The effect of the central bank’s standing facilities on interbank lending and bank liquidity holding

Abstract

How does the central bank influence interbank lending? The central bank’s policy rates determine the attractiveness of the standing facilities compared with the interbank market. Therefore, by choosing the policy rates the central bank affects the number of banks using the standing facilities and the number of banks using the interbank market. There is also a second channel. The policy rates may influence bank liquidity holding and thus the chances that interbank lending occurs. To address both channels, bank liquidity holding is endogenous in the presented model. The results show that liquidity is not held to insure against idiosyncratic risk but to lend to the interbank market in case counterparty risk is not too high. If banks expect interbank lending to be sufficiently likely and profitable, a smooth liquidity transfer at the interbank market is guaranteed. The central bank can create such a situation under the constraint that counterparty risk is moderate and counterparty risk perceptions are not too distorted. If, however, counterparty risk is perceived to be too large, this may result in liquidity hoarding.

Appendices

Appendix

Proof of Lemma 1

Bank 1 is able to repay the repayment claim in case of a successful investment if

$$\begin{aligned} r_{IB}\le \overline{r}_{IB}&=\frac{1}{L_{1}}\left( \alpha _{1}R_{1}-r_{l}\left( -1+\alpha _{1}-\lambda _{1}-L_{1}\right) \right) \end{aligned}$$

(11)

$$\begin{aligned}&=r_{l}+\frac{1}{L_{1}}\left( \alpha _{1}R_{1}-r_{l}\left( -1+\alpha _{1}-\lambda _{1}\right) \right) . \end{aligned}$$

(12)

Bank 1 accepts the loan contract if

$$\begin{aligned}&E^{t=1}\varPi _{1}\ge {\textit{TP}}_{1}\\&\quad \iff p_{1}\left( \alpha _{1}R_{1}-r_{IB}L_{1}-r_{l}\left( -1+\alpha _{1}-\lambda _{1}-L_{1}\right) \right) ^{+} \\&\quad \ge p_{1}\left( \alpha _{1}R_{1}-r_{l}\left( -1+\alpha _{1}-\lambda _{1}\right) \right) ^{+} \end{aligned}$$

This yields the condition

$$\begin{aligned} r_{IB}{\left\{ \begin{array}{ll} \in \mathbb {R}_{+}, &{} r_{l}>\frac{\alpha _{1}R_{1}}{-1+\alpha _{1}-\lambda _{1}},\\ \le r_{l}, &{} r_{l}\le \frac{\alpha _{1}R_{1}}{-1+\alpha _{1}-\lambda _{1}}. \end{array}\right. } \end{aligned}$$

(13)

Bank 2 accepts the loan contract if

$$\begin{aligned} E^{t=1}\varPi _{2}\ge & {} {\textit{TP}}_{2}\nonumber \\\iff & {} r_{d}\left( 1-\alpha _{2}-\lambda _{2}-L_{1}\right) +p_{2}\alpha _{2}R_{2}+p_{1}r_{IB}L_{1}\ge r_{d}\left( 1-\alpha _{2}+\lambda _{2}\right) +p_{2}\alpha _{2}R_{2}\nonumber \\\iff & {} r_{IB}\ge \frac{r_{d}}{p_{1}}. \end{aligned}$$

(14)

Interbank lending is not taking place if all three conditions (11), (13) and (14) cannot be satisfied at the same time, i.e., if

• $\frac{r_{d}}{p_{1}}>\frac{1}{L_{1}}\left( \alpha _{1}R_{1}-r_{l}\left( -1+\alpha _{1}-\lambda _{1}-L_{1}\right) \right)$ and $r_{l}>\frac{\alpha _{1}R_{1}}{-1+\alpha _{1}-\lambda _{1}},$ or

• $\frac{r_{d}}{p_{1}}>r_{l}$ and $r_{l}\le \frac{\alpha _{1}R_{1}}{-1+\alpha _{1}-\lambda _{1}}.$

These conditions are equivalent to

$$\begin{aligned} \frac{r_{d}}{p_{1}}>r_{l}-\frac{1}{L_{1}}\left( r_{l}\left( -1+\alpha _{1}-\lambda _{1}\right) -\alpha _{1}R_{1}\right) ^{+} \end{aligned}$$

which proves the claim.

Proof of Corollary 1

Denote $LHS=\frac{r_{d}}{p_{1}}$ and $RHS=r_{l}-\frac{1}{L_{1}}\left( r_{l}\left( -1+\alpha _{1}-\lambda _{1}\right) -\alpha _{1}R_{1}\right) ^{+}$. Note that LHS depends on $r_{d}$ but not on $r_{l}$, and RHS depends on $r_{l}$ but not on $r_{d}$. LHS is increasing in $r_{d}$ which proves (a). If $r_{l}\le \frac{\alpha _{1}R_{1}}{-1+\alpha _{1}-\lambda _{1}}$, then $RHS=r_{l}$ is increasing in $r_{l}$ which proves (b1). If $r_{l}>\frac{\alpha _{1}R_{1}}{-1+\alpha _{1}-\lambda _{1}}$ and $L_{1}=1-\alpha _{2}+\lambda _{2}$, then $RHS=r_{l}-r_{l}\frac{-1+\alpha _{1}-\lambda _{1}}{1-\alpha _{2}+\lambda _{2}}+\frac{\alpha _{1}R_{1}}{L_{1}}$ is decreasing in $r_{l}$ which proves (b2). If $r_{l}>\frac{\alpha _{1}R_{1}}{-1+\alpha _{1}-\lambda _{1}}$ and $L_{1}=-1+\alpha _{1}-\lambda _{1}$, then $RHS=\frac{\alpha _{1}R_{1}}{L_{1}}$ does not depend on $r_{l}$ which proves (b3).

Proof of Lemma 2

The lemma is proven by case distinction.

• $r_{l}\le \frac{\alpha _{1}R_{1}}{-1+\alpha _{1}-\lambda _{1}}:$

  Problem (7) can be described by

  $$\begin{aligned} \max _{r_{l}\ge r_{IB}\ge \frac{r_{d}}{p_{1}}}p_{1}L_{1}^{2}\left( r_{l}-r_{IB}\right) \left( p_{1}r_{IB}-r_{d}\right) . \end{aligned}$$

  Solving this problem yields the interbank rate

  $$\begin{aligned} r_{IB}^{*}=\frac{1}{2p_{1}}\left( r_{d}+p_{1}r_{l}\right) . \end{aligned}$$

• $r_{l}>\frac{\alpha _{1}R_{1}}{-1+\alpha _{1}-\lambda _{1}}:$

  According to (7), the interbank rate is the solution to

  $$\begin{aligned} \max _{r_{IB}}p_{1}\left( \alpha _{1}R_{1}-r_{IB}L_{1}-r_{l}\left( -1+\alpha _{1}-\lambda _{1}-L_{1}\right) \right) \left( p_{1}r_{IB}-r_{d}\right) L_{1} \end{aligned}$$

  subject to

  $$\begin{aligned} \frac{1}{L_{1}}\left( \alpha _{1}R_{1}-r_{l}\left( -1+\alpha _{1}-\lambda _{1}-L_{1}\right) \right) \ge r_{IB}\ge \frac{r_{d}}{p_{1}}. \end{aligned}$$

  We obtain

  $$\begin{aligned} r_{IB}^{*}=\frac{1}{2p_{1}}r_{d}-\frac{1}{2L_{1}}\left( r_{l}\left( -1+\alpha _{1}-\lambda _{1}-L_{1}\right) -\alpha _{1}R_{1}\right) . \end{aligned}$$

Problem (7) is therefore solved by the interbank rate

$$\begin{aligned} r_{IB}^{*}=\frac{1}{2p_{1}}\left( r_{d}+p_{1}r_{l}\right) -\frac{1}{2L_{1}}\left( r_{l}\left( -1+\alpha _{1}-\lambda _{1}\right) -\alpha _{1}R_{1}\right) ^{+}. \end{aligned}$$

Proof of Proposition 1

The proof proceeds in the following way: First, banks’ expected profits at $t=0$ is derived. The second step is the calculation of the best-response functions and liquidity holding in equilibrium.

The expected profit at time $t=1$ is defined piecewise by

$$\begin{aligned} E^{t=0}\varPi _{i}=\frac{1}{4}\int _{-1}^{1}\int _{-1}^{1}E^{t=1}\varPi _{i}\left( \lambda _{1},\lambda _{2}\right) \,\,d\lambda _{1}\,d\lambda _{2}. \end{aligned}$$

Depending on the liquidity shocks, banks face a different expected profit. The different sections are illustrated in Fig. 2. According to this case distinction, we split the integral

$$\begin{aligned} E^{t=0}\varPi _{i}=E^{t=0}\left( \varPi _{i}\wedge I.\right) +E^{t=0}\left( \varPi _{i}\wedge II.\right) +E^{t=0}\left( \varPi _{i}\wedge III.\right) +E^{t=0}\left( \varPi _{i}\wedge IV.\right) \end{aligned}$$

and derive each of the summands separately. Note that $E^{t=0}\left( \varPi _{i}\wedge II.\right)$ and $E^{t=0}\left( \varPi _{i}\wedge III.\right)$ can again be split into several summands. For reasons of symmetry, it is sufficient to consider bank 1, only.

