Decoupled Liability and Efficiency

Abstract

A basic feature of tort law is that of coupled liability. Liability is said to be ‘coupled’ if the liability imposed on the injurer equals payment to the victim and to be ‘decoupled’ if the two amounts are unequal. The ‘coupled liability’ feature of tort law is incorporated in the very definition of a liability rule by having the shares of loss borne by the two parties sum to one. In this chapter, the relationship between the ‘coupled liability’ feature of tort law and efficiency is investigated. It is shown that coupled liability is necessary for efficiency, i.e. if a rule is such that it invariably gives rise to efficient outcomes then it must be the case that under it the liability is coupled. In other words, no rule with decoupled liability can be such that it invariably yields efficient outcomes. A corollary of the result establishing necessity of coupled liability for efficiency is that the rule under which the injurer pays tax equal to the harm and the victim bears his loss is not an efficient rule. This rule, however, has an interesting property, namely, that under it the configuration of due care levels is always a Nash equilibrium. A rule under which the configuration of due care levels is always a Nash equilibrium is termed a quasi-efficient rule. A complete characterization of quasi-efficient rules is also obtained in this chapter.

Appendix

Lemma 4.1.

Let f be a h-liability rule; $f: [0,1]^{2}\mapsto [0,\infty )^{2}$ . If f is efficient for every application belonging to $\mathcal{A}$ , then $[\forall p,q \in [0,1)][x(p,1) \geq 1 \wedge y(1,q) \geq 1]$ .

Proof.

Let f be a h-liability rule; $f: [0,1]^{2}\mapsto [0,\infty )^{2}$. Suppose [∃q ∈ [0, 1)][y(1, q) < 1].

Let y(1, q ₀) < 1.

Let t > 0. Choose r such that 0 ≤ y(1, q ₀)t < r < t. Let $d_{0} = \frac{r} {1-q_{0}}$. Let c ₀ > 0 and ε > 0.

Let C and D be specified as follows:

$C =\{ 0,c_{0}\},D =\{ 0,q_{0}d_{0},d_{0}\}.$

For (c, d) ∈ C × D, let L(c, d) be as given in the following array³:

$$\displaystyle\begin{array}{rcl} L(c,d)& & {}\\ \begin{array}{|c|c|c|c|}\hline & d = 0 &d = q_{0}d_{0} & d = d_{0} \\\hline c = 0 &c_{0} +\epsilon +t + q_{0}d_{0} & c_{0} +\epsilon +t & c_{0}+\epsilon \\\hline c = c_{0} & t + q_{0}d_{0} & t & 0 \\\hline \end{array} & & {}\\ \end{array}$$

ε > 0 and $t > r = (1 - q_{0})d_{0}$ imply that M = { (c ₀, d ₀)}.

Let (c ^∗, d ^∗) = (c ₀, d ₀).

Now,

$$\displaystyle\begin{array}{rcl} & & EC_{2}(c_{0},d_{0}) = d_{0} {}\\ & & EC_{2}(c_{0},q_{0}d_{0}) {}\\ & & \!= q_{0}d_{0} + y(1,q_{0})L(c_{0},q_{0}d_{0}) {}\\ & & \!= q_{0}d_{0} + y(1,q_{0})t {}\\ & & EC_{2}(c_{0},d_{0}) - EC_{2}(c_{0},q_{0}d_{0}) {}\\ & & \!= d_{0} - q_{0}d_{0} - y(1,q_{0})t {}\\ & & \!= (1 - q_{0})d_{0} - y(1,q_{0})t {}\\ & & \!= r - y(1,q_{0})t {}\\ & & \!> 0. {}\\ \end{array}$$

This implies that the unique TSC-minimizing configuration (c ₀, d ₀) is not a Nash equilibrium. f is therefore not efficient for the application under consideration.

If [∃p ∈ [0, 1)][x(p, 1) < 1] holds, then by an analogous argument one can demonstrate the existence of an application belonging to $\mathcal{A}$ for which f is not efficient.

