There exists a unique x1 , where 0 ? x1 ? 1 , given that Vt(q1, q2) is an increasing function in q1 .
Proof. When the strategic RA (RA1) gets a bad project, it will get pay-off if it gives the project a GR, and if it refuses rating. 1 and is increasing in x1 . Given that Vt(q1, q2) is increasing in q1 , it is easy to see that ?(lie) is decreasing in x1 and that ?(honest) is increasing in x1 . Thus, if we define x1 such that
dos Proof Proposition 2
Proof. Suppose that the strategic RA (RA1) gets a good project and that its strategy is x1 . Let us examine whether RA1 wants to deviate:
•if x1 = 1 , we have ?(lie) ? ?(honest) , or . If the RA1 gives NR to the good project, it will get and otherwise. Since RA1 does not want to deviate.
•if x1 = 0 , , hence reputation becomes dating positivesingles irrelevant and the RA does not have an incentive to give NR to the good project.
•if 0 < x1 < 1 , we have ?(lie) = ?(honest) , so , and hence RA1 does not want to deviate.
3 Proof Corollary step 1
Proof. Suppose that the equilibrium strategy is x1 = 0 . Then and we must have I + ?Vt(q1, q2) ? ?Vt(q1, q2) . This is impossible as long as I > 0 . Hence, x1 = 0 cannot be an equilibrium strategy.
cuatro Proof Corollary 2
Suppose the model ends in period T. Then the equilibrium strategy of the strategic RA is xt = 1 at t = T ? step one, T .
Proof. At t = T , the strategic RA does not have any reputational concerns. This implies that the strategy of strategic RA will be to always give GR if the project is bad, that is, xT = 1 .
Also, within t = T ? step one , the fresh strategic RA will always rest. Imagine that a bad venture involves proper RA, say RA1. The newest questioned pay-away from RA1 was
Although in this case RA1 does have reputational concerns, these are not sufficient to prevent RA1 from being lax and not giving GR to bad projects. Since by being honest RA1 is giving up I today, in exchange for having a higher chance of getting I in the next period, it is not optimal for RA1 to be honest, given that RA1 is impatient (i.e., ? < 1 ). Hence, the optimal strategy of RA1 is to always lie, that is, xT ? 1 = 1 .
5 Evidence of Proposal 4
Proof. Since pG = 1 , the reputation of RA1 (i.e., the strategic RA) will go to zero if it gives a GR to a bad project since now every good project succeeds and every bad project fails. So the expected pay-off from giving a GR to a bad project is I . This simplifies expressions (9) and (10) and allows us to derive RA1’s equilibrium strategy.
Having fun with Equations (6) and you can (7) and you can listing you to RA1 will always be rest into the attacks t = dos, 3 , this might be rewritten just like the
We look for an equilibrium of the game by examining RA1’s trade-off between lying and not lying. If the pay-off from lying is greater when x1 = 1 , we have a pure-strategy equilibrium in which RA1 always lies; if the pay-off from not lying is greater when x1 = 0 , we have a pure-strategy equilibrium in which RA1 never lies; otherwise we have a mixed-strategy equilibrium in which RA1 is indifferent between lying or not given some prior beliefs about its strategy, that is, 0 < x1 < 1 .