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The result has several avenues for further research. First, there may exist other inter-mediaries. Competition between intermediaries may occur and can be further studied.

Second, in our current study, the intermediary cannot influence the demand of customers.

In reality, intermediary might provide mechanism that affect customers’ valuation on the product such as advertisement, data technology, better services, etc. It is interesting how these options may influence the decisions of the players in the environment.

To highlight the impact of industry structure and product similarity on the pricing decisions, we omit demand uncertainty in our model; otherwise, the intermediary would have a strong incentive favoring the platform model. We also assume that there is no

information asymmetry in the market. If the product quality is hidden to consumers, quality risk may force the intermediary to choose the merchant model. Nevertheless, it would contribute more to the literature if we further study the joint impact of these factors.

Appendix A

Proofs of lemmas and propositions

Proof of Lemma 1. We first verify the concavity of the objective function. Since

O2πIW W(p1, p2) =

, and its leading principals are −2 < 0 and will be necessary and sufficient for an optimal solution. Let’s omit the constraints for a while. The first-order condition gives us δπδpW WI

1 = (1 − 2p1+ bp2 + w1) + b(p2− w2) = 0, i.e., p1 = 1+2bp2+w2 1−bw2. Similarly, we have p2 = 1+2bp1+w2 1−bw2. Solving the two equations leads to p1 = 1+(1−b)w2(1−b) 1 and p2 = 1+(1−b)w2(1−b) 2.

Based on the response of the intermediary, manufacturer i maximizes his profit πiW W(wi) = (wi − ci)1−wi+bw2 3−i, i = 1, 2. As πW Wi (wi) is concave (d2πW Widw2(wi)

i

= −1 < 0), the first-order condition requires the optimal wholesale price wito satisfy wi = 1+bw3−i2 +ci, i = 1, 2, where c1 = c and c2 = 0. Solving the two equations for the two manufacturers results in w1 = (2+b)+2c4−b21+bc2 and w2 = (2+b)+2c4−b22+bc1. By plugging in w1 and w2 to the expressions of pi, πiW W, and πW WI , we may obtain the equilibrium retail prices and profits

given in the lemma. applying the first-order condition dpRRi

i = 0 for manufacturer i and yield pi = 1+bp3−i2+ci+r. Solving the two equations results in pi(r) = (2+b)(1+r)+2ci+bc3−i

(4−b2) , which is the optimal price responding to r. Note that whatever r is, p1(r) ≥ p2(r) owing to the fact that c1 ≥ c2.

Proof of Lemma 3. While our main focus is the intermediary’s profit, it can be obtained as a function of r by plugging in pi into her profit function. Afterwards, the in-termediary decides r to maximize her profit πIRR(r) = maxr 2−br (2 − 2(1 − b)r − (1 − b)c). there are two possible equilibrium profit functions in the platform model. That is,

πIRR =

where the former is the first-order condition (FOC) case and the latter is the boundary

case. We would discuss the two cases separately.

Taking account of the FOC case first, after tedious calculations we get πRRI − πIW W =

G

−16 < 0 ). Second, we can prove that the first-order condition point of T , ˙c, re-sides at the left side of ¯c, the point that distinguishes the FOC case and boundary case. More precisely, ¯c − ˙c = −3b2(2+b)2+b+6(4b3−14b2−2b+20)(2+b)

−U = −U (−3b(2+b)H2+b+6) ≥ 0, where H = −4b5+ 2b4+ 28b3− 26b2− 24b + 24 = (b − 1)(b + 1)(b − 2)(−4b2− 6b + 12) ≥ 0.

Proof of Proposition 2. To compare the system profits of the two models, we have to derive the system profit of each model first. For the merchant model part, according to Lemma 1, the system profit of the merchant model is

πW Ws = πIW W + π1W W + π2W W

can derive

Now we are ready to compare the system profits.

πsRR− πW Ws = (1−b)(8+16b−6b2−10b3+4b4)(c21+c22)−4(1−b)(4+8b−3b2−5b3+2b4)c1c2

Proof of Lemma 4. Applying backward induction, we start from the optimal prices.

First, since πIW R and πW R2 are concave ( d2dpπIW R2

1 = d2dpπW R22

2 = −2 ), we get the optimal pi by applying the first-order condition dpIRR

1 = 0 and dp2RR

2 = 0, and yield p1 = 1+bp2+w2 1+br and p2 = 1+bp12+c2+r. Solving the two equations results in p1 = (2+b)+3br+2w1+bc2

4−b2 and

p2 = (2+b)(3−b2)+(2+b4−b22)r+bw1+2c2.

