Conjectures and underpricing in repeated mass disputes with heterogeneous plaintiffs

Abstract

The present Bayes-conjectural model focuses on a single defendant repeatedly formulating settlement offers when facing a heterogeneous population of victims. As usual, victims are risk-averse. However, the defendant cannot know the victims’ characteristics or their distribution. The defendant is endowed with an initial conjecture on the victims’ aggregate behavior. On each date, it formulates a settlement offer and updates the conjecture based on what it observes. We show that there is a multiplicity of conjectural equilibria and that the one reached depends on the initial conjecture. Independent of the degree of victims’ risk aversion, an equilibrium settlement can underprice the correct value of the claim. The results show that higher defendant court fees and aggregate procedures may alleviate the underpricing problem.

Appendices

Appendix

A.1

From Assumption 1 it follows that, for given${\gamma }_{i}\delta$,${\gamma }_{i}\delta {r}_{i,t}\sim N\left({\gamma }_{i}\delta \rho ,{\gamma }_{i}^{2}{\delta }^{2}{\sigma }^{2}\right)$. On the other hand, it is well-known that, if r is a normal variable with mean ρ and variance σ², then ${e}^{-r}$ is log-normal with mean${e}^{-\rho +{\sigma }^{2}/2}$. Therefore, we conclude that i’s expected utility of filing a lawsuit at time t is$E\left(A-{e}^{-{\gamma }_{i}\delta {r}_{i,t}}\right)={A-e}^{-{\gamma }_{i}\delta \rho +\frac{{\gamma }_{i}^{2}{\delta }^{2}}{2}{\sigma }^{2}}$.

2.1 A.2 (RESULT 1, part b)

${q}_{s}(s)$ Is studied over the interval $\rho -\frac{\delta {\sigma }^{2}}{2}\le {s}_{t}\le \rho$:

$$\frac{{dq}_{s}(s)}{ds}=\frac{{2}^{g}gh{\left(1-{\left(\frac{2\left(\rho -s\right)}{\delta {\sigma }^{2}}\right)}^{g}\right)}^{h-1}{\left(\frac{\rho -s}{\delta {\sigma }^{2}}\right)}^{g-1}}{\delta {\sigma }^{2}}>0.Hence:$$

$$(1) \frac{{dq}_{s}(s)}{d\rho }=-\frac{{dq}_{s}\left(s\right)}{ds}<0$$

$$(2) \frac{{dq}_{s}(s)}{d{\sigma }^{2}}=\left(\rho -s\right)\frac{{dq}_{s}\left(s\right)}{ds}\ge 0$$

A.3 (RESULT 2)

• (a) ${\mathrm{min}}_{{s}_{t}}{ E}_{A,t}\left[{O}_{t}\left(s\right)\right]$First order condition: ${\beta }_{t-1}s+\left({\alpha }_{t-1}+{\beta }_{t-1}s\right)-{\beta }_{t-1}{r}_{D}=0 \to$ $s=\frac{{\beta }_{t-1}{r}_{D}-{\alpha }_{t-1}}{2{\beta }_{t-1}}$Second order condition (${2\beta }_{t-1}>0$) is verified.

• (b) ${s}_{t}<{r}_{D}$: to verify that, remember that Assumption 7.3 about $a+b{r}_{D}>0$ implies $-\frac{\mathrm{\alpha }}{\upbeta }<{r}_{D}$.

• (c) ${s}_{t}>0$: to verify that, remember the part $b{r}_{D}>a$ of Assumption 7.3 that implies $\frac{\mathrm{\alpha }}{\upbeta }<{r}_{D}$.

• (d) By replacing the optimal settlement offer provided in Eq. (2) into the defendant’s forecast about the share of individuals who will accept the settlement at t, we obtain ${q}_{S,t}^{e}={\alpha }_{t-1}+{\beta }_{t-1}{s}_{t}=\frac{{\alpha }_{t-1}}{2}+\frac{{\beta }_{t-1}{r}_{D}}{2}$.

