Government Guarantees, Transparency, and Bank Risk Taking

Abstract

We present a model of bank risk taking and government guarantees. Levered banks take excessive risk as their actions are not fully priced at the margin by debt holders. The impact of government guarantees on bank risk taking depends critically on the portion of bank investors that can observe bank behavior and hence price debt at the margin. Greater guarantees increase risk taking (moral hazard) when informed investors hold a sufficiently large fraction of liabilities. But, otherwise, they reduce risk taking by increasing the profits of the bank (franchise value effect). The results extend to the case in which information disclosure and, thus, the portion of informed investors is endogenous but costly. The model also shows that, when bank capital is endogenous, public guarantees lead unequivocally to an increase in bank leverage and an associated increase in risk taking. The analysis points to a complex relationship between prudential policy and the institutional framework governing bank resolution and bailouts.

Appendix: Proofs

1.1 Proposition 1

Proof

Differentiating (5) with respect to $\gamma$, we have that

$$\begin{aligned} \frac{\partial q^{*}}{\partial \gamma }=\frac{(1-k)\overline{r}}{2c\Phi }(-R+(1-k)\overline{r}\gamma +2c(1-\theta )-\Phi ), \end{aligned}$$

(16)

with $\Phi \equiv \sqrt{\left( R-\gamma (1-k)r\right) ^{2}-4cr\left( 1-\gamma \right) \left( 1-\theta \right) \left( 1-k\right) }\ge 0$. At $\theta =0$,

$$\begin{aligned} (-R+(1-k)\overline{r}\gamma +2c(1-\theta )-\Phi )=2c\left( 1-q^{*}\right) , \end{aligned}$$

which is positive when parameters are such that q admits an internal solution so that (16) is always positive. At $\theta =1$, (A.1), instead, implies that (16) is always negative. Thus, to complete the proof, it is enough to show that $\frac{\partial ^{2}q^{*}}{\partial \theta \partial \gamma }<0$. Differentiating (16) with respect to $\gamma$, we obtain:

$$\begin{aligned} \frac{\partial ^{2}q^{*}}{\partial \theta \partial \gamma }=\frac{-(1-k)\overline{r}\Psi }{\left( \left( R-\gamma (1-k)\overline{r}\right) ^{2}-4c\overline{r}(1-k)(1-\theta )\left( 1-\gamma \right) \right) ^{\frac{3}{2}}}, \end{aligned}$$

(17)

with

$$\begin{aligned} \Psi \equiv (R-(1-k)\overline{r})(R-(1-k)\overline{r}\gamma )-2c\overline{r}(1-\gamma )(1-k)(1-\theta ). \end{aligned}$$

(18)

To show that $\frac{\partial ^{2}q^{*}}{\partial \theta \partial \gamma }<0,$ it is enough to show that $\Psi >0$. Since $\Psi$ is a linear function of $\gamma$, to prove that it is positive for any $\gamma$, it is enough to show that it is positive at $\gamma =0$, and $\gamma =1$ (for any k and $\theta$). When $\gamma =0$, we have that

$$\begin{aligned} \Psi =R(R-(1-k)r)-2c\overline{r}(1-k)(1-\theta ). \end{aligned}$$

(19)

Since (19) is an increasing function of k and $\theta$, if we know that if it is positive at $k=0$ and $\theta =0$, then it is positive for all k and $\theta$. We thus have that

$$\begin{aligned} \Psi \left| _{\gamma =0;\theta =0}\right. =R^{2}-Rr-2c\overline{r}>0 \end{aligned}$$

However, since from (A.1), we know that $R<c$ a sufficient condition for the inequality to hold is that

$$\begin{aligned} R^{2}-3\overline{r}c>0 \end{aligned}$$

or

$$\begin{aligned} c<\frac{R^{2}}{3\overline{r}}\text {,} \end{aligned}$$

which is always satisfied if (A.1) holds. This, in turn, implies that a single $\widehat{\theta }\in (0,1)$ exists such that $\frac{\partial q^{*} }{\partial \gamma }>0$, if $\theta<$ $\widehat{\theta }$; and $\frac{\partial q^{*}}{\partial \gamma }<0$, if $\theta >\widehat{\theta }$. $\square$

1.2 Lemma 1

Proof

Define

$$\begin{aligned} H\equiv \frac{\hbox {d}\Pi }{\hbox {d}\theta }=\left( 1-\theta \right) \frac{\overline{r} ^{2}\left( 1-\gamma \right) ^{2}\left( 1-k\right) ^{2}}{q\sqrt{\left( R-\gamma (1-k)\overline{r}\right) ^{2}-4c\overline{r}\left( 1-\gamma \right) \left( 1-\theta \right) \left( 1-k\right) }}-\theta \varphi . \end{aligned}$$

(20)

