Internet channel entry: retail coverage and entry cost advantage

Abstract

In this research we study how existing market coverage affects the outcome of the Internet channel entry game between an existing retailer and a new entrant. A market is not covered when some consumers with low reservation prices are priced out by existing retailers and do not purchase. In a model with multiple existing retailers and a potential new entrant, we demonstrate that when entry costs are equal, one of the existing retailers enters the Internet channel first. However, if the market is covered by existing retailers before entry, then because of the threat of Internet channel entry by the potential new entrant, retailer entry cannibalizes existing retail profits—cannibalizing at a loss. In addition, if a potential new entrant has a slight advantage in Internet channel entry costs and the market is not covered by existing retailers, then the new entrant enters the Internet channel first. If the market is covered by existing retailers, then the new entrant must have a larger Internet channel entry cost advantage to be first to enter the Internet channel.

Appendix of mathematical proofs

Proof of Lemma 1

Referring to Salop’s [19] classical circular model with two firms (e.g., [22]), the equilibrium prices are ${p_{a}^{s1} = p_{b}^{s1} = t/2}$. In order for all the consumers on the circle to be served—that is, for the market to be covered—in this symmetric model the most distant consumer from a retailer is x = 1/4 and the equilibrium prices have to satisfy ${p_{a}^{s1} + t/4 < R}$ and ${p_{b}^{s1} + t/4 < R}$, which requires that R > 3t/4.

When R ≤ 3t/4, the reservation price is sufficiently low that the retailers no longer compete with each other. In this case, taking retailer A as an example, x is given by ${p_{a}^{s1} + tx = R }$. Retailer A’s maximization problem is

$$ \max_{p_{a}^{s1}} \pi_{a}^{s1} = \max_{p_{a}^{s1}} \left\{ 2 p_{a}^{s1} \left( \frac { R - p_{a}^{s1}} {t}\right) \right\} \;\; \ni \;\; p_{a}^{s1} + t/4 \geq R \; , \;\; p_{a}^{s1} \geq 0. $$

Solving this constrained maximization problem, we get that, if R ≤ t/2, x < 1/4 (which means the market is not covered), ${p_{a}^{s1} = p_{b}^{s1} = R/2 }$, and ${ \pi_{a}^{s1} = \pi_{b}^{s1} = R^{2}/2t }$. If ${ t/2 \leq R \leq 3t/4 }$, x = 1/4 (the constraint is binding), ${ p_{a}^{s1} = p_{b}^{s1} = R - t/4 }$, and the profits are stated in Lemma 1. □

Proof of Lemma 2

Using (1), we can write the Lagrangian function for retailer A as

$$ \begin{aligned} L=2p_{a}^{s2}\left(\frac {p_{d}^{s2}-p_{a}^{s2}+\mu}{t}\right) + 2p_{d}^{s2} \left(\frac{1}{2}-\frac {p_{d}^{s2} - p_{a}^{s2} + \mu}{t}-\frac {p_{d}^{s2}-p_{b}^{s2}+\mu}{t}\right) \\ - \lambda_{1} (p_{d}^{s2}+\mu-R) - \lambda_{2} \left(\frac {p_{d}^{s2} - p_{a}^{s2} + \mu}{t} + \frac {p_{d}^{s2}-p_{b}^{s2}+\mu}{t} - \frac {1}{2}\right). \end{aligned} $$

The resulting Kuhn-Tucker conditions for retailer A are

$$ \frac {\partial L}{\partial p_{a}^{s2}} = \frac {2}{t}(2p_{d}^{s2}-2p_{a}^{s2}+\mu) + \frac {\lambda_{2}} {t} \leq 0, $$

(7)

$$ \frac {\partial L}{\partial p_{d}^{s2}} = \frac {2}{t}(2p_{a}^{s2}+p_{b}^{s2}-4p_{d}^{s2}-2\mu) + 1 - \lambda_{1} - \frac {2\lambda_{2}} {t} \leq 0, $$

(8)

$$ \frac {\partial L}{\partial \lambda_{1}} = - p_{d}^{s2} - \mu + R \geq 0, $$

(9)

$$ \frac {\partial L}{\partial \lambda_{2}} = - x - y + 1/2 \geq 0, $$

(10)

$$ p_{a}^{s2} \geq 0, \;\; p_{d}^{s2} \geq 0, \;\; \lambda_{1} \geq 0, \;\; \lambda_{2} \geq 0, $$

$$ p_{a}^{s2} \frac {\partial L}{\partial p_{a}^{s2}} = 0, \;\; p_{d}^{s2} \frac {\partial L}{\partial p_{d}^{s2}} = 0, \;\; \lambda_{1} \frac {\partial L}{\partial \lambda_{1}} = 0, \;\; \lambda_{2} \frac {\partial L}{\partial \lambda_{2}} = 0. $$

Using (2) the Kuhn-Tucker conditions for retailer B are

$$ \frac {\partial \pi_{b}^{s2}}{\partial p_{b}^{s2}} = \frac{2}{t} (p_{d}^{s2}-2p_{b}^{s2}+\mu) \leq 0, $$

(11)

$$ p_{b}^{s2} \geq 0, \;\; p_{b}^{s2}\frac {\partial \pi_{b}^{s2}}{\partial p_{b}^{s2}} = 0. $$

Excluding the non-negativity constraints, there are the two possible binding constraints for retailer A’s problem, giving rise to four cases which we discuss in turn.

Interior solution. In this case the shadow prices are zero, λ₁ = λ₂ = 0. Assume the three prices are strictly positive so that (7), (8), and (11) hold with equality. From these equations we get the candidate Nash equilibrium prices:

$$ p_{a}^{s2} = \frac {2t + \mu} {6}, \;\; p_{d}^{s2} = \frac {t-\mu} {3}, \;\; p_{b}^{s2} = \frac {t + 2\mu} {6}. $$

From the market comparison condition, ${\mu/t < 1/2}$, the candidate Nash equilibrium prices are strictly positive. Using the strict inequalities from our constraints, ${p_{d}^{s2} + \mu < R}$ and x + y < 1/2, and our market coverage condition R ≥ t/2, we derive the additional conditions on μ/t and R: ${\frac{\mu}{t}<\frac{1}{4}}$ and ${R \geq \frac{t}{2}}$, or ${\frac{1}{4} \leq \frac{\mu}{t} < \frac{2}{5}}$ and ${R > \frac{t + 2 \mu}{3}}$. And the candidate Nash equilibrium profits are ${\pi_{a}^{s2} = \frac{13 \mu^{2} - 8 \mu t + 4 t^{2}}{18t}}$ and ${\pi_{b}^{s2} = \frac{(2 \mu + t)^{2}}{18t}}$.

