Optimal bailouts, bank’s incentive and risk

Abstract

We show how the impact of a government bailout in the form of liquidity assistance on the ex ante effort of a representative bank depends on the volatility of its investment. The bank’s investment delivers a cashflow that follows a geometric Brownian motion and the government guarantees the bank’s liabilities. To counter the bank’s expectations of a bailout, the government may choose a tighter liquidity policy when the bank’s effort is not observable. This tighter liquidity induces more prudent ex ante behavior by the bank, but it may have the opposite effect when investment volatility is high. This novel effect arises because the bank could be discouraged to be prudent precisely because the chances of receiving liquidity assistance are low.

Appendix

1.1 Liquidity buffers

We can model liquidity buffers introducing a second process $Y_{t}$ that represents the bank’s payouts strategy. Defining $\tilde{v}_{t}$ a version of the process $v_{t}$ with the liquid reserves, this is given by:

$$\begin{aligned} \tilde{v}_{t}=v_{t}+L_{t}-Y_{t} \end{aligned}$$

(33)

where $Y_{t}$ is an adapted, left-continuous and non-decreasing process with initial value $Y_{0}=0$ which represents the bank’s cumulative payouts to its shareholders. Now the two cumulative processes become:

$$\begin{aligned} Y_{t}=\max _{0\le s\le t}\left[ b-v_{s}-L_{s}\right] ^{-}, \end{aligned}$$

(34)

and

$$\begin{aligned} L_{t}=\max _{0\le s\le t}\left[ v_{s}-Y_{s}-r(1-k)\right] ^{-} \end{aligned}$$

(35)

where b is the upper level of v above which the bank distributes dividends. That is, the payout policy consists in distributing dividends to maintain liquid reserves at or below the target level b. With the process (33), the government intervenes only when $L_{t}-Y_{t}\ge 0,$ i.e. when the liquidity buffer generated by $Y_{t}$ is not sufficient to keep $\tilde{v}_{t}$ above $r(1-k).$ Under these assumptions the objective function of the bank is:

$$\begin{aligned} V=\max _{q}q\left[ \max _{b}\left( \mathbb {E}\left[ \int _{0}^{\tau }e^{-rt}(\tilde{v} _{t}-b)dt-\left( 1-e^{-r\tau }\left( 1-k\right) \right) \right] \right) \right] - \frac{a}{2}q^{2}, \end{aligned}$$

(36)

subject to (34) and (35).

In (36), the bank chooses both q and b,  while the government chooses $\tau$. If reserves do not pay interests (e.g. as in Hugonnier and Morellec 2017) it is easy to conjecture that the bank chooses $b=r(1-k)$. Indeed the government’s liquidity support is akin to raise equity in the market; there is no reason to hold liquidity buffers if the bank can count on liquidity support for a while. In general, however, with a $b>r(1-k)$ the value of the project (36) is lowered, although qualitatively, nothing changes regarding the choice of q.

1.2 Proof of Lemma 1

Defining:

$$\begin{aligned} Q^{B}(v_{0})\equiv \mathbb {E}\int _{0}^{\infty }e^{-rt}e^{-\psi L_{t}}(\tilde{ v}_{t}-r\left( 1-k\right) )dt+(1-k)\mathbb {E}\int _{0}^{\infty }e^{-rt}\psi e^{-\psi L_{t}}{} { dL}_{t} \end{aligned}$$

the Bellman equation is:

$$\begin{aligned} rQ^{B}=v_{0}-r\left( 1-k\right) +lim_{dt\rightarrow 0}\frac{1}{dt}\mathbb {E} [dQ^{B}]. \end{aligned}$$

(37)

Using the stochastic process (1) and Ito’s Lemma on $\mathbb {E}[dQ^{B}(v_{0})]$, we obtain the following partial differential equation:

$$\begin{aligned} \frac{1}{2}\sigma ^{2}v^{2}Q^{B\prime \prime }-rQ^{B}=-(v_{0}-r(1-k))\,\,\,\,\,\,\,\,\,\text {for}\quad v_{0}\in [r(1-k),\infty ), \end{aligned}$$

(38)

with boundary conditions:

$$\begin{aligned}&\lim _{v_{0}\rightarrow \infty }\left[ Q^{B}-\frac{v_{0}-r\left( 1-k\right) }{r}\right] =0, \end{aligned}$$