The proposition is therefore established. □ 

The following corollary follows immediately from Lemma 4.1.

Corollary 4.1.

Let f be a h-liability rule; $f: [0,1]^{2}\mapsto [0,\infty )^{2}$ . If f is efficient for every application belonging to $\mathcal{A}$ , then $\overline{s} \geq 1$ .

For every h-liability rule such that $[\exists p \in [0,1)][x(p,1) < 1] \vee [\exists q \in [0,1)][y(1,q) < 1]$, the proof of Lemma 4.1 shows the existence of an application belonging to $\mathcal{A}$ for which the configuration of costs of due care levels (c ^∗, d ^∗) ∈ M is not a Nash equilibrium. Thus from the proof of Lemma 4.1, it follows that the following lemma holds.

Lemma 4.2.

Let f be a h-liability rule; $f: [0,1]^{2}\mapsto [0,\infty )^{2}$ . If f is such that for every application belonging to $\mathcal{A}$ the configuration of costs of due care levels (c ^∗ ,d ^∗ ) ∈ M is a Nash equilibrium, then $[\forall p,q \in [0,1)][x(p,1) \geq 1 \wedge y(1,q) \geq 1]$ .

The following corollary follows immediately from Lemma 4.2.

Corollary 4.2.

Let f be a h-liability rule; $f: [0,1]^{2}\mapsto [0,\infty )^{2}$ . If f is such that for every application belonging to $\mathcal{A}$ the configuration of costs of due care levels (c ^∗ ,d ^∗ ) ∈ M is a Nash equilibrium, then $\overline{s} \geq 1$ .

Lemma 4.3.

Let f be a h-liability rule; $f: [0,1]^{2}\mapsto [0,\infty )^{2}$ . If f is efficient for every application belonging to $\mathcal{A}$ , then $[x^{{\ast}}\leq 1 \wedge y^{{\ast}}\leq 1]$ .

Proof.

Suppose x ^∗ > 1.

Choose $0 < \frac{1} {x^{{\ast}}} < t < 1$.

Choose $c_{0},d_{0},\epsilon,\delta _{1},\delta _{2} > 0$.

Let C and D be specified as follows:

$C =\{ 0,c_{0},c_{0}+\epsilon \},D =\{ 0,d_{0}\}.$

For (c, d) ∈ C × D, let L(c, d) be as given in the following array:

$$\displaystyle\begin{array}{rcl} L(c,d)& & {}\\ \begin{array}{|c|c|c|}\hline & d = 0 & d = d_{0} \\\hline c = 0 &c_{0} +\delta _{1} + d_{0} +\delta _{2} + t\epsilon &c_{0} +\delta _{1} + t\epsilon \\\hline c = c_{0} & d_{0} +\delta _{2} + t\epsilon & t\epsilon \\\hline c = c_{0}+\epsilon & d_{0} +\delta _{2} & 0 \\\hline \end{array} & & {}\\ \end{array}$$

δ ₁ > 0, δ ₂ > 0, and 0 < t < 1 imply that (c ₀, d ₀) is the unique TSC-minimizing configuration, i.e. M = { (c ₀, d ₀)}.

Let (c ^∗, d ^∗) = (c ₀, d ₀).

Now,

$$\displaystyle\begin{array}{rcl} & & EC_{1}(c_{0},d_{0}) = c_{0} + x^{{\ast}}t\epsilon {}\\ & & EC_{1}(c_{0}+\epsilon,d_{0}) = c_{0} +\epsilon {}\\ & & EC_{1}(c_{0},d_{0}) - EC_{1}(c_{0}+\epsilon,d_{0}) = c_{0} + x^{{\ast}}t\epsilon - c_{ 0} -\epsilon {}\\ & &\!=\epsilon [x^{{\ast}}t - 1] {}\\ & & \!> 0. {}\\ \end{array}$$

Thus the unique TSC-minimizing configuration of costs of care is not a Nash equilibrium, establishing that the rule is not efficient for the application in question.

