Attorney fees in repeated relationships

Abstract

We investigate contracts between a law firm and a corporate client involved in a repeated relationship. In contrast to the previous literature pertaining to one-time interactions between clients and attorneys, we find that the contingent fee is not the best arrangement. Rather, the contingent fee is dominated by a contract which, we argue, an outside observer could not distinguish from simple hourly fee contract. This contract includes an hourly fee equal to the law firm’s opportunity cost, a lump sum, and a retention function. The lump sum payment is independent of the number of hours worked by the law firm and the outcome of the case. The repeated nature of the relationship allows the client to create a contract where the desire to maintain the relationship induces the law firm to exert the optimal level of effort in the current case.

Appendices

A Appendix

Before we proceed with the Proofs of Proposition 1 and Theorem 1, we verify the claims regarding Assumption 1. Let $F_i(\xi )$ and $f_i(\xi )$ denote the CDF and density of the noise term $\xi (i,\theta )$. Clearly $F_i(\xi (i,\theta )) =1/2 + \theta$. Taking a derivative with respect to $\theta$ yields $f_i(\xi ) \cdot \xi _{\theta } = 1$ or

$$\begin{aligned} f_i(\xi ) = \frac{1}{\xi _{\theta }} \end{aligned}$$

Now clearly $\xi$ is symmetrically distributed around zero if and only if $f_i(\xi ) = f_i(-\xi )$, which (from the above) holds if and only if $\xi _{\theta }(i,-\theta ) = \xi _{\theta }(i,\theta )$. Hence, $\xi$ is symmetrically distributed if and only if $\xi (i,-\theta ) = -\xi (i,\theta )$, which is Assumption 1.2.

Taking a derivative of $f_i(\xi )$ yields

$$\begin{aligned} \frac{d f_i}{d \xi } = \frac{-\xi _{\theta \theta }}{[\xi _{\theta }]^3} \end{aligned}$$

So Assumption 1.1 is the same as assuming that the density on $\xi$ is weakly increasing for $\xi <0$ and weakly decreasing for $\xi >0$. If there are no point masses, then this is a necessary and sufficient condition for the distribution to be uni-modal.

The above Equations are not defined if $\xi _{\theta }=0$. If $\xi _{\theta }=0$ over some range $\theta \in [a,b]$, then there is a point mass at $\xi (i,a)$, and $F_i$ takes a discrete jump at this point. However, notices that Assumption 1.1 assures that $\xi _{\theta } \ge 0$ achieves its minimum at zero. Hence the only possible point mass is at $\xi (i,0) =0$. Again, we have a uni-modal distribution.

B Proof of Proposition 1

Let $\mathcal {Q}= z P + (1-z)$ for $z \in (0,1]$. Clearly $\mathcal {Q}$ is a more general retention probability function. Rather than proving Proposition 1, we prove a more general result. We use the following Assumption to make the statement of the more general result easier.

Assumption 2

$\Gamma$ is sufficiently small and one of the following holds:

• 1. $z=1$, or

• 2. $\delta$ is sufficiently large.

Proposition 1 uses Assumption 2.1.

Proposition 3

Let $\alpha =0$, $w=c$ and $R_i=h_i^*$. Fix $z \in (0,1]$. Fix a value i and then fix $b_j$ and $\Delta _j$ for $j \not =i$. Further, assume that for $j \not =i$, the law firm must set $h_j=r_j =h_j^*$. If Assumptions 1 and 2 hold, then there exist L, $\Delta _i$, and $b_i$ such that in any period t with $i_t=i$:

• 1. The client prefers the long term contract to any one shot contract.

• 2. The long term contract induces the law firm to set hours efficiently and report hours truthfully.

The firm acts to maximize it’s discounted expected stream of profits. Hence, he acts to satisfy first order conditions of his Bellman Equation

$$\begin{aligned} {\hat{V}}(i) = \max _{h,r} \left\{ L + \alpha {\hat{A}} -c\cdot h + w \cdot r + \delta {\hat{\mathcal {Q}}} {\tilde{V}} \right\} \end{aligned}$$

(14)

The firm’s first order condition with respect to h is

$$\begin{aligned} \frac{d {\hat{\mathcal {Q}}}}{dh} = \frac{c-\alpha {\hat{A}}_h}{\delta {\tilde{V}}} \end{aligned}$$

(15)

while the first order condition with respect to r is

$$\begin{aligned} \frac{d {\hat{\mathcal {Q}}}}{dr} = \frac{-w}{\delta {\tilde{V}}} \end{aligned}$$

(16)

Our next step is to replace the derivatives of ${\hat{\mathcal {Q}}}$ in Eqs. 15 and 16 with more explicit expressions. We recall that the expectation of P is

Taking a derivative of Eq. 11 with respect to h yields

$$\begin{aligned} \frac{d {\hat{P}}}{d h}= & {} \tau ({\overline{\theta }}_i) \frac{d\overline{\theta }_i}{dh} -\tau ({\underline{\theta }}_i) \frac{d\underline{\theta }_i}{dh} +\int _{{\underline{\theta }}_i}^{{\overline{\theta }}_i} \frac{d\tau }{dh} d\theta -\frac{d{\overline{\theta }}_i}{dh}\\= & {} -\tau ({\underline{\theta }}_i) \frac{d{\underline{\theta }}_i}{dh} + \left[ \tau ({\overline{\theta }}_i)-1\right] \frac{d{\overline{\theta }}_i}{dh} +\int _{{\underline{\theta }}_i}^{{\overline{\theta }}_i} \frac{d\tau }{dh} d\theta \\= & {} \int _{{\underline{\theta }}_i}^{{\overline{\theta }}_i} \frac{d\tau }{dh} d\theta . \end{aligned}$$

