The role of the private sector under insecure property rights

Abstract

It is well known that the so-called voracity effect can be observed in an economy with a weak property rights system. Voracious behavior is regarded as one of the excess uses of the common assets. In this paper, we seek to examine voracious behavior from a different perspective by introducing a new direction of capital flow: from the private sector to the common sector. A government mandates that all competing interest groups invest their private capital in the common sector to mitigate the effects of excess use of the commons. In this situation, we study how this capital flow affects the voracious behavior of the groups and the growth rate of the economy. The main findings are that, while there is no standard voracity effect, an increase in the contribution of the private sector into the common sector causes more voracious behavior and thus reduces economic growth. This suggests that policies designed to preserve the commons can lead to a harmful effect on the economy.

Appendices

Appendix 1: Proof of Lemma 1

In the case where the value function is independent of the opponents’ private capital stocks, (11), we obtain

$$\begin{aligned} \frac{\partial V_{i}}{\partial h_{j}} = \frac{\partial ^{2} V_{i}}{\partial h_{j} \partial K} = \frac{\partial ^{2} V_{i}}{\partial h_{j} \partial h_{i}} = \frac{\partial ^{2} V_{i}}{\partial h_{j}^{2}} = 0. \end{aligned}$$

Substituting these appropriation strategies into (10) yields $\frac{\partial V}{\partial K} \left( Au - (n - 1)\delta \right) = 0$. Since $\frac{\partial V}{\partial K} \ne 0$, for the equation to be satisfied, $Au - (n - 1)\delta$ must be zero and thus $\delta = \frac{Au}{n - 1}$. The optimal condition (7) requires that $\alpha = 1$ and that the conjectured value function must hold

$$\begin{aligned} \frac{\partial ^{2} V}{\partial K^{2}} = \frac{\partial ^{2} V}{\partial h_{i} \partial K} = \frac{\partial ^{2} V}{\partial h_{i}^{2}} = -\theta \xi (K + \alpha h_{i})^{-\theta - 1}. \end{aligned}$$

These conditions mean that (8) is equivalent to (9), which leads to $\gamma = \frac{A - B(1 - u)}{n - 1}$.

Next, from the optimal condition (6), the value function (11), and the consumption strategy, we confirm that

$$\begin{aligned} \left( a' + aK + eh_{i} + bZ_{i} \right) ^{-\theta } = c_{i}^{-\theta } = \frac{\partial V_{i}}{\partial K} = \left( \xi ^{-\frac{1}{\theta }}K + \xi ^{-\frac{1}{\theta }}h_{i} \right) ^{-\theta }. \end{aligned}$$

For the condition to be satisfied, $a' = b = 0$ and $a = e = \xi ^{-\frac{1}{\theta }}$ are required.

Since we focus on symmetric equilibrium, $h_{i}$ is equivalent to $h_{j}$ for $j \ne i$ at equilibrium. Using this and the results obtained above, we can arrange (8) as follows.

$$\begin{aligned}{}[\rho - B(1 - u)]\frac{\partial V_{i}}{\partial K} = \frac{\partial ^{2}V_{i}}{\partial K^{2}}[B(1 - u)(K + h_{i}) - c_{i}] \iff \rho - B(1 - u) = -\theta B(1 - u) + \theta \xi ^{-\frac{1}{\theta }}. \end{aligned}$$

This leads to

$$\begin{aligned} a = e = \xi ^{-\frac{1}{\theta }} = \left( \frac{\theta - 1}{\theta } \right) B(1 - u) + \frac{\rho }{\theta }. \end{aligned}$$

Appendix 2: Proof of Lemma 3

First, we confirm that, in the case of (18), the optimal condition (7) requires $\alpha = 1$ and thus the following relations are obtained.

