Governance efficiency with and without government

Abstract

This paper explores the social interactions between public and private agents through a comparative institutional approach to consider the roles of community and government in societies with and without State. Using a theoretical framework where the private agents have different political power and they are, or are not, able to efficiently coordinate their actions, we study how public, private, and self-governance affect the level of welfare and capacity in each society. In particular, assuming two alternative private agent motivations (self-interested or other-regarding preferences) and community behaviors (collectivistic and individualistic societies), and a public agent as a bureaucracy with coercive power, that could either be partisan or bipartisan if it can, or cannot, be captured by private agents, we find that governance efficiency and capacity in the societies with State are lower when the government is partisan rather than when it is bipartisan. Moreover, society rankings for welfare and governance capacity are the same; thus, the welfare of a society is higher when the governance capacity is higher.

Appendix A: Proofs of Proposition 2

The proof of Proposition 2 consists of 2 parts, one for the ranking of W and one for those of $\gamma$.

We first present the proof for $W_{i,P}<W_{i,N}$, $i=I,C$. Substituting the results from (12) and (14), we have:

$$\begin{aligned}&(\theta _G^2(1-\pi ^2) )^{\theta _A+\theta _B} ((1-\theta _G)(1+\pi ) )^{1-\theta _A} ((1-\theta _G)(1-\pi ) )^{1-\theta _B} \\&\quad < (\theta _G^2)^{\theta _A+\theta _B} (1-\theta _G)^{1-\theta _A} (1-\theta _G)^{1-\theta _B} \end{aligned}$$

We can rewrite it as:

$$\begin{aligned} (1-\pi ^2 )^{\theta _A+\theta _B} \left( 1+\pi \right) ^{1-\theta _A}\left( 1-\pi \right) ^{1-\theta _B}<1 \end{aligned}$$

That becomes:

$$\begin{aligned} \left( 1-\pi \right) ^{1+\theta _A} \left( 1+\pi \right) ^{1+\theta _B}<1 \end{aligned}$$

(A.1)

Given that $(1-\pi )^{1+\theta _A}(1+\pi )^{1+\theta _A}>(1-\pi )^{1+\theta _A}(1+\pi )^{1+\theta _B}$ is always true given Assumption 1, (A.1) will always hold since $(1-\pi )^{1+\theta _A}(1+\pi )^{1+\theta _A}=(1-\pi ^2)^{1+\theta _A}<1$.

We now present the proof for $W_{i,N}<W_{I,S}$, with $i=I,C$. Substituting the results from (7) and (12), we have:

$$\begin{aligned}&(\theta _G^2)^{\theta _A} (\theta _G^2)^{\theta _B} (1-\theta _G)^{1-\theta _A} (1-\theta _G)^{1-\theta _B} \\&\quad < (\theta _A \theta _B)^{\theta _A} (\theta _A \theta _B)^{\theta _B} (1-\theta _A)^{1-\theta _A} (1-\theta _B)^{1-\theta _B} \end{aligned}$$

We first prove that $(\theta _G^2)^{\theta _A} (1-\theta _G)^{1-\theta _A} < (\theta _A \theta _B)^{\theta _A} (1-\theta _A)^{1-\theta _A}$. We can rewrite this inequality as:

$$\begin{aligned} \left( \frac{\theta _A \theta _B}{\theta _G^2}\right) ^{\theta _A} \left( \frac{1-\theta _A}{1-\theta _G}\right) ^{1-\theta _A}>1 \end{aligned}$$

We can rewrite this as follows:

$$\begin{aligned} (\theta _A \theta _B)^{\theta _A}>\left( \frac{1-\theta _G}{1-\theta _A}\right) ^{1-\theta _A} (\theta _G )^{2 \theta _A} \end{aligned}$$

(A.2)

Notice that the right-hand side of (A.2) is strictly increasing in $\theta _G$. Given Assumption 1, the maximum level $\theta _G$ can reach is $\theta _B$. Hence, studying (A.2) when $\theta _G\rightarrow \theta _B$ leads to:

$$\begin{aligned} (\theta _A \theta _B )^{\theta _A}>\left( \frac{1-\theta _B}{1-\theta _A}\right) ^{1-\theta _A} (\theta _B )^{2 \theta _A} \end{aligned}$$

that we can simplify as follows:

$$\begin{aligned} (\theta _A )^{\theta _A}>\left( \frac{1-\theta _B}{1-\theta _A}\right) ^{1-\theta _A} (\theta _B )^{\theta _A} \end{aligned}$$

(A.3)

Again, right-hand side of (A.3) is strictly increasing in $\theta _B$, so we can study how it behaves as $\theta _B\rightarrow \theta _A$. The right-hand side of (A.3) will always be lower than $(\theta _A )^{\theta _A}$, since $\lim \nolimits _{\theta _B\rightarrow \theta _A} \left( \frac{1-\theta _B}{1-\theta _A} \right) ^{1-\theta _A} (\theta _B )^{\theta _A} = (\theta _A )^{\theta _A}$ and $\theta _B<\theta _A$.