The final equality follows from the relationship between $\tau$ and ${\underline{\theta }}_i$ and ${\overline{\theta }}_i$. In particular, if $\tau ({\overline{\theta }}_i) <1$, then ${\overline{\theta }}_i =1/2$ and $\frac{d{\overline{\theta }}_i}{dh} =0$. Likewise, if $\tau ({\underline{\theta }}_i) >0$, then ${\underline{\theta }}_i =-1/2$ and $\frac{d{\underline{\theta }}_i}{dh} =0$. Since $\frac{d {\hat{\mathcal {Q}}}}{d h} = \frac{d {\hat{\mathcal {Q}}}}{d {\hat{P}}} \cdot \frac{d {\hat{P}}}{d h}$ and $\tau (i,h,r,\theta ) = \frac{A-c\cdot r -b_i}{\Delta _i}$ we have

$$\begin{aligned} \frac{d {\hat{\mathcal {Q}}}}{d h} = z\int _{{\underline{\theta }}_i}^{\overline{\theta }_i} \frac{A_h}{\Delta _i} d\theta \end{aligned}$$

(17)

Combining Eqs. 15 and 17 yields Eq. 18.

$$\begin{aligned} \frac{c - \alpha {\hat{A}}_h}{\delta {\tilde{V}}} = z\int _{\underline{\theta }_i}^{{\overline{\theta }}_i} \frac{A_h}{\Delta _i} d\theta \end{aligned}$$

(18)

By arguments analogous to those used in the derivation of Eq. 17, the derivative of ${\hat{\mathcal {Q}}}$ with respect to r is

$$\begin{aligned} \frac{d {\hat{\mathcal {Q}}}}{d r} = z\int _{{\underline{\theta }}_i}^{\overline{\theta }_i} \frac{-c}{\Delta _i} d\theta = \frac{-c(\overline{\theta }_i-{\underline{\theta }}_i)z}{\Delta _i} \end{aligned}$$

(19)

Combining Eqs. 16 and 19 yields Eq. 20 below.

$$\begin{aligned} \frac{\Delta _i}{z} = \left( \frac{c}{w} \right) ({\overline{\theta }}_i-{\underline{\theta }}_i) \delta {\tilde{V}} \end{aligned}$$

(20)

Equations 18 and 20 both include ${\tilde{V}}$ which is endogenous. This motivates us to divide Eq. 18 by Eq. 20 to remove common terms including ${\tilde{V}}$. Doing this yields Eq. 21.

$$\begin{aligned} \int _{{\underline{\theta }}_i}^{{\overline{\theta }}_i} A_h d\theta = \left( \frac{c}{w}\right) (c-\alpha {\hat{A}}_h) \cdot \left( \overline{\theta }_i-{\underline{\theta }}_i\right) \end{aligned}$$

(21)

Equation 21 is a necessary condition for both first order conditions to hold.

Proof of Lemma 1

By definition ${\hat{A}}_h^* =c$. If $A_{h,\theta }^*=0$, then $A_h^*$ is constant in $\theta$. That is to say $A_h^*={\hat{A}}_h^* =c$ within the integral in Eq. 21. This reduces Eq. 21 to $w=(1-\alpha ) c$.

$\square$

As in the body, we set $\alpha =0$ and $w=c$ henceforth. Doing this, and requiring that Eq. 21 holds when $h=r=h_i^*$ yields the first order condition for h presented in the body

If $\alpha =0$, $w=c$, and Eq. 21 holds, then Eqs. 18 and 20 both reduce to

$$\begin{aligned} \frac{\Delta _i}{z} = ({\overline{\theta }}_i-{\underline{\theta }}_i) \delta {\tilde{V}} \end{aligned}$$

(22)

We convert Eq. 22 into Eq. 24 by replacing ${\tilde{V}}$ with an endogenously determined value. If $\alpha =0$, $w=c$, and the first order conditions hold so that $h=r=h^*_i$, then the firm’s Bellman Equation 5 becomes

$$\begin{aligned} {\hat{V}}(i) = L + \delta {\hat{\mathcal {Q}}}_i^* {\tilde{V}} \end{aligned}$$

Taking expectations over i, and recalling that, e.g., ${\tilde{\mathcal {Q}}}^* = E_i({\hat{\mathcal {Q}}}_i^*)$, we have

$$\begin{aligned} {\tilde{V}} = \frac{L}{1- \delta {\tilde{\mathcal {Q}}}^*} \end{aligned}$$

(23)

Plugging Eq. 23 into Eq. 22 and setting $r=h=h_i^*$ yields

$$\begin{aligned} 1=\left( \frac{{\overline{\theta }}_i^*-{\underline{\theta }}_i^*}{\Delta _i}\right) \left( \frac{z\delta L}{1- \delta {\tilde{\mathcal {Q}}}^*}\right) \end{aligned}$$

(24)

Setting $z=1$ makes ${\tilde{\mathcal {Q}}}^* = {\tilde{P}}^*$, and turns Eq. 24 into Eq. 12.

