Proof of Proposition 5

By definition of δ, some ECli (i ≥ 1) will bind in some left-neighborhood of δ, while ECl0 remains slack. In this neighborhood, the profit-maximizing n^l_i (i ≥ 1) are obtained by maximizing Π^l₁.

Thus, we maximize

Π^l₁ =

∞

X

τ=1

(δ(1 − q))^{τ −1} θ^lg(n^l_τ) − n^l_τc
+ δqΠ^h 1 1 − δ(1 − q) subject to

−n^l_ic+δqΠ^h(1 + δ(1 − q))+δ θ^l− qθ^h
g(n^l_i+1) − (1 − q)n^l_i+1c
+δ²(1−q)²Π^l_i+2 ≥ 0

13One shows that n^l< n^{F B}l (n^h< n^{F B}h ) by showing that the partial derivative of (ECh) with respect to n^l0(n^h) is always strictly negative at n^l0= n^{F B}l (n^h= n^{F B}h ).

14As the only exception, there is a direct impact of ˆn^l0 in (ECl0). Yet, as ˆn^l0 ≤ n^{F B}l , (ECl0) is slacker than the other (ECli) constraints, and thus continues to hold as well.

for all i ≥ 1. We proceed in several steps.

Lemma 10 For any i ≥ 1, Π^l₁ ≥ Π^l_i.

Proof. Suppose to the contrary that Π^l_j > Π^l₁, for some j > 1. For all i ≥ 1, replace ni by nj+i−1. (This operation is feasible because all (ECli) were satisfied by assumption.) Thus, our previous Π^l₁ cannot solve our

maximiza-tion problem. 

Lemma 11 n^l₁ ≥ n^l_i∀i

Proof.Suppose to the contrary that there is a j > 1 with n^l_j > n^l₁. Re-place n^l_j with n^l₁ and the continuation play following n^l_j with the continuation play following n^l₁. This is clearly feasible and (weakly) profitable (as Π^l_j ≤ Π^l₁

by Lemma 10). 

Lemma 12 For all odd i ≥ 1, Π^l_i ≥ Π^l_i+2 and n^l_i ≥ n^l_i+2.

Proof. We proceed by induction over i. That Π^l₃ ≤ Π^l₁ follows from Lemma 10. For the induction step, suppose that Π^l_j ≥ Π^l_j+2, for some odd integer j. We have to show that Π^l_j+2 ≥ Π^l_j+4. Suppose to the contrary that Π^l_j+2 < Π^l_j+4. Now, for all i ≥ j + 2, replace n^l_i by n^l_i+2. This is feasible if n^l_j+1 ≤ n^l_j+3. Therefore, our operation increases Π^l_j+2 and hence Π^l₁. If n^l_j+1 > n^l_j+3, by contrast, we distinguish two cases: (1.) If Π^l_j+1 ≤ Π^l_j+3, we can replace n^l_i by n^l_i+2 for all i ≥ j + 1. This replacement is feasible and weakly increases Π^l₁. (2.) If, however, Π^l_j+1 > Π^l_j+3, we replace n^l_i+2 by n^l_i for all i ≥ j + 1. This is feasible if n^l_j ≥ n^l_j+2. If, however, n^l_j < n^l_j+2, we can replace n^l_i+2 by n^l_i for all i ≥ j. Because, by the induction hypothesis, Π^l_j ≥ Π^l_j+2, this increases Π^l₁.

Suppose that n^l_j < n^l_j+2 for some odd integer j. Replace all n^l_i+2 by n^l_i for all i ≥ j. This is clearly feasible and (weakly) profitable (as Π^l_j ≥ Π^l_j+2).

Lemma 13 For all even i ≥ 2, Π^l_i ≤ Π^l_i+2 and n^l_i ≤ n^l_i+2.

