Does collusive advertising facilitate collusive pricing? Evidence from experimental duopolies

Abstract

This article analyzes experimentally whether the degree of collusion for one dimension of duopolists’ interactions influences the degree of collusion for another dimension. More precisely, I will explore whether a high degree of collusion for advertisement expenditures facilitates tacit price collusion. Two environments are tested, in which the size of the spillover between advertising expenditures is varied. The results indicate that both degrees of collusion are correlated: a high degree of collusion on advertising functions as a signalling device triggering a significantly higher degree of price collusion by the opponent. Thus advertising expenditures seem to be a useful indicator for market regulators to detect non-competitive pricing.

Appendices

Author’s translation of the German instructions for the low condition

Appendix A: Instructions

Author’s translation of the German instructions for the low condition

Thank you for participating in our experiment. We kindly ask you to refrain from any public statements and attempts to communicate directly with other participants. If you violate this rule, we have to exclude you from the experiment. If you have any questions, please raise your hand, and one of the persons who runs the experiment will come to your place and answer your questions. Please read these instructions carefully. In this experiment, you will earn money based on repeated decisions. How much you will earn depends on your decisions as well as the decisions of another participant.

You will repeatedly interact with another, anonymous participant for 60 periods. The instructions are identical for all participants. The other participant will be randomly assigned to you and will remain with you for the entire experiment.

In this experiment, you as well as the other participant have to sell a product on a market. Your profit equals the number of sold entities of your product multiplied by the price, minus production costs. You have to decide on the price of the product. The price can range between 0 and 25 ECU, with 0.01 as the smallest incremental step. The higher the price you choose, the smaller the number of sold entities per period. Whenever the price exceeds a certain level, you cannot sell any entity at all. However, the number of sold entities increases if your competitor increases the price for her product. In summary, your number of sold entities (q _own ) equals

$$ q_{own}=6-p_{own}+0.5p_{other}, $$

where p _own denotes the price you choose and p _other the price the other participant chooses. Note that for each entity you sell, there are production costs of 1 ECU. Therefore, your profit (G _own ) is

$$ G_{own}=q_{own}\times p_{own}-1\times q_{own}. $$

In every fifth period, you have the opportunity to invest in your product. We denote your investments as m _own . Investments can range between 0 and 9, with 0.01 as the smallest incremental step. Although you can only change your investments in every fifth period, they increase the number of sold entities in every period. Additionally, the investments of the other participant increase the number of sold entities of your product. The number of sold entities of your product (q _own ) equals

$$ q_{own}=6+m_{own}+0.3m_{other}-p_{own}+0.5p_{other}, $$

where m _other denotes the investments of the other participant.²⁷ However, in every period, investments also cost 2(m _own )² (not only in every fifth period when you can change them). You participate in the investments of the other participant without any costs. Thus, your profit equals

$$ G_{own}=q_{own}\times p_{own}-1\times q_{own}-2(m_{own})^{2}. $$

Please consider that you may accumulate losses due to unfavorable investment choices. If you earn a negative total profit throughout the entire experiment, you will be asked to pay back this amount by doing clerical work at our institute (120 ECU = 1 h). If you do not accept this rule, please leave the experiment now.

In every first out of five periods (i.e., in periods 1, 6, 11,...), you will be asked to specify your investments. You will not be asked for this in the subsequent 4 periods. Please note that the level of investments cannot be changed for these 4 periods, though you have to carry the costs for them. In each period, you are then informed on the level of investments you chose as well as that chosen by the other participant. Additionally, you are informed on the investments and prices chosen in the previous period. Then you have to choose the price for your product in this period. Finally, we inform you on your profit in this period, the profit of the other participant, and the accumulated profits. At the end of the experiment, we will exchange all ECUs you earned in the 60 periods at a rate of 120 ECU = 1 Euro.

Before the first round starts, we will ask you several questions concerning the rules of this experiment in a questionnaire. Please answer them correctly. One of the persons who runs the experiment will come to your place and clarify incorrect answers.

Appendix B: Proofs of propositions 1–4

Of course, Proposition 1 is simple to prove. The first derivative of π _i with respect to p _i leads to i’s best-response function $2^{-1}[A^{t}_{i}+\alpha p_{-i}+c]=p_{i}$. Applying the corresponding first derivative of π_−i with respect to p _−i leads to −i’s best-response function. Applying simple math yields $p^{N}_{i}$.

