Over-the-counter trade and the value of assets as collateral

Abstract

We study asset pricing within a general equilibrium model where unsecured credit is ruled out, and a real asset helps agents carry out mutually beneficial transactions by serving as collateral. A unique feature of our model is that the agent who provides the loan might have a low valuation for the collateral asset. Nevertheless, the lender rationally chooses to accept the collateral because she can access a secondary asset market where she can sell the asset. Following a recent strand of the finance literature, based on the influential work of Duffie et al. (Econometrica 73(6):1815–1847, 2005), we model this secondary asset market as an over-the-counter market characterized by search and bargaining frictions. We study how the asset’s property to serve as collateral affects its equilibrium price, and how the asset price and the economy’s welfare are affected by the degree of liquidity in the secondary asset market.

Appendix

Proof of Lemma 3.2

We define the Langrangian function for the LW bargaining problem as

$$\begin{aligned} L= & {} u(q) - (1-l) \varphi d - (1-l)b - \lambda _1 ( d - m) - \lambda _2 ( - b) - \lambda _3 ( b - a)\\&-\,\lambda _4 [q - \varphi d - (1-l) b - l(1-\delta + \alpha _\mathrm{_S} \lambda \delta ) a ], \end{aligned}$$

where $\lambda _1$, $\lambda _2$, $\lambda _3$, and $\lambda _4$ are Lagrange multipliers. The FOCs are

$$\begin{aligned}&\{ q \} : u'(q) - \lambda _4 = 0, \end{aligned}$$

(6.1)

$$\begin{aligned}&\{ d \} : -(1-l)\varphi - \lambda _1 + \lambda _4 \varphi = 0, \end{aligned}$$

(6.2)

$$\begin{aligned}&\{ b \} : -(1-l) + \lambda _2 -\lambda _3 + \lambda _4 (1-l) = 0, \end{aligned}$$

(6.3)

$$\begin{aligned}&\{ \lambda _1 \} : d - m \le 0, \ \ \lambda _1 \ge 0, \end{aligned}$$

(6.4)

$$\begin{aligned}&\{ \lambda _2 \} : - b \le 0, \ \ \ \ \ \ \lambda _2 \ge 0, \end{aligned}$$

(6.5)

$$\begin{aligned}&\{ \lambda _3 \} : b - a \le 0, \ \ \ \lambda _3 \ge 0, \end{aligned}$$

(6.6)

$$\begin{aligned}&\{ \lambda _4 \} : q - \varphi d - (1-l) b - l(1-\delta + \alpha _\mathrm{_S} \lambda \delta ) a = 0. \end{aligned}$$

(6.7)

• Case 1: $\lambda _1 = 0$, $\lambda _2 = 0$, $\lambda _3 = 0$ ($d < m$, $0 < b < a$)

  Equations (6.1) and (6.2) imply $u'(q) = 1-l$. On the other hand, Eqs. (6.1) and (6.3) imply $u'(q) = 1$. These contradict each other.

• Case 2: $\lambda _1 = 0$, $\lambda _2 = 0$, $\lambda _3 > 0$ ($d < m$, $0 < b = a$)

  Equations (6.1) and (6.2) imply $u'(q) = 1-l$. On the other hand, Eqs. (6.1) and (6.3) imply $\lambda _3 = (1-l)[ u'(q) - 1]$. Since $\lambda _3 > 0$, this implies $u'(q) > 1$. Again, these contradict each other.

• Case 3: $\lambda _1 = 0$, $\lambda _2 > 0$, $\lambda _3 = 0$ ($d < m$, $0 = b < a$)

  Equations (6.1) and (6.2) imply $u'(q) = 1-l$. Hence, $q=q^{**}$. Equations (6.1) and (6.3) imply $\lambda _2 = (1-l)[ 1 - u'(q^{**})]$, which is consistent with $\lambda _2>0$. Equation (6.5) implies $b=0$. Then, from Eq. (6.7), we get $d = [q^{**} - l(1-\delta +\alpha _\mathrm{_S} \lambda \delta )a]/\varphi$, i.e., $d=(q^{**}-xa)/\varphi$. To sum up, in Case 3 the solution is $q= q^{**}$, $d = (q^{**} - xa)/\varphi$, and $b= 0$. Notice that in this case we have $\pi (m,a) > q^{**}$, which is consistent with the fact that $d<m$.