If y ^∗ > 1, an analogous argument shows that there exists an application belonging to $\mathcal{A}$ for which the rule is not efficient.

The lemma is therefore established. □ 

The following corollary follows immediately from Lemma 4.3.

Corollary 4.3.

Let f be a h-liability rule; $f: [0,1]^{2}\mapsto [0,\infty )^{2}$ . If f is efficient for every application belonging to $\mathcal{A}$ , then $\overline{s} \leq 2$ .

For every h-liability rule such that $[x^{{\ast}} > 1 \vee y^{{\ast}} > 1]$, the proof of Lemma 4.3 shows the existence of an application belonging to $\mathcal{A}$ for which the configuration of costs of due care levels (c ^∗, d ^∗) ∈ M is not a Nash equilibrium. Thus from the proof of Lemma 4.3, it follows that the following lemma holds.

Lemma 4.4.

Let f be a h-liability rule; $f: [0,1]^{2}\mapsto [0,\infty )^{2}$ . If f is such that for every application belonging to $\mathcal{A}$ , the configuration of costs of due care levels (c ^∗ ,d ^∗ ) ∈ M is a Nash equilibrium, then $[x^{{\ast}}\leq 1 \wedge y^{{\ast}}\leq 1]$ .

The following corollary follows immediately from Lemma 4.4.

Corollary 4.4.

Let f be a h-liability rule; $f: [0,1]^{2}\mapsto [0,\infty )^{2}$ . If f is such that for every application belonging to $\mathcal{A}$ , the configuration of costs of due care levels (c ^∗ ,d ^∗ ) ∈ M is a Nash equilibrium, then $\overline{s} \leq 2$ .

Theorem 4.1.

Let f be a h-liability rule; $f: [0,1]^{2}\mapsto [0,\infty )^{2}$ . If f is efficient for every application belonging to $\mathcal{A}$ , then $\overline{s} = 1$ .

Proof.

Let h-liability rule $f: [0,1]^{2}\mapsto [0,\infty )^{2}$ be efficient for every application belonging to $\mathcal{A}$. Then from Corollaries 4.1 and 4.3, we have $1 \leq \overline{s} \leq 2$.

Suppose $1 < \overline{s} \leq 2$.

We have x ^∗ + y ^∗ > 1 as $\overline{s} > 1$.

Therefore, we obtain: [$1 \geq x^{{\ast}} > 0 \wedge 1 \geq y^{{\ast}} > 0$], in view of Lemma 4.3.

Let:

• (i) $c_{0},d_{0},\epsilon _{1},\epsilon _{2} > 0$

• (ii) β > 0

• (iii) $\frac{x^{{\ast}}} {x^{{\ast}}+y^{{\ast}}}\beta <\alpha _{1} <\theta _{1} < x^{{\ast}}\beta$

• (iv) $\frac{y^{{\ast}}} {x^{{\ast}}+y^{{\ast}}}\beta <\alpha _{2} <\theta _{2} < y^{{\ast}}\beta$.

Let C and D be specified as follows:

$C =\{ 0,c_{0},c_{0} +\theta _{1}\},D =\{ 0,d_{0},d_{0} +\theta _{2}\}.$

For (c, d) ∈ C × D, let L(c, d) be as given in the following array:

$$\displaystyle\begin{array}{rcl} L(c,d)& & {}\\ \begin{array}{|c|c|c|c|}\hline & d = 0 & d = d_{0} & d = d_{0} +\theta _{2} \\\hline c = 0 &c_{0} +\epsilon _{1} + d_{0} +\epsilon _{2} +\theta _{1} +\theta _{2} & c_{0} +\epsilon _{1} +\theta _{1} +\theta _{2} & c_{0} +\epsilon _{1} +\theta _{1} +\alpha _{2} \\\hline c = c_{0} & d_{0} +\epsilon _{2} +\theta _{1} +\theta _{2} & \theta _{1} +\theta _{2} & \theta _{1} +\alpha _{2} \\\hline c = c_{0} +\theta _{1} & d_{0} +\epsilon _{2} +\alpha _{1} +\theta _{2} & \alpha _{1} +\theta _{2} & \alpha _{1} +\alpha _{2}-\beta \\\hline \end{array} & & {}\\ \end{array}$$

