Scarce human resources and equilibrium industry structure

Abstract

We investigate how the initial distribution of scarce human resources determines the equilibrium industry structure. Using a two-firm competition model with resource-dependent production technology, we examine the flow of these resources in gradual trading, one-off trading, and Walrasian trading. We find that the evolution of an industry’s structure depends on the total amount of resources available relative to the market size. When resources are plentiful, a symmetric duopoly emerges through the trading of the resources. In all three models, a monopoly emerges when the resources are sufficiently scarce. This is in contrast to the results obtained by many existing models of capacity found in the literature.

Appendix: Proof of Proposition 3

Proof

We shall prove the first half of this proposition in 2 directions.

(i) If $w^{*}$, $N_{i}(w^{*})$ and $N_{j}(w^{*})$ constitute a Walrasian equilibrium, then for any $N_{i}+N_{j}=N$, we have

$$\begin{aligned} \pi _{i}(N_{i}(w^{*}),N_{j}(w^{*}))+w^{*}(N_{i}^{0}-N_{i}(w^{*}))\ge & {} \pi _{i}(N_{i},N_{j})+w^{*}(N_{i}^{0}-N_{i}), \\ \pi _{j}(N_{i}(w^{*}),N_{j}(w^{*}))+w^{*}(N_{j}^{0}-N_{j}(w^{*}))\ge & {} \pi _{j}(N_{i},N_{j})+w^{*}(N_{j}^{0}-N_{j}). \end{aligned}$$

Noting that $N_{i}^{0}+N_{j}^{0}=N$ and that $N_{i}(w^{*})+N_{j}(w^{*})=N$, the sum of the above 2 inequalities leads to

$$\begin{aligned} \pi _{i}(N_{i}(w^{*}),N_{j}(w^{*}))+\pi _{j}(N_{i}(w^{*}),N_{j}(w^{*})) \ge \pi _{i}(N_{i},N_{j})+\pi _{j}(N_{i},N_{j}), \end{aligned}$$

for any $N_{i}+N_{j}=N$. In this case, $N_{i}^{*}=N_{i}(w^{*})$ and $N_{j}^{*}=N_{j}(w^{*})$.

(ii) For any $N_{i}^{*}$ and $N_{j}^{*}$ that maximize (13), we need to show that there exists a $w^{*}$, such that $N_{i}(w^{*})=N_{i}^{*}$ and $N_{j}(w^{*})=N_{j}^{*}$.

We need to discuss 2 cases. In case 1, $N_{i}^{*}$ and $N_{j}^{*}$ form a corner solution. Without a loss of generality, let $N_{i}^{*}=N$ and $N_{j}^{*}=0$. Then we must have

$$\begin{aligned} \frac{d(\pi _{i}+\pi _{j})}{d N_{i}}= \frac{\partial \pi _{i}}{\partial N_{i}}-\frac{\partial \pi _{i}}{\partial N_{j}} + \frac{\partial \pi _{j}}{\partial N_{i}}-\frac{\partial \pi _{j}}{\partial N_{j}} =V_i-V_j \ge 0. \end{aligned}$$

Any $w^{*} \in [V_j,V_i]$ would induce the corner solution $N_{i}(w^{*})=N$ and $N_{j}(w^{*})=0$. This configuration constitutes a Walrasian equilibrium.

In case 2, $N_{i}^{*}$ and $N_{j}^{*}$ form an interior solution. In this case,

$$\begin{aligned} \frac{d(\pi _{i}+\pi _{j})}{d N_{i}} =V_i-V_j = 0. \end{aligned}$$

Let $w^{*} = V_i =V_j$. Then the first order condition (15) is satisfied for firm i and for firm j. From (6), we have

$$\begin{aligned} V_{i}= & {} \frac{\Omega }{\gamma }\left[ (1+\beta )A-(2+\beta )c_{i}+c_{j}\right] \left[ (2+\beta )c_{i}^{2}+c_{j}^{2}\right] = w^{*}, \end{aligned}$$

(16)

$$\begin{aligned} V_{j}= & {} \frac{\Omega }{\gamma }\left[ (1+\beta )A-(2+\beta )c_{j}+c_{i}\right] \left[ (2+\beta )c_{j}^{2}+c_{i}^{2}\right] = w^{*}. \end{aligned}$$

(17)

By subtracting the first equation from the second, we obtain [c.f. inequality (8)]

$$\begin{aligned} (c_{j}-c_{i})\left[ \Xi \cdot (c_{j}+c_{i})A-(c_{i}^{2}+c_{j}^{2})-\Xi \cdot c_{i}c_{j}\right] =0. \end{aligned}$$

(18)

From (18), we have 2 solutions. Solution 1 gives $c_{i}=c_{j}$. Solution 2 gives

$$\begin{aligned} \Xi \cdot (c_{j}+c_{i})A-(c_{i}^{2}+c_{j}^{2})-\Xi \cdot c_{i}c_{j}=0. \end{aligned}$$

(19)

The second order condition for maximization (13) implies that

$$\begin{aligned} dV_{i}(N_{i}^{*},N_{j}^{*})/dN_{i} +dV_{j}(N_{i}^{*},N_{j}^{*})/dN_{j}\le 0. \end{aligned}$$

Note that

$$\begin{aligned} \frac{dV_{i}(N_{i}^{*},N_{j}^{*})}{dN_{i}}= & {} \frac{\Omega }{\gamma ^{2}}\left[ (2+\beta )c_{i}^{2}+c_{j}^{2}\right] ^{2}\\&-\frac{2\Omega }{\gamma ^{2}}\left[ (2+\beta )c_{i}^{3}-c_{j}^{3}\right] \left[ (1+\beta )A-(2+\beta )c_{i}+c_{j}\right] , \\ \frac{dV_{j}(N_{i}^{*},N_{j}^{*})}{dN_{j}}= & {} \frac{\Omega }{\gamma ^{2}}\left[ (2+\beta )c_{j}^{2}+c_{i}^{2}\right] ^{2}\\&-\frac{2\Omega }{\gamma ^{2}}\left[ (2+\beta )c_{j}^{3}-c_{i}^{3}\right] \left[ (1+\beta )A-(2+\beta )c_{j}+c_{i}\right] . \end{aligned}$$

Combining the above 2 equations, we obtain

$$\begin{aligned}&\frac{dV_{i}(N_{i}^{*},N_{j}^{*})}{dN_{i}} +\frac{dV_{j}(N_{i}^{*},N_{j}^{*})}{dN_{j}} \nonumber \\&\quad = -2c_{i}c_{j}\left[ \Xi \cdot (c_{j}+c_{i})A-(c_{i}^{2}+c_{j}^{2})-\Xi \cdot c_{i}c_{j}\right] +(c_{i}-c_{j})^{2}(c_{i}+c_{j})^{2} \nonumber \\&\quad \le 0, \end{aligned}$$

(20)

as required by the second order condition of the joint profit maximization.

Solution 2 [Eq. (19) above] from the first order condition obviously does not satisfy this second order condition. Solution 1 (i.e., $c_{i}=c_{j}$) does satisfy this second order condition. This is because at $c_{i}=c_{j}$, ${dV_{i}(N_{i}^{*},N_{j}^{*})}/{dN_{i}}= {dV_{j}(N_{i}^{*},N_{j}^{*})}/{dN_{j}}$, and therefore, (20) implies that both ${dV_{i}}/{dN_{i}}$ and ${dV_{j}}/{dN_{j}}$ are non-positive. This concludes the proof for the first half of this proposition. The second half of the proposition is immediate from the proof. $\square$