Considering the interaction between manufacturer 2 and the intermediary, manufac-turer 1 sets w1 to maximize his profit π1W R = (w1 − c1)(2+b)+b(b2−1)r+(b4−b2 2−2)w1+bc2. Due to the concavity of the profit function (d2dwπW R12

1

= −2(2−b4−b22) < 0 as b ∈ [0, 1)), he opti-mally sets the wholesale price by applying the first-order condition W R1 = 0 and derives

w1 = (2+b)−b(1−b2)r+(2−b2)c1+bc2

Proof of Lemma 5. Following Lemma 4, we first check the concavity of πIW R(r).

Since d2πdrW RI 2(r) = −2b24(4−b(1−b22)(7−b)2 2) + (1−b2(4−b2)(−16+6b2)(2−b2)2) < 0, πIW R(r) is concave in r. There-fore, the first-order condition IW Rdr(r) = 0 results to the optimal transaction fee r =

2(2+b)(8−3b−3b2+b3)−2b(1−b)(2−b2)c1+2(1−b)(−16+9b2−b4)c2) (64−26b2−3b4+b6)(1−b) .

We have to verify the non-negative demand constraints in the mixed model (WR).

Starting from the demand of product 1, since DW R1 = (2+b)B0+(2−b2(4−b22)C)A00c1+bD0c2 where A0 = DW R2 (r) > 0, r is feasible considering the non-negative demand constraints.

Now we are able to compare pW R1 and pW R2 . First, since pW R1 −pW R2 = (2+b)B+(2−b2)Cc1+Dc2 optimal pi by applying the first-order condition dpRRi

i = 0 for manufacturer i and yield pi = (1−φ)(1+bp2(1−φ)3−i)+ci, where c1 = c and c2 = 0. Solving the two equations results to pi = (2+b)(1−φ)+2ci+bc3−i

(4−b2)(1−φ) . The equilibrium demands and the intermediary’s profit as a function of φ can then be obtained by plugging in pi into their formulas.

Proof of Lemma 7. Through several steps of arithmetic, we obtain h(φ) = −2(2 + b)2φ3+ 6(2 + b)2φ2+ (−6(2 + b)2− (2 + b)2bc + (3b2− 4)c2)φ + (2(2 + b)2+ (2 + b)2bc + monotonically decreases as φ goes up, and eventually reaches h(1) < 0. In other words, ΠRRI (φ) is quasi-concave. Therefore, there exists a unique first-order solution ˜φ ∈ (0, 1) that satisfies RRI(φ) = 0.

Now we take the constraints into consideration. First, D2 = (2+b)(1−φ)+bc

(4−b2)(1−φ) ≥ 0. How-ever, D1 = (2+b)(1−φ)+(−2+b2)c

(4−b2)(1−φ) ≥ 0 if c ≤ (1−φ)(2+b)(2−b2 , i.e., φ ≤ ˆφ = 1 − (2−b2+b2)c. Therefore, if

φ ≤ ˆ˜ φ, ˜φ is optimal; if not, then ˆφ is optimal.

Proof of Proposition 3. When c = 0 and b = 0, ˜φ = maxφφ2 = 1 = ˆφ. Therefore, we have φ = 2 and ΠRRI = 12. By plugging c = 0 and b = 0 into Lemma 1, we derive πIW W = 18. The facts that ΠRRI |b=c=0 > πIW W|b=c=0 and both profit functions are continuous result in our first conclusion regarding the existence of ˆb1 ∈ (0, 1) and ˆ

c1 ∈ (0, 1). On the contrary, when c and b both approach 1, ˆφ approaches 23 and ˜φ approaches to a value around 0.81, which is greater than ˆφ. Therefore, φ = ˆφ = 23 and ΠRRI = 83. In this case, πIW W approaches infinity. Again, because all functions are continuous, we obtain our second conclusion regarding the existence of ˆb2 ∈ (0, 1) and ˆ

c2 ∈ (0, 1).

Proof of Proposition 4. Consider the case that c = 45. Through several steps of arith-metic, we obtain h( ˆφ)|c=4 models can be conducted by investigating the sign of

g(b) = ΠRRI)|c=4 proposition then follows due to the continuity of all profit functions.