Constraint C1 is binding. In this case the shadow price of C2 is zero, λ₂ = 0. Assume the three prices are strictly positive so that (7), (8), (11) and (9) hold with equality, where the latter equation implies ${p_{d}^{s2} = R - \mu}$. From these equations we get the remaining candidate Nash equilibrium prices and the shadow price for C1:

$$ p_{a}^{s2} = R - \mu/2, \;\; p_{b}^{s2} = R/2, \;\; \lambda_{1} = \frac {2\mu - 3R + t} {t}. $$

From the market comparison condition, ${\mu/t < 1/2}$, and the market coverage condition, R ≥ t/2, the candidate Nash equilibrium prices are strictly positive. Using the strict inequality from C2, x + y < 1/2, our market coverage condition, and the non-negative shadow price for C1, λ₁ ≥ 0, we derive the additional conditions on ${\mu/t}$ and R: ${\frac{1}{4} \leq \frac{\mu}{t} < \frac{2}{5} }$ and ${ \frac{t}{2} \leq R \leq \frac{t + 2 \mu}{3} }$, or ${ \frac{2}{5} \leq \frac{\mu}{t} < \frac{1}{2} }$ and ${ \frac{t}{2} \leq R < t - \mu }$. And the candidate Nash equilibrium profits are ${\pi_{a}^{s2} = \frac{\mu^{2} + 2 \mu R - 2 R^{2}}{2t} + R - \mu}$ and ${\pi_{b}^{s2} = \frac{R^{2}}{2t}}$.

Constraint C2 is binding. In this case the shadow price of C1 is zero, λ₁ = 0. Assume the three prices are strictly positive so that (7), (8), (9) and (10) hold with equality. From these equations we get candidate Nash equilibrium prices and the shadow price for C2:

$$ p_{a}^{s2} = \frac {2t} {5}, \;\; p_{d}^{s2} = \frac {3t - 5\mu} {5}, \;\; p_{b}^{s2} = \frac {3t} {10}, \;\; \lambda_{2} = \frac { 2 ( 5 \mu - 2 t ) } {5}. $$

From the market comparison condition, ${\mu/t < 1/2}$, all candidate Nash equilibrium prices are strictly positive. Using the strict inequality from C1, ${p_{d}^{s2} + \mu < R}$, our market coverage condition R ≥ t/2, and the non-negative shadow price for C2, λ₂ ≥ 0, we derive the additional conditions on ${\mu/t}$ and R: ${\frac{2}{5} \leq \frac{\mu}{t} < \frac{1}{2}}$ and ${R > \frac{3t}{5}}$. And the candidate Nash equilibrium profits are ${\pi_{a}^{s2} = \frac{4t}{25}}$ and ${\pi_{b}^{s2} = \frac{9t}{50}}$.

Constraints C1 and C2 are binding. Assume the three prices are strictly positive so that (7), (8), (11), (9) and (10) hold with equality. From (9) we get ${p_{d}^{s2} = R - \mu}$. From the remaining equations we get other candidate Nash equilibrium prices and the shadow prices for C1 and C2:

$$ p_{a}^{s2} = \frac {3R-t} {2}, \;\; p_{b}^{s2} = \frac {R} {2}, \;\; \lambda_{1} = \frac { 3 t - 5 R } {t}, \;\; \lambda_{2} = 2 ( \mu + R - t ). $$

From our market coverage condition R ≥ t/2 all prices are strictly positive. Using this condition, together with the two non-negative shadow prices, λ₁, λ₂ ≥ 0, we derive the additional conditions on ${\mu/t}$ and R: ${\frac{2}{5} \leq \frac{\mu}{t} < \frac{1}{2}}$ and ${t-\mu \leq R \leq \frac{3t}{5}}$. And the candidate Nash equilibrium profits are ${\pi_{a}^{s2} = \frac{(t-R)(3R-t)}{2t}}$ and ${\pi_{b}^{s2} = \frac{R^{2}}{2t}}$.

Our second step is to verify that the Kuhn-Tucker conditions are also the sufficient condition for the maximization problem. It is easy to verify that both the objective functions are concave (the Hessian matrix of the objective function of retailer A is negative definite, and the second derivative of the objective function of retailer B is negative), and the constraints are linear and therefore concave. Thus, what we derive above is the optimal solution to the constrained profit maximization problem.

Next since retailer A always has the option to charge higher than the reservation price and not sell in the Internet channel, we need to verify when profits for retailer A are greater in State 2 than in State 1. By comparing the profits of retailer A in these two states, we derive the conditions under which retailer A chooses to sell in the Internet channel and under which he chooses not to sell in the Internet channel. The conditions are given in the lemma.

Lastly, in order to prove that the optimal solution to the constrained optimization problem is indeed the unique Nash equilibrium, following Friedman [6, p. 152], Gruca and Sudharshan [9], four conditions have to satisfy: (G1) the number of players is finite, (G2) the strategy space of every player is compact and convex, (G3) the payoff functions are continuous and bounded, and (G4) the payoff functions are quasi-concave. It is easy to verify that in our case, the number of players is finite (G1), the constraints insure that the strategy space of every player is compact and convex (G2), the payoff functions are continuous and bounded (G3), and the payoff functions are concave and thus quasi-concave (G4).□

Proof of Lemma 4

The Lagrangian function for (3) is:

$$ L = 2 p_{a}^{s4} \left( \frac { \mu - p_{a}^{s4} } {t} \right) - \lambda \left(\frac {\mu - p_{a}^{s4}}{t} + \frac {\mu -p_{b}^{s4}}{t} - \frac{1}{2}\right) $$

The resulting Kuhn-Tucker conditions are

$$ \frac {\partial L}{\partial p_{a}^{s4}} = \frac {2}{t}( \mu - 2 p_{a}^{s4}) + \frac {\lambda} {t} \leq 0, $$

$$ \frac {\partial L}{\partial \lambda} = - x - y + 1/2 \geq 0, $$

$$ p_{a}^{s4} \geq 0,\;\; \lambda \geq 0, $$

$$ p_{a}^{s4} \frac {\partial L}{\partial p_{a}^{s4}} = 0, \;\; \lambda \frac {\partial L}{\partial \lambda} = 0. $$

Excluding the non-negativity constraints, there is the one possible binding constraint for retailer A’s problem, giving rise to two cases which we discuss in turn.

Interior solution. In this case the shadow prices are zero, λ = 0. Following the same analysis as Lemma 2 and the assumption of identical firms, we get candidate Nash equilibrium prices: ${p_{a}^{s4} = p_{b}^{s4} = \mu/2}$. Using the strict inequality from the constraint, x + y < 1/2, we get the condition for the candidate Nash equilibrium prices to exist: ${\mu/t < 1/2}$, which is our market comparison condition.

Constraint C2 is binding. From x + y = 1/2, we get ${p_{a}^{s4} = \mu - t/4}$. The first Kuhn-Tucker condition holds with equality, since p ^s4 _a is positive because of the market comparison condition. Plugging p ^s4 _a into the first condition, we get ${\lambda = 2 ( \mu - t/2 )}$. From λ ≥ 0, we get ${\mu/t \geq 1/2}$, which is contradictory to our market comparison condition. Therefore, in State 4, C2 is never binding.

It is easy to verify that both the objective functions are concave (the second derivative is negative), and the constraints are linear and therefore concave. Thus, what we derive above is the optimal solution to the constrained profit maximization problem. Similar as in the proof of Lemma 2, it is easy to verify that in our case, the number of players is finite (G1), the constraints insure that the strategy space of every player is compact and convex (G2), the payoff functions are continuous and bounded (G3), and the payoff functions are concave and thus quasi-concave (G4). Therefore, the optimal solution to the constrained maximization problem is indeed the unique Nash equilibrium.□

Proof of Corollary 1

We first prove the dominance of State 4 by State 1, then by State 2 and State 3.

