Bank Competition, Risk, and Subordinated Debt

Abstract

This paper studies a dynamic model of banking in which banks compete for insured deposits, issue subordinated debt, and invest in either a prudent or a gambling asset. The model allows banks to choose their level of risk after the interest rate on subordinated debt is contracted. We show that requiring banks to issue a small amount of subordinated debt can reduce their gambling incentives. Moreover, when equity capital is more expensive than subordinated debt, adding a subordinated debt requirement to a policy regime that only uses equity capital requirements is Pareto improving.

Appendix

Lemma 1

The first no-gambling condition (9) is equivalent to

$$\left( {1 - \pi } \right)\left[ {\gamma - \alpha - 2\left( {h - 1} \right)\frac{\mu }{n}} \right] - \left[ {\pi \left( {1 + \rho } \right) - \left( {1 - \pi } \right)\left( {\gamma - \alpha } \right)} \right]k - \left[ {\pi \left( {1 + \lambda } \right) - \left( {1 - \pi } \right)\left( {\gamma - \alpha } \right)} \right]s \leqslant 0,$$

(18)

where h is given by Eq. 12.

Proof

Differentiating the left-hand side of Eq. 9 with respect to k _j , and using assumption (1) and (2), yields

$$ \left( -1+\frac{1-\pi }{1+\rho }\left( 1+\gamma \right) \right) D\left( r_{j},r_{P}\right) <\left( -1+\frac{1+\alpha }{1+\rho }\right) D\left( r_{j},r_{P}\right) <0. $$

Hence the shareholders choose k _j  = k.

Similarly, differentiating the left-hand side of Eq. 9 with respect to s _j , and using assumption (1) and (3), yields

$$ \frac{1-\pi }{1+\rho }\left( 1+\gamma -\frac{1+\lambda }{1-\pi }\right) D\left( r_{j},r_{P}\right) <\frac{1-\pi }{1+\rho }\left( 1+\gamma -\frac{ 1+\alpha }{1-\pi }\right) D\left( r_{j},r_{P}\right) <0. $$

Hence the shareholders choose s _j  = s.

Finally, substituting k _j  = k and s _j  = s into Eq. 9, and then differentiating the left-hand side of it with respect to r _j , yields the following first-order condition:

$$ r_{j}=\frac{r_{P}+r_{G}}{2}, $$

where

$$ r_{G}=\gamma -\frac{\mu }{n}-\left( \frac{1+\rho }{1-\pi }-1-\gamma \right) k-\left( \frac{1+\lambda }{1-\pi }-1-\gamma \right) s. \label{r_G} $$

(19)

Therefore

$$ D\left( r_{j},r_{P}\right) =\frac{1}{n}+\frac{r_{G}-r_{P}}{2\mu }. $$

Substituting r _j and $D\left( r_{j},r_{P}\right)$ back into Eq. 9 and simplifying terms yields

$$ \left( \frac{\mu }{n}+\frac{r_{G}-r_{P}}{2}\right) ^{2}\leq h^{2}\left( \frac{\mu }{n}\right) ^{2}. $$

When $D\left( r_{j},r_{P}\right) \geq 0$, above inequality is equivalent to

$$ r_{G}-r_{P}\leq 2\left( h-1\right) \frac{\mu }{n}. $$

Substituting r _G from Eq. 19 and r _P from Eq. 7 into this expression and simplifying terms yields the desired result. □

Lemma 2

The second no-gambling condition Eq. 10 is equivalent to

$$ \left( 1-\pi \right) \left[ \left( \gamma -\alpha \right) \left( 1+s\right) -\left( h^{2}-1\right) \frac{\mu }{n}\right] -\left[ \pi \left( 1+\rho \right) -\left( 1-\pi \right) \left( \gamma -\alpha \right) \right] k\leq 0. \label{no_gambling_2} $$

(20)

Proof

Substitute $D\left( r_{P},r_{P}\right) =1/n$ and V _P from Eqs. 8 into (10). Simple calculations yield the desired result. □