ε ₁ > 0, ε ₂ > 0, α ₁ > 0, α ₂ > 0 and $\alpha _{1} +\alpha _{2}-\beta > 0$ imply⁴ that M = { (c ₀, d ₀)}.

Let (c ^∗, d ^∗) = (c ₀, d ₀).

Now,

$$\displaystyle{ EC_{1}(c_{0} +\theta _{1},d_{0} +\theta _{2}) = c_{0} +\theta _{1} + x^{{\ast}}(\alpha _{ 1} +\alpha _{2}-\beta ) }$$

(4.1)

$$\displaystyle{ EC_{1}(c_{0},d_{0} +\theta _{2}) = c_{0} + x^{{\ast}}(\theta _{ 1} +\alpha _{2}) }$$

(4.2)

$$\displaystyle{ EC_{1}(0,d_{0} +\theta _{2}) = x(0,1)(c_{0} +\epsilon _{1} +\theta _{1} +\alpha _{2}) }$$

(4.3)

$$\displaystyle{ EC_{2}(c_{0} +\theta _{1},d_{0} +\theta _{2}) = d_{0} +\theta _{2} + y^{{\ast}}(\alpha _{ 1} +\alpha _{2}-\beta ) }$$

(4.4)

$$\displaystyle{ EC_{2}(c_{0} +\theta _{1},d_{0}) = d_{0} + y^{{\ast}}(\alpha _{ 1} +\theta _{2}) }$$

(4.5)

$$\displaystyle{ EC_{2}(c_{0} +\theta _{1},0) = y(1,0)(d_{0} +\epsilon _{2} +\alpha _{1} +\theta _{2}) }$$

(4.6)

$$\displaystyle\begin{array}{rcl} & & EC_{1}(c_{0} +\theta _{1},d_{0} +\theta _{2}) - EC_{1}(c_{0},d_{0} +\theta _{2}) = [c_{0} +\theta _{1} + x^{{\ast}}(\alpha _{ 1} +\alpha _{2}-\beta )] \\ & & \!-[c_{0} + x^{{\ast}}(\theta _{ 1} +\alpha _{2})] \\ & & = (1 - x^{{\ast}})\theta _{ 1} + x^{{\ast}}\alpha _{ 1} - x^{{\ast}}\beta \\ & &\!< (1 - x^{{\ast}})\theta _{ 1} + x^{{\ast}}\theta _{ 1} - x^{{\ast}}\beta,\;\mathrm{as}\;\theta _{ 1} >\alpha _{1} > 0\;\mathrm{and}\;x^{{\ast}} > 0 \\ & & \!=\theta _{1} - x^{{\ast}}\beta \\ & &\!< 0 {}\end{array}$$

(4.7)

$$\displaystyle\begin{array}{rcl} & & EC_{1}(c_{0} +\theta _{1},d_{0} +\theta _{2}) - EC_{1}(0,d_{0} +\theta _{2}) \\ & & \!= [c_{0}\! +\!\theta _{1}\! +\! x^{{\ast}}(\alpha _{ 1}\! +\!\alpha _{2}\!-\!\beta )]\! -\! x(0,1)[c_{0}\! +\!\epsilon _{1}\! +\!\theta _{1}\! +\!\alpha _{2}] \\ & & \!\leq [c_{0}\! +\!\theta _{1}\! +\! x^{{\ast}}(\alpha _{ 1}\! +\!\alpha _{2}\!-\!\beta )]\! -\! [c_{0}\! +\!\epsilon _{1}\! +\!\theta _{1}\! +\!\alpha _{2}],\text{ as }x(0,1) \geq 1\text{ by Lemma 4.1} \\ & & \!= x^{{\ast}}(\alpha _{ 1}-\beta ) - (1 - x^{{\ast}})\alpha _{ 2} -\epsilon _{1} \\ & & \!< 0 {}\end{array}$$