We now argue that it is possible to solve Eqs. 13 and 24 simultaneously. The first step to this is to show that we have some freedom in choosing pairs $(b_i,\Delta _i)$ to solve Eq. 13. There are two simple cases which we dispense with first. If $A_{h,\theta }^* =B_h^* \cdot \xi _{\theta }=0$, then Eq. 13 holds automatically. Likewise, if $b_i$ and $\Delta _i$ are such that $-1/2 = {\underline{\theta }}^*_i$ and ${\overline{\theta }}^*_i= 1/2$, then Eq. 13 holds by the definition of $h_i^*$.

Equations 8 and 9 imply that if $-1/2 = {\underline{\theta }}^*_i$ and ${\overline{\theta }}^*_i= 1/2$, then $\Delta _i \ge A(i,h^*_i,{\overline{\theta }}_i^*) - A(i,h^*_i,{\underline{\theta }}_i^*)$, while if $-1/2< {\underline{\theta }}^*_i< {\overline{\theta }}^*_i< 1/2$, then

$$\begin{aligned} \Delta _i = A\left( i,h^*_i,{\overline{\theta }}_i^*\right) - A\left( i,h^*_i,\underline{\theta }_i^*\right) . \end{aligned}$$

(25)

Lemma B.1

Set $\alpha =0$ and $w=c$. Let Assumption 1 hold, fix i and assume that $B_h(i,h^*_i) \not =0$. If $0 < \Delta _i \le A(i,h^*_i,1/2) - A(i,h^*_i,-1/2)$, then $\exists !b_i(\Delta _i)$ with $b_i +c\cdot h^*_i \in [A(i,h^*_i,-1/2), A(i,h^*_i,0))$ such that Eqs. 13 and 25 hold for ${\underline{\theta }}^*_i(b_i)$ and ${\overline{\theta }}^*_i(b_i,\Delta _i)$.

Furthermore, $b_i(\Delta _i)$ is strictly decreasing in $\Delta _i$, and ${\underline{\theta }}^*_i(b_i) = -{\overline{\theta }}^*_i(b_i,\Delta _i)$.

Before we prove Lemma B.1, there are some preliminaries to attend. We consider the comparative statics from changing $b_i$ and $\Delta _i$. Let ${\underline{A}}^*= A(i, h^*_i, {\underline{\theta }}^*_i)$ and ${\overline{A}}^*= A(i, h^*_i, {\overline{\theta }}^*_i)$. Since it is the case for Lemma B.1, assume that $B_h^* \not =0$ and $-1/2< {\underline{\theta }}^*_i< {\overline{\theta }}^*_i<1/2$. In this case ${\underline{\theta }}^*_i$ is defined by ${\underline{A}}^*-ch^*_i-b_i =0$. It follows that

$$\begin{aligned} \frac{d {\underline{\theta }}^*_i}{d b_i} = \frac{1}{{\underline{A}}^*_{\theta }} >0 \end{aligned}$$

Given ${\underline{\theta }}^*_i$, we determine ${\overline{\theta }}^*_i$ by $\int _{{\underline{\theta }}^*_i}^{{\overline{\theta }}^*_i} A_h d \theta = c({\overline{\theta }}^*_i- {\underline{\theta }}^*_i)$. It follows that

$$\begin{aligned} \frac{d {\overline{\theta }}^*_i}{d {\underline{\theta }}^*_i} = \frac{{\underline{A}}^*_h -c}{{\overline{A}}^*_h -c} <0 \end{aligned}$$

Using the functional form $A={\hat{A}} +B\cdot \xi$ and Assumption 1, we have

$$\begin{aligned} \frac{d {\overline{\theta }}^*_i}{d {\underline{\theta }}^*_i} =\frac{{\hat{A}}^*_h -B^*_h\cdot \xi ({\underline{\theta }}^*_i)-c}{{\hat{A}}^*_h -B^*_h\cdot \xi ({\overline{\theta }}^*_i)-c} =\frac{\xi ({\underline{\theta }}^*_i)}{\xi ({\overline{\theta }}^*_i)}= -1 \end{aligned}$$

(26)

The final equality derives from the following. We know that when ${\underline{\theta }}^*_i= -\frac{1}{2}$, then ${\overline{\theta }}^*_i= \frac{1}{2}= - {\underline{\theta }}^*_i$. By Assumption 1.2, the final equality must then hold at ${\underline{\theta }}^*_i= -\frac{1}{2}$. Furthermore, as long as ${\underline{\theta }}^*_i=- {\overline{\theta }}^*_i$, the equality will hold, and as long as the equality holds along the path from $-\frac{1}{2}$ to ${\underline{\theta }}^*_i$, then ${\underline{\theta }}^*_i=- {\overline{\theta }}^*_i$. Hence, Eq. 26 holds, as does the following Lemma.

Lemma B.2

If $\alpha =0$, $w=c$ and Assumption 1 holds, then ${\overline{\theta }}^*_i= - {\underline{\theta }}^*_i$.

