Second-Degree Price Discrimination on Two-Sided Markets

Appendix: Proofs

Proof of Lemma 1:

The proof can be divided into two steps. At first, we show that λ₁=1, λ₂=0, and λ₃=(1−μ₁). Here, we know that the relevant Kuhn-Tucker conditions are given by

λ1[θ1Lu1(q1L)+α1LQ2−t1L]=0, λ1, [θ1Lu1(q1L)+α1LQ2−t1L]≥0,λ2[θ1Lu1(q1L)−t1L−θ1Lu1(q1H)+t1H]=0, λ2, [θ1Lu1(q1L)−t1L−θ1Lu1(q1H)+t1H]≥0,λ3[θ1Hu1(q1H)−t1H−θ1Hu1(q1L)+t1L]=0, λ3, [θ1Hu1(q1H)−t1H−θ1Hu1(q1L)+t1L]≥0.

Respecting these conditions, we are now checking for potential solutions. First, suppose that λ₁=0 and solve (11) for λ₃, which yields λ₃=(1−μ₁)+λ₂. Then, we find that (10) becomes μ₁−λ₂+(1−μ₁)+λ₂=0⇔1=0, which produces a contradiction. Hence, we must have λ_1>0. Now suppose that λ₃=0. In this case, (11) becomes (1−μ₁)+λ₂=0, which also yields a contradiction, because λ₂≥0 and 1−μ₁>0. Therefore, it must be that λ₃>0. Finally, suppose that λ₁>0, λ₂>0, and λ₃>0. This implies that IC1L=0 as well as IC1H=0. Solving IC1L=0 for t1L, plugging the result into IC1H=0, and rearranging terms gives (θ1H−θ1L)(u1(q1H)−u1(q1L))=0. Since θ1H−θ1L>0, this only holds for q1L=q1H, which is not a profit-maximizing solution. Hence, we must have that λ₂=0, which easily yields λ₁=1 and λ₃=(1−μ₁). Showing that λ₄=1, λ₅=0 as well as λ₆=(1−μ₂) follows the same approach.

Proof of Proposition 1:

In a first step, we can show that

Then, we use (16) and (30) and compare Equations (2) and (14) to find that qiL∗∗<qiL∗. Using (3), (15), and (30), we can show that qiH∗∗<qiH∗.

(q.e.d.)

Proof of Proposition 2:

Suppose αik>0. As we know that θiH>θiL,αiH>αiL, and q˜iH∗>q˜iL∗, we can immediately conclude from (20) that

t˜iH∗−t˜iL∗=θiHui(q˜iH∗)−θiLui(q˜iL∗)︸>0+Qj(q˜iH∗αiH−q˜iL∗αiL)︸>0>0.

For αik=0, (20) becomes

t˜iH∗−t˜iL∗=θiHu(q˜iH∗)−θiHu(q˜iH∗)︸>0>0,

which closes the proof for the first part of Proposition 2.

In case of a negative indirect network effect, i.e. for αik<0, we find by using (20) that

t˜iH∗−t˜iL∗=θiHu(q˜iH∗)−θiLu(q˜iL∗)︸>0+Qj︸>0(q˜iH∗αiH−q˜iL∗αiL)︸<0 or ≥0,

which obviously implies that t˜iH∗−t˜iL∗ can either be positive, negative or equal to zero. In order to prove that t˜iH∗−t˜iL∗<0 is feasible in the optimum, we have to show that

θiHu(q˜iH∗)−θiLu(q˜iL∗)︸>0+Qj︸>0(q˜iH∗αiH−q˜iL∗αiL)︸<0 or ≥0<0

is, at least for some parameter constellations, in line with the first-order conditions. Suppose that θiH≈θiL and αiH≈αiL. Then we have in the optimum that q˜iL∗=q˜iH∗−ε, where ε>0 is assumed to be very small. In this case, it approximately holds that

θiHui(q˜iH*)−θiLui(q˜iL*)=θiH∂ui∂qiH|q˜iH∗|q˜iL∗−q˜iH∗|+θiL∂ui∂qiL|q˜iL∗(q˜iH∗−q˜iL∗) =θiH∂ui∂qiH|q˜iH∗|−ε|+θiL∂ui∂qiL|q˜iL∗ε=ε(θiH∂ui∂qiH|q˜iH∗+θiL∂ui∂qiL|q˜iL∗).

