Tasks, technology, and factor prices in the neoclassical production sector

Abstract

This paper introduces tasks into the neoclassical production sector. Competitive firms choose the profit-maximizing amounts of factor-specific tasks that determine their factor demands and output supplies. We show that the effect of factor-augmenting technical change on relative and absolute factor prices can be decomposed into a productivity effect and a task-demand effect of opposite sign. These effects appear since the novel task-based approach distinguishes between the demands for tasks and the demands for factors. This perspective provides a new intuition for the emergence of relative and absolute factor biases and the role of the elasticity of substitution.

Appendix: Proofs

1.1 Proof of Proposition 1

Consider firm i’s demand for capital. With (2.5), we have $K(i)=M(i)/b$. Hence, for $R>0$, (2.6) delivers $\int _0^1 K(i)di=\int _0^1 M(i)/b=K$. Hence, $\int _0^1 M(i)di=bK=M^*$. Similarly, one derives for labor and $w>0$ that $\int _0^1 N(i)di=aL=N^*$. Next consider the first-order conditions (3.2). Since the derivatives $\partial F\left( M(i),N(i)\right) /\partial M(i)$ and $\partial F\left( M(i),N(i)\right) /\partial N(i)$ are homogeneous of degree zero, these conditions imply $M(i)/N(i)=M(i')/N(i')$ for all $i,i'\in [0,1]$ and $i\ne i'$. Therefore, it holds that $M(i)/N(i)=\left[ M(i)+M(i')\right] /\left[ N(i)+N(i')\right]$ and

$$\begin{aligned} \left( \frac{M(i)}{N(i)}\right) ^*=\frac{\int _0^1 M(i)di}{\int _0^1 N(i)di}=\frac{bK}{aL}=\frac{M^*}{N^*}. \end{aligned}$$

(6.1)

Accordingly, the equilibrium task intensity, the aggregate equilibrium output, and the equilibrium factor prices are as stated in Proposition 1. Since $M^*$ and $N^*$ are positive and finite, we have $R^*>0$ and $w^*>0$. Since $\pi (i)$ is homogenous of degree one in $\left( M(i),N(i)\right)$, the absolute levels $M^*(i)$ and $N(i)^*$ remain indeterminate. $\square$

1.2 Proof of Proposition 2

In equilibrium (3.6) becomes

$$\begin{aligned} \ln \left( \frac{R^*}{w^*}\right) =\ln \gamma +\ln \left( \frac{b}{a}\right) - \frac{1}{\sigma }\ln \left( \frac{M^*}{N^*}\right) . \end{aligned}$$

In light of (3.8) the proposition follows. $\square$

1.3 Proof of Proposition 3

We start proving that $M(i)=M(j)=M^*$ and $N(i)=N(j)=N^*$ for all $i,j\in [0,1]$ and $i\ne j$ holds in equilibrium. Since $C(i)=C\left( M(i),N(i)\right)$ is strictly convex, profits $\pi (i)=\pi \left( M(i),N(i)\right)$ are strictly concave. Assume to the contrary that two firms i and j choose $\left( M(i),N(i)\right) \ne \left( M(j),N(j)\right)$. This is only possible if $\pi \left( M(i),N(i)\right) =\pi \left( M(j),N(j)\right)$. Now, consider $\left( M,N\right) =\mu \left( M(i),N(i)\right) +(1-\mu )\left( M(j),N(j)\right)$ for some $\mu \in (0,1)$. Since profits are strictly concave we have $\pi \left( M,N\right) >\mu \pi \left( M(i),N(i)\right) +(1-\mu )\pi \left( M(j),N(j)\right)$. Therefore, $\pi \left( M,N\right) >\pi \left( M(i),N(i)\right) =\pi \left( M(j),N(j)\right)$ which is a contradiction. Hence, for all $i\in [0,1]$ we must have $\left( M(i),N(i)\right) =\left( M,N\right)$.

As a consequence, the aggregate demand for capital of all firms is $\int _0^M e^{\delta m}dm/b=\left( e^{\delta M}-1\right) /(\delta b)$. Hence, for $R>0$, (2.6) delivers $M^*$ as indicated in the proposition. Similarly, one derives $N^*$. The first-order conditions (4.2) deliver $R^*$ and $w^*$ as indicated. Since $F(\cdot )$ has constant returns to scale, $F\left( M^*,N^*\right) =F\left( \ln (1+\delta bK),\ln (1+\delta a L)\right) /\delta$. $\square$

1.4 Proof of Proposition 4

The statements of Eq. (4.6) are immediate from (4.5). Computing the respective effects delivers with $z\equiv \delta b K$

$$\begin{aligned} \delta M^* \frac{\partial \ln M^*}{\partial \ln b}=\frac{z}{1+z}\quad \text{ and }\quad \frac{1}{\sigma }\frac{\partial \ln M^*}{\partial \ln b}=\frac{1}{\sigma }\left( \frac{z}{1+z}\right) \left( \frac{1}{\ln (1+z)}\right) . \end{aligned}$$

(6.2)

It follows that ${\bar{\sigma }}_b=z/(\ln (1+z))>1$ which is increasing in z, hence, also in $\delta$. As to labor, let $x\equiv \delta a L$. Then,

$$\begin{aligned} \delta N^* \frac{\partial \ln N^*}{\partial \ln a}=\frac{x}{1+x}\quad \text{ and }\quad \frac{1}{\sigma }\frac{\partial \ln N^*}{\partial \ln a}=\frac{1}{\sigma }\left( \frac{x}{1+x}\right) \left( \frac{1}{\ln (1+x)}\right) . \end{aligned}$$

(6.3)

It follows that ${\bar{\sigma }}_a=x/(\ln (1+x))>1$ which is increasing in x, hence, also in $\delta$. $\square$