Proof. Suppose to the contrary that Π^l_j+4 < Π^l_j+2 for some even integer j. Then, we can replace all n^l_i+2 by n^l_i for all i ≥ j + 2. This is feasible as n^l_j+1 ≥ n^l_j+3 by Lemma 12. Suppose that n^l_j > n^l_j+2 for some even integer j. Replace all n^l_i by n^l_i+2 for all i ≥ j. This is clearly feasible and (weakly)

profitable (as Π^l_j+2 ≥ Π^l_j). 

Lemma 14 n^l_i 6= n^l_i+2 ⇒ n^l_j 6= n^l_j+2∀j ≤ i.

Proof. Suppose to the contrary that n^l_i 6= n^l_i+2 but n^l_j = n^l_j+2 for some integer j < i. Consider the biggest such integer j, i.e., n^l_j+1 6= n^l_j+3. First, assume that j is even, i.e., j + 1 is odd and, by Lemma 12, n^l_j+1 > n^l_j+3. Replace all n^l_ι+2 by n^l_ι for all ι ≥ j + 1. This is feasible as n^l_j = n^l_j+2 and (weakly) profitable (as Π^l_j+1 ≥ Π^l_j+3). Second, assume that j is odd, i.e., j + 1 is even and, by Lemma 13, n^l_j+1 < n^l_j+3. Replace all n^l_ι by n^l_ι+2for all ι ≥ j +1.

This is feasible as n^l_j = n^l_j+2 and (weakly) profitable (as Π^l_j+1 ≤ Π^l_j+3).  Lemma 15 n^l_i = n^l_i+2 ⇒ n^l_j = n^l_j+2∀j ≥ i.

Proof. Suppose to the contrary that n^l_i = n^l_i+2 but n^l_j 6= n^l_j+2 for some integer j > i. Consider the smallest such integer j, i.e., n^l_j−1 = n^l_j+1. First, assume that j − 1 is even, i.e., j is odd and, by Lemma 12, n^l_j > n^l_j+2. Replace all n^l_ι+2 by n^l_ι for all ι ≥ j. This is feasible as n^l_j−1 = n^l_j+1 and (weakly) profitable (as Π^l_j ≥ Π^l_j+2). Second, assume that j − 1 is odd, i.e., j is even and, by Lemma 13, n^l_j < n^l_j+2. Replace all n^l_ι by n^l_ι+2 for all ι ≥ j. This is feasible as n^l_j−1 = n^l_j+1 and (weakly) profitable (as Π^l_j1 ≤ Π^l_j+2).  Lemma 16 n^l₁ = n^l₂ ⇒ n^l_i = n^l₁∀i ≥ 1.

Proof. By Lemma 13, n^l₁ = n^l₂ ⇒ n^l_j = n^l₁ for all even j. Hence, by Lemma 15, n^l_ι = n^l_ι+2 for all odd ι ≥ 3. Suppose to the contrary that n^l₁ > n^l₃. Replace n^l₃ with n^l₁ and the continuation play following n^l₃ with the continuation play following n^l₁. This is feasible and (weakly) profitable (as

Π^l_j ≤ Π^l₁ by Lemma 10). 

Lemma 17 Assume there is one i for which the (ECli) constraint is slack.

Then, the (ECli+1) constraint binds.

Proof. To the contrary, assume that the (ECli+1) constraint is slack.

Increase n^l_i+1 by a small ε > 0. This is feasible and increases Π^l₁.  Lemma 18 Assume there is one odd i > 1 for which the (ECli) constraint is slack. Then, n^l_j = n^l_{F B}∀j ≥ 1.

Proof. Suppose (ECli) is slack for i odd, with i > 1. Then, there must exist an optimum with ni+j = nj∀j ≥ 1. This implies that Π^l_i+1 = Π^l₁, Π^l_i+2 =

Π^l₂, ..., Π^l_2i−1 = Π^l_i−1. By Lemma 13, Π^l₂ ≤ Π^l₄ ≤ ... ≤ Π^l_i+1 ≤ Π^l_i+3 ≤ ....