Proof of Proposition 1

Substituting s = 0 and k = 0 into Eq. 18, and rearranging terms, the first no-gambling condition becomes

$$ \frac{\mu }{n}\geq \frac{\gamma -\alpha }{2\left( h-1\right) }. \label{c_1} $$

(21)

Similarly, substituting s = 0 and k = 0 into Eq. 20, and rearranging terms, the second no-gambling condition becomes

$$ \frac{\mu }{n}\geq \frac{\gamma -\alpha }{h^{2}-1}. \label{c_2} $$

(22)

Because h > 1, condition (22) is satisfied as long as Eq. 21 is satisfied. Therefore, no prudent equilibrium can exist if

$$ \frac{\mu }{n}<\frac{\gamma -\alpha }{2\left( h-1\right) }. $$

□

Proof of Proposition 2

Substituting k = 0 into Eq. 18, the first no-gambling condition becomes

$$ \left( 1-\pi \right) \left[ \gamma -\alpha -2\left( h-1\right) \frac{\mu }{n} \right] -\left[ \pi \left( 1+\lambda \right) -\left( 1-\pi \right) \left( \gamma -\alpha \right) \right] s\leq 0. \label{s_c_1} $$

(23)

It then follows from Eq. 14 that Eq. 23 can be rewritten as

$$ s\geq \frac{\left( 1-\pi \right) \left[ \gamma -\alpha -2\left( h-1\right) \frac{\mu }{n}\right] }{\pi \left( 1+\lambda \right) -\left( 1-\pi \right) \left( \gamma -\alpha \right) }. \label{s_bigger} $$

(24)

Substituting k = 0 into Eq. 20, and rearranging terms, the second no-gambling condition becomes

$$ s\leq \left( \frac{h^{2}-1}{\gamma -\alpha }\right) \frac{\mu }{n}-1. \label{s_smaller} $$

(25)

The bank will have no incentive to deviate when both conditions (24) and (25) are satisfied. For this to be possible, we need

$$ \frac{\left( 1-\pi \right) \left[ \gamma -\alpha -2\left( h-1\right) \frac{ \mu }{n}\right] }{\pi \left( 1+\lambda \right) -\left( 1-\pi \right) \left( \gamma -\alpha \right) }\leq \left( \frac{h^{2}-1}{\gamma -\alpha }\right) \frac{\mu }{n}-1, $$

or

$$ {\kern65pt} \frac{\mu }{n}\geq \frac{\pi \left( 1+\lambda \right) \left( \gamma -\alpha \right) }{\pi \left( 1+\lambda \right) \left( h^{2}-1\right) -\left( 1-\pi \right) \left( \gamma -\alpha \right) \left( h-1\right) ^{2}}. $$

□

Proof of Proposition 3

Assumption (1) and (2) imply that

$$ \pi \left( 1+\rho \right) -\left( 1-\pi \right) \left( \gamma -\alpha \right) =\left( \rho -\alpha \right) \pi +\alpha -\left( 1-\pi \right) \gamma +\pi >0. \label{before_k_positive} $$

(26)

Therefore the first no-gambling condition, Eq. 18, can be rewritten as k ≥ k ₁, where the parameter k ₁ is given by Eq. 15. Similarly, the second no-gambling condition, Eq. 20, can be rewritten as k ≥ k ₂, where k ₂ is given by Eq. 16.

To satisfy both conditions, the regulator sets $k\geq \max \left( k_{1},k_{2}\right)$. □

Proof of Proposition 4

Some calculations show that k ₁ = k ₂ when s = s ^∗, where s ^∗ is given by Eq. 17. Because h > 1, we have s ^∗ > 0.

