Optimal capital structures for private firms

Abstract

This article shows how to construct an optimal capital structure for a private firm. Since the agents who supply the firm’s capital are risk averse, they diversify by holding both debt and equity. This can mitigate, or even eliminate, the classical risk shifting problem. There is a wealth effect since the optimal capital structure, which can involve multiple types of debt, depends on the amount of wealth that each agent contributes to the firm. However, it is shown that the agents’ equity holdings do not depend on the contributed amounts of wealth. Thus the model can produce a wedge between ownership rights and equity cashflow rights. These features are illustrated in a firm with three agents.

Appendix

Proof of Proposition 1

Note that $K_{1}$ and $K_{2}$ in (13)–(14) and D in (11) depend on $\{\lambda _{i}\}_{i=1}^{3}$, which in turn depend on the initial wealth levels $\{W_{0i}\}_{i=1}^{3}$. For some parameter values and initial wealth levels we have $D>0$, while for other parameter values and initial wealth levels we have $D\le 0$. For each sign of D, $K_{1}$ and $K_{2}$ may satisfy either $K_{1}\le K_{2}\le 0$, $K_{1}\le 0<K_{2}$, or $0<K_{1}\le K_{2}$. Thus to prove the proposition, we must analyze six cases.

The agents’ personal loan positions are described first. The ith agent’s loan position is $d_{i}({\lambda })=$ $m_{i}({\lambda })/g_{0}( {\lambda })$, where $m_{i}({\lambda })$ is given below and $g_{0}({\lambda })$ is the after-tax payoff on the loan. From the discussion in Sect. 2.5, $g_{0}({\lambda })=P_{r}+\left( 1-t_{d}\right) \left[ 1-P_{r}\right]$, where $P_{r}$ is from (8). Define the expressions $w_{i}({\lambda })$ for $i=1,2,3$ as

$$\begin{aligned} w_{1}({\lambda })= & {} \rho \log \left[ \frac{\lambda _{1}}{\varDelta } \right] -\frac{\rho \left[ 1-\left( 1-t_{e}\right) \left( 1-t_{c}\right) \right] P_{e}}{\rho +\tau +\nu }, \\ w_{2}({\lambda })= & {} \phi \log \left[ \frac{\lambda _{2}}{\varDelta } \right] -\frac{\tau \left[ 1-\left( 1-t_{e}\right) \left( 1-t_{c}\right) \right] P_{e}}{\rho +\tau +\nu }, \\ w_{3}({\lambda })= & {} \alpha \log \left[ \frac{\lambda _{3}}{\varDelta } \right] -\frac{\nu \left[ 1-\left( 1-t_{e}\right) \left( 1-t_{c}\right) \right] P_{e}}{\rho +\tau +\nu }, \end{aligned}$$

where $\varDelta$ is from (9). First consider $K_{1}\le K_{2}\le 0$. In this case if $D>0$, the expressions for $\{m_{i}( {\lambda })\}_{i=1}^{3}$ are

$$\begin{aligned} m_{1}({\lambda })= & {} w_{1}({\lambda })-K_{1}\left[ \frac{\rho \left( 1-t_{d}\right) }{\rho +\tau +\alpha }-\frac{\rho \left( 1-t_{d}\right) }{\rho +\phi +\alpha }\right] -K_{2}\left[ \frac{\rho \left( 1-t_{d}\right) }{\rho +\tau +\nu }-\frac{\rho \left( 1-t_{d}\right) }{\rho +\tau +\alpha }\right] , \\ m_{2}({\lambda })= & {} w_{2}({\lambda })-K_{1}\left[ \frac{\tau \left( 1-t_{d}\right) }{\rho +\tau +\alpha }-\frac{\phi \left( 1-t_{d}\right) }{\rho +\phi +\alpha }\right] -K_{2}\left[ \frac{\tau \left( 1-t_{d}\right) }{\rho +\tau +\nu }-\frac{\tau \left( 1-t_{d}\right) }{\rho +\tau +\alpha }\right] , \\ m_{3}({\lambda })= & {} w_{3}({\lambda })-K_{1}\left[ \frac{ \alpha \left( 1-t_{d}\right) }{\rho +\tau +\alpha }-\frac{\alpha \left( 1-t_{d}\right) }{\rho +\phi +\alpha }\right] -K_{2}\left[ \frac{\nu \left( 1-t_{d}\right) }{\rho +\tau +\nu }-\frac{\alpha \left( 1-t_{d}\right) }{\rho +\tau +\alpha }\right] . \end{aligned}$$

