Can underemployment persist in an expanding economy? Clues from a non-Walrasian OLG model with endogenous longevity

Abstract

This paper aims at casting a new light on the persistence of underemployment in emerging economies, by examining the relationship between labour market imperfections and longevity changes. For that purpose, we develop a two-period OLG model where longevity depends positively on the real wage, but negatively on the underemployment level, which both result from wage negotiations between a trade-union, representing workers (i.e. young generation), and the management, representing capital-holders (i.e. old generation). The existence, uniqueness and stability of a non-trivial steady-state equilibrium are studied. The distribution of bargaining power is shown to be a major determinant of the short run and long run dynamics of employment, production and longevity. The dynamics is also shown to be significantly sensitive to the precise form under which job quality affects longevity.

Appendix

Existence of a non-trivial steady-state The case where $\delta > \frac{(1-\alpha)(1-\beta)}{\alpha \beta +(1-\beta)}$ can be treated as follows.

Substituting the kk locus within the hh locus allows us to derive an expression giving us h _t+1 as a function G(h _t ):

$$ \begin{aligned} h_{t+1}=&Z\left(A(1-\alpha)\left(\left(\frac{A(1-\alpha)(1-\beta )h_{t} }{\beta +(1-\beta)h_{t}}\hbox{min} \left(\left(\frac{\left[ \delta \beta +(1-\beta)h_{t}\right] }{(1-\alpha)^{-1}\delta \rho }\right) ^{\frac{1}{ \rho }},\left(\frac{\beta +(1-\beta )h_{t}}{\rho }\right) ^{\frac{1}{\rho} }\right) \right) ^{\frac{1}{1-\alpha }}\right) ^{\alpha }\right) ^{\psi}\\ &\left(\hbox{min} \left(\Uplambda,\Uptheta \right) \right)^{ \frac{\psi }{1-\alpha }}\left(\hbox{min} \left(\frac{\left(\frac{(1-\alpha) \left[ \delta \beta +(1-\beta )h_{t}\right] }{\delta \rho }\right) ^{\frac{1 }{\rho }}}{\left(\frac{\beta +(1-\beta)h_{t}}{\rho }\right) ^{\frac{1}{ \rho }}},1\right) \right) ^{\phi} \end{aligned} $$

where $\Uplambda \equiv \left(\frac{(1-\alpha)\left(\delta \beta +(1-\beta)h_{t}\right) }{\delta \rho }\right) ^{\frac{-\alpha }{\rho }}$ and $\Uptheta \equiv \frac{\delta \left(\frac{\beta +(1-\beta)h_{t}}{\rho }\right) ^{\frac{-\alpha }{\rho }}\left(\beta +(1-\beta)h_{t}\right) }{(1-\alpha)(\delta \beta +(1-\beta)h_{t})}.$

Under $\delta > \frac{(1-\alpha)(1-\beta)}{\alpha \beta +(1-\beta)},$ we know that Λ < Θ, and that:

$$ \hbox{min} \left(\left(\frac{(1-\alpha)\left[ \delta \beta +(1-\beta)h_{t} \right] }{\delta \rho }\right) ^{\frac{1}{\rho }},\left(\frac{\beta +(1-\beta)h_{t}}{\rho }\right) ^{\frac{1}{\rho }}\right) =\left(\frac{ (1-\alpha)\left[ \delta \beta +(1-\beta)h_{t}\right] }{\delta \rho} \right) ^{\frac{1}{\rho}} $$

so that we can rewrite G(h _t ) as:

$$ h_{t+1}=Z\left(\left[ A(1-\alpha)\right] ^{\frac{1}{1-\alpha }}\left(\frac{(1-\beta )h_{t}}{\beta +(1-\beta)h_{t}}\left(\frac{\left[ \delta \beta +(1-\beta)h_{t}\right] }{(1-\alpha)^{-1}\delta \rho }\right) ^{\frac{ 1}{\rho }}\right) ^{\frac{\alpha }{1-\alpha }}\left(\Uplambda \right) ^{\frac{ 1}{1-\alpha }}\right) ^{\psi }\left(\frac{\left(\frac{\left[ \delta \beta +(1-\beta )h_{t}\right] }{(1-\alpha)^{-1}\delta \rho }\right) ^{\frac{1}{ \rho }}}{\left(\frac{\beta +(1-\beta)h_{t}}{\rho }\right) ^{\frac{1}{\rho} }}\right) ^{\phi} $$