Given the negative symmetry of $\xi$ it follows that $\xi _{\theta }(-\theta ) = \xi _{\theta }(\theta )$ so that ${\underline{A}}^*_{\theta } = {\overline{A}}^*_{\theta }$. Using the above results, we have that

$$\begin{aligned} \frac{d {\overline{\theta }}^*_i}{d b_i} = \frac{d {\overline{\theta }}^*_i}{d {\underline{\theta }}^*_i} \cdot \frac{d {\underline{\theta }}^*_i}{d b_i} =\frac{-1}{{\underline{A}}^*_{\theta }} <0 \end{aligned}$$

Finally, $\Delta _i ={\overline{A}}^*-{\underline{A}}^*$, which leads us to

$$\begin{aligned} \frac{d \Delta _i}{d {\underline{\theta }}^*_i} = {\overline{A}}^*_{\theta } \cdot \frac{d {\overline{\theta }}^*_i}{d {\underline{\theta }}^*_i} - {\underline{A}}^*_{\theta } = -2 {\underline{A}}^*_{\theta } \end{aligned}$$

and

$$\begin{aligned} \frac{d \Delta _i}{d b_i} = \frac{d \Delta _i}{d {\underline{\theta }}^*_i} \cdot \frac{d {\underline{\theta }}^*_i}{d b_i} = -2 \end{aligned}$$

Proof of Lemma B.1

Given the above, we can define $b_i(\Delta _i) = {\hat{A}}^*_i -ch_i^* - \frac{1}{2}\Delta _i$ for $0 < \Delta _i \le A(i,h^*_i, \frac{1}{2}) - A(i,h^*_i, -\frac{1}{2})$. It remains only to verify uniqueness. To this end we first consider if, given ${\underline{\theta }}^*_i$ it is possible to choose ${\overline{\theta }}^*_i\not = -{\underline{\theta }}^*_i$. Equation 13 can be written as

$$\begin{aligned} \int _{{\underline{\theta }}^*_i}^{{\overline{\theta }}^*_i} A_h^* d \theta - c({\overline{\theta }}^*_i-{\underline{\theta }}^*_i)=0. \end{aligned}$$

Now if $B^*_h >0$ (resp. $<0$) then we know that $A_h^* < c$ (resp. $>c$) if and only if $\theta <0$. Hence, ${\underline{\theta }}^*_i<0 < {\overline{\theta }}^*_i$. The derivative of the LHS of the above equation w.r.t. ${\overline{\theta }}^*_i$ is ${\overline{A}}^*_h -c = {\hat{A}}^*_h +B_h^* \cdot \xi ({\overline{\theta }}^*_i) -c = B_h^* \cdot \xi ({\overline{\theta }}^*_i)$ which does not change sign for ${\overline{\theta }}^*_i>0$. Hence as one moves away from ${\overline{\theta }}^*_i=-{\underline{\theta }}^*_i$ the difference between the LHS and 0 only gets larger. This verifies that ${\overline{\theta }}^*_i= - {\underline{\theta }}^*_i$ is the only possibility. Now $\Delta _i = A(i,h^*_i,{\overline{\theta }}^*_i) - A(i,h^*_i,{\underline{\theta }}^*_i) = A(i,h^*_i,-{\underline{\theta }}^*_i) - A(i,h^*_i,{\underline{\theta }}^*_i)$. The RHS is monotonically decreasing in ${\underline{\theta }}^*_i$. Hence there is a unique value of ${\underline{\theta }}^*_i$ for a given $\Delta _i$. It is trivial from the definition of ${\underline{\theta }}^*_i$ that there is a unique $b_i$ for each value of ${\underline{\theta }}^*_i$ which establishes uniqueness. $\square$

Lemma B.1 establishes the relationship $b_i(\Delta _i)$ for $0 < \Delta _i \le A(i,h^*_i, \frac{1}{2}) - A(i,h^*_i, -\frac{1}{2})$ and $B_h^* \not =0$. It is convenient to have $b_i(\Delta _i)$ defined in all cases. To that end, if $\Delta _i > A(i,h^*_i, 1/2) - A(i,h^*_i, -1/2)$, then we set $b_i(\Delta _i) = A(i,h^*_i, -1/2) - c\cdot h^*_i$. Finally, if $0 < \Delta _i \le A(i,h^*_i, \frac{1}{2}) - A(i,h^*_i, -\frac{1}{2})$ and $B_h^*=0$, then we set $b_i(\Delta _i)$ so that ${\underline{\theta }}^*_i(b_i) = -{\overline{\theta }}^*_i(b_i,\Delta _i)$.

So long as $b_i=b_i(\Delta _i)$, then we know that $(b_i,\Delta _i)$ solves Eq. 13. Now in trying to choose $\Delta _i$ to satisfy Eq. 24, we see that both $({\overline{\theta }}^*_i-{\underline{\theta }}^*_i)$ and ${\tilde{\mathcal {Q}}}^* = z \cdot \sum _j q_j \cdot P^*_j + (1-z)$ depends upon $\Delta _i$. However, this dependence is well behaved.

Lemma B.3

If $\alpha =0$, $w=c$, and Assumption 1 holds, then $\frac{\Delta _i}{{\overline{\theta }}^*_i-{\underline{\theta }}^*_i}$ is weakly increasing in $\Delta _i$ and unbounded above.

