Information Acquisition Behavior: An Evolutionary Game Theory Perspective

Jialu Li; Meiying Yang; Wei Xing; Xuan Zhao

doi:10.1007/s13235-017-0224-8

Dynamic Games and Applications

June 2018, Volume 8, Issue 2, pp 434–455 | Cite as

Information Acquisition Behavior: An Evolutionary Game Theory Perspective

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Jialu LiEmail author
Meiying Yang
Wei Xing
Xuan Zhao

Article

First Online: 29 May 2017

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Abstract

This paper explores the retailers’ information acquisition behavior under horizontal competition and investigates the impact of information leakage from an evolutionary game theory perspective. When information leakage does not occur, we find that firms’ acquisition behavior differ under different types of competition. Under Cournot competition, rivals prefer the competitor not to acquire information. Under Bertrand competition, however, rivals prefer the competitor to acquire information. As a result, at a moderate level of acquisition cost mixed strategies (acquire and not acquire) coexist under Cournot competition, whereas all retailers adopt the same (pure) strategy under Bertrand competition. With information leakage, two effects (collaborative forecasting and free-riding) influence firms’ decision making. When the collaborative forecasting effect dominates, all retailers would choose the acquisition strategy in the equilibrium. However, if the free-riding effect is strong, no retailer would acquire information in the equilibrium. With a moderate acquisition cost, there are still retailers willing to acquire information and mixed strategies coexist, regardless of the type of competition. Interestingly, no matter information is leaked or not, we find that Cournot competition in general better accommodates information acquisition than Bertrand competition. As a result, Cournot competition can lead to a higher total welfare than Bertrand competition when the information acquisition cost is moderate and the demand uncertainty is large, in contrast to the standard result of duopoly models.

Keywords

Information acquisition Leakage Evolutionary game

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Notes

Acknowledgements

The authors are grateful to the associate editor and the anonymous referees for their constructive feedback. This research is supported by China Scholarship Council (Grant No. 201506020047), National Natural Science Foundation of China (Grant Nos. 71632003, 71471101), Natural Sciences and Engineering Research Council of Canada (Grant No. 312572) and Social Sciences and Humanities Research Council of Canada (Grant No. 00657).

Compliance with Ethical Standards

Conflict of interest

The authors declare that they have no conflict of interest.

APPENDIX

Proof of Lemma 1

See [24]. $\square $

Proof of Proposition 1

From the replicated dynamic Eq. (6), it is easy to find that the equation has three possible stable state at most, ${x^*} = 0$, ${x^*} = 1$, $x^{*}=-\frac{\alpha }{\beta }$. Because ${F}'\left( x \right) = \left( {1 - 2x} \right) \left( {\alpha + \beta x} \right) + \left( {x - {x^2}} \right) \beta $, we have: (1) When ${x^*} = 0$, ${F}'\left( {{x^*}} \right) = \alpha $; (2) When ${x^*} = 1$, ${F}'\left( {{x^*}} \right) = - \left( {\alpha + \beta } \right) $; (3) When $x^{*}=-\frac{\alpha }{\beta }$, ${F}'\left( {{x^*}} \right) = {x^*}\left( {1 - {x^*}} \right) B$. Because ${x^*} \in \left[ {0,1} \right] $ is an stable point if and only if ${F}'\left( {{x^*}} \right) < 0$. Under Cournot competition and if information leakage does not occur, we have the following:

(a)

if $c > \frac{{{\sigma ^\mathrm{{2}}}}}{{\mathrm{{4}}\left( {\mathrm{{1 + }}s} \right) }}$, when ${x^*} = 0$, ${F}'\left( {{x^*}} \right) < 0$. So ${x^*} = 0$ is a stable point; when ${x^*} = 1$, ${F}'\left( {{x^*}} \right) > 0$, so ${x^*} = 1$ is not a stable state; when $x^{*}=-\frac{\alpha }{\beta }$, ${x^*} \notin \left( {0,1} \right) $, so it is not a stable state.
(b)

if $c < \frac{{\left( {1 + s} \right) {\sigma ^\mathrm{{2}}}}}{{{{\left( {\mathrm{{2 + }}\gamma \mathrm{{ + 2}}s} \right) }^2}}}$, when ${x^*} = 0$, ${F}'\left( {{x^*}} \right) > 0$, so ${x^*} = 0$ is not a stable point; when ${x^*} = 1$, ${F}'\left( {{x^*}} \right) < 0$, so ${x^*} = 1$ is a stable point; when $x^{*}=-\frac{\alpha }{\beta }$, ${x^*} \notin \left( {0,1} \right) $, so it is not a stable state.
(c)

if $\frac{{\left( {1 + s} \right) {\sigma ^2}}}{{{{\left( {2 + \gamma + 2s} \right) }^2}}}< c < \frac{{{\sigma ^2}}}{{4\left( {1 + s} \right) }}$, when ${x^*} = 0$, ${F}'\left( {{x^*}} \right) > 0$, So ${x^*} = 0$ is not a stable state; when ${x^*} = 1$, ${F}'\left( {{x^*}} \right) > 0$, so ${x^*} = 1$ is not a stable state; when $x^{*}=-\frac{\alpha }{\beta }$, ${x^*} \in \left( {0,1} \right) $, ${F}'\left( {{x^*}} \right) < 0$, so it is a stable state.