Given that G(h _t ) ≥ 0, it can be shown that G(h _t ) has necessarily a fixed point, that is, a value of h _t such that G(h _t ) = h _t , if the following three conditions are satisfied:

• (i) G(0) = 0

• (ii) $\lim_{h\rightarrow 1}G(h_{t})/h_{t} < 1$

• (iii) $\lim_{h\rightarrow 0}G^{\prime }(h_{t})=+\infty$

Regarding (i), it is easy to check that:

$$ G(0)=Z\left(\left[ A(1-\alpha)\right] ^{\frac{1}{1-\alpha }}\left(\frac{0 }{\beta }\left(\frac{(1-\alpha)\left[ \delta \beta \right] }{\delta \rho} \right) ^{\frac{1}{\rho }}\right) ^{\frac{\alpha }{1-\alpha }}\left(\Uplambda \right) ^{\frac{1}{1-\alpha }}\right) ^{\psi }\frac{\left(\frac{(1-\alpha) \left[ \delta \beta \right] }{\delta \rho }\right) ^{\frac{\phi }{\rho }}}{ \left(\frac{\beta }{\rho }\right) ^{\frac{\phi }{\rho }}}=0 $$

Regarding (ii),

$$ \lim_{h\rightarrow 1}G(h_{t})/h_{t}=Z\left(\left[ A(1-\alpha)\right] ^{\frac{1}{1-\alpha }}\left(\frac{ (1-\beta)}{\beta +(1-\beta )}\left(\frac{\left[ \delta \beta +(1-\beta) \right] }{(1-\alpha )^{-1}\delta \rho }\right) ^{\frac{1}{\rho }}\right)^{ \frac{\alpha }{1-\alpha }}\left(\Uplambda \right) ^{\frac{1}{1-\alpha} }\right) ^{\psi }\left(\frac{\left(\frac{\left[ \delta \beta +(1-\beta) \right] }{(1-\alpha)^{-1}\delta \rho }\right) ^{\frac{1}{\rho }}}{\left(\frac{\beta +(1-\beta)}{\rho }\right) ^{\frac{1}{\rho }}}\right) ^{\phi} $$

so that the constraint imposed is:

$$ Z\left[ A(1-\alpha )\right] ^{\frac{\psi }{1-\alpha }}\left(1-\beta \right) ^{\frac{\alpha \psi }{1-\alpha }}\left(\frac{(1-\alpha)\left[ \delta \beta +(1-\beta)\right] }{\delta }\right) ^{\frac{\phi }{\rho }} < 1 $$

Regarding (iii), G(h _t ) can be rewritten as:

$$ G(h_{t})=Z\left[ A(1-\alpha)\right]^{\frac{\psi }{1-\alpha}}\left((1-\beta)h_{t}\right)^{\frac{\psi \alpha }{1-\alpha}}\left(\beta +(1-\beta)h_{t}\right)^{\frac{-\psi \alpha \rho-\phi(1-\alpha)}{ (1-\alpha)\rho}}\left(\frac{\left[\delta\beta+(1-\beta)h_{t}\right]}{(1-\alpha )^{-1}\delta }\right) ^{\frac{\phi }{\rho}} $$

the derivative G′(h _t ) is:

$$ \begin{array}{l} Z\left[ A(1-\alpha)\right] ^{\frac{\psi }{1-\alpha }}\frac{\psi \alpha}{ 1-\alpha }\left((1-\beta)h_{t}\right) ^{\frac{\psi \alpha }{1-\alpha} -1}(1-\beta)\left[ \left(\beta +(1-\beta)h_{t}\right) ^{\frac{-\psi \alpha \rho -\phi (1-\alpha)}{(1-\alpha)\rho }}\left(\frac{\delta \beta +(1-\beta)h_{t}}{(1-\alpha)^{-1}\delta }\right) ^{\frac{\phi }{\rho }} \right]\\ +\,Z\left[ A(1-\alpha )\right] ^{\frac{\psi }{1-\alpha }}\left((1-\beta)h_{t}\right) ^{\frac{\psi \alpha }{1-\alpha }}\left[ \frac{-\psi \alpha \rho -\phi (1-\alpha)}{(1-\alpha)\rho (1-\beta)^{-1}}\left(\beta +(1-\beta)h_{t}\right) ^{\frac{-\psi \alpha \rho -\phi (1-\alpha )}{ (1-\alpha)\rho }-1}\left(\frac{\delta \beta +(1-\beta )h_{t}}{(1-\alpha)^{-1}\delta }\right) ^{\frac{\phi }{\rho }}\right]\\ +\,Z\left[ A(1-\alpha)\right] ^{\frac{\psi }{1-\alpha }}\left((1-\beta)h_{t}\right) ^{\frac{\psi \alpha }{1-\alpha }}\left[ \left(\beta +(1-\beta)h_{t}\right) ^{\frac{-\psi \alpha \rho -\phi (1-\alpha)}{(1-\alpha)\rho }} \frac{\phi }{\rho }\left(\frac{\delta \beta +(1-\beta)h_{t}}{(1-\alpha)^{-1}\delta }\right) ^{\frac{\phi }{\rho }-1}\frac{(1-\alpha)(1-\beta)}{ \delta }\right] \end{array} $$

Thus, $\lim_{h\rightarrow 0}G^{\prime }(h)$ is:

$$ Z\left[ A(1-\alpha )\right] ^{\frac{\psi }{1-\alpha }}\frac{\psi \alpha}{ 1-\alpha }\left(0\right) ^{\frac{\psi \alpha }{1-\alpha }-1}(1-\beta )\left[ \left(\beta \right) ^{\frac{-\psi \alpha \rho -\phi (1-\alpha)}{(1-\alpha)\rho }}\left((1-\alpha)\beta \right) ^{\frac{\phi }{\rho }}\right]+0 $$

which tends to +∞ when $\frac{\alpha \varphi }{1-\alpha } < 1$ .

The three conditions garantee that the function G(h _t ) will cross the 45degree line at least once on the interval [0,1]: actually, G(0) is 0, but for higher values of h _t , the G(h _t ) curve starts above the 45 line by condition (iii), but must be lower than 1 when h _t tends to 1 by condition (ii) so that it must cross the 45line somewhere.

The case where $0 < \delta\leq\frac{(1-\alpha)(1-\beta)}{\alpha \beta+(1-\beta)}$ can be discussed by following those lines of arguments.

Actually, in that general case, (i) is still true, as:

$$ \begin{aligned} G(0)=\,&Z\left(\left[ A(1-\alpha )\right]^{\frac{1}{1-\alpha}}\left(\frac{0}{\beta}\hbox{min}\left(\left(\frac{(1-\alpha)\left[ \delta\beta\right]}{\delta\rho}\right)^{\frac{1}{\rho}},\left(\frac{\beta}{\rho}\right)^{\frac{1}{\rho }}\right)\right)^{\frac{\alpha }{1-\alpha}}\right)^{\psi}\\ &\left(\hbox{min}\left(\left(\frac{(1-\alpha)\left(\delta\beta \right) }{\delta \rho }\right) ^{\frac{-\alpha }{\rho }},\frac{ \left(\frac{\beta }{\rho }\right) ^{\frac{-\alpha }{\rho }}}{(1-\alpha)} \right) \right) ^{\frac{\psi }{1-\alpha }}\left(\hbox{min} \left(\frac{\left(\frac{(1-\alpha)\left[ \delta \beta \right] }{\delta \rho }\right) ^{\frac{1 }{\rho }}}{\left(\frac{\beta}{\rho}\right) ^{\frac{1}{\rho }}},1\right) \right) ^{\phi}=0 \end{aligned} $$