• Case 4: $\lambda _1 = 0$, $\lambda _2 > 0$, $\lambda _3 > 0$ ($d < m$, $0 = b = a$)

  This case is similar to Case 3, with the exception that here $a=0$. It is easy to check that the solution is $q = q^{**}$, $d=q^{**}/\varphi$, and $b = 0$. Since here $d<m$ and $a=0$, we have $\pi (m,0)>q^{**}$.

• Case 5: $\lambda _1 > 0$, $\lambda _2 = 0$, $\lambda _3 = 0$ ($d = m$, $0 < b < a$)

  Equations (6.1) and (6.3) imply $u'(q) = 1$, and so $q=q^*$. Notice that $q^{**}>q^{*}$ because $u'' <0$. Equations (6.1) and (6.2) imply $\lambda _1 = \varphi [ u'(q^*) - (1-l) ]$, which is consistent with $\lambda _1>0$. Equation (6.4) implies $d=m$. Then, from Eq. (6.7), we get $b = \big [q^{*} -\varphi m - l(1-\delta +\alpha _\mathrm{_S} \lambda \delta )a\big ]/(1-l)$, i.e., $b=\big [q^*-\pi (m,a)\big ]/(1-l)$. To sum up, the solution is $q = q^{*}$, $d= m$, and $b=[q^*-\pi (m,a)]/(1-l)$. Since here $0 < b < a$, we have $q^*-(1-l)a<\pi (m,a)< q^*$.

• Case 6: $\lambda _1 > 0$, $\lambda _2 = 0$, $\lambda _3 > 0$ ($d = m$, $0 < b = a$)

  Equations (6.1) and (6.3) imply $u'(q) > 1$. Equations (6.1) and (6.2) imply $u'(q) > 1 - l$. Therefore, the solution of q must satisfy $q < q^{*}$. Equation (6.4) implies $d=m$. Equation (6.6) implies $b=a$. Then, from Eq. (6.7), we get $q = \varphi m + [1-l+l(1-\delta +\alpha _\mathrm{_S} \lambda \delta )] a$, i.e., $q=\pi (m,a)+(1-l)a$. To sum up, in this case the solution is $q = \pi (m,a) + (1-l)a$, $d= m$, and $b = a$. Notice that since here $q<q^*$, we have $\pi (m,a) < q^*-(1-l)a$.

• Case 7: $\lambda _1 > 0$, $\lambda _2 > 0$, $\lambda _3 = 0$ ($d = m$, $0 = b < a$)

  Equations (6.1) and (6.3) imply $u'(q) < 1$. Equations (6.1) and (6.2) imply $u'(q) > 1 - l$. Thus, $1-l < u'(q) < 1$, which means that $q^{*} < q < q^{**}$. Equation (6.4) implies $d=m$. Equation (6.5) implies $b=0$. Then, from Eq. (6.7), we get $q = \varphi m + l(1-\delta +\alpha _\mathrm{_S} \lambda \delta ) a$, i.e., $q=\pi (m,a)$. To sum up, in this case the solution is $q= \pi (m,a)$, $d = m$, and $b = 0$. Since here $q^*< q < q^{**}$, we have $q^* < \pi (m,a) < q^{**}$.

• Case 8: $\lambda _1 > 0$, $\lambda _2 > 0$, $\lambda _3 > 0$ ($d = m$, $0 = b = a$)

  Equations (6.1) and (6.2) imply $u'(q) > 1 - l$. Thus, $q < q^{**}$. Equation (6.4) implies $d=m$. Equation (6.5) implies $b=0$. Then, from Eq. (6.7), we get $q = \varphi m$. Summing up, we have $q = \varphi m$, $d= m$, and $b = 0$. Since here $q < q^{**}$, we have $q^{**} > \pi (m,0)$.

