The Effects of Introducing Advertising in Pay TV: A Model of Asymmetric Competition between Pay TV and Free TV

A.1 Proof of Lemma 1

First, we derive the second-order conditions. By noting that ∂ n p ∂ s p = − 1 2 t , ∂ 2 n p ∂ s p 2 = 0 , ∂ n p ∂ a p = − γ 2 t , ∂ 2 n p ∂ a p 2 = 0 , ∂ c ∂ a p > 0 and ∂ 2 c ∂ a p 2 > 0 , the second-order conditions for a maximum require

It is easy to see that (a) is satisfied by definition. To ensure that (b) is satisfied, we must assume that t > k − γ 2 4 c ′ ′ a p * , where c ′ ′ ( a p ) = ∂ 2 c ∂ a p 2 .

Second, we derive the optimality conditions in Part (i). By solving ∂ π p ∂ s p = 0 for s _p with ∂ n p ∂ s p = − 1 2 t , we obtain s p * = 1 2 ( γ a f * − ( k + γ ) a p * + t ) . Plugging s p * into ∂ π p ∂ a p = k n p * s p * , a p * + s p * + k a p * ∂ n p * ∂ a p − c ′ a p * = 0 and ∂ π f ∂ a f = k n f * a f * + k a f * ∂ n f * ∂ a f − c ′ a f * = 0 and rearranging these equations produces the optimality conditions for PTV and FTV.

Part (ii). To prove the existence and uniqueness of the equilibrium s p A , a p A , a f A , we proceed as follows. First, we show that the reaction function R _f (a _p ) of FTV is a monotonous decreasing function in a _p and the reaction function R _p (a _f ) of PTV is a monotonous increasing function in a _f . We rearrange the optimality conditions of PTV and FTV and we define

With the implicit function theorem, we derive

Hence, PTV reacts with a higher advertising level to an increase in the FTV advertising level, while the opposite is true for the reaction of FTV. We illustrate the reaction functions in Figure 2.

Figure 2:

Reaction functions

We define a ̄ p as the advertising level, which PTV would have to set such that the FTV broadcaster has no incentives to places advertising on its channel, i.e. R f ( a ̄ p ) = 0 . Moreover, a ̲ p is the PTV advertising level if FTV refrains from advertising, i.e. a ̲ p = R p ( 0 ) . Now, we have to show that a ̄ p > a ̲ p . This guarantees that the monotonous reaction function have exactly one intersection point, which characterizes the unique equilibrium. Throughout the paper, we assume that the viewers’ disutility from advertising does not exceed the marginal return on advertising γ ≤ k.

Because c ′ a f A = k 4 t ( − 3 γ a f A − ( k − γ ) a p A + 3 t ) is the optimality condition for FTV, we derive that if a p = 3 t k − γ then a _f = 0. Hence, a ̄ p ( t ) = 3 t k − γ and thus ∂ a ̄ p ∂ t > 0 . Moreover, a ̲ p is implicitly characterized by the PTV’s optimality condition c ′ ( a ̲ p ) = k − γ 4 t ( ( k − γ ) a ̲ p + t ) . With the implicit function theorem, we derive

for t sufficiently large. Because a ̄ p ( t ) is continuous and a monotonously increasing function in t and a ̲ p ( t ) is continuous and a monotonously decreasing function in t, there exists a value t _eq such that a ̄ p ( t ) > a ̲ p ( t ) for all t > t _eq. Hence, we have shown that for a sufficiently large differentiation parameter, i.e. t > t _eq, an unique equilibrium a p A , a f A exists, which is implicitly defined by c ′ a p A = k − γ 4 t ( γ a f A + ( k − γ ) a p A + t ) and c ′ a f A = k 4 t ( − 3 γ a f A − ( k − γ ) a p A + 3 t ) . Note that a p A is always larger than zero, while t has to be sufficiently large to ensure that a f A is larger than zero, i.e. a f A > 0 ⇔ t > t a f ≡ γ a f A + k − γ 3 a p A .

In a next step, we will derive the conditions under which a positive equilibrium subscription fee s p A exists and is unique. According to the first-order conditions, the subscription fee s p A is implicitly defined by s p A = 1 2 ( γ a f A − ( k + γ ) a p A + t ) . Hence, s p A > 0 ⇔ t > t s p ≡ k a p A + γ a p A − a f A . Because for t > max { t eq , t a f , t S O C } , a unique equilibrium a p A , a f A exists and is positive, we conclude that t > t s p ≡ k a p A + γ a p A − a f A ensures that a unique subscription fee exists and is positive.