$$\begin{aligned} \frac{\partial V_{i}}{\partial h_{j}}&= \beta \frac{\partial V_{i}}{\partial K}, \quad \frac{\partial ^{2}V_{i}}{\partial K^{2}} = \frac{\partial ^{2}V_{i}}{\partial h_{i}\partial K} = \frac{\partial ^{2}V_{i}}{\partial h_{i}^{2}}, \\ \frac{\partial ^{2}V_{i}}{\partial h_{j}\partial K}&= \frac{\partial ^{2}V_{i}}{\partial h_{j}\partial h_{i}} = \beta \frac{\partial ^{2}V_{i}}{\partial K^{2}}, \quad {\text{and}} \quad \frac{\partial ^{2}V_{i}}{\partial h_{j}^{2}} = \beta ^{2} \frac{\partial ^{2}V_{i}}{\partial K^{2}}. \end{aligned}$$

Next, substituting these and the strategies into Eqs. (8)–(10), we obtain

$$\begin{aligned} \frac{\partial ^{2} V_{i}}{\partial K^{2}}\times F(K, h)&= \{ \rho - A + (1 - \beta )(n - 1)\gamma + a\beta (n -1)\} \frac{\partial V_{i}}{\partial K},\nonumber \\ \frac{\partial ^{2} V_{i}}{\partial K^{2}}\times F(K, h)&= \{ \rho - u[ A - (1 - \beta )(n - 1)\gamma ] - B(1 - u) + a\beta ^{2}(n - 1) \} \frac{\partial V_{i}}{\partial K}, \end{aligned}$$

(23)

and

$$\begin{aligned} \beta \frac{\partial ^{2} V_{i}}{\partial K^{2}}\times F(K, h)&= \{ \beta \rho - u[ A - (1 - \beta )(n - 1)\gamma ] - \beta B(1 - u) + a\beta [ 1 + \beta (n - 2)] \} \frac{\partial V_{i}}{\partial K}, \end{aligned}$$

where the function F(K, h) represents

$$\begin{aligned} F(K, h)&= \{ A - (1 - \beta )(n - 1)\gamma - a[ 1 + \beta (n - 1)] \} K \\&\quad + \{ u[A - (1 - \beta )(n - 1)\gamma ] + B(1 - u) - a[ 1 + \beta ^{2}(n - 1)] \} h_{i} \\&\quad + \{ u[A - (1 - \beta )(n - 1)\gamma ] + \beta B(1 - u) - a\beta [ 2 + \beta (n - 2)] \} Z_{i}. \end{aligned}$$

We can summarize the three equations as follows:

$$\begin{aligned} (1 - \beta )(n - 1)(1 - u)\gamma&= (A - B)(1 - u) - a\beta (n - 1)(1 - \beta ), \end{aligned}$$

(24)

$$\begin{aligned} (1 - \beta )(n - 1)(\beta - u)\gamma&= A(\beta - u) - \beta [B(1 - u) - a(1 - \beta )]. \end{aligned}$$

(25)

The unknown parameters, a, $\beta$, and $\gamma$, must satisfy both of the above equations simultaneously. First, if $\beta = 1$, the above conditions require that the contribution rate u must be a unity because of the assumption $A > B$. This contradicts the assumption $u \in (0, 1)$, and thus this is not an equilibrium. Second, we consider the possibility that $\beta$ is zero. Substituting $\beta = 0$ into (24) and (25), we get two equations, $(n - 1)\gamma = A - B$ and $(n - 1)\gamma = A$. For the two equations to be satisfied simultaneously, B must be zero, which contradicts the positivity of B. Therefore, $\beta = 0$ is not an equilibrium. Finally, we consider the case $\beta \ne 0, 1$. Substituting (24) into (25), we obtain $a \beta ^{2}(n- 1) -u a \beta (n - 1) + a \beta (1 - u) - uB(1 - u) = 0$. We solve this for a,

$$\begin{aligned} a = \frac{uB(1 - u)}{\beta [(n - 1)\beta + 1 - un]}. \end{aligned}$$