• Dominance of State 4 by State 1: From Lemma 4, ${\pi_{a}^{s4} = \pi_{b}^{s4} < t/8}$ from the market comparison constraint ${\mu/t < 1/2}$ . From Lemma 1, if R >  3t/4, then ${\pi_{a}^{s1} = \pi_{b}^{s1} > \pi_{a}^{s4} = \pi_{b}^{s4}}$ . If t/2 ≤  R ≤  3t/4, then from the market coverage constraint, R ≥  t/2, ${\pi_{a}^{s1} = \pi_{b}^{s1} \geq t/8 > \pi_{a}^{s4} = \pi_{b}^{s4}}$.

• Dominance of State 4 by State 2: Since the profit of retailer A in State 2 is at least as large as that in State 1 (Lemma 2), ${\pi_{a}^{s2} > \pi_{a}^{s1} > \pi_{a}^{s4}}$ . It is also easy to verify that ${\pi_{b}^{s2} > \pi_{b}^{s4}}$.

• Dominance of State 4 by State 3: From Lemma 3 and Lemma 4, ${\pi_{a}^{s3} - \pi_{a}^{s4} = \pi_{b}^{s3} - \pi_{b}^{s4} = \frac{(t+4\mu)^2}{72t} - \frac{\mu^2}{2t} = \frac{(t+10\mu)(t-2\mu)}{72t} > 0}$. □

Proof of Theorem 1

In order to compare the stand-alone incentives and preemption incentives, we summarize the equilibrium profits under different parameter ranges in State 1 through 3 in Table 3. In this proof, row numbers refer to Table 3.

Table 3 Equilibrium profits when the market is covered

• Row 1: If ${ \mu / t \in ( 0, 1/4 ) }$ and ${R \in [ \frac{t}{2}, \frac{52 \mu^{2} - 32 \mu t + 25 t^{2} }{36t})},$

  $$ \begin{aligned} (\pi_{a}^{s2}-\pi_{a}^{s1})-\pi_{e}^{s3}&= \frac{(13\mu^{2}-8\mu t+4t^{2})}{18t}-\frac{(R-\frac{t}{4})}{2}-\frac{(t-2\mu)^{2}}{9t}\\ &=\frac{1}{2}\left(\frac{20\mu^{2}+17t^{2}}{36t}-R\right). \end{aligned} $$

  Since R has to be in the range of ${[ \frac{t}{2},\frac{52 \mu^{2} - 32 \mu t + 25 t^{2} }{36t} )}$, if ${ \mu/t\geq 1/\sqrt {20} }$, then ${ t/2 \leq ( 20 \mu^{2} + 17 t^{2})/36t < 3t/4 }$, so if ${ R \leq(20 \mu^{2} + 17 t^{2} ) / 36t }$, then ${ ( \pi_{a}^{s2} -\pi_{a}^{s1}) - \pi_{e}^{s3} \geq 0 }$, and if ${ R > ( 20 \mu^{2} +17 t^{2})/36t}$, then ${ ( \pi_{a}^{s2} - \pi_{a}^{s1} ) - \pi_{e}^{s3} < 0 }$. If ${ \mu / t < 1 / \sqrt {20} }$, then ${ ( 20 \mu^{2}+17 t^{2} ) / 36t < t/2 \leq R \leq 3t/4}$, and ${ ( \pi_{a}^{s2} - \pi_{a}^{s1} ) - \pi_{e}^{s3} < 0 }$.

• Row 2: If ${ \mu / t \in ( 0, 1/4 ) }$ and ${ R \in [ \frac{52 \mu^{2} - 32 \mu t + 25 t^{2} }{36t}, \frac{3t}{4} )}$,

  $$ ( \pi_{a}^{s2} - \pi_{a}^{s1} ) = 0 < \pi_{e}^{s3}. $$

• Row 3: If ${ \mu / t \in ( 0, 1/4 ) }$ and ${ R \in ( \frac{3t}{4}, \infty ) }$,

  $$ ( \pi_{a}^{s2} - \pi_{a}^{s1} ) = 0 < \pi_{e}^{s3}. $$

• Row 4: If ${ \mu / t \in [ \frac{1}{4}, \frac{14 - 3\sqrt{3}}{26} ) }$ and ${ R \in [ t/2, ( t + 2 \mu ) / 3 ] }$,

  $$ \begin{aligned} ( \pi_{a}^{s2} - \pi_{a}^{s1} ) - \pi_{e}^{s3}&= \frac{1}{2t} ( \mu^{2} + 2\mu R - 2R^{2} ) + R - \mu - \frac{(R - \frac{t}{4})}{2} - \frac{(t-2\mu)^{2}}{9t} \\ &= \frac {1}{72t} ( -72 R^{2} + 36tR + 72 \mu R + 4 \mu ^{2} - 40 \mu t + t^{2} ) \\ &= \frac {1}{t} \left( - \left( R - \frac {1}{4} ( t + 2 \mu )\right ) ^ {2} + \frac {11}{144} ( t - 2 \mu ) ^{2} \right) \\ &= \frac {1}{t} \left( R - \frac { 3 ( t + 2 \mu ) - \sqrt{11} ( t - 2\mu ) }{12} \right) \left(\frac { 3 ( t + 2 \mu ) + \sqrt{11} ( t - 2\mu )}{12} - R \right). \end{aligned} $$

  Note that ${ ( 3 ( t + 2 \mu ) - \sqrt{11} ( t - 2\mu ) ) / 12 < t/2 \leq R}$. Since ${ R \in [ t/2, ( t + 2 \mu ) / 3 ] }$, if ${ \mu / t \geq ( 6 - \sqrt {11} ) / 10 }$, then ${ t/2 < ( 3 ( t + 2 \mu ) + \sqrt{11} ( t - 2\mu ) ) / 12 \leq ( t + 2 \mu ) / 3}$, so if ${ R > ( 3 ( t + 2 \mu ) + \sqrt{11} ( t - 2\mu ) ) / 12 }$, then ${ ( \pi_{a}^{s2}-\pi_{a}^{s1} ) - \pi_{e}^{s3} < 0 }$, and if ${ R \leq ( 3 ( t + 2\mu)+\sqrt{11} ( t - 2\mu ) ) / 12 }$, then ${ ( \pi_{a}^{s2}-\pi_{a}^{s1} ) - \pi_{e}^{s3} \geq 0 }$. If ${ \mu / t < ( 6 - \sqrt {11} ) / 10 }$, then ${ ( 3 ( t + 2 \mu ) + \sqrt{11} ( t-2\mu)) / 12 > ( t + 2\mu ) / 3 \geq R }$, and ${ ( \pi_{a}^{s2} - \pi_{a}^{s1} ) - \pi_{e}^{s3} > 0 }$.

