Banking Regulation and Collateral Screening in a Model of Information Asymmetry

Abstract

This paper explores the impact of banking regulation on a competitive credit market with ex-ante asymmetric information and aggregate uncertainty. I construct a model where the government imposes a regulatory constraint that limits the losses banks make in the event of their default. I show that the addition of banking regulation results in three deviations from the standard theory. First, collateral is demanded of both high and low risk firms, even in the absence of asymmetric information. Second, if banking regulation is sufficiently strict, there may not exist an adverse selection problem. Third, a pooling Nash equilibrium can exist.

Appendix

1.1 Proof of Proposition 1

In this proof it is assumed that W is sufficiently high that any quantity of collateral can be implemented in equilibrium. The Lagrangian that solves for the competitive equilibrium contract for an single firm type is

$$ {\displaystyle \begin{array}{c}{\mathcal{L}}_i={p}_i\left(\varphi {k}_i^{\alpha }-{R}_i{k}_i\right)-\left(1-{p}_i\right){C}_i+W\\ {}+q{\lambda}_B\left[{p}_i\left({z}_G\right){R}_i{k}_i+\delta \left(1-{p}_i\left({z}_G\right)\right){C}_i-{k}_i\right]\\ {}\begin{array}{c}+{\lambda}_S\left[{p}_i\left({z}_B\right){R}_i{k}_i+\delta \left(1-{p}_i\left({z}_B\right)\right){C}_i-\gamma {k}_i\right]\\ {}+{\lambda}_C^{-}{C}_i,\end{array}\end{array}} $$

(A.45)

where λ_S, λ_B and ${\lambda}_C^{-}$ are the multipliers on Eqs. (7), (8) and the non-negativity constraint on collateral. The first order conditions are

$$ {p}_i={\lambda}_Bq{p}_i\left({z}_G\right)+{\lambda}_S{p}_i\left({z}_B\right), $$

(A.46)

$$ \left(1-{p}_i\right)={\lambda}_B q\delta \left(1-{p}_i\left({z}_G\right)\right)+{\lambda}_S\delta \left(1-{p}_i\left({z}_B\right)\right)+{\lambda}_C^{-}, $$

(A.47)

$$ {p}_i\left(\alpha \varphi {k}_i^{\alpha -1}-{R}_i\right)+{\lambda}_Bq\left[{p}_i\left({z}_G\right){R}_i-1\right]+{\lambda}_S\left[{p}_i\left({z}_B\right){R}_i-\gamma \right]=0. $$

(A.48)

First note that if the stress-test condition does not bind, λ_S = 0 and the contract terms that solve the first order conditions are given by C_i = 0, R_i = 1 and ${k}_i={\left(\alpha {p}_i\left({z}_G\right)\right)}^{\frac{1}{1-\alpha }}$. Plugging these equations into Eq. (7), the regulatory constraint will be satisfied only if γ ≤ ξ. Next, consider when λ_B will be strictly positive. To do this, suppose instead that λ_B = 0, then Eq. (A.46) implies that ${\lambda}_S=\frac{p_i}{p_i\left({z}_B\right)}$. Substituting this into Eq. (A.47) and rearranging yields

$$ \left(1-{p}_i\right)=\delta {p}_i\left(\frac{1-{p}_i\left({z}_B\right)}{p_i\left({z}_B\right)}\right)+{\lambda}_C^{-}. $$

(A.49)

This yields a contradiction whenever $\delta >\left(\frac{p_i\left({z}_B\right)}{1-{p}_i\left({z}_B\right)}\right)\left(\frac{1-{p}_i}{p_i}\right)$. Thus it follows that if γ > ξ and $\delta >\left(\frac{p_i\left({z}_B\right)}{1-{p}_i\left({z}_B\right)}\right)\left(\frac{1-{p}_i}{p_i}\right)$ then λ_S > 0, λ_B > 0 and ${\lambda}_C^{-}=0$. Thus Eqs. (7) and (8) will bind in equilibrium and the equilibrium will feature a strictly positive amount of collateral. Solving the system of first order conditions then yields the equilibrium loan size and collateral size set out in Eqs. (16) and (17). The interest rate charged to the firm follows from substituting Eqs. (16) and (17) into Eq. (8) and the payoff ${U}_i^{\ast }$ follows from substituting the contract terms into

$$ {U}_i\left(k,R,C\right)={p}_i\left(\varphi {k}_i^{\alpha }-{R}_i{k}_i\right)-\left(1-{p}_i\right){C}_i. $$

1.1.1 Properties of the Equilibrium with identical customers

The derivative of ${k}_i^{\ast }$ with respect to γ is

$$ \frac{d{k}_i^{\ast }}{d\gamma}=-\left(\frac{1}{1-\alpha}\right)\left(\frac{\left(1-q\right)+\left(\frac{1-\delta }{\delta}\right)\left(1-{p}_i\right)\left(\frac{1}{1-\xi}\right)}{\left(q+\left(1-q\right)\gamma \right)+\left(\frac{1-\delta }{\delta}\right)\left(1-{p}_i\right)\left(\frac{\gamma -\xi }{1-\xi}\right)}\right){k}_i^{\ast }, $$

(A.50)

which is strictly decreasing.

The derivative of ${U}_i^{\ast }$ with respect to γ is

$$ \frac{{\partial U}_i^{\ast }}{\partial \gamma }=-\left[\left(1-q\right)+\left(\frac{1-\delta }{\delta}\right)\left(1-{p}_i\right)\left(\frac{1}{1-\xi}\right)\right]{k}_i^{\ast }, $$

(A.51)

which is strictly decreasing.