Applying a similar reasoning to $(\theta _G^2)^{\theta _B} (1-\theta _G)^{1-\theta _B} < (\theta _A \theta _B)^{\theta _B} (1-\theta _B)^{1-\theta _B}$, we obtain:

$$\begin{aligned} (\theta _A \theta _B )^{\theta _B}>\left( \frac{1-\theta _G}{1-\theta _B}\right) ^{1-\theta _B} (\theta _G )^{2 \theta _B} \end{aligned}$$

(A.4)

The right-hand side of (A.4) is strictly increasing in $\theta _G$, so we can study what happens when $\theta _G\rightarrow \theta _B$, obtaining:

$$\begin{aligned} (\theta _A )^{\theta _B}> (\theta _B)^{\theta _B} \end{aligned}$$

which is always true, given Assumption 1.

Hence, $W_{i,N}<W_{I,S}$, with $i=I,C$.

We now present the proof for $W_{I,S}<W_{C,S}$. Substituting the value in (7) and (9) in the condition $W_{I,S}<W_{C,S}$, we have:

$$\begin{aligned}&(\theta _A \theta _B)^{\theta _A+\theta _B} (1-\theta _A)^{1-\theta _A} (1-\theta _B)^{1-\theta _B}\\&\quad <\left( \frac{ (\theta _A+\theta _B )^2}{\left( 1+\theta _A\right) \left( 1+\theta _B\right) }\right) ^{\theta _A+\theta _B} \left( \frac{1-\theta _A}{1+\theta _B}\right) ^{1-\theta _A} \left( \frac{1-\theta _B}{1+\theta _A}\right) ^{1-\theta _B} \end{aligned}$$

We can rewrite this inequality as:

$$\begin{aligned} \frac{ (\theta _A+\theta _B )^2}{\theta _A \theta _B}> (1+\theta _A )^{\frac{1+\theta _A}{\theta _A+\theta _B}} (1+\theta _B )^{\frac{1+\theta _B}{\theta _A+\theta _B}} \end{aligned}$$

(A.5)

Notice that the left-hand side of (A.5) is always larger than 4, since it can be rewritten as $2+\frac{\theta _A}{\theta _B}+\frac{\theta _B}{\theta _A}$, which is larger than 4 if $\frac{ (\theta _A-\theta _B )^2}{\theta _A \theta _B}>0$, which is always true.

The right-hand side of (A.5) is increasing in $\theta _A$, since its derivative with respect to $\theta _A$ is equal to:

$$\begin{aligned} \frac{(1+\theta _A)^{\frac{1+\theta _A}{\theta _A+\theta _B}} (1+\theta _B)^{\frac{1+\theta _B}{\theta _A+\theta _B}} \left[ \theta _A+\theta _B-(1-\theta _B) \ln (1+\theta _A)-(1+\theta _B) \ln (1+\theta _B)\right] }{(\theta _A+\theta _B)^2} \end{aligned}$$

and it is positive when the expression in square brackets is positive, that is when $\theta _A+\theta _B-(1-\theta _B)\ln (1+\theta _A)-(1+\theta _B) \ln (1+\theta _B)>0$, which is always true since $\theta _A>\ln (1+\theta _A)$, $\theta _B>\ln (1+\theta _B)$, and $\theta _B\left[ \ln (1+\theta _A) -\ln (1+\theta _B)\right] >0$ given Assumption 1. Hence, the right-hand side of (A.5), as $\theta _A\rightarrow 1$, will be equal to:

$$\begin{aligned} 2^{\frac{2}{1+\theta _B}} (1+\theta _B) \end{aligned}$$

(A.6)

The expression in (A.6) is convex in (0, 1), so we can study how it behaves as $\theta _B$ approaches either 0 or 1:

$$\begin{aligned}&\lim _{\theta _B\rightarrow 1} 2^{\frac{2}{1+\theta _B}} (1+\theta _B) = 4\\&\lim _{\theta _B\rightarrow 0} 2^{\frac{2}{1+\theta _B}} (1+\theta _B) = 4 \end{aligned}$$

This implies that the right-hand side of (A.5) is always smaller than 4 and, thus, of the left-hand side of (A.5).

This completes the first part of the proof.

We now present the proof for the ranking of $\gamma$.

$\gamma _{i,P}<\gamma _{i,N}$, $i=I,C$, is already proved, as it is the Lemma 1.

$\gamma _{i,N}<\gamma _{I,S}$, with $i=I,C$, is true when $\theta _G^2<\theta _A \theta _B$. Recall that $\theta _G<\theta _B<\theta _A$. Hence, is always verified.

$\gamma _{I,S}<\gamma _{C,S}$ is already proved, as it is the Corollary 1.

This completes the proof.