If instead $D\le 0$, the expressions for $\{m_{i}({\lambda })\}_{i=1}^{3}$ are

$$\begin{aligned} m_{1}({\lambda })= & {} w_{1}({\lambda })-K_{1}\left[ \frac{\rho \left( 1-t_{d}\right) }{\rho +\phi +\nu }-\frac{\rho \left( 1-t_{d}\right) }{ \rho +\phi +\alpha }\right] -K_{2}\left[ \frac{\rho \left( 1-t_{d}\right) }{ \rho +\tau +\nu }-\frac{\rho \left( 1-t_{d}\right) }{\rho +\phi +\nu }\right] , \\ m_{2}({\lambda })= & {} w_{2}({\lambda })-K_{1}\left[ \frac{\phi \left( 1-t_{d}\right) }{\rho +\phi +\nu }-\frac{\phi \left( 1-t_{d}\right) }{ \rho +\phi +\alpha }\right] -K_{2}\left[ \frac{\tau \left( 1-t_{d}\right) }{ \rho +\tau +\nu }-\frac{\phi \left( 1-t_{d}\right) }{\rho +\phi +\nu }\right] , \\ m_{3}({\lambda })= & {} w_{3}({\lambda })-K_{1}\left[ \frac{\nu \left( 1-t_{d}\right) }{\rho +\phi +\nu }-\frac{\alpha \left( 1-t_{d}\right) }{\rho +\phi +\alpha }\right] -K_{2}\left[ \frac{\nu \left( 1-t_{d}\right) }{ \rho +\tau +\nu }-\frac{\nu \left( 1-t_{d}\right) }{\rho +\phi +\nu }\right] . \end{aligned}$$

Next consider $K_{1}\le 0<K_{2}$. If $D>0$, the expressions for $\{m_{i}({\lambda })\}_{i=1}^{3}$ are

$$\begin{aligned} m_{1}({\lambda })= & {} w_{1}({\lambda })-\frac{\rho t_{d}P_{s}}{ \rho +\tau +\alpha }-K_{1}\left[ \frac{\rho \left( 1-t_{d}\right) }{\rho +\tau +\alpha }-\frac{\rho \left( 1-t_{d}\right) }{\rho +\phi +\alpha } \right] , \\ m_{2}({\lambda })= & {} w_{2}({\lambda })-\frac{\tau t_{d}P_{s}}{ \rho +\tau +\alpha }-K_{1}\left[ \frac{\tau \left( 1-t_{d}\right) }{\rho +\tau +\alpha }-\frac{\phi \left( 1-t_{d}\right) }{\rho +\phi +\alpha } \right] , \\ m_{3}({\lambda })= & {} w_{3}({\lambda })-\frac{\alpha t_{d}P_{s} }{\rho +\tau +\alpha }-K_{1}\left[ \frac{\alpha \left( 1-t_{d}\right) }{\rho +\tau +\alpha }-\frac{\alpha \left( 1-t_{d}\right) }{\rho +\phi +\alpha } \right] . \end{aligned}$$

If instead $D\le 0$, the expressions for $\{m_{i}({\lambda })\}_{i=1}^{3}$ are

$$\begin{aligned} m_{1}({\lambda })= & {} w_{1}({\lambda })-\frac{\rho t_{d}P_{s}}{ \rho +\phi +\nu }-K_{1}\left[ \frac{\rho \left( 1-t_{d}\right) }{\rho +\phi +\nu }-\frac{\rho \left( 1-t_{d}\right) }{\rho +\phi +\alpha }\right] , \\ m_{2}({\lambda })= & {} w_{2}({\lambda })-\frac{\phi t_{d}P_{s}}{ \rho +\phi +\nu }-K_{1}\left[ \frac{\phi \left( 1-t_{d}\right) }{\rho +\phi +\nu }-\frac{\phi \left( 1-t_{d}\right) }{\rho +\phi +\alpha }\right] , \\ m_{3}({\lambda })= & {} w_{3}({\lambda })-\frac{\nu t_{d}P_{s}}{ \rho +\phi +\nu }-K_{1}\left[ \frac{\nu \left( 1-t_{d}\right) }{\rho +\phi +\nu }-\frac{\alpha \left( 1-t_{d}\right) }{\rho +\phi +\alpha }\right] . \end{aligned}$$