The derivative of ${C}_i^{\ast }$ with respect to γ is

$$ \frac{d{C}_i^{\ast }}{d\gamma}=\frac{1}{\delta}\left(\frac{1}{1-\xi}\right)\left[1-\left(\frac{1}{1-\alpha}\right)\left(\frac{\left(1-q\right)\left(\gamma -\xi \right)+\left(\frac{1-\delta }{\delta}\right)\left(1-{p}_i\right)\left(\frac{\gamma -\xi }{1-\xi}\right)}{\left(q+\left(1-q\right)\gamma \right)+\left(\frac{1-\delta }{\delta}\right)\left(1-{p}_i\right)\left(\frac{\gamma -\xi }{1-\xi}\right)}\right)\right]{k}_i^{\ast }. $$

(A.52)

The derivative of this is strictly positive if

$$ \left(\frac{1}{1-\alpha}\right)\left(\frac{\left(1-q\right)\left(\gamma -\xi \right)+\left(\frac{1-\delta }{\delta}\right)\left(1-{p}_i\right)\left(\frac{\gamma -\xi }{1-\xi}\right)}{\left(q+\left(1-q\right)\gamma \right)+\left(\frac{1-\delta }{\delta}\right)\left(1-{p}_i\right)\left(\frac{\gamma -\xi }{1-\xi}\right)}\right)<1. $$

This holds at sufficiently low values of γ and the left hand side of the above is zero when γ = ξ. However, at high values of γ, this inequality may not hold, especially if α is high. In this case, the derivative will be negative and the level of collateral will fall in response to an increase in regulation. As discussed in the text, the reason for this is that ${k}_i^{\ast }$ falls in response to γ and at sufficiently high levels of α and γ, this effect dominates the increased collateral ratio and the level of collateral required falls.

1.2 Proof of Proposition 2

First, if $\gamma \ge \xi, \delta \ge \left(\frac{p_H\left({z}_B\right)}{1-{p}_H\left({z}_B\right)}\right)\left(\frac{1-{p}_H}{p_H}\right)$ and $W\ge {C}_L^{\ast }$ then the equilibrium contracts if the incentive compatibility constraint does not bind are given by ${\left\{\left({k}_i^{\ast },{R}_i^{\ast },{C}_i^{\ast}\right)\right\}}_{i=\in \left\{L,H\right\}}$. Then Eq. (21) can be derived by substituting these contract terms into Eq. (19) and rearranging. If for some value of γ Eq. (21) is weakly positive, then the incentive compatibility will not bind at γ, otherwise, it will. As discussed in the text, the limit of Λ_IC(γ) as γ → ξ is negative and the incentive compatibility constraint will always bind for γ ≤ ξ. On the other hand, as discussed in the text it is clear that the incentive compatibility constraint is slack at the upper-limit as γ → 1.

The rest off the proposition follows so long as $\frac{\partial {\Lambda}_{IC}\left(\gamma \right)}{\partial \gamma }>0$ and there exists a threshold γ^∗ ∈ [ξ, 1) such that for any γ > γ^∗ Λ_IC(γ) > 0.

To show that $\frac{\partial {\Lambda}_{IC}\left(\gamma \right)}{\partial \gamma }>0$ note that the derivative of Eq. (21) with respect to γ can be written as

$$ {\displaystyle \begin{array}{c}\frac{\partial {\Lambda}_{IC}\left(\gamma \right)}{\partial \gamma }=\alpha {\left(\frac{p_H}{p_L}\right)}^{\frac{\alpha }{1-\alpha }}\left({p}_L-{p}_H\right)\left(\frac{q+\left(1-q\right)\xi }{1-\xi}\right)\\ {}\times {\left(\frac{\left[q+\left(1-q\right)\upgamma \right]+\left(\frac{1-\delta }{\delta}\right)\left(1-{p}_L\right)\left(\frac{\gamma -\xi }{1-\xi}\right)}{\left[q+\left(1-q\right)\upgamma \right]+\left(\frac{1-\delta }{\delta}\right)\left(1-{p}_H\right)\left(\frac{\gamma -\xi }{1-\xi}\right)}\right)}^{\frac{1}{1-\alpha }}\\ {}\begin{array}{c}+\alpha \left(\frac{\left(q+\left(1-q\right)\xi \right)\frac{1}{\delta}\left(\frac{p_L}{p_H}-1\right)\left(\frac{1}{1-\xi}\right)}{{\left[\left[q+\left(1-q\right)\upgamma \right]+\left(\frac{1-\delta }{\delta}\right)\left(1-{p}_L\right)\left(\frac{\gamma -\xi }{1-\xi}\right)\right]}^2}\right)\\ {}\times \left[1-{p}_H{\left(\frac{p_H}{p_L}\right)}^{\frac{\alpha }{1-\alpha }}{\left(\frac{\left[q+\left(1-q\right)\gamma \right]+\left(\frac{1-\delta }{\delta}\right)\left(1-{p}_L\right)\left(\frac{\gamma -\xi }{1-\xi}\right)}{\left[q+\left(1-q\right)\gamma \right]+\left(\frac{1-\delta }{\delta}\right)\left(1-{p}_H\right)\left(\frac{\gamma -\xi }{1-\xi}\right)}\right)}^{\frac{1}{1-\alpha }}\right]\end{array}\end{array}} $$

(A.53)

A sufficient condition for $\frac{\partial {\Lambda}_{IC}\left(\gamma \right)}{\partial \gamma }>0$ is

$$ {p}_H{\left(\frac{p_H}{p_L}\right)}^{\frac{\alpha }{1-\alpha }}{\left(\frac{\left[q+\left(1-q\right)\gamma \right]+\left(\frac{1-\delta }{\delta}\right)\left(1-{p}_L\right)\left(\frac{\gamma -\xi }{1-\xi}\right)}{\left[q+\left(1-q\right)\gamma \right]+\left(\frac{1-\delta }{\delta}\right)\left(1-{p}_H\right)\left(\frac{\gamma -\xi }{1-\xi}\right)}\right)}^{\frac{1}{1-\alpha }}\le 1. $$