• Row 5: If ${ \mu / t \in [ \frac{1}{4}, \frac{14 - 3\sqrt{3}}{26} ) }$ and ${ R \in [\frac {t + 2\mu} {3}, \frac{52 \mu^{2} - 32 \mu t + 25 t^{2} }{36t} ) }$,

  $$ \begin{aligned} ( \pi_{a}^{s2} - \pi_{a}^{s1} ) - \pi_{e}^{s3} &= \frac{(13 \mu^{2} - 8 \mu t + 4 t^{2})}{18t} - \frac{(R - \frac{t}{4})}{2} - \frac{(t-2\mu)^{2}}{9t} \\ & = \frac {1}{2} \left(\frac { 20 \mu^{2} +17 t^{2} }{ 36t } - R \right). \end{aligned} $$

  Since ${ R \in [\frac {t + 2\mu} {3}, \frac{52 \mu^{2} - 32 \mu t + 25 t^{2} }{36t} ) }$, if ${\mu / t \leq ( 6 - \sqrt {11} ) / 10 }$, then ${ ( t + 2\mu ) / 3 \leq ( 20 \mu^{2} +17 t^{2} ) / 36t < \frac{52 \mu^{2} - 32 \mu t + 25 t^{2} }{36t} }$, so if ${ R \leq ( 20 \mu^{2} +17 t^{2} ) / 36t }$, then ${ ( \pi_{a}^{s2} - \pi_{a}^{s1} ) - \pi_{e}^{s3} \geq 0 }$, and if ${ R > ( 20 \mu^{2} +17 t^{2} ) / 36t }$, then ${ ( \pi_{a}^{s2} - \pi_{a}^{s1} ) - \pi_{e}^{s3} < 0 }$. If ${\mu / t > ( 6 - \sqrt {11} ) / 10 }$, then ${ ( 20 \mu^{2} +17 t^{2} ) / 36t < ( t + 2\mu ) / 3 < R }$, and ${ ( \pi_{a}^{s2} - \pi_{a}^{s1} ) -\pi_{e}^{s3} < 0}$.

• Row 6: If ${ \mu / t \in [ 1/4, \frac{14 - 3\sqrt{3}}{26}) }$ and ${ R \in [\frac{52 \mu^{2} - 32 \mu t + 25 t^{2} }{36t}, 3t/4) }$, the analysis is the same as that of Row 2.

• Row 7: If ${ \mu / t \in [ 1/4, \frac{14 - 3\sqrt{3}}{26}) }$ and ${ R \in ( 3t/4, \infty ) }$, the analysis is the same as that of Row 3.

• Row 8: If ${ \mu / t \in [\frac{14 - 3\sqrt{3}}{26}, 1/2 ) }$ and ${R \in [ t/2, \frac{(\sqrt{3} + 1) t - 2 (\sqrt{3} - 1) \mu }{4} )}$,

  $$ \begin{aligned} ( \pi_{a}^{s2} - \pi_{a}^{s1} ) - \pi_{e}^{s3} &= \frac{1}{2t} ( \mu^{2} + 2\mu R - 2R^{2} ) + R - \mu - \frac{(R - \frac{t}{4})}{2} - \frac{(t-2\mu)^{2}}{9t}\\ &= \frac {1}{t} \left( R - \frac { 3 ( t + 2 \mu ) - \sqrt{11} ( t - 2\mu ) }{12} \right) \left(\frac { 3 ( t + 2 \mu ) + \sqrt{11} ( t - 2\mu) }{12} - R \right). \end{aligned} $$

  Note that ${ ( 3 ( t + 2 \mu ) - \sqrt{11} ( t - 2\mu ) ) / 12 < t/2 \leq R}$, and ${ t/2 < ( 3 ( t + 2 \mu ) + \sqrt{11} ( t - 2\mu ) ) / 12 < \frac{(\sqrt{3} + 1) t - 2 (\sqrt{3} - 1) \mu }{4} }$. So if ${ R \leq ( 3 ( t + 2 \mu ) + \sqrt{11} ( t - 2\mu ) ) / 12 }$, then ${ ( \pi_{a}^{s2} - \pi_{a}^{s1} ) \geq \pi_{e}^{s3} }$. Otherwise, ${ ( \pi_{a}^{s2} - \pi_{a}^{s1} ) < \pi_{e}^{s3} }$.

• Row 9: If ${ \mu / t \in [\frac{14 - 3\sqrt{3}}{26}, 1/2 ) }$ and ${ R \in [ \frac{(\sqrt{3} + 1) t - 2 (\sqrt{3} - 1) \mu }{4}, 3t/4 ) }$, the analysis is the same as that of Row 2.

• Row 10: If ${ \mu / t \in [\frac{14 - 3\sqrt{3}}{26}, 1/2 ) }$ and ${ R \in ( 3t/4, \infty ) }$, the analysis is the same as that of Row 3.

In sum, ${ ( \pi_{a}^{s2} - \pi_{a}^{s1} ) < \pi_{e}^{s3} }$, except for the following parameter combinations:

• (i) ${1 / \sqrt{20} \leq \mu / t < 1/4 }$ and ${ t/2 \leq R \leq \frac {20 \mu^{2} + 17 t^{2} } {36t} }$, or,

• (ii) ${ \frac { 6 - \sqrt {11} }{10} \leq \mu / t < 2/5 }$ and ${ t/2 \leq R \leq \frac { 3 ( t + 2 \mu ) + \sqrt{11} ( t - 2\mu ) }{12} }$, or,

• (iii) ${ 1/4 \leq \mu / t < \frac { 6 - \sqrt {11} }{10} }$ and ${ t/2 \leq R \leq \frac { t + 2 \mu } {3} }$, or,

• (iv) ${ 1/4 \leq \mu / t \leq \frac { 6 - \sqrt {11} }{10} }$ and ${ \frac { t + 2 \mu }{ 3 } < R \leq \frac { 20 \mu^{2} +17 t^{2} }{ 36t } }$, or,

• (v) ${ 2/5 \leq \mu / t \leq 1/2 }$ and ${ t/2 \leq R \leq \frac { 3 ( t + 2 \mu ) + \sqrt{11} ( t - 2\mu ) }{12} }$.

Combining (iii) and (iv), we get ${ 1/4 \leq \mu / t \leq \frac { 6 - \sqrt {11} }{10} }$ and ${ t/2 \leq R \leq \frac { 20 \mu^{2} +17 t^{2} }{ 36t } }$, which we then combine with (i), and we get the first parameter combination in Theorem 1. Combining (ii) and (v) results in the second parameter combination in Theorem 1. □

Proof of Theorem 2

Again, in this proof, row numbers refer to Table 3.

• Row 1: If ${ \mu / t \in ( 0, 1/4 )}$ and ${R \in [ \frac{t}{2}, \frac{52 \mu^{2} - 32 \mu t + 25 t^{2} }{36t})}$,

  $$ \begin{aligned} (\pi_{a}^{s2} - \pi_{a}^{s3} ) - \pi_{e}^{s3} &= \frac{(13 \mu^{2} - 8 \mu t + 4 t^{2})}{18t} - \frac{(t + 4 \mu)^2}{72t} - \frac{(t-2\mu)^2}{9t} \\ & = \frac{1}{72t} (4 \mu ^{2} - 8 \mu t + 7 t^{2} ) \\ &= \frac{1}{72t} (4 ( \mu - t ) ^ {2} + 3 t^{2}) \\ & > 0. \end{aligned} $$

• Row 2: If ${ \mu / t \in ( 0, 1/4 ) }$ and ${ R \in [ \frac{52 \mu^{2} - 32 \mu t + 25 t^{2} }{36t}, \frac{3t}{4} )}$,

  $$ \begin{aligned} (\pi_{a}^{s2} - \pi_{a}^{s3} ) - \pi_{e}^{s3} &= \frac{R - \frac{t}{4}}{2} - \frac{ (t + 4 \mu) ^{2} }{72t} - \frac{(t-2\mu)^{2}}{9t} \\ & = \frac{1}{2}\left(R - \frac{ 8 \mu ^{2} - 4 \mu t + 3 t^{2} }{6t}\right) \\ &> \frac{1}{2}\left(R - \frac{t}{2}\right)\\ &\geq 0. \end{aligned} $$