This completes the Proof of Proposition 1. And the Proof of Proposition 2 is similar to the above. $\square $

Proof of Proposition 3

Under Cournot competition and when information leakage occurs, we have the following:

(a)

if $c > \frac{{{\sigma ^2}}}{{4\left( {1 + s} \right) {{\left( {2 + \gamma } \right) }^2}}}$, when ${x^*} = 0$, ${F}'\left( {{x^*}} \right) < 0$. So ${x^*} = 0$ is a stable state; when ${x^*} = 1$, ${F}'\left( {{x^*}} \right) > 0$, so ${x^*} = 1$ is not a stable state; when $x^{*}=-\frac{\alpha }{\beta }$, ${x^*} < 0$, so it is not a stable state.
(b)

if $c < \frac{{s{\sigma ^2}}}{{4\left( {1 + s} \right) \left( {2 + s} \right) {{\left( {2 + \gamma } \right) }^2}}}$, when $x^{*}=0$, ${F}'\left( {{x^*}} \right) > 0$, so ${x^*} = 0$ is not a stable state; when ${x^*} = 1$, ${F}'\left( {{x^*}} \right) < 0$, so ${x^*} = 1$ is a stable state; when $x^{*}=-\frac{\alpha }{\beta }$, ${x^*} > 1$, so it is not a stable state.
(c)

if $\frac{{s{\sigma ^2}}}{{4\left( {1 + s} \right) \left( {2 + s} \right) {{\left( {2 + \gamma } \right) }^2}}}< c < \frac{{{\sigma ^2}}}{{4\left( {1 + s} \right) {{\left( {2 + \gamma } \right) }^2}}}$, when ${x^*} = 0$, ${F}'\left( {{x^*}} \right) > 0$, So ${x^*} = 0$ is not a stable state; when ${x^*} = 1$, $\dot{F}'\left( {{x^*}} \right) > 0$, so ${x^*} = 1$ is not a stable state; when $x^{*}=-\frac{\alpha }{\beta }$, ${F}'\left( {{x^*}} \right) < 0$, so it is a stable state.

This completes the Proof of Proposition 3. And the Proof of Proposition 4 is similar to the above. $\square $

Proof of Theorem 3

(i) When information leakage occurs and $\gamma > \gamma ^{*}$, we have:

(a)

$c < B_{2}^{I}$. ETS for the population under Bertrand and Cournot competition is $T_{B}^{(2)}$ and $T_{C}^{(2)}$, respectively. In this situation, Bertrand is better than Cournot since $T_{B}^{(2)}>T_{C}^{(2)}$.
(b)

$B_{2}^{I}\le c<B_{1}^{I}$. ETS for the population under Bertrand and Cournot competition is ${(p_{B}^{I})^2} \cdot T_B^{(2)} + {\left( {1 - p_{B}^{I}} \right) ^2}T_B^{(0)} + 2p_{B}^{I}\left( {1 - p_{B}^{I}} \right) T_B^{(1)}$ and $T_C^{(2)}$, respectively. It can be proved that ETS under Bertrand competition goes from $T_{B}^{(2)}$ to $T_{B}^{(0)}$ as c changes from $B_{2}^{I}$ to $B_{1}^{I}$ in this situation. Therefore, Cournot competition can give higher total surplus than Bertrand competition if
$$\begin{aligned} T_B^{(0)} < T_C^{(2)}\left( c=B_{1}^{I}\right) . \end{aligned}$$

(A1)
(c)

$B_{1}^{I} \le c<C_{2}^{I}$. ETS for the population under Bertrand and Cournot competition is $T_{B}^{(0)}$ and $T_{C}^{(2)}$, respectively. In this situation, therefore, Cournot competition can give higher total surplus than Bertrand competition if the condition (A1) holds.
(d)

$C_{2}^{I} \le c \le C_{1}^{I}$. ETS for the population under Bertrand and Cournot competition is $T_{B}^{(0)}$ and ${(p_{C}^{I})^2} \cdot T_C^{(2)} + {\left( {1 - p_{C}^{I}} \right) ^2}T_C^{(0)} + 2p_{C}^{I}\left( {1 - p_{C}^{I}} \right) T_C^{(1)}$, respectively. It can be proved that ETS under Cournot competition goes from $T_{C}^{(2)}$ to $T_{C}^{(0)}$ as c changes from $C_{2}^{I}$ to $C_{1}^{I}$ in this situation. Therefore, Cournot competition can give higher total surplus than Bertrand competition if
$$\begin{aligned} T_B^{(0)} < T_C^{(2)}\left( c=C_{2}^{I}\right) . \end{aligned}$$

(A2)
(e)

$c> C_{1}^{I}$. ETS for the population under Bertrand and Cournot competition is $T_{B}^{(0)}$ and $T_{C}^{(0)}$, respectively. In this situation, Bertrand is better than Cournot since $T_{B}^{(0)}>T_{C}^{(0)}$.

It can be verified that the condition (A1) and (A2) hold if $\sigma ^{2}$ is enough large. This completes the proof of part (i). When information leakage does not occur, the required result can be obtained in similar ways of part (i). $\square $

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Authors and Affiliations

Jialu Li
- 1
Email authorView author's OrcID profile
Meiying Yang
- 1
Wei Xing
- 2
Xuan Zhao
- 3

1.School of Economics and ManagementBeihang UniversityBeijingChina
2.School of ManagementQufu Normal UniversityRizhaoChina
3.Lazaridis School of Business and EconomicsWilfrid Laurier UniversityWaterlooCanada

Information Acquisition Behavior: An Evolutionary Game Theory Perspective

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