(A.54)

To show this, note that from the definition of ${k}_i^{\ast }$ the following is true $\alpha \varphi {p}_i{\left({k}_i^{\ast}\right)}^{\alpha -1}=\left[q+\left(1-q\right)\gamma \right]+\left(\frac{1-\delta }{\delta}\right)\left(1-{p}_i\right)\left(\frac{\gamma -\xi }{1-\xi}\right)$. Thus Eq. (A.54) can be rewritten as

$$ {p}_H{\left(\frac{p_H}{p_L}\right)}^{\frac{\alpha }{1-\alpha }}{\left(\frac{\alpha \varphi {p}_L{\left({k}_L^{\ast}\right)}^{\alpha -1}}{\varphi {p}_H{\left({k}_H^{\ast}\right)}^{\alpha -1}}\right)}^{\frac{1}{1-\alpha }}\le 1, $$

(A.55)

which simplifies to

$$ {\left(\frac{p_H}{p_L}\right)}^{\frac{\alpha }{1-\alpha }}\frac{k_H^{\ast }}{k_L^{\ast }}\le 1, $$

(A.56)

which will always be satisfied as ${k}_H^{\ast }<{k}_L^{\ast }$ and p_L > p_H.

1.3 Proof of Proposition 3

First, note that the cutoff $\overline{\gamma}$ as defined in Proposition 3 is simply the point where ${\hat{C}}_L={\overline{C}}_L$ which is the point where the following equation must hold

$$ {\Lambda}_{\overline{C}}\left(\gamma \right)\equiv \frac{1}{\delta}\left(\frac{\gamma -\xi }{1-\xi}\right){\overline{k}}_L-\frac{1}{\delta}\left(\frac{\frac{p_H}{p_L}\left({p}_L\varphi {\overline{k}}_L^{\alpha -1}-\left(q+\left(1-q\right)\xi \right)\right){\overline{k}}_L-{U}_H^{\ast }}{1-\frac{p_H}{p_L}\left(q+\left(1-q\right)\xi \right)+\left(\frac{1-\delta }{\delta}\right)\left(1-{p}_H\right)}\right)=0. $$

(A.57)

First note that when γ = ξ, ${\overline{C}}_L=0$ and ${\hat{C}}_L>0$ implies that ${\Lambda}_{\overline{C}}\left(\xi \right)<0$. Next, note that when γ = γ^*, ${C}_L={C}_L^{\ast }<{\overline{C}}_L$ and thus ${\Lambda}_{\overline{C}}\left({\gamma}^{\ast}\right)>0$. It follows from this that for there to exists a unique $\overline{\gamma}\in \left(\xi, {\gamma}^{\ast}\right)$ such that ${C}_L={\overline{C}}_L$ it is sufficient to show that $\frac{{\mathrm{\partial \Lambda}}_{\overline{C}}}{\mathrm{\partial \upgamma }}>0$. To prove this note that differentiating Eq. (A.57) yields

$$ \frac{{\mathrm{\partial \Lambda}}_{\overline{C}}}{\mathrm{\partial \upgamma }}=\left(\frac{1}{1-\xi}\right){\overline{k}}_L-\left(\frac{\left(1-q\right)\left(1-\xi \right)+\left(\frac{1-\delta }{\delta}\right)\left(1-{p}_H\right)}{1-\frac{p_H}{p_L}\left(q+\left(1-q\right)\xi \right)+\left(\frac{1-\delta }{\delta}\right)\left(1-{p}_H\right)}\right){k}_H^{\ast }. $$

(A.58)

Now note that as $\left(\frac{\left(1-q\right)\left(1-\xi \right)+\left(\frac{1-\delta }{\delta}\right)\left(1-{p}_H\right)}{1-\frac{p_H}{p_L}\left(q+\left(1-q\right)\xi \right)+\left(\frac{1-\delta }{\delta}\right)\left(1-{p}_H\right)}\right)$ and ${\overline{k}}_L>{k}_H^{\ast }$ it follows that $\frac{{\mathrm{\partial \Lambda}}_{\overline{C}}}{\mathrm{\partial \upgamma }}>0$.

1.3.1 Analysis of optimal policy with single firm type

Planner’s First Order Condition

Equation (28) yields a first order condition for the government’s optimal policy problem when there is a single firm of type i. From Eqs. (16) and (17) the derivatives of the equilibrium contracts with respect to the parameter γ are as follows

$$ \frac{d{k}_i^{\ast }}{d\gamma}=-\left(\frac{1}{1-\alpha}\right)\left(\frac{\left(1-q\right)+\left(\frac{1-\delta }{\delta}\right)\left(1-{p}_i\right)\left(\frac{1}{1-\xi}\right)}{p_i\varphi \alpha {\left[{k}_i^{\ast}\right]}^{\alpha -1}}\right){k}_i^{\ast }, $$

(A.59)

$$ \frac{d{C}_i^{\ast }}{d\gamma}=\frac{1}{\delta}\left(\frac{1}{1-\xi}\right)\left(\frac{\alpha }{1-\alpha}\right)\left(\frac{\frac{1}{\alpha}\left(q+\left(1-q\right)\xi \right)-{p}_i\varphi \alpha {\left[{k}_i^{\ast}\right]}^{\alpha -1}}{p_i\varphi \alpha {\left[{k}_i^{\ast}\right]}^{\alpha -1}}\right){k}_i^{\ast }. $$

