Rent-Seeking and Litigation: The Hidden Virtues of Limited Fee Shifting

B.1 Proof of impossibility of Y≤d<X

We show the impossibility by contradiction. For this, we first assume that both inequalities hold true. Based on the consequential definitions of the parties’ payoffs, we then show that these investments cannot be a Nash equilibrium.

Suppose Y≤d<X. Then

With these payoffs, we get the following first order condition for the defendant:

which simplifies to

It is useful to write this as:

The defendant’s second order condition is:

If we replace μXr as suggested by eq. [36] inside the brackets, this turns into:

Hence, Y≤d<X can only be an equilibrium if r<1.

We now turn to the plaintiff’s foc. It is given by:

Inserting eq. [36] yields:

Then by our initial assumption we have:

which implies

As we already know that r<1, inequality [38] also implies

or

which contradicts the initial assumption. Hence, there is no equilibrium with Y≤d<X. By symmetry, we also cannot have an equilibrium with X≤d<Y. q.e.d.

B.2 Equilibrium with X,Y>d

We first check whether investment levels Xˆ,Yˆ>d defined by the first-order conditions of maximizing ΠP(X,Y) and ΠD(X,Y) as defined by eqs [13] and [14] constitute a Nash equilibrium of the litigation game. We know from standard rent-seeking theory that it is not sufficient to consider the first-order conditions but that investing nothing may be a better alternative. However, the plaintiff is not willing to deviate if exerting effort X˜ when the defendant invests Y˜ yields a higher payoff than exerting no effort (X=0). This happens when ΠP(Xˆ,Yˆ)>−d, that is, when 1+μ−r>0. The latter inequality is always satisfied for r≤1. For r>1, it is required that the case presents μ>r−1.

Similarly, the defendant does not deviate if and only is ΠD(Yˆ,Xˆ)>−V−d, which implies 1+μ(1−r)>0, which is always satisfied for r<1 and requires μ<1r−1 if r>1.

Knowing that his payoff will be ΠD(Yˆ,Xˆ) in the litigation stage, the defendant is willing to defend in court if and only if ΠD(Xˆ,Yˆ)>−V, i.e., if and only if d<d˜D which we can invert to

The plaintiff, on the other hand, is willing to file if and only if the defendant will not defend her case in court or ΠP(Xˆ,Yˆ)>0, that is d<dˆP which we can invert to

Equations [39] and [40] represent respectively the defendant’s and the plaintiff’s participation constraints. When they are satisfied, investment levels Xˆ,Yˆ>d constitute a subgame-perfect Nash equilibrium of the litigation game.^[28]

To see that for r≤1/2 we have dˆD>dˆ we note that the denominator of dˆ is always positive and the same is true for the denominator of dˆD if and only if μ>1+r2−r<1. Hence, if the latter condition is violated, dˆD=∞>dˆ. If the condition holds true, dˆD>dˆ is equivalent to

which after some algebra reduces to r<12+12μ which is satisfied by assumption. Since dˆP is the same as dˆD with μ replaced by 1/μ and dˆ does not change when we replace μ by 1/μ, exactly the same argument also proves that dˆP>dˆ if r≤1/2. Obviously, if r>1/2, the argument fails if μ or, respectively 1/μ is large enough. And if r>1, the argument for dˆD is reversed for all μ≥1 and the argument for dˆP is reversed for all μ≤1, whence min[dˆD,dˆP]<dˆ for all μ.

B.3 Equilibrium with X,Y<d

We prove first that, for r<1, investment levels X˜ and Y˜ obtained in [24] are always Nash equilibria of the litigation game. Following the same procedure adopted for the American rule in Appendix A, the plaintiff does not deviate from the equilibrium X˜,Y˜ if and only if ΠP(X˜,Y˜)>−Y˜, which follows from the fact that the plaintiff has to pay the defendant’s legal fees even if her own expenditures are X=0. This implies that the plaintiff does not deviate if and only if μ11−r(1−r)>0, which is always true if r<1. Similarly, the defendant does not deviate if and only is ΠD(Y˜,X˜)>−V−X˜, which implies μ11−r<1+μ11−r, which is always satisfied for r<1. However, as shown in the text, the equilibrium fails to exist if r≥1.

The rest of the argument is in the text.

B.4 Equilibrium with X=Y=d

To prove the conditions under which X=Y=d is an equilibrium, we first show that increasing the expenditures beyond d reduces the plaintiff’s payoff if d≥dˆ. We then turn to the condition d≤dˉ. The argument for the defendant is symmetric and need not be made explicit here.

To see that the plaintiff will not increase X above d when d≥dˆ, we first consider her payoff

for X≥d and its first derivative of at X=Y=d:

with equality only for d=dˆ. The plaintiff will thus not increase his expenditures marginally above d.

