The Effects of Corruption in a Monetary Union

Abstract

72% of Europeans in European Monetary Union countries think that corruption is widespread in their country. Corruption can affect the individual country’s performance and, in turn, impact on the performance of the other members of the monetary union. In this article, we will present a monetary union composed of two countries and we will study how the presence of corruption in one member country will affect its economy and the neighbour’s economy. Additionally, we will analyse what measures could reduce the negative spillover effects generated by the presence of corruption.

Appendix

We initially derive the expressions for output and public spending for country 1. To ease the notation we omit the subscript 1 in the following two proofs.

Derivation of Expression (1).

Output of a representative firm is given by X = L^λ, where X denotes the real output, L represents labour and λ indicates the output elasticity. Distortionary taxes are levied on production. The firm maximises profit, given by: (1 − τ)PL^λ − WL, where τ denotes the tax rate on total revenue of firms, P represents the price level and W is the nominal wage. Solving for the firm’s labour demand, assuming it can hire the labour it demands at the given nominal wage, taking logs, we have x = a(p − τ − w) + b, where lower-case letters denote logs of nominal variables, a = λ/(1 − λ), b = λ ln λ/(1 − λ), and ln (1 − τ) ≈  − τ. Following Debelle and Fischer (1994), for simplicity, we set λ = 0.5, so that a = 1 and we approximate ln λ to 0. Hence, x = p − τ − w.

In addition, following Alesina and Tabellini (1987), Debelle and Fischer (1994) and Beetsma and Bovenberg (2001), among others, we assume that workers are represented by a centralised trade union, which sets nominal wage (in logs) and it has the following objective function: E((w − p)²)/2. The first order condition of this optimisation problem immediately yields w = p^e. Therefore, x = p − p^e − τ. Finally, approximating p − p^e by π − π^e, where π represents the inflation rate and π^e the expected inflation rate, we get the aggregate supply equation of the model, i.e., x = π − π^e − τ.

Derivation of Expression (2).

The government budget constraint in nominal terms is given by P_tG_t = ϕτ_tP_tX_t, where G_t denotes public spending and ϕ the degree of institutional quality. Dividing the government budget constraint by nominal income, P_tX_t, yields g_t = ϕτ_t, where g_t is public spending as a share of X_t.

Proof of Proposition 1.

Substituting Expressions (1), (2) and (3) into (4) and (5), it follows that

$$ {\displaystyle \begin{array}{c}{L}_1=\frac{1}{2}\left[{\pi}^2+{\delta}_1{\left(\pi -{\pi}^e-{\tau}_1-{\overline{x}}_1\right)}^2+{\gamma}_1{\left({\phi \tau}_1-{\overline{g}}_1\right)}^2\right]\\ {}{L}_2=\frac{1}{2}\left[{\pi}^2+{\delta}_2{\left(\pi -{\pi}^e-{\tau}_2-{\overline{x}}_2\right)}^2+{\gamma}_2{\left({\tau}_2-{\overline{g}}_2\right)}^2\right],\mathrm{and}\\ {}{L}_{CCB}=\frac{1}{2}\left[{\pi}^2+{\delta}_{CCB}{\left(z\left(\pi -{\pi}^e-{\tau}_1\right)+\left(1-z\right)\left(\pi -{\pi}^e-{\tau}_2\right)-\left(z{\overline{x}}_1+\left(1-z\right){\overline{x}}_2\right)\right)}^2\right].\end{array}} $$

The first-order conditions of the fiscal authorities’ optimisation problems are given by

$$ {\displaystyle \begin{array}{c}\frac{{\partial L}_1}{{\partial \tau}_1}=-{\delta}_1\left(\pi -{\pi}^e-{\tau}_1-{\overline{x}}_1\right)+{\phi \gamma}_1\left({\phi \tau}_1-{\overline{g}}_1\right)=0,\mathrm{and}\ \\ {}\frac{{\partial L}_2}{{\partial \tau}_2}=-{\delta}_2\left(\pi -{\pi}^e-{\tau}_2-{\overline{x}}_2\right)+{\gamma}_2\left({\tau}_2-{\overline{g}}_2\right)=0.\end{array}} $$

