Online Shopping Intermediaries: The Strategic Design of Search Environments

Transcription

1 Online Supplemental Appendix to Online Shopping Intermediaries: The Strategic Design of Search Environments Anthony Dukes University of Southern California Lin Liu University of Central Florida February 2, 2015 In this online supplemental appendix, we provide arguments for two claims made in the main article. SA.1. Evaluation Costs Quadratic in Breadth The consumer s evaluation cost function in section 2 specified that search costs are linear in the breadth but quadratic in depth. We can show that the main result in Proposition 1 can continue to hold even for a search cost specification that is quadratic in both dimensions. Proposition SA.1 Suppose the consumer s evaluations cost is, ;. Let \. If then 1 1. Proof: Let 0,1 be an arbitrary search environment and the consumer s evaluation objective be a modified version of (B1). Then the consumer s optimal evaluation plan is if if and if 1 if where 1. Equilibrium prices are clearly decreasing in and therefore cannot exceed. When, equilibrium prices increasing in. This implies that the optimal level of search aids. are

2 SA.2. Competing Intermediaries The two intermediaries are located at the two ends of a Hotelling line. We show that the main result in Proposition 1 remains when the transportation cost () is sufficiently large. However, when the transportation cost is small, we show that it is optimal for the intermediary to provide full aids ( 1). Proposition SA. 2 (i) When intermediaries are relatively differentiated ( ln ), the equilibrium outcomes are the same as in Proposition 1. (ii) Otherwise, in equilibrium, the intermediary minimizes search costs in the search environment 1. The symmetric equilibrium price is. Proof: We claim that there is a symmetric equilibrium in which both intermediaries set 1 1 0, where ln. To prove this claim we demonstrate directly that one intermediary, intermediary 1, cannot be more profitable by deviating from given that the other intermediary, intermediary 2, chooses. (i) Suppose and consider any deviation 1, with the corresponding profits denoted by. Any deviation 0, 1 leads to profits, which we show is increasing on this interval. Specifically, 0 as long as for all. We have,. ln where the first inequality holds by assumption, the second since, and the third for 0, 1. Therefore, any deviation 1 is not profitable.. Any deviation 1,1 leads to ln.

3 This deviation is not profitable if 0, which requires ln Since ln 1, profits are decreasing near (and to the right of). Note that the function is strictly increasing in. Therefore, the condition at 1, the right endpoint of the interval, is sufficient for at any 1,1. This condition is. ln 2 1, which holds by our assumption ln Any deviation 1,1 leads to a profit of ln 1., which is increasing in and therefore bounded above by 1. The condition ln directly implies that ln 1, for all 1,1. (ii) Now suppose ln 0,1 e. This leads to a deviation profit of. We first consider any deviation ln. Characterizing the shape of this deviation profit function depends on the level of. We argue that for different three levels of. For 0ln, the expression for the demand at intermediary 1, ln 0. So any deviation under this condition is not profitable. For ln ln, the derivative / has the following property:

4 0. Since the derivative is continuous, it means that any maximizer,, of in 0,1 e must solve 0. This solution is expressed 1 1 ln and leads to profits ln 1 ln. Under the condition that ln, /, where the last term is the profit the intermediary earns by sticking to 1. Hence, no deviation 0,1 is profitable. For [ln 0,1 e. Thus, ln, we have 0 for all 1 ln, for all 0,1. Hence, for any ln deviation for any 0,1., there is no profitable Now consider deviations 1,1 when 1. Intermediary 1 s profit is given by It can be shown that 0 ln ln where 0 is increasing on 1,1. Suppose 0 1,1. Thus, 1ln. ln. ln1. Then 0 for all 1 1

5 for all 1,1. However, by choosing 1, intermediary 1 earns which exceeds 1. Suppose ln ln /,. Then 0 near 1. In this case, is bounded by either 1 or 1. We know from above that both of these values are exceeded by the profit. Hence there is no profitable deviation 1,1. Finally consider any deviation 1,1. This leads to profits given by 1, which is obviously increasing in. Therefore, choosing 1 gives intermediary 1 more profit than any in 1,1.