Observe that and you may , that’s, are coming down during the x

1 Evidence of Suggestion step one

There exists a unique x₁ , where 0 ? x₁ ? 1 , given that V_t(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 GramsR, and if it refuses rating. ₁ and is increasing in x₁ . Given that V_t(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 of Suggestion dos

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 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.

step 3 Proof of Corollary step one

Proof. Suppose that the equilibrium strategy is x₁ = 0 http://datingranking.net/plenty-of-fish-review . Then and we must have I + ?V_t(q₁, q₂) ? ?V_t(q₁, q₂) . This is impossible as long as I > 0 . Hence, x₁ = 0 cannot be an equilibrium strategy.

4 Evidence of Corollary 2

Suppose the model ends in period T. Then the equilibrium strategy of the strategic RA is x_t = 1 at t = T ? 1, 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, x_T = 1 .

Also, within t = T ? step one , the latest strategic RA will always rest. That is amazing a bad endeavor involves strategic RA, state RA1. The expected pay-off 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, x_{T ? 1} = 1 .

5 Proof Offer cuatro

Proof. Since p_G = 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.

Playing with Equations (6) and you can (7) and noting one RA1 will always lay when you look at the episodes t = dos, 3 , this is exactly rewritten once 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 .