Online Appendix for. Breakthroughs, Deadlines, and Self-Reported Progress: Contracting for Multistage Projects. American Economic Review, forthcoming

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1 Online Appendix for Breakthroughs, Deadlines, and Self-Reported Progress: Contracting for Multistage Projects American Economic Review, forthcoming by Brett Green and Curtis R. Taylor Overview This supplemental appendix contains proofs of formal results in Section IV of Green and Taylor In this appendix, equations are numbered S.1, S.2, S.3,... References to equations, propositions, lemmas etc. not prefaced with S corresponds to ones in Green and Taylor Omitted Proofs Proof of Proposition 3. The proof takes several steps. In Steps 1 through 3 we assume that the principal must use the formal channel, but that she does so optimally. In Step 4 we prove that, for ρ sufficiently small, the principal indeed benefits by using the formal channel. Step 1. First observe that the only possible reason for requiring a costly report is to relax the no-false-progress constraint so as to add more time to the clock following the first reported breakthrough. This implies that the formal communication channel should only be used to report progress not lack of progress. Next, observe that it cannot be optimal for the principal to require the costly formal report at date t and take no action until date t > t, because this is dominated by waiting until t to require the report. The project might be completed between t and t. Thus, it is optimal to putoff requiring a formal report as long as possible. Let y be the highest value of the low type s continuation utility at which the principal requires a formal report, and let F 2 u 1, u 2 ; ρ denote the principal s value function in the second stage. This value function is given by F 2 u 1, u 2 ; ρ = c 1 e u 1 +ρ/ u 2 1 {u1 y}ρe u 1 y/. S.1 This is established using the same method of proof as for Proposition A.2 with two straightforward alterations. First, to satisfy B9, the termination policy must be such that 1

2 ET ] u 1 + ρ/. Second, for u 1 y, there is a chance that the high type will have to pay the reporting cost. Therefore, the promise keeping constraint necessitates u 2 + ρe u 1 y/ = E a=1 T τ2 0 ] dy t s = 1, Because F 2 u 1, u 2 ; ρ is concave in u 1 and linear in u 2, σ = 0 is optimal. Step 2. We solve for the first-stage value function assuming σ = 0 and check for concavity. The HJB is F 1 u 1 ; ρ = max y 0 F 2u 1, u 1 + /; ρ c df 1 du 1. Because the principal prefers to delay formal reports as long as possible, y = 0 is optimal. Therefore we have F 1u 1 ; ρ + F 1u 1 ; ρ = 2c u 1 ρ + c ] e ρ/ e u 1/. This has a solution of the form F1 c u 1 ; ρ = 2c u 1 ρ + c e ρ/ u1 e u 1/ + Ke u1/. S.2 Using the boundary condition F 1 0; ρ = 0 gives F c 1 u 1 ; ρ = 2c 1 e u 1 / u 1 ρ + c e ρ/ u1 e u 1/. For small ρ, this is convex for u 1 sufficiently close to zero, necessitating random termination at some point u s ρ. Step 3. Using the same analysis as in the proof of Lemma A.4 yields F 1 u 1 ; ρ = 2c u1 u s ρ u s ρ + 2 2c ] u 1 u u s ρ where u s ρ is implicitly defined by e u 1 u sρ 2c ρ + c ρ/ 2 + u sρ e usρ/ 0. ] u 1 u 1 u s ρ, S.3 u 1 0, u s ρ. S.4 Step 4. We wish to show that for ρ sufficiently small and any u 1 > 0, F 1 u 1 ; ρ > F 1 u 1 ; 0 = 2

3 F 1 u 1. The claim will follow if we show F 1u;0 ρ u s0 = c c > 0 for all u > 0. Note first that 2 + us 1 + us < 0, and F 1 u 1 ; ρ depends on ρ only through u s ρ. Therefore, for all u > 0, F1 u; 0 F1 u; 0 sign = sign > 0, ρ u s where the inequality follows from observing that both expressions in S.3-S.4 are strictly decreasing in u s ρ for all u 1 > 0. Proof of Proposition 4. First, we extend the analysis from Appendix A to characterize F α 1 for an arbitrary α. Using the same arguments as in Lemma A.2, we have that F α 2 u 1, u 2 = 1 e u 1 1 α c1 α u 2 Replacing F 2 with F2 α in HJB and the appropriately modified binding constraints, we find that F1 α has the form ˆF α 1 u 1 = 2c u α 2α 1 αc e u 1 1 α + C α 1 e u 1 1+α. S.5 If the terminal boundary condition is imposed ˆF 1 α 0 = 0 then C1 α = 1+αΠ c1+α 2α α and ˆF 1 0 = 2 Π > 0. Hence, there exists some u 1 α 2 sα such that random termination is optimal for u 0, u s α]. Let c α u denote the constant in the principal s value function that satisfies the smooth-pasting condition i.e., A9 at an arbitrary u > 0. That is, c α u α+1e 2αu α 2 1 c α 2 2α+4αe α u +1 +u α α+2αe α u +u+ 2αα+u+. S.6 Twice differentiability at u s α i.e., A10 is equivalent to u s α = max u c α u, which requires the first-order condition e usα/1 α u s α + 2 = c1 α 1 α 2c. S.7 The right-hand side of the above expression is strictly greater than 1/2 for all α 1, 1. The left-hand side is equal to 1/2 at u s α = 0, strictly increasing in u s α and unbounded. 3