• Row 3: If ${ \mu / t \in ( 0, 1/4 ) }$ and ${ R \in ( \frac{3t}{4}, \infty)}$,

  $$ \begin{aligned} (\pi_{a}^{s2} - \pi_{a}^{s3} ) - \pi_{e}^{s3}& = \frac{t}{4} - \frac{ (t + 4 \mu) ^{2} }{72t} - \frac{(t-2\mu)^{2}}{9t}\\ & = \frac{ \mu (t - 2\mu) }{3t} + \frac{t}{8}\\ &> 0. \end{aligned} $$

• Row 4: If ${ \mu / t \in [ \frac{1}{4}, \frac{14 - 3\sqrt{3}}{26} ) }$ and ${ R \in [ t/2, ( t + 2 \mu ) / 3 ] }$,

  $$ \begin{array}{lcl} ( \pi_{a}^{s2} - \pi_{a}^{s3} ) - \pi_{e}^{s3} &=& \frac{1}{2t} ( \mu^{2} + 2\mu R - 2R^{2} ) + R - \mu - \frac{ (t + 4 \mu) ^{2} }{72t} - \frac{(t-2\mu)^{2}}{9t}\\ & =& \frac {1}{24t} ( - 24 R^{2} + 24 \mu R + 24 t R - 4 \mu ^{2} -16 \mu t - 3 t^{2})\\ &=& \frac {1}{24t} \left( - 24 \left( R - \frac { \mu + t }{2}\right ) ^ {2} + 2 \mu ^ {2} - 4 \mu t + 3 t^{2}\right)\\ & \stackrel { t/2 \leq R \leq \frac { t + 2 \mu }{3} < \frac { \mu + t }{2} }{ > }& \frac {1}{24t} \left( - 24 \left( \frac{t}{2} - \frac { \mu + t }{2} \right) ^ {2} + 2 \mu ^ {2} - 4 \mu t + 3 t^{2}\right)\\ &=& \frac {1} {24t} ( - 4 \mu ^{2} - 4 \mu t + 3 t^{2})\\ &= & \frac {1} {24t} ( t - 2 \mu ) ( 3 t + 2 \mu)\\ & \stackrel { \mu / t < 1/2 }{ > }& 0. \end{array} $$

• Row 5: If ${ \mu / t \in [ \frac{1}{4}, \frac{14 - 3\sqrt{3}}{26} ) }$ and ${ R \in [\frac {t + 2\mu} {3}, \frac{52 \mu^{2} - 32 \mu t + 25 t^{2} }{36t} ) }$, the analysis is the same as that of Row 1.

• Row 6: If ${ \mu / t \in [ 1/4, \frac{14 - 3\sqrt{3}}{26}) }$ and ${ R \in [\frac{52 \mu^{2} - 32 \mu t + 25 t^{2} }{36t}, 3t/4) }$, the analysis is the same as that of Row 2.

• Row 7: If ${ \mu / t \in [ 1/4, \frac{14 - 3\sqrt{3}}{26}) }$ and ${ R \in ( 3t/4, \infty ) }$, the analysis is the same as that of Row 3.

• Row 8: If ${ \mu / t \in [\frac{14 - 3\sqrt{3}}{26}, 1/2 ) }$ and ${ R \in [ t/2, \frac{(\sqrt{3} + 1) t - 2 (\sqrt{3} - 1) \mu }{4})}$,

  $$ \begin{array}{lcl} (\pi_{a}^{s2} - \pi_{a}^{s3} ) - \pi_{e}^{s3} & =& \frac{1}{2t} ( \mu^{2} + 2\mu R - 2R^{2} ) + R - \mu - \frac{ (t + 4 \mu) ^{2} }{72t} - \frac{(t-2\mu)^{2}}{9t}\\ & = & \frac {1}{24t} ( - 24 R^{2} + 24 \mu R + 24 t R - 4 \mu ^{2} -16 \mu t - 3 t^{2})\\ & = & \frac {1}{24t} ( - 24 ( R - \frac { \mu + t }{2} ) ^ {2} + 2 \mu ^ {2} - 4 \mu t + 3 t^{2}) \\ & \stackrel { t/2 \leq R \leq \frac{(\sqrt{3} + 1) t - 2 (\sqrt{3} - 1) \mu }{4} < \frac { \mu + t }{2} }{ > } & \frac {1}{24t} ( - 24 ( \frac{t}{2} - \frac { \mu + t }{2} ) ^ {2} + 2 \mu ^ {2} - 4 \mu t + 3 t^{2} )\\ & = & \frac {1} {24t} ( - 4 \mu ^{2} - 4 \mu t + 3 t^{2} )\\ & = & \frac {1} {24t} ( t - 2 \mu ) ( 3 t + 2 \mu)\\ & \stackrel { \mu / t < 1/2 }{ > } & 0. \end{array} $$

• Row 9: If ${ \mu / t \in [\frac{14 - 3\sqrt{3}}{26}, 1/2 ) }$ and ${R \in [ \frac{(\sqrt{3} + 1) t - 2 (\sqrt{3} - 1) \mu }{4}, 3t/4)}$, the analysis is the same as that of Row 2.

• Row 10: If ${ \mu / t \in [\frac{14 - 3\sqrt{3}}{26}, 1/2 ) }$ and ${ R \in ( 3t/4, \infty ) }$, the analysis is the same as that of Row 3. □

Proof of Lemma 6

The Kuhn-Tucker conditions of retailers A and B are the same as in the proof of Lemma 2.

Interior solution. Following the same analysis as what we have in proof of Lemma 2, we get candidate Nash equilibrium prices:

$$ p_{a}^{s2} = \frac {2t + \mu} {6}, \;\; p_{d}^{s2} = \frac {t-\mu} {3}, \;\; p_{b}^{s2} = \frac {t + 2\mu} {6}. $$

Using the strict inequalities from our constraints, ${p_{d}^{s2} + \mu < R}$ and x + y < 1/2, and our market coverage condition R <  t/2, we derive the additional conditions on ${\mu/t}$ and R in Lemma 6.

Constraint C1 is binding. Following the same analysis as what we have in proof of Lemma 2, we get candidate Nash equilibrium prices and the shadow price for C1:

$$ p_{a}^{s2} = R - \mu/2, \;\; p_{d}^{s2} = R - \mu , \;\; p_{b}^{s2} = R/2, \;\; \lambda_{1} = \frac {2\mu - 3R + t} {t}. $$

Using the strict inequality from C2, x + y < 1/2, our market coverage condition, R < t/2, and the non-negative shadow price for C1, λ₁ ≥ 0, we derive the additional conditions on ${\mu/t}$ and R in Lemma 6.