Substituting it into (24), we obtain the appropriation rate $\gamma$:

$$\begin{aligned} \gamma&= \frac{A - B}{(1 - \beta )(n - 1)} - \frac{uB}{(n - 1)\beta + 1 - un} \nonumber \\&= \frac{A [(n - 1)\beta + 1 - un] -B(1 - u)[(n - 1)\beta + 1]}{(1 - \beta )(n - 1)[ (n - 1)\beta + 1 - un ] }. \end{aligned}$$

(26)

Next, from the optimal condition (6) and (18), and the consumption strategy, we confirm that

$$\begin{aligned} (a' + aK + eh_{i} + bZ_{i}) ^{-\theta } = c_{i}^{-\theta } = \frac{\partial V_{i}}{\partial K} = \left( \xi ^{-\frac{1}{\theta }}K + \xi ^{-\frac{1}{\theta }}h_{i} + \xi ^{-\frac{1}{\theta }}\beta Z_{i} \right) ^{-\theta }, \end{aligned}$$

which leads to $a' = 0$, $a = e = \xi ^{-\frac{1}{\theta }}$, and $b = a\beta$.

Finally, since we focus on symmetric equilibrium, $h_{i}$ is equivalent to $h_{j}$ for $j \ne i$ at equilibrium. We substitute the above results into (23), and after some manipulation, we obtain the following equation:

$$\begin{aligned}&\left[ \rho - B + \frac{(n - 1)uB(\beta - u)}{(n - 1)\beta + 1 -un} \right] \xi (K + h_{i} + \beta Z_{i}) ^{-\theta } \\&\quad = -\theta \xi (K + h_{i} + \beta Z_{i})^{-1-\theta } \left[ \frac{B(1 - u)(\beta - u)[\beta (n - 1) + 1]}{\beta [(n - 1)\beta + 1 - un]} \right] (K + h_{i} + \beta Z_{i}). \end{aligned}$$

It is rewritten as $(n - 1)[\rho + (1 - u)(\theta - 1)B] \beta ^{2} -y \beta - \theta uB(1 - u) = 0$, where $y \equiv (un - 1)\rho + (1 - u)[n(1+u) + 1] (\theta - 1)B$. Solving the quadratic equation for $\beta$,

$$\begin{aligned} \beta = \frac{y \pm \sqrt{y^{2} + 4(n - 1)[\rho + (1 - u)(\theta - 1)B]\theta uB(1 - u)}}{2(n - 1)[\rho + (1 - u)(\theta - 1)B]}. \end{aligned}$$

This implies that if the quadratic equation has two different real roots, one is negative and the other is positive.

Appendix 3: Proof of Lemma 4

Let us derive the growth rate of consumption. The consumption of group i is represented by $c_{i} = \psi ^{*}_{i} = a (K + h_{i} + \beta Z_{i})$. Differentiating this with respect to t and dividing it by $c_{i}$ yields

$$\begin{aligned} \frac{{\dot{c}}_{i}}{c_{i}} = \frac{{\dot{K}} + {\dot{h}}_{i} + \beta {\dot{Z}}_{i}}{K + h_{i} + \beta Z_{i}}. \end{aligned}$$

The state dynamics of the model are represented as follows.

$$\begin{aligned} {\dot{K}}&= (A - n\gamma )K + (Au - n\gamma u)h_{i} + (Au - n\gamma u)Z_{i}, \\ {\dot{h}}_{i}&= (\gamma - a)K + (B(1 - u) +\gamma u - a)h_{i} + (\gamma u - a\beta )Z_{i}, \\ {\dot{h}}_{j}&= (\gamma - a)K + (\gamma u - a\beta )h_{i} + (B(1 - u) + \gamma u - a)h_{j} + (\gamma u - a\beta )\sum _{k \ne i, j}h_{k}. \end{aligned}$$

