SUPPLEMENT TO Random Authority

Transcription

1 SUPPLEMENT TO Random Authority Siguang Li Xi Weng September 15, 2015 In this Online Appendix, we provide a complete proof of Proposition 2, and extend the analysis of optimal authority allocation to asymmetric organizations. Throughout the Online Appendix, we refer often to results in the main text and its Appendix using the numbering established there e.g., Proposition 2. The numbers of equations and figures in this Online Appendix are all prefixed by A to distinguish them from those in the main text and its Appendix, and we always use this prefix in referring to them. 1 Proof of Proposition 2 Proposition 2 is proved by showing the following three lemmas: Lemma A.1 In symmetric organizations, the optimal authority allocation ρ must be symmetric or be a corner solution i.e., if it is optimal to have ρ 3 > 0, then the optimal authority allocation must satisfy ρ 4 ρ 3 or ρ 4 0. Lemma A.2 Symmetric mixed authority is suboptimal in symmetric organizations i.e., if ρ 3 ρ 4, then it is optimal to set ρ 3 ρ 4 0. Lemma A.3 In symmetric organizations, authority allocation with ρ 3 > 0 and ρ 4 0 or ρ 4 > 0 and ρ 3 0 is always suboptimal. These lemmas imply Proposition 2 naturally. From Lemma A.1, there are two possibilities for the optimal random authority structure in symmetric organizations: i it must be symmetric when introducing both mixed authority structures, {M 1} and {M 2}; and ii there exists only one mixed authority structure. Lemma A.2 rules out symmetric mixed authority arrangements in symmetric organizations. Lemma A.3 implies that introducing only M 1 or M 2 will be suboptimal as well. 1 It is immediately clear that including mixed authority is always suboptimal in symmetric organizations. 1 Lemma A.3 has already been proved numerically by [1. We provide a formal proof here. 1

2 PROOF OF LEMMA A.1 Proof. From Proposition 1, by plugging ρ 4 1 ρ 1 ρ 2 ρ 3 into π RA and taking the partial derivative over ρ 3, we obtain π RA 3δ δ2λ 1 2 λ2δ + λ 2 1 ρ 1 ρ 2 2ρ 3 h 1 ρ 1, ρ 2, λ, δ ρ 3 λ + δ + 2λδh 2 2 ρ. A.1 1, ρ 2, ρ 3, λ, δ Since h 1 is independent of ρ 3, we have: i when h 1 0, then the maximum of π RA is achieved at ρ 3 1 ρ 1 ρ 2 2 ; and ii when h 1 < 0, then π RA is maximized at a corner solution i.e. ρ 3 0 or ρ 3 1 ρ 1 ρ 2. Thus if mixed authority is included in the optimal authority allocation, it must be symmetric i.e., ρ 3 ρ 4 1 ρ 1 ρ 2 2 or be a corner solution i.e., ρ 3 0 or ρ 4 0. PROOF OF LEMMA A.2 Proof. Suppose by contradiction that there exists some λ, δ such that the optimal random authority is given by ρ ρ 1, ρ 2, ρ 3, ρ 4, which satisfies ρ 3 ρ 4 > 0. Consider another randomized structure with a new probability vector ρ ρ 1 +, ρ 2 +, ρ 3, ρ 4, > 0 is sufficiently small. Under symmetric mixed authority i.e., ρ 3 ρ 4, S 1 RA S2 RA and let S RA S 1 RA S2 RA. From Proposition 1, we obtain: 4 4 π RA ρ i a i S RA + ρ i b i, A δ a δ, a 2 2δ2 2δ 2 + 6λ 2 δ + 4λ 1λ 2 λ + δ 2 λ + 2δ 2, a 3 a 4 λ λ λλ 2 δ + λ5 + 8λ + 4λ 2 δ λ + 4λ 3 δ 3 + 8δ 4 λ + δ 2 λ + δ + 2λδ 2, b 1 4δ 1 + 4δ, b 2 4δδ + 4δ2 + λ 2 + 4δλ δλ + 2δ 2, b 3 b 4 2δ δ + 2λ δλ 2 + 4δ λ δλ + δ + 2λδ 2, Similarly, the total expected profits π RA under ρ can be rewritten as: π RA [ ρ 1 + a 1 + ρ 2 + a 2 + ρ 3 a 3 + ρ 4 a 4 S RA + [ ρ 1 + b 1 + ρ 2 + b 2 + ρ 3 b 3 + ρ 4 b 4, A.3 2