Constraint C2 is binding. Following the same analysis as what we have in proof of Lemma 2, we get candidate Nash equilibrium prices and the shadow price for C2:

$$ p_{a}^{s2} = \frac {2t} {5}, \;\; p_{d}^{s2} = \frac {3t - 5\mu} {5}, \;\; p_{b}^{s2} = \frac {3t} {10}, \;\; \lambda_{2} = \frac { 2 ( 5 \mu - 2 t ) } {5}. $$

Using the strict inequality from C1, ${p_{d}^{s2} + \mu < R}$, we get R > 3t/5, which contradicts our market coverage condition R < t/2. Therefore, in State 2, when the market is not covered, C2 is never binding.

Constraints C1 and C2 are binding. Following the same analysis as what we have in proof of Lemma 2, we get candidate Nash equilibrium prices and the shadow prices for C1 and C2:

$$ p_{a}^{s2} = \frac {3R-t} {2}, \;\; p_{d}^{s2} = R - \mu, \;\; p_{b}^{s2} = \frac {R} {2}, \;\; \lambda_{1} = \frac { 3 t - 5 R } {t}, \;\; \lambda_{2} = 2 ( \mu + R - t ). $$

From λ₂ ≥ 0, we get ${R \geq t - \mu}$, which is contradictory to our market coverage condition R < t/2, since ${t/2 < t - \mu}$, because of our market comparison condition. Therefore, in State 2, when the market is not covered, C1 and C2 are never binding simultaneously.

Our second step is to verify that the Kuhn-Tucker conditions are also the sufficient condition for the maximization problem. It is easy to verify that both the objective functions are concave (the Hessian matrix of the objective function of retailer A is negative definite, and the second derivative of the objective function of retailer B is negative), and the constraints are linear and therefore concave. Thus, what we derive above is the optimal solution to the constrained profit maximization problem.

Unlike Lemma 2 for the covered market, retailer A always chooses to sell in the Internet channel in the uncovered market. Similar as in the proof of Lemma 2, it is easy to verify that in our case, the number of players is finite (G1), the constraints insure that the strategy space of every player is compact and convex (G2), the payoff functions are continuous and bounded (G3), and the payoff functions are concave and thus quasi-concave (G4). Therefore, the optimal solution to the constrained maximization problem is indeed the unique Nash equilibrium. □

Proof of Lemma 7

From (4) we can write the Lagrangian function for the new entrant as

$$ \begin{aligned} L & = 2p_{d}^{s3} \left(\frac{1}{2}-\frac {p_{d}^{s3} - p_{a}^{s3} + \mu}{t}-\frac {p_{d}^{s3}-p_{b}^{s3}+\mu}{t}\right) \\ & \quad - \lambda_{1} (p_{d}^{s3}+\mu-R) - \lambda_{2} \left(\frac {p_{d}^{s3} - p_{a}^{s3} + \mu}{t} + \frac {p_{d}^{s3}-p_{b}^{s3}+\mu}{t} - \frac {1}{2}\right). \end{aligned} $$

The resulting Kuhn-Tucker conditions are

$$ \frac {\partial L}{\partial p_{d}^{s3}} = \frac {2}{t}(p_{a}^{s3}+p_{b}^{s3}-4p_{d}^{s3}-2\mu) + 1 - \lambda_{1} - \frac {2\lambda_{2}} {t} \leq 0, $$

$$ \frac {\partial L}{\partial \lambda_{1}} = - p_{d}^{s3} - \mu + R \geq 0, $$

$$ \frac {\partial L}{\partial \lambda_{2}} = - x - y + 1/2 \geq 0, $$

$$ p_{d}^{s3} \geq 0, \;\; \lambda_{1} \geq 0, \;\; \lambda_{2} \geq 0, $$

$$ p_{d}^{s3} \frac {\partial L}{\partial p_{d}^{s3}} = 0, \;\; \lambda_{1} \frac {\partial L}{\partial \lambda_{1}} = 0, \;\; \lambda_{2} \frac {\partial L}{\partial \lambda_{2}} = 0. $$

From (5) the Kuhn-Tucker conditions for retailer A are

$$ \frac {\partial \pi_{a}^{s3}}{\partial p_{a}^{s3}} = \frac{2}{t} (p_{d}^{s3}-2p_{a}^{s3}+\mu) \leq 0, $$

$$ p_{a}^{s3} \geq 0, \;\; p_{a}^{s3} \frac {\partial \pi_{a}^{s3}}{\partial p_{a}^{s3}} = 0. $$

From (6) the Kuhn-Tucker conditions for retailer B are

$$ \frac {\partial \pi_{b}^{s3}}{\partial p_{b}^{s3}} = \frac{2}{t} (p_{d}^{s3}-2p_{b}^{s3}+\mu) \leq 0, $$

$$ p_{b}^{s3} \geq 0, \;\; p_{b}^{s3} \frac {\partial \pi_{b}^{s3}}{\partial p_{b}^{s3}} = 0. $$

Excluding the non-negativity constraints, there are the two possible binding constraints for the new entrant’s problem, giving rise to four cases which we discuss in turn.

Interior solution. Following similar analysis as proof of Lemma 2, we get candidate Nash equilibrium prices:

$$ p_{a}^{s3} = p_{b}^{s3} = \frac {t + 4 \mu} {12}, \;\; p_{d}^{s3} = \frac {t - 2 \mu} {6}. $$

From the market comparison condition, ${\mu/t < 1/2}$ , the candidate Nash equilibrium prices are strictly positive. Using the strict inequalities from our constraints, ${p_{d}^{s3} + \mu < R}$ and x + y < 1/2, and our market coverage condition R < t/2, we derive the additional conditions on ${\mu/t}$ and R in Lemma 7.

Constraint C1 is binding. Following similar analysis as proof of Lemma 2, we get candidate Nash equilibrium prices and the shadow price for C1:

$$ p_{a}^{s3} = p_{b}^{s3} = R/2, \;\; p_{d}^{s3} = R - \mu, \;\; \lambda_{1} = \frac {t - 2 (3R - 2 \mu)} {t}. $$

From ${R>\mu}$, the candidate Nash equilibrium prices are strictly positive. Using the strict inequality from C2, x + y < 1/2, our market coverage condition, R < t/2, and the non-negative shadow price for C1, λ₁ ≥ 0, we derive the additional conditions on ${\mu/t}$ and R in Lemma 7.

Constraint C2 is binding. Following similar analysis as proof of Lemma 2, we get candidate Nash equilibrium prices and the shadow price for C2:

$$ p_{a}^{s3} = p_{b}^{s3} = \frac {t} {4}, \;\; p_{d}^{s3} = \frac {t - 2\mu} {2}, \;\; \lambda_{2} = 2 \mu - t. $$

From the non-negative shadow price for C2, λ₂ ≥ 0, we get ${\mu/t \geq 1/2}$, which is contradictory to our market comparison condition. Therefore, in State 3, when the market is not covered, C2 is never binding.

Constraints C1 and C2 are binding. Directly following from the binding constraint C1, ${p_{d}^{s3} = R - \mu}$. Plugging p ^s3 _d into ${\partial \pi_{a}^{s3} / \partial p_{a}^{s3} = \partial \pi_{b}^{s3} / \partial p_{b}^{s3} = 0}$, we get p ^s3 _a = p ^s3 _b = R/2. Plugging p ^s3 _a , p ^s3 _b , and p ^s3 _d into the binding constraint C2, x + y = 1/2, we get R = t/2, which contradicts our market coverage condition, R < t/2. Therefore, in State 3, when the market is not covered, C1 and C2 are never binding simultaneously.