A.4 (RESULT 4)

A learning dynamical system can be described through the notation ${\mathbf{v}}_{t}=L\left({\mathbf{v}}_{t-1}\right)$ whose state-variables are described by a vector v. The stationary states of this dynamical system will be characterized by configurations of its variables such that the system stays there forever. The relevant state-variables of our learning dynamical system are ${q}_{S}\left({s}_{t}\right)$ plus the defendant’s hyper-parameters: these variables are described by the row-vector ${\mathbf{v}}_{t}=[\begin{array}{ccc}{\alpha }_{t}& {\beta }_{t}& {s}_{t}\end{array}$ ${q}_{S}\left({s}_{t}\right)$]. All this given, we posit the following:

Definition 1

A Conjectural Equilibrium (CE) at date t–1 is a vector ${\mathbf{v}}_{t-1}$ such that ${\mathbf{v}}_{t}={\mathbf{v}}_{t-1}$ under the operation of the learning dynamical system. In other terms, it is a fixed point of the learning dynamical system ${\mathbf{v}}_{t}=L\left({\mathbf{v}}_{t-1}\right)$. Every CE is characterized by constant values of the variables: the defendant’s belief, its settlement offer, and the acceptance rate.

• (a) We have a single equation in two variables, α and β: hence, we have a one-dimensional set (or, technically, manifold) of couples $\left(\alpha ,\beta \right)$ solving condition (5).

• (b) Taking advantage of some results obtained until now, write condition (5) as $H\left(\alpha ,\beta \right)=\frac{\alpha }{2}+\frac{\beta {r}_{D}}{2}-G\left(\frac{2\rho -{r}_{D}+\frac{\alpha }{\beta }}{\delta {\sigma }^{2}}\right)=0$, where $G\left(\cdot \right)=1-F\left(\cdot \right)$. This implicit equation defines the CE set, and the relationship between α and β in the set derives from the Implicit Function Theorem.

  The partial derivatives of H with respect to α and β are $\frac{\partial H}{\partial \alpha }=\frac{1}{2}+F^{\prime}\frac{1}{\beta \delta {\sigma }^{2}}$, and $\frac{\partial H}{\partial \beta }={\frac{{r}_{D}}{2}-F}^{^{\prime}}\cdot \frac{\alpha }{{\beta }^{2}\delta {\sigma }^{2}}$. Hence, in the CE set, we have ${\frac{d\alpha }{d\beta }}_{|CE}=-\frac{\partial H/\partial \beta }{\partial H/\partial \alpha }=\frac{{F}^{^{\prime}}\cdot \frac{\alpha }{{\beta }^{2}\delta {\sigma }^{2}}-\frac{{r}_{D}}{2}}{F^{\prime}\frac{1}{\beta \delta {\sigma }^{2}}+\frac{1}{2}}$. Recalling that in a CE $s=\frac{{r}_{D}}{2}-\frac{1}{2}{\frac{\alpha }{\beta }}_{|CE}$, to understand the behaviour of the equilibrium settlement amount s in the set as a function of β, we compute ${\frac{ds}{d\beta }}_{|CE}=\frac{\partial s}{\partial \beta }+\frac{\partial s}{\partial \alpha }\cdot {\frac{d\alpha }{d\beta }}_{|CE}=\frac{\alpha }{2{\beta }^{2}}-\frac{1}{2\beta }\cdot \frac{{F}^{^{\prime}}\cdot \frac{\alpha }{{\beta }^{2}\delta {\sigma }^{2}}-\frac{{r}_{D}}{2}}{F^{\prime}\frac{1}{\beta \delta {\sigma }^{2}}+\frac{1}{2}}=\frac{1}{2\beta }\left(\frac{\alpha }{\beta }-\frac{{\frac{\alpha }{\beta }F}^{^{\prime}}\cdot \frac{1}{\beta \delta {\sigma }^{2}}-\frac{{r}_{D}}{2}}{F^{\prime}\frac{1}{\beta \delta {\sigma }^{2}}+\frac{1}{2}}\right)$. After some algebra, we observe that ${\frac{ds}{d\beta }}_{|CE}>0$ if $\frac{\alpha }{\beta }+{r}_{D}>0$. Given the Assumption 7.3 we conclude that ${\frac{ds}{d\beta }}_{|CE}>0$.