(A.60)

Substituting these derivatives into Eq. (28) yields the following

$$ {\displaystyle \begin{array}{c}\frac{\partial {\mathcal{U}}_i}{\partial \gamma }=-\left(\left(\frac{1-q}{1-\alpha}\right)+\left(\frac{1-\delta }{\delta}\right)\left(1-{p}_i\right)\left(\frac{1}{1-\xi}\right)\right){k}_i^{\ast}\\ {}+\left(\frac{1-q}{1-\alpha}\right)\left(\frac{1-\delta }{\delta}\right)\left(1-{p}_i\right)\frac{k_i^{\ast }}{p_i\varphi \alpha {\left[{k}_i^{\ast}\right]}^{\alpha -1}}.\end{array}} $$

(A.61)

Thus the value of γ that sets the government’s first order condition to zero is such that the marginal product of the project is given by the following equation

$$ {p}_i\varphi \alpha {\left[{k}_i^{\ast}\right]}^{\alpha -1}=\left(\frac{1+\left(\frac{1-\delta }{\delta}\right)\left(1-{p}_i\right)}{1+\left(\frac{1-\delta }{\delta}\right)\left(1-{p}_i\right)\left(\frac{1-\alpha }{1-q}\right)\left(\frac{1}{1-\xi}\right)}\right). $$

(A.62)

As the marginal product at the first best loan size is 1, it follows that the first best loan size is achieved only if this will only occur if (q + (1 − q) ξ) = α. Substituting Eq. (17) into Eq. (A.62) and rearranging the optimal γ can be written as

$$ \gamma =\xi +\left(\frac{\left(1-q\right)\left(1-\xi \right)+\left[1-\left(1-\alpha \right)\left(\frac{q+\left(1-q\right)\xi }{\left(1-q\right)\left(1-\xi \right)}\right)\right]\left(\frac{1-\delta }{\delta}\right)\left(1-{p}_i\right)}{\left[\left(1-q\right)+\left(\frac{1-\eta }{\eta}\right)\left(1-{p}_i\right)\left(\frac{1}{1-\xi}\right)\right]\left[1+\left(\frac{1-\delta }{\delta}\right)\left(1-{p}_i\right)\left(\frac{1-\alpha }{1-q}\right)\left(\frac{1}{1-\xi}\right)\right]}\right). $$

(A.63)

It follows from the above that a sufficient condition for the government to impose a binding regulatory constraint such that γ > ξ is the following

$$ \left(\frac{1-\delta }{\delta}\right)\left[1-\left(1-\alpha \right)\left(\frac{q+\left(1-q\right)\xi }{\left(1-q\right)\left(1-\xi \right)}\right)\right]>0. $$

(A.64)

Proof of $\frac{d{\gamma}_i^{OPT}}{d{p}_i}>0$

The implicit function theorem can be used to show that $\frac{d{\gamma}_i^{OPT}}{d{p}_i}>0$. First, consider Λ_{F, i}, a simple transformation of Eq. (A.61) defined as

$$ {\Lambda}_{F,i}\equiv \left(\frac{1-\alpha }{1-q}\right)\frac{\partial {\mathcal{U}}_i}{\partial \gamma}\frac{1}{k_i^{\ast }}. $$

(A.65)

Now note the partial derivatives of this function with respect to γ and p_i are

$$ \frac{\partial {\Lambda}_{F,i}}{\partial \gamma }=-\left(\frac{\left(1+\left(\frac{1-\delta }{\delta}\right)\left(1-{p}_i\right)\right)\left(\left(1-q\right)+\left(\frac{1-\delta }{\delta}\right)\left(1-{p}_i\right)\left(\frac{1}{1-\xi}\right)\right)}{{\left[\left(q+\left(1-q\right)\gamma \right)+\left(\frac{1-\delta }{\delta}\right)\left(1-{p}_i\right)\left(\frac{\gamma -\xi }{1-\xi}\right)\right]}^2}\right)<0, $$

(A.66)

and

$$ \frac{\partial {\Lambda}_{F,i}}{\partial {p}_i}=\left[\left(\frac{1-\alpha }{1-q}\right)-\left(\frac{\left(q+\left(1-q\right)\xi \right)\left(\left(1-\gamma \right)\right)}{{\left[\left(q+\left(1-a\right)\gamma \right)+\left(\frac{1-\delta }{\delta}\right)\left(1-{p}_i\right)\left(\frac{\gamma -\xi }{1-\xi}\right)\right]}^2}\right)\right]\left(\frac{1-\delta }{\delta}\right)\left(\frac{1}{1-\xi}\right). $$

(A.67)

The latter derivative is positive around the optimum, that is around the point where Λ_{F, i} = 0. To see this note that at Λ_{F, i} the following must hold

$$ \left(\frac{1-\alpha }{1-q}\right)\left(\frac{1}{1-\xi}\right)\left(\frac{1-\delta }{\delta}\right)=\left(\frac{1}{1-{p}_i}\right)\left(\frac{\left(1-q\right)\left(1-\gamma \right)+\left(\frac{1-\delta }{\delta}\right)\left(1-{p}_i\right)\left(\frac{1-\gamma }{1-\xi}\right)}{\left(q+\left(1-q\right)\gamma \right)+\left(1-{p}_i\right)\left(\frac{1-\delta }{\delta}\right)\left(\frac{\gamma -\xi }{1-\xi}\right)}\right). $$

(A.68)

Substituting this into the above equation and rearranging shows that $\frac{\partial {\Lambda}_{F,i}}{\partial {p}_i}>0$. It follows immediately from the implicit function theorem that $\frac{d{\gamma}_i^{OPT}}{d{p}_i}>0$ and thus ${\gamma}_H^{OPT}<{\gamma}_L^{OPT}$.