Note that

$$\begin{aligned} \frac{\hbox {d}\theta }{\hbox {d}\gamma }= -\frac{\frac{\hbox {d}H}{\hbox {d}\gamma }}{\frac{\hbox {d}H}{\hbox {d}\theta }}, \end{aligned}$$

where $\frac{\hbox {d}H}{\hbox {d}\theta }<0$ from the second-order conditions. Then, we have that $\frac{\hbox {d}\theta }{\hbox {d}\gamma }$ will have the same sign as $\frac{\hbox {d}H}{\hbox {d}\gamma }$. We can write:

$$\begin{aligned} \frac{\hbox {d}H}{\hbox {d}\gamma }=-\left( 1-\theta \right) r^{2}\left( 1-\gamma \right) \left( 1-k\right) ^{2}\frac{2A+\frac{\partial A}{\partial \gamma }\left( 1-\gamma \right) }{A^{2}}, \end{aligned}$$

where $A=q\sqrt{\left( R-\gamma (1-k)\overline{r}\right) ^{2}-4c\overline{r}\left( 1-\gamma \right) \left( 1-\theta \right) \left( 1-k\right) }.$ Now:

$$\begin{aligned} \frac{\partial A}{\partial \gamma }=-qr(1-k)\frac{R-r\gamma (1-k)-2c\left( 1-\theta \right) }{\sqrt{\left( R-\gamma (1-k)r\right) ^{2}-4cr\left( 1-\gamma \right) \left( 1-\theta \right) \left( 1-k\right) }}, \end{aligned}$$

which, after substituting, gives us

$$\begin{aligned} \frac{\hbox {d}H}{\hbox {d}\gamma }=-\left( 1-\theta \right) r^{2}\left( 1-\gamma \right) \left( 1-k\right) ^{2}\frac{\frac{2q^{2}\left( \left( R-\gamma (1-k)r\right) ^{2}-4cr\left( 1-\gamma \right) \left( 1-\theta \right) \left( 1-k\right) \right) }{A}-\frac{q^{2}r(1-k)\left( 1-\gamma \right) \left( R-r\gamma (1-k)-2c\left( 1-\theta \right) \right) }{A}}{A^{2}}. \end{aligned}$$

So $\frac{\hbox {d}H}{\hbox {d}\gamma }$ will have the opposite sign of

$$\begin{aligned} B&\equiv \frac{2q^{2}\left( \left( R-\gamma (1-k)r\right) ^{2}-4cr\left( 1-\gamma \right) \left( 1-\theta \right) \left( 1-k\right) \right) }{A}+\\ &\quad -\frac{q^{2}r(1-k)\left( 1-\gamma \right) \left( R-r\gamma (1-k)-2c\left( 1-\theta \right) \right) }{A}, \end{aligned}$$

expression that we can rewrite as

$$\begin{aligned} \frac{BA}{q^{2}}&=\left( R-\gamma (1-k)r\right) ^{2}-4cr\left( 1-\gamma \right) \left( 1-\theta \right) \left( 1-k\right) +\\ &\quad +\left( R-(1-k)r\right) \left( R-r(1-k)\gamma \right) -2cr\left( 1-\gamma \right) \left( 1-\theta \right) \left( 1-k\right) . \end{aligned}$$

The first term is just $A^{2}/q>0.$ A sufficient condition for the second term to be positive is that $R>2r$, which is always verified from our other restrictions. It follows that $B>0$ and that $\frac{\hbox {d}H}{\hbox {d}\gamma }<0$, which, in turn, implies that $\frac{\hbox {d}\theta }{\hbox {d}\gamma }<0.$ $\square$

1.3 Proposition 2

Proof

From (5), we have that

$$\begin{aligned} q^{*}\left| _{\gamma =0}\right.\equiv & {} q_{0}^{*}=\frac{1}{2c}(R+\sqrt{R^{2}-4c\overline{r}(1-k)(1-\theta )}, \end{aligned}$$

(21)

$$\begin{aligned} q^{*}\left| _{\gamma =1}\right.\equiv & {} q_{1}^{*}=\frac{R-(1-k)\overline{r}}{c}. \end{aligned}$$

(22)

Let now define $\overline{\theta }$ as the value of $\theta$ such that $q_{0}^{*}=q_{1}^{*}$, or

$$\begin{aligned} \frac{1}{2c}\left( R+\sqrt{R^{2}-4c\overline{r}\left( 1-\theta \right) \left( 1-k\right) }\right) =\frac{R-(1-k)\overline{r}}{c}, \end{aligned}$$

(23)

which gives

$$\begin{aligned} \overline{\theta }=1-\frac{R-(1-k)\overline{r}}{c}=1-q_{1}^{*}. \end{aligned}$$

(24)

The first-order condition of (7) with respect to $\theta$ can be written as

$$\begin{aligned} \frac{\hbox {d}\Pi }{\hbox {d}\theta }=\left( 1-\theta \right) \frac{\overline{r}^{2}\left( 1-\gamma \right) ^{2}\left( 1-k\right) ^{2}}{q^{*}\sqrt{\left( R-\gamma (1-k)\overline{r}\right) ^{2}-4c\overline{r}\left( 1-\gamma \right) \left( 1-\theta \right) \left( 1-k\right) }}-\theta \varphi =0, \end{aligned}$$