For the PTV channel, the subscription fee s p A and advertising level a p A in equilibrium are implicitly given by

For example, for a quadratic cost function c(a) = 1/2a ², the advertising level a ̲ p ( t ) is given by

and thus

A.2 Proof of Proposition 1

To prove Part (i), we differentiate three cases:

1. Suppose that γ = k. In this case, we derive c ′ a p A = 0 and hence a p A = 0 . The subscription fee is then given by s p A = 1 2 k a f A + t and a f A is implicitly defined by c ′ a f A = 3 k 4 t − k a f A + t . To ensure that the FTV broadcaster sets a positive advertising level, it must hold t > k a f A and hence a f A > a p A for γ = k.

2. Suppose that γ = 0. In this case, the subscription fee is given by s p A = 1 2 − k a p A + t . To ensure a positive fee, we assume that t > k a p A . The advertising level a p A , a f A are then implicitly defined by c ′ a p A = k 4 t k a p A + t and c ′ a f A = k 4 t − k a p A + 3 t . Since t > k a p A , the FTV broadcaster sets a positive advertising level because − k a p A + 3 t > 0 . Next, we derive

   c ′ a p A − c ′ a f A = k k a p A − t 2 t < 0 ,

   because t > k a p A . It follows that a f A > a p A for γ = 0.

3. Suppose that γ ∈ (0, k). We can show that a p A and a f A will never coincide in the interval γ ∈ (0, k). Suppose that a p A = a f A = a A and hence c a p A = c a f A such that k − γ 4 t ( k a A + t ) = k 4 t − ( 2 γ + k ) a A + 3 t . This equality is satisfied if and only if γ = − 2k, which is not in the interval of feasible γ.

From 1.–3., it follows that a f A > a p A for γ ∈ (0, k].

Moreover, we can rule out the case γ > k. To show this claim, we provide a proof by contradiction. We know that c ′ a p A = k − γ 4 t ( γ a f A + ( k − γ ) a p A + t ) ≥ 0 . Now, suppose that γ > k. Hence, it must be the case that t + γ a f A ≤ ( γ − k ) a p A . However, a positive subscription fee s p A > 0 implies t + γ a f A > ( γ + k ) a p A . That is, t + γ a f A ∈ ( ( γ + k ) a p A , ( γ − k ) a p A ] , which cannot be satisfied in equilibrium. Therefore, our assumption was wrong and it must hold γ ≤ k.

To prove the comparative statics result in Part (ii), we rearrange (15) and (14) and define F p γ , a p A = k − γ 4 t ( γ a f A + ( k − γ ) a p A + t ) − c ′ a p A = 0 and F f γ , a f A = k 4 t ( − 3 γ a f A − ( k − γ ) a p A + 3 t ) − c ′ a f A = 0 . With the implicit function theorem, we derive

Hence, the advertising level on the FTV channel is always decreasing in γ. Regarding the advertising level on PTV, we derive

The second-order conditions require 4 t c ′ ′ a p A > ( k − γ ) 2 , which implies that the denominator is negative. Moreover, it is always the case that a f A > a p A . As a result, we derive that for γ < γ ̂ , the numerator will be negative, and hence,

A.3 Proof of Proposition 2

To show that n f A ≥ n p A ∀ γ ∈ [ 0 , k ] , we differentiate three cases:

1. Suppose that γ = k. In this case, we derive n f A − n p A = 1 2 t t − k a f A > 0 because t > k a f A .

2. Suppose that γ = 0. In this case, we derive n f A − n p A = 1 2 t t − k a p A > 0 because t > k a p A .

3. Suppose that γ ∈ (0, k). To show that n f A ≥ n p A ∀ γ ∈ ( 0 , k ) , we provide a proof by contradiction. Suppose that n f A can fall below n p A in the interval γ ∈ (0, k). Note that n f A ( γ ) and n p A ( γ ) are both continuous functions in γ. Hence, it must be the case that n f A and n p A intersect twice in the interval γ ∈ (0, k) because n f A ( 0 ) > n p A ( 0 ) and n f A ( k ) > n p A ( k ) . However, only one point of intersection exists because

It follows that the assumption was wrong and n f A ≥ n p A ∀γ ∈ (0, k).