3 From Eq. A.2 and A.3, we can get: 4 πra ρ i a i SRA S RA + { a 1 + a 2 a 3 a 4 SRA + b 1 + b 2 b 3 b 4 }, A.4 Obviously, as 0, we have S RA S RA. Define by S RA lim 0 S RA S RA. Rearranging Eq. A.4 yields πra lim 0 4 SRA ρ i a i lim S { } RA + a 1 + a 2 a 3 a 4 lim SRA + b 1 + b 2 b 3 b ρ i a i S RA + { a 1 + a 2 a 3 a 4 S RA + b 1 + b 2 b 3 b 4 } A.5 Thus, to verify that ρ strictly dominates ρ, it suffices to show that lim 0 π RA π RA > 0. From the definition of S RA, we have S RA δλ + δ2λ 1 3 ρ iω i 3 ρ iω i+3, A.6 ω 1 λ + 2δ 2 λ + δ + 2λδ 2, ω 2 δ1 + 4δλ + δ + 2λδ 2, ω δλ + 2δ 2 λ + δλ + λ 2 + 2δλ δλ + 2δ 2, ω 4 λ + 2δ 2 λ + δ + 2λδ 2 3λ 2 + 2λδ1 + 4λ + 8λ 1δ 2, A.7 ω 5 δ δλ + δ + 2λδ 2 5λ 1λ + 8λ 1δ, ω 6 4λ 2 δ δλ + 2δ 2 λ + δ + 4λδ + λ δλ + 2δ 2 3λ 4 + 8δλ λ + δ 2 λ 1 + 9λ + 16λ 2 + δ λ + 8λ 2, From the definition of S RA, we obtain 3λ1 2λ 2 δ δλ + δ λ λλδ λδ S RA ρ iϖ i 3 2, A.8 ρ iϖ i+3 3

4 ϖ 1 λ 4 + λ 2 δ 2 + 5λ + 2λ 2 + λδ λ + 8λ 2 + δ 3 3 5λ + 6λ 2 4δ 4, ϖ 2 δ 2[ λ3λ 1 + δ 2 + λ + 6λ 2 4δ 2, ϖ 3 λ 4 + δλ λ + 2λ 2 + 2δ 2 λ 3 + 3λ + 4λ 2 + δ 3 5 4λ + 12λ 2 8δ 4 ϖ 4 λ + 2δ 2 λ + δ + 2λδ 2[ 3λ 2 + 2δλ1 + 4λ + δ 2 8λ 1, ϖ 5 δ δλ + δ + 2λδ 2[ λ5λ 1 + δ8λ 1, ϖ 6 λ + 2δ δ [ 3λ 4 + 8δλ λ + δ 2 λ 1 + 9λ + 20λ 2 + 4λ 3 + δ λ + 12λ λ 3, A.9 Plugging Eq. A.6, A.7, A.8 and A.9 into Eq. A.5, we obtain πra lim 0 χ 3 ρ 1, ρ 2, ρ 3 χ 4 ρ 1, ρ 2, ρ δλ + δλ + 2δ 2 λ + δ + 2λδ 2 3 ρ iϖ i ρ iω i+3,a.10 χ 3 3λδ 2λδ 2 { ρ 1 λ + δ + 2λδλ + 2δ 2[ 3λ 2 + 2λ1 + 4λδ + δ 2 8λ 1 + ρ 2 δ δλ + δ + 2λδ 2[ λ5λ 1 + δ8λ 1 + ρ 3 λ + 2δ δ [ 3λ λλ 3 δ + δ 2 λ 1 + 9λ + 20λ 2 + 4λ 3 + δ λ + 12λ λ 3 }, and χ δ [ λ 2 + λδ3 + 2λ λδ 2 2 K1 K 2 + 4δλ + δλ + 2δλ + δ + 2λδ [ 21 λλ + δ3 + 4λ 4λ 2 + 8δ 2 K 3 K 4. A.11 4