Hence,

$$ {\tau}_1=\frac{{\overline{g}}_1}{\phi }-\frac{\frac{\delta_1}{\gamma_1}}{\phi \left({\phi}^2+\frac{\delta_1}{\gamma_1}\right)}\left(\phi {\overline{x}}_1+{\overline{g}}_1\right)+\frac{\frac{\delta_1}{\gamma_1}}{\phi^2+\frac{\delta_1}{\gamma_1}}\left(\pi -{\pi}^e\right),\mathrm{and} $$

(17)

$$ {\tau}_2={\overline{g}}_2-\frac{\frac{\delta_2}{\gamma_2}}{1+\frac{\delta_2}{\gamma_2}}\left({\overline{x}}_2+{\overline{g}}_2\right)+\frac{\frac{\delta_2}{\gamma_2}}{1+\frac{\delta_2}{\gamma_2}}\left(\pi -{\pi}^e\right). $$

(18)

For the central bank, the first-order condition of its optimisation problem implies that

$$ \frac{\partial {L}_{CCB}}{\partial \pi }=\pi +{\delta}_{CCB}\left(z\left(\pi -{\pi}^e-{\tau}_1\right)+\left(1-z\right)\left(\pi -{\pi}^e-{\tau}_2\right)-\left(z{\overline{x}}_1+\left(1-z\right){\overline{x}}_2\right)\right)=0. $$

Thus,

$$ \pi =\frac{\delta_{CCB}}{1+{\delta}_{CCB}}\left({\pi}^e+z\left({\tau}_1+{\overline{x}}_1\right)+\left(1-z\right)\left({\tau}_2+{\overline{x}}_2\right)\right). $$

(19)

Plugging Expressions (17) and (18) into (19), it follows that

$$ \pi =\frac{z\frac{\phi^2}{\phi^2+\frac{\delta_1}{\gamma_1}}+\left(1-z\right)\frac{1}{1+\frac{\delta_2}{\gamma_2}}}{\Delta}{\pi}^e+z\frac{\phi }{\left({\phi}^2+\frac{\delta_1}{\gamma_1}\right)\Delta}\left(\phi {\overline{x}}_1+{\overline{g}}_1\right)+\left(1-z\right)\frac{1}{\left(1+\frac{\delta_2}{\gamma_2}\right)\Delta}\left({\overline{x}}_2+{\overline{g}}_2\right), $$

(20)

where $\Delta =\frac{1}{\delta_{CCB}}+z\frac{\phi^2}{\phi^2+\frac{\delta_1}{\gamma_1}}+\left(1-z\right)\frac{1}{1+\frac{\delta_2}{\gamma_2}}$. Using the rational expectation hypothesis, we know that π  =  π^e. Then, from (20), (17) and (18), we obtain Expressions (7)–(9).

Proof of Corollary 2.

We differentiate the expressions for output (Eqs. 10 and 11), public spending (Eqs. 12 and 13) and inflation (Eq. 9) with respect to ϕ. Therefore, we obtain the following expressions:

$$ {\displaystyle \begin{array}{c}\frac{\partial }{\partial \phi }{x}_1^{\ast }=-\frac{2\phi \frac{\delta_1}{\gamma_1}{\overline{x}}_1+\left(\frac{\delta_1}{\gamma_1}-{\phi}^2\right){\overline{g}}_1}{{\left({\phi}^2+\frac{\delta_1}{\gamma_1}\right)}^2},\frac{\partial }{\partial \phi }{g}_1^{\ast }=\frac{\delta_1}{\gamma_1}\frac{\left({\phi}^2-\frac{\delta_1}{\gamma_1}\right){\overline{x}}_1+2\phi {\overline{g}}_1}{{\left({\phi}^2+\frac{\delta_1}{\gamma_1}\right)}^2},\\ {}\frac{\partial }{\partial \phi }{\pi}^{\ast }=z{\delta}_{CCB}\frac{2\phi \frac{\delta_1}{\gamma_1}{\overline{x}}_1+\left(\frac{\delta_1}{\gamma_1}-{\phi}^2\right){\overline{g}}_1}{{\left({\phi}^2+\frac{\delta_1}{\gamma_1}\right)}^2},\mathrm{and}\\ {}\frac{\partial }{\partial \phi }{x}_2^{\ast }=\frac{\partial }{\partial \phi }{g}_2^{\ast }=0.\end{array}} $$