• I. $1-\alpha _{i}+\lambda _{i}\ge 0$:

  $$\begin{aligned} 4E^{t=0}\left( \varPi _{1}\wedge I.\right) =&\int _{-1+\alpha _{1}}^{1}\int _{-1+\alpha _{2}}^{1}r_{d}\left( 1-\alpha _{1}+\lambda _{1}\right) +p_{1}\alpha _{1}R_{1}\\&-\left( 1-\lambda _{1}\right) \text { }d\lambda _{2}\,d\lambda _{1}\\ =&\left( 2-\alpha _{2}\right) \left( 2-\alpha _{1}\right) \left[ r_{d}\left( 1-\frac{1}{2}\alpha _{1}\right) +p_{1}\alpha _{1}R_{1}\right] \\&-\int _{-1+\alpha _{1}}^{1}\int _{-1+\alpha _{2}}^{1}\left( 1-\lambda _{1}\right) \text { }d\lambda _{2}\,d\lambda _{1} \end{aligned}$$

• IV. $1-\alpha _{i}+\lambda _{i}<0$:

  $$\begin{aligned} 4E^{t=0}\left( \varPi _{1}\wedge IV.\right)= & {} \int _{-1}^{-1+\alpha _{1}}\int _{-1}^{-1+\alpha _{2}}p_{1}\left( \alpha _{1}R_{1}-r_{l}\left( -1+\alpha _{1}-\lambda _{1}\right) \right) \\&-\left( 1-\lambda _{1}\right) \text { }d\lambda _{2}\,d\lambda _{1}\\= & {} \alpha _{1}^{2}\alpha _{2}p_{1}\left( R_{1}-\frac{r_{l}}{2}\right) -\int _{-1}^{-1+\alpha _{1}}\int _{-1}^{-1+\alpha _{2}}\left( 1-\lambda _{1}\right) \text { }d\lambda _{2}\,d\lambda _{1} \end{aligned}$$

• II. $1-\alpha _{1}+\lambda _{1}<0$, $1-\alpha _{2}+\lambda _{2}\ge 0$:

• a. $r_{l}<\frac{r_{d}}{p_{1}}$:

  $$\begin{aligned} 4E^{t=0}\left( \varPi _{1}\wedge II.\wedge a\right)= & {} \int _{-1}^{-1+\alpha _{1}}\int _{-1+\alpha _{2}}^{1}p_{1}\left( \alpha _{1}R_{1}-r_{l}\left( -1+\alpha _{1}-\lambda _{1}\right) \right) \\&-\left( 1-\lambda _{1}\right) \text { }d\lambda _{2}\,d\lambda _{1}\\= & {} \alpha _{1}^{2}\left( 2-\alpha _{2}\right) p_{1}\left( R_{1}-\frac{r_{l}}{2}\right) \\&-\int _{-1}^{-1+\alpha _{1}}\int _{-1+\alpha _{2}}^{1}\left( 1-\lambda _{1}\right) \text { }d\lambda _{2}\,d\lambda _{1} \end{aligned}$$

• b. $r_{l}\ge \frac{r_{d}}{p_{1}}$:

• II.1. $1-\alpha _{2}+\lambda _{2}\ge -1+\alpha _{1}-\lambda _{1}$:

  $$\begin{aligned} 4E^{t=0}\left( \varPi _{1}\wedge II.\wedge b\wedge II.1.\right)= & {} \int _{-1}^{-1+\alpha _{1}}\int _{-2+\alpha _{1}+\alpha _{2}-\lambda _{1}}^{1}-\left( 1-\lambda _{1}\right) +p_{1}\left( \alpha _{1}R_{1}\right. \\&\left. -\frac{1}{2p_{1}}\left( r_{d}+p_{1}r_{l}\right) \left( -1+\alpha _{1}-\lambda _{1}\right) \right) \,\,d\lambda _{2}\,d\lambda _{1}\\= & {} p_{1}\alpha _{1}^{2}\left[ R_{1}\left( 2-\frac{1}{2}\alpha _{1}-\alpha _{2}\right) \right. \\&\left. +\frac{1}{2p_{1}}\left( r_{d}+p_{1}r_{l}\right) \left( \frac{1}{3}\alpha _{1}+\frac{1}{2}\alpha _{2}-1\right) \right] \\&-\int _{-1}^{-1+\alpha _{1}}\int _{-2+\alpha _{1}+\alpha _{2}-\lambda _{1}}^{1}\left( 1-\lambda _{1}\right) \text { }d\lambda _{2}\,d\lambda _{1} \end{aligned}$$

• II.2. $1-\alpha _{2}+\lambda _{2}<-1+\alpha _{1}-\lambda _{1}$:

  $$\begin{aligned}&4E^{t=0}\left( \varPi _{1}\wedge II.\wedge b\wedge II.2.\right) \\&\quad = \int _{-1}^{-1+\alpha _{1}}\int _{-1+\alpha _{2}}^{-2+\alpha _{1}+\alpha _{2}-\lambda _{1}}p_{1}\left( \alpha _{1}R_{1}-r_{l}\left( -2+\alpha _{1}+\alpha _{2}-\lambda _{1}-\lambda _{2}\right) \right. \\&\qquad \left. -\frac{1}{2p_{1}}\left( r_{d}+p_{1}r_{l}\right) \left( 1-\alpha _{2}+\lambda _{2}\right) \right) -\left( 1-\lambda _{1}\right) \,\,d\lambda _{2}\,d\lambda _{1}\\&\quad = \frac{1}{2}p_{1}\alpha _{1}^{3}\left( R_{1}-\frac{1}{6p_{1}}\left( r_{d}+p_{1}r_{l}\right) -\frac{1}{3}r_{l}\right) -\int _{-1}^{-1+\alpha _{1}}\int _{-1+\alpha _{2}}^{-2+\alpha _{1}+\alpha _{2}-\lambda _{1}}\left( 1-\lambda _{1}\right) \text { }d\lambda _{2}\,d\lambda _{1} \end{aligned}$$

• III. $1-\alpha _{1}+\lambda _{1}\ge 0$, $1-\alpha _{2}+\lambda _{2}<0$:

• a. $r_{l}<\frac{r_{d}}{p_{2}}$:

  $$\begin{aligned} 4E^{t=0}\left( \varPi _{1}\wedge III.\wedge a\right)= & {} \int _{-1}^{-1+\alpha _{2}}\int _{-1+\alpha _{1}}^{1}r_{d}\left( 1-\alpha _{1}+\lambda _{1}\right) +p_{1}\alpha _{1}R_{1}\\&-\left( 1-\lambda _{1}\right) \,\,d\lambda _{1}\,d\lambda _{2}\\= & {} \alpha _{2}\left( 2-\alpha _{1}\right) \left( r_{d}\left( 1-\frac{1}{2}\alpha _{1}\right) +p_{1}\alpha _{1}R_{1}\right) \\&-\int _{-1}^{-1+\alpha _{2}}\int _{-1+\alpha _{1}}^{1}\left( 1-\lambda _{1}\right) \,\,d\lambda _{1}\,d\lambda _{2} \end{aligned}$$

• b. $r_{l}\ge \frac{r_{d}}{p_{2}}$:

• III.1. $1-\alpha _{1}+\lambda _{1}\ge -1+\alpha _{2}-\lambda _{2}$:

  $$\begin{aligned}&4E^{t=0}\left( \varPi _{1}\wedge III.\wedge b\wedge III.1.\right) \\&\quad = \int _{-1}^{-1+\alpha _{2}}\int _{-2+\alpha _{1}+\alpha _{2}-\lambda _{2}}^{1}r_{d}\left( 2-\alpha _{1}-\alpha _{2}+\lambda _{1}+\lambda _{2}\right) \\&\qquad +p_{1}\alpha _{1}R_{1}+\frac{1}{2}\left( r_{d}+p_{2}r_{l}\right) \left( -1+\alpha _{2}-\lambda _{2}\right) \\&\qquad -\left( 1-\lambda _{1}\right) \,\,d\lambda _{1}\,d\lambda _{2}\\&\quad = \frac{1}{6}r_{d}\alpha _{2}\left( \alpha _{2}^{2}+3\alpha _{1}\alpha _{2}-6\alpha _{2}+3\alpha _{1}^{2}-12\alpha _{1}+12\right) \\&\qquad +p_{1}\alpha _{2}\alpha _{1}R_{1}\left( 2-\frac{1}{2}\alpha _{2}-\alpha _{1}\right) \\&\qquad +\frac{1}{2}\left( r_{d}+p_{2}r_{l}\right) \alpha _{2}^{2}\left( -\frac{1}{3}\alpha _{2}-\frac{1}{2}\alpha _{1}+1\right) \\&\qquad -\int _{-1}^{-1+\alpha _{2}}\int _{-2+\alpha _{1}+\alpha _{2}-\lambda _{2}}^{1}\left( 1-\lambda _{1}\right) \,\,d\lambda _{1}\,d\lambda _{2} \end{aligned}$$

• III.2. $1-\alpha _{1}+\lambda _{1}<-1+\alpha _{2}-\lambda _{2}$:

  $$\begin{aligned}&4E^{t=0}\left( \varPi _{1}\wedge III.\wedge b\wedge III.2.\right) \\&\quad = \int _{-1}^{-1+\alpha _{2}}\int _{-1+\alpha _{1}}^{-2+\alpha _{1}+\alpha _{2}-\lambda _{2}}\frac{1}{2}\left( r_{d}+p_{2}r_{l}\right) \left( 1-\alpha _{1}+\lambda _{1}\right) \\&\qquad +p_{1}\alpha _{1}R_{1}-\left( 1-\lambda _{1}\right) \,\,d\lambda _{1}\,d\lambda _{2}\\&\quad = \frac{1}{2}\alpha _{2}^{2}\left( p_{1}\alpha _{1}R_{1}+\frac{1}{6}\left( r_{d}+p_{2}r_{l}\right) \alpha _{2}\right) \\&\qquad -\int _{-1}^{-1+\alpha _{2}}\int _{-2+\alpha _{1}+\alpha _{2}-\lambda _{2}}^{1}\left( 1-\lambda _{1}\right) \,\,d\lambda _{1}\,d\lambda _{2} \end{aligned}$$

When adding up, four situations need to be distinguished:

• $p_{1}r_{l},\,p_{2}r_{l}<r_{d}$:

  $$\begin{aligned} E^{t=0}\varPi _{1}= & {} E^{t=0}\left( \varPi _{i}\wedge I.\right) +E^{t=0}\left( \varPi _{i}\wedge II.\wedge a\right) \\&+\,E^{t=0}\left( \varPi _{i}\wedge III.\wedge a\right) +E^{t=0}\left( \varPi _{i}\wedge IV.\right) \end{aligned}$$