Proof of Lemma 8. Applying backward induction, we start from the optimal prices.

First, since πiERR is concave ( d2πdpiERR2 i

= −2 ) where i = 1, 2, we get the optimal pi by applying the first-order condition dpRRi

i = 0 and yield pi = 1+bp3−i2+ci+ri. Solving the two equations results in pi(r1, r2) = (2+b)+2(ri+c4−bi)+b(r2 3−i+c3−i), which is the optimal price responding to r.

Considering the interaction between manufacturers, the intermediary sets r1 and r2

to maximize his profit

πIERR(r1, r2) = maxr1,r2r1D1+ r2D2

= maxr1,r2(r1+ r2) − (r1− br2)p1(r) − (r2− br2)p2(r)

= maxr1,r2 ((2+b)+bc2−(2−b2)(c1+r1))r1+((2+b)+bc4−b2 1−(2−b2)(c2+r2))r2+2br1r2. We first check the concavity of the profit function, since

O2πIERR(r1, r2) = 4−b12

given b ∈ [0, 1), πERRI (r1, r2) is negative semi-definite. Therefore, the intermediary op-timally sets the transaction fees by applying the first-order condition. Let’s omit the constraints for a while. The first-order condition gives us δπδrIERR

1 = 0, which implies that r2 = −(2+b)+bc2−(2−b2)(c1+2r1). Similarly, we have r1 = −(2+b)+bc2−(2−b2)(c1+2r2). Solving the

two equations leads to r1 = 2(1−b)112c1 and r2 = 2(1−b)112c2. Furthermore, we can derive

Proof of Proposition 5. We may show that the intermediary can well simulate the product prices in the mixed model by operating the transaction fees in the enhanced plat-form model. According to Lemma 4, we have pW R1 (r) = (2+b)(3−b2)+b(5−2b(4−b22)r+(2−b)(2−b2)2)c1+b(3−b2)c2

and pW R2 (r) = (2+b)(4+b−2b2)+(8−b2−b4)r+b(2−b2)c1+(8−3b2)c2

2(4−b2)(2−b2) . In contrast, according to Lemma 8, we have pERRi (r1, r2) = (2+b)+2(ri+c4−bi)+b(r2 3−i+c3−i), where i = 1, 2. Solving the two equa-tions pERR1 = pW R1 and pERR2 = pW R2 leads to r1 = (2+b)+(3−b2(2−b2)br−(2−b2) 2)c1+bc2 and r2 = r.

In addition, we have to prove that the profit of the intermediary is higher in the enhanced platform model than that in the mixed model. Since the product prices are the same, the demands of both products are also the same. Consequently, we only have to compare the margin profit of each products. For the product 2 part, the mar-gin profits are the same in both models, i.e., r2 = r. For the product 1, the margin profit is r1 in the enhanced platform model and the margin profit is pW R1 − w1W R in the mixed model. The difference between the margin profits in two models is D(r) = r1 − (pW R1 − w1W R) = (2+b)+(3−b2(2−b2)br−(2−b2) 2)c1+bc2 − ((2+b)(3−b2)+b(5−2b(4−b22)r+(2−b)(2−b2)2)c1+b(3−b2)c2

(2+b)−(1−b2)br+(2−b2)c1+bc2

2 ) = (2+b)−b(1−b2)r−(2−b2)c1+bc2

2 2 .

It can be observed that D(r) is decreasing in r. Therefore, if we can find a r0 greater than the optimal transaction fee rW R such that D(r0) ≥ 0, we show that D(rW R) ≥ 0.

By plugging r0 = 2(2+b)(8−3b−3b2+b3)

(1−b)(62−29b)2(2+b)(8−3b−3b2+b3)−2b(1−b)(2−b2)c1+2(1−b)(−16+9b2−b4)c2

(64−26b2−3b4+b6)(1−b) = rW R into D(r), we have D(r0) = (2+b)(64−45b−10b2+12b3+4b4−2b5)+(64−29b)(bc2−(2−b2)c1)

(4−b2)(2−b2)(64−29b) . It can be prove that D(r0) ≥ 0, since D(r0) is increasing in c2 but decreasing in c1, and D(r0)|c1=1,c2=0= b(32−b−15b(4−b2)(2−b22+20b)(64−29b)3−4b5) ≥ 0 for all b ∈ [0, 1), c1 ∈ [0, 1] and c2 ∈ [0, 1].

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