1.3.2 Analysis of optimal policy with two firm types

Equation (32) is the first order condition for the government’s optimal policy problem when there are two firm types. As the high-risk firm will receive $\left({k}_H^{\ast },{R}_H^{\ast },{C}_H^{\ast}\right)$, the derivatives $\frac{d{k}_H^{\ast }}{d\gamma}$ and $\frac{d{C}_H^{\ast }}{d\gamma}$ are given by Eqs. (A.59) and (A.60) respectively.

The value of ${\hat{C}}_L$ can be found from combining Eqs. (8) and (19) with the value of ${\hat{k}}_L$ defined in Eq. (31) such that ${\hat{C}}_L$ must satisfy the following equation

$$ {\hat{C}}_L=\frac{1}{\delta}\left(\frac{\frac{p_H}{p_L}\left({p}_L\varphi {\left({\hat{k}}_L\right)}^{\alpha -1}-\Big(q+\left(1-q\right)\xi \right){\hat{k}}_L-{\pi}_H^{\ast }}{1-\frac{p_H}{p_L}\left(q+\left(1-q\right)\xi \right)+\left(\frac{1-\delta }{\delta}\right)\left(1-{p}_H\right)}\right). $$

(A.69)

The derivative $\frac{d{C}_H^{\ast }}{d\gamma}$ can then be found by totally differentiating the above equation with respect to γ.

In the first case where ${\hat{C}}_L\ge {\overline{C}}_L$, the derivative is simply

$$ \frac{d{\hat{C}}_L}{d\gamma}=\frac{1}{\delta}\left(\frac{\left(1-q\right)+\left(\frac{1-\delta }{\delta}\right)\left(1-{P}_H\right)\left(\frac{1}{1-\xi}\right)}{1-\left(\frac{p_H}{p_L}\right)\left(q+\left(1-q\right)\xi \right)+\left(\frac{1-\delta }{\delta}\right)\left(1-{P}_H\right)}\right){k}_H^{\ast }>0. $$

(A.70)

It follows that when ${\hat{C}}_L\ge {\overline{C}}_L,\frac{d{\hat{C}}_L}{d\gamma}>0$ and an increase in regulation results in a larger collateral requirement.

Next consider the case where ${\hat{C}}_L<{\overline{C}}_L$. The derivative becomes

$$ \frac{d{\hat{C}}_L}{d\gamma}=\frac{1}{\delta}\left(\frac{1}{1-\xi}\right)\left(\frac{\left(1-q\right)\left(1-\xi \right)+\left(\frac{1-\delta }{\delta}\right)\left(1-{P}_H\right)\left(\frac{1}{1-\xi}\right)\iota }{1-\left(\frac{p_H}{p_L}\right)\left(q+\left(1-q\right)\xi \right)+\left(\frac{1-\delta }{\delta}\right)\left(1-{P}_H\right)-\frac{{\hat{k}}_L}{k_H^{\ast }}\iota}\right){k}_H^{\ast }, $$

(A.71)

where

$$ \iota \equiv \left(\frac{1-\xi }{\gamma -\xi}\right)\frac{p_H}{p_L}\left(\alpha {p}_L\varphi {\left({k}_L\right)}^{\alpha -1}-\left(q+\left(1-q\right)\xi \right)\right)\ge 0. $$

(A.72)

From Eq. (31) it follows that for any γ ∈ [1, ξ) that $\alpha {p}_L\varphi {\left({\hat{k}}_L\right)}^{\alpha -1}\ge \left(q+\left(1-q\right)\xi \right)$ and thus ι > 0 and is increasing in γ above $\overline{\upgamma}$. As $\frac{d{\hat{C}}_L}{d\gamma}$ is decreasing in ι, it follows that the derivative will be decreasing in γ.

1.3.3 Analysis of Returns to scale

Impact of α on ${\hat{C}}_L$ and ${\hat{k}}_L$

First, note that the incentive compatibility constraint binds so long as F = 0 where the function F is defined below as

$$ {\displaystyle \begin{array}{c}F\equiv \frac{p_H}{p_L}\left({p}_L\varphi {\hat{k}}_L^{\alpha }-\left(q+\left(1-q\right)\xi \right){\hat{k}}_L\right)\\ {}-\delta \left[1-\frac{p_H}{p_L}\left(q+\left(1-q\right)\xi \right)\left(\frac{1-\delta }{\delta}\right)\left(1-{P}_H\right)\right]{\hat{C}}_L-{U}_H^{\ast }=0\end{array}} $$

(A.73)

When ${C}_L\ge {\overline{C}}_L$ the derivative for ${\hat{k}}_L$ with respect to α is

$$ \frac{\partial {\hat{k}}_L}{\partial \alpha }=\frac{1}{{\left(1-\alpha \right)}^2\alpha}\left(\alpha \ln \left(\frac{\alpha {p}_L\varphi }{q+\left(1-q\right)\xi}\right)+\left(1-\alpha \right)\right){\hat{k}}_L, $$

(A.74)

thus given the assumption that αp_Hφ ≥ 1 the derivative is strictly positive. The derivative with respect to ${\hat{C}}_L$ can be found through application of the implicit function theorem to the function F. The partial derivative of F with respect to ${\hat{C}}_L$ is

$$ \frac{\partial F}{\partial {\hat{C}}_L}=-\delta \left[1-\frac{p_H}{p_L}\left(q+\left(1-q\right)\xi \right)+\left(\frac{1-\delta }{\delta}\right)\left(1-{P}_H\right)\right]<0. $$

