Product Differentiation and Trade

Appendix A: The Optimal Problem of Consumers

From Eq. (2), the first-order conditions for the representative consumer at home is given by:

Integrating Eq. (A1) with respect to i for each given z and rearranging the equation, we obtain:

Combining Eqs. (A1) and (A2), we have:

Then, integrating Eq. (A3) with respect to i and z and rearranging the equation, we have:

where ∫ 0 1 ∫ 0 N p i y i d i d z is exactly equal to income I.

Thus, the marginal utility of income in the home country can be rewritten as:

where μ 1 P and μ 2 P are given by:

Appendix B: Equilibrium Conditions of Production

In symmetric equilibrium, the first-order conditions for domestic firms (i.e. Eqs. (10) and (12)) can be rewritten as:

Integrating Eq. (B1) with respect to i in 0 , δ j , we obtain:

With δ _j = δ and ∫ 0 δ j c j i d i = c z δ j + 1 2 γ δ j 2 , Eq. (B3) can be rewritten as:

Then, with c j δ j = c z − w δ j and Eq. (B3), we have:

Combining Eqs. (B4) and (B5), we obtain:

Similarly, for the foreign country, we have:

Furthermore, combining Eqs. (B5)–(B8) and eliminating X and X*, we obtain:

Appendix C: Impacts of Product Differentiation on Extensive Margins

In industry z ̃ ,with X = 0 and δ = 0, Eqs. (B9) and (B10) can be rewritten as:

Differentiating Eqs. (C1) and (C2) with respect to δ*, z ̃ , and e and solving the equations, we obtain:

Symmetrically, in industry z ̃ * , we have:

Appendix D: Impacts of Product Differentiation on the Scopes of Variety

For z ∈ z ̃ * , z ̃ , differentiating Eqs. (B9) and (B10) with respect to δ, δ*, and e, and solving the equations, we obtain:

where Δ = 4 1 − e 2 + 2 1 − e e 1 + n δ + 1 + n * δ * + e 2 1 + n + n * δ δ * .

For z ∈ 0 , z ̃ * ∪ z ̃ , 1 , although the first-order conditions are not differentiable at z ̃ and z ̃ * , Eqs. (B9) and (B10) are continuous at z ̃ and z ̃ * . Thus, ∂ δ ∂ e and ∂ δ * ∂ e have the same sign as Eqs. (D1) and (D2). Thus, we know that ∂ δ ∂ e < 0 and ∂ δ * ∂ e < 0 in all industries.

Appendix E: Impacts of Product Differentiation on the Wage in the General Equilibrium

Totally differentiating Eqs. (15) and (17), we have:

Denote x i as d x i = A 1 d e + A 2 d w , where A ₁ and A ₂ will be calculated later. As dL = 0, combining Eqs. (E1) and (E2), we have:

Then, we have:

Furthermore, we calculate the numerator and the denominator of Eq. (E4) in general equilibrium. In symmetric equilibrium, the equilibrium scopes and aggregate outputs of firms are determined by:

Totally differentiating Eqs. (E5)E8)–(E8) and simplifying the equations, we obtain:

Solving Eq. (E9), we have:

where Δ 0 = b ̄ 2 e 2 δ δ * 2 n + 1 + 2 1 − e e δ + δ * n + 1 + 4 1 − e 2 > 0 , B 1 = b ̄ e δ δ * β + 1 2 δ n + 1 − n β * + 1 2 δ * + 2 b ̄ δ β + 1 2 δ 2 1 − e , B 2 = b ̄ 2 δ n + 1 − 2 X + δ n X * e δ * n + 1 + 2 1 − e − b ̄ 2 e δ n δ * n X + δ * n + 1 − 2 X * , B 3 = b ̄ e δ n + 1 + 2 1 − e δ * β * + 1 2 ( δ * ) 2 − b ̄ e δ * n δ β + 1 2 δ 2 , and B 4 = b ̄ 2 e δ n + 1 + 2 1 − e δ * n X + δ * n * + 1 − 2 X * − b ̄ 2 e δ * n δ n + 1 − 2 X + δ n X * .

Recall that the output of a single product is given by:

Totally differentiating Eq. (E12), we have:

Combining Eqs. (E10), (E11), and (E13), we obtain:

Combining Eqs. (E10), (E11) and (E14), we have:

where A 3 = 2 β δ + δ 2 2 2 1 − e X + n X + n X * + e 1 + 2 n X δ * Δ / b ̄ 2 .

Then, we have:

Since ∫ 0 δ z i − δ / 2 d i = 0 and x i is decreasing in i, Eq. (E17) is negative unambiguously. A ₃ − X/δ > 0 is a sufficient condition for ∫ z ̃ * z ̃ ∫ 0 δ z β z + i A 1 d i d z < 0 .

Combining Eqs. (E5) and (E6) and denoting N ₀ = N/n = δ + δ*, we have:

From Eq. (E18), it is clear that A ₃ − X/δ > 0 holds if N ₀ is large enough. N ₀ is a measure of the aggregate scope of variety in an industry. Therefore, if the aggregate scope of variety is large enough in all industries, ∫ 0 z ̃ ∫ 0 δ z β z + i A 1 d i d z < 0 holds.

Furthermore, from Eqs. (E7) and (E8), a ̄ is essential in the determination of the scales of δ and δ*. If a ̄ tends to be zero, δ and δ* tend to be zero, too. If a ̄ increases, the scales of δ and δ* also increase with X and X*.

However, ∫ 0 z ̃ ∫ 0 δ z β z + i A 2 d i d z is always negative. We calculate the interval z ̃ * , z ̃ first. From Eq. (E15), we have:

Combining Eqs. (E7) and (E8), we have:

In symmetric equilibrium, we have:

As β < β* and δ > δ* hold when z ∈ z ̃ * , z ̃ , with Eq. (E21), we obtain:

Combining Eqs. (E20) and (E22), we can find that ∫ z ̃ * z ̃ ∫ 0 δ z β z + i A 2 d i d z < 0 is proved.

Then, in the interval 0 , z ̃ * , we have δ* = 0 and X* = 0. Thus, we have:

Thus, ∫ 0 z ̃ ∫ 0 δ z β z + i A 2 d i d z < 0 always holds. Therefore, with Eq. (E4), it is clear that d w d e < 0 if and only if ∫ 0 z ̃ ∫ 0 δ z β z + i A 1 d i d z < 0 . Proposition 4 is proved.

Appendix F: The Case with Single-Product Firms

For home and foreign single-product firms, the first-order conditions can be obtained from Eqs. (10) and (13) immediately:

Solving Eqs. (F1) and (F2), we obtain the equilibrium outputs:

Furthermore, differentiating s with respect to e and w, we have:

In general equilibrium, the market-clearing condition in the home labor market is given by:

Similar to Appendix E, from Eq. (F7), the impact of product differentiation on the wage is determined by:

We consider the symmetric case where n = n*, L = L*, and w = w* hold. In the interval z ∈ 0 , z ̃ * , as the outputs of foreign firms are zero, it is equivalent to the case where n* = 0. Thus, Eqs. (F5) and (F6) are negative unambiguously in the interval z ∈ 0 , z ̃ * . In the interval z ∈ z ̃ * , z ̃ , we have:

Thus, in the interval z ∈ z ̃ * , z ̃ Eqs. (F5) and (F6) are both negative. Then, it is clear that Eq. (F8) is negative unambiguously in symmetric equilibrium. In the case with single-product firms, an increment of product differentiation increases the wage.