Hence, $\frac{\partial }{\partial \phi }{x}_1^{\ast }>0\ \mathrm{and}\ \frac{\partial }{\partial \phi }{\pi}^{\ast }<0$ if and only if ${\gamma}_1>{\overline{\gamma}}_1$, where the expression of ${\overline{\gamma}}_1$ is given in the statement of this corollary. Finally, taking into account the assumption that $\frac{{\overline{x}}_1}{{\overline{g}}_1}<\frac{\phi {\gamma}_1}{\delta_1}$, we can conclude that $\frac{\partial }{\partial \phi }{g}_1^{\ast }>0$.

Proof of Corollary 3.

a) Substituting Expressions (9), (10) and (12) into (4) for country 1 and differentiating the resulting expression with respect to ϕ, we have

$$ \frac{\partial {L}_1^{\ast }}{\partial \phi }={\gamma}_1\frac{p\left({\gamma}_1\right)}{{\left({\delta}_1+{\phi}^2{\gamma}_1\right)}^3\left(\frac{\delta_2}{\gamma_2}+1\right)}, $$

where $p\left({\gamma}_1\right)={p}_2{\gamma}_1^2+{p}_1{\gamma}_1+{p}_0$, with

$$ {\displaystyle \begin{array}{c}{p}_2=-\left(z\left(1-z\right){\phi}^4{\delta}_{CCB}^2{\overline{g}}_1\left({\overline{x}}_2+{\overline{g}}_2\right)+{\phi}^3\left({z}^2{\delta}_{CCB}^2+{\delta}_1\right)\left(1+\frac{\delta_2}{\gamma_2}\right)\left(\phi {\overline{x}}_1+{\overline{g}}_1\right){\overline{g}}_1\right),\\ {}{p}_1=2z\left(1-z\right){\phi}^3{\delta}_1{\delta}_{CCB}^2{\overline{x}}_1\left({\overline{x}}_2+{\overline{g}}_2\right)+\phi {\updelta}_1\left({\updelta}_1\left(\phi {\overline{x}}_1-{\overline{g}}_1\right)+{z}^2{\delta}_{CCB}^2\left(2\phi {\overline{x}}_1+{\overline{g}}_1\right)\right)\left(1+\frac{\delta_2}{\gamma_2}\right)\left(\phi {\overline{x}}_1+{\overline{g}}_1\right),\mathrm{and}\ \\ {}{p}_0=z\left(1-z\right){\delta}_1^2{\delta}_{CCB}^2\left(2\phi {\overline{x}}_1+{\overline{g}}_1\right)\left({\overline{x}}_2+{\overline{g}}_2\right)+{\delta}_1^3\left(1+\frac{\delta_2}{\gamma_2}\right)\left(\phi {\overline{x}}_1+{\overline{g}}_1\right){\overline{x}}_1.\end{array}} $$

Notice that p₂ < 0 and p₀ > 0 . This allows us to guarantee that there exists a unique positive root of the polynomial p(γ₁), denoted by ${\overset{\sim }{\gamma}}_1$. Hence, we can conclude that $\frac{\partial {L}_1^{\ast }}{\partial \phi }<0$ if and only if ${\gamma}_1>{\overset{\sim }{\gamma}}_1$. Moreover, in order to show that

$$ {\overline{\gamma}}_1>{\overset{\sim }{\gamma}}_1 $$

(21)

it suffices to prove that $p\left({\overline{\gamma}}_1\right)<0$. Direct computations yield $p\left({\overline{\gamma}}_1\right)=-2{\delta}_1^3\frac{1+\frac{\delta_2}{\gamma_2}}{\phi {\overline{g}}_1}{\left(\phi {\overline{x}}_1+{\overline{g}}_1\right)}^3<0\ \mathrm{and}$, hence, (21) is satisfied.