Substituting these into the numerator, we obtain

$$\begin{aligned} {\dot{K}} + {\dot{h}}_{i} + \beta {\dot{Z}}_{i}&= [A - \gamma (n - 1) - a + \beta (n - 1)(\gamma - a)]K \\&\quad + [Au - \gamma u (n - 1) + B(1 - u) - a \beta (n - 1)(\gamma u - a\beta )]h_{i} \\&\quad + [Au - \gamma u(n - 1) - a\beta + \beta (B(1 - u) + \gamma u - a) + \beta (n - 2)(\gamma u - a\beta )]Z_{i}, \\&= \frac{B(1 - u)(\beta - u)[\beta (n - 1) + 1]}{\beta [\beta (n - 1) + 1 - un]} (K + h_{i} + \beta Z_{i}). \end{aligned}$$

Therefore, we obtain the following growth rate of consumption,

$$\begin{aligned} \frac{{\dot{c}}_{i}}{c_{i}} = \frac{B(1 - u)(\beta - u)[\beta (n - 1) + 1]}{\beta [\beta (n - 1) + 1 - un]}. \end{aligned}$$

Appendix 4: The sign of $\gamma - a$

According to Lemma 3, we can represent $\gamma - a$ as follows:

$$\begin{aligned} \gamma - a = \frac{A\beta [(n - 1)\beta + 1 - un] - B(1 - u)[nu + (\beta - u)(1 + (n - 1)\beta )]}{(n - 1)(1 - \beta )\beta [(n - 1)\beta + 1 - un]}. \end{aligned}$$

(27)

Similarly, we can represent $A - n\gamma$ as follows:

$$\begin{aligned} A - n\gamma = \frac{[\beta (n - 1) + 1]\{ nB(1 - u) - A[\beta (n - 1) + 1 - un] \} }{(1 - \beta )(n - 1)[\beta (n - 1) + 1 - un]}. \end{aligned}$$

We subtract $A - n\gamma$ from the growth rate of consumption, g:

$$\begin{aligned} g - (A - n\gamma ) = \frac{\begin{matrix} [(n - 1)\beta + 1]\{ A\beta [(n - 1)\beta + 1 - un] \\ - B(1 - u)[nu + (\beta - u)(1 + (n - 1)\beta )] \} \end{matrix} }{(n - 1)(1 - \beta )\beta [(n - 1)\beta + 1 - un]}. \end{aligned}$$

(28)

Substituting (28) into (27), we obtain

$$\begin{aligned} \gamma - a = \frac{g - (A - n\gamma )}{(n - 1)\beta + 1}. \end{aligned}$$

(29)

If $\beta$ is positive, the denominator in the right-hand side is positive and if $\beta$ is negative, it must be negative because of the need to assure positive growth of consumption.

Furthermore, on the balanced growth path of the model, the growth rate of consumption and common capital must coincide, that is, from (19) and (20),

$$\begin{aligned} g = \frac{{\dot{K}}}{K} = (A - n\gamma ) + n u (A - n\gamma )\frac{h_{i}}{K}, \end{aligned}$$

and therefore we obtain the ratio of private capital to common capital:

$$\begin{aligned} \chi = \frac{h_{i}}{K} = \frac{g - (A - n\gamma ) }{nu(A - n\gamma )}. \end{aligned}$$

(30)

Since $\chi$ must be positive, the numerator $g - (A - n\gamma )$ must be also positive. Therefore, from (29), if $\beta$ is positive, $\gamma - a$ is positive and if $\beta$ is negative, $\gamma - a$ is negative.

Appendix 5: Proof of Proposition 4

First, we derive the symmetric MPE strategies. As discussed in Sect. 4.1, $\beta$ is positive. Substituting $\beta$ into the parameters obtained in Lemma 3 yields the MPE parameters. Therefore, the optimal strategies are $\psi _{i}^{*} = aK + ah_{i} + bZ_{i}$ and $\phi _{i}^{*} = \gamma \left[ K + uh_{i} + uZ_{i} \right]$.

