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On Lawyer Compensation When Appeals Are Possible

Christian At, Tim Friehe and Yannick Gabuthy

Abstract

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.

JEL Classification: D82; K41

Appendix

A Proof of Lemma 1

Using the t2(y1,1) implied by (9) in (4) yields:

L2(e2(y1))=e2(y1)ce2y1(e2(y1))c2y1(e2(y1))

Since L2(0)=0 and L2(e2(y1))=e2(y1)ce2′′(e2(y1),y1)0, we get L2(e2(y1))0 such that (4) is satisfied.

Restating (8) yields a(1)=a(0)+c1(e1), which allows to state (3) as follows:

L1(e1)=a(0)+e1c1(e1)c1(e1)0.

Since (4) is satisfied, we have a(0)0. By noting f(e1)=e1c1(e1)c1(e1), we have f(0) = 0 and f(e1)=e1c1′′(e1)0 since c1′′(e1)0 by assumption. As a result, (3) is satisfied.

(2) is induced by (9) since ce2(e2(y1),y1)0.

B Proof of Proposition 1

Denote by x(y1)=qy1+1y1 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 t2(y1,1)=c2y1(e2(y1)), we consider:

P(e2(y1),y1)=e2(y1)(Jc2y1(e2(y1)))and A(e2(y1),y1)=e2(y1)c2y1(e2(y1))c2y1(e2(y1))

as the expected payoff of the plaintiff and the lawyer, respectively, from an appeal. The expected social surplus is thus given by:

S(e2(y1),y1)=e2(y1)Jc2y1(e2(y1)).

Hereafter, we omit y1 when the computations are true whatever the value of y1.

Using a restatement of (5), a=t1+xA(e2), the plaintiff’s program to determine the optimal continuation payoff π(a) may be written:

maxt1,e2xP(e2)t1s.t.t1=axA(e2)0

Replacing t1=axA(e2) in the plaintiff’s objective function, this becomes:

maxe2xS(e2)as.t.axA(e2)0

With λ0 as the Lagrange multiplier, the Lagrangian is:

L=xS(e2)a+λ(axA(e2))

The first-order conditions are then given by:

(10)S(e2)λA(e2)=0
(11)λ(axA(e2))=0

and allow us to reason about the plaintiff’s desired level of appeals court effort as a function of the outcome y0.

Step1. Proof of e20,e2FB

We first show that e2=0 cannot be a solution. We know from the proof of Lemma 1 that a(0)0. From (8) (i.e. a(1)a(0)=c1(e1)), and c1(e1)0, we deduce that a(1)a(0). Hence, since a0 and A(0) = 0, we get λ = 0 and Sʹ(0) = 0, which is impossible since Sʹ(0) = J.

Next, we show that any e2>e2FB cannot be a solution. For e2>e2FB, we have S(e2)<0, which violates (10) since λA(e2)0.

Step2. Proof of e2(0)0,e2FB(0) and e2(1)0,e2FB(1)

We know that t1(0)=0 is optimal. The plaintiff’s program with outcome y1=0 is thus:

maxe2(0)P(e2(0))

as x is a constant.

Since P(e2FB)=0 and P(e2FB)=e2FBc′′(e2FB)<0, we know that e2(0)<e2FB(0).

For the case with outcome y1=1 such that x(1) = q, we can state the following lemma.

Lemma 2

For a given lawyer’s continuation payoff a(1), we have:

  1. e2(1)=e2FB(1) and t1(1)=a(1)qAe2FB(1)>0 if q<a(1)Ae2FB(1)=qˉ ;

  2. e2(1)=A1a(1)q<e2FB(1) and t1(1)=0 if qqˉ.

Proof 4

Consider the first-order conditions (10) and (11).

If λ = 0, then S(e2)=0 and, thus, e2=e2FB, implying π(a,e2FB)=qS(e2FB)a0 since aqA(e2FB)=qS(e2FB). Furthermore, we get t1(1)=a(1)qAe2FB(1)>0.