$\square$

Proof of Lemma 3.3

As a first step, we highlight some important properties of $J: \mathbf {R}^2_+ \rightarrow \mathbf {R}$. First, this function is continuous throughout its domain. To see why this is the case, recall that the solution to the LW market bargaining problem is continuous, and so is u. Hence, J is also continuous.

Second, the function J is differentiable within each of the four regions defined in Lemma 3.2. To see this point, notice that the solution to the bargaining problem is differentiable within each of the four regions defined in Lemma 3.2, and u is also (twice continuously) differentiable. Therefore, J is differentiable within each of the four regions.

Third, J is weakly concave in both arguments everywhere. To see this point, recall that J is continuous everywhere and differentiable within each region, thus $J_1$ and $J_2$ are well defined within each region. Moreover, $J_1$ and $J_2$ are constant in Regions 1 and 3, and strictly decreasing in Regions 2 and 4 in $\hat{m}$ and $\hat{a}$ because $u''<0$. Thus, J is weakly concave everywhere.

Given the properties of J, we have the following:

• (a) Since J is weakly concave and differentiable (with the exception of the boundary points), the optimal choice ($\hat{m}$, $\hat{a}$) in each region satisfies $\nabla J(\hat{m},\hat{a})=\mathbf {0}$.

• (b) If the asset price is equal to the fundamental value, i.e., $\psi =\beta (1-l)$, Regions 2, 3, and 4 are ruled out, because $\psi = \beta (1-l)$ implies that $J_2^1=0$, while $J_2^2$, $J_2^3$ and $J_2^4 > 0$ for any bundle ($\hat{m}$, $\hat{a}$). There is no reason to choose less $\hat{a}$ than $q^{**}/x$ in this case. On the other hand, the fact that $J_1^1 < 0$ in the case where $\varphi > \beta \hat{\varphi } (1-l)$, i.e., the cost of holding money is positive means that $\hat{m}=0$ because a buyer can purchase the amount $q^{**}$ without using any money. Hence, any bundle ($\hat{m}$, $\hat{a}$) with $\hat{m}=0$ and $\hat{a}\ge q^{**}/x$ is optimal.

• (c) When $\psi > \beta (1-l)$, any choice within Region 1 is ruled out because $J^1_2 < 0$. In Regions 2 and 4, $J_1$ and $J_2$ are strictly decreasing in $\hat{m}$ and $\hat{a}$, therefore the optimal choice ($\hat{m}$, $\hat{a}$) satisfying $J_1=0$ and $J_2=0$ is unique (given that $\varphi > \beta \hat{\varphi } (1-l)$ and $\psi > \beta (1-l)$). In Region 3, since any bundle ($\hat{m}$, $\hat{a}$) satisfies $J^3_1 = 0$ and $J^3_2 = 0$, the optimal choice is indeterminate.

$\square$

Proof of Lemma 4.2

From Lemma 3.1, it is straightforward to show that $\chi$ and c exist and are unique, given A, in the equilibrium. From Lemma 3.2 and 3.3, we have the following equations on each region in the steady state where $\varphi /\hat{\varphi }=1+\mu$.

• $1+\mu =\beta (1-l)$,   $\psi =\beta (1-l)$, $q=q^{**}$    and   $b=0$, in Region 1,

• $1+\mu =\beta u'(\tilde{q}_2(z))$,   $\psi =\beta \big [u'(\tilde{q}_2(z))x+1-l]$   and    $b=0$, in Region 2,

• $1+\mu =\beta$,   $\psi =\beta (1-l+x)$,   $q=q^{*}$,   and    $b=\tilde{b}(z)$, in Region 3,

• $1+\mu =\beta u'(\tilde{q}_4(z))$,    $\psi =\beta u'(\tilde{q}_4(z)) (1-l+x)$    and    $b=A$, in Region 4.