• $p_{1}r_{l}<r_{d}\le p_{2}r_{l}$:

  $$\begin{aligned} E^{t=0}\varPi _{1}= & {} E^{t=0}\left( \varPi _{i}\wedge I.\right) +E^{t=0}\left( \varPi _{i}\wedge II.\wedge a\right) \\&+\,E^{t=0}\left( \varPi _{i}\wedge III.\wedge b\wedge III.1.\right) +E^{t=0}\left( \varPi _{i}\wedge III.\wedge b\wedge III.2.\right) \\&+\,E^{t=0}\left( \varPi _{i}\wedge IV.\right) \end{aligned}$$

• $p_{2}r_{l}<r_{d}\le p_{1}r_{l}$:

  $$\begin{aligned} E^{t=0}\varPi _{1}= & {} E^{t=0}\left( \varPi _{i}\wedge I.\right) +E^{t=0}\left( \varPi _{i}\wedge II.\wedge b\wedge II.1.\right) \\&+\,E^{t=0}\left( \varPi _{i}\wedge II.\wedge b\wedge II.2.\right) +E^{t=0}\left( \varPi _{i}\wedge III.\wedge a\right) \\&+\,E^{t=0}\left( \varPi _{i}\wedge IV.\right) \end{aligned}$$

• $r_{d}\le p_{1}r_{l},\,p_{2}r_{l}$:

  $$\begin{aligned} E^{t=0}\varPi _{1}= & {} E^{t=0}\left( \varPi _{i}\wedge I.\right) +E^{t=0}\left( \varPi _{i}\wedge II.\wedge b\wedge II.1.\right) \\&+\,E^{t=0}\left( \varPi _{i}\wedge II.\wedge b\wedge II.2.\right) +E^{t=0}\left( \varPi _{i}\wedge III.\wedge b\wedge III.1.\right) \\&+\,E^{t=0}\left( \varPi _{i}\wedge III.\wedge b\wedge III.2.\right) +E^{t=0}\left( \varPi _{i}\wedge IV.\right) \end{aligned}$$

We obtain expected profit at time $t=0$:

• If $p_{1}r_{l}<r_{d},\,p_{2}r_{l}<r_{d},$

  $$\begin{aligned} E^{t=0}\varPi _{1}=p_{1}\alpha _{1}R_{1}+\frac{1}{4}r_{d}\left( 2-\alpha _{1}\right) ^{2}-\frac{1}{4}p_{1}r_{l}\alpha _{1}^{2}-1. \end{aligned}$$

• If $p_{1}r_{l}\ge r_{d}>p_{2}r_{l},\,$

  $$\begin{aligned} E^{t=0}\varPi _{1}= & {} -1+p_{1}\alpha _{1}R_{1}+r_{d}\left( 1-\alpha _{1}\right) \\&-\left( p_{1}r_{l}-r_{d}\right) \frac{1}{8}\alpha _{1}^{2}\left( \frac{1}{6}\alpha _{1}+\frac{1}{2}\alpha _{2}+1\right) . \end{aligned}$$

• If $p_{2}r_{l}\ge r_{d}>p_{1}r_{l},$

  $$\begin{aligned} E^{t=0}\varPi _{1}= & {} -1+p_{1}\alpha _{1}R_{1}+\frac{1}{4}r_{d}\left( 2-\alpha _{1}\right) ^{2}-\frac{1}{4}p_{1}r_{l}\alpha _{1}^{2}\\&+\left( p_{2}r_{l}-r_{d}\right) \frac{1}{8}\alpha _{2}^{2}\left( 1-\frac{1}{2}\alpha _{1}-\frac{1}{6}\alpha _{2}\right) . \end{aligned}$$

• If $p_{1}r_{l}\ge r_{d},p_{2}r_{l}\ge r_{d}$,

  $$\begin{aligned} E^{t=0}\varPi _{1}= & {} -1+p_{1}\alpha _{1}R_{1}+r_{d}\left( 1-\alpha _{1}\right) \\&-\left( p_{1}r_{l}-r_{d}\right) \frac{1}{8}\alpha _{1}^{2}\left( \frac{1}{6}\alpha _{1}+\frac{1}{2}\alpha _{2}+1\right) \\&+\left( p_{2}r_{l}-r_{d}\right) \frac{1}{8}\alpha _{2}^{2}\left( 1-\frac{1}{2}\alpha _{1}-\frac{1}{6}\alpha _{2}\right) . \end{aligned}$$

  A symmetric result holds for bank 2.

Now, the best-response functions and liquidity holding in equilibrium is derived:

• First, consider $r_{d}>p_{1}r_{l},\,p_{2}r_{l}$, i.e., there is no interbank lending at $t=1$. Expected profit at $t=0$ of bank 1 is given by

  $$\begin{aligned} E^{t=0}\varPi _{1}=p_{1}\alpha _{1}R_{1}+\frac{1}{4}r_{d}\left( 2-\alpha _{1}\right) ^{2}-\frac{1}{4}p_{1}r_{l}\alpha _{1}^{2}-1. \end{aligned}$$

  To derive the best-response function, we form the derivative with respect to $\alpha _{1}$

  $$\begin{aligned} \frac{\partial E^{t=0}\varPi _{1}}{\partial \alpha _{1}}= & {} p_{1}R_{1}-\frac{1}{2}r_{d}\left( 2-\alpha _{1}\right) -\frac{1}{2}p_{1}r_{l}\alpha _{1}\\= & {} p_{1}R_{1}-r_{d}+\frac{1}{2}\alpha _{1}\left( r_{d}-p_{1}r_{l}\right) >0 \end{aligned}$$

  which yields

  $$\begin{aligned} \alpha _{1}^{*}=1. \end{aligned}$$

  For reasons of symmetry we also get

  $$\begin{aligned} \alpha _{2}^{*}=1. \end{aligned}$$

• Second, consider $p_{1}r_{l}\ge r_{d}>p_{2}r_{l}$. Expected profit at $t=0$ of bank 1 is given by

  $$\begin{aligned} E^{t=0}\varPi _{1}= & {} -1+p_{1}\alpha _{1}R_{1}+r_{d}\left( 1-\alpha _{1}\right) \\&-\left( p_{1}r_{l}-r_{d}\right) \frac{1}{8}\alpha _{1}^{2}\left( \frac{1}{6}\alpha _{1}+\frac{1}{2}\alpha _{2}+1\right) . \end{aligned}$$

  Differentiating yields

  $$\begin{aligned} \frac{\partial E^{t=0}\varPi _{1}}{\partial \alpha _{1}}= & {} p_{1}R_{1}-r_{d}-\left( p_{1}r_{l}-r_{d}\right) \frac{1}{4}\alpha _{1}\left( \frac{1}{4}\alpha _{1}+\frac{1}{2}\alpha _{2}+1\right) \\= & {} -\frac{1}{16}\left( p_{1}r_{l}-r_{d}\right) \alpha _{1}^{2}-\frac{1}{4}\left( p_{1}r_{l}-r_{d}\right) \alpha _{1}\left( \frac{1}{2}\alpha _{2}+1\right) \\&+\left( p_{1}R_{1}-r_{d}\right) . \end{aligned}$$

  If $p_{1}r_{l}=r_{d},$ the derivative equals $p_{1}R_{1}-r_{d}>0$, so $\alpha _{1}^{*}=1$ holds. Otherwise, the derivative describes a quadratic function with two zeros $\alpha _{1}^{\min }$ and $\alpha _{1}^{\max }$. Since $p_{1}r_{l}>r_{d}$, we know from the expected profit function that $\alpha _{1}^{\min }<\alpha _{1}^{\max }.$ Thus,

  $$\begin{aligned} \alpha _{1}^{\min }= & {} -\left( 2+\alpha _{2}\right) -\sqrt{\left( 2+\alpha _{2}\right) ^{2}+16\frac{p_{1}R_{1}-r_{d}}{p_{1}r_{l}-r_{d}}}<0,\\ \alpha _{1}^{\max }= & {} -\left( 2+\alpha _{2}\right) +\sqrt{\left( 2+\alpha _{2}\right) ^{2}+16\frac{p_{1}R_{1}-r_{d}}{p_{1}r_{l}-r_{d}}}. \end{aligned}$$

  Since $R_{1}>r_{l}$, we have

  $$\begin{aligned} \alpha _{1}^{\max }> & {} -\left( 2+\alpha _{2}\right) +\sqrt{\left( 2+\alpha _{2}\right) ^{2}+16}\\> & {} -\left( 2+\alpha _{2}\right) +\sqrt{\left( 2+\alpha _{2}\right) ^{2}+5+2\alpha _{2}}\\= & {} -\left( 2+\alpha _{2}\right) +\sqrt{\left( 3+\alpha _{2}\right) ^{2}}=1. \end{aligned}$$

  From the now obtained inequalities $\alpha _{1}^{\min }<0<1<\alpha _{1}^{\max }$ we know that the expected profit of bank 1 at $t=0,\ E^{t=0}\varPi _{1},$ must be increasing with respect to $\alpha _{1}$. Thus,

  $$\begin{aligned} \alpha _{1}^{*}=1 \end{aligned}$$

  holds. For reasons of symmetry we can derive $\alpha _{2}^{*}$ from the case $p_{2}r_{l}\ge r_{d}>p_{1}r_{l}$ which is considered next.

