The Role of Informative Advertising in Aligning Preferences Over Product Design

Proof of Proposition (1)

We proceed to show that firms have no profitable deviation form the advertising reach and prices established in the proposition.

• Partially-informed market, ϕ* < 1:

  The equilibrium price is p * = 2 / α ( 1 + z ) . Setting any price other than p* in the vicinity is unprofitable, as it will not be in the best response. Setting a price p′ = (V − 2)/2 such that only captive consumers purchase is not profitable whenever:

  ϕ * p * + ϕ * ( 1 − ϕ * ) 2 p * > 2 ϕ * ( 1 − ϕ * ) ( V − 2 ) / 2 .

  Rearranging terms, this is equivalent to p*(2 − ϕ*)/(1 − ϕ*) + 2 > V, and substituting for the equilibrium price and advertising reach gives V < ( 4 + 2 2 α ( 1 + z ) − 2 α ) / ( 2 α ( 1 + z ) − α ) ≔ V ̄ .

  A fully-covered market implies that V ≥ 2p* + 2. Substituting for the equilibrium prices gives V ≥ ( 2 2 + 2 α ( 1 + z ) ) / ( α ( 1 + z ) ) ≔ V ̲ .

  It is easy to show that for any α > 0, we get V ̄ > V ̲ since

  ( 4 + 2 2 α ( 1 + z ) − 2 α ) / ( 2 α ( 1 + z ) − α ) > ( 2 2 + 2 α ( 1 + z ) ) / ( α ( 1 + z ) ) ⟺ 4 + 2 2 η − 2 α ) / ( 2 η − α ) > ( 2 2 + 2 η ) / η ⟺ 0 > − 2 2 α ,

  where the second lines comes from setting η ≔ α ( 1 + z ) . Hence, there is no profitable deviation in prices as long as V < V ̄ .

  A deviation in adverting in the neighborhood of ϕ*, will not be profitable as it will not belong to the firm’s best response, and hence, we consider the extremes. A deviation consisting in setting ϕ′ = 0 is clearly not profitable as no consumer will be aware of the product. Setting ϕ′ = 1 is not profitable whenever:

  p * ( ϕ * ) 2 + 2 ϕ * ( 1 − ϕ * ) − ϕ * − 2 ( 1 − ϕ * ) ≥ E ( ϕ * ) − E ( ϕ ′ = 1 ) ⟺ 1 α − ( ϕ * ) 2 α ≥ p * 2 + ( ϕ * ) 2 − 3 ϕ * ⟺ 1 − 2 α α + 2 η α ≥ 2 η 2 + 4 α 2 ( α + 2 η ) 2 − 6 α α + 2 η ,

  where the second line comes from introducing the equilibrium advertising reach and setting η ≔ α ( 1 + z ) . The previous inequality reduces to 4 η 2 + 3 α 2 ≥ 4 2 α η . Then, since η ≔ α ( 1 + z ) and arranging terms gives 4 ( 1 + z ) + 3 α 2 4 2 2 ≥ α ( 1 + z ) , which is equivalent to 16(1 + z)² + 24(1 + z)α ² + 9α ⁴ ≥ 32α(1 + z), and this is fulfilled for any z and α.

• Fully-informed market, ϕ* = 1:

  The equilibrium price is p* = 1/(1 + z). Setting any price other than p* in the vicinity is unprofitable, as it will not be in the best response. Moreover, pricing at p′ = (V − 2)/2 is also unprofitable due to the absence of captive consumers. The market is fully-covered whenever V ≥ 1 + 2p* which gives V ≥ (3 + z)/(1 + z), and this is fulfilled for any V ≥ 3.

  A deviation in advertising reach, ϕ′ = 1 − ϵ for ϵ ∈ (0, 1], results in profit loss of 1/2(1 − ϵ)2p* − (1/2)2p* = −ϵp*, while gains from lower advertising costs are ϵ ²/α. A deviation is unprofitable if ϵp* > ϵ ²/α which is equivalent to α > ϵ(1 + z), which is always true for ϵ ∈ (0, 1] given that in a fully-informed market α > 2(1 + z).

Proof of Proposition (2).

To establish that an intermediate level of compatibility maximizes consumer surplus, I differentiate CS(⋅) with respect to z to show that CS(⋅) is a quasi-concave function in z. Then:

Evaluating the function at both extremes gives d C S ( z = 0 , α ) d z = 2 α 8 2 + 3 2 α − α ( 10 + α ) ( 2 + α ) 3 , and d C S ( z = 1 , α ) d z = 2 α 8 2 + 2 α − 2 α ( 10 + 2 α ) ( 2 + α ) 3 . The derivative evaluated at z = 0 is positive if 8 2 + 3 2 α − α ( 10 + α ) > 0 The derivative evaluated at z = 1 is negative if 8 2 + 2 α − 2 α ( 10 + 2 α ) < 0 . Define α′ as the solution of 8 2 = α ′ ( 10 + α ′ ) − 3 2 α ′ , and α″ as the solution of 8 = α ′ ′ ( 10 + 2 α ′ ′ ) − α ′ ′ . Since 2 α ( 10 + 2 α ) − 2 α > α ( 10 + α ) − 3 2 α , it implies that α″ < α′, and there will be a maximum at an intermediate level of product compatibility for α ∈ (α″, α′). Then, it is only left to show that this range for α is included in the case of a partially-informed market, or equivalently, α < α ̄ ( z ) .

We first show that α ′ > α ̄ ( z ) . To see this, use 8 2 = α ( 10 + α ) − 3 2 α and define RHS ( α ) ≔ α ( 10 + α ) − 3 2 α . Since ∂ RHS ( α ) / ∂ α = 5 / α + 3 / 2 α − 3 / 2 > 0 , then, replacing for the highest possible α which is α ̄ ( z = 1 ) = 4 , we get 8 2 > 4 ( 10 + 4 ) − 3 2 × 4 . This implies that the solution α′ must be larger than α ̄ ( z ) .

To show that α ′ ′ < α ̄ ( z ) use 8 = α ( 10 + 2 α ) − α . Define RHS ( α ) ≔ α ( 10 + 2 α ) − α . Since ∂ RHS ( α ) / ∂ α = 10 / α + 3 α − 1 > 0 , then, replacing for the highest possible α which is α ̄ ( z = 1 ) = 4 , we get 8 < 4 ( 10 + 8 ) − 4 . Then the solution α″ must be smaller than α ̄ ( z ) .

This demonstrates that there exists an ϵ > 0 such that for α ∈ α ′ ′ , α ′ ′ + ϵ , consumer welfare is maximized at an intermediate level of product compatibility as stated in the proposition.