(A.75)

The partial derivative of F with respect to α is

$$ \frac{\partial F}{\partial \alpha }=\frac{p_H}{p_L}\left(\alpha {p}_L\varphi {\hat{k}}_L^{\alpha -1}-\left(q+\left(1-q\right)\xi \right)\right)\frac{\partial {\hat{k}}_L}{\partial \alpha }-\frac{\partial {U}_H^{\ast }}{\partial \alpha }. $$

(A.76)

Note that when ${C}_L\ge {\overline{C}}_L$ then $\alpha {p}_L\varphi {\hat{k}}_L^{\alpha -1}=\left(q+\left(1-q\right)\xi \right)$ and thus,

$$ \frac{\partial F}{\partial \alpha }=\frac{\partial {U}_H^{\ast }}{\partial \alpha }, $$

(A.77)

where

$$ \frac{\partial {U}_H^{\ast }}{\partial \alpha }={p}_H\varphi\ \ln \left({k}_H^{\ast}\right){\left({k}_H^{\ast}\right)}^{\alpha }, $$

(A.78)

is strictly positive due to the assumption that αp_Hφ ≥ 1 . Thus from the implicit function theorem it follows that $\frac{d{\hat{C}}_L}{d\alpha}<0$.

When ${C}_L<{\overline{C}}_L$ the derivative for ${\hat{k}}_L$ with respect to α becomes

$$ {\displaystyle \begin{array}{c}\frac{\partial F}{\partial {\hat{C}}_L}=\frac{p_H}{p_L}\left(\alpha {p}_L\varphi {\hat{k}}_L^{\alpha -1}-\left(q+\left(1-q\right)\xi \right)\right){\hat{k}}_L^{\alpha -1}\left(\frac{1-\xi }{\gamma -\xi}\right)\\ {}-\delta \left[1-\frac{p_H}{p_L}\left(q+\left(1-q\right)\xi \right)+\left(\frac{1-\delta }{\delta}\right)\left(1-{P}_H\right)\right].\end{array}} $$

(A.79)

The sign of the above equation can be found by noting that ${\hat{k}}_L\ge {k}_L^{\ast }$ and thus

$$ \alpha {p}_L\varphi {\hat{k}}_L^{\alpha -1}\le \left[q+\left(1-q\right)\gamma \right]+\left(\frac{1-\delta }{\delta}\right)\left(1-{P}_L\right)\left(\frac{\gamma -\xi }{1-\xi}\right). $$

(A.80)

Substituting this into inequality into the above derivative is enough to show that $\frac{\partial F}{\partial {\hat{C}}_L}<0$ as before.

Next, note that the derivative of F with respect to α is

$$ \frac{\partial F}{\partial \alpha }=\frac{p_H}{p_L}\left({p}_L\varphi {k}_L^{\alpha }\ \ln \left({k}_L\right)\right)-\frac{\partial {U}_H^{\ast }}{\partial \alpha }, $$

(A.81)

where

$$ \frac{\partial {U}_H^{\ast }}{\partial \alpha }={p}_H\varphi\ \ln \left({k}_H^{\ast}\right){\left({k}_H^{\ast}\right)}^{\alpha }. $$

(A.82)

Thus the derivative can be written as

$$ \frac{\partial F}{\partial \alpha }={p}_H\varphi \left[{k}_L^{\alpha }\ \ln \left({k}_L\right)-{k}_H^{\alpha }\ \ln \left({k}_H\right)\right], $$

(A.83)

which is strictly positive. Hence through application of the implicit function theorem $\frac{d{\hat{C}}_L}{d\alpha}>0$.

Proof that $\frac{d\overline{\gamma}}{d\alpha}<0$

The derivative $\frac{d\overline{\gamma}}{d\alpha}$ can be found by applying the implicit function theorem to the function ${\Lambda}_{\overline{C}}$ as defined in Eq. (A.57). The partial derivative of ${\Lambda}_{\overline{C}}$ with respect to α is

$$ \frac{\partial {\Lambda}_{\overline{C}}}{\partial \alpha }=\frac{1}{\delta }{p}_H\varphi \left(\frac{\ln \left({\overline{k}}_L\right){\left({\overline{k}}_L\right)}^{\alpha }-\ln \left({k}_H^{\ast}\right){\left({k}_H^{\ast}\right)}^{\alpha }}{1-\frac{p_H}{p_L}\left(q+\left(1-q\right)\xi \right)+\left(\frac{1-\delta }{\delta}\right)\left(1-{p}_H\right)}\right)>0. $$

As this partial derivative is strictly positive, and it has already been found that $\frac{\partial {\Lambda}_{\overline{C}}}{\partial \gamma }>0$, it follows from the implicit function theorem that $\frac{d\overline{\gamma}}{d\alpha}<0$.

