The Long Italian Stagnation and the Welfare Effects of Outsourcing

Abstract

The stagnation of the Italian economy over the last two decades is widely documented. During this period, the world economy has become highly integrated, and foreign outsourcing has become a standard practice for firms. While trade theory predicts benefits from the internationalization of production, Italy seems to have gained negligibly from it, or, rather to have lost. In a simple model, we show that this may be the case when markets are overregulated and competition policies are weak. We study a small open economy with one oligopolistic and one competitive sector, which outsources part of its production process abroad. Advances in globalization entail lower tariff rates of outsourcing. Contrary to the common wisdom, we show that national welfare is an inverted U-shaped function of tariffs. There exists a tariff threshold, below which the economy loses from globalization because the competitive sector overproduces and the oligopolistic underproduces (the oligopolistic good has a higher marginal effect on welfare). Competition policies that target the competitive sector lower the threshold and allow the economy to benefit from increased openness.

Appendix: Proof of Propositions

The proof of both propositions is based on utility function (25) and on the model solutions:

$$ {X}^D=\frac{\varphi }{1-\varphi }{A}^V{\left(\frac{\alpha }{\varOmega}\overline{K}\right)}^{\alpha }{\left(\frac{1-\alpha }{\varPsi}\overline{L}\right)}^{1-\alpha}\frac{1}{{\overline{P}}_X}\cdot {P}_{V,x}\left(\tau \right)\cdot T\left(\tau \right) $$

(29)

$$ Y=\frac{N-1}{N}{A}^Y{\left(\frac{\beta }{\varOmega}\overline{K}\right)}^{\beta }{\left(\frac{1-\beta }{\varPsi}\overline{L}\right)}^{1-\beta}\cdot T\left(\tau \right) $$

(30)

where $\varOmega \left(\tau, N\right):=\alpha +\beta \cdot T\left(\tau \right)\cdot \left(N-1/N\right)$, and $\varPsi \left(\tau, N\right):=\left(1-\alpha \right)+\left(1-\beta \right)\cdot T\left(\tau \right)\cdot$ $\cdot \left(N-1/N\right)$. The conditions $N>1$ and $\tau \ge 0>-\eta$ ($\tau >-1+{A}^X{\left(1-a\right)}^{1/\eta}\left({\overline{P}}_X/{\overline{P}}_O\right)$) guarantee positive solutions in the Cobb–Douglas (CES) case.

Proposition 1

We first show that utility (1) is continuous in τ for $\tau \ge 0$. This is immediately seen from the fact that Ω(τ, N) and Ψ(τ, N) are continuous in τ and strictly positive since $T\left(\tau \right)>0$ for any $\tau \ge 0$. Hence, X ^D and Y are also continuous in τ. Differentiating Eq. (1) with respect to the tariff rate yields

$$ \frac{\partial U}{\partial \tau }=\left[\varphi \frac{1}{X^D}\frac{\partial {X}^D}{\partial \tau }+\left(1-\varphi \right)\frac{1}{Y}\frac{\partial Y}{\partial \tau}\right]\cdot U. $$

(31)

If τ goes to infinity, utility is zero since Ω(τ, N) and Ψ(τ, N) are finite and X ^D collapses to zero (see Eq. (27)). For $\tau \ge 0$ $0<U\left(\tau, N\right)<\infty, \forall N>1$. Thus, ${U}_{\tau}^{\prime }=0$ if and only if the term in square brackets in Eq. (31) is zero. Its opposite is equivalent to the following cubic equation in the level of tariffs:

$$ {\tau}^3+a\cdot {\tau}^2+b\cdot \tau +c=0 $$

(32)

where

$$ {\fontsize{8.5}{10.5}\selectfont{\begin{array}{l}a:=\dfrac{1}{d}\left\{\dfrac{1}{\eta^2}\dfrac{N-1}{N}\left[\dfrac{\beta \left(1-\beta \right)\left(1-\varphi \right)}{\varphi}\left(3\dfrac{N-1}{N}-\eta \right)+\left[\alpha \left(1-\beta \right)+\left(1-\alpha \right)\beta \right]\left(1+\eta \right)\right]+\right.\\[9pt] \qquad\qquad\left.+\alpha \left(1-\alpha \right)\left(\dfrac{1}{\eta}\dfrac{N-1}{N}+2\dfrac{\varphi }{\left(1-\varphi \right)\eta }-1\right)\right\}\end{array} }}$$

