Strategic Interactions in a One-Sector Growth Model

Abstract

We study the effect of dynamic and investment externalities in a one-sector growth model. In our model, two agents interact strategically in the utilization of capital for consumption, savings, and investment in technical progress. We consider two types of investment choices: complements and substitutes. For each case, we derive the equilibrium and provide the corresponding stationary distribution. We then compare the equilibrium with the social planner’s solution.

Appendix: Solution of the Social Planner

In this Appendix, we derive the social planner’s solution in the case of complementary and substitutionary investment choices. We consider the infinite horizon by conjecturing that the value function is of the form $W^{\infty }(y) = \kappa _{\infty } \ln y + \Psi _{\infty }$. As noted, the linear conjecture can be inferred by solving the problem recursively.

Given $W^{\infty }(y) = \kappa _{\infty } \ln y + \Psi _{\infty }$, (62) is rewritten as

$$\begin{aligned} W^{\infty }(y) =&~ \max _{\{e_{j},i_{j}\}_{j=1}^{2}} \left\{ \ln (e_{1}-i_{1}) + \ln (e_{2}-i_{2}) + \delta \kappa _{\infty } \overline{\eta }_{1} \ln i_{1} + \delta \kappa _{\infty } \overline{\eta }_{2} \ln i_{2} \right. \nonumber \\&~ \left. + \delta \kappa _{\infty } \overline{\alpha } \ln (y-e_{1}-e_{2}) + \delta \Psi _{\infty } \right\} \end{aligned}$$

(76)

if investment choices are complementary and

$$\begin{aligned} W^{\infty }(y) =&~ \max _{\{e_{j},i_{j}\}_{j=1}^{2}} \left\{ \ln (e_{1}-i_{1}) + \ln (e_{2}-i_{2}) + \delta \kappa _{\infty } \overline{\eta } \ln (i_{1}+i_{2}) \right. \nonumber \\&~ \left. + \delta \kappa _{\infty } \overline{\alpha } \ln (y-e_{1}-e_{2}) + \delta \Psi _{\infty } \right\} \end{aligned}$$

(77)

if investment choices are substitutionary.

For complements, for $j=1,2$, the first-order conditions corresponding to (76) are

$$\begin{aligned} e_{j}&: \frac{1}{e_{j}-i_{j}} = \frac{\delta \kappa _{\infty } \overline{\alpha }}{y-e_{1}-e_{2}}, \end{aligned}$$

(78)

$$\begin{aligned} i_{j}&: \frac{1}{e_{j}-i_{j}} = \frac{\delta \kappa _{\infty } \overline{\eta }_{j}}{i_{j}}, \end{aligned}$$

(79)

which yields

$$\begin{aligned} E_{j}^{*\infty }(y)&= \frac{1 + \delta \kappa _{\infty } \overline{\eta }_{j}}{2 + \delta \kappa _{\infty } (\overline{\eta }_{1}+\overline{\eta }_{2}+\overline{\alpha })} y, \end{aligned}$$

(80)

$$\begin{aligned} I_{j}^{*\infty }(y)&= \frac{\delta \kappa _{\infty } \overline{\eta }_{j}}{2 + \delta \kappa _{\infty } (\overline{\eta }_{1}+\overline{\eta }_{2}+\overline{\alpha })} y. \end{aligned}$$

(81)

Plugging (80) and (81) back into (76) implies that

$$\begin{aligned} \kappa _{\infty } = \frac{2}{1-\delta (\overline{\eta }_{1}+\overline{\eta }_{2}+\overline{\alpha })}. \end{aligned}$$

(82)

Plugging (82) into (80) and (81) and summing over $j$ yields (63) and (65). Plugging (63) and (65) into $\sum \nolimits _{j=1}^{2} C^{*\infty }_{j}(y) = \sum \nolimits _{j=1}^{2} (E^{*\infty }_{j}(y) - I^{*\infty }_{j}(y))$ yields (64).

For substitutes, for $j=1,2$, the first-order conditions corresponding to (77) are

$$\begin{aligned} e_{j}&: \frac{1}{e_{j}-i_{j}} = \frac{\delta \kappa _{\infty } \overline{\alpha }}{y-e_{1}-e_{2}} \end{aligned}$$

(83)

$$\begin{aligned} i_{j}&: \frac{1}{e_{j}-i_{j}} = \frac{\delta \kappa _{\infty } \overline{\eta }}{i_{1}+i_{2}}, \end{aligned}$$

(84)

which yields

$$\begin{aligned} E_{j}^{*\infty }(y)&= \frac{2 + \delta \kappa _{\infty } \overline{\eta }}{4 + 2 \delta \kappa _{\infty } (\overline{\eta } + \overline{\alpha })} y \end{aligned}$$

(85)

$$\begin{aligned} I_{1}^{*\infty }(y)+I_{2}^{*\infty }(y)&= \frac{\delta \kappa _{\infty } \overline{\eta }}{2 + \delta \kappa _{\infty } (\overline{\eta } + \overline{\alpha }) } y \end{aligned}$$

(86)

since the social planner only needs to solve for total investment. Plugging (85) and (86) back into (77) yields

$$\begin{aligned} \kappa _{\infty } = \frac{2}{1-\delta (\overline{\eta } + \overline{\alpha })}. \end{aligned}$$

(87)

Plugging (87) into (85) and summing over $j$ yields (66). Plugging (87) into (86) yields (68). Plugging (66) and (68) into $\sum \nolimits _{j=1}^{2} C^{*\infty }_{j}(y) = \sum \nolimits _{j=1}^{2} (E^{*\infty }_{j}(y) - I^{*\infty }_{j}(y))$ yields (67).