5 The coefficients K i i 1, 2, 3, 4 in χ 4 can be explicitly expressed as K 1 [ ρ 1 λ 4 + δλ λ + 2λ 2 + δ 2 λ 5 + 3λ + 8λ 2 + δ 3 3 5λ + 6λ 2 4δ 4 + ρ 2 δ 2[ λ3λ 1 + δ 2 + λ + 6λ 2 4δ 2 + [ ρ 3 λ 4 + λ 2 δ 2 + 5λ + 2λ 2 + 2λδ λ + 4λ 2 + δ 3 5 4λ + 12λ 2 8δ 4 K 2 ρ δ [ λ 3 + 2δλ λ + δ λ + δ 2 λ5 + 6λ ρ δδ 2 λ + δ + 2λδ 2[ λ 2 4λ 1 + 6δλ 2 + 2δ 2 ρ 3 λ1 + 4δλ + 2δ 2 λ 3 + 2δλ λ + δ 2 λ5 + 8λ + 4λ 2 + 2δ λ + 4λ 3 + 8δ 4 K 3 ρ δ [ λ 3 + 2δλ λ + δ 2 λ5 + 6λ + δ λ + ρ δδ 2 λ + δ + 2λδ + ρ δ λ 3 + 4δλ 2 + δ 2 λ5 + 2λ + δ λ and K 4 ρ 1 λ + 2δ 2 λ + δ + 2λδ 2[ 3λ 2 + 2δλ1 + 4λ + δ 2 8λ 1 + ρ 2 δ δλ + δ + 2λδ [ λ5λ 1 + δ8λ 1 + ρ δ [ 3λ 4 + 8δλ λ + δ 2 λ 1 + 9λ + 20λ 2 + 4λ 3 + δ λ + 12λ λ 3. Note that given λ 1/2, 1 and δ > 0, ω 4, ω 5, ω 6 and χ 3 are strictly positive. Hence, the sign πra of lim π RA 0 is the same as the sign of χ 4. Finally, we need to check the sign of χ 4. Plugging ρ 2 1 ρ 1 2ρ 3, we can see that χ 4 ρ 1, ρ 2, ρ 3 > 0. 2 Therefore, the new authority allocation ρ ρ 1 +, ρ 2 +, ρ 3, ρ 4 always outperforms the original one for any sufficiently small > 0, yielding a contradiction! PROOF OF LEMMA A.3 Proof. Suppose by contradiction that only introducing some possibility of M1 can be optimal. 3 In other words, there exists some λ, δ such that the optimal authority allocation ρ ρ 1, ρ 2, ρ 3, ρ 4 satisfies ρ 3 > 0, ρ 4 0. Consider another authority allocation with a new probability vector ˆρ ρ 1 +, ρ 2 +, ρ 3 2, 0 for sufficiently small > 0. From Proposition 1, we obtain: π RA ρ i a i SRA 1 + ρ i b i SRA 2 + ρ i c i, A.12 2 The details of calculations can be obtained upon request. 3 The case for M2 is analogous and thus omitted. 5