If λ > 0 or t1(1)=0, then the second-best effort e2 is the solution of a=qA(e2). We have π(a,e2)=qP(e2). Since P(e2FB)=0 and P(e2FB)=e2FBc′′(e2FB)<0, we get P(e2)>P(e2FB)=0 and, thus, π(a,e2)>π(a,e2FB).

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, q<qˉ, since e2FB(1)=supe2(1)0,e2FB(1)A(e2(1)).   □

Step3. Analysis of the optimal continuation payoff π(a).

Consider, first, the case qqˉ. We know from Step 2 that e20,e2FB and a=xA(e2). We also know that a(0,xA(e2FB)) since A(e2) is continuous and increasing in e2 and A(0) = 0. Furthermore, the derivative of the function π(a)=xP(e2)=xP(A1(a/x)) is π(a)=P(e2)A(e2). When axA(e2FB) (i.e. when e2e2FB), we have:

π(a)=P(e2)A(e2)P(e2FB)A(e2FB)=1<0

While, when a0 (i.e. when e20), we get:

π(a)P(0)A(0)+

Some algebra gives:

π′′(a)=P′′(e2)A(e2)P(e2)A′′(e2)A(e2)2=S′′(e2)A(e2)S(e2)A′′(e2)A(e2)2<0

implying that π(a) is concave. Therefore, the continuation payoff reaches a maximum in the interval (0,xA(e2FB)).

Second, if q<qˉ, then the continuation payoff is π(a(1))=qS(e2FB(1))a(1).

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:

maxe1,a(0),a(1)e1[(1q)J+π(a(1))]+(1e1)π(a(0))

Replacing a(1) by a(0)+c1(e1) using (8), this problem becomes:

maxe1,a(0)e1[(1q)J+π(a(0)+c1(e1))]+(1e1)π(a(0))

The first-order conditions, which are sufficient since the objective function is concave, are given by:

(12)(1q)J+π(a(1))π(a(0))+π(a(1))e1c1′′(e1)=0
(13)e1π(a(1))+(1e1)π(a(0))=0

Consider that e1=0. 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. e1>0).

Note that qa(1)Ae2FB(1)>0 is satisfied since e1>0 (implying that a(1) > 0).

Step5. Assessing payments after the appeals court’s judgment.

Since e1>0, from (8) we deduce a(1) > a(0) or qA(e2(1),1)>A(e2(0),0)qA(e2(0),0). Remember A(e2,y1)=e2c2y1(e2)c2y1(e2). Our assumptions ensure that Ae(e,y1)=e2c2y1′′(e2)0 and Ae′′(e,y1)=c2y1′′(e2)+ec2y1′′′(e2)0. We have two cases to consider.

Case (a): Suppose that c21(e2)c20(e2)<c21(e2)c20(e2)e2<0e2(0,e2FB), implying A(e2(0),1)<A(e2(0),0). We must have A(e2(1),1)>A(e2(0),0) at the optimum, which can be satisfied only for e2(1)>e2(0). From (9), we know that t2(0,1)=c20(e2(0)) and t2(1,1)=c21(e2(1)). Furthermore, there exists eˆ<e2(1) such that c21(e2(1))=c20(eˆ) since c21(e2)<c20(e2), c2y1′′(e2)0 and e2(1)>e2(0). We deduce that t2(1,1)=c21(e2(1))<t2(0,1)=c20(e2(0)) if e2(0)>eˆ, and t2(1,1)>t2(0,1) otherwise.

Figure 2: e2(1)e_{2}(1)must be greater than e2(1)_>e2(0)\underline{e_{2}(1)}>e_{2}(0) to ensure A(e2(1),1)>A(e2(0),0)A(e_{2}(1),1)>A(e_{2}(0),0).

Figure 2:

e2(1)

must be greater than e2(1)_>e2(0) to ensure A(e2(1),1)>A(e2(0),0).

Figure 3:

Figure 3:

Case (b): If c21(e2)c20(e2)e2<c21(e2)c20(e2)<0, then A(e2,1)>A(e2,0)e(0,e2FB) and the signs of both e2(1)e2(0) and t2(1,1)t2(0,1) are ambiguous.