Next, the ratio of private capital stock to common capital stock is (30) obtained in Appendix 4. To derive the common and private capital growth rates, we divide (20) and (21) by K and $h_{i}$, respectively.

$$\begin{aligned} \frac{{\dot{K}}}{K}&= (A - n\gamma ) + n u (A - n\gamma )\frac{h_{i}}{K}, \end{aligned}$$

(31)

$$\begin{aligned} \frac{{\dot{h}}_{i}}{h_{i}}&= (\gamma - a)\frac{K}{h_{i}} + B(1 - u) +n\gamma u - a[ (n - 1)\beta + 1]. \end{aligned}$$

(32)

Substituting (30) into (31), we obtain

$$\begin{aligned} \frac{{\dot{K}}}{K} = \frac{B(1 - u)(\beta - u)[\beta (n - 1) + 1]}{\beta [\beta (n - 1) + 1 - un]}. \end{aligned}$$

Also, substituting (30) into (32), we obtain

$$\begin{aligned} \frac{{\dot{h}}_{i}}{h_{i}}&= \frac{1}{(n - 1)\beta + 1} \left[ \{ g -(A - n\gamma ) \} \frac{K}{h_{i}} - nu(A - n\gamma ) \right] + g\\&= \frac{B(1 - u)(\beta - u)[\beta (n - 1) + 1]}{\beta [\beta (n - 1) + 1 - un]}. \end{aligned}$$

Third, we derive the growth rate of appropriation. The appropriation of group i is represented by $d_{i} = \gamma [K + uh_{i} + uZ_{i}]$. As in the discussion above, since we focus on the symmetric MPE, $h_{i} = h_{j}$ for all $j \ne i$. Therefore, differentiating this with respect to t yields

$$\begin{aligned} \frac{{\dot{d}}_{i}}{d_{i}} = \frac{{\dot{K}} + un{\dot{h}}_{i}}{K + un h_{i}}. \end{aligned}$$

On the balanced growth path, the growth rate of the common capital is equivalent to that of the private capital, ${\dot{K}}/K = {\dot{h}}_{i}/h_{i}$, and thus

$$\begin{aligned} \frac{{\dot{d}}_{i}}{d_{i}} = \frac{\frac{{\dot{K}}}{K}\left( 1 + nu\frac{h_{i}}{K}\right) }{1 + nu\frac{h_{i}}{K}} = \frac{{\dot{K}}}{K} = \frac{B(1 - u)(\beta - u)[\beta (n - 1) + 1]}{\beta [\beta (n - 1) + 1 - un]}. \end{aligned}$$

Finally, we check the boundary condition. Note that since the value function $V_{i}(K, h)$ has the properties $V_{i}(0, 0) = 0$ and strict concavity, holding the boundary condition (5) guarantees that the transversality conditions are satisfied. In the same manner as the previous section, for the boundary condition to be satisfied,

$$\begin{aligned}&\frac{\xi (K_{0} + h_{i0} + \beta Z_{i0})^{1 - \theta }}{1 - \theta } \lim _{t \rightarrow \infty } \exp \left( \left[ \frac{Bu(1 - u)(1 - \theta )(\beta - 1) }{\beta [(n - 1)\beta + 1 - un]}\right. \right. \\&\quad \left. \left. -\, \{ \rho + (\theta -1)(1 - u)B \} t \right] \right) . \end{aligned}$$

must converge to zero. If $\theta > 1$, it is easy to verify that this is satisfied. Let us consider the case where $0< \theta < 1$. According to (26) in Appendix 2, $\beta$ must be smaller than one to assure positive appropriation, which means that the first term is negative. Under Assumption 3, the equation $\rho + (\theta - 1)(1 - u)B$ is positive so that the second term is negative. Therefore, the boundary condition is satisfied.