6 a δ 1 + 4δ, a 2 δ2 2δ 2 + λ 2 6δλ 2 + 4λ δλ2δ + λ δ + λ 2 2δ + λ 2, a 3 δ + λ + 2δλ 2, b δ 1 + 4δ, b 2 δ2 2δ 2 + λ 2 6δλ 2 + 4λ 3 δ + λ 2 2δ + λ 2, b 3 4δ2 λδ 2 + λ 2 + δλ + 2δλ 3 δ + λ 2 δ + λ + 2δλ 2, c 1 4δ 1 + 4δ, c 2 4δδ + λ2 2δ + λ 2, and c 3 2δδ + 2λ2 + 4δλ 2 δ + λ + 2δλ 2. A.13 Denote S 1, RA S1 RA ρˆρ and S 2, RA S2 RA ρˆρ, and from Eq. A.12 we can get: π RA [ ρ 1 + a 1 + ρ 2 + a 2 + ρ 3 2a 3 S 1, RA + [ ρ 1 + b 1 + ρ 2 + b 2 + ρ 3 2b 3 S 2, RA + [ ρ 1 + c 1 + ρ 2 + c 2 + ρ 3 2c 3. A.14 Hence, 3 3 πra ρ i a i S 1, RA S RA + ρ i b i S 2, RA S RA { } + a 1 + a 2 2a 3 S 1, RA + b 1 + b 2 2b 3 S 2, RA + c 1 + c 2 2c 3. A.15 Obviously, as 0, S 1, RA S1 RA and S2, RA S2 RA. Denote the limits by SRA/ 1 S 1, lim 0 Rearranging Eq. A.15 yields πra lim 0 3 ρ i a i RA S1 RA lim 0, SRA/ 2 S 2, lim 0 S 1, 3 RA S RA + ρ i b i RA S2 RA lim 0. S 2, RA S RA { } + lim a 1 + a 2 2a 3 S 1, 0 RA + b 1 + b 2 2b 3 S 2, RA + c 1 + c 2 2c ρ i a i SRA/ 1 + ρ i b i SRA/ 2 + { a 1 + a 2 2a 3 S 1 RA + b 1 + b 2 2b 3 S 2 RA + c 1 + c 2 2c 3 }. A.16 6

7 From the definition of S j RA, we have S 1 RA δλ + δ2λ 1 ρ 1 w 1 + ρ 2 w 2 + ρ 3 w 3 ρ1 w 5 + ρ 2 w 6 + ρ 3 w 7, S 2 RA δλ + δ2λ 1 ρ 1 w 1 + ρ 2 w 2 + ρ 3 w 4 ρ1 w 5 + ρ 2 w 6 + ρ 3 w 8, A.17 w 1 λ + 2δ 2 λ + δ + 2λδ 2, w 2 δ1 + 4δλ + δ + 2λδ 2, w δλ + 2δ 2 λ + δλ + λ 2, w 4 2δ1 + 4δλ 2 2δ + λ 2, w 5 δ + λλ + 2δ 2 λ + δ + 2λδ 2 [3λ + 8λ 1δ, A.18 w 6 δ δλ + δ + 2λδ 2 [5λ 1λ + 8λ 1δ, w δλ + δλ + 2δ 2[ 3λ 3 + δλ λ + δ λ + 8λ 2, w 8 4λ 2 δ δλ + 2δ 2 λ + δ + 4λδ + λ 2. From the definition of SRA 1 / and S2 RA /, we obtain S 1 RA/ υ 0ρ 1 υ 1 + ρ 2 υ 2 + ρ 3 υ 3 ρ 1 w 5 + ρ 2 w 6 + ρ 3 w 7 2, S2 RA/ υ 0ρ 1 υ 4 + ρ 2 υ 5 + ρ 3 υ 6 ρ 1 w 5 + ρ 2 w 6 + ρ 3 w 8 2, A.19 υ 0 3δ δλδ + λ2λ 1[λ 2 + δλ3 + 2λ + δ λ 2, υ 1 λ 3 2λ 1 + δ13λ 7λ 2 + δ 2 λ λ + 8λ 2 + δ 3 7 3λ + 8λ 2 + 4λ 3 8δ 4, υ 2 λ 3 + δλ λ + δ 2 λ1 + 16λ + δ λ 4λ 2 8δ 4, υ 3 λ 4 + δλ λ + 2δ 2 λ 3 + 8λ + 2λ 2 + δ λ + 4λ 2 8δ 4, υ 4 λ 3 4λ 5 δλ λ 16λ 2 8λ 3 + δ 2 λ3 + 10λ 4λ 2 32λ 3 + δ λ + 24λ 2 28λ δ 4 λ, υ 5 λ 3 + δλ λ + 3δ 2 λ1 + 2λ + 4λ 2 + δ λ 8λ λ 3 16δ 4 λ, υ 6 2λ 4 + 4δλ λ + 2δ 2 λ 1 + 4λ + 8λ 2 + 2δ 3 1 8λ + 12λ 2 16δ 4. A.20 Plugging Eq. A.13, A.17, A.18, A.19 and A.20 into Eq. A.16, we obtain πra lim 0 3λδ 2 2λ 1 2 χ 5 ρ 1, ρ 2, ρ δ[λ 3 + 2δλ λ + δ λ + δ 2 λ5 + 6λχ 2 6 ρ 1, ρ 2, ρ 3, A.21 χ 6 ρ 1 w 5 + ρ 2 w 6 + ρ 3 w 7 ρ 1 w 5 + ρ 2 w 6 + ρ 3 w 8. 7