• Consider $p_{2}r_{l}\ge r_{d}>p_{1}r_{l}$. Expected profit of bank 1 at $t=0$ is given by

  $$\begin{aligned} E^{t=0}\varPi _{1}= & {} -1+p_{1}\alpha _{1}R_{1}+\frac{1}{4}r_{d}\left( 2-\alpha _{1}\right) ^{2}-\frac{1}{4}p_{1}r_{l}\alpha _{1}^{2}\\&+\left( p_{2}r_{l}-r_{d}\right) \frac{1}{8}\alpha _{2}^{2}\left( 1-\frac{1}{2}\alpha _{1}-\frac{1}{6}\alpha _{2}\right) \end{aligned}$$

  which is a quadratic function in $\alpha _{1}$. Since $r_{d}>p_{1}r_{l}$, the expected profit is convex in $\alpha _{1}$ which yields that the share of the funds invested by bank 1 is equal to zero or equal to one. Moreover, for reasons of symmetry we know from the above case that $\alpha _{2}^{*}=1$. Comparing expected profit at $\alpha _{1}=0$ and $\alpha _{1}=1$ yields:

  $$\begin{aligned} E^{t=0}\varPi _{1}\left( 0,1\right)= & {} -1+r_{d}+\left( p_{2}r_{l}-r_{d}\right) \frac{5}{48}\\ E^{t=0}\varPi _{1}\left( 1,1\right)= & {} -1+p_{1}R_{1}+\frac{1}{4}\left( r_{d}-p_{1}r_{l}\right) +\frac{1}{24}\left( p_{2}r_{l}-r_{d}\right) . \end{aligned}$$

  Therefore,

  $$\begin{aligned} \alpha _{1}^{*}=\left\{ \begin{array}{cc} 0, &{}\quad 16p_{1}R_{1}-11r_{d}-4p_{1}r_{l}-p_{2}r_{l}<0\\ \left\{ 0,1\right\} , &{}\quad 16p_{1}R_{1}-11r_{d}-4p_{1}r_{l}-p_{2}r_{l}=0\\ 1, &{}\quad 16p_{1}R_{1}-11r_{d}-4p_{1}r_{l}-p_{2}r_{l}>0 \end{array}\right. \end{aligned}$$

  and

  $$\begin{aligned} \alpha _{2}^{*}=1. \end{aligned}$$

• Last, we have to consider the case $p_{1}r_{l},\,p_{2}r_{l}\ge r_{d}$. Bank 1 expects the profit

  $$\begin{aligned} E^{t=0}\varPi _{1}= & {} -1+p_{1}\alpha _{1}R_{1}+r_{d}\left( 1-\alpha _{1}\right) \\&-\left( p_{1}r_{l}-r_{d}\right) \frac{1}{8}\alpha _{1}^{2}\left( \frac{1}{6}\alpha _{1}+\frac{1}{2}\alpha _{2}+1\right) \\&+\left( p_{2}r_{l}-r_{d}\right) \frac{1}{8}\alpha _{2}^{2}\left( 1-\frac{1}{2}\alpha _{1}-\frac{1}{6}\alpha _{2}\right) . \end{aligned}$$

  Differentiating with respect to $\alpha _{1}$ yields

  $$\begin{aligned} \frac{\partial E^{t=0}\varPi _{1}}{\partial \alpha _{1}}= & {} p_{1}R_{1}-r_{d}-\left( p_{2}r_{l}-r_{d}\right) \frac{1}{16}\alpha _{2}^{2}\\&-\left( p_{1}r_{l}-r_{d}\right) \frac{1}{4}\alpha _{1}\left( \frac{1}{4}\alpha _{1}+\frac{1}{2}\alpha _{2}+1\right) . \end{aligned}$$

  Assume without loss of generality $p_{1}\ge p_{2}$. Then:

  $$\begin{aligned} \frac{\partial E^{t=0}\varPi _{1}}{\partial \alpha _{1}}\ge & {} p_{1}R_{1}-r_{d}-\left( p_{1}r_{l}-r_{d}\right) \frac{1}{4}\left( \frac{1}{4}\alpha _{1}^{2}+\frac{1}{2}\alpha _{1}\alpha _{2}+\alpha _{1}+\frac{1}{4}\alpha _{2}^{2}\right) \\\ge & {} p_{1}R_{1}-r_{d}-\frac{1}{2}\left( p_{1}r_{l}-r_{d}\right) >0 \end{aligned}$$

  and thus

  $$\begin{aligned} \alpha _{1}^{*}=1. \end{aligned}$$

  Symmetry yields

  $$\begin{aligned} \frac{\partial E^{t=0}\varPi _{2}}{\partial \alpha _{2}}= & {} p_{2}R_{2}-r_{d}-\left( p_{1}r_{l}-r_{d}\right) \frac{1}{16}\alpha _{1}^{2}\\&-\left( p_{2}r_{l}-r_{d}\right) \frac{1}{4}\alpha _{2}\left( \frac{1}{4}\alpha _{2}+\frac{1}{2}\alpha _{1}+1\right) . \end{aligned}$$

  If $p_{2}r_{l}=r_{d}$,

  $$\begin{aligned} \frac{\partial E^{t=0}\varPi _{2}}{\partial \alpha _{2}}=p_{2}R_{2}-r_{d}-\left( p_{1}r_{l}-r_{d}\right) \frac{1}{16}\alpha _{1}^{2} \end{aligned}$$

  and together with $\alpha _{1}^{*}=1$

  $$\begin{aligned} \alpha _{2}^{*}=\left\{ \begin{array}{cc} 0, &{}\quad 16p_{2}R_{2}-15r_{d}-p_{1}r_{l}<0\\ \left[ 0,1\right] , &{}\quad 16p_{2}R_{2}-15r_{d}-p_{1}r_{l}=0\\ 1, &{}\quad 16p_{2}R_{2}-15r_{d}-p_{1}r_{l}>0. \end{array}\right. \end{aligned}$$

  Now, consider $p_{2}r_{l}>r_{d}$. The expected profit of bank 2 at $t=0$ is a cubical function in $\alpha _{2}$. We use $\alpha _{1}^{*}=1$. The first-order condition is equivalent to

  $$\begin{aligned} \alpha _{2}^{2}+6\alpha _{2}-16\frac{\left( p_{2}R_{2}-r_{d}\right) }{\left( p_{2}r_{l}-r_{d}\right) }+\frac{\left( p_{1}r_{l}-r_{d}\right) }{\left( p_{2}r_{l}-r_{d}\right) }\overset{!}{=}0. \end{aligned}$$

  Any solution of the first-order condition is a maximum because

  $$\begin{aligned} \frac{\partial ^{2}E^{t=0}\varPi _{2}}{\partial \alpha _{2}^{2}}=-\left( p_{2}r_{l}-r_{d}\right) \frac{1}{4}\left( \frac{1}{2}\alpha _{2}+\frac{3}{2}\right) <0. \end{aligned}$$

  The profit function has no maximum in $\left( 0,1\right)$ if

  $$\begin{aligned} 9+\frac{1}{p_{2}r_{l}-r_{d}}\left( 16p_{2}R_{2}-15r_{d}-p_{1}r_{l}\right) \le 0. \end{aligned}$$

  In this case the profit function is monotonically decreasing and thus

  $$\begin{aligned} \alpha _{2}^{*}=0. \end{aligned}$$

  Otherwise, the solutions to the first-order condition are given by

  $$\begin{aligned} \alpha _{2}^{\min }= & {} -3-\sqrt{9+\frac{1}{p_{2}r_{l}-r_{d}}\left( 16p_{2}R_{2}-15r_{d}-p_{1}r_{l}\right) }<0\\ \alpha _{2}^{\max }= & {} -3+\sqrt{9+\frac{1}{p_{2}r_{l}-r_{d}}\left( 16p_{2}R_{2}-15r_{d}-p_{1}r_{l}\right) }. \end{aligned}$$

  We check the conditions for $\alpha _{2}^{\max }\in \left[ 0,1\right]$ and obtain, with eliminating the redundant inequalities,

  $$\begin{aligned} \alpha _{2}^{*}=\left\{ \begin{array}{ll} 0, &{}\quad 16p_{2}R_{2}-15r_{d}-p_{1}r_{l}<0\\ 1, &{}\quad 16p_{2}R_{2}-7p_{2}r_{l}-8r_{d}-p_{1}r_{l}>0\\ \alpha _{2}^{\max }, &{}\quad \text {otherwise} \end{array}\right. \end{aligned}$$

  which proofs the claim.

Proof of Corollary 2

Since $\alpha _{2}^{*}=1$ if $r_{d}>p_{1}r_{l}$, (ii) is clear. Therefore, now concentrate on $r_{d}\le p_{1}r_{l}$. Note that liquidity holding in equilibrium $1-\alpha _{2}^{*}$ is increasing and convex in $r_{l}$ if $\alpha _{2}^{*}$ is decreasing and concave in $r_{l}.$

First, consider the case that liquidity holding of bank 2 in equilibrium is constantly equal to zero if $r_{d}<p_{2}r_{l}$, i.e.,

$$\begin{aligned} \frac{1}{8}\left( 16p_{2}R_{2}-7p_{2}r_{l}-p_{1}r_{l}\right) \ge p_{2}r_{l}. \end{aligned}$$

A short calculation shows that this condition is equivalent to

$$\begin{aligned} p_{2}r_{l}\le & {} \frac{1}{15}\left( 16p_{2}R_{2}-p_{1}r_{l}\right) \text { and to}\\ p_{2}r_{l}\le & {} \frac{1}{11}\left( 16p_{2}R_{2}-p_{1}r_{l}-4p_{2}r_{l}\right) . \end{aligned}$$

Thus, in case equality holds in the above conditions we get

$$\begin{aligned} \alpha _{2}^{*}=\left\{ \begin{array}{cc} 1 &{}\quad \text {if }p_{2}r_{l}>r_{d}\\ \left[ 0,1\right] , &{}\quad \text {if }p_{2}r_{l}=r_{d}\\ 0, &{}\quad \text {if }p_{2}r_{l}<r_{d}, \end{array}\right. \end{aligned}$$

and in case equality does not hold we have $\alpha _{2}^{*}=1$ if $p_{2}r_{l}=r_{d}$ which proves the claim.