b) Taking into account that the output and public spending of country 2 are not affected by the level of corruption in country 1, we have that $\frac{\partial {L}_2^{\ast }}{\partial \phi }={\pi}^{\ast}\frac{\partial {\pi}^{\ast }}{\partial \phi }$. Combining the positiveness of π^∗ and Corollary 2, it follows that $\frac{\partial {L}_2^{\ast }}{\partial \phi }<0$ if and only if ${\gamma}_1>{\overline{\gamma}}_1$.

Proof of Corollary 4.

Note that country 2 has the same losses whether there is corruption in country 1 or not if ${L}_2^{\ast}\left(\phi \right)-{L}_2^{\ast }(1)=0.$ Using Expression (4) for country 2 and Corollary 2c, the previous equality is equivalent to

$$ \frac{1}{2}{\left({\pi}^{\ast}\left(\phi \right)\right)}^2-\frac{1}{2}{\left({\pi}^{\ast }(1)\right)}^2=0, $$

and, hence, π^∗(ϕ) = π^∗(1). Using Expression (9), the previous equality can be rewritten as

$$ z\frac{\phi {\delta}_{CCB}}{\phi^2+\frac{\delta_1}{\gamma_1}}\left(\phi {\overline{x}}_1+{\overline{g}}_1^R\right)+\left(1-z\right)\frac{\delta_{CCB}}{1+\frac{\delta_2}{\gamma_2}}\left({\overline{x}}_2+{\overline{g}}_2\right)=z\frac{\delta_{CCB}}{1+\frac{\delta_1}{\gamma_1}}\left({\overline{x}}_1+{\overline{g}}_1\right)+\left(1-z\right)\frac{\delta_{CCB}}{1+\frac{\delta_2}{\gamma_2}}\left({\overline{x}}_2+{\overline{g}}_2\right). $$

Isolating the value of ${\overline{g}}_1^R$, it follows that ${\overline{g}}_1^R=\frac{\phi^2+\frac{\delta_1}{\gamma_1}}{\phi \left(1+\frac{\delta_1}{\gamma_1}\right)}\left({\overline{x}}_1+{\overline{g}}_1\right)-\phi {\overline{x}}_1$, which can be rewritten as ${\overline{g}}_1^R={\overline{g}}_1-\Psi$, where the expression of Ψ is given in the statement of this corollary.

Proof of Corollary 5.

Using the expression of Ψ, we get that

$$ {L}_1^{\ast}\left(\Psi \right)-{L}_1^{\ast }(0)=\frac{\phi {\gamma}_1\Psi}{2\left({\delta}_1+{\gamma}_1\right){\left({\delta}_1+{\phi}^2{\gamma}_1\right)}^2\left({\delta}_2+{\gamma}_2\right)}l\left({\gamma}_1\right), $$

where $l\left({\gamma}_1\right)={l}_3{\gamma}_1^3+{l}_2{\gamma}_1^2+{l}_1{\gamma}_1+{l}_0,$ with