Proof

Let $y = \frac{\Delta _i}{{\overline{\theta }}^*_i- {\underline{\theta }}^*_i}$. If $\Delta _i > A(i,h^*_i, \frac{1}{2}) -A(i,h^*_i, -\frac{1}{2})$, then $y = \Delta _i$. This establishes both that the derivative is positive in this range and that the function is unbounded above. Now consider $\Delta _i < A(i,h^*_i, \frac{1}{2}) -A(i,h^*_i, -\frac{1}{2})$. Let $x = ({\overline{\theta }}^*_i-{\underline{\theta }}^*_i)^2 \cdot \frac{dy}{d \Delta _i}$. Clearly x and $\frac{dy}{d \Delta _i}$ have the same sign. $x= ({\overline{\theta }}^*_i-{\underline{\theta }}^*_i) - \Delta _i\left( \frac{d {\overline{\theta }}^*_i}{d \Delta _i} - \frac{d {\underline{\theta }}^*_i}{d \Delta _i}\right) = ({\overline{\theta }}^*_i-{\underline{\theta }}^*_i) - \Delta _i\left( \frac{d {\overline{\theta }}^*_i}{d b_i} - \frac{d {\underline{\theta }}^*_i}{d b_i} \right) \cdot \frac{d b_i}{d \Delta _i} =({\overline{\theta }}^*_i-{\underline{\theta }}^*_i) - \frac{\Delta _i}{{\underline{A}}^*_\theta }$. We observe that $\frac{\Delta _i}{{\overline{\theta }}^*_i-{\underline{\theta }}^*_i} \le {\underline{A}}^*_\theta$ because the first term is the average slope which by Assumption 1.1 is weakly less than the slope at the edge, ${\underline{A}}^*_\theta$. Hence $x >0$. Finally, there is a kink at $\Delta _i = A(i,h^*_i, \frac{1}{2}) -A(i,h^*_i, -\frac{1}{2})$. However, at a kink a function is inarguably increasing if both its left and right hand derivatives are positive. $\square$

Lemma B.4

Let $\alpha =0$, $w=c$, and $b_i = b_i(\Delta _i)$ as defined above. If Assumption 1 holds, then ${\hat{P}}^*_i=\frac{1}{2}$ for $\Delta _i \le A(i,h^*_i, \frac{1}{2}) -A(i,h^*_i, -\frac{1}{2})$ and is decreasing in $\Delta _i$ for $\Delta _i > A(i,h^*_i, \frac{1}{2}) -A(i,h^*_i, -\frac{1}{2})$.

Proof

We first consider the case in which $\Delta _i \le A(i,h^*_i, \frac{1}{2}) -A(i,h^*_i, -\frac{1}{2})$. In this case $b_i = A^*(i,{\underline{\theta }}^*_i) -c\cdot h_i^*$ so that

$$\begin{aligned} A^*-c\cdot h^*_i -b_i= & {} A^* - {\underline{A}}^*= \left[ {\hat{A}}^* +B^*\cdot \xi (\theta )\right] - \left[ {\hat{A}}^* +B^*\cdot \xi ({\underline{\theta }}^*_i)\right] \\= & {} B^*\left[ \xi (\theta ) -\xi ({\underline{\theta }}^*_i)\right] \end{aligned}$$

Also

$$\begin{aligned} \Delta _i= & {} {\overline{A}}^*-{\underline{A}}^*= \left[ {\hat{A}}^* + B^*\cdot \xi (i,{\overline{\theta }}^*_i)\right] - \left[ {\hat{A}}^* + B^*\cdot \xi (i,{\underline{\theta }}^*_i)\right] \\= & {} B^*\left[ \xi (i,{\overline{\theta }}^*_i)-\xi (i,{\underline{\theta }}^*_i)\right] = 2B^* \cdot \xi ({\overline{\theta }}^*_i) \end{aligned}$$

Hence we have that

$$\begin{aligned} {\hat{P}}^*_i= & {} \int _{{\underline{\theta }}^*_i}^{{\overline{\theta }}^*_i} \frac{B^* \left[ \xi (\theta ) -\xi ({\underline{\theta }}^*_i)\right] }{2B^* \cdot \xi ({\overline{\theta }}^*_i)} d \theta +\left( \frac{1}{2}-{\overline{\theta }}^*_i\right) \\= & {} \int _{{\underline{\theta }}^*_i}^{{\overline{\theta }}^*_i}\frac{\xi (\theta ) }{2\cdot \xi ({\overline{\theta }}^*_i)} d \theta + \int _{{\underline{\theta }}^*_i}^{{\overline{\theta }}^*_i}\frac{\xi ({\overline{\theta }}^*_i) }{2\cdot \xi ({\overline{\theta }}^*_i)} d \theta +\left( \frac{1}{2}-{\overline{\theta }}^*_i\right) \\= & {} 0 + \frac{1}{2}\left( {\overline{\theta }}^*_i-{\underline{\theta }}^*_i\right) + \frac{1}{2}-{\overline{\theta }}^*_i= \frac{1}{2}\end{aligned}$$

On the other hand, if $\Delta _i > A(i,h^*_i, \frac{1}{2}) -A(i,h^*_i, -\frac{1}{2})$ then $b_i = A(i,h^*_i, -\frac{1}{2}) - c\cdot h^*_i$, ${\underline{\theta }}^*_i= -\frac{1}{2}$ and ${\overline{\theta }}^*_i= \frac{1}{2}$. That is, none of these parameters depend upon $\Delta _i$. Hence

$$\begin{aligned} \frac{d {\hat{P}}^*_i}{d \Delta _i} = -\left( \frac{1}{\Delta _i} \right) ^2 \int _{-\frac{1}{2}}^{\frac{1}{2}} \left( A^* - c \cdot h^*_i - b_i\right) d \theta <0 \end{aligned}$$

$\square$

Lemma B.5

Let Assumption 1 hold, and set $\alpha =0$, $w=c$ and $b_i =b_i(\Delta _i)$ as defined above.