Second, consider

$$\begin{aligned} \frac{1}{8}\left( 16p_{2}R_{2}-7p_{2}r_{l}-p_{1}r_{l}\right) <p_{2}r_{l}, \end{aligned}$$

meaning that liquidity holding is not constantly equal to zero if $r_{d}<p_{2}r_{l}$. Again the inequality is equivalent to

$$\begin{aligned} p_{2}r_{l}> & {} \frac{1}{15}\left( 16p_{2}R_{2}-p_{1}r_{l}\right) \text { and to}\\ p_{2}r_{l}> & {} \frac{1}{11}\left( 16p_{2}R_{2}-p_{1}r_{l}-4p_{2}r_{l}\right) . \end{aligned}$$

This immediately yields $\alpha _{2}^{*}=0$ in case $r_{d}\ge p_{2}r_{l}$. It remains to be shown that $\alpha _{2}^{*}$ is decreasing and concave if $p_{2}r_{l}>r_{d},\,\ 16p_{2}R_{2}-15r_{d}-p_{1}r_{l}\ge 0$ and $16p_{2}R_{2}-8r_{d}-p_{1}r_{l}-7p_{2}r_{l}\le 0$. In this case

$$\begin{aligned} \alpha _{2}^{*}=-3+\sqrt{9+\frac{1}{p_{2}r_{l}-r_{d}}\left( 16p_{2}R_{2}-15r_{d}-p_{1}r_{l}\right) }=:-3+\sqrt{x}. \end{aligned}$$

The derivative with respect to $r_{d}$ is given by

$$\begin{aligned} \frac{\partial \alpha _{2}^{*}}{\partial r_{d}}=\frac{16p_{2}R_{2}-p_{1}r_{l}-15p_{2}r_{l}}{2\sqrt{x}\left( p_{2}r_{l}-r_{d}\right) ^{2}}. \end{aligned}$$

The conditions $16p_{2}R_{2}-8r_{d}-p_{1}r_{l}-7p_{2}r_{l}\le 0$ and $p_{2}r_{l}>r_{d}$ imply that the numerator and thus the whole derivative is negative. The share of the funds that is invested in equilibrium is monotonically decreasing with respect to $r_{d}$. We form the second derivative:

$$\begin{aligned} \frac{\partial ^{2}\alpha _{2}^{*}}{\partial r_{d}}= & {} \frac{-\left( 16p_{2}R_{2}-p_{1}r_{l}-15p_{2}r_{l}\right) }{2x\left( p_{2}r_{l}-r_{d}\right) ^{4}}\\&\cdot \left( \frac{\partial \alpha _{2}^{*}}{\partial r_{d}}\left( p_{2}r_{l}-r_{d}\right) ^{2}-2\sqrt{x}\left( p_{2}r_{l}-r_{d}\right) \right) \\< & {} 0 \end{aligned}$$

and obtain that $\alpha _{2}^{*}$ is concave with respect to $r_{d}$.

Proof of Corollary 3

If $r_{l}$ is small ($p_{1}r_{l}<r_{d}$), then $\alpha _{2}^{*}=1$ and thus liquidity holding is equal to zero. It is therefore sufficient to concentrate on the section $p_{1}r_{l}\ge r_{d}$.

First, assume that $\alpha _{2}^{*}$ is constantly equal to zero in case $p_{2}r_{l}>r_{d},$ i.e.,

$$\begin{aligned} \frac{16p_{2}R_{2}-15r_{d}}{p_{1}}\le \frac{r_{d}}{p_{2}}. \end{aligned}$$

A short calculation shows that this condition is equivalent to

$$\begin{aligned} \frac{16p_{2}R_{2}-11r_{d}}{p_{1}+4p_{2}}\le & {} \frac{r_{d}}{p_{2}}\text {.} \end{aligned}$$

Thus, in case equality holds in the above conditions we get

$$\begin{aligned} \alpha _{2}^{*}=\left\{ \begin{array}{cc} \left[ 0,1\right] , &{}\quad \text {if }p_{2}r_{l}=r_{d}\\ 1, &{}\quad \text {if }p_{2}r_{l}<r_{d}, \end{array}\right. \end{aligned}$$

and in case equality does not hold we have $\alpha _{2}^{*}=0$ if $p_{2}r_{l}=r_{d}$ which proves the claim.

Second, consider

$$\begin{aligned} \frac{16p_{2}R_{2}-15r_{d}}{p_{1}}>\frac{r_{d}}{p_{2}}, \end{aligned}$$

meaning that liquidity holding is not constantly equal to zero if $r_{d}<p_{2}r_{l}$. Again the inequality is equivalent to

$$\begin{aligned} \frac{16p_{2}R_{2}-11r_{d}}{p_{1}+4p_{2}}> & {} \frac{r_{d}}{p_{2}}. \end{aligned}$$

This immediately yields $\alpha _{2}^{*}=1$ in case $r_{d}\ge p_{2}r_{l}$. It remains to be shown that $\alpha _{2}^{*}$ is decreasing if $p_{2}r_{l}>r_{d},\,\ 16p_{2}R_{2}-15r_{d}-p_{1}r_{l}\ge 0$ and $16p_{2}R_{2}-8r_{d}-p_{1}r_{l}-7p_{2}r_{l}\le 0$. In this case we have

$$\begin{aligned} \alpha _{2}^{*}=-3+\sqrt{9+\frac{1}{p_{2}r_{l}-r_{d}}\left( 16p_{2}R_{2}-15r_{d}-p_{1}r_{l}\right) }=:-3+\sqrt{x}. \end{aligned}$$

The derivative with respect to $r_{l}$ is given by

$$\begin{aligned} \frac{\partial \alpha _{2}^{*}}{\partial r_{l}}=\frac{-p_{1}\left( p_{2}r_{l}-r_{d}\right) -\left( 16p_{2}R_{2}-15r_{d}-p_{1}r_{l}\right) p_{2}}{2\sqrt{x}\left( p_{2}r_{l}-r_{d}\right) ^{2}}. \end{aligned}$$

The conditions $16p_{2}R_{2}-15r_{d}-p_{1}r_{l}\ge 0$ and $p_{2}r_{l}>r_{d}$ imply that the numerator and thus the whole derivative is negative.

Proof of Corollary 4

First, consider changes in $r_{d}$. If liquidity holding of bank 2 in equilibrium is equal to zero for all $r_{d}$ satisfying $p_{2}r_{l}<r_{d}\le p_{1}r_{l}$, liquidity holding is constantly equal to zero for all $r_{d}$ according to Corollary 2. Therefore, it is sufficient to concentrate on the case $p_{2}r_{l}<r_{d}\le p_{1}r_{l}$. A necessary and sufficient condition for liquidity holding being constantly equal to zero for all $p_{2}r_{l}<r_{d}\le p_{1}r_{l}$ is given by

$$\begin{aligned} 16p_{2}R_{2}-12p_{1}r_{l}-4p_{2}r_{l}>0 \end{aligned}$$

which is equivalent to

$$\begin{aligned} 16p_{2}\underset{>0}{\underbrace{\left( R_{2}-r_{l}\right) }}+12r_{l}\underset{\le 0}{\underbrace{\left( p_{2}-p_{1}\right) }}>0. \end{aligned}$$

Therefore, liquidity holding is constantly equal to zero if $p_{1}$ and $p_{2}$ are sufficiently close, meaning that the second summand is not too small.

Second, consider changes in $r_{l}$. If liquidity holding of bank 2 in equilibrium is equal to zero for all $r_{l}$ satisfying $r_{d}<p_{2}r_{l}\le p_{1}r_{l}$, liquidity holding is constantly equal to zero for all $r_{l}$ according to Corollary 3. Therefore, it is sufficient to concentrate on the case $r_{d}<p_{2}r_{l}\le p_{1}r_{l}$. A necessary and sufficient condition for liquidity holding being constantly equal to zero is given by

$$\begin{aligned} 16p_{2}R_{2}-8r_{d}-7p_{2}r_{l}-p_{1}r_{l}>0\text { for all } r_{d}<p_{2}r_{l}. \end{aligned}$$

We use $r_{l}<R_{2}$ and get the condition

$$\begin{aligned} 8\underbrace{\left( p_{2}R_{2}-r_{d}\right) }_{>0}+R_{2}\underbrace{\left( p_{2}-p_{1}\right) }_{\le 0}>0. \end{aligned}$$

Therefore, liquidity holding is constantly equal to zero if $p_{1}$ and $p_{2}$ are sufficiently close, meaning that the second summand is not too small.

Robustness check: interbank rate and ex ante contracting

In our model the interbank rate is determined by bilateral bargaining at $t=1$. More precisely, $r_{IB}$ is determined by the Nash bargaining solution. While it is consistent with empirical evidence that the interbank rate is determined bilaterally, one may question the use of the Nash bargaining solution. In particular, it is also plausible that the lender makes a “take it or leave it” offer. This yields an interbank rate that maximizes the lender’s expected profit. More generally, the banks’ bargaining power must not necessarily be equal as previously assumed.

Note that Lemma 1 holds also if the Nash bargaining solution is not used. Assume again without loss of generality that bank 1 needs liquidity, $1-\alpha _{1}+\lambda _{1}<0$, while bank 2 does not, $1-\alpha _{2}+\lambda _{2}\ge 0$. Any bilateral process that results in an agreement to engage in interbank lending yields an interbank rate satisfying

$$\begin{aligned} \frac{r_{d}}{p_{1}}\le r_{IB}\le r_{l}-\frac{1}{L_{1}}\left( r_{l}\left( -1+\alpha _{1}-\lambda _{1}\right) -\alpha _{1}R_{1}\right) ^{+} \end{aligned}$$

If $\frac{r_{d}}{p_{1}}>r_{IB}$ holds, bank 2’s expected return from supplying liquidity at the interbank market, $p_{1}r_{IB}$, is smaller than bank 2’s expected return from depositing liquidity at the central bank, $r_{d}$. Hence, bank 2 does not supply liquidity at the interbank market. If $r_{IB}>r_{l}$ holds, bank 1’s repayment rate for borrowing from the interbank market, $r_{IB}$, is larger than the repayment rate for borrowing from the central bank, $r_{l}$, and bank 1 does not demand liquidity from the interbank market. If $r_{IB}>r_{l}-\frac{1}{L_{1}}\left( r_{l}\left( -1+\alpha _{1}-\lambda _{1}\right) -\alpha _{1}R_{1}\right)$ with $r_{l}\left( -1+\alpha _{1}-\lambda _{1}\right) >\alpha _{1}R_{1}$ the de facto interbank rate is still $r_{l}-\frac{1}{L_{1}}\left( r_{l}\left( -1+\alpha _{1}-\lambda _{1}\right) -\alpha _{1}R_{1}\right)$ because bank 1 is not able to repay a higher amount. Therefore, lending volume on the interbank market between bank 1 and 2 is given by

$$\begin{aligned} {\left\{ \begin{array}{ll} \min \left\{ -1+\alpha _{1}-\lambda _{1},1-\alpha _{2}+\lambda _{2}\right\} , &{} \frac{r_{d}}{p_{1}}\le r_{IB}\le r_{l}-\frac{1}{L_{1}}\left( r_{l}\left( -1+\alpha _{1}-\lambda _{1}\right) -\alpha _{1}R_{1}\right) ^{+}\\ 0, &{} \text {otherwise}. \end{array}\right. } \end{aligned}$$