To show that the plaintiff will not invest any much higher amount, we consider the second derivative of her payoff:

Obviously, this is strictly negative if r≤1. If r>1, the second derivative is positive for X=0 and eventually becomes negative as X increases. Leaving aside the restriction that the definition in eq. [42] only is valid for X≥d, one can easily check that the right-hand side of [42] becomes −d when X=0. Hence this right-hand side must have been increasing in some range before X reaches d. Since the first derivative is negative at X=d, the derivative must have been decreasing, i.e. the second derivative must have been negative somewhere in the range 0<X<d. However, a negative second derivative for any value of X implies that the second derivative is negative for all larger values of X too. As a consequence, (∂2ΠP(X,d))/(∂X2)<0 for all X≥d which also implies that (∂ΠP(X,d))/(∂X)<0 for all X≥d. Hence ΠP(X,d)<ΠP(d,d) for all X>d whence the plaintiff will never increase his investments above d when d≥dˆ. We can make a corresponding argument for the defendant to show that he will never increase her investments above d when d≥dˆ.

We now turn to the claim that neither of the parties will reduce expenditures below d if d≤dˉ. For this proof it is helpful to transform the plaintiff’s first-order condition for a payoff maximum with X,Y<d

into

Inserting this into the second derivative

implies

Hence, for r<1, there is no payoff minimum for the plaintiff in 0≤X<d if Y≤d.

Since for equal investments the derivative

is positive for Y≤d≤dˉ and strictly so, if Y<d, the derivative is also strictly positive for all X<Y and thus the plaintiff will always invest more than the defendant if Y<d and not less than the defendant if Y=d if r<1.

If r≥1, any extremum in 0≤X<d cannot be a maximum. Due to Y≤d≤dˉ we get after some algebra ΠP(0,Y)=−Y<ΠP(Y,Y)=μ1+μ(V+2Y)−2Y. Hence the plaintiff will again invest not less than the defendant and due to inequality [47] she will invest more unless Y=d.

By a symmetric argument for the defendant, we find that he also invests more than the plaintiff if X<d and the same amount if X=d.

Suppose that the defendant invests Y≤d. Then for X≤d we have

which for X=Y simplifies to

which is obviously positive if 1+1/μ−2r≤0 and positive due to Y≤d≤dˉ if 1+1/μ−2r>0 by the following argument:

μ ≥ max [ 1 , 2 r − 1 ] implies that Y≤rV1+μ−2r≤rV1+1/μ−2r which implies that the derivative is positive and strictly so, if Y<d.

2 r − 1 > μ > 1 2 r − 1 implies that 1+1/μ−2r<0 whence the derivative is positive.

μ ≤ min 1 , 1 2 r − 1 implies that Y≤rV1+1/μ−2r which implies that the derivative is positive and strictly so, if Y<d.

By a corresponding argument, we also get

with strict inequality for X<d.

B.5 Equilibrium with X<Y=d for μ<1 or Y<X=d for μ>1

To prove the various claims of Section 4.4, we first recall that according to eq. [48] (∂ΠP(X,d))/(∂X)|X=d is positive if d≤dˆ. This sufficient condition was also necessary, if μ<1, i.e. if the plaintiff is the weaker party (cf. the last alternative in the discussion of eq. [48]). Hence if d>dˆ and the plaintiff is the weaker party, we have (∂ΠP(X,d))/(∂X)|X=d<0 and thus the best reply of the plaintiff to the defendant’s choice of d is some X<d (X>d is excluded by Appendix B.1). The first-order condition for the optimal X is given by

We call this solution X˜(d).

Since (∂ΠP(X,d))/(∂X)|X=d<0 and limX→0(∂ΠP(X,d))/(∂X)=+∞ and the derivative does not display any discontinuity between zero and d, X˜(d) exists. Since the second derivative at X=X˜(d) can easily be shown to be negative for r<1 the solution X˜(d) is unique.

To make sure that (X˜(d),d) is a Nash equilibrium, we have to prove that d is the defendant’s best reply to (X˜(d). We know from the previous appendix, that μ<1 implies

By continuity, the inequality prevails if d slightly increases above dˆ. We thus have

Hence we have Xr<rYr−1(V+X+Y)−Yr/μ. Inserting this into the second derivative

which strictly increases in Xr, yields

Hence, reducing Y slightly results in a larger and thus still positive first derivative. Reducing Y further step by step always results in ever larger first derivatives and thus negative second derivatives by exactly the argument of eqs [53] through [54]. Hence, the defendant always gains by increasing his investment until he invests d, which completes the proof of d being the best reply to X˜(d).

The symmetric argument works for μ>1 and thus the plaintiff being the stronger party.

To prove the payoffs of eqs [30] and [31], note that eq. [51] implies dr+μX˜(d)r=μrX˜(d)r−1(V+X˜(d)+d). Inserting this into the plaintiff’s payoff for μ<1, yields:

The other expressions in eqs [30] and [31] can be derived accordingly.

Finally, we show that for r≤1/2, μ<1 implies ΠD(d,X˜(d))>−V. Suppose the reverse were true, i.e. −1rX˜(d)d≤−Vd. We could then insert this into

which is but a rewritten form of eq. [51] and obviously declines in X˜(d)d. Hence, this would imply

and thus

where the last inequality follows from the fact that for r≤1/2 the equilibrium with (X˜(d),d) may only occur for d<rV. However we know from d>dˉ that μ<1/(rVd+2r−1). Some simple algebra shows that this is compatible with inequality [58] only if d>rV. Hence ΠD(d,X˜(d))>−V must be true. The proofs for the other claims in the same paragraph follow the same structure and are omitted here.