$$ {\fontsize{8.5}{10.5}\selectfont{\begin{array}{l}b:=\dfrac{1}{d}\left\{\left[\left(1-\alpha \right)\left(\alpha \left(1+\eta \right)+\beta \right)-\dfrac{\beta \left(1-\beta \right)\left(1-\varphi \right)}{\varphi}\left(1+\eta -3\dfrac{N-1}{N}\right)+\alpha \left(1-\beta \right)\right]\right.\\[12pt] \qquad\qquad\left.\dfrac{N-1}{N}\dfrac{1}{\eta }-\alpha \left(1-\alpha \right)\left[2-\dfrac{\varphi }{\left(1-\varphi \right)\eta}\right]\right\}\end{array}}} $$

$$ c:=-\frac{1}{d}\left\{\frac{1}{N}\left[\alpha \left(1-\alpha \right)+\beta \left(1-\beta \right)\frac{1-\varphi }{\varphi}\frac{N-1}{N}\right]\right\} $$

where

$$ d:=\frac{1}{\eta}\left\{\!\frac{1}{\eta}\frac{N{-}1}{N}\left[\alpha \left(1{-}\beta \right)+\left(1{-}\alpha \right)\beta \right]+\frac{\alpha \left(1{-}\alpha \right)\varphi }{1{-}\varphi }+\frac{\beta \left(1{-}\beta \right)\left(1{-}\varphi \right)}{\varphi }{\left(\!\frac{1}{\eta}\frac{N{-}1}{N}\!\right)}^2\!\right\} $$

Note first that $a>0$, $d>0$, and $c<0$, which ensure two negative and one positive solution. (The sign of b is irrelevant.) Let ${\tau}^{\ast }$ be the positive solution. In order to prove that the positive solution is a maximum observe that ${U}_{\tau}^{\prime}\left(0,N\right)>0$ because $c<0$ and Eq. (32) is the opposite of the term in square brackets in (31). Since U(τ, N) is continuous, and the other roots of Eq. (32) are negative, it follows that ${U}_{\tau}^{\prime}\left(0,N\right)>0$ in $\left[0,{\tau}^{\ast}\right)$. The fact that ${\tau}^{\ast }$ is a root of a cubic equation with at least two distinct solutions ensures that ${U}_{\tau}^{\prime}\left(0,N\right)<0$ if $\tau >{\tau}^{\ast }$. Thus, ${\tau}^{\ast }$ is a utility maximum. This proves Proposition 1.

Proposition 2

We show first that utility function (1) is continuous in N for $N>1$. This is immediately from the fact that Ω(τ, N) and Ψ(τ, N) are continuous in N and strictly positive for any $N>1$ and so are X ^D and Y. Differentiating the utility Eq. (25) and setting U ^′ _N (τ, N) equal to zero yields the following quadratic equation in $M:=\left(N-1\right)/N$:

$$ A\cdot {M}^2+B\cdot M+C=0 $$

(33)

with

$$ \begin{array}{c}A:=-\beta \left(1-\beta \right)\varphi \cdot {\left[T\left(\tau \right)\right]}^2\\ {}B:=-\left[\alpha \varphi \left(1-\alpha \right)-\beta \left(1-\varphi \right)\left(1-\beta \right)\right]\cdot T\left(\tau \right)\\ {}C:=\alpha \left(1-\alpha \right)\left(1-\varphi \right)\end{array} $$

(34)

Since $A<0$ and $C>0$ for all feasible model parameters, $\left({B}^2-4 AC\right)$ is strictly positive. This ensures the existence of two real and distinct solutions, which are discordant in sign. Since ${N}_{1,2}=1/\left(1-{M}_{1,2}\right)$, the negative solution ${M}_2=\left(-B-\sqrt{B^2-4 AC}\right)/2A$ is unfeasible because $N>1$ must hold. The positive solution is feasible only if ${M}_1=\left(-B+\sqrt{B^2-4 AC}\right)/2A<1$, which is equivalent to $\left(A+B+C\right)>0$. Replace A, B, C by their definitions and verify that this is a product of positive terms. Since $A<0$ and $\left({B}^2-4 AC\right)>0$, $\left(A\cdot {M}^2+B\cdot M+C\right)$ is positive (negative) for $M<{M}_1$ $\left(M>{M}_1\right)$ which proves that ${N}_1=1/\left(1-{M}_1\right)$ is a utility maximum. Use definitions (34) and (26) to verify that the optimal N is

$$ {N}^{\ast }=\frac{1}{1-\eta}\cdot \left(1+\frac{\eta }{\tau}\right) $$

(35)

Observe that if τ becomes zero, ${N}^{\ast }$ is infinite. This proves Proposition 2.