In this case $\frac{\Delta _i (1-\delta {\tilde{\mathcal {Q}}}^*)}{{\overline{\theta }}^*_i-{\underline{\theta }}^*_i}$ is weakly increasing in $\Delta _i$ and unbounded above.

Proof

We note that if $j \not =i$, then $P_j^*$ does not depend upon $\Delta _i$, hence by Lemma B.4${\tilde{\mathcal {Q}}}^*= z\sum _j P_j^*+ (1-z)$ is weakly decreasing in $\Delta _i$. The result then follows from an application of Lemma B.3. $\square$

With Lemma B.5 in hand, there are no difficulties if $\delta L$ is large enough. In particular, if $\delta L$ is large enough that the LHS of Eq. 24 is larger than the RHS when $\Delta _i = A(i,h^*_i, 1/2) - A(i,h^*_i, -1/2)$, then we can simply keep increasing $\Delta _i$ until Eq. 24 holds. We now address the case in which $\Delta _i < A(i,h^*_i, 1/2) - A(i,h^*_i, -1/2)$.

Lemma B.6

Let $b_i = b_i(\Delta _i)$ as defined above. $\lim _{\Delta _i \rightarrow 0} \frac{\Delta _i}{{\overline{\theta }}^*_i- {\underline{\theta }}^*_i} = A_{\theta } (i,h^*_i,0)$.

Proof

$\lim _{\Delta _i \rightarrow 0} \frac{\Delta _i}{{\overline{\theta }}^*_i- {\underline{\theta }}^*_i} =\lim _{\Delta _i \rightarrow 0} \frac{A(i,h^*_i,\overline{\theta }_i^*) - A(i,h^*_i,{\underline{\theta }}_i^*)}{\overline{\theta }_i^*-{\underline{\theta }}_i^*}= A_{\theta } (i,h^*_i,0)$. The first equality follows from Eq. 25 which holds for $\Delta _i$ sufficiently small. The second equality is just the definition of a derivative. $\square$

We rewrite Eq. 24 as

$$\begin{aligned} z \delta L = \left( \frac{\Delta _i}{{\overline{\theta }}_i^*-{\underline{\theta }}_i^*} \right) \left( 1- \delta {\tilde{\mathcal {Q}}}^*\right) \end{aligned}$$

(27)

We use $1-\delta {\tilde{\mathcal {Q}}}^* = 1-\delta (1-z) -z \delta {\tilde{P}}^*$ and ${\tilde{P}}^* = q_i \cdot {\hat{P}}_i^* + \sum _{j \not = i} q_j \cdot {\hat{P}}_j^*$ to transform Eq. 27 into

$$\begin{aligned} z \delta L = \left( \frac{\Delta _i}{{\overline{\theta }}^*_i-{\underline{\theta }}^*_i} \right) \left( 1-\delta (1-z)- z \delta \left( \sum _{j \not = i} q_j \cdot {\hat{P}}_j^* \right) - z \delta \cdot q_i \cdot {\hat{P}}_i^* \right) \end{aligned}$$

(28)

Let $K_i = 1-\delta (1-z) -\delta z \left( \sum _{j \not = i} q_j {\hat{P}}^*_j\right) - \frac{q_i z \delta }{2}$. We note that $K_i \in (0,1)$ and is constant in $\Delta _i$ and $b_i$. We use Eq. 11 to transform Eq. 28 to

$$\begin{aligned} z\delta L = \left( \frac{\Delta _i}{{\overline{\theta }}^*_i-{\underline{\theta }}^*_i} \right) \left( K_i +z \delta \cdot q_i \cdot {\overline{\theta }}^*_i- \delta z\cdot q_i \int _{{\underline{\theta }}^*_i}^{{\overline{\theta }}^*_i} \left( \frac{A^*-c\cdot h^*-b}{\Delta _i}\right) d\theta \right) \nonumber \\ \end{aligned}$$

(29)

We break Eq. 29 into two pieces, divide by z, and factor $\frac{\Delta _i}{{\overline{\theta }}^*_i-{\underline{\theta }}^*_i}$ out of the second piece to arrive at

$$\begin{aligned} \delta L = \left( \frac{\Delta _i}{{\overline{\theta }}^*- {\underline{\theta }}^*}\right) \left( \frac{ K_i + \delta z q_i {\overline{\theta }}^*_i}{z}\right) - \delta q_i \int _{{\underline{\theta }}^*}^{{\overline{\theta }}^*} \frac{A*-ch^*-b_i}{{\overline{\theta }}^*-{\underline{\theta }}^*} d \theta \end{aligned}$$

(30)