• (c) To prove this part, consider ${\frac{d\alpha }{d\beta }}_{|CE}=\frac{{F}^{^{\prime}}\cdot \frac{\alpha }{{\beta }^{2}\delta {\sigma }^{2}}-\frac{{r}_{D}}{2}}{F^{\prime}\frac{1}{\beta \delta {\sigma }^{2}}+\frac{1}{2}}$ as derived above. The denominator is always positive, so we concentrate on the numerator: its sign is clearly equal to the sign of $\frac{\alpha }{\beta }\frac{1}{\beta }{F}^{^{\prime}}\cdot \frac{2}{\delta {\sigma }^{2}}-{r}_{D}$. As long as $\alpha$ is negative, we have ${\frac{d\alpha }{d\beta }}_{|CE}<0$. Now, observe that ${F}^{^{\prime}}\cdot \frac{2}{\delta {\sigma }^{2}}$ is the derivative of the actual relation between settlement offer and acceptance share ${q}_{s}\left(s\right)=1-F\left(\frac{2\left(\rho -s\right)}{\delta {\sigma }^{2}};g,h\right)$ (given that $\rho -\frac{\delta {\sigma }^{2}}{2}\le s\le \rho$), while $\beta$ is the derivative of the conjectured relation ${q}_{S}^{e}=\alpha +\beta s$.

  Therefore, for positive $\alpha$, as long as the conjectured relation is steeper than the actual one at a conjectural equilibrium, we still have ${\frac{d\alpha }{d\beta }}_{|CE}<0$, since Assumption 7.3 guarantees that $\frac{\alpha }{\beta }<{r}_{D}$. It follows that as $\alpha$ increases $\beta$ keeps decreasing in the CE set, and the equilibrium s decreases as well, as we proved in part (b) of this proof. This implies that the equilibrium point keeps shifting to South-West in the (${s,q}_{s}$) as showed in Fig. 3.

  Suppose now that, following this path, a certain point is reached such that ${F}^{^{\prime}}\cdot \frac{2}{\delta {\sigma }^{2}}$ becomes large enough to turn ${\frac{d\alpha }{d\beta }}_{|CE}$ positive. Now, as $\alpha$ keeps increasing, $\beta$ should start increasing, and the CE point should move to North-East in the (${s,q}_{s}$) space. Therefore, if this were the case, one would observe a CE point already observed in the first phase (when $\beta$ was decreasing), however characterized by a higher $\alpha$ and a higher $\beta$. Call (${\alpha }_{1},{\beta }_{1}$) the parameter pair of the first observation and (${\alpha }_{2},{\beta }_{2}$) that of the second observation of the same CE point, with ${\alpha }_{2}>{\alpha }_{1}$. Since the CE point is the same in the two cases, one must have that ${q}_{S}^{e}\left({\alpha }_{1},{\beta }_{1}\right)={q}_{S}^{e}\left({\alpha }_{2},{\beta }_{2}\right)$ and $s\left({\alpha }_{1},{\beta }_{1}\right)=s\left({\alpha }_{2},{\beta }_{2}\right)$. Given (3) and (2), these latter equalities correspond to $\frac{{\alpha }_{1}}{2}+\frac{{\beta }_{1}{r}_{D}}{2}=\frac{{\alpha }_{2}}{2}+\frac{{\beta }_{2}{r}_{D}}{2}$and $\frac{{r}_{A}}{2}-\frac{{\alpha }_{1}}{{2\beta }_{1}}=\frac{{r}_{A}}{2}-\frac{{\alpha }_{2}}{2{\beta }_{2}}$, respectively. The second equality implies that ${\alpha }_{2}={\alpha }_{1}\frac{{\beta }_{2}}{{\beta }_{1}}$; since we assumed ${\alpha }_{2}>{\alpha }_{1}>0$, this would indeed imply that ${\beta }_{2}>{\beta }_{1}$. However, the first equality, coupled with the second one and after some passages, implies ${\alpha }_{1}\left(1-\frac{{\beta }_{2}}{{\beta }_{1}}\right)={r}_{D}\left({\beta }_{2}-{\beta }_{1}\right)$: but this impossible. Therefore, as $\alpha$ keeps increasing, ${\frac{d\alpha }{d\beta }}_{|CE}$ must be always negative.