Expression (14) implies that k ₁ is a linearly decreasing function of s, while assumption (1) implies that k ₂ is a linearly increasing function of s. Therefore, for s < s ^∗, we have k = k ₁, or dk/ds < 0. Similarly, for s ≥ s ^∗, we have k = k ₂, or dk/ds > 0. □

Proof of Proposition 5

From the proof of Proposition 4, we know that for s < s ^∗, we have$\ k=k_{1}$, thus

$$\begin{array}{*{20}c} {\frac{{dr_P }}{{d_S }}}{ = \frac{{\partial r_P }}{{\partial _S }} + \frac{{\partial r_P }}{{\partial k}}\frac{{dk}}{{ds}}} \\ {}{ = - \left( {\lambda - \alpha } \right) + \left( {\rho - \alpha } \right)\frac{{\pi \left( {1 + \lambda } \right) - \left( {1 - \pi } \right)\left( {\gamma - \alpha } \right)}}{{\pi \left( {1 + \rho } \right) - \left( {1 - \pi } \right)\left( {\gamma - \alpha } \right)}}} \\ {}{ = \frac{{\left( {\rho - \lambda } \right)\left[ {\alpha - \left( {1 - \pi } \right)\gamma + \pi } \right]}}{{\alpha + \left( {\rho - \alpha } \right)\pi - \left( {1 - \pi } \right)\gamma + \pi }}.} \\ \end{array} $$

It follows from assumption (1) and (2) that for s < s ^∗, we have dr _P /ds > 0 if and only if ρ > λ . This implies that when equity capital is more expensive than subordinated debt (i.e., when ρ > λ) and for s < s ^∗, an increase of s will increase the returns of the depositors. At the same time, the bank’s expected returns do not change, and the regulator (as the deposit insurance provider) does not incur any insurance outflow. Hence the policy is Pareto improving. □

Proof

that k ₁ is decreasing in π

Rewrite Eq. 12 as $\pi =\rho \left( h^{2}-1\right) /\left( \rho h^{2}+1\right)$. Substituting it into Eq. 15 and simplifying terms yields

$$ k_{1}=\frac{\left( 1+\rho \right) \left[ \gamma -\alpha -2(h-1)\frac{\mu }{n} \right] -\left[ \rho \left( h^{2}-1\right) \left( 1+\lambda \right) -\left( 1+\rho \right) \left( \gamma -\alpha \right) \right] s}{\rho \left( h^{2}-1\right) \left( 1+\rho \right) -\left( 1+\rho \right) \left( \gamma -\alpha \right) }. $$

It is easy to verify that $\partial k_{1}/\partial h<0$. But $\partial h/\partial \pi >0$, therefore $\partial k_{1}/\partial \pi <0$. □

Proof

that k ₂ is decreasing in π

Substituting Eq. 12 into Eq. 16 and simplifying terms yields

$$\begin{array}{*{20}c} {k_2 = }{\frac{{\left( {1 - \pi } \right)\left( {\gamma - \alpha } \right)\left( {1 + s} \right) - \pi \left( {1 + {1 \mathord{\left/ {\vphantom {1 \rho }} \right. \kern-\nulldelimiterspace} \rho }} \right)\tfrac{\mu }{n}}}{{\pi \left( {1 + \rho } \right) - \left( {1 - \pi } \right)\left( {\gamma - \alpha } \right)}}} \\ = {\frac{{\left( {\gamma - \alpha } \right)\left( {1 + s} \right) - \pi \left[ {\left( {\gamma - \alpha } \right)\left( {1 + s} \right) + \left( {1 + {1 \mathord{\left/ {\vphantom {1 \rho }} \right. \kern-\nulldelimiterspace} \rho }} \right)\tfrac{\mu }{n}} \right]}}{{\pi \left( {1 + \rho + \gamma - \alpha } \right) - \gamma + \alpha }}.} \\ \end{array} $$

Because $\left( \gamma -\alpha \right) \left( 1+s\right) +\left( 1+1/\rho \right) \frac{\mu }{n}>0$, and 1 + ρ + γ − α > 0, we have $\partial k_{2}/\partial \pi <0$. □

Proof

that s ^∗ is decreasing in π

Rewrite Eq. 12 as $\pi =\rho \left( h^{2}-1\right) /\left( \rho h^{2}+1\right)$. Substituting it into Eq. 17 and simplifying terms yields

$$ s^{\ast }=\frac{\left( 1+\rho \right) \mu }{\rho \left( 1+\lambda \right) n} \left( \frac{h-1}{h+1}\right) . $$

It is easy to verify that $\partial s^{\ast }/\partial h>0$. But $\partial h/\partial \pi >0$, therefore $\partial s^{\ast }/\partial \pi >0$. □