Recall that ${\bar{X}}$ is the minimum (over i) efficiency loss from a one shot contract. We notice that $\lim _{\Delta _i \rightarrow 0} b_i(\Delta _i) =\beta _i \equiv A(i,h_i^*,0)-c\cdot h_i^*$. We have that

$$\begin{aligned} \lim _{\Delta _i \rightarrow 0} \left( \frac{\Delta _i}{{\overline{\theta }}^*_i-{\underline{\theta }}^*_i}\right) \left( \frac{ K_i + \delta z q_i {\overline{\theta }}^*_i}{z}\right)= & {} A_{\theta }(i,h_i^*,0) \left( \frac{ K_i + \delta z q_i/2 }{z}\right) \nonumber \\< & {} \delta {\bar{X}} \left( \frac{ K_i + \delta z q_i/2 }{z}\right) \end{aligned}$$

(31)

$$\begin{aligned} \lim _{\Delta _i \rightarrow 0} \int _{{\underline{\theta }}^*_i}^{{\overline{\theta }}^*_i} \left( \frac{A^*-c\cdot h_i^*-b_i}{{\overline{\theta }}^*_i-{\underline{\theta }}^*_i}\right) d\theta= & {} A(i,h_i^*,0)-c\cdot h_i^* -\beta _i =0 \end{aligned}$$

(32)

We must now consider which part of Assumption 2 holds. If $z=1$, then we have

$$\begin{aligned} \frac{ K_i + \delta z q_i/2 }{z} = 1 - \delta \left( \sum _{j\not =i} q_j \cdot {\hat{P}}_j^*\right) <1. \end{aligned}$$

Consequently, the limit as $\Delta _i \rightarrow 0$ of the RHS of Eq. 30 is less than $\delta {\bar{X}} (K_i + \delta q_i {\overline{\theta }}^*_i) < \delta {\bar{X}}$. Hence, $\exists \epsilon >0$ such that if $L = {\bar{X}} -\epsilon$, then the contract is feasible and the client strictly prefers the current expected payoff from this contract to any one shot contract.

We next turn to the case in which $\delta$ is sufficiently close to one. We can see that

$$\begin{aligned} \lim _{\delta \rightarrow 1} \frac{ K_i + \delta z q_i/2 }{z}= & {} lim_{\delta \rightarrow 1} \frac{1-\delta (1-z) -\delta z \left( \sum _{j \not = i} q_j {\hat{P}}^*_j\right) }{z}\\= & {} \frac{z - z\left( \sum _{j \not = i} q_j {\hat{P}}^*_j\right) }{z} = 1- \sum _{j \not = i} q_j {\hat{P}}^*_j <1 \end{aligned}$$

Hence for $\delta$ sufficiently close to 1, it is again the case that the limit as $\Delta _i \rightarrow 0$ of the RHS of Eq. 30 is less than $\delta {\bar{X}}$.

It remains only to establish that the firm cannot make itself better off by attempting to defraud the client. The choice of $r=h=h_i^*$ is locally optimal. However, choosing h and r so as to defraud the client is a non-local alternative. Being a non-local alternative, it must lead to a discrete drop in the probability of being retained. Since being retained has a strictly positive value, this leads to discrete drop in the expected present value of future payoffs. For $\Gamma$ sufficiently small, this lowers the payoff to the law firm. This concludes the proof of Proposition 1.

C Proof of Theorem 1 and Proposition 2

We note that the difference between these results is that in Theorem 1 Assumption 2.1 holds, while in Proposition 2 Assumption 2.2 hold.

We proceed with $\Delta _i$ determining $b_i$, ${\underline{\theta }}^*_i$, and ${\overline{\theta }}^*_i$. This assures that Eq. 13 holds for each i. It remains to show that we can also simultaneously satisfy Eq. 24 for each i. Let $\rho _i(\Delta _1, \ldots \Delta _n) = \left( \frac{\Delta _i}{{\overline{\theta }}^*_i-{\underline{\theta }}^*_i} \right) \left( 1 - \delta {\tilde{\mathcal {Q}}}^* \right)$. Equation 24 can be written as $z \delta L = \rho _i$. Let $\Delta _i^e$ denote the solution to Eq. 24 given $\Delta = (\Delta _1 \ldots \Delta _n)$.

Lemma C.1

Let $\alpha =0$, $w=c$, and $b_i = b_i(\Delta _i)$ as defined above. $\Delta _i^e$ is continuous in and weakly decreasing in $\Delta _j$.

Proof

We notice that $\Delta _i^e$ acts to set $z\delta L = \rho _i$. The function $\rho _i$ is: continuous in both $\Delta _i$ and $\Delta _j$, strictly monotonically increasing in $\Delta _i$, and weakly monotonically increasing in $\Delta _j$. Hence an infinitesimal increase in $\Delta _j$ must be met by no more than an infinitesimal decrease in $\Delta _i^e$.$\square$

Let $\Delta ^e \equiv (\Delta _1^e, \ldots \Delta _n^e)$. The Brouwer Fixed Point Theorem states that there is a fixed point if $\Delta ^e$ is a continuous function on a compact and convex domain. Lemma C.1 states that $\Delta ^e$ is continuous. However, the domain of $\Delta$ is $\mathfrak {R}^n_{++}$ which is not compact. We now demonstrate that there is a compact and convex sub-domain of $\mathfrak {R}^n_{++}$ which $\Delta ^e$ maps into itself. This will suffice, since there must be a fixed point on this sub-domain.