(39)

$$\begin{aligned}&Q^{B\prime }(r(1-k))-\psi [Q^{B}(r(1-k))-(1-k)]=0, \end{aligned}$$

(40)

where $Q^{B\prime }$ and $Q^{B\prime \prime }$ represent the first and the second derivative of $Q^{B}\left( v_{0}\right)$ w.r.t. v. Equation (39) states that, when cash flows go to infinity the effort must be bounded. In fact, the second term in (39) represents the discounted present value of excess returns over an infinite horizon starting from $v_{0}$. The boundary condition (40) means that when the cash flows reach the lower boundary $r(1-k)$, to continue to keep the bank open the marginal value of one extra unit of liquidity must not fall below the bank’s cost to increase effort by one unit represented by $\psi [Q^{B}(r(1-k))-(1-k)]$.¹⁶ By the linearity of the differential equation (38) and making use of (39), the general solution of (38) takes the form:

$$\begin{aligned} Q^{B}=\frac{v_{0}-r(1-k)}{r}+A(v_{0})^{\beta }, \end{aligned}$$

(41)

where A is a constant to be determined and $\beta ,$ with $-\infty<\beta <0,$ is the negative root of the characteristic equation $\frac{1}{2}\sigma ^{2}\beta (\beta -1)-r=0.$ The boundary condition (40), yields the value of the constant A:

$$\begin{aligned}&\left[ \frac{1}{r}+\beta A(r(1-k)^{\beta -1}\right] -\psi A[r(1-k)]^{\beta }+\psi (1-k)=0\nonumber \\&A=-\frac{(1-k)(1+\psi r(1-k))}{\underset{<0}{\underbrace{\beta -\psi r(1-k)}} }[r\left( 1-k\right) ]^{-\beta }>0. \end{aligned}$$

(42)

Then, the expected present value of the total liquidity supplied is equal to:

$$\begin{aligned} Av_{0}^{\beta }=-\left( \frac{v_{0}}{r(1-k)}\right) ^{\beta }\frac{ (1-k)(1+\psi r(1-k))}{\beta -\psi r(1-k)}>0. \end{aligned}$$

(43)

Finally, substituting (43) in (41), we are able to write the effort as in the text.

1.3 Proof or Proposition 1

Taking the derivative of the effort level (10) with respect to $\psi ,$ we obtain:

$$\begin{aligned} \frac{\partial q^{B}}{\partial \psi }&\!=\!-\left( \frac{v_{0}}{r(1\!-\!k)}\right) ^{\beta }(1\!-\!k)\frac{r(1-k)(\beta -\psi r(1-k))+(1+\psi r(1-k))r(1-k)}{(\beta -\psi r(1-k))^{2}} \nonumber \\&=-\Gamma (\beta )r(1-k)\left( \beta +1\right) , \end{aligned}$$

(44)

where

$$\begin{aligned} \Gamma (\beta )\equiv \left( \frac{v_{0}}{r(1-k)}\right) ^{\beta }\frac{1-k}{ (\beta -\psi r(1-k))^{2}}>0. \end{aligned}$$

(45)

By (44) it is easy to show that:

$$\begin{aligned} \frac{\partial q^{B}}{\partial \psi }>0 \ \ \ \text { if }\ \ \ \beta<-1; { \ \ \ \ \ \ \ \ \ \ }\frac{\partial q^{B}}{\partial \psi }<0\ \ \ \text { if }\ \ \ \beta >-1. \end{aligned}$$

Moreover, taking the limits:

$$\begin{aligned} \lim _{\beta \rightarrow 0}\frac{\partial q^{B}}{\partial \psi }=-\frac{1}{ (\psi r)^{2}}r<0;{ \ \ \ \ \ \ \ \ \ }\lim _{\beta \rightarrow -\infty } \frac{\partial q^{B}}{\partial \psi }=0. \end{aligned}$$

