This paper describes how plaintiff should compensate lawyers, who choose unobservable effort, when litigation may proceed from the trial to the appeals court. We find that, when it is very likely that the defendant will appeal, transfers made to the lawyer only after an appeals court’s ruling are key instruments in incentivizing both trial and appeal court effort. Indeed, the lawyer may not receive any transfer after the trial court’s ruling. In contrast, when reaching the appeals stage is unlikely, a favorable trial court ruling triggers a positive transfer to the lawyer and first-best appeals effort. In our setup, the lawyer may receive a lower transfer after winning in both the trial and the appeals court as compared to the scenario in which the first-instance court ruled against the plaintiff and the appeals court reversed that ruling.
A Proof of Lemma 1
Using the implied by (9) in (4) yields:
Since and , we get such that (4) is satisfied.
Restating (8) yields , which allows to state (3) as follows:
Since (4) is satisfied, we have . By noting , we have f(0) = 0 and since by assumption. As a result, (3) is satisfied.
(2) is induced by (9) since .
B Proof of Proposition 1
Denote by the probability that there will be an appeal as a function of the trial court’s judgment. Moreover, define π(a) as the plaintiff’s maximum continuation payoff when she implements a lawyer continuation payoff a.
Using , we consider:
as the expected payoff of the plaintiff and the lawyer, respectively, from an appeal. The expected social surplus is thus given by:
Hereafter, we omit when the computations are true whatever the value of .
Using a restatement of (5), , the plaintiff’s program to determine the optimal continuation payoff π(a) may be written:
Replacing in the plaintiff’s objective function, this becomes:
With as the Lagrange multiplier, the Lagrangian is:
The first-order conditions are then given by:
and allow us to reason about the plaintiff’s desired level of appeals court effort as a function of the outcome .
Step1. Proof of
We first show that cannot be a solution. We know from the proof of Lemma 1 that . From (8) (i.e. ), and , we deduce that . Hence, since and A(0) = 0, we get λ = 0 and Sʹ(0) = 0, which is impossible since Sʹ(0) = J.
Next, we show that any cannot be a solution. For , we have , which violates (10) since .
Step2. Proof of and
We know that is optimal. The plaintiff’s program with outcome is thus:
as x is a constant.
Since and , we know that .
For the case with outcome such that x(1) = q, we can state the following lemma.
For a given lawyer’s continuation payoff a(1), we have:
and if ;
and if .
Consider the first-order conditions (10) and (11).
If λ = 0, then and, thus, , implying since . Furthermore, we get .
If λ > 0 or , then the second-best effort is the solution of . We have . Since and , we get and, thus, .
Consequently, it is always optimal to make the constraint binding. However, the constraint cannot be bind if the probability that the defendant will appeal is too small, , since . □
Step3. Analysis of the optimal continuation payoff π(a).
Consider, first, the case . We know from Step 2 that and . We also know that since is continuous and increasing in and A(0) = 0. Furthermore, the derivative of the function is . When (i.e. when ), we have:
While, when (i.e. when ), we get:
Some algebra gives:
implying that π(a) is concave. Therefore, the continuation payoff reaches a maximum in the interval .
Second, if , then the continuation payoff is .
Step4. Determination of the second-best effort level at the trial stage.
The plaintiff can choose any pair of non-negative a(1) and a(0) to solve the following maximization problem:
Replacing a(1) by using (8), this problem becomes:
The first-order conditions, which are sufficient since the objective function is concave, are given by:
Consider that . From (8), we deduce that a(1) = a(0) since cʹ(0) = 0 by assumption. From (13), we get πʹ(a(0)) = 0 = πʹ(a(1)) and π(a(1)) = π(a(0)), which violates (12). Therefore, the lawyer exerts a positive effort at the trial stage (i.e. ).
Note that is satisfied since (implying that a(1) > 0).
Step5. Assessing payments after the appeals court’s judgment.
Since , from (8) we deduce a(1) > a(0) or . Remember . Our assumptions ensure that and . We have two cases to consider.
Case (a): Suppose that , implying . We must have at the optimum, which can be satisfied only for . From (9), we know that and . Furthermore, there exists such that since , and . We deduce that if , and otherwise.
Case (b): If , then and the signs of both and are ambiguous.
C Proof of Proposition 2
Step1. Proof of and
Consider, first, the case where . We know that and, thus, . For an optimal choice of a(1), we have , since . From (3), we deduce that .
Consider now the case where . We have , implying that .
Step2. Proof of , and
Consider the first-order conditions (12) and (13), and note for simplicity:
By totally differentiating the system, we get:
Given that , we get , since the program is concave. We deduce that:
where , and (since, from (13), and πʹ(a(q)) have opposite signs, and π(.) is concave). Therefore, we get .
We also have . We know that and, thus, , implying .
Step3. Determination of the effect of q on
Only the case where matters as otherwise. We know that and obtain:
which exhibits an ambiguous sign.
D Endogenous Probability
Suppose that the plaintiff lost the case in trial court (i.e. we have ) and now considers setting the probability of appeal, p ∈ [0,1]. The plaintiff would at this stage want to
where a(0) and were defined above. With as the Lagrange multiplier, the Lagrangian is (omitting from the parentheses):
The first-order conditions are then given by:
It is straightforward from (15) that λ = 0 cannot be a solution, hence using (16), we can replace in the program that simplifies to:
We deduce that the optimal choice of the probability of an appeal is p = 1. This is consistent with the result of Ohlendorf and Schmitz (2012) when the continuation costs are equal to 0.
We are grateful for the helpful comments received from two anonymous reviewers.
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