Figure 4: e2(0)e_{2}(0)must be smaller than e2(0)‾>e2(1)\overline{e_{2}(0)}>e_{2}(1) to ensure A(e2(1),1)>A(e2(0),0)A(e_{2}(1),1)>A(e_{2}(0),0); therefore, we can have e2(0)>e2(1)e_{2}(0)>e_{2}(1) or e2(0)<e2(1)e_{2}(0)<e_{2}(1) as a solution.

Figure 4:

e2(0)

must be smaller than e2(0)>e2(1) to ensure A(e2(1),1)>A(e2(0),0); therefore, we can have e2(0)>e2(1) or e2(0)<e2(1) as a solution.

C Proof of Proposition 2

Step1. Proof of de2(1)dq0 and dt2(1,1)dq0

Consider, first, the case where q>q. We know that t1(1)=0 and, thus, a(1)=qA(e2(1),1). For an optimal choice of a(1), we have de2(1)dq=a(1)q2A(e2(1),1)<0, since e2(1)=A1a(1)q. From (3), we deduce that dt2(1,1)dq<0.

Consider now the case where qq. We have e2(1)=e2FB(1), implying that de2(1)dq=dt2(1,1)dq=0.

Step2. Proof of de1dq<0, de2(0)dq>0 and dt2(0,1)dq>0

Consider the first-order conditions (12) and (13), and note for simplicity:

E(a(q),e1(q),q)=(1q)J+π(a(q)+c1(e1(q))π(a(q))+π(a(q)+c1(e1(q))e1(q)c1′′(e1(q))=0and A(a(q),e1(q),q)=e1(q)π(a(q)+c1(e1(q))+(1e(q))π(a(q))=0

By totally differentiating the system, we get:

Eadadq+Eededq+Eq=0Aadadq+Aededq+Aq=0

Given that Aq=0, we get EeAa(EaAe)2>0, since the program is concave. We deduce that:

sign dedq=sign EqAasign dadq=sign EqAe

where Eq=J<0, Aa=e(q)π′′(a(q)+c1(e(q))+(1e(q))π′′(a(q))<0 and Ae=π(a(q)+c1(e(q))π(a(q))+π′′(a(q)+c1(e(q))e(q)c1′′(e(q))<0 (since, from (13), π(a(q)+c1(e(q)) and πʹ(a(q)) have opposite signs, and π(.) is concave). Therefore, we get de1dq<0.

We also have da(0)dq>0. We know that e2(0)=A1(a(0)) and, thus, de2(0)dq=1A(e2(0),0)da(0)dq>0, implying dt2(0,1)dq>0.

Step3. Determination of the effect of q on t1(1)

Only the case where q<q matters as t1(1)=0 otherwise. We know that t1(1)=a(1)qAe2FB(1) and obtain:

dt1(1)dq=da(0)dq+c1′′(e1)de1dqAe2FB(1)

which exhibits an ambiguous sign.

D Endogenous Probability

Suppose that the plaintiff lost the case in trial court (i.e. we have y1=0) and now considers setting the probability of appeal, p ∈ [0,1]. The plaintiff would at this stage want to

maxe2(0),ppS(e2(0))a(0)s.t.a(0)pA(e2(0))0

where a(0) and A(e2(0)) were defined above. With λq0 as the Lagrange multiplier, the Lagrangian is (omitting y1=0 from the parentheses):

L=pS(e2)a+λ(apA(e2))

The first-order conditions are then given by:

(14)S(e2)λA(e2)=0
(15)S(e2)λA(e2)=0
(16)λ(apA(e2))=0

It is straightforward from (15) that λ = 0 cannot be a solution, hence using (16), we can replace a=pA(e2) in the program that simplifies to:

maxe2,ppS(e2)pA(e2)=pP(e2)

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.

Acknowledgements

We are grateful for the helpful comments received from two anonymous reviewers.

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Published Online: 2019-03-23

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