Proof that $\frac{d{\gamma}^{\ast }}{d\alpha}<0$

The derivative $\frac{d{\gamma}^{\ast }}{d\alpha}$ can be found by applying the implicit function theorem to the function Λ_IC as defined in Eq. (21). The partial derivative of Λ_IC with respect to α is

$$ {\displaystyle \begin{array}{c}\frac{{\mathrm{\partial \Lambda}}_{IC}}{\partial \alpha }={\left(\frac{1}{1-\alpha}\right)}^2{\left(\frac{p_{H\left[\left(q+\left(1-q\right)\gamma \right)+\left(\frac{1-\delta }{\delta}\right)\left(1-{p}_L\right)\left(\frac{\gamma -\xi }{1-\xi}\right)\right]}}{p_{L\left[\left(q+\left(1-q\right)\gamma \right)+\left(\frac{1-\delta }{\delta}\right)\left(1-{p}_H\right)\left(\frac{\gamma -\xi }{1-\xi}\right)\right]}}\right)}^{\frac{\alpha }{1-\alpha }}\\ {}\times \ln \left(\frac{p_H\left[\left(q+\left(1-q\right)\gamma \right)+\left(\frac{1-\delta }{\delta}\right)\left(1-{p}_L\right)\left(\frac{\gamma -\xi }{1-\xi}\right)\right]}{p_L\left[\left(q+\left(1-q\right)\gamma \right)+\left(\frac{1-\delta }{\delta}\right)\left(1-{p}_H\right)\left(\frac{\gamma -\xi }{1-\xi}\right)\right]}\right)\\ {}+{\left(\frac{1}{1-\alpha}\right)}^2\left(\frac{\frac{1}{\delta}\left(\frac{p_H}{p_L}-1\right)\left(\frac{\gamma -\xi }{1-\xi}\right)}{\left(q+\left(1-q\right)\gamma \right)+\left(\frac{1-\delta }{\delta}\right)\left(1-{p}_L\right)\left(\frac{\gamma -\xi }{1-\xi}\right)}\right).\end{array}} $$

(A.84)

Now note that around Λ_IC = 0, that is around the point where γ = γ^*, the following must.hold

$$ {\displaystyle \begin{array}{c}{\left(\frac{p_H\left[\left(q+\left(1-q\right)\gamma \right)+\left(\frac{1-\delta }{\delta}\right)\left(1-{p}_L\right)\left(\frac{\gamma -\xi }{1-\xi}\right)\right]}{p_L\left[\left(q+\left(1-q\right)\gamma \right)+\left(\frac{1-\delta }{\delta}\right)\left(1-{p}_H\right)\left(\frac{\gamma -\xi }{1-\xi}\right)\right]}\right)}^{\frac{\alpha }{1-\alpha }}\\ {}=\left[1-\left(\frac{\alpha }{1-\alpha}\right)\left(\frac{\frac{1}{\delta}\left(\frac{p_H}{p_L}-1\right)\left(\frac{\gamma -\xi }{1-\xi}\right)}{\left(q+\left(1-q\right)\gamma \right)+\left(\frac{1-\delta }{\delta}\right)\left(1-{p}_L\right)\left(\frac{\gamma -\xi }{1-\xi}\right)}\right)\right].\end{array}} $$

(A.85)

Thus by substituting this equation into the partial derivative above, at the point γ = γ^*, the partial derivative can be written as

$$ \frac{{\mathrm{\partial \Lambda}}_{IC}}{\partial \alpha }=\left(\frac{1-\alpha }{\alpha}\right){\left(\frac{1}{1-\alpha}\right)}^2\left[\left(1-x\right)\ln \left(1-x\right)+x\right], $$

(A.86)

where

$$ x=\left(\frac{\alpha }{1-\alpha}\right)\left(\frac{\frac{1}{\delta}\left(\frac{p_L}{p_H}-1\right)\left(\frac{\gamma -\xi }{1-\xi}\right)}{\left(q+\left(1-q\right)\gamma \right)+\left(\frac{1-\delta }{\delta}\right)\left(1-{p}_L\right)\left(\frac{\gamma -\xi }{1-\xi}\right)}\right)>0. $$

(A.87)

Thus it follows that $\frac{{\mathrm{\partial \Lambda}}_{IC}}{\partial \alpha }>0$ and as it has already been established that $\frac{{\mathrm{\partial \Lambda}}_{IC}}{\partial \gamma }>0$ from application of the implicit function theorem it follows that $\frac{d{\upgamma}^{\ast }}{d\alpha}<0$.

1.3.4 Analysis of Collateral

Impact of δ on ${\hat{C}}_L$ and ${\hat{k}}_L$

When ${C}_L\ge {\overline{C}}_L$ the derivative for ${\hat{k}}_L$ with respect to δ is equal to zero and thus the partial derivative of F, as defined by Eq. (A.73) can be written as

$$ \frac{\partial F}{\partial \delta }=-\frac{\partial {U}_H^{\ast }}{\partial \delta }, $$

(A.88)

where

$$ \frac{\partial {U}_H^{\ast }}{\partial \delta }={\left(\frac{1}{\delta}\right)}^2\left(1-{p}_H\right)\left(\frac{\gamma -\xi }{1-\xi}\right){k}_H^{\ast}\ge 0. $$

(A.89)

Thus it follows that $\frac{\partial F}{\partial \delta }<0$ and through Eq. (A.75) which states that $\frac{\partial F}{\partial {\hat{C}}_L}<0$ and application of the implicit function theorem it follows that $\frac{d{\hat{C}}_L}{d\delta}>0$. As $\frac{\partial {\hat{k}}_L}{\partial \delta }=0$ this implies that the collateral ratio is increasing in δ when ${C}_L\ge {\overline{C}}_L$.

When ${C}_L\ge {\overline{C}}_L$ the collateral ratio is determined by the following relationship

$$ \frac{{\hat{C}}_L}{{\hat{k}}_L}=\frac{1}{\varepsilon}\left(\frac{\gamma -\xi }{1-\xi}\right), $$

(A.90)

and thus the collateral ratio is decreasing in δ when ${C}_L\ge {\overline{C}}_L$.