(25)

and, using (5):

$$\begin{aligned} \frac{\hbox {d}\Pi }{\hbox {d}\theta }=\left( 1-\theta \right) \frac{\overline{r}^{2}\left( 1-\gamma \right) ^{2}\left( 1-k\right) ^{2}}{q^{*}(2cq^{*}-R+(1-k)\overline{r}\gamma )}-\theta \varphi =0. \end{aligned}$$

(26)

When $\gamma =0$, the expression becomes:

$$\begin{aligned} \frac{\hbox {d}\Pi }{\hbox {d}\theta }=\left( 1-\theta \right) \frac{\overline{r}^{2}\left( 1-k\right) ^{2}}{q_{0}^{*}(2cq^{*}-R)}-\theta \varphi =0. \end{aligned}$$

(27)

If now we substitute $\overline{\theta }$ from (24) into (27), we obtain:

$$\begin{aligned} \frac{\hbox {d}\Pi }{\hbox {d}\theta }=q_{1}^{*}\frac{\overline{r}^{2}\left( 1-k\right) ^{2}}{q_{0}^{*}(2cq^{*}-R)}-(1-q_{1}^{*})\varphi =0. \end{aligned}$$

(28)

However, at $\theta =\overline{\theta }$, $q_{0}^{*}=q_{1}^{*} =\frac{R-(1-k)r}{c}$ so that we can rewrite (28) as

$$\begin{aligned} \frac{\hbox {d}\Pi }{\hbox {d}\theta }=\frac{\overline{r}^{2}\left( 1-k\right) ^{2} }{(2cq^{*}-R)}-\left( \frac{c-R+(1-k)\overline{r}}{c}\right) \varphi =0; \end{aligned}$$

(29)

Solving for $\varphi$, we obtain

$$\begin{aligned} \overline{\varphi }=\frac{c\overline{r}^{2}\left( 1-k\right) ^{2}}{\left( c-R+(1-k)\overline{r}\right) \left( R-2(1-k)\overline{r}\right) }. \end{aligned}$$

Notice that, by construction, $\overline{\varphi }$ is the cost of transparency such that, after having chosen $\theta$ optimally, the bank chooses the same portfolio riskiness when its deposit are fully insured ($\gamma =1$) and when they are not ($\gamma =0$). Formally, this means $\widehat{\theta }\left( \overline{\varphi }\right) =$ $\overline{\theta }.$ Now, note that $q_{1} ^{*}$ is invariant in $\theta ,$ but $\frac{\partial q_{0}^{*}}{\partial \theta }>0.$ Then, since from (24), $\frac{\partial \widehat{\theta }}{\partial \varphi }>0,$ we have that $q_{0}^{*}>q_{1}^{*}\Longleftrightarrow \varphi <\overline{\varphi }$. Finally, numerical simulations indicate that $\max \limits _{\gamma }q^{*}\left( \gamma \right)$ never admits an interior solution, so that $\arg \max \limits _{\gamma }q^{*}\left( \gamma \right) \in \left\{ 0,1\right\}$. (See Cordella et al. 2017 for numerical examples.) $\square$

1.4 Proposition 3

Proof

From 5, it is immediate that $\frac{\partial q}{\partial k}>0$ (k enters always with a positive sign, actually two negative signs) and $\frac{\partial ^{2}q}{\partial k\partial \theta }<0.$ It follows that

$$\begin{aligned} \frac{\hbox {d}\widehat{k}}{\hbox {d}\theta }=-\frac{\frac{\partial ^{2}\Pi }{\partial k\partial \theta }}{\frac{\partial ^{2}\Pi }{\partial k^{2}}}<0, \end{aligned}$$

since $\frac{{\text {d}}^{2}\Pi }{{\text{d}}k^{2}}<0$ from the second-order conditions. Note that for $\theta =1,$ we have

$$\begin{aligned} \frac{\hbox {d}\Pi }{\hbox {d}k}=-\left( 1-\frac{R-(1-k)\overline{r}\gamma }{c}\right) \gamma \overline{r}-\xi <0,\text{for}\,\text{all }\,k , \end{aligned}$$

and the first-order conditions for k can never be satisfied. Then, the model will only admit a corner solution with $\widehat{k}=0.$ $\square$

1.5 Proposition 4

Proof

From 15, we have that $\frac{\partial ^{2}\Pi }{\partial k\partial \gamma }<0,$ which, together with the second-order conditions with respect to k,  implies that

$$\begin{aligned} \frac{\hbox {d}\widehat{k}}{\hbox {d}\gamma }=-\frac{\frac{\partial ^{2}\Pi }{\partial k\partial \gamma }}{\frac{\partial ^{2}\Pi }{\partial k^{2}}}<0, \end{aligned}$$

that is, as government guarantees increase, banks find it optimal to reduce their capitalization (increase their leverage). $\square$