Entrepreneurs, legal institutions and firm dynamics

Abstract

This paper assesses the impact of legal institutions on firm dynamics in a model where entrepreneurs have heterogeneous risk aversion, credit constraints and may default. Entrepreneurs choose firm size, capital structure, consumption, default and whether to incorporate. We find that less risk-averse entrepreneurs tend to incorporate while more risk-averse entrepreneurs do not; this occurs because leaving some personal assets exposed by not incorporating allows more risk-averse borrowers to credibly commit to lower default rates. We show that incorporation is determined by two effects: the standard effect that bankruptcy insures low firm returns and a new “scale effect”—more risk-averse entrepreneurs run smaller firms and default more often. The more risk-averse choose to leave some personal assets unshielded in bankruptcy due to a commitment problem that dominates the value of insurance. The less risk-averse run larger firms, default less and incorporate.

Appendix

We now show that problem 1 is equivalent to problem 3. As a first step, we show that the value function has the form $V_S(w)=w^{1-\rho }v_S$ and $V_B(w)=w^{1-\rho } v_B$.

In order to see this, substitute these functional forms into problem 1. Thus,

$$\begin{aligned} \begin{aligned} V_S(w)&=\max _{c,A,\epsilon , \bar{v} ,\gamma } u(c) + \beta \biggl [ \int _{\mathfrak {B}} ((1-\gamma )(1+r)(w-\epsilon A-c))^{1-\rho }v_B\,\mathrm{d}F(x)\\&\quad + \int _{\mathfrak {B}^c} (A(x-\bar{v})+(1+r)(w-\epsilon A-c))v_S \,\mathrm{d}F(x)\biggr ] \end{aligned} \end{aligned}$$

(10)

Subject to: (1), (3), (4) and

$$\begin{aligned}&x\in \mathfrak {B}\text { iff } v_B\left( (1-\gamma )(1+r)(w-\epsilon A-c)\right) ^{1-\rho }>v_S\left( A(x-\bar{v})\right. \nonumber \\&\quad \quad \left. +\,(1+r)(w-\epsilon A-c)\right) ^{1-\rho } \end{aligned}$$

(11)

Note that $v_S=V_S(1)$ is the continuation utility from starting with an initial net-worth of one unit. Since utility is of the form $u(x)=x^{1-\rho }/(1-\rho )$ it follows that $v_S>0$ if $\rho <1$ and $v_S<0$ if $\rho >1$. The right-hand side of (11) is therefore strictly increasing in x. Thus, $\mathfrak {B}$ must be a lower interval, i.e., there exists $x^*$ such that $\mathfrak {B}=\left\{ x \bigm | x <x^*\right\}$. Further, for $x=x^*$ Eq. (11) must hold with equality, i.e.,

$$\begin{aligned} v_B\left( (1-\gamma )(1+r)(w-\epsilon A-c)\right) ^{1-\rho }=v_S\left( A(x^*-\bar{v})+(1+r)(w-\epsilon A-c)\right) ^{1-\rho } \end{aligned}$$

(12)

Solving (12) for $x^*$ yields constraint (6) of Problem 3.

Now let $\lambda >0$. We show that $V_S(\lambda w) = \lambda ^{1-\rho } V_S(w)$. Note that if we replace w by $\lambda w$, A by $\lambda A$ and c by $\lambda c$ then the right-hand side of (10) becomes

$$\begin{aligned}&\lambda ^{1-\rho }\left( \int _{\mathfrak {B}} ((1-\gamma )(1+r)(w-\epsilon A-c))^{1-\rho }v_B\,\mathrm{d}F(x) \right. \nonumber \\&\quad \left. +\,\int _{\mathfrak {B}^c} (A(x-\bar{v})+(1+r)(w-\epsilon A-c))v_S \,\mathrm{d}F(x)\right) \end{aligned}$$

(13)

Similarly, this substitution multiplies both sides of Eq. (12) by $\lambda$, and hence $x^*$ and the bankruptcy set $\mathfrak {B}$ do not change. Constraint (1) does not change, since w, c and A enter the expression only as ratios w / A and c / A and are therefore unchanged. Replacing A by $\lambda A$ and w by $\lambda w$ in (3) multiplies both sides by $\lambda >0$, and hence this equations are also unchanged. Finally, the non-negativity constraints in (4) also remain valid.

Thus, $\lambda c$, $\lambda A$, $\epsilon$ and $\bar{v}$ satisfy the constraints of our optimization problem at net-worth level $\lambda w$. Equation (13) therefore implies that $V_S(\lambda w)\le \lambda ^{1-\rho } V_S(w)$ for all $\lambda ,w>0$. Now let $\tilde{w}=w/\lambda$. Then

$$\begin{aligned} \lambda ^{1-\rho } V_S(\tilde{w})= \lambda ^{1-\rho }V_S\left( \textstyle {\frac{1}{\lambda }} \lambda \tilde{w}\right) =\lambda ^{1-\rho }V_S\left( \textstyle {\frac{1}{\lambda }} w\right) \ge \lambda ^{1-\rho }\left( \textstyle {\frac{1}{\lambda }}\right) ^{1-\rho } V_S(w) = V_S(\lambda \tilde{w}) \end{aligned}$$

Hence $V_S(w)=w^{1-\rho }v_S$. The argument for $\rho =1$, i.e., for utility $u(x)=\log (x)$ is similar.

We can apply the same argument to Problem 2 to show that $V_B(w)=w^{1-\rho }v_S$. It is now immediate that Problem 1 is equivalent to Problem 3.