Since an optimal solution requires

θiH∂ui∂qiH|q˜iH∗+αiHQj=θiL∂ui∂qiL|q˜iL∗+αiLQj   ⇔   θiL∂ui∂qiL|q˜iL∗=θiH∂ui∂qiH|q˜iH∗+Qj(αiH−αiL),

we find by using (20) that t˜iH∗−t˜iL∗ can be expressed as

t˜iH∗−t˜iL∗=ε(2θiH∂ui∂qiH|q˜iH∗+Qj(αiH−αiL))+Qj(q˜iH∗αiH−q˜iL∗αiL).

Since αiH≈αiL and q˜iL∗=q˜iH∗−ε, we know that

q˜iH∗αiH−q˜iL∗αiL≈(q˜iH∗−q˜iL∗)(αiH+αiL)=ε(αiH+αiL),

so that we can write (20) as

t˜iH∗−t˜iL∗=ε(2θiH∂ui∂qiH|q˜iH∗+Qj(αiH−αiL))+Qjε(αiH+αiL)=ε(2θiH∂ui∂qiH|q˜iH∗+2QjαiH).

Hence, we have that

t˜iH∗−t˜iL∗<0   ⇔   θiH∂ui∂qiH|q˜iH∗+QjαiH<0   ⇔   θiH∂ui∂qiH|q˜iH∗<−QjαiH.

By analyzing the first-order conditions, in particular Equation (19), we can conclude that this inequality is satisfied in the optimum, if and only if it is true that

μjqjLαjL+(1−μj)qjHαjH−ci>0,

which only holds for αjk>0.

(q.e.d.)

Proof of Proposition 3:

First, we consider the case of αik>0. Then, respecting q˜iH∗∗>q˜iL∗∗, we find by using Equation (29) that

t˜iH∗∗−t˜iL∗∗=θiHui(q˜iH∗∗)−θiHui(q˜iL∗∗)︸>0+αiH︸>0Qj︸>0(q˜iH∗∗−q˜iL∗∗)︸>0>0.

For αik=0, Equation (29) simplifies to

t˜iH∗∗−t˜iL∗∗=θiHui(q˜iH∗∗)−θiHui(q˜iL∗∗)︸>0>0,

which proves Proposition 3’s first statement.

Now, suppose that αik<0. In this case, we find from (29) that

t˜iH∗∗−t˜iL∗∗=θiHui(q˜iH∗∗)−θiHui(q˜iL∗∗)︸>0+αiH︸<0Qj︸>0(q˜iH∗∗−q˜iL∗∗)︸>0,

yielding ambiguous results, i.e. t˜iH∗∗−t˜iL∗∗ can have any sign or be equal to zero. Showing that t˜iH∗∗−t˜iL∗∗<0 is feasible in the optimum, requires to verify that

θiHui(q˜iH∗∗)−θiHui(q˜iL∗∗)︸>0+αiH︸<0Qj︸>0(q˜iH∗∗−q˜iL∗∗)︸>0<0

is covered by the first-order conditions. We start the proof by assuming that θiH≈θiL. Hence, we have in the optimum that q˜iL∗∗=q˜iH∗∗−ε with ε>0 being very small. By approximation, it holds that

θiHui(q˜iH∗∗)−θiHui(q˜iL∗∗)=θiH∂u∂qiH|q˜iH∗∗(q˜iH∗∗−q˜iL∗∗)=θiH∂u∂qiH|q˜iH∗∗ε.

Then, using (29) we find that t˜iH**−t˜iL** can be described by

t˜iH∗∗−t˜iL∗∗=θiH∂u∂qiH|q˜1H∗∗ε+αiHQjε=ε(θiH∂u∂qiH|q˜1H∗∗+αiHQj),

which allows us to conclude that

t˜iH∗∗−t˜iL∗∗<0   ⇔   θiH∂u∂qiH|q˜1H∗∗+αiHQj<0   ⇔   θiH∂u∂qiH|q˜1H∗∗<−αiHQj.

By analyzing (28) we find that this inequality is satisfied in the optimum if it holds that

μjqjLαjL+(1−μj)(qjLαjL+(qjH−qjL)αjH)−ci>0,

which requires positive indirect network effects on market side j, i.e. αjk>0.

(q.e.d.)

Proof of Proposition 4:

First, we can verify that

because αjH>αjL. Then, respecting θiH>θiL,αiH>αiL, and (31), we can show by comparing Equations (18) and (27) that q˜iL∗∗<q˜iL∗. Using (31), we compare (19) and (28), which yields q˜iH∗∗<q˜iH∗.

(q.e.d.)