Since Π^l_i+1 = Π^l₁ ≤ Π^l_i+3 = Π^l₃ ≤ Π^l_i+5 = Π^l₅ ≤ ... ≤ Π^l_2i = Π^l_i ≤ ... ≤ Π^l₁, Π^l_j = Π^l₁ for all even j.

By Lemma 12, we have Π^l₁ ≥ Π^l₃ ≥ ... ≥ Π^l_i = Π^l_2i = Π^l₁, and hence Π^l_j = Π^l₁ for all odd j. Thus, n^l_j = n^l₁ for all j ≥ 1. Therefore, the Lagrange parameters satisfy λj = λj+1 = 0 for all j, and n^l₁ = n^l_{F B}.  Lemma 18 implies that, in our left-neighborhood of δ, all odd-numbered constraints will bind, i.e. the Lagrange parameters satisfyλj > 0 for all odd integers j.

Lemma 19 Assume there is one even i for which the (ECli) constraint is slack. Then, the (EClj) constraints are slack for any even j. Moreover, n^l_j = n^l_j+2= ... = n^l₁ for all odd j, and n^l_ι = n^l_ι+2 = ... = n^l₂ for any even ι.

Proof. Suppose (ECli) is slack for i even. Then, there must exist an optimum with ni+j = nj∀j ≥ 1. This implies that Π^l_i+1 = Π^l₁, Π^l_i+2 = Π^l₂, ..., Π^l_2i−1 = Π^l_i−1. By Lemma 13, Π^l₂ ≤ Π^l₄ ≤ ... ≤ Π^l_i ≤ Π^l_i+2 = Π^l₂, and hence Πι = Π^l₂ for all even ι. It follows that (EClι) is slack for all even ι.

Thus, n^l_ι = n^l₂ for all even ι.

By Lemma 12, we have Π^l₁ ≥ Π^l₃ ≥ ... ≥ Π^l_i+1 = Π^l₁, and hence Π^l_j = Π^l₁ for all odd j. Thus, n^l_j = n^l₁ for all odd j. Therefore, the Lagrange parameters

λι = λι+2 = 0 for all even ι. 

The previous lemmata imply that there are two possibilities for an op-timum. Either, all even (ECli) constraints are slack, in which case n^l_j = n^l₁ for all odd j and n^l_ι = n^l₂ for all even ι. Otherwise, all (ECli) constraints will bind. In the following, we characterize effort levels n^l_i (i ≥ 1) for the latter possibility.

Lemma 20 Assume all (ECli) constraints bind. Then, either n^l₁ = n^l₃ = n^l₅ = ... and n^l₂ = n^l₄ = n^l₆ = ..., or n^l₁ > n^l₃ > n^l₅ > ... and n^l₂ < n^l₄ < n^l₆ < ....

Proof. To the contrary, assume that n^l_j+2 > n^l_j for j even, but that n^l_j+3 = n^l_j+1. By Lemma 15, this implies that n^l_j+2 = n^l_j+4 = ... and n^l_j+3 = n^l_j+5= ..., and in particular also that Π^l_j+2 = Π^l_j+4. But then, (EClj) can not bind, a contradiction. The same logic can be applied to show that n^l_j+2 < n^l_j for j odd, but that n^l_j+3 = n^l_j+1, is not feasible.  Lemma 21 n^l_i > n^l_j ⇒ Π^l_i ≥ Π^l_j.

Proof. Suppose to the contrary that there exist integers i and j such that n^l_i > n^l_j yet Π^l_i < Π^l_j. Then,

Π^l_j−Π^l_i = θ^lg(n^l_j) − cn^l_j
+δqΠ^h+δ(1−q)Π^l_j+1−

θ^lg(n^l_i) − cn^l_i
+ δqΠ^h+ δ(1 − q)Π^l_i+1

= 

θ^lg(n^l_j) − cn^l_j
− θ^lg(n^l_i) − cn^l_i
 + δ(1 − q) Π^l_j+1− Π^l_i+1
≥ 0. Be-cause n^l_i > n^l_j, 

θ^lg(n^l_j) − cn^l_j
− θ^lg(n^l_i) − cn^l_i


< 0. Therefore, Π^l_j+1 − Π^l_i+1 > 0. Hence, replacing the history n^l_i by n^l_j and the continuation play af-ter n^l_i by the continuation play after n^l_j is feasible, and also strictly profitable.



