Finance and Economic Growth in a Dynamic Game

Abstract

We investigate how the relaxation of financial constraints affects economic growth in a dynamic game of the tragedy of the commons by introducing an imperfect financial market into Tornell and Velasco’s (J Polit Econ 100(6):1208–1231, 1992) model. It is shown that whereas the relaxation of financial constraints enhances economic growth if agents have access only to a common asset whose property rights are not secure, the relaxation of financial constraints reduces economic growth if agents can have access not only to a common asset but also to a private asset whose property rights are secure.

Appendix

1.1 Proof of Lemma 2

We set up the Lagrangian as follows:

$$\begin{aligned} L_t^{i,h}:&=E_t\left[ \Sigma _{\tau =t}^\infty \beta ^{\tau -t} \ln c_\tau ^{i,h}\right] \nonumber \\&\quad +E_t\left[ \Sigma _{\tau =t}^\infty \beta ^{\tau -t}\psi _\tau ^{i,h}\left( (1-\theta _\tau (N-1))R_\tau ^h s_{\tau -1}^h-d_\tau ^{i,h}-s_\tau ^h\right) \right] \nonumber \\&\quad +E_t\left[ \Sigma _{\tau =t}^\infty \beta ^{\tau -t}\zeta _\tau ^{i,h}\left( \rho f_{\tau -1}+d_\tau ^{i,h}-f_\tau ^{i,h}-c_\tau ^{i,h}\right) \right] . \end{aligned}$$

(B.1)

The first-order conditions are given by

$$\begin{aligned} \frac{1}{c_t^{i,h}}-\zeta _t^{i,h}= & {} 0, \end{aligned}$$

(B.2)

$$\begin{aligned} -\psi _t^{i,h}+\beta (1-\theta _{t+1}(N-1))R_{t+1}^hE_t[\psi _{t+1}^{i,h}]= & {} 0, \end{aligned}$$

(B.3)

$$\begin{aligned} -\psi _t^{i,h}+\zeta _t^{i,h}= & {} 0, \end{aligned}$$

(B.4)

and

$$\begin{aligned} -\zeta _t^{i,h}+\beta \rho E_t[\zeta _{t+1}^{i,h}]=0. \end{aligned}$$

(B.5)

Under the Markov perfect equilibrium, the optimal control path, $\{\tilde{d}_t^{i,h}, c_t^{i,h}\}$, exists if and only if $d_t^{i,h}=\tilde{d}_t^{i,h}=\theta _t R_t^hs_{t-1}^h$ in (21) and (23), and the first-order conditions, (B.2)-(B.5) and the transversality conditions, $\lim _{s\rightarrow \infty }\beta ^sE_t\left[ s_{t+s}^h/c_{t+s}^{i,h}\right] =0$ and $\lim _{s\rightarrow \infty }\beta ^sE_t\left[ f_{t+s}^{i,h}/c_{t+s}^{i,h}\right] =0$, are satisfied.

From (B.3)-(B.5), we have

$$\begin{aligned} (1-\theta _{t+1}(N-1))R_{t+1}^h=\rho \end{aligned}$$

(B.6)

or, equivalently,

$$\begin{aligned} \theta _{t+1}=\frac{R_{t+1}^h-\rho }{R_{t+1}^h(N-1)}. \end{aligned}$$

(B.7)

From (B.7), we obtain

$$\begin{aligned} \tilde{d}_t^{i,h}(I_t^h)=\frac{R_t^h-\rho }{R_t^h(N-1)}I_t^h=:\theta _t I_t^h. \end{aligned}$$

(B.8)

This is equation (25). From (23), with $d_t^{i,h}=\theta _{t}I_t^{i,h}=\theta _{t}R_t^hs_{t-1}^h$, it follows that $s_t^h=(1-\theta _t N)R_t^hs_{t-1}^h$. From this equation and (B.7), we obtain

$$\begin{aligned} s_t^h=\frac{\rho N-R_t^h}{R_t^h(N-1)}I_t^h. \end{aligned}$$

(B.9)