Assuming that the two banks interact at the interbank market, expected profits are given by (4) and (5), respectively. Hence, the lender’s profit is maximized by the highest possible interbank rate that does not prevent interaction, i.e.,

$$\begin{aligned} r_{IB}=r_{l}-\frac{1}{L_{1}}\left( r_{l}\left( -1+\alpha _{1}-\lambda _{1}\right) -\alpha _{1}R_{1}\right) ^{+}. \end{aligned}$$

The borrower’s profit is maximized by the lowest possible interbank rate that does not prevent interaction, $r_{IB}=\frac{r_{d}}{p_{1}}$. More generally, any interbank rate that does not prevent interaction on the interbank market is given by

$$\begin{aligned} \tilde{r}_{IB}:=\gamma \frac{r_{d}}{p_{1}}+\left( 1-\gamma \right) \left[ r_{l}-\frac{1}{L_{1}}\left( r_{l}\left( -1+\alpha _{1}-\lambda _{1}\right) -\alpha _{1}R_{1}\right) ^{+}\right] \end{aligned}$$

for some $\gamma \in \left[ 0,1\right]$. We show that Proposition 1 and Corollaries 2, 3 and 4 can be transferred to this more general case. The proofs proceed exactly analogous. As before, we assume $r_{l}<R_{i},i\in \left\{ 1,2\right\}$ which guarantees that bank i’s project can be financed completely through the central bank. Let us start with Proposition 1. If the interbank rate is given by $\tilde{r}_{IB}$, the equivalent of Proposition 1 is as follows:

Assume$p_{1}\ge p_{2}$holds. The bank with the less risky investment holds no liquidity, i.e.,$\alpha _{1}^{*}=1$. The liquidity holding if the bank with the riskier project can be described in the following way:

(a):

If$r_{d}>p_{1}r_{l}\ge p_{2}r_{l}$,

$$\begin{aligned} \alpha _{2}^{*}=1. \end{aligned}$$

(b):

If$p_{1}r_{l}\ge r_{d}>p_{2}r_{l}$,

$$\begin{aligned} \alpha _{2}^{*}={\left\{ \begin{array}{ll} 0, &{} \text {if }8p_{2}R_{2}-\left( 5+\gamma \right) r_{d}-2p_{2}r_{l}-\left( 1-\gamma \right) p_{1}r_{l}<0\\ \left\{ 0,1\right\} , &{} \text {if }8p_{2}R_{2}-\left( 5+\gamma \right) r_{d}-2p_{2}r_{l}-\left( 1-\gamma \right) p_{1}r_{l}=0\\ 1, &{} \text {if }8p_{2}R_{2}-\left( 5+\gamma \right) r_{d}-2p_{2}r_{l}-\left( 1-\gamma \right) p_{1}r_{l}>0 \end{array}\right. } \end{aligned}$$

(c):

If$p_{1}r_{l}\ge p_{2}r_{l}=r_{d}$,

$$\begin{aligned} \alpha _{2}^{*}={\left\{ \begin{array}{ll} 0, &{} 8p_{2}R_{2}-\left( 7+\gamma \right) r_{d}-\left( 1-\gamma \right) p_{1}r_{l}<0,\\ \left[ 0,1\right] , &{} 8p_{2}R_{2}-\left( 7+\gamma \right) r_{d}-\left( 1-\gamma \right) p_{1}r_{l}=0,\\ 1, &{} 8p_{2}R_{2}-\left( 7+\gamma \right) r_{d}-\left( 1-\gamma \right) p_{1}r_{l}>0. \end{array}\right. } \end{aligned}$$

(d):

If$p_{1}r_{l}\ge p_{2}r_{l}>r_{d}$,

$$\begin{aligned} \alpha _{2}^{*}={\left\{ \begin{array}{ll} 0, &{} 8p_{2}R_{2}-7r_{d}-p_{1}r_{l}<0,\\ 1, &{} 8p_{2}R_{2}-3r_{d}-p_{1}r_{l}-4p_{2}r_{l}>0,\\ \frac{8p_{2}R_{2}-7r_{d}-p_{1}r_{l}}{4\left( p_{2}r_{l}-r_{d}\right) }, &{} \text {otherwise} \end{array}\right. } \end{aligned}$$

for$\gamma =0$and

$$\begin{aligned} \alpha _{2}^{*}=\left\{ \begin{array}{l} 0,\qquad \qquad \qquad \ \ 8p_{2}R_{2}-\left( 7+\gamma \right) r_{d}-\left( 1-\gamma \right) p_{1}r_{l}<0\\ 1,\qquad \qquad \qquad \ \ 8p_{2}R_{2}-\left( 3+2\gamma \right) r_{d}-\left( 1-\gamma \right) p_{1}r_{l}-\left( 4-\gamma \right) p_{2}r_{l}>0\\ -\frac{2-\gamma }{\gamma }+\sqrt{\frac{\left( 2-\gamma \right) ^{2}}{\gamma ^{2}}+\frac{8}{\gamma }\frac{p_{2}R_{2}-r_{d}}{p_{2}r_{l}-r_{d}}-\frac{1-\gamma }{\gamma }\frac{p_{1}r_{l}-r_{d}}{p_{2}r_{l}-r_{d}}},\,\text {otherwise} \end{array}\right. \end{aligned}$$

for$\gamma \ne 0$.

Proof

We distinguish four different situations

• 1. $p_{1}r_{l},p_{2}r_{l}<r_{d}:$ Expected profit of bank 1 is given by

  $$\begin{aligned} E^{t=0}\varPi _{1}=-1+p_{1}\alpha _{1}R_{1}+r_{d}\left( 1-\alpha _{1}+\frac{1}{4}\alpha _{1}^{2}\right) -\frac{1}{4}\alpha _{1}^{2}p_{1}r_{l}. \end{aligned}$$

  Hence, we have

  $$\begin{aligned} \frac{\partial E^{t=0}\varPi _{1}}{\partial \alpha _{1}}=p_{1}R_{1}-r_{d}+\frac{1}{2}\alpha _{1}\left( r_{d}-p_{1}r_{l}\right) >0 \end{aligned}$$

  which yields $\alpha _{1}^{*}=1$ and for reasons of symmetry also $\alpha _{2}^{*}=1$.

• 2. $p_{1}r_{l}\ge r_{d}>p_{2}r_{l}$: Expected profit of bank 1 is given by

  $$\begin{aligned} E^{t=0}\varPi _{1}=-1+p_{1}\alpha _{1}R_{1}+r_{d}\left( 1-\alpha _{1}\right) -\left( p_{1}r_{l}-r_{d}\right) \frac{1}{4}\alpha _{1}^{2}\left( \left( 1-\gamma \right) +\gamma \left( \frac{\alpha _{1}}{6}+\frac{\alpha _{2}}{2}\right) \right) . \end{aligned}$$

  Hence, we have

  $$\begin{aligned} \frac{\partial E^{t=0}\varPi _{1}}{\partial \alpha _{1}}&=p_{1}R_{1}-r_{d}-\left( p_{1}r_{l}-r_{d}\right) \underbrace{\left( \left( 1-\gamma \right) \frac{\alpha _{1}}{2}+\gamma \left( \frac{\alpha _{1}^{2}}{8}+\frac{\alpha _{1}\alpha _{2}}{4}\right) \right) }_{<1}\\&>p_{1}R_{1}-p_{1}r_{l}>0 \end{aligned}$$

  which yields $\alpha _{1}^{*}=1$. For reasons of symmetry we can derive $\alpha _{2}^{*}$ from the case $p_{2}r_{l}\ge r_{d}>p_{1}r_{l}$ which is considered next.

• 3. $p_{2}r_{l}\ge r_{d}>p_{1}r_{l}$: Expected profit of bank 1 is given by

  $$\begin{aligned} E^{t=0}\varPi _{1}= & {} -1+p_{1}\alpha _{1}R_{1}+\frac{r_{d}}{4}\left( 2-\alpha _{1}\right) ^{2}\\&+\,\frac{1-\gamma }{4}\alpha _{2}^{2}\left( \frac{\alpha _{2}}{6}+\frac{\alpha _{1}}{2}-1\right) \left( r_{d}-p_{2}r_{l}\right) -p_{1}r_{l}\frac{\alpha _{1}^{2}}{4}. \end{aligned}$$

  Hence, we have

  $$\begin{aligned} \frac{\partial E^{t=0}\varPi _{1}}{\partial \alpha _{1}}=p_{1}R_{1}-\frac{r_{d}}{2}\left( 2-\alpha _{1}\right) +\frac{1-\gamma }{8}\alpha _{2}^{2}\left( r_{d}-p_{2}r_{l}\right) -p_{1}r_{l}\frac{\alpha _{1}}{2} \end{aligned}$$

  and

  $$\begin{aligned} \frac{\partial ^{2}E^{t=0}\varPi _{1}}{\partial \alpha _{1}^{2}}=\frac{1}{2}\left( r_{d}-p_{1}r_{l}\right) >0. \end{aligned}$$

  The maximum $\alpha _{1}^{*}$ is either 0 or 1. Using $\alpha _{2}^{*}=1$ (obtained from the previous case), we get

  $$\begin{aligned} E^{t=0}\varPi _{1}\left( 0,1\right)&=-1+r_{d}+\frac{5}{24}\left( 1-\gamma \right) \left( p_{2}r_{l}-r_{d}\right) ,\\ E^{t=0}\varPi _{1}\left( 1,1\right)&=-1+p_{1}R_{1}+\frac{1}{4}\left( r_{d}-p_{1}r_{l}\right) +\frac{1}{12}\left( 1-\gamma \right) \left( p_{2}r_{l}-r_{d}\right) . \end{aligned}$$