Proof that $\frac{d\overline{\gamma}}{d\delta}>0$

The derivative $\frac{d\overline{\gamma}}{d\delta}$ can be found by applying the implicit function theorem to the function ${\Lambda}_{\overline{C}}$ as defined in Eq. (A.57). The partial derivative of ${\Lambda}_{\overline{C}}$ with respect to δ can be written as

$$ {\displaystyle \begin{array}{c}\frac{{\mathrm{\partial \Lambda}}_{\overline{C}}}{\mathrm{\partial \updelta }}=-\frac{1}{\delta }{\Lambda}_{\overline{C}}-\frac{1}{\delta^3}\left(1-{p}_H\right)\left(\frac{\frac{p_H}{p_L}\left({p}_L\varphi {\overline{k}}_L^{\alpha -1}-\left(q+\left(1-q\right)\xi \right)\right){\overline{k}}_L-{U}_H^{\ast }}{{\left(1-\frac{p_H}{p_L}\left(q+\left(1-q\right)\xi \right)+\left(\frac{1-\delta }{\delta}\right)\left(1-{p}_H\right)\right)}^2}\right)\\ {}+\frac{1}{\delta}\left(\frac{\frac{\partial {U}_H^{\ast }}{\mathrm{\partial \updelta }}}{1-\frac{p_H}{p_L}\left(q+\left(1-q\right)\xi \right)+\left(\frac{1-\delta }{\delta}\right)\left(1-{p}_H\right)}\right),\end{array}} $$

(A.91)

where $\frac{{\partial U}_H^{\ast }}{\partial \delta }={\left(\frac{1}{\delta}\right)}^2\left(1-{p}_H\right)\left(\frac{\gamma -\xi }{1-\xi}\right){k}_H^{\ast }>0$.

Note that at the point where $\upgamma =\overline{\upgamma}$ it must be the case that ${\Lambda}_{\overline{C}}=0$ and thus the above equation can be written as

$$ \frac{{\mathrm{\partial \Lambda}}_{\overline{C}}}{\partial \delta }=-\frac{1}{\delta^3}\left(1-{p}_H\right)\left(\frac{\gamma -\xi }{1-\xi}\right)\left(\frac{{\overline{k}}_L-{k}_H^{\ast }}{1-\frac{p_H}{p_L}\left(q+\left(1-q\right)\xi \right)\left(\frac{1-\delta }{\delta}\right)\left(1-{p}_H\right)}\right)<0. $$

(A.92)

As this partial derivative is strictly negative, and it has already been found that $\frac{{\mathrm{\partial \Lambda}}_{\overline{C}}}{\partial \gamma }>0$, it follows from the implicit function theorem that $\frac{d\overline{\gamma}}{d\delta}>0$.

Proof that $\frac{d{\gamma}^{\ast }}{d\delta}>0$

The derivative $\frac{d{\gamma}^{\ast }}{d\delta}$ can be found by applying the implicit function theorem to the function Λ_IC as defined in Eq. (21). The partial derivative of Λ_IC with respect to δ is

$$ {\displaystyle \begin{array}{c}\frac{{\mathrm{\partial \Lambda}}_{\overline{C}}}{\partial \delta }=B\left(q+\left(1-q\right)\gamma \right){\left(\frac{p_H\left[\left(q+\left(1-q\right)\gamma \right)+\left(\frac{1-\delta }{\delta}\right)\left(1-{p}_L\right)\left(\frac{\gamma -\xi }{1-\xi}\right)\right]}{p_L\left[\left(q+\left(1-q\right)\gamma \right)+\left(\frac{1-\delta }{\delta}\right)\left(1-{p}_H\right)\left(\frac{\gamma -\xi }{1-\xi}\right)\right]}\right)}^{\frac{1}{1-\alpha }}\\ {}-B\frac{1}{p_L}\left[\left(q+\left(1-q\right)\gamma \right)-\left(1-{p}_L\right)\left(\frac{\gamma -\xi }{1-\xi}\right)\right],\end{array}} $$

(A.93)

where

$$ B=\frac{\frac{1}{\delta^2}\left(\frac{\alpha }{1-\alpha}\right)\left(\frac{p_L}{p_H}-1\right)\left(\frac{\gamma -\xi }{1-\xi}\right){p}_L}{{\left[\left(q+\left(1-q\right)\gamma \right)+\left(\frac{1-\delta }{\delta}\right)\left(1-{p}_L\right)\left(\frac{\gamma -\xi }{1-\xi}\right)\right]}^2}>0. $$

(A.94)

Now note that around ${\Lambda}_{\overline{C}}=0$, that is around the point where γ = γ^*, Eq. (A.85) must hold and the partial derivative can be written as

$$ {\displaystyle \begin{array}{c}\frac{{\mathrm{\partial \Lambda}}_{\overline{C}}}{\partial \delta }=-B\left(\frac{1}{p_L}-1\right)\left(q+\left(1-q\right)\xi \right)\left(\frac{1-\gamma }{1-\xi}\right)\\ {}-B\left(q+\left(1-q\right)\gamma \right)\left[1-{\left(\frac{p_H\left[\left(q+\left(1-q\right)\gamma \right)+\left(\frac{1-\delta }{\delta}\right)\left(1-{p}_L\right)\left(\frac{\gamma -\xi }{1-\xi}\right)\right]}{p_L\left[\left(q+\left(1-q\right)\gamma \right)+\left(\frac{1-\delta }{\delta}\right)\left(1-{p}_H\right)\left(\frac{\gamma -\xi }{1-\xi}\right)\right]}\right)}^{\frac{1}{1-\alpha }}\right],\end{array}} $$

which is strictly negative. Thus it follows that $\frac{{\mathrm{\partial \Lambda}}_{IC}}{\partial \delta }<0$ and as it has already been established that $\frac{{\mathrm{\partial \Lambda}}_{IC}}{\partial \gamma }>0$ from application of the implicit function theorem it follows that $\frac{d{\gamma}^{\ast }}{d\delta}>0$.