This is equation (26). We must verify that $\tilde{d}_t^{i,h}(I_t^h)>0$ and $s_t^h>0$. Because $R_{t}^h-\rho \ge r_{t}-\rho =A\mu -\rho >0$, from Assumption 2, we obtain $\tilde{d}_t^{i,h}(I_t^h)>0$. Moreover, from Assumption 1, it follows that $R_t^h\le (A-r_t \mu )/(1-\mu )=A(1+\mu )$. Therefore, we obtain $\rho N-R_t^h\ge \rho N-A(1+\mu )>0$ from Assumption 2. Finally, we derive agent i’s consumption schedule in equilibrium. From (21), (23), and (B.6), it follows that

$$\begin{aligned} \frac{f_{t+1}^{i,h}+s_{t+1}^h}{c_{t+1}^{i,h}}+1=\rho \frac{f_{t}^{i,h}+s_{t}^h}{c_{t+1}^{i,h}}. \end{aligned}$$

(B.10)

Taking the expectation operator $E_t(.)$ for both sides of (B.10) yields

$$\begin{aligned} E_t\left[ \frac{f_{t+1}^{i,h}+s_{t+1}^h}{c_{t+1}^{i,h}}\right] +1=(f_{t}^{i,h}+s_{t}^h)\rho E_t\left[ \frac{1}{c_{t+1}^{i,h}}\right] . \end{aligned}$$

(B.11)

By applying (B.2) and (B.5) to the right-hand side of (B.11), we obtain

$$\begin{aligned} \frac{f_{t}^{i,h}+s_{t}^h}{c_{t}^{i,h}}=\beta +\beta E_t\left[ \frac{f_{t+1}^{i,h}+s_{t+1}^h}{c_{t+1}^{i,h}}\right] . \end{aligned}$$

(B.12)

From (B.12), it follows that

$$\begin{aligned} \frac{f_{t}^{i,h}+s_{t}^h}{c_{t}^{i,h}}&=\beta +\beta ^2+\cdots +\beta ^s+\beta ^s E_t\left[ \frac{f_{t+s}^{i,h}+s_{t+s}^h}{c_{t+s}^{i,h}}\right] \nonumber \\&=\frac{\beta }{1-\beta }+\lim _{s\rightarrow \infty }\beta ^s E_t\left[ \frac{f_{t+s}^{i,h}+s_{t+s}^h}{c_{t+s}^{i,h}}\right] \end{aligned}$$

(B.13)

From the transversality conditions, we have $\lim _{s\rightarrow \infty }\beta ^s E_t\left[ (f_{t+s}^{i,h}+s_{t+s}^h)/c_{t+s}^{i,h}\right] =0$, and thus, $f_{t}^{i,h}+s_{t}^h=\beta c_{t}^{i,h}/(1-\beta )$. This equation and (B.10) yield

$$\begin{aligned} c_t^{i,h}=\rho (1-\beta )(f_{t-1}^{i,h}+s_{t-1}^h). \end{aligned}$$

(B.14)

This is agent i’s consumption schedule given by (27). Conversely, we verify that the transversality conditions hold. Suppose that agent i consumes following the schedule given by (B.14). Then, from (B.10) and (B.14), we obtain $f_{t}^{i,h}+s_{t}^h=\beta c_{t}^{i,h}/(1-\beta )$. From the last equation and (B.13), it follows that

$$\begin{aligned} \lim _{s\rightarrow \infty }\beta ^s E_t\left[ \frac{f_{t+s}^{i,h}+s_{t+s}^h}{c_{t+s}^{i,h}}\right] =0. \end{aligned}$$

(B.15)

In (B.15), because $f_{t+s}^{i,h}\ge 0$ and $s_{t+s}^h>0$, it holds that $\lim _{s\rightarrow \infty }\beta ^sE_t\left[ s_{t+s}^h/c_{t+s}^{i,h}\right] =0$ and $\lim _{s\rightarrow \infty }\beta ^sE_t\left[ f_{t+s}^{i,h}/c_{t+s}^{i,h}\right] =0$. $\square$

1.2 Proof of Theorem 2

The second claim is obvious. For the first claim, define a function such that $f(x)=(2\rho x-A(1+\mu ))/(x-1)$ in $x\in (0, \infty )$. Then, it follows from Assumption 2 that ${{\mathrm{sign}}}(f^\prime (x))={{\mathrm{sign}}}(A(1+\mu )-2\rho )>0$. The first claim follows from the last inequality. $\square$