Social Efficiency of Free Entry in a Vertically Related Industry with Cost and Technology Asymmetry

Proof of Lemma 1

We derive q _i and q _j with respect to m:

Recall that all firms’ outputs should be positive. Thus, from Equation (4) and q j = a − w − c e 1 + 2 d + c i − c e n 1 + 2 d 1 + 2 d + m + n > 0 . Due to 1 + 2 d 1 + 2 d + m + n > 0 , we have a − w − c e 1 + 2 d + c i − c e n > 0 and then ∂ q i ∂ m = ∂ q j ∂ m < 0 .

Proof of Lemma 2

We derive w ^*ES with respect to m:

Letting ∂ w * ∂ m = 0 , it yields c _i = c _e .

Proof of ∂ S W 1 N E S * ∂ m

where f ₃ = −12n ³ − 20n ² − 11n − 2, f ₂ = 6a − 6c + 22an − 22cn + 20an ² − 20cn ² + c _i n(11 + 40n + 36n ²), f ₁ = 12ac − 11a ² n − 11c ² n − 6a ² − 6c ² + 22acn + c _i n(−40an − 22a + 40n + 22cn − 36c _i n ² − 20c _i n), and f 0 = 2 a 3 − 6 a 2 c + 6 a c 2 − 2 c 3 + c i n 11 a 2 − 22 ac + 20 a c i n + 11 c 2 − 20 c c i n + 12 c i 2 n 2 .

Proof of Proposition 1

Firstly, we evaluate the numerator of ∂ S W 1 NES * ∂ m m = m NES * . Solving f 3 c e 3 + f 2 c e 2 + f 1 c e + f 0 = 0 with respect to c _e yields three roots as follows: c e 1 NES * = c e 2 NES * = a − c + 2 c i n 2 n + 1 , c e 3 NES * = 2 a − 2 c + 3 c i n 3 n + 2 . Given that a − c > c _i , we have c e 3 NES * > a − c + 2 c i n 2 n + 1 and then f 3 c e 3 + f 2 c e 2 + f 1 c e + f 0 > 0 .

Secondly, we evaluate the denominator of ∂ S W 1 NES * ∂ m m = m NES * Solving 2(a − c) − 2c _e (n + 1) + 2c _i n = 0 with respect to c _e yields c e 4 NES * = a − c + c i n n + 1 . Given that a − c > c _i , we have c e 4 NES * > a − c + 2 c i n 2 n + 1 and then 2(a − c) − 2c _e (n + 1) + 2c _i n > s0.

Obviously, because f 3 c e 3 + f 2 c e 2 + f 1 c e + f 0 > 0 and 2(a − c) − 2c _e (n + 1) + 2c _i n > 0, we have ∂ S W 1 NES * ∂ m m = m NES * > 0 and then entry in the final goods market is always insufficient.

Proof of ∂ S W D E S * ∂ m

where f ₃ = −20n ³ − 108n ² − 189n − 108, f ₂ = 324a − 324c + 378an − 378cn + 108an ² − 108cn ² + 189c _i n + 216c _i n ² + 60c _i n ³, f 1 = 648 a c − 189 a 2 n − 189 c 2 n + 324 a 2 − 324 c 2 + 378 a c n − 216 a c i n 2 − 378 a c i n + 216 c c i n 2 − 60 c i 2 2 n 3 − 108 c i 2 n 2 , and f 0 = 108 a 3 − 324 a 2 c + 324 a c 2 − 108 c 3 + 189 a 2 c i n − 378 ac c i n + 108 a c i 2 n 2 + 189 c 2 c i n − 108 c c i 2 n 2 + 20 c i 3 n 3 .

Proof of Proposition 2

Firstly, we evaluate the numerator of ∂ S W 1 D E S * ∂ m m = m D E S * . Solving f 3 c e 3 + f 2 c e 2 + f 1 c e + f 0 = 0 with respect to c _e yields three roots as follows: c e 1 D E S * = c e 2 D E S * = 3 a − c + 2 c i n 2 n + 3 , c e 3 D E S * = 12 a − c + 5 c i n 5 n + 12 . Given that a − c > c _i , we have c e 3 NES * > 3 a − c + 2 c i n 2 n + 3 and then f 3 c e 3 + f 2 c e 2 + f 1 c e + f 0 > 0 .

Secondly, we evaluate the denominator of ∂ S W 1 D E S * ∂ m m = m NES * Solving 3(a − c) − c _e (n + 3) + c _i n = 0 with respect to c _e yields c e 4 NES * = 3 a − c + c i n n + 3 . Given that a − c > c _i , we have c e 4 NES * > 3 a − c + 2 c i n 2 n + 3 and then 3(a − c) − c _e (n + 3) + c _i n > 0.

Obviously, because f 3 c e 3 + f 2 c e 2 + f 1 c e + f 0 > 0 and 3(a − c) − c(n + 3) + c _i n > 0, we have ∂ S W 1 D E S * ∂ m m = m D E S * > 0 and then entry in the final goods market is always insufficient.

Proof of Proposition 3

Substituting m PTA * = n a 1 + n λ − 1 − 1 − c n − 1 + n λ a n 2 − λ 1 − λ + λ 2 − c n λ − 1 + λ n 1 − λ 2 + λ 2 into ∂ S W PTA * ∂ m , we have ∂ S W PTA * ∂ m m = m PTA * = f 4 f 5 2 a λ − c n λ − 1 + λ n λ − 1 + 2 λ a − c n λ − 1 + λ , where f 4 = 1 2 n λ − 1 + λ 5 > 0 and f 5 = a n λ − 2 λ − 1 + λ 2 − c n λ − 1 + λ n λ − 1 2 + λ 2 2 > 0 . We consider when the market size is large enough, i.e. a ⇒ +∞, 2 a λ − c n λ − 1 + λ n λ − 1 + 2 λ a − c n λ − 1 + λ ⇒ 2 λ > 0 , and then entry in the final goods market is insufficient.