1.4 Equity and liquidity policy

In the model we have assumed that the project is financed with the exogenous equity of the bank’s shareholder k and deposits $1-k.$ In what follows we explore the effectiveness of the liquidity policy in inducing effort if the project is financed with more equity. Recall that the impact of liquidity on effort is given by

$$\begin{aligned} \frac{\partial q^{B}}{\partial \psi }=-\Gamma (\beta )r(1-k)\left( \beta +1\right) , \end{aligned}$$

where recall $\Gamma (\beta )$ is defined in (45). Taking the derivative of $\Gamma (\beta )$ with respect to k we obtain:

$$\begin{aligned} \frac{\partial \Gamma (\beta )}{\partial k}=\Gamma (\beta )\left[ \frac{ \beta -1}{(1-k)}+\frac{2\psi r}{(\beta -\psi r(1-k))}\right] <0. \end{aligned}$$

Therefore

$$\begin{aligned} \frac{\partial }{\partial k}\left( \frac{\partial q^{B}}{\partial \psi } \right) =r\left( \beta +1\right) \underset{>0}{\underbrace{\Gamma (\beta )}} \underset{>0}{\underbrace{\left[ -\beta +2-\frac{2\psi r(1-k)}{(\beta -\psi r(1-k))}\right] }} \end{aligned}$$

(46)

whose sign depends on the sign of $\beta +1.$ If the project’s volatility is high, i.e. $\mid \beta \mid$ is close to zero such that $\beta +1>0$, then the sign of (46) is positive. Intuitively, in Proposition 1 we have established that if volatility is high the decision of the government to provide a more generous liquidity support (that is to lower $\psi$) induces the bank to increase effort $(\frac{\partial q^{B}}{ \partial \psi }<0).$ Then if k increases, the marginal value of liquidity to induce the bank to provide effort declines, i.e. $\frac{\partial }{ \partial k}\left( \frac{\partial q^{B}}{\partial \psi }\right) >0.$

On the contrary if the project volatility is low, when the government increases liquidity it induces the bank to lower effort $(\frac{\partial q^{B}}{\partial \psi }>0).$ In this case, since if $\beta \rightarrow -\infty$ we have $\beta +1<0$, then the larger is the equity, the lower will be the reduction of effort, $\frac{\partial }{\partial k}\left( \frac{ \partial q^{B}}{\partial \psi }\right) <0$. That is, again the marginal value of liquidity on the bank’s effort declines. To sum up the effectiveness of a more generous liquidity policy in inducing effort declines with effort.

1.5 Proof of Lemma 2

Defining

$$\begin{aligned} Q^{W}(v_{0})\equiv \mathbb {E}\int _{0}^{\infty }e^{-rt}e^{-\psi L_{t}}\left[ ( \tilde{v}_{t}-r\left( 1-k\right) )dt-{ dC}_{t}\right] -Z\mathbb {E}\int _{0}^{\infty }e^{-rt}\psi e^{-\psi L_{t}}{} { dL}_{t}, \end{aligned}$$

the solution for $Q^{W}$ is obtained by solving the following Bellman equation:

$$\begin{aligned} \frac{1}{2}\sigma ^{2}v^{2}Q^{W\prime \prime }-rQ^{W}=-(v_{0}-r(1-k))\,\,\,\,\,\,\,\,\,\text {for }\quad v_{0}\in [r(1-k),\infty ), \end{aligned}$$

(47)

with boundary conditions:

$$\begin{aligned}&\lim _{v_{0}\rightarrow \infty }\left[ Q^{W}-\frac{v_{0}-r(1-k)}{r}\right] =0, \end{aligned}$$

(48)

$$\begin{aligned}&Q^{W\prime }(r(1-k))-\psi [Q^{W}(r(1-k))+Z]=c\left( \psi \right) . \end{aligned}$$

(49)

While (48) is equal to (39), and has the same meaning, condition (49) replaces the boundary condition (40). In fact, since liquidity is costly for the government, at each liquidity injection the marginal value of continuing to keep the bank open must not fall below the marginal cost, that now includes both the cost of liquidity $c\left( \psi \right)$ as well as the deadweight cost of bank closure Z, i.e.:

$$\begin{aligned} c\left( \psi \right) +\psi [Q^{W}(r(1-k))+Z]. \end{aligned}$$

Again, by the linearity of the differential equation (47) and making use of (48), the general solution takes the form:

$$\begin{aligned} Q^{W}=\frac{v_{0}-r(1-k)}{r}+Bv_{0}^{\beta }, \end{aligned}$$

(50)