  This yields

  $$\begin{aligned} \alpha _{1}^{*}={\left\{ \begin{array}{ll} 0, &{} \text {if }8p_{1}R_{1}-\left( 5+\gamma \right) r_{d}-2p_{1}r_{l}-\left( 1-\gamma \right) p_{2}r_{l}<0,\\ \left\{ 0,1\right\} , &{} \text {if }8p_{1}R_{1}-\left( 5+\gamma \right) r_{d}-2p_{1}r_{l}-\left( 1-\gamma \right) p_{2}r_{l}=0,\\ 1, &{} \text {if }8p_{1}R_{1}-\left( 5+\gamma \right) r_{d}-2p_{1}r_{l}-\left( 1-\gamma \right) p_{2}r_{l}>0. \end{array}\right. } \end{aligned}$$

• 5. $p_{1}r_{l},p_{2}r_{l}\ge r_{d}$: Without loss of generality assume $p_{1}\ge p_{2}$. Expected profit of bank 1 is given by

  $$\begin{aligned} E^{t=0}\varPi _{1}=&-1+p_{1}\alpha _{1}R_{1}+r_{d}\left( 1-\alpha _{1}\right) +\left( 1-\gamma \right) \left( r_{d}-p_{1}r_{l}\right) \frac{\alpha _{1}^{2}}{4}\\&+\gamma \left( r_{d}-p_{1}r_{l}\right) \frac{\alpha _{1}^{2}}{4}\left( \frac{\alpha _{1}}{6}+\frac{\alpha _{2}}{2}\right) +\left( p_{2}r_{l}-r_{d}\right) \left( 1-\frac{\alpha _{2}}{6}-\frac{\alpha _{1}}{2}\right) \left( 1-\gamma \right) \frac{\alpha _{2}^{2}}{4}. \end{aligned}$$

  Hence, we have

  $$\begin{aligned} \frac{\partial E^{t=0}\varPi _{1}}{\partial \alpha _{1}}&= p_{1}R_{1}-r_{d}-\left( 1-\gamma \right) \left( p_{1}r_{l}-r_{d}\right) \frac{\alpha _{1}}{2}-\frac{\gamma }{4}\left( \frac{\alpha _{1}^{2}}{2}+\alpha _{1}\alpha _{2}\right) \left( p_{1}r_{l}-r_{d}\right) \\&\quad -\left( p_{2}r_{l}-r_{d}\right) \left( 1-\gamma \right) \frac{\alpha _{2}^{2}}{8}\\&> \left( p_{1}R_{1}-r_{d}\right) -\left( p_{1}r_{l}-r_{d}\right) >0 \end{aligned}$$

  which yields $\alpha _{1}^{*}=1$. Using $\alpha _{1}^{*}=1$, we obtain for bank 2

  $$\begin{aligned} \frac{\partial E^{t=0}\varPi _{2}}{\partial \alpha _{2}}&= p_{2}R_{2}-r_{d}-\left( 1-\gamma \right) \left( p_{2}r_{l}-r_{d}\right) \frac{\alpha _{2}}{2}-\frac{\gamma }{4}\left( \frac{\alpha _{2}^{2}}{2}+\alpha _{2}\right) \left( p_{2}r_{l}-r_{d}\right) \\&\quad -\left( p_{1}r_{l}-r_{d}\right) \frac{\left( 1-\gamma \right) }{8} \end{aligned}$$

  and

  $$\begin{aligned} \frac{\partial ^{2}E^{t=0}\varPi _{2}}{\partial \alpha _{2}^{2}}=-\frac{\gamma }{4}\left( 1+\alpha _{2}\right) \left( p_{2}r_{l}-r_{d}\right) -\frac{1-\gamma }{2}\left( p_{2}r_{l}-r_{d}\right) <0. \end{aligned}$$

  If $p_{2}r_{l}=r_{d}$, we have

  $$\begin{aligned} \frac{\partial E^{t=0}\varPi _{2}}{\partial \alpha _{2}}&= p_{2}R_{2}-r_{d}-\left( p_{1}r_{l}-r_{d}\right) \frac{\left( 1-\gamma \right) }{8} \end{aligned}$$

  and hence

  $$\begin{aligned} \alpha _{2}^{*}={\left\{ \begin{array}{ll} 0, &{} 8p_{2}R_{2}-\left( 7+\gamma \right) r_{d}-\left( 1-\gamma \right) p_{1}r_{l}<0,\\ \left[ 0,1\right] , &{} 8p_{2}R_{2}-\left( 7+\gamma \right) r_{d}-\left( 1-\gamma \right) p_{1}r_{l}=0,\\ 1, &{} 8p_{2}R_{2}-\left( 7+\gamma \right) r_{d}-\left( 1-\gamma \right) p_{1}r_{l}>0. \end{array}\right. } \end{aligned}$$

  If $p_{2}r_{l}>r_{d}$ and $\gamma =0$, we have

  $$\begin{aligned} \frac{\partial E^{t=0}\varPi _{2}}{\partial \alpha _{2}}=&p_{2}R_{2}-r_{d}-\left( p_{2}r_{l}-r_{d}\right) \frac{\alpha _{2}}{2}-\left( p_{1}r_{l}-r_{d}\right) \frac{1}{8} \end{aligned}$$

  and hence

  $$\begin{aligned} \alpha _{2}^{*}={\left\{ \begin{array}{ll} 0, &{} 8p_{2}R_{2}-7r_{d}-p_{1}r_{l}<0,\\ 1, &{} 8p_{2}R_{2}-3r_{d}-p_{1}r_{l}-4p_{2}r_{l}>0,\\ \frac{8p_{2}R_{2}-7r_{d}-p_{1}r_{l}}{4\left( p_{2}r_{l}-r_{d}\right) }, &{} \text {otherwise}. \end{array}\right. } \end{aligned}$$

  If otherwise $p_{2}r_{l}>r_{d}$ and $\gamma >0$, the first-order condition is given by

  $$\begin{aligned} \alpha _{2}^{2}+\alpha _{2}\frac{2\left( 2-\gamma \right) }{\gamma }-\frac{8}{\gamma }\frac{p_{2}R_{2}-r_{d}}{p_{2}r_{l}-r_{d}}+\frac{1-\gamma }{\gamma }\frac{p_{1}r_{l}-r_{d}}{p_{2}r_{l}-r_{d}}=0 \end{aligned}$$

  This equation has no solution if

  $$\begin{aligned} \left( 2-\gamma \right) ^{2}\left( p_{2}r_{l}-r_{d}\right) +\gamma \left[ 8p_{2}R_{2}-\left( 7+\gamma \right) r_{d}-\left( 1-\gamma \right) p_{1}r_{l}\right] <0, \end{aligned}$$

  in this case expected profit is decreasing in $\alpha _{2}$ and $\alpha _{2}^{*}=0$. Otherwise, solutions to the first-order condition are given by $\alpha _{2}^{\min }<\alpha _{2}^{\max }$:

  $$\begin{aligned} \alpha _{2}^{\min }&=-\frac{2-\gamma }{\gamma }-\sqrt{\frac{\left( 2-\gamma \right) ^{2}}{\gamma ^{2}}+\frac{8}{\gamma }\frac{p_{2}R_{2}-r_{d}}{p_{2}r_{l}-r_{d}}-\frac{1-\gamma }{\gamma }\frac{p_{1}r_{l}-r_{d}}{p_{2}r_{l}-r_{d}}}<0\\ \alpha _{2}^{\max }&=-\frac{2-\gamma }{\gamma }+\sqrt{\frac{\left( 2-\gamma \right) ^{2}}{\gamma ^{2}}+\frac{8}{\gamma }\frac{p_{2}R_{2}-r_{d}}{p_{2}r_{l}-r_{d}}-\frac{1-\gamma }{\gamma }\frac{p_{1}r_{l}-r_{d}}{p_{2}r_{l}-r_{d}}} \end{aligned}$$

  We check the conditions for $\alpha _{2}^{\max }\in \left[ 0,1\right]$ and obtain, with eliminating the redundant inequalities,

  $$\begin{aligned} \alpha _{2}^{*}=\left\{ \begin{array}{l} 0,\qquad {\qquad \qquad \ \ }8p_{2}R_{2}-\left( 7+\gamma \right) r_{d}-\left( 1-\gamma \right) p_{1}r_{l}<0\\ 1,\qquad {\qquad \qquad \ \ }8p_{2}R_{2}-\left( 3+2\gamma \right) r_{d}-\left( 1-\gamma \right) p_{1}r_{l}-\left( 4-\gamma \right) p_{2}r_{l}>0\\ -\frac{2-\gamma }{\gamma }-\sqrt{\frac{\left( 2-\gamma \right) ^{2}}{\gamma ^{2}}+\frac{8}{\gamma }\frac{p_{2}R_{2}-r_{d}}{p_{2}r_{l}-r_{d}}-\frac{1-\gamma }{\gamma }\frac{p_{1}r_{l}-r_{d}}{p_{2}r_{l}-r_{d}}},\,\text {otherwise} \end{array}\right. \end{aligned}$$

  which proves the claim.

$\square$

We proceed with proving Corollaries 2, 3 and 4 that can be transferred one to one.

Proof of Corollary 2

Since $\alpha _{2}^{*}=1$ if $r_{d}>p_{1}r_{l}$, (ii) is clear. Therefore, concentrate on $r_{d}\le p_{1}r_{l}$.