Lastly consider $0<K_{1}\le K_{2}$. If $D>0$, the expressions for $\{m_{i}({\lambda })\}_{i=1}^{3}$ are

$$\begin{aligned} m_{1}({\lambda })= & {} w_{1}({\lambda })-\frac{\rho t_{d}P_{s}}{ \rho +\phi +\alpha }-\frac{\rho t_{d}P_{j}}{\rho +\tau +\alpha }, \\ m_{2}({\lambda })= & {} w_{2}({\lambda })-\frac{\phi t_{d}P_{s}}{ \rho +\phi +\alpha }-\frac{\tau t_{d}P_{j}}{\rho +\tau +\alpha }, \\ m_{3}({\lambda })= & {} w_{3}({\lambda })-\frac{\alpha t_{d}P_{s} }{\rho +\phi +\alpha }-\frac{\alpha t_{d}P_{j}}{\rho +\tau +\alpha }. \end{aligned}$$

If instead $D\le 0$, the expressions for $\{m_{i}({\lambda })\}_{i=1}^{3}$ are

$$\begin{aligned} m_{1}({\lambda })= & {} w_{1}({\lambda })-\frac{\rho t_{d}P_{s}}{ \rho +\phi +\alpha }-\frac{\rho t_{d}P_{j}}{\rho +\phi +\nu }, \\ m_{2}({\lambda })= & {} w_{2}({\lambda })-\frac{\phi t_{d}P_{s}}{ \rho +\phi +\alpha }-\frac{\phi t_{d}P_{j}}{\rho +\phi +\nu }, \\ m_{3}({\lambda })= & {} w_{3}({\lambda })-\frac{\alpha t_{d}P_{s} }{\rho +\phi +\alpha }-\frac{\nu t_{d}P_{j}}{\rho +\phi +\nu }. \end{aligned}$$

For each of the six cases, it is easy to verify that $d_{1}( {\lambda })+d_{2}({\lambda })+d_{3}({\lambda })=0$. Parts (ii)–(iv) of Proposition 1 give the fraction of equity, senior debt, and junior debt that is held by each agent in each of the six cases. For all six cases, the fractions of each security sum to one. This verifies condition (v) of Definition 1.

If $K_{1}\le K_{2}\le 0$, then $N({\lambda })=1$ and there is only equity in the capital structure. The after-tax equity payoff is

$$\begin{aligned} g_{1}(Y,{\lambda })=P_{e}+(1-t_{e})(1-t_{c})(Y-P_{e}), \end{aligned}$$

which coincides with the total after-tax payoff $f(Y,{\lambda })$. See item 12 of Table 1. If $K_{1}\le 0<K_{2}$, then $N({\lambda })=2$ and the capital structure contains equity and senior debt with face value $K_{2}$. In this case the after-tax equity payoff is

$$\begin{aligned} g_{1}(Y,{\lambda })=P_{e}+(1-t_{e})(1-t_{c})(\max [0,Y-K_{2}]-P_{e}), \end{aligned}$$

(15)

while the after-tax senior debt payoff is $g_{2}(Y,{\lambda })=P_{s}+(1-t_{d})(\min [Y,K_{2}]-P_{s})$.

The total after-tax payoff is $f(Y,{\lambda })=g_{1}(Y,{\lambda })+g_{2}(Y,{\lambda })$. Again, see item 12 of Table 1. Lastly, if $0<K_{1}\le K_{2}$, then $N({\lambda })=3$ and the capital structure contains equity, senior debt with face value $K_{1}$, and junior debt with face value $K_{2}-K_{1}$. The after-tax equity payoff is the same as (15), but the after-tax senior debt payoff is $g_{2}(Y,{\lambda } )=P_{s}+(1-t_{d})(\min [Y,K_{1}]-P_{s})$

and the after-tax junior debt payoff is

$$\begin{aligned} g_{3}(Y,{\lambda })=P_{j}+(1-t_{d})(\max [0,Y-K_{1}]-\max [0,Y-K_{2}]-P_{j}). \end{aligned}$$