where B is a constant to be determined and $\beta <0$ is still the negative root of the characteristic equation $\frac{1}{2}\sigma ^{2}\beta (\beta -1)-r=0.$ Using (49) we obtain:

$$\begin{aligned} B=\frac{(1-k)[(c(\psi )r-1)+\psi rZ]}{\underset{<0}{\underbrace{\beta -\psi r(1-k)}}}(r(1-k))^{-\beta }<0, \end{aligned}$$

from which:

$$\begin{aligned} Bv_{0}{}^{\beta }=\left( \frac{v_{0}}{r(1-k)}\right) ^{\beta }\frac{ (1-k)[(c(\psi )r-1)+\psi rZ]}{\beta -\psi r(1-k)}<0, \end{aligned}$$

(51)

which is H, the loss of project value due to government’s intervention. Since (51) is negative it concurs to lower the level of effort. Finally, substituting (51) in (50) we are able to obtain the expression in the text.

1.6 Proof of Propositions 2 and 3

To prove Proposition 2, recall that from (18) maximizing W is equivalent to maximize $q^{W},$ and that from (19) the probability of success is

$$\begin{aligned} aq^{W}=\frac{v_{0}-r(1-k)}{r}+H(\psi ,\beta )-1, \end{aligned}$$

where, recall,

$$\begin{aligned} H(\psi ,\beta )=\left( \frac{v_{0}}{r(1-k)}\right) ^{\beta }(1-k)\frac{ (c(\psi )r-1)+\psi rZ}{\beta -\psi r(1-k)}<0. \end{aligned}$$

(52)

To maximize $q^{W}$ we look for a $\psi ^{W}$ that minimizes H. Let us consider the F.O.C.:

$$\begin{aligned} \frac{\partial H}{\partial \psi }&=\left( \frac{v_{0}}{r(1-k)}\right) ^{\beta }(1-k)\frac{(c^{\prime }(\psi )r+rZ)(\beta -\psi r(1-k))+[c(\psi )r-1+\psi rZ]r(1-k)}{(\beta -\psi r(1-k))^{2}}\nonumber \\&=\Gamma (\beta )\left[ c^{\prime }(\psi )r\beta +rZ\beta -\psi r(1-k)c^{\prime }(\psi )r+(c(\psi )r-1)r(1-k)\right] =0, \end{aligned}$$

(53)

where recall $\Gamma (\beta )$ is defined in (45). Hence we obtain:

$$\begin{aligned} \frac{\partial H}{\partial \psi }=\Gamma (\beta )r\left[ c^{\prime }(\psi )(\beta -\psi r(1-k))+(c(\psi )r-1)(1-k)+Z\beta \right] =0. \end{aligned}$$

(54)

In addition the S.O.S.C. is always satisfied, i.e.:

$$\begin{aligned} \frac{\partial ^{2}H}{\partial \psi ^{2}}&=\frac{\partial \Gamma (\beta )}{ \partial \psi }\underset{=0\text { by F.O.C.}}{\underbrace{\left[ \ldots \right] } }+\Gamma (\beta )\left[ c^{\prime \prime }(\psi )r\beta -r(1-k)c^{\prime }(\psi )r-\psi r(1-k)c^{\prime \prime }(\psi )r\right] \nonumber \\&=\underset{\textit{SOSC}}{\underbrace{\Gamma (\beta )\left[ rc^{\prime \prime }(\psi )(\beta -\psi r(1-k))-r(1-k)c^{\prime }(\psi )r\right] }}>0. \end{aligned}$$

(55)

This proves Proposition 2.

To prove Proposition 3 we first analyze the effect of $\sigma$ on the level of effort $q^{W}.$ In particular, taking the derivative of (10), i.e. (52), with respect to $\sigma$ we obtain:

$$\begin{aligned} \frac{\partial H}{\partial \sigma }=\frac{\partial H}{\partial \psi ^{W}} \frac{\partial \psi ^{W}}{\partial \sigma }+\frac{\partial H}{\partial \beta }\frac{\partial \beta }{\partial \sigma }. \end{aligned}$$