A.4 Proof of Proposition 3

For Proposition 3, we calculate the equilibrium solutions in Regimes A and B for the quadratic cost function c ( a ) = θ 2 a 2 .

In Regime A, the maximization problems for the PTV and FTV broadcasters, respectively, are given by

By deriving the corresponding first-order conditions and solving the system of equations, it is straightforward to calculate the subscription fee s p A and the level of advertising a p A for the PTV broadcaster in Regime A as

with λ = 8 θ 2 t 2 − k k − γ 2 γ − 2 θ t k 2 − 5 k γ + γ 2 .

For the FTV broadcaster, the level of advertising a f A in Regime A is

The viewer demands on the PTV channel n p A and FTV channel n f A are

As in the general model, we assume that the differentiation parameter is sufficiently large with t > t ^A . This implies

In Regime B, the maximizations problems for the PTV and FTV broadcasters, respectively, are given by

By deriving the corresponding first-order conditions and solving the system of equations, we calculate the subscription fee s p B for the PTV broadcaster in Regime B as

For the FTV broadcaster the level of advertising a f B in Regime B is

so that the viewer demands in Regime B are given by

To prove Part (i) of Proposition 3, we must show that

Rearranging of the inequality yields t 2 θ t + 3 k γ 2 θ t − k 2 + k γ 4 θ t + 3 k γ − λ 2 θ t 2 + 3 k t γ < 0 . We further simplify the inequality and obtain

which proves the claim in Part (i).

To prove Part (ii) of Proposition 3, we must show that

Rearranging of the inequality yields [ 2 k t ( 3 θ t − k − γ 2 ) ] 4 θ t + 3 k γ − λ 3 k t < 0 . We further simplify the inequality and obtain

which proves the claim in Part (ii).

To prove Part (iii) of Proposition 3, we must show that

Rearranging of the inequality yields θ t 2 θ t + 3 k γ 2 ( 4 θ t + 3 k γ ) − λ ( 2 θ t + 3 k γ ) > 0 . We further simplify the inequality and obtain

Because n _f = 1 − n _p it follows that n f A < n f B , which proves Part (iii) and thus completes the proof of the proposition.

A.5 Proof of Proposition 4

First, we will prove Part (i) of the proposition which states that aggregate consumer surplus is higher in Regime A than in Regime B, i.e. CS p A + CS f A > CS p B + CS f B .

In Regime A, we compute the aggregate consumer surplus CS A = CS p A + CS f A as

with

In Regime B, the aggregate consumer surplus CS B = CS p B + CS f B is given by

To prove Part (i) of the proposition, we must show that CS ^A > CS ^B or equivalently

We further rearrange this inequality as

with

Because of k > γ the above inequality simplifies to τ ₁ + τ ₂ < 0. After rearrangements and simplifications we obtain β ₁ + β ₂ + β ₃ > 0 with

and η ₁ = (k ² + γ ²) and η 2 = k − γ 2 . From our assumption t > t ^A it follows 3θt > η ₂ (see Eq. (25)) and thus it holds β ₁ + β ₂ + β ₃ > 0, which proves Part (i) of the proposition.

Second, we will prove Part (ii) of the proposition which states that fhe profit of the FTV (PTV) broadcaster is lower (higher) in Regime A than in Regime B, i.e. π f A < π f B and π p A > π p B .

In Regime A, we derive the profits of the PTV and FTV broadcasters, respectively, as

with

In Regime B, the profits of the PTV and FTV broadcasters, respectively, are

To prove the claim that the profits of the PTV broadcaster is higher in Regime A than in Regime B, we show

After rearranging the inequality we get

After rearrangements

with

Because k > γ the proof further simplifies to ξ < 0. After further rearrangements and by denoting η 2 = k − γ 2 we obtain

Again, we rearrange and simplify the inequality to

From our assumption t > t ^A it follows 3θt > η ₂ (see (25)) and thus the above inequality is larger than zero, which proves the claim.

Next, we show that the profits of the FTV broadcaster is lower in Regime A than in Regime B, i.e.

After rearranging the inequality we get

We further simplify the inequality and obtain

Thus, we only need to show that

From our assumption t > t ^A it follows 3θt > η ₂ (see (25)) and thus the above inequality is larger than zero, which proves the claim. This completes the proof of Proposition 4.