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Cheating in Ranking Systems

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Abstract

Consider a software application that pays a commission fee to be sold on an on-line platform (e.g., Google Play). The sales depend on the applications’ customer rankings. Therefore, developers have an incentive to dishonestly promote their application’s ranking, e.g., by faking positive customer reviews. The platform detects dishonest behavior (cheating) with some probability, and then decides whether to ban the application. We provide an analysis and find the equilibrium behaviors of both the application (cheat or not) and the platform (setting the commission fee). We provide insights into how the platform’s detection accuracy affects the incentives of the application’s developers.

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Notes

  1. Statista: The Statistics Portal https://www.statista.com/statistics/276623/number-of-apps-available-in-leading-app-stores/.

  2. An application is an experience good. The users assume that high-rated apps are ones with which other users have had a positive experience. If an app receives a high rank by cheating, the user might be disappointed from his experience with the app.

  3. The platform maximizes its expected utility. The technicalities are characterized in Proposition 3 in the “Appendix”.

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Acknowledgements

We wish to thank Christopher Thomas Ryan, Yair Tauman, Richard Zeckhauser, the journal editor—Lawrence J. White—and the anonymous reviewers for their helpful suggestions.

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Correspondence to Artyom Jelnov.

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Appendix

Appendix

Proof of Proposition 1

Consider first an equilibrium where P chooses \(\bar{r}=1\). Then c is a superior action of A. If P chooses \({\hat{b}}\), its payoff is therefore \(\gamma f r-w\). If P chooses b, its payoff is 0; and the platform prefers \({\hat{b}}\) to b for each \(d \le r \le 1\) if \(w<\gamma f d\).

Next, consider \(\bar{r}=d\). Observe, that for \(\alpha >0\), pure \({\hat{c}}\) is not an equilibrium in this case. By contrary, if A chooses pure \({\hat{c}}\), with positive probability it obtains a rating above d, and a false signal s is sent to P; therefore, b is not the best reply of the platform.

If A chooses c and P (following s) chooses pure b, A’s expected utility is \(\frac{\gamma (1-f)\beta (1+d)}{2}\). If A does not cheat and if P (following s and \(r>d\)) chooses pure b, A’s expected utility is \(\frac{\gamma (1-f)}{2}[d+(1-\alpha )(1+d)]\). Thus, A prefers c to \({\hat{c}}\) for \(\beta >\frac{d+(1-\alpha )(1+d)}{1+d}\). If A cheats with certainty, P prefers to ban for each \(d \le r \le 1\) if \(\gamma f <w\).

Consider next \(d<\bar{r}<1\). A is indifferent between c and \({\hat{c}}\) for

$$\begin{aligned} \frac{\gamma (1-f)}{1-d}\left[\int _{d}^{\bar{r}} r \,dr+\beta \int _{\bar{r}}^1 r \,dr]=\gamma (1-f)[\int _0^{\bar{r}} r \,dr+(1-\alpha )\int _{\bar{r}}^1 r \,dr\right], \end{aligned}$$

which simplifies to

$$\begin{aligned} \bar{r}=\sqrt{\frac{(1-\alpha )(1-d)-\beta +d^2}{d-\beta +(1-\alpha )(1-d)}}. \end{aligned}$$
(1)

By (1), for \(\beta <\frac{(1-\alpha )(1-d^2)+d^2}{1+d}\), \(d<\bar{r}<1\).

Let A choose c with probability \(P_c\). Given alert s and rating \(r>d\), let P(c|s) be belief of P that A cheats:

$$\begin{aligned} P(c|s)=\frac{P_c(1-\beta )\frac{r-d}{1-d}}{P_c(1-\beta )\frac{r-d}{1-d}+(1-P_c)\alpha r}. \end{aligned}$$
(2)

Following alert s, P is indifferent between b and \({\hat{b}}\) for the threshold rating r of A iff

$$\begin{aligned} \gamma f\bar{r}-wP(c|s)+v(1-P(c|s))=0, \end{aligned}$$
(3)

and by (2), this is equivalent to

$$\begin{aligned} P_c=\frac{\alpha \bar{r}(\gamma f \bar{r}+v)(1-d)}{\alpha \bar{r}(\gamma f \bar{r}+v)(1-d)+(1-\beta )(w-\gamma f \bar{r})(\bar{r}-d)}. \end{aligned}$$
(4)

By (4), \(0<P_c<1\) for \(w>\gamma f \bar{r}\).