$$ {\displaystyle \begin{array}{c}{l}_3=\left(1-\phi \right){\phi}^3\left({\delta}_2+{\gamma}_2\right){\overline{g}}_1,\\ {}{l}_2=\phi \left[\left({\left(1-\phi \right)}^2{\delta}_1{\overline{g}}_1-\left(1+\phi \right){z}^2{\delta}_{CCB}^2{\overline{g}}_1-\left(\left(1-{\phi}^2\right){\delta}_1+2{z}^2{\delta}_{CCB}^2\right)\phi {\overline{x}}_1\right)\left({\delta}_2+{\gamma}_2\right)-2\phi z\left(1-z\right){\gamma}_2{\delta}_{CCB}^2\left({\overline{x}}_2+{\overline{g}}_2\right)\right],\\ {}\begin{array}{c}{l}_1=-{\delta}_1\Big[\left(\left(1-\phi \right)\left(\left(1+\phi \right){\overline{x}}_1+{\overline{g}}_1\right){\delta}_1+{z}^2{\delta}_{CCB}^2\left(\left(1+{\phi}^2\right){\overline{x}}_1+\left(1+\phi \right){\overline{g}}_1\right)\right)\left({\delta}_2+{\gamma}_2\right)\\ {}\begin{array}{c}+2\left(1+{\phi}^2\right)z\left(1-z\right){\gamma}_2{\updelta}_{CCB}^2\left({\overline{x}}_2+{\overline{g}}_2\right)\Big],\mathrm{and}\\ {}{l}_0=-2z\left(1-z\right){\delta}_1^2{\gamma}_2{\delta}_{CCB}^2\left({\overline{x}}_2+{\overline{g}}_2\right).\end{array}\end{array}\end{array}} $$

Note that $l\left({\delta}_1\frac{\left(1+\phi \right){\overline{x}}_1+{\overline{g}}_1}{\phi {\overline{g}}_1}\right)<0$ and limγ_{1 ⟶ ∞}l(γ₁) > 0. Then, applying the Descartes’ rule, we can conclude that there exists a unique value of γ₁, denoted by $\hat{\gamma_1}$, that satisfies ${L}_1^{\ast}\left(\Psi \right)-{L}_1^{\ast }(0)=0.$ Moreover, we know that $\hat{\gamma_1}\in \left({\delta}_1\frac{\left(1+\phi \right){\overline{x}}_1+{\overline{g}}_1}{\phi {\overline{g}}_1},\infty \right)$. Therefore, if ${\gamma}_1<\hat{\gamma_1}$, country 1 is better off with its required public spending target. Otherwise, if ${\gamma}_1>\hat{\gamma_1}$, country 1 is worse off with its required target.

Proof of Corollary 6.

Using the expression of Ψ, it follows that the inequality ${L}_1^{\ast}\left(\Psi \right)>{L}_1^{NM\ast }$ is equivalent to $\frac{\phi^2{\left(1-\phi \right)}^2{\gamma}_1^2{\overline{g}}_1^2}{\left({\delta}_1+{\phi}^2{\gamma}_1\right){\left({\delta}_1+{\gamma}_1\right)}^2}{\left({\gamma}_1-{\delta}_1\frac{\left(1+\phi \right){\overline{x}}_1+{\overline{g}}_1}{\phi {\overline{g}}_1}\right)}^2>\frac{\phi^2{\delta}_{CCB}^2}{{\left({\phi}^2+\frac{\delta_1}{\gamma_1}\right)}^2}{\left(\phi {\overline{x}}_1+{\overline{g}}_1\right)}^2-{\left(z\frac{\delta_{CCB}}{1+\frac{\delta_1}{\gamma_1}}\left({\overline{x}}_1+{\overline{g}}_1\right)+\left(1-z\right){\delta}_{CCB}\frac{{\overline{x}}_2+{\overline{g}}_2}{1+\frac{\delta_2}{\gamma_2}}\right)}^2.$

Note that if γ₁ is high enough, the previous inequality is satisfied and, consequently, in this case country 1 prefers to leave the monetary union. By contrast, if ${\gamma}_1=\hat{\gamma_1}$, then ${L}_1^{\ast}\left(\Psi \right)={L}_1^{\ast }(0)$. It is easy to see that this value is lower than ${L}_1^{NM\ast }$ whenever $\frac{{\overline{x}}_2+{\overline{g}}_2}{1+\frac{\delta_2}{\gamma_2}}<\frac{\phi }{\phi^2+\frac{\delta_1}{\gamma_1}}\left(\phi {\overline{x}}_1+{\overline{g}}_1\right).$