Lemma 22 Suppose all (ECli) constraints bind. Then, sup_j∈Nn^l_2j ≤ infj∈Nn^l_2j−1. Proof. Suppose to the contrary that sup_j∈Nn^l_2j > infj∈Nn^l_2j−1. Then, by Lemmata 12 and 13, this implies lim sup_j∈Nn^l_2j > lim infj∈Nn^l_2j−1. Therefore, there exists an integer i such that n^l_2i > n^l_2i−1 ≥ n^l_2i+1. By Lemma 21, this implies that Π^l_2i > Π^l_2i+1. Yet, as all constraints (EClι), and in particular (ECl2i − 2), are binding, n^l_2i > n^l_2i−1 and Π^l_2i > Π^l_2i+1 implies that n^l_2i−2 >

n^l_2i−1, which, by Lemma 21, implies Π^l_2i−2 > Π^l_2i−1. As furthermore Π^l_2i ≥ Π^l_2i−2 by Lemma 13 and all constraints, in particular (ECli−2) and (ECli−3), are binding, we can conclude that n^l_2i−2 > n^l_2i−3 and thus, by Lemma 21, Π^l_2i−2 > Π^l_2i−3. Iterating this argument finally yields n^l₂ > n^l₁, a contradiction

to Lemma 11. 

Lemma 23 n^l₁ = n^l₂ ⇔ qθ^h = θ^l.

Proof. Recall that for δ < δ, the values n^l_i, i ≥ 1, can be obtained by maximimizing

Π^l₁ =

∞

P

τ=1

(δ(1 − q))^{τ −1} θ^lg(n^l_τ) − n^l_τc
+ δqΠ^h_{1−δ(1−q)}¹ , s.t. (ECli) con-straints for i ≥ 1, and treating Π^h as a constant. The Lagrange function of

this problem is first assume that λ2 = 0 and verify later that it holds.

If λ2 = 0, the condition gives n^l₁ = n^l₂. Furthermore, if λ2 = 0, Lemma 19

implies that n^l₁ = n^l₃ = ... and n^l₂ = n^l₄ = .... Then, Π^l₁ = (^θlg(nl1)−n^l₁c)^+δ(1−q)(^θlg(nl2)−n^l₂c)

1−(δ(1−q))² +

δqΠ^h_{1−δ(1−q)}¹ , and the (binding) (ECl1) constraint equals

−n^l₁c + _{1−δ(1−q)}^δq Π^h+ δ θ^l− qθ^h
g(n^l₂) − (1 − q)n^l₂c
+_{1−(δ(1−q))}^(δ(1−q))22

 θ^lg(n^l₁) − n^l₁c
+ δ(1 − q) θ^lg(n^l₂) − n^l₂c
 = 0 Plugging

δq 1−δ(1−q)Π^h

= n^l₁c − δ θ^l− qθ^h
g(n^l₂) − (1 − q)n^l₂c

−_{1−(δ(1−q))}^(δ(1−q))22

 θ^lg(n^l₁) − n^l₁c
+ δ(1 − q) θ^lg(n^l₂) − n^l₂c
 into (ECl2) gives

g(n^l₁) − g(n^l₂)
δ θ^l− qθ^h− qδ(1 − q) θ^h− θ^l

+ c n^l₁− n^l₂
≥ 0.

For qθ^h = θ^l and n^l₁ = n^l₂, the left hand side equals zero, hence (ECl2) is

satisfied. 

This concludes the proof of Proposition 5.