Taking the derivative of H with respect to $\beta ,$ and observing that $\frac{\partial H}{\partial \psi ^{W}}=0$ by (54), we obtain:

$$\begin{aligned}&\frac{\partial H}{\partial \beta }\frac{\partial \beta }{\partial \sigma }\\&\quad =\left( \frac{v_{0}}{r(1-k)}\right) ^{\beta }(1-k)\frac{c(\psi )r-1+\psi rZ }{\beta -\psi r(1-k)}\\&\qquad \times \left[ \log \frac{v_{0}}{r(1-k)}-\frac{1}{\beta -\psi r(1-k)}\right] \frac{\partial \beta }{\partial \sigma }<0, \end{aligned}$$

from which it follows that $\frac{\partial q^{W}}{\partial \sigma }<0.$ Now, totally differentiating (54) with respect to $\sigma$ and using (55) we obtain:

$$\begin{aligned} \frac{\partial \psi ^{W}}{\partial \sigma }= & {} -\frac{\frac{\partial (\cdot )}{ \partial \sigma }}{\frac{\partial (\cdot )}{\partial \psi }}=-\frac{r[c^{\prime }(\psi )+Z]\frac{\partial \beta }{\partial \sigma }}{\textit{SOSC}}\nonumber \\= & {} -\frac{\overset{>0 }{\overbrace{\left[ \frac{-(c(\psi )r-1)r(1-k)-rZ\psi r(1-k)}{\beta -\psi r(1-k)}\right] } }\overset{>0}{\overbrace{\frac{\partial \beta }{\partial \sigma }}}}{\textit{SOSC}>0} <0. \end{aligned}$$

(56)

To gain the intuition for why $\frac{\partial \psi ^{W}}{\partial \sigma }<0$ observe that the regulator maximizes W w.r.t. $\psi$ by maximizing $q^{W}.$ So, at the maximum we have:

$$\begin{aligned} \frac{\partial q^{W}}{\partial \psi }(\psi (\sigma ),\sigma )=0. \end{aligned}$$

(57)

Taking the total differential of (57) w.r.t. $\sigma$, we obtain:

$$\begin{aligned} \frac{\partial ^{2}q^{W}}{\partial \psi ^{2}}(\psi (\sigma ),\sigma )\frac{ \partial \psi ^{W}}{\partial \sigma }+\frac{\partial ^{2}q^{W}}{\partial \psi \partial \sigma }(\psi (\sigma ),\sigma )=0, \end{aligned}$$

from which

$$\begin{aligned} \frac{\partial \psi ^{W}}{\partial \sigma }=-\frac{\frac{\partial ^{2}q^{W}}{ \partial \psi \partial \sigma }(\psi (\sigma ),\sigma )}{\frac{\partial ^{2}q^{W}}{\partial \psi ^{2}}(\psi (\sigma ),\sigma )}<0. \end{aligned}$$

(58)

Now, since $\frac{\partial ^{2}q^{W}}{\partial \psi ^{2}}(\psi (\sigma ),\sigma )<0,$ from (58) we have $\frac{ \partial \psi ^{W}}{\partial \sigma }<0$ if $\frac{\partial ^{2}q^{W}}{ \partial \psi \partial \sigma }(\psi (\sigma ),\sigma )<0.$ If we interpret $\frac{\partial q^{W}}{\partial \psi }$ as the marginal productivity of a restrictive liquidity policy, then $\frac{\partial ^{2}q^{W}}{\partial \psi \partial \sigma }(\psi (\sigma ),\sigma )<0$ denotes that an increase in output volatility lowers the marginal productivity of a restrictive liquidity policy. Therefore, as output volatility increases, if the government wants that the productivity of liquidity increases, it has to increase the liquidity, i.e. $\frac{\partial \psi ^{W}}{\partial \sigma }<0.$

1.7 Proof of Proposition 4

We prove $\psi ^{h}\ge \psi ^{W}$ for high volatility projects first, and then we add a sufficient condition for $\psi ^{h}\ge \psi ^{W}$ for low volatility projects. We proceed in steps.