By substitution of (2) into (3) one can verify that for sufficient high w the left hand side of (3) decreases in \(\bar{r}\). For \(r<\bar{r}\) the platform does not ban; and for \(r>\bar{r}\), it bans, as is required by a threshold strategy. \(\square \)

Proof of Corollary 1

Since conditions of part 3 of proposition 1 hold, probabilities of cheating and the threshold of banning are given by (4) and (1), respectively. Results follow directly by (4) and (1). \(\square \)

Proposition 3

  1. 1.

    If \(\frac{2w}{\gamma (1+d)}< f\), then in equilibrium the expected utility of P is \(\frac{\gamma f(1+d)}{2}-w\).

  2. 2.

    If \(\frac{2w}{\gamma (1+d)}> f\) and \(\beta >\frac{d+(1-\alpha )(1+d)}{1+d}\), then in equilibrium the expected utility of P is \(\beta (\frac{\gamma f(1+d)}{2}-w)\).

  3. 3.

    If w are sufficiently high and \(\beta <\frac{(1-\alpha )(1-d^2)+d^2}{1+d}\), then in equilibrium the expected utility of P is

    $$\begin{aligned}&(1-P_c)[\frac{\gamma f}{2}[\bar{r}^2+(1-\alpha )(1-\bar{r}^2)]+v[\bar{r}+(1-\alpha )(1-\bar{r})]]\\&\qquad +\frac{P_c}{1-d}[\frac{\gamma f}{2} [\bar{r}^2-d^2+\beta (1-\bar{r}^2)]-w[\bar{r}-d+\beta (1-\bar{r})]], \end{aligned}$$

    where \(P_c\) and \(\bar{r}\) are given by (4) and (1).

Proof

Directly from Proposition 1 and Fig. 1. \(\square \)

Proof of Proposition 2

Consider first an equilibrium where P chooses pure \({\hat{b}}\). Then c is a superior action of A. The expected utility of P in this case is \(\gamma f-w\). If P chooses b, its payoff is 0, and the platform prefers \({\hat{b}}\) to b for \(w< \gamma f\).

Next, consider that P, following s, chooses pure b. Observe, that for \(\alpha (\rho )>0\), pure \({\hat{c}}\) is not an equilibrium in this case. By contrary, if A chooses pure \({\hat{c}}\), with positive probability it obtains the rating 1 and a false signal s is sent to P; therefore, b is not the best reply of the platform.

If A chooses c and if P (following s) chooses pure b, A’s expected utility is \(\gamma (1-f)\beta (\rho )\). If A does not cheat and if P (following s) chooses pure b, A’s expected utility is \(\gamma (1-f)[\rho (1-l(\rho ))+l(\rho )(1-\alpha (\rho ))]\). Thus, A prefers c to \({\hat{c}}\) for \(\beta (\rho )>r-rl(\rho )+l(\rho )-\alpha (\rho )l(\rho )\).

Let A choose c with probability \(P_c\). Similar to the proof of Proposition 1,

$$\begin{aligned} P_c=\frac{\alpha (\rho )l(\rho )(\gamma f+v)}{\alpha (\rho )l(\rho )(\gamma f+v)+(1-\beta (\rho ))(w-\gamma f)}. \end{aligned}$$
(5)

By (5), \(0<P_c<1\) for \(w>\gamma f\).

Let \(P_b\) be a probability with which P bans the application, following alert s. A is indifferent between c and \({\hat{c}}\) for

$$\begin{aligned} \gamma (1-f)[1-(1-\beta (\rho ))P_b]=\gamma (1-f)[(1-l(\rho )) \rho +l(\rho )(1-\alpha (\rho )P_b)], \end{aligned}$$

which simplifies to

$$\begin{aligned} P_b=\frac{(1-\rho )(1-l(\rho ))}{1-\beta (\rho )-\alpha (\rho )l(\rho )}. \end{aligned}$$
(6)

By (6), \(0<P_b<1\) for \(l(\rho )-\alpha (\rho )l(\rho )+\rho - \rho l(\rho )>\beta (\rho )\). \(\square \)

Proof of Corollary 2

Since the conditions of part 3 of Proposition 2 hold, probabilities of cheating and of banning are given by (5) and (6), respectively. Results follow directly by (5) and (6). \(\square \)

Proof of Corollary 3

Since the conditions of part 3 of Proposition 2 hold, probabilities of cheating is given by (5). The result follows directly by (5) and , by \(\frac{\partial l(\rho )}{\partial \rho }>0\) and \(\frac{\partial \beta (\rho )}{\partial \rho }\ge 0\). \(\square \)

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Dery, L., Hermel, D. & Jelnov, A. Cheating in Ranking Systems. Rev Ind Organ 58, 303–320 (2021). https://doi.org/10.1007/s11151-020-09754-2

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