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
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1.
If \(\frac{2w}{\gamma (1+d)}< f\), then in equilibrium the expected utility of P is \(\frac{\gamma f(1+d)}{2}-w\).
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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 \(\beta (\frac{\gamma f(1+d)}{2}-w)\).
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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 \(\frac{\partial \beta (\rho )}{\partial \rho }\ge 0\). \(\square \)