First step. Recall that $W^{h}=W-I.$ The F.O.C. is:

$$\begin{aligned} \frac{\partial W^{h}}{\partial \psi }=\frac{\partial W}{\partial \psi }- \frac{\partial I}{\partial \psi }=0. \end{aligned}$$

(59)

As W is concave with $\frac{\partial W}{\partial \psi }_{\mid \psi =\psi ^{W}}=0,$ if a value $\psi ^{h}$ that satisfies (59) exists, for $\psi ^{h}>\psi ^{W}$ it must be that $\frac{\partial I}{ \partial \psi }<0$. In particular, since $a[q^{B}-q^{W}]=P+D>0,$ we have $\frac{\partial I}{\partial \psi }=a[q^{B}-q^{W}]\frac{\partial (P+D)}{ \partial \psi }.$ Then, comparing $\psi ^{h}$ with $\psi ^{W},$ we obtain $\psi ^{h}\ge \psi ^{W}$ if $\frac{\partial (P+D)}{\partial \psi }\le 0.$

Second step. The sign of $\frac{\partial (P+D)}{\partial \psi }=a\frac{ \partial [q^{B}-q^{W}]}{\partial \psi }$ is given by:

$$\begin{aligned} \frac{\partial (P+D)}{\partial \psi }= & {} -\Gamma (\beta )r\left\{ [1-k+c^{\prime }(\psi )+Z](\beta -\psi r(1-k))\right. \nonumber \\&\left. +\,[\psi r(1-k)+c(\psi )r+\psi rZ](1-k)\right\} , \end{aligned}$$

(60)

where $\Gamma (\beta )>0$ is defined in (45). Expression (60) is continuous in $\sigma$ (i.e. $\beta$). In addition, $\frac{\partial (P+D)}{\partial \psi }$ from (60) is $\le 0$ if:

$$\begin{aligned} -[c^{\prime }(\psi )+((1-k)+Z)]\beta \le [c(\psi )-c^{\prime }(\psi )\psi ]r(1-k)). \end{aligned}$$

(61)

Since the R.H.S. of (61) is always positive, there exists a value of $\sigma$ in the region $[0,\infty )$ that satisfies (61). In the specific, if the project is high volatility (i.e. $\mid \beta \mid$ is close to 0), (61) is easily satisfied for any acceptable range of $\psi$.

Third step. Since $q^{B}-q^{W}>0$, combining with $\frac{\partial (P+D)}{ \partial \psi }\le 0,$ for high volatility projects we obtain $\psi ^{h}\ge \psi ^{W}.$

Fourth step. For a low volatility project (i.e. when $\mid \beta \mid$ is high),  the sign of $\frac{\partial (P+D)}{\partial \psi }$ is harder to determine. However, from (61) a sufficient condition for $\frac{\partial (P+D)}{\partial \psi }\le 0$ regardless of the project’s volatility is (30), that is

$$\begin{aligned} c^{\prime }(\psi )+(1-k)+Z<0\Leftrightarrow -rc^{\prime }(\psi )>r(1-k)+rZ. \end{aligned}$$

(62)

Thus, if (62) holds then $\psi ^{h}>\psi ^{W}$ for all value of $\sigma .$ This proves Proposition 4.

1.8 Analysis of Eq. (32)

Replacing the expressions for $q^{B}(\psi ^{W}),$$q^{W}(\psi ^{W})$ and $\frac{\partial q(\psi ^{W})}{\partial \psi }$ into (31), we obtain:

$$\begin{aligned} \Delta q= & {} {-\Gamma (\beta )}\left[ \psi ^{W}r(1-k))+c(\psi ^{W})r+\psi ^{W}rZ\right] \nonumber \\&-\,\Gamma (\beta )\frac{r(1-k)(\beta -\psi ^{W}r(1-k))+(1+\psi ^{W}r(1-k))r(1-k)}{(\beta -\psi ^{W}r(1-k))}\Delta \psi .\nonumber \\ \end{aligned}$$

(63)

Rearranging (63) it is easy to see that for the difference $\Delta q$ to be positive it must be:

$$\begin{aligned} \Delta \psi =\psi ^{h}-\psi ^{W}<-\frac{(\beta -\psi ^{W}r(1-k))[\psi ^{W}r(1-k))+c(\psi ^{W})r+\psi ^{W}rZ]}{r(1-k)\left( \beta +1\right) }, \end{aligned}$$

where we assume, like in Proposition 1, that if the project is high volatility, i.e. $\mid \beta \mid$ is close to zero, then $\beta +1>0.$