• (d) This particular equilibrium requires ${q}_{S}^{e}\left(\alpha ,\beta \right)=1$ and $s\left(\alpha ,\beta \right)=\rho$. Given (3) and (2), these latter equalities correspond to $\frac{\alpha }{2}+\frac{\beta {r}_{D}}{2}=1$ and $\frac{{r}_{D}}{2}-\frac{\alpha }{2\beta }=\rho$. After some algebra, these equalities imply $\alpha =\frac{{r}_{D}-2\rho }{{r}_{D}-\rho }$, which is negative for $\frac{1}{2}{r}_{D}\le \rho \le {r}_{D}$, as guaranteed by Assumptions 5a and 5b (see the comments after those assumptions).

A.5 (RESULT 5)

We are interested in analyzing the behavior of ${q}_{S}$ with respect to c, provided that we focus on CE.

$${\left.\frac{\mathrm{d}{q}_{S}}{\mathrm{d}c}\right|}_{CE}={\left.\frac{\mathrm{d}{q}_{\mathrm{S}}}{\mathrm{d}\left({r}_{D}-\rho \right)}\right|}_{CE}=\left(\frac{\partial {q}_{S}}{\partial \left({r}_{D}-\rho \right)}+\frac{\partial {q}_{S}}{\partial \left(\frac{\alpha }{\beta }\right)}{\left.\frac{\mathrm{d}\left(\frac{\alpha }{\beta }\right)}{\mathrm{d}\left({r}_{D}-\rho \right)}\right|}_{CE}\right)$$

Since ${q}_{S}=1-F={\left(1-{\left(\frac{\rho -\left({r}_{D}-\rho \right)+\frac{\alpha }{\beta }}{\delta {\sigma }^{2}}\right)}^{g}\right)}^{h}$, then $\frac{\partial {q}_{S}}{\partial \left({r}_{D}-\rho \right)}=F^{\prime}\frac{1}{\delta {\sigma }^{2}}$ and $\frac{\partial {q}_{S}}{\partial \left(\frac{\alpha }{\beta }\right)}=-F^{\prime}\frac{1}{\delta {\sigma }^{2}}$. Now, focus on ${\left.\frac{\mathrm{d}\left(\frac{\alpha }{\beta }\right)}{\mathrm{d}\left({r}_{A}-\rho \right)}\right|}_{CE}$. We rewrite the CE condition presented in (5) as following:

$$\frac{\beta }{2}\left({r}_{D}+\frac{\alpha }{\beta }+\rho -\rho \right)={\left(1-{\left(\frac{\left(2\rho -\left[{r}_{D}-\frac{\alpha }{\beta }\right]\right)}{\delta {\sigma }^{2}}\right)}^{g}\right)}^{h}$$

$$\frac{\beta }{2}\left(\left({r}_{D}-\rho \right)+\frac{\alpha }{\beta }+\rho \right)={\left(1-{\left(\frac{\left(\rho -\left({r}_{D}-\rho \right)+\frac{\alpha }{\beta }\right)}{\delta {\sigma }^{2}}\right)}^{g}\right)}^{h}$$

As we did in Part b of A.4., to exploit the Implicit Function Theorem, the previous condition is rewritten as:

$$H\left(\frac{\alpha }{\beta },\left({r}_{D}-\rho \right)\right)\equiv \frac{\beta }{2}\left(\frac{\alpha }{\beta }+\left({r}_{D}-\rho \right)+\rho \right)-{\left(1-{\left(\frac{\rho -\left({r}_{D}-\rho \right)+\frac{\alpha }{\beta }}{\delta {\sigma }^{2}}\right)}^{g}\right)}^{h}$$

In order to be in the CE set (along the CE manifold) where condition (5) is verified, the following must hold: $H\left(\frac{\alpha }{\beta },\left({r}_{D}-\rho \right)\right)=\frac{\beta }{2}\left(\frac{\alpha }{\beta }+\left({r}_{D}-\rho \right)+\rho \right)-1+F\left(\frac{\rho -\left({r}_{D}-\rho \right)+\frac{\alpha }{\beta }}{\delta {\sigma }^{2}}\right)=0$.