Regarding (ii), one should notice that, if δ ≤ $\frac{(1-\alpha)(1-\beta )}{\alpha \beta +(1-\beta)},$ the economy cannot remain in underemployment when h _t tends to 1. Therefore, the condition under which $\lim_{h_{t}\rightarrow 1}G(h_{t})/h_{t} < 1$ becomes:

$$ Z\left[ A(1-\alpha )\right] ^{\frac{\psi }{1-\alpha }}(1-\beta)^{\frac{ \alpha \psi }{1-\alpha }}\left(\frac{(1-\alpha)[\delta \beta +(1-\beta )]}{\delta}\right)^{\frac{-\psi }{1-\alpha }} < 1 $$

Regarding (iii), one should notice that, when h _t tends to 0, the economy must, given δ > 0, tend to excess labour supply, so that the above condition $\frac{\alpha \psi }{1-\alpha } < 1$ must remain true.

The case where δ is zero is trivial. Under δ = 0, the negotiated wage is zero, so that workers will not be able to consume nor to save anything. Hence, given the full depreciation of capital, the capital stock at the next period will be zero, so that the production can no longer take place, whatever initial conditions are.

Uniqueness If the steady-state exhibits no underemployment, G(h _t ) becomes:

$$ h_{t+1}=G(h_{t})=Z\left[ A(1-\alpha)\right] ^{\frac{\psi }{1-\alpha }}\left(\frac{\delta }{1-\alpha }\right) ^{\frac{\psi }{1-\alpha }} \left[ (1-\beta )h_{t}\right] ^{\frac{\alpha \psi }{1-\alpha }}\left(\beta +(1-\beta)h_{t}\right) ^{\psi }(\delta \beta +(1-\beta )h_{t})^{\frac{-\psi }{1-\alpha }} $$

However, it follows from this that the derivative of G(h _t ) at such a steady-state must be non-positive. Indeed, G′(h _t ) can be rewritten as the product:

$$ \begin{array}{l} \left[ Z\left[ A(1-\alpha)\right] ^{\frac{\psi }{1-\alpha}}\left(\frac{\delta }{1-\alpha }\right) ^{\frac{\psi }{1-\alpha}}\left[ (1-\beta)h_{t}\right] ^{\frac{\alpha \psi }{1-\alpha}}\left(\beta +(1-\beta)h_{t}\right) ^{\psi }(\delta \beta +(1-\beta)h_{t})^{\frac{-\psi }{1-\alpha }}(1-\beta )\right]\\ \left[ \frac{\alpha \psi }{1-\alpha }\left[ (1-\beta )h_{t}\right] ^{-1}+\psi \left(\beta +(1-\beta)h_{t}\right) ^{-1}+\left(\frac{-\psi}{ 1-\alpha }\right) (\delta \beta +(1-\beta)h_{t})^{-1}\right] \end{array} $$

whose first factor is positive, whereas the second factor is necessarily non-positive.

Actually, a non-positive second factor is equivalent to:

$$ \frac{\alpha }{1-\alpha }\frac{1}{(1-\beta)h}+\frac{1}{\beta +(1-\beta)h} \leq \frac{1}{1-\alpha }\frac{1}{\delta \beta +(1-\beta)h} $$

which, in terms of δ, is equivalent to: $\delta \leq \frac{(1-\beta)(1-\alpha)h}{\alpha \beta +h(1-\beta)}.$

But there is no underemployment here, so that $\delta \leq \frac{(1-\beta)(1-\alpha)h}{\alpha \beta +h(1-\beta)}$ is necessarily satisfied.

Hence, under that constraint, G′(h) ≤ 0. In other words, the slope of G(h) in the full employment area (i.e. on the right of the vertical dotted line) is non-positive. It follows from that inequality that the non-trivial steady-state must here be unique. Actually, given that G(h) is decreasing with h for h > h*, it is hard to see how G(h) could cross the 45° line another time on the interval [h*,1], so that the non-trivial steady-state must be unique.