Clearly, $\psi$, z, q, and b exist and are unique in Region 1. In Regions 2 and 4, $\mu$ uniquely pins down z, q, and $\psi$. In addition, b is unique in all regions. Hence, equilibrium exists and is unique in Regions 2 and 4. In Region 3, z can obtain any value in the interval of $[q^*-(1-l+x)A, q^*-xA]$, even though $\psi$ and q are unique. It follows that b is also indeterminate because it is a function of z. Hence, they all exist, but z and b are not unique in Region 3. $\square$

Proof

Derivation of expressions $\bar{\mu }_i(A)$, $i=2,4$, in Sect. 4.2.

Consider Region 2. For any given A, the term $\bar{\mu }_2$ indicates the level of inflation that would lead to the same q as in the case where $z=0$. The latter is simply given by $q_{2,n}=xA$ (the second index, n, stands for “non-monetary” equilibrium). To identify the relationship between $\mu$ and q in an equilibrium where money is valued, set the derivative in (3.11) equal to zero, and evaluate it in steady state. This yields $1+\mu =\beta u'(q_{2,m}),$ where $q_{2,m}=z+xA$ (in the same spirit as above, the second index, m, stands for “monetary” equilibrium). Hence, $\bar{\mu }_2$ solves $\bar{\mu }_2 = \{\mu : q_{2,m}=q_{2,n}\}$, which implies that $\bar{\mu }_2=\beta u'(xA)-1$. The derivation of $\bar{\mu }_4$ follows identical steps. $\square$

Proof of Proposition 4.5

Consider first equilibria in Region 4. From the definition of steady state equilibrium, we know that in this region $b=A$. Also, from Proposition 4.3, we know that $\psi =\psi _4(\mu )$. Hence, in this region the haircut is given by ${\varXi }=1-1/\psi _4$. Consequently, we have

$$\begin{aligned} \frac{\partial {\varXi }}{\partial \mu }=&\frac{\partial {\varXi }}{\partial \psi _4}\, \frac{\partial \psi _4}{\partial \mu }=\frac{1}{\psi _4^2}\, \frac{\partial \psi _4}{\partial \mu },\\ \frac{\partial {\varXi }}{\partial \alpha _\mathrm{_S}}=&\frac{\partial H}{\partial \psi _4}\, \frac{\partial \psi _4}{\partial \alpha _\mathrm{_S}}=\frac{1}{\psi _4^2}\, \frac{\partial \psi _4}{\partial \alpha _\mathrm{_S}}, \end{aligned}$$

and since we have established that $\psi _4$ is increasing in both $\mu$ and $\alpha _\mathrm{_S}$, the same is true about ${\varXi }$.

In Region 3, it is impossible to obtain an expression for $\partial {\varXi }/\partial \mu$, since being in this region requires $\mu =\beta -1$. However, we can study how ${\varXi }$ depends on $\alpha _\mathrm{_S}$. To that end, recall from the definition of equilibrium that in Region 3 we have $b=\tilde{b}(z)=(1-l)^{-1}(q^*-z-xA)$. Also, from the discussion that follows Proposition 4.3, recall that the asset price is given by $\psi =\psi _4(\beta -1)=\beta (1-l+x)$. Hence, in Region 3 the haircut is given by

$$\begin{aligned} {\varXi }=1-\frac{q^*-z-xA}{(1-l)\beta (1-l+x)A.} \end{aligned}$$

It is now straightforward to verify that

$$\begin{aligned} \frac{\partial {\varXi }}{\partial \alpha _\mathrm{_S}}=\frac{\partial {\varXi }}{\partial x}\, \frac{\partial x}{\partial \alpha _\mathrm{_S}}=\frac{A(1-l+x)+\tilde{b}(z)(1-l)}{(1-l)\beta (1-l+x)^2A} \, l\lambda \delta , \end{aligned}$$

which is clearly strictly positive.$\square$