It is easy to verify that all the objective functions are concave (the second derivative is negative), and the constraints are linear and therefore concave. Thus, what we derive above is the optimal solution to the constrained profit maximization problem. Similar as in the proof of Lemma 2, it is easy to verify that in our case, the number of players is finite (G1), the constraints insure that the strategy space of every player is compact and convex (G2), the payoff functions are continuous and bounded (G3), and the payoff functions are concave and thus quasi-concave (G4). Therefore, the optimal solution to the constrained maximization problem is indeed the unique Nash equilibrium. □

Proof of Corollary 2

From Lemma 5 and Lemma 8, ${ \pi_{a}^{s1} = \pi_{b}^{s1} = R^{2}/2t \geq \pi_{a}^{s4} = \pi_{b}^{s4} = \mu^{2}/2t }$, because ${R > \mu}$. Proof of dominance of State 4 by State 2 and State 3 is similar to that in Corollary 1. □

Proof of Theorem 3

Same as in the covered market, we summarize the equilibrium profits under different parameter ranges in States 1 through 3 in Table 4. In this proof, row numbers refer to Table 4.

Table 4 Equilibrium profits when the market is not covered

• Row 1: If ${ \mu / t \in ( 0, 1/4 ) }$ and ${ R \in [ \mu, ( t + 4 \mu ) / 6 ]}$,

  $$ \begin{aligned} ( \pi_{a}^{s2} - \pi_{a}^{s1} ) - \pi_{e}^{s3} & = \frac{1}{2t} ( \mu^{2} + 2 \mu R - 2 R^{2} ) + R - \mu - \frac{R^{2}}{2t} - \frac {(R - \mu)(t - 2 R)} {t} \\ & = \frac { ( R - \mu ) ^{2} } { 2t }\\ & > 0. \end{aligned} $$

• Row 2: If ${\mu/t \in ( 0, 1/4 )}$ and ${R \in ( ( t + 4 \mu) / 6, ( t + 2 \mu )/3]}$,

  $$ (\pi_{a}^{s2} - \pi_{a}^{s1} ) - \pi_{e}^{s3} = \frac{1}{2t} ( \mu^{2} + 2 \mu R - 2R^{2} ) + R - \mu - \frac{R ^ {2}}{2t} - \frac{(t-2\mu)^{2}}{9t} = \frac {1}{18t} (-27 R^{2} + 18 \mu R + 18tR + \mu ^{2} - 10 \mu t - 2 t^{2} ) = \frac {1}{18t} ( -27 ( R - \frac {1}{3} ( \mu + t) ) ^ {2} + ( t - 2 \mu ) ^ {2} ) \stackrel { ( t + 4 \mu ) / 6 < R < t/2 } {>} 0.$$

• Row 3: If ${ \mu / t \in ( 0, 1/4 ) }$ and ${ R \in ( ( t + 2 \mu ) / 3, t/2 ) }$,

  $$ \begin{aligned} ( \pi_{a}^{s2} - \pi_{a}^{s1} ) - \pi_{e}^{s3} & = \frac{( 13 \mu^{2} - 8 \mu t + 4t^{2})}{18t} - \frac{R^{2}}{2t} - \frac{(t-2\mu)^{2}}{9t}\\ & = \frac {1}{18t} ( 5 \mu ^ {2} + 2 t ^ { 2 } - 9 R ^ { 2} ). \end{aligned} $$

  Since ${R \in ( ( t + 2 \mu ) / 3, t/2 ) }$, if ${ \mu / t \leq 1 / \sqrt{20}}$, then ${ ( t + 2 \mu ) / 3 < \sqrt { 5 \mu ^ {2} + 2 t ^ {2} } / 3 \leq t/2}$, so if ${ R \geq \sqrt { 5 \mu ^ {2} + 2 t ^ {2} } / 3 }$, then ${ ( \pi_{a}^{s2} - \pi_{a}^{s1} ) - \pi_{e}^{s3} \leq 0 }$, and if ${ R < \sqrt { 5 \mu ^ {2} + 2 t ^ {2} } / 3 }$, then ${ ( \pi_{a}^{s2} - \pi_{a}^{s1} ) - \pi_{e}^{s3} > 0 }$. If ${ \mu / t > 1 / \sqrt {20} }$, then ${ \sqrt { 5 \mu ^ {2} + 2 t ^ {2} } / 3 > t/2 > R }$, and ${ ( \pi_{a}^{s2} - \pi_{a}^{s1} ) - \pi_{e}^{s3} > 0 }$.

• Row 4: If ${ \mu / t \in [ 1/4, 1/2 ) }$ and ${ R \in [ \mu, ( t + 4 \mu ) / 6 ] }$, the analysis is the same as that of Row 1.

• Row 5: If ${ \mu / t \in [ 1/4, 1/2 ) }$ and ${ R \in ( ( t + 4 \mu ) / 6, t/2 )}$, the analysis is the same as that of Row 2. □

Proof of Theorem 4

Again, in this proof, row numbers refer to Table 4.

• Row 1: If ${\mu / t \in ( 0, 1/4 )}$ and ${R \in [ \mu, ( t + 4 \mu ) / 6 ]}$,

  $$ \begin{array}{ll} ( \pi_{a}^{s2} - \pi_{a}^{s3} ) - &\pi_{e}^{s3} = ( \pi_{a}^{s2} - \pi_{a}^{s1} ) - \pi_{e}^{s3} \\ & \stackrel{{\rm Theorem\;3}}{>}0. \end{array} $$

• Row 2: If ${ \mu / t \in ( 0, 1/4 ) }$ and ${ R \in ( ( t + 4 \mu ) / 6, ( t + 2 \mu ) / 3 ] }$,

  $$ \begin{array}{ll} ( \pi_{a}^{s2} - \pi_{a}^{s3} ) - \pi_{e}^{s3} &\stackrel{ R > ( t + 4 \mu ) / 6 }{>} ( \pi_{a}^{s2} - \pi_{a}^{s1} ) - \pi_{e}^{s3} \cr & \stackrel{\rm Theorem\;3}{>}0. \end{array} $$

• Row 3: If ${ \mu / t \in ( 0, 1/4 ) }$ and ${ R \in ( ( t + 2 \mu ) / 3, t/2 ) }$, the analysis is the same as that of Row 1 in the proof of Theorem 2.

• Row 4: If ${\mu / t \in [ 1/4, 1/2 )}$ and ${R \in [ \mu, ( t + 4 \mu ) / 6 ]}$, the analysis is the same as that of Row 1.

• Row 5: If ${\mu / t \in [ 1/4, 1/2 )}$ and ${R \in ( ( t + 4 \mu ) / 6, t/2 )}$, the analysis is the same as that of Row 2. □

Proof of Corollary 3

The minimum entry cost advantage for the entrant is given by:

$$ K_{a}(T) - K_{e}(T) > \frac{e^{ -r T }}{r} [ ( \pi_{a}^{s2} - \pi_{a}^{s3} ) - \pi_{e}^{s3} ]. $$

So if ${ ( \pi_{a}^{s2} - \pi_{a}^{s3} ) - \pi_{e}^{s3} }$ is smaller in the uncovered market than in the covered market, we can prove that K _a (T) − K _e (T) requires to be smaller in the uncovered market than in the covered market. Note that ${ ( \pi_{a}^{s2} - \pi_{a}^{s3} ) - \pi_{e}^{s3} }$ is actually the difference between retailer A’s preemption incentive and the new entrant’s. The following proves that ${ ( \pi_{a}^{s2} - \pi_{a}^{s3} ) - \pi_{e}^{s3} }$ is smaller in the uncovered market than in the covered market.