First case: $\alpha _{2}^{*}=1$ for all $r_{d}<p_{2}r_{l}$, i.e.,

$$\begin{aligned} \frac{1}{3+2\gamma }\left( 8p_{2}R_{2}-\left( 4-\gamma \right) p_{2}r_{l}-\left( 1-\gamma \right) p_{1}r_{l}\right) \ge p_{2}r_{l}. \end{aligned}$$

A short calculation shows that this condition is equivalent to

$$\begin{aligned} p_{2}r_{l}\le & {} \frac{1}{7+\gamma }\left( 8p_{2}R_{2}-\left( 1-\gamma \right) p_{1}r_{l}\right) \text { and to}\\ p_{2}r_{l}\le & {} \frac{1}{5+\gamma }\left( 8p_{2}R_{2}-\left( 1-\gamma \right) p_{1}r_{l}-2p_{2}r_{l}\right) . \end{aligned}$$

Thus, in case equality holds in the above conditions, we get

$$\begin{aligned} \alpha _{2}^{*}=\left\{ \begin{array}{cc} 1 &{}\quad \text {if }p_{2}r_{l}>r_{d}\\ \left[ 0,1\right] , &{}\quad \text {if }p_{2}r_{l}=r_{d}\\ 0, &{}\quad \text {if }p_{2}r_{l}<r_{d}, \end{array}\right. \end{aligned}$$

and, in case the equality does not hold, we have $\alpha _{2}^{*}=1$ if $p_{2}r_{l}=r_{d}$ which proves the claim.

Second case:

$$\begin{aligned} \frac{1}{3+2\gamma }\left( 8p_{2}R_{2}-\left( 4-\gamma \right) p_{2}r_{l}-\left( 1-\gamma \right) p_{1}r_{l}\right) <p_{2}r_{l}, \end{aligned}$$

Again the inequality is equivalent to

$$\begin{aligned} p_{2}r_{l}> & {} \frac{1}{7+\gamma }\left( 8p_{2}R_{2}-\left( 1-\gamma \right) p_{1}r_{l}\right) \text { and to}\\ p_{2}r_{l}> & {} \frac{1}{5+\gamma }\left( 8p_{2}R_{2}-\left( 1-\gamma \right) p_{1}r_{l}-2p_{2}r_{l}\right) . \end{aligned}$$

This immediately yields $\alpha _{2}^{*}=0$ in case $r_{d}\ge p_{2}r_{l}$. It remains to be shown that $\alpha _{2}^{*}$ is decreasing and concave if $p_{2}r_{l}>r_{d},\,\ 8p_{2}R_{2}-\left( 7+\gamma \right) r_{d}-\left( 1-\gamma \right) p_{1}r_{l}\ge 0$ and $8p_{2}R_{2}-\left( 3+2\gamma \right) r_{d}-\left( 1-\gamma \right) p_{1}r_{l}-\left( 4-\gamma \right) p_{2}r_{l}\le 0$. This is easily verified for $\gamma =0$ as well as for $\gamma >0$ by considering the first and the second derivative of $\alpha _{2}^{*}$ with respect to $r_{d}$. $\square$

Proof of Corollary 3

If $r_{l}$ is small ($p_{1}r_{l}<r_{d}$), then $\alpha _{2}^{*}=1$ and thus liquidity holding is equal to zero. Thus, concentrate on $p_{1}r_{l}\ge r_{d}$.

First case: $\alpha _{2}^{*}=0$ for all $p_{2}r_{l}>r_{d},$ i.e.

$$\begin{aligned} \frac{8p_{2}R_{2}-\left( 7+\gamma \right) r_{d}}{\left( 1-\gamma \right) p_{1}}\le \frac{r_{d}}{p_{2}}. \end{aligned}$$

This condition is equivalent to

$$\begin{aligned} \frac{8p_{2}R_{2}-\left( 5+\gamma \right) r_{d}}{\left( 1-\gamma \right) p_{1}+2p_{2}}\le & {} \frac{r_{d}}{p_{2}}\text {.} \end{aligned}$$

Thus, in case equality holds in the above conditions, we get

$$\begin{aligned} \alpha _{2}^{*}=\left\{ \begin{array}{cc} \left[ 0,1\right] , &{}\quad \text {if }p_{2}r_{l}=r_{d}\\ 1, &{}\quad \text {if }p_{2}r_{l}<r_{d}, \end{array}\right. \end{aligned}$$

and, in case the equality does not hold, we have $\alpha _{2}^{*}=0$ if $p_{2}r_{l}=r_{d}$ which proves the claim.

Second case:

$$\begin{aligned} \frac{8p_{2}R_{2}-\left( 7+\gamma \right) r_{d}}{\left( 1-\gamma \right) p_{1}}>\frac{r_{d}}{p_{2}}, \end{aligned}$$

Again the inequality is equivalent to

$$\begin{aligned} \frac{8p_{2}R_{2}-\left( 5+\gamma \right) r_{d}}{\left( 1-\gamma \right) p_{1}+2p_{2}}> & {} \frac{r_{d}}{p_{2}}. \end{aligned}$$

This immediately yields $\alpha _{2}^{*}=1$ in case $r_{d}\ge p_{2}r_{l}$. It remains to be shown that $\alpha _{2}^{*}$ is decreasing if $p_{2}r_{l}>r_{d},\,\ 8p_{2}R_{2}-\left( 7+\gamma \right) r_{d}-\left( 1-\gamma \right) p_{1}r_{l}\ge 0$ and $8p_{2}R_{2}-\left( 3+2\gamma \right) r_{d}-\left( 1-\gamma \right) p_{1}r_{l}-\left( 4-\gamma \right) p_{2}r_{l}\le 0$. This is easily verified for $\gamma =0$ as well as for $\gamma >0$ by considering the first and the second derivative of $\alpha _{2}^{*}$ with respect to $r_{l}$. $\square$

Proof of Corollary 4

First, consider changes in $r_{d}$. If liquidity holding in equilibrium of bank 2 is equal to zero for all $r_{d}$ satisfying $p_{2}r_{l}<r_{d}\le p_{1}r_{l}$, liquidity holding is constantly equal to zero for all $r_{d}$ according to Corollary 2. Therefore, it is sufficient to concentrate on the case $p_{2}r_{l}<r_{d}\le p_{1}r_{l}$. A necessary and sufficient condition for liquidity holding being constantly equal to zero for all $p_{2}r_{l}<r_{d}\le p_{1}r_{l}$ is given by

$$\begin{aligned} 8p_{2}R_{2}-6p_{1}r_{l}-2p_{2}r_{l}>0 \end{aligned}$$

which is equivalent to

$$\begin{aligned} 8p_{2}\underset{>0}{\underbrace{\left( R_{2}-r_{l}\right) }}+6\underset{\le 0}{\underbrace{r_{l}\left( p_{2}-p_{1}\right) }}>0. \end{aligned}$$

Therefore, liquidity holding is constantly equal to zero if $p_{1}$ and $p_{2}$ are sufficiently close, meaning that the second summand is not too small.

Second, consider changes in $r_{l}$. If liquidity holding in equilibrium of bank 2 is equal to zero for all $r_{l}$ satisfying $r_{d}<p_{2}r_{l}\le p_{1}r_{l}$, liquidity holding is constantly equal to zero for all $r_{l}$ according to Corollary 3. Therefore, it is sufficient to concentrate on the case $r_{d}<p_{2}r_{l}\le p_{1}r_{l}$. A necessary and sufficient condition for liquidity holding being constantly equal to zero is given by

$$\begin{aligned} 8p_{2}R_{2}-\left( 3+2\gamma \right) r_{d}-\left( 4-\gamma \right) p_{2}r_{l}-\left( 1-\gamma \right) p_{1}r_{l}>0\text { for all }{r_{d}}<{p_{2}r_{l}}. \end{aligned}$$

We use $r_{l}<R_{2}$ and get the condition

$$\begin{aligned} \left( 3+2\gamma \right) \underbrace{\left( p_{2}R_{2}-r_{d}\right) }_{>0}+\left( 1-\gamma \right) R_{2}\underbrace{\left( p_{2}-p_{1}\right) }_{\le 0}>0. \end{aligned}$$

Therefore, liquidity holding is constantly equal to zero if $p_{1}$ and $p_{2}$ are sufficiently close, meaning that the second summand is not too small. $\square$

Another possibility of extending the model is allowing for ex ante contracts. The idea is that banks agree on an interbank rate at $t=0$, i.e., before liquidity shocks realize. At $t=1$ banks decide on interbank lending volumes based on the interbank rate that was fixed at $t=0$. Note that also ex ante, i.e., at $t=0$, banks do not agree on an interbank rate larger than $r_{l}$ or smaller than $\frac{r_{d}}{p_{i}},i\in \left\{ 1,2\right\}$ because otherwise using the central bank’s lending or deposit facility is more beneficial than using the interbank market.¹⁷ Since lending volumes are unknown at $t=0$, banks may ex ante agree on an interbank rate that ex post, i.e., at $t=1$, is larger than $r_{l}-\frac{1}{L_{i}}\left( r_{l}\left( -1+\alpha _{i}-\lambda _{i}\right) -\alpha _{i}R_{i}\right) ^{+}$ where i denotes the bank that ex post is in need of liquidity and j the bank that ex post has excess liquidity. In this case the lender anticipates that the borrower is not able to repay the full interbank rate and decides on interbank lending at $t=1$ based on the truncated interbank rate $r_{l}-\frac{1}{L_{i}}\left( r_{l}\left( -1+\alpha _{i}-\lambda _{i}\right) -\alpha _{i}R_{i}\right) ^{+}$ instead of the interbank rate fixed at $t=0$. Hence, any interbank rate that banks agree on ex ante is ex post equal to

$$\begin{aligned} \gamma \frac{r_{d}}{p_{1}}+\left( 1-\gamma \right) \left[ r_{l}-\frac{1}{L_{1}}\left( r_{l}\left( -1+\alpha _{1}-\lambda _{1}\right) -\alpha _{1}R_{1}\right) ^{+}\right] \end{aligned}$$

for some $\gamma \in \left[ 0,1\right]$. Therefore, introducing ex ante contracts does not change our results. A potential difference between ex ante and ex post interbank contracts is not whether interbank lending takes place or not but the actual value of the interbank rate. For example it could be the case that banks ex ante agree on a lower interbank rate than the interbank rate resulting from ex post bargaining. In this case, the lender is worse off due to the ex ante agreement.