(4.8)

$$\displaystyle\begin{array}{rcl} & & EC_{2}(c_{0} +\theta _{1},d_{0} +\theta _{2}) - EC_{2}(c_{0} +\theta _{1},d_{0}) = [d_{0} +\theta _{2} + y^{{\ast}}(\alpha _{ 1} +\alpha _{2}-\beta )] \\ & & \!-[d_{0} + y^{{\ast}}(\alpha _{ 1} +\theta _{2})] \\ & & \!= (1 - y^{{\ast}})\theta _{ 2} + y^{{\ast}}\alpha _{ 2} - y^{{\ast}}\beta \\ & &\!< (1 - y^{{\ast}})\theta _{ 2} + y^{{\ast}}\theta _{ 2} - y^{{\ast}}\beta,\;\mathrm{as}\;\theta _{ 2} >\alpha _{2} > 0\;\mathrm{and}\;y^{{\ast}} > 0 \\ & & \!=\theta _{2} - y^{{\ast}}\beta \\ & &\!< 0 {}\end{array}$$

(4.9)

$$\displaystyle\begin{array}{rcl} & & EC_{2}(c_{0} +\theta _{1},d_{0} +\theta _{2}) - EC_{2}(c_{0} +\theta _{1},0) \\ & & \!= [d_{0}\! +\!\theta _{2}\! +\! y^{{\ast}}(\alpha _{ 1}\! +\!\alpha _{2}\!-\!\beta )]\! -\! y(1,0)[d_{0}\! +\!\epsilon _{2}\! +\!\alpha _{1}\! +\!\theta _{2}] \\ & & \!\leq [d_{0}\! +\!\theta _{2}\! +\! y^{{\ast}}(\alpha _{ 1}\! +\!\alpha _{2}\!-\!\beta )]\! -\! [d_{0}\! +\!\epsilon _{2}\! +\!\alpha _{1}\! +\!\theta _{2}],\text{ as }y(1,0) \geq 1\text{ by Lemma 4.1} \\ & & \!= y^{{\ast}}(\alpha _{ 2}-\beta ) - (1 - y^{{\ast}})\alpha _{ 1} -\epsilon _{2} \\ & & \!< 0 {}\end{array}$$

(4.10)

(4.7)–(4.10) establish that $(c_{0} +\theta _{1},d_{0} +\theta _{2})$ is a Nash equilibrium. But $(c_{0} +\theta _{1},d_{0} +\theta _{2})\notin M$, which implies that f is not efficient for the application under consideration.

This contradiction establishes the theorem. □ 

As a h-liability rule $f: [0,1]^{2}\mapsto [0,\infty )^{2}$ exhibits decoupled liability iff $\overline{s}\neq 1$, Theorem 4.1 can also be stated as follows.

Theorem 4.2.

If a h-liability rule $f: [0,1]^{2}\mapsto [0,\infty )^{2}$ exhibits decoupled liability, then it is not the case that f is efficient for every application belonging to $\mathcal{A}$ .

Theorem 4.3.

Let f be a h-liability rule; $f: [0,1]^{2}\mapsto [0,\infty )^{2}$ . f has the property that for every application belonging to $\mathcal{A}$ the configuration of costs of due care levels (c ^∗ ,d ^∗ ) ∈ M is a Nash equilibrium iff it satisfies the following two conditions:

• (i) $[\forall p,q \in [0,1)][x(p,1) \geq 1 \wedge y(1,q) \geq 1]$

• (ii) $x^{{\ast}}\leq 1 \wedge y^{{\ast}}\leq 1$ .

Proof.