Again, the proof of Proposition 2 is simple, but demanding in notation. Partial derivation of π _i with respect to m _i defines i’s best-response function implicitly as $(2k)^{-1}[(p_{i}-c)A^{\prime t}_{i} +(A^{t}_{i}-2p_{i}+\alpha p_{-i}+c)p^{\prime}_{i}-\alpha cp^{\prime}_{-i}]=m_{i}$, where $A^{\prime t}_{i}$ ($p^{\prime}_{i}$) denotes the first derivative of $A^{t}_{i}$ (p _i ) with respect to m _i . I substitute on the left hand side p _i and p _−i ($A_{i}^{t}$) by investments according to Eq. 4 (by Eq. 2) and first derivatives of prices according to first derivatives of Eq. 4 with respect to m _i : $p^{\prime}_{i}=(2+\alpha\beta)/(4-\alpha^{2})$ and $p^{\prime}_{-i}=(2\beta+\alpha)/(4-\alpha^{2})$. Finally, the first derivative of Eq. 2 with respect to m _i , $A^{\prime t}_{i}=1$, helps us to obtain the explicit best-response function

$$ m_{i}=\frac{\gamma_{1}A_{0}+\gamma_{3}m_{-i}+\gamma_{4}c}{1-\gamma_{2}} $$

with $\gamma_{1}=\frac{2+\alpha}{8k-2k\alpha^{2}}$, $\gamma_{2}=\frac{2+\alpha\beta}{8k-2k\alpha^{2}}$ $\gamma_{3}=\frac{2\beta+\alpha}{8k-2k\alpha^{2}}$, and $\gamma_{4}=\frac{2\alpha^{3}\beta-8\alpha\beta-\alpha^{3}+4\alpha-8+2\alpha^{2}} {2k(4-\alpha^{2})^{2}}$. Applying the corresponding technique to π_−i leads to −i’s best-response function. Simple math leads to

$$ m^{N}=\frac{(\gamma_{2}-1-\gamma_{3})(\gamma_{1}A_{0}+\gamma_{4}c)} {\gamma_{3}^{2}-\gamma_{2}^{2}+2\gamma_{2}-1}. $$

Concerning the proof of Proposition 3, the first derivative of π _i  + π_−i with respect to p _i leads to i’s best-response function $2^{-1}[A^{t}_{i}+2\alpha p_{-i}+(1-\alpha)c]=p_{i}$. Applying the corresponding technique to the first deviation of π _i  + π_−i with respect to p _−i yields −i’s best-response function. Again, simple math yields $p^{C}_{i}$.

Finally, the proof of Proposition 4 is simple, but demanding in notation. Partial derivation of π _i  + π_−i with respect to m _i defines i’s best-response function implicitly as $(2k)^{-1}[(A^{\prime t}_{i}-2p^{\prime}_{i}+2\alpha p^{\prime}_{-i})p_{i} +p^{\prime}_{i}A^{t}_{i}+(A^{\prime t}_{-i}-2p^{\prime}_{-i}+2\alpha p^{\prime}_{i})p_{-i}+p^{\prime}_{-i}A^{t}_{-i}+((1-\alpha)(p^{\prime}_{i} +p^{\prime}_{-i})-A^{\prime t}_{i} -A^{\prime t}_{-i})c]=m_{i}$. I substitute on the left hand side p _i and p _−i (A ^t _i ) by investments according to Eq. 5 (by Eq. 2) and first derivatives of prices according to first derivatives of Eq. 5 with respect to m _i : $p^{\prime}_{i}=(1+\alpha\beta)/(2-2\alpha^{2})$ and $p^{\prime}_{-i}=(\beta+\alpha)/(2-2\alpha^{2})$. Finally, the first derivative of Eq. 2 with respect to m _i , $A^{\prime t}_{i}=1$ and $A^{\prime t}_{-i}=\beta$, helps us to obtain the explicit best-response function

$$ m_{i}=\frac{(\gamma_{5}+\gamma_{6})A_{0}+(\beta\gamma_{5}+\gamma_{6})m_{-i}+\gamma_{7}c} {1+\gamma_{5}-\beta\gamma_{6}} $$

with $\gamma_{5}=\frac{1+\alpha\beta}{4k-4k\alpha^{2}}$, $\gamma_{6}=\frac{\alpha+\beta}{4k-4k\alpha^{2}}$, $\gamma_{7}=\frac{\alpha^{2}(1+\beta)-\beta-1}{4k-4k\alpha^{2}}$. Applying the corresponding technique to π _i  + π_−i with respect to m _−i leads to −i’s best-response function. Simple math leads to

$$ m^{C}=\frac{(\beta\gamma_{5}+\gamma_{6}+1-\gamma_{5}+\beta\gamma_{6})((\gamma_{5}+\gamma_{6})A_{0}+\gamma_{7}c)} {(1-\gamma_{5}+\beta\gamma_{6})^{2}-(\beta\gamma_{5}+\gamma_{6})^{2}}. $$