4 This guarantees the existence of a unique u s α satisfying S.7, which completes the characterization of F α 1. To summarize, For u u s α, F1 α is of the form given in S.5 with C1 α = c α u s α, where u α s is the unique solution to S.7. For u 0, u s α, F α 1 u = u F α u sα 1 u s α. To prove i, first note that by the envelope theorem d dα cα u s α = α cα u s α. Using this fact, evaluating the derivative and taking the it as α 0, we get that for u u s 0 = u s, d α 0 dα F 1 α e u uus + u 2 u = s 2 u s + 2 Π 2 u 2s 2 e us 1 + u s c 2 u 2s 2 2 e us 1 + 2u s. The first-term on the right hand side is clearly positive for u u s. Using S.7, the second term reduces to u s + cu s Π, which is also clearly positive if u s / > Π/c Π/c 1. We now claim that if u s solves S.7 for α = 0, then this latter inequality must hold. Let x u s / 0, y Π/c 2 > 0, and α = 0. The claim is that e x x + 2 = y + 1 y = x > y + 2 y + 1. To see that this is true, suppose that increasing. Therefore, e x x + 2 e x = y+1 and x y+2. Note that e x x+2 y y+1 x+2 e y+2 y+1 y < y + 1, y y+1 is strictly which gives the contradiction. We have thus shown that at α = 0, the derivative of F α 1 u w.r.t. α is strictly positive for all u u s α. That the same statement is true for u 0, u s is immediate by the linearity of the value function below u s. Since F1 α is also continuously differentiable in both of its arguments, it must be strictly increasing in a neighborhood around α = 0 for all u > 0, which completes the proof of i. We prove ii and iii for the case of α 1, the proof for α 1 follows a similar argument. We first show that α 1 u s α = 0. To do so, rewrite S.7 as 1 αe usα/1 α = c1 αu sα c Suppose u s 1 α 1 u s α 0,. Then α 1 1 αe usα/1 α = > Π c1 αu s1+2 Π 2c, a contradiction. Also, clearly u s 1 < otherwise the principal s 4

5 value function would be arbitrarily negative. The only remaining possibility is u s 1 = 0. From S.5, we have that for u u α s, d α 1 dα F 1 α u = e u 2 u α 1 4 cα u s α + α cα u s α. Notice from S.6 that α 1 u 0 c α u = u 0 α 1 c α u = 2c/. Therefore, we can conclude that α 1 c α u s α = Π 2c/. Hence, to prove ii, it is sufficient to show that α 1 α cα u s α = 0, for this implies d F α dα 1 u e u 2 2c/ u < 0 for 4 u u s α and α sufficiently close to 1. To see that α 1 α cα u s α = 0, first notice from S.7 that e usα 1 α 0,, S.8 α 1 1 α implying that u s α is O1 α ln1 α as α 1. Differentiating S.6 with respect to α and omitting the argument of u s α, we get that α cα u s α = e 2αus α α 2 α 2 α + 1α + u s + Π 1 α 2 α u 3sα α ] + 2 u 2 s 3α + 2α 4 e us 1 α + α 3 5 4e us 1 α + α 2 2e us 1 α αα u s 2 1 αc α 1 2 α α α u 3 s 2 u 2 s α 4 4α + 4α 3 e us 1 α 1 4α 2 e us 1 α αα u s 2 ] ]. Using S.8, we know that e 2αusα α 2 1 is O1 α as α 1. Therefore, any term inside the outermost brackets that goes to zero faster than O1 α will converge to zero when scaled by the fraction outside the brackets. By inspection, the only terms that do not clearly go to zero faster than O1 α are ] 2 u s α 2 2α 4 e usα 1 α 4α 3 e usα 1 α + 2α 2 e usα 1 α. 5