We use monotonicity to establish an upper bound for $\Delta _i^e$ which we denote as ${\bar{\Delta }}_i$. Let us assume for the moment that $\bar{\Delta }_i > A(i,h_i^*,\frac{1}{2}) - A(i,h_i^*,-\frac{1}{2})$. In this case ${\overline{\theta }}^*_i-{\underline{\theta }}^*_i=1$, and

$$\begin{aligned} \Delta _i^e = \frac{z\delta L}{(1-\delta {\tilde{\mathcal {Q}}}^*)} \le \frac{\delta L}{(1-\delta )} \end{aligned}$$

Hence $\Delta _i^e \le {\bar{\Delta }}_i \equiv \max \{ \frac{\delta L}{(1-\delta )}, A(i,h_i^*,\frac{1}{2}) - A(i,h_i^*,-\frac{1}{2}),\}$.

We now establish a lower bound for $\Delta _i^e$ which we denote as ${\underline{\Delta }}_i$. Clearly 0 is a lower bound for $\Delta _i^e$. However, neither $\tau$ nor $\mathcal {Q}$ are defined for $\Delta _i =0$. Hence, what we need is to establish a lower bound ${\underline{\Delta }}_i >0$. We work with $\Delta _j \le {\bar{\Delta }}_j$ in which case we have

$$\begin{aligned} \frac{\Delta _i^e}{{\overline{\theta }}^*_i- {\underline{\theta }}^*_i} = \frac{z\delta L}{1-{\tilde{\mathcal {Q}}}^*} > z\delta L \end{aligned}$$

(33)

The strict inequality follows since ${\hat{\mathcal {Q}}}^*_j =0$ only if $\Delta _j = \infty$ and $z=1$. We know from Lemmas B.3 and B.6 that we may set the LHS of Eq. 33 to any value strictly greater than $A_{\theta }(i,h_i^*,0)$ which itself is strictly less than $\delta {\bar{X}}$. Hence if $z=1$, then we can fix $\epsilon$ with $0< \epsilon < X - \frac{A_{\theta }(i,h^*_i, 0)}{\delta }$. Set $L= {\bar{X}} - \epsilon$. Now since $\delta L > A_{\theta }(i,h^*_i, 0)$ and $\frac{\Delta _i}{{\overline{\theta }}^*_i- {\underline{\theta }}^*_i}$ is increasing, it follows that there exists a ${\underline{\Delta }}_i >0$ such that $\frac{{\underline{\Delta }}_i}{{\overline{\theta }}^*_i- {\underline{\theta }}^*_i} =\delta L$. From Eq. 33 it follows that $\Delta _i^e > \underline{\Delta }_i$ as long as $\Delta _j \le {\bar{\Delta }}_j$.

Now on the other hand, suppose that $z<1$. In this case we note that

$$\begin{aligned} \lim _{\delta \rightarrow 1}\frac{z\delta L}{1-{\tilde{\mathcal {Q}}}^*} = \lim _{\delta \rightarrow 1}\frac{z\delta L}{1 -\delta (1-z) - {\tilde{P}}^*} = \frac{L}{1-{\tilde{P}}^*} > L \end{aligned}$$

Hence, for $\delta$ sufficiently close to one, we can for identical reasons find a ${\underline{\Delta }}_i >0$ such that $\frac{\underline{\Delta }_i}{{\overline{\theta }}^*_i- {\underline{\theta }}^*_i} =\delta L$. Again, from Eq. 33 it follows that $\Delta _i^e > {\underline{\Delta }}_i$ as long as $\Delta _j \le {\bar{\Delta }}_j$.

With this lower bound established, we may apply the Brouwer Fixed Point Theorem. A fixed point to $\Delta ^e$ is a simultaneous solution to Eq. 24 for each value of i. Hence, the firm voluntarily sets $h_j=r_j=h_j^*$ for each value of j. This renders moot the fact that $\Delta ^e$ was defined as the solution for Eq. 24 with $h_j=r_j=h_j^*$ for $j\not =i$. That is, define ${\bar{\Delta }}^e$ as we defined $\Delta ^e$, but with the requirement that each $h_j$ and $r_j$ are chosen to maximize the discounted present value of payments to the firm. The fixed point to $\Delta ^e$ must also be a fixed point for ${\bar{\Delta }}^e$. Hence, at this fixed point the law firm is choosing $h_i=r_i=h_i^*$ for each value of i absent any assumptions concerning how other $h_j$ and $r_j$ are chosen. Finally, we note that L was set less than ${\bar{X}}$. Hence, the client prefers the long term contract for each value of i.

As in the proof of Proposition 3, the law firm has no desire to defraud the client. This follows for the exact same reasons. This concludes the proof of Theorem 1.

To complete the proof of Proposition 2 we need to show that we can choose z to set ${\tilde{\mathcal {Q}}}^*$ to any value in [1 / 2, 1). We first note that ${\tilde{\mathcal {Q}}}^* = {\tilde{P}}^* \le 1/2$ when $z=1$ and ${\tilde{\mathcal {Q}}}^* \rightarrow 1$ as $z \rightarrow 0$. Hence, it remains only to show that ${\tilde{\mathcal {Q}}}^*$ is continuous. We note that an infinitesimal change in z creates an infinitesimal change in the law firm’s incentives which can be rebalanced with an infinitesimal change in L, and $\{\Delta _i\}_i$. Hence, if we think of L and $\{\Delta _i\}_i$ as functions of z, then ${\tilde{\mathcal {Q}}}^*$ is continuous in z. This completes the proof of Proposition 2.