According to the Implicit Function Theorem, we compute ${\left.\frac{\mathrm{d}\left(\frac{\alpha }{\beta }\right)}{\mathrm{d}\left({r}_{D}-\rho \right)}\right|}_{CE}=-\frac{\frac{\partial H}{\partial \left({r}_{D}-\rho \right)}}{\frac{\partial H}{\partial \left(\frac{\alpha }{\beta }\right)}}$. Since $\frac{\partial H}{\partial \left(\frac{\alpha }{\beta }\right)}=\frac{\beta }{2}+F^{\prime}\frac{1}{\delta {\sigma }^{2}}$ and$\frac{\partial H}{\partial \left({r}_{D}-\rho \right)}=\frac{\beta }{2}-F^{\prime}\frac{1}{\delta {\sigma }^{2}}$, we conclude that: ${\left.\frac{\mathrm{d}\left(\frac{\alpha }{\beta }\right)}{\mathrm{d}\left({r}_{D}-\rho \right)}\right|}_{CE}=\left(\frac{{F}^{^{\prime}}\frac{1}{\delta {\sigma }^{2}}-\frac{\beta }{2}}{{F}^{^{\prime}}\frac{1}{\delta {\sigma }^{2}}+\frac{\beta }{2}}\right)$

Therefore,

$${\left.\frac{\mathrm{d}{q}_{S}}{\mathrm{d}c}\right|}_{CE}=F^{\prime}\frac{1}{\delta {\sigma }^{2}}-F^{\prime}\frac{1}{\delta {\sigma }^{2}}\frac{F^{\prime}\frac{1}{\delta {\sigma }^{2}}-\frac{\beta }{2}}{F^{\prime}\frac{1}{\delta {\sigma }^{2}}+\frac{\beta }{2}}=F^{\prime}\frac{1}{\delta {\sigma }^{2}}\cdot \left[1-\frac{F^{\prime}\frac{1}{\delta {\sigma }^{2}}-\frac{\beta }{2}}{F^{\prime}\frac{1}{\delta {\sigma }^{2}}+\frac{\beta }{2}}\right]=\frac{F^{\prime}\frac{2}{\delta {\sigma }^{2}}\beta }{F^{\prime}\frac{2}{\delta {\sigma }^{2}}+\beta }>0$$

5.1 A.6 (RESULT 6)

We are interested in analyzing K, provided that we focus on CE.

$$\begin{aligned} K & \equiv \left( {1 - q_{S} } \right)\left( {r_{D} - \rho } \right) + q_{s} \left( {\rho - s} \right) \\ & = \left( {1 - q_{S} } \right)\left( {r_{D} - \rho } \right) + q_{s} \left( {\rho - \frac{{r_{D} }}{2} + \frac{\alpha }{2\beta }} \right) \\ & = \left( {1 - q_{S} } \right)\left( {r_{D} - \rho } \right) + \frac{1}{2}q_{s} \left( {\rho - \left( {r_{D} - \rho } \right) + \frac{\alpha }{\beta }} \right) \\ \end{aligned}$$

In particular, we analyze its relationship with c.

$${\left.\frac{\mathrm{d}K}{\mathrm{d}c}\right|}_{CE}={\left.\frac{\mathrm{d}K}{\mathrm{d}\left({r}_{D}-\rho \right)}\right|}_{CE}=\frac{\partial K}{\partial \left({r}_{D}-\rho \right)}+\frac{\partial K}{\partial {q}_{S}}{\left.\frac{\mathrm{d}{q}_{S}}{\mathrm{d}\left({r}_{D}-\rho \right)}\right|}_{CE}+\frac{\partial K}{\partial \frac{\alpha }{\beta }}{\left.\frac{\mathrm{d}\left(\frac{\alpha }{\beta }\right)}{\mathrm{d}\left({r}_{D}-\rho \right)}\right|}_{CE}$$