6 Thus, we get have that α 1 α cα u s α = e 2αusα α 2 1 e usα 1 α α 1 1 α 2 α + 1α + u s α + u sα 2 α 2 2α + 1 ] 2 e usα 1+α = α 1 α + 1α + u s α + u sα 2 2 = 2 4 u sα 2 α 1 = 0, which completes the proof of ii. For iii, we have ˆF 1 α u V 1 α 2 u = 2α 1 α 2α 1 α 2α 1 αc 1 αc Π + C α 1 2c/ + e u 1 α + C α 1 e u 1+α 2c + Cα 1 2c + C1 α e x α 2 e x α 1+α, Cα 1 e u 2 e u 1+α e u 2 where the first inequality uses the triangle inequality and e x 1 and the second uses the fact that e x 2 e x 1+α is hump-shaped in x and achieves its maximum at x α 21 + α ln 1+α /α 1. Clearly all three terms converge to 0 as α 1. 2 Proof of Proposition 5. Consider first any simple contract with deadline T. If the agent does not shirk, 1 then the probability that the project succeeds at t 0, T ] is given by 2 te t and the probability that the project does not succeed prior to T is given by e T 1 + T. Therefore, the total surplus is given by ST T 0 2 te t ctdt + e T 1 + T ct. In order to induce the agent to work, he must be given rents in the amount of at least u = T, otherwise he can do better by shirking. Making the change of variables from T to u, we have that the principal s ex-ante expected payoff under a simple contract with deadline T = u/ is bounded above by Gu Su/ u = 1 e u/ 1 + u/ 2c 1 e u/ 1 + u/2 u. S.9 1 Arguments similar to those made for a single-stage project can be used to confirm shirking is suboptimal. 6

7 Note that Gu is not the principal s value function under the optimal contract, since her belief about the project stage changes over time. We will construct the value function shortly. To prove the proposition, it suffices to show that i The principal s ex-ante payoff for a project with unobservable progress under any contract is bounded above by max u Gu. ii There exists an incentive-compatible simple contract under which the principal s ex-ante expected payoff is max u Gu. iii For all u > 0, Gu < F 1 u. Therefore, the principal does strictly better with intangible progress than she does with unobservable progress. For i, let w = arg max u Gu, which is generically unique. We have already argued that the principal s maximal payoff under a simple contract is bounded above by Gw. Given that neither player has any information about the status of the project, the only possibility is that the principal randomizes over the termination date. It therefore suffices to show that the principal cannot benefit from such randomization, or equivalently, that the principal s value function under this simple contract with the optimally chosen deadline is globally concave in the agent s continuation value. Suppose that the principal can implement a simple contract in which the incentive compatibility condition holds with equality for all t this is the best possible case for the principal. It is most intuitive to construct this value function from the pair of value functions that are conditional on s i.e., whether a breakthrough has been made and weight them appropriately by the probability that the principal assigns to each. Given u, the principal s payoff conditional on being in the first stage s = 1 is Gu; i.e., F unobs u s = 1 = u/ 0 2 te t ctdt e u/ 1 + u/cu/ u = 1 e u/ 2c/ u e u/ c/ u. Conditional on being in the second stage, the principal s value function is the benchmark payoff V u: F unobs u s = 2 = u/ 0 e t ct + e u/ cu/ u = 1 e u/ c/ u. Over time, the principal s beliefs will evolve about the state of the project. Conditional on reaching state u < w prior to project success, a period of time of length tu; w = w u has 7

8 elapsed. Therefore, the principal s beliefs are given by µu; w = Prτ 1 tu; w τ 2 > tu; w = 1 + w u w u. The principal s value function for u w is therefore given by F unobs u; w = µu; w F unobs u s = µu; w F unobs u s = 1. We will now verify this value function is concave for all u w. Using the functional forms given above and twice differentiating F unobs u; w, we get that d 2 du 2 F unobs u; w = e u/ 2 w u w w u 2 + w 2 c + 2 w u 2 2c + 3 Π + 2e u 1 ]. All three terms inside the brackets are clearly positive, implying the value function is concave in u for all u w, which completes the proof of i. We prove ii by showing that for any T, there exists a w : 0, T ] R + such that 1 it is incentive compatible for the agent to work for all t 0, T ], and 2 the agent s continuation utility at date t is ut = T t. Let u 2 t be the promised continuation value conditional on being in the second stage and ut be the unconditional continuation value at t. Promise keeping requires that ut = µtwt + 1 µtu 2 t + u t. Conditional on progress, the evolution of u 2 is given by u 2 t = wt + u 2t. S.10 We want to find wt such that ut = T t for all t 0, T ]. Note that this implies that u t =. Using the promise keeping condition, T t + 1/ = 1 µtu 2 t + µtwt. 8

9 Substituting for wt from S.10, we get that T t + 1 = 1 µtu 2 t + µt u 2 t u 2t = u 2 t t 1 + t u 2t. Imposing the boundary condition u 2 T = 0, we arrive at a unique solution for u 2 t, which we can then substitute back into S.10, to arrive at t 1 e T wt = T t et q T q t, 2 T where qz = z e x /xdx. It is straightforward to check that wt > 0 for all t 0, T ], which completes the proof of ii. For iii, note that Gu has the same form as F 1 u see A8 with H 1 = 2c Π. The result then follows from the fact that F 1 has a constant strictly larger than 2c Π. References Green, Brett, and Curtis Taylor Breakthroughs, Deadlines, and Self-Reported Progress: Contracting for Multistage Projects. American Economic Review forthcoming. 9