The total after-tax payoff is $f(Y,{\lambda })=g_{1}(Y, {\lambda })+g_{2}(Y,{\lambda })+g_{3}(Y,{\lambda })$. Once again, see item 12 of Table 1. Using the above values of $\{d_{i}( {\lambda })\}_{i=1}^{3}$ and using the security fractions from parts (ii)–(iv) of Proposition 1, the agents’ sharing rules $\{s_{i}(Y,{\lambda })\}_{i=1}^{3}$ are constructed from condition (iii) of Definition 1. Note that for each of the six cases, $s_{1}(Y,{\lambda })+s_{2}(Y, {\lambda })+s_{3}(Y,{\lambda })=f(Y,{\lambda })$, which verifies condition (ii) of Definition 1.

For each of the six cases, it is now possible to verify by direct substitution that

$$\begin{aligned} \lambda _{1}U_{1}^{^{\prime }}(s_{1}(Y,{\lambda }))=\lambda _{2}U_{2}^{^{\prime }}(s_{2}(Y,{\lambda }))=\lambda _{3}U_{3}^{^{\prime }}(s_{3}(Y,{\lambda }))=\xi (Y,{\lambda }), \end{aligned}$$

(16)

where $\xi (Y,{\lambda })$ is given in part (v) of Proposition 1. Recall that the sign of D in (11) is related to the agents’ preferences in (2)–(3). If (11) is positive, the events $\left\{ Y>K_{1}\right\}$ and $\left\{ W_{2}>\theta \right\}$ are equivalent, as are the events $\left\{ Y>K_{2}\right\}$ and $\left\{ W_{3}>\eta \right\}$. On the other hand, if (11) is negative, the events $\left\{ Y>K_{1}\right\}$ and $\left\{ W_{3}>\eta \right\}$ are equivalent, as are the events $\left\{ Y>K_{2}\right\}$ and $\left\{ W_{2}>\theta \right\}$. First suppose $K_{1}\le K_{2}\le 0$. Since $Y>0$, we know that $Y>K_{1}$ and $Y>K_{2}$. Thus regardless of the sign of D, we know that $W_{2}>\theta$ and $W_{3}>\eta$. To evaluate (16), we use only the piece of $U_{2}$ in (2) that corresponds to $W_{2}>\theta$ and only the piece of $U_{3}$ in (3) that corresponds to $W_{3}>\eta$. Next suppose $K_{1}\le 0<K_{2}$. In this case we know that $Y>K_{1}$. To evaluate (16) when $D>0$, we use only the piece of $U_{2}$ in (2) that corresponds to $W_{2}>\theta$, but we use both pieces of $U_{3}$ in (3). On the other hand, to evaluate (16) when $D\le 0$, we use both pieces of $U_{2}$ in (2), but we use only the piece of $U_{3}$ in (3) that corresponds to $W_{3}>\eta$. Lastly, suppose that $0<K_{1}\le K_{2}$. In this case we use both pieces of $U_{2}$ and both pieces of $U_{3}$. To evaluate (16) when $D>0$, we rely on the fact that the events $\left\{ Y>K_{1}\right\}$ and $\left\{ W_{2}>\theta \right\}$ are equivalent, as are the events $\left\{ Y>K_{2}\right\}$ and $\left\{ W_{3}>\eta \right\}$ . On the other hand, when $D\le 0$, we rely on the fact that the events $\left\{ Y>K_{1}\right\}$ and $\left\{ W_{3}>\eta \right\}$ are equivalent, as are the events $\left\{ Y>K_{2}\right\}$ and $\left\{ W_{2}>\theta \right\}$. Thus for each of the six cases, it can be verified that condition (i) of Definition 1 is satisfied.

Once a probability measure $\mathbb {P}$ is specified, the values for $\lambda _{1}$, $\lambda _{2}$, and $\lambda _{3}$ can be extracted from the following system of equations

$$\begin{aligned} {\mathbb {E}}[\xi (Y,{\lambda })\,s_{i}(Y,{\lambda })]=W_{0i}\,, \end{aligned}$$

(17)

for $i=1,2,3$, which is the same as condition (iv) of Definition 1. This completes the proof. $\square$