$${\left.\frac{\mathrm{d}K}{\mathrm{d}c}\right|}_{CE}={\left.\frac{\mathrm{d}K}{\mathrm{d}\left({r}_{D}-\rho \right)}\right|}_{CE}=\left(1-{q}_{S}-\frac{1}{2}{q}_{S}\right)+{\left.\frac{\mathrm{d}{q}_{S}}{\mathrm{d}\left({r}_{D}-\rho \right)}\right|}_{CE}\frac{1}{2}\left(\rho -3\left({r}_{A}-\rho \right)+\frac{\alpha }{\beta }\right)+\frac{1}{2}{q}_{S}{\left.\frac{\mathrm{d}\left(\frac{\alpha }{\beta }\right)}{\mathrm{d}\left({r}_{D}-\rho \right)}\right|}_{CE}$$

$$=\left(1-\frac{3}{2}{q}_{S}\right)+{\left.\frac{\mathrm{d}{q}_{S}}{\mathrm{d}\left({r}_{D}-\rho \right)}\right|}_{CE}\frac{1}{2}\left(2\rho -2\left({r}_{D}-\rho \right)-{r}_{D}+\frac{\alpha }{\beta }\right)+\frac{1}{2}{q}_{S}{\left.\frac{\mathrm{d}\left(\frac{\alpha }{\beta }\right)}{\mathrm{d}\left({r}_{D}-\rho \right)}\right|}_{CE}$$

$$=\left(1-\frac{3}{2}{q}_{S}\right)+{\left.\frac{\mathrm{d}{q}_{S}}{\mathrm{d}\left({r}_{D}-\rho \right)}\right|}_{CE}\left(\rho -\left({r}_{D}-\rho \right)-\frac{1}{2}\left({r}_{A}-\frac{\alpha }{\beta }\right)\right)+\frac{1}{2}{q}_{S}{\left.\frac{\mathrm{d}\left(\frac{\alpha }{\beta }\right)}{\mathrm{d}\left({r}_{D}-\rho \right)}\right|}_{CE}$$

Recalling that that $s=\frac{1}{2}\left({r}_{A}-\frac{\alpha }{\beta }\right)$, we obtain:

$${\left.\frac{\mathrm{d}K}{\mathrm{d}c}\right|}_{CE}=\left(1-\frac{3}{2}{q}_{S}\right)+{\left.\frac{\mathrm{d}{q}_{S}}{\mathrm{d}\left({r}_{D}-\rho \right)}\right|}_{CE}\left(\rho -{r}_{D}+\rho -s\right)+\frac{1}{2}{q}_{S}{\left.\frac{\mathrm{d}\left(\frac{\alpha }{\beta }\right)}{\mathrm{d}\left({r}_{D}-\rho \right)}\right|}_{CE}$$

From A.5. we already know that ${\left.\frac{\mathrm{d}{q}_{S}}{\mathrm{d}c}\right|}_{CE}=\frac{{F}^{^{\prime}}\frac{2}{\delta {\sigma }^{2}}\beta }{{F}^{^{\prime}}\frac{2}{\delta {\sigma }^{2}}+\beta }$ and ${\left.\frac{\mathrm{d}\left(\frac{\alpha }{\beta }\right)}{\mathrm{d}\left({r}_{D}-\rho \right)}\right|}_{CE}=\frac{{F}^{^{\prime}}\frac{1}{\delta {\sigma }^{2}}-\frac{\beta }{2}}{{F}^{^{\prime}}\frac{1}{\delta {\sigma }^{2}}+\frac{\beta }{2}}$.

