2}-\frac{1}{2}\frac{ \gamma ^{2}}{\mu }\varepsilon _{1}^{2} \\\le & {} 0 \end{aligned}$$

Since the banks are almost symmetric, we have a similar welfare for the case $\varepsilon _{1}\le 0$ and $\varepsilon _{2}>0$.

$$\begin{aligned} L(d_{1}^{**},d_{2}^{**},c,\varepsilon _{1},\varepsilon _{2})= \frac{\gamma ^{2}}{\mu }\varepsilon _{1}\varepsilon _{2}-\frac{1}{2}\frac{ \gamma ^{2}}{\mu }\varepsilon _{2}^{2}\le 0 \end{aligned}$$

If $\varepsilon _{1}\le 0$ and $\varepsilon _{2}\le 0$, $L(d_{1}^{**},d_{2}^{**},c,\varepsilon _{1},\varepsilon _{2})$ is trivially zero.

Substituting these into the Eq. 7:

$$\begin{aligned}&E\left[ L(d_{1}^{**},d_{2}^{**},c,\varepsilon _{1},\varepsilon _{2})\right] \\&\quad =\int _{0}^{\bar{\varepsilon }}\int _{0}^{\bar{ \varepsilon }}\left\{ 2\gamma ^{2}\frac{\beta +\mu }{\mu ^{2}}\varepsilon _{1}\varepsilon _{2}-\frac{1}{2}\frac{\gamma ^{2}}{\mu }\varepsilon _{1}^{2}- \frac{1}{2}\frac{\gamma ^{2}}{\mu }\varepsilon _{2}^{2}\right\} \psi (\varepsilon _{1})\psi (\varepsilon _{2})d\varepsilon _{1}d\varepsilon _{2}\\&\qquad +\int _{0}^{\bar{\varepsilon }}\int _{-\bar{\varepsilon }}^{0}\left\{ \frac{ \gamma ^{2}}{\mu }\varepsilon _{1}\varepsilon _{2}-\frac{1}{2}\frac{\gamma ^{2}}{\mu }\varepsilon _{1}^{2}\right\} \psi (\varepsilon _{1})\psi (\varepsilon _{2})d\varepsilon _{1}d\varepsilon _{2} \\&\qquad +\int _{-\bar{\varepsilon }}^{0}\int _{0}^{\bar{\varepsilon }}\left\{ \frac{ \gamma ^{2}}{\mu }\varepsilon _{1}\varepsilon _{2}-\frac{1}{2}\frac{\gamma ^{2}}{\mu }\varepsilon _{1}^{2}\right\} \psi (\varepsilon _{1})\psi (\varepsilon _{2})d\varepsilon _{1}d\varepsilon _{2} \end{aligned}$$

The first term can be written as follows:

$$\begin{aligned}&\int _{0}^{\bar{\varepsilon }}\int _{0}^{\bar{\varepsilon }}\left\{ 2\gamma ^{2}\frac{\beta +\mu }{\mu ^{2}}\varepsilon _{1}\varepsilon _{2}-\frac{1}{2} \frac{\gamma ^{2}}{\mu }\varepsilon _{1}^{2}-\frac{1}{2}\frac{\gamma ^{2}}{ \mu }\varepsilon _{2}^{2}\right\} \frac{1}{4\bar{\varepsilon }^{2}} d\varepsilon _{1}d\varepsilon _{2} \\&\quad =\int _{0}^{\bar{\varepsilon }}\left[ \gamma ^{2}\frac{\beta +\mu }{\mu ^{2}} \varepsilon _{1}\varepsilon _{2}^{2}-\frac{1}{2}\frac{\gamma ^{2}}{\mu } \varepsilon _{1}^{2}\varepsilon _{2}-\frac{1}{6}\frac{\gamma ^{2}}{\mu } \varepsilon _{2}^{3}\right] _{0}^{\bar{\varepsilon }}\frac{1}{4\bar{ \varepsilon }^{2}}d\varepsilon _{1} \\&\quad = \frac{1}{4\bar{\varepsilon }^{2}}\left[ \frac{1}{2}\gamma ^{2}\frac{\beta +\mu }{\mu ^{2}}\varepsilon _{1}^{2}\bar{\varepsilon }^{2}-\frac{1}{6}\frac{ \gamma ^{2}}{\mu }\varepsilon _{1}^{3}\bar{\varepsilon }-\frac{1}{6}\frac{ \gamma ^{2}}{\mu }\varepsilon _{1}\bar{\varepsilon }^{3}\right] _{0}^{\bar{ \varepsilon }} \\&\quad =\frac{1}{24}\frac{\gamma ^{2}}{\mu }\left\{ 3\frac{\beta +\mu }{\mu } -2\right\} \bar{\varepsilon }^{2} \end{aligned}$$

The second term is:

$$\begin{aligned}&\int _{0}^{\bar{\varepsilon }}\left[ \frac{1}{2}\frac{\gamma ^{2}}{\mu } \varepsilon _{1}\varepsilon _{2}^{2}-\frac{1}{2}\frac{\gamma ^{2}}{\mu } \varepsilon _{1}^{2}\varepsilon _{2}\right] _{\bar{\varepsilon }}^{0}\frac{1}{ 4\bar{\varepsilon }^{2}}d\varepsilon _{1} \\&\quad =\frac{1}{4\bar{\varepsilon }^{2}}\left[ \frac{1}{6}\frac{\gamma ^{2}}{\mu } \varepsilon _{1}^{3}\bar{\varepsilon }-\frac{1}{4}\frac{\gamma ^{2}}{\mu } \varepsilon _{1}^{2}\bar{\varepsilon }^{2}\right] _{0}^{\bar{\varepsilon }}= \frac{1}{48}\frac{\gamma ^{2}}{\mu }\left( -1\right) \bar{\varepsilon }^{2} \end{aligned}$$

The third term is equivalent to the second one. Now we have:

$$\begin{aligned} E\left[ L(d_{1}^{**},d_{2}^{**},c,\varepsilon _{1},\varepsilon _{2})\right]= & {} \frac{1}{24}\frac{\gamma ^{2}}{\mu }\left\{ 3\frac{\beta +\mu }{\mu }-2\right\} \bar{\varepsilon }^{2}+2\cdot \frac{1}{48} \frac{\gamma ^{2}}{\mu }\left( -1\right) \bar{\varepsilon }^{2} \\= & {} \frac{1}{24}\frac{\gamma ^{2}}{\mu }\bar{\varepsilon }^{2}\left( 3\frac{ \beta +\mu }{\mu }-2-1\right) \end{aligned}$$

Since $\beta <0$, $\left( \beta +\mu \right) /\mu <1$ and it is sufficient to show the following:

$$\begin{aligned} E\left[ L(d_{1}^{**},d_{2}^{**},c,\varepsilon _{1},\varepsilon _{2})\right] <0 \end{aligned}$$

Since the aggregate loss under the throughput guidelines, $E\left[ L(0,0,c,\varepsilon _{1},\varepsilon _{2})\right]$ is obviously zero ($d_{1}=d_{2}=0$ for any cases), now we complete proving that the aggregate loss under the tariff is smaller than the loss under the throughput guidelines. $\square$