In the covered market (R ≥ t/2):

If ${\mu / t \in ( 0, 1/4 )}$,

$$ ( \pi_{a}^{s2} - \pi_{a}^{s3} ) - \pi_{e}^{s3} = \left\{ \begin{array}{ll} \frac { 1 } { 72t } ( 4 ( \mu - t ) ^ {2} + 3 t^{2} ) & R \in [ \frac{t}{2}, \frac{52 \mu^{2} - 32 \mu t + 25 t^{2} }{36t} ) \\ \frac{1}{2}(R - \frac{t}{2}) & R \in [ \frac{52 \mu^{2} - 32 \mu t + 25 t^{2} }{36t}, \frac{3t}{4} ) \\ \frac{ \mu (t - 2\mu) }{3t} + \frac{t}{8} & R \in ( \frac{3t}{4}, \infty ) \end{array} \right. $$

It can be shown that as R varies, ${( \pi_{a}^{s2} - \pi_{a}^{s1} )-\pi_{e}^{s3}}$ is non-decreasing. So the minimum ${ ( \pi_{a}^{s2} - \pi_{a}^{s1} ) - \pi_{e}^{s3}}$ takes is ${\frac { 1 } { 72t } ( 4 ( \mu - t ) ^ {2} + 3 t^{2} )}.$

If ${ \mu / t \in [ 1/4, \frac{14 - 3\sqrt{3}}{26})}$ ,

$$ (\pi_{a}^{s2} - \pi_{a}^{s3} ) - \pi_{e}^{s3} = \left\{ \begin{array}{ll} \frac {1}{24t} ( - 24 ( R - \frac { \mu + t }{2} ) ^ {2} + 2 \mu ^ {2} - 4 \mu t + 3 t^{2} ) & R \in [ t/2, ( t + 2 \mu ) / 3 ] \\ \frac { 1 } { 72t } ( 4 ( \mu - t ) ^ {2} + 3 t^{2} ) & R \in [\frac {t + 2\mu} {3}, \frac{52 \mu^{2} - 32 \mu t + 25 t^{2} }{36t} ) \\ \frac{1}{2}(R - \frac{t}{2}) & R \in [ \frac{52 \mu^{2} - 32 \mu t + 25 t^{2} }{36t}, \frac{3t}{4} ) \\ \frac{ \mu (t - 2\mu) }{3t} + \frac{t}{8} & R \in ( \frac{3t}{4}, \infty ) \end{array} \right. $$

It can be shown that as R varies, ${ ( \pi_{a}^{s2} - \pi_{a}^{s1} ) - \pi_{e}^{s3} }$ is non-decreasing. So the minimum ${ ( \pi_{a}^{s2} - \pi_{a}^{s1} ) - \pi_{e}^{s3} }$ takes is ${ \frac { 1 } { 24t } ( -4 \mu^{2} - 4 \mu t + 3 t^{2})}.$

If ${\mu / t \in [\frac{14 - 3\sqrt{3}}{26}, 1/2 )}$ ,

$$ ( \pi_{a}^{s2} - \pi_{a}^{s3} ) - \pi_{e}^{s3} = \left\{ \begin{array}{ll} \frac {1}{24t} ( - 24 ( R - \frac { \mu + t }{2} ) ^ {2} + 2 \mu ^ {2} - 4 \mu t + 3 t^{2} ) & R \in [ t/2, \frac{(\sqrt{3} + 1) t - 2 (\sqrt{3} - 1) \mu }{4} ) \\ \frac{1}{2}(R - \frac{t}{2}) & R \in [\frac{(\sqrt{3} + 1) t - 2 (\sqrt{3} - 1) \mu }{4}, \frac{3t}{4} ) \\ \frac{ \mu (t - 2\mu) }{3t} + \frac{t}{8} & R \in ( \frac{3t}{4}, \infty ) \end{array} \right. $$

It can be shown that as R varies, ${ ( \pi_{a}^{s2} - \pi_{a}^{s1} )-\pi_{e}^{s3} }$ is non-decreasing. So the minimum ${(\pi_{a}^{s2} -\pi_{a}^{s1} ) - \pi_{e}^{s3}}$ takes is ${\frac{1 } { 24t } ( -4 \mu^{2} - 4 \mu t + 3 t^{2})}.$

In the uncovered market ${(\mu \leq R < t/2)}:$

If ${\mu / t \in ( 0, 1/4 )}$,

$$ ( \pi_{a}^{s2} - \pi_{a}^{s3} ) - \pi_{e}^{s3} = \left\{ \begin{array}{ll} \frac { ( R - \mu ) ^{2} } { 2t } & R \in [ \mu, ( t + 4 \mu ) / 6 ] \\ \frac {1}{24t} ( - 24 ( R - \frac { \mu + t }{2} ) ^ {2} + 2 \mu ^ {2} - 4 \mu t + 3 t^{2} ) & R \in ( ( t + 4 \mu ) / 6, ( t + 2 \mu ) / 3 ] \\ \frac { 1 } { 72t } ( 4 ( \mu - t ) ^ {2} + 3 t^{2} ) & R \in ( ( t + 2 \mu ) / 3, t/2 ) \end{array} \right. $$

It can be shown that as R varies, ${ ( \pi_{a}^{s2} - \pi_{a}^{s1} ) - \pi_{e}^{s3} }$ is non-decreasing. So the maximum ${ ( \pi_{a}^{s2} - \pi_{a}^{s1} ) - \pi_{e}^{s3} }$ takes is ${ \frac { 1 } { 72t } ( 4 ( \mu - t ) ^ {2} + 3 t^{2} )}.$

If ${ \mu / t \in [ 1/4, 1/2 ) }$,

$$ ( \pi_{a}^{s2} - \pi_{a}^{s3} ) - \pi_{e}^{s3} = \left\{ \begin{array}{ll} \frac { ( R - \mu ) ^{2} } { 2t } & R \in [ \mu, ( t + 4 \mu ) / 6 ] \\ \frac {1}{24t} ( - 24 ( R - \frac { \mu + t }{2} ) ^ {2} + 2 \mu ^ {2} - 4 \mu t + 3 t^{2} ) & R \in ( ( t + 4 \mu ) / 6, t/2 ] \end{array} \right. $$

It can be shown that as R varies, ${( \pi_{a}^{s2} - \pi_{a}^{s1} ) - \pi_{e}^{s3}}$ is increasing. So the maximum ${ ( \pi_{a}^{s2} - \pi_{a}^{s1} ) - \pi_{e}^{s3}}$ takes is ${\frac { 1 } { 24t } (-4 \mu^{2} - 4 \mu t + 3 t^{2})}.$

By comparing the minimum that ${ ( \pi_{a}^{s2} - \pi_{a}^{s1} ) - \pi_{e}^{s3} }$ takes in the covered market and the maximum in the uncovered market, it is easy to show that ${ ( \pi_{a}^{s2} - \pi_{a}^{s1} ) - \pi_{e}^{s3}}$ is smaller in the uncovered market than in the covered market. □