Mixed Bundling and Mergers

Proof of Result 4

For a fixed vector of prices, p 1 A , p 2 A , P A , p 1 B , p 2 B , P B , assuming all consumers buy all bundles, the market is partitioned as given in Lemma 1. The uniform distribution then gives the measures of the four market segments as the area of the sets AB, AA, BB, BA:

These definitions yield the following:

Using the fact that

we can also write

Firm A’s profit function is

Thus the first order condition for p 1 A is

A symmetric condition holds for p 2 A . The first order condition for P ^A is

Assuming a symmetric solution, p j A = p j B , P A = P B , x ̲ j = 1 − x ̄ j , μ(AB) = μ(BA) and Δ = p 1 A + p 2 A − P A . Summing the first order conditions for p 1 A , p 2 A then eliminates the p j A s yielding an expression only in terms of Δ:

This is a convex quadratic in Δ (and therefore p j A ) so necessary second order conditions imply selecting the smallest of the roots. Substituting into (10) for p 1 A and (11) for P ^A , yields necessary conditions for a symmetric equilibrium. If t₁ = t₂ = t, then the solution for Δ = t/2. This then gives p j = 11 t / 12 + c j , P = 8 t / 6 + c 1 + c 2 , x ̲ j = 1 − x ̄ j ′ = 1 / 4 and firm profits are (2/3 + 1/32)t = 0.698t.

Proof of Theorem 1

Suppose that Firm A offers its product solely as a bundle at price P ^A and no synthetic bundling occurs.^[18] Define P B = p 1 B + p 2 B and δ = P ^B − P ^A . This implies that the market is partitioned by the partition of the unit square determined by intersection of the line

and the unit square.

The partition could occur in three ways: (a) the dividing manifold could intersect the left edge and the bottom of the square (δ < t₂ − t₁); (b) the manifold could intersect the top and bottom of the square, (δ ∈ [t₂ − t₁, t₁ − t₂]), or; (c) the top of the square and the right edge of the square (δ ≥ t₁ − t₂).

Case (a) can be shown never to be an equilibrium since in that case, Firm A’s bundle price exceeds the sum of the prices of the independent firms and Firm A always prefers to lower its price to capture more market.

In Case (b), suppose δ = t₁ − t₂ − γ, γ > 0. The firms’ first order conditions can be solved directly to yield

and this case can only arise for t 2 ∈ ( 0 , 3 4 t 1 ] .

Suppose instead of offering only the bundle price Firm A offers the equilibrium bundle price and a stand-alone price for product 2A of

At this price, consumers in the lower right corner of the unit square will choose to form the synthetic bundle BA since a consumer of type ( t 1 + t 2 + δ 2 t 1 , 0 ) was previously just indifferent between buying the two goods from A and buying the two goods from B. This segment of the market is the rectangle x 1 ϵ , 1 × [ 0 , x ̲ 2 ] where

and, from Lemma 1,

since P B = p 1 B + p 2 B . This rectangle has area ϵ 1 − x 1 ϵ = γ ϵ + 2 t 2 ϵ 2 2 t 1 The intersection of this segment with the segment Firm A was selling under pure bundling is a triangle with vertices

where t 1 + t 2 + Δ 2 t 1 − x 1 ϵ = t 2 t 1 ϵ so its area is t 2 2 t 1 ϵ 2 . Thus, the deviation generates a gross gain of

and a gross loss from cannibalized sales of the bundle of

For small ϵ, the gain dominates the loss and this deviation is profitable.

Case (c) occurs for t 2 ∈ [ 3 4 t 1 , t 1 ] . In this case, BB is a triangle in the upper right corner of the unit square with vertices

where

and

The first order conditions for 1B, 2B, A imply

and

Combining the two equations from the FOCs gives

which implies

The candidate equilibrium price P ^B is increasing in t₂ so it reaches its maximum at t₂ = t₁. At that value, P ^B = 1.76t₁ and assuming symmetry, p 1 B = . 86 t 1 . Thus, in this region, any optimal p 1 B < t 1 .

Suppose Firm A offers instead of the pure bundling solution, the bundle price P ^A and a stand-alone price, p 2 A = P A (or equivalently, suppose a consumer can purchase the bundle, AA and discard 1A costlessly.) Applying the definitions in Lemma 1, if x ̄ 1 < 1 , the market segment BA will have positive measure.

Since this implies (using Lemma 1)

and p 2 A = P A , Firm A obtains the same revenue per sale in market segment BA and the union of the new segments AA and BA strictly contains the old AA. Thus, this deviation is profitable and breaks the candidate equilibrium.

Proof of Theorem 2

The market partitions are as outlined in Lemma 1, however, since the price of bundle BB is the sum of the standalone prices, p j B , the definitions of the boundaries between BB and AB or BA are simplified somewhat to

The upper boundary, x ̄ j is the same as before. For convenience, set Δ A = p 1 A + p 2 A − P A and note that

Given the uniform distribution, the measures of the sets are computed in the same way as for the mixed bundling two-firm market. Now, however, the sales of (say) product 2B, then is the sum μ(BA) + μ(BB). This yields

Only the first term depends on p 2 B so

Firm 2B then selects its best response in price, p 2 B holding fixed the three prices of Firm A and the price of Firm 1B. The first order condition yields a unique solution

Observe this is decreasing in p 2 B so yields a unique best response. Rewriting, yields

A similar condition, replacing 1 with 2 and BA for AB, holds for the other independent firm. These equations form a linear system in p 1 B , p 2 B and yield unique solutions as functions of P 1 A , p 2 A , p 1 A .

Writing p i B = A i − Δ A 4 t 1 p 1 B we get

So,

Thus

The derivatives of the profit function for the integrated firm are the same as given in (10) and (11). These solutions are then inserted in (12) to obtain the prices for the single-good firms.