8 It is easy to observe that πra sign{lim } sign{χ 5 ρ 1, ρ 2, ρ 3 }. 0 Therefore, we only need to verify that χ 5 ρ 1, ρ 2, ρ 3 > 0, which is true. 4 structure ˆρ outperforms the original one ρ for any 0. Contradiction! Hence, the new 2 Optimal Authority Allocation in Asymmetric Organizations In asymmetric organizations, the optimal authority structure may involve randomization among all possible deterministic authority allocations, making it difficult to explicitly characterize the optimal authority allocation. However, the intuition in symmetric organizations should still apply, and thus random authority can be strictly optimal under some parameter values. There are two major findings in this section. First, we show that mixed authority can be included as part of the optimal authority allocation in asymmetric organizations. Second, due to the existence of mixed authority, we show by example that the optimal degree of delegation may not be monotonic in incentive misalignment. Observation A.1 In asymmetric organizations, for any σ1 2 σ2 2, then there exists λ, δ such that the optimal authority allocation satisfies either ρ 3 > 0 or ρ 4 > 0. 5 Observation A.1 reinforces the result in [1, which numerically claims that mixed authority i.e., partial centralization can be optimal when the degree of information asymmetry is sufficiently large. We theoretically prove that mixed authority can be part of the optimal authority allocation even under tiny asymmetry. We have already proved that the optimal degree of delegation can change non-monotonically with the coordination need in symmetric organizations. However, in symmetric organizations, the optimal degree of delegation monotonically decreases with the incentive misalignment. asymmetric organizations, the optimal degree of delegation may be non-monotonic in the incentive misalignment as illustrated by the following numerical example. Denote α σ2 2/σ2 1. Then the optimal authority allocation satisfies: In C α,λ,µ1.5,0.56, {P } {M1}, C α,λ,µ1.5,0.58, {P } {M1}, C α,λ,µ1.5,0.9,0.6 {F }. 4 The exact expressions of χ 5 can be obtained upon request. 5 The proofs are available upon request. 8

9 In this example, the coordination need is quite large, and there exists some degree of information distribution asymmetry in organizations. When the incentives are sufficiently aligned, including mixed authority can improve organizational performance, mainly due to the fact that mixed authority is a better fit to the information structure. However, as λ increases, the benefits from including mixed authority to enhance communication will be offset by the adaptation loss caused by the functional manager. This makes project authority more attractive. Finally, when incentive misalignment is sufficiently severe λ very large, enforcing functional authority will become necessary in order to implement effective coordination. References [1 Rantakari, H. 2008: Governing Adaption, Review of Economic Studies, 75,