Then, we can conclude that:

$${\left.\frac{\mathrm{d}K}{\mathrm{d}c}\right|}_{CE}=\left(1-\frac{3}{2}{q}_{S}\right)+\frac{{F}^{^{\prime}}\frac{2}{\delta {\sigma }^{2}}\beta }{{F}^{^{\prime}}\frac{2}{\delta {\sigma }^{2}}+\beta }\left(\rho -{r}_{D}+\rho -s\right)+\frac{1}{2}{q}_{S}\frac{{F}^{^{\prime}}\frac{1}{\delta {\sigma }^{2}}-\frac{\beta }{2}}{{F}^{^{\prime}}\frac{1}{\delta {\sigma }^{2}}+\frac{\beta }{2}}$$

By inspecting ${\left.\frac{\mathrm{d}K}{\mathrm{d}c}\right|}_{CE}$, we observe that

• i. the first addend is negative for $q_{S} > 2/3$;

• ii. the sign of the second addend depends on its second term since the first term is always positive. The second term is negative when ${s}_{|CE}$ in not too much smaller than $\rho$; increasing litigation costs $c=\left({r}_{D}-\rho \right)$ guarantee that the term is even more negative;

• iii. the third addend is negative when the slope of the sigmoid function describing the probability of settlement acceptance (${F}^{^{\prime}}\frac{1}{\delta {\sigma }^{2}}$) is smaller than the slope of the function describing the conjecture of the defendant about the probability that its offer will be accepted ($\frac{\beta }{2}$). This happens more and more easily when ${s}_{|CE}$ increases and approaches $\rho$ from the left.

Given this analysis and recalling that in CE a large $s$ implies that both ${q}_{S}$ and $\frac{\beta }{2}$ are relatively large (Result 4), we conclude that when $s$ is sufficiently large then the overall cost of dispute resolution K is decreasing in $s$ (Result 6a). Furthermore, given the observation ii. and the fact that $s$ is increasing in c we also conclude that when $s$ is sufficiently large, then K is decreasing in c (Result 6b).

A.7 (Aggregate procedure)

$${E}_{D,t}\left[\left.{O}_{t}\left(s\right)\right|\theta \right]=\left\{\begin{array}{*{20}l}\left({\alpha }_{t-1}+{\beta }_{t-1}s\right)\left(1-\theta \right)s+\left[1-\left({\alpha }_{t-1}+{\beta }_{t-1}s\right)\left(1-\theta \right)\right]\cdot {r}_{D} for s>\theta \rho \\ {r}_{D} for s\le \theta \rho \end{array}\right.$$

We wish to solve ${\mathrm{min}}_{s}{E}_{D,t}\left[\left.{O}_{t}\left(s\right)\right|\theta \right]$. The derivative of $\left({\alpha }_{t-1}+{\beta }_{t-1}s\right)\left(1-\theta \right)s+\left[1-\left({\alpha }_{t-1}+{\beta }_{t-1}s\right)\left(1-\theta \right)\right]\cdot {r}_{D}$ with respect to s is ${\alpha }_{t-1}\left(1-\theta \right)+2{\beta }_{t-1}\left(1-\theta \right)s-{\beta }_{t-1}\left(1-\theta \right)\cdot {r}_{D}$. Equating this derivative to zero yields ${s}_{t}=\frac{{r}_{D}}{2}-\frac{{\alpha }_{t-1}}{2{\beta }_{t-1}}, \mathrm{provided that} \frac{{r}_{D}}{2}-\frac{{\alpha }_{t-1}}{2{\beta }_{t-1}}>\theta \rho$.

If $\frac{{r}_{D}}{2}-\frac{{\alpha }_{t-1}}{2{\beta }_{t-1}}\le \theta \rho$, then the defendant should fix $\widetilde{{s}_{t}}=\theta \rho +\varepsilon <\rho$ (small ε): this allows him to spend less than fixing ${s}_{t}\le \theta \rho$. Therefore, we conclude that the optimal settlement amount is:

$${s}_{t}=\left\{\begin{array}{*{20}l}\frac{{r}_{D}}{2}-\frac{{\alpha }_{t-1}}{2{\beta }_{t-1}} & \mathrm{if} \frac{{r}_{D}}{2}-\frac{{\alpha }_{t-1}}{2{\beta }_{t-1}}>\theta \rho \\ \theta \rho +\varepsilon <\rho , (small \varepsilon )& \mathrm{if} \frac{{r}_{D}}{2}-\frac{{\alpha }_{t-1}}{2{\beta }_{t-1}}\le \theta \rho \end{array}\right.$$