Suppose f has the property that for every application belonging to $\mathcal{A}$ the configuration of costs of due care levels (c ^∗, d ^∗) ∈ M is a Nash equilibrium. Then (i) holds by Lemma 4.2 and (ii) holds by Lemma 4.4.

Next, assume that f satisfies conditions (i) and (ii). Take any application < C, D, π, H, (c ^∗, d ^∗) ∈ M > belonging to $\mathcal{A}$. Suppose (c ^∗, d ^∗) is not a Nash equilibrium. This implies:

$$\displaystyle\begin{array}{rcl} & & (\exists c^{{\prime}}\in C)[c^{{\prime}} + x[p(c^{{\prime}}),q(d^{{\ast}})]L(c^{{\prime}},d^{{\ast}}) < c^{{\ast}} + x^{{\ast}}L^{{\ast}}]\; \vee \; (\exists d^{{\prime}}\in D) \\ & & [d^{{\prime}} + y[p(c^{{\ast}}),q(d^{{\prime}})]L(c^{{\ast}},d^{{\prime}}) < d^{{\ast}} + y^{{\ast}}L^{{\ast}}]. {}\end{array}$$

(4.11)

$$\displaystyle\begin{array}{rcl} & & \text{Suppose }(\exists c^{{\prime}}\in C)[c^{{\prime}} + x[p(c^{{\prime}}),q(d^{{\ast}})]L(c^{{\prime}},d^{{\ast}}) < c^{{\ast}} + x^{{\ast}}L^{{\ast}}]\text{ holds. } \\ & & c^{{\prime}} < c^{{\ast}}\wedge \mbox{ (4.12)} \rightarrow c^{{\prime}} + L(c^{{\prime}},d^{{\ast}}) < c^{{\ast}} + x^{{\ast}}L^{{\ast}},\text{ as }x[p(c^{{\prime}}),q(d^{{\ast}})] \\ & & \geq 1\text{ by condition (i) } \\ & & \rightarrow c^{{\prime}} + L(c^{{\prime}},d^{{\ast}}) < c^{{\ast}} + L^{{\ast}},\text{ as }x^{{\ast}}\leq 1\text{ by condition (ii) } \\ & & \rightarrow c^{{\prime}} + d^{{\ast}} + L(c^{{\prime}},d^{{\ast}}) < c^{{\ast}} + d^{{\ast}} + L^{{\ast}} \\ & &\rightarrow \mathrm{TSC}(c^{{\prime}},d^{{\ast}}) <\mathrm{ TSC}(c^{{\ast}},d^{{\ast}}). {}\end{array}$$

(4.12)

This is a contradiction as TSC are minimum at (c ^∗, d ^∗). Therefore we conclude:

$$\displaystyle{ c^{{\prime}} < c^{{\ast}}\rightarrow \mbox{ (4.12)}\text{ cannot hold. } }$$

(4.13)

For c ^′ > c ^∗, we have: $x[p(c^{{\prime}}),q(d^{{\ast}})] = x(1,1) = x^{{\ast}}$.

Consequently,

$$\displaystyle\begin{array}{rcl} & & c^{{\prime}} > c^{{\ast}}\wedge (4.12) \rightarrow c^{{\prime}} + x^{{\ast}}L(c^{{\prime}},d^{{\ast}}) < c^{{\ast}} + x^{{\ast}}L^{{\ast}} \\ & &\rightarrow (1 - x^{{\ast}})c^{{\prime}} + x^{{\ast}}[c^{{\prime}} + d^{{\ast}} + L(c^{{\prime}},d^{{\ast}})] < (1 - x^{{\ast}})c^{{\ast}} + x^{{\ast}}[c^{{\ast}} + d^{{\ast}} + L^{{\ast}}] \\ & & \rightarrow (1 - x^{{\ast}})c^{{\prime}} < (1 - x^{{\ast}})c^{{\ast}},\text{ as }\mathrm{TSC}(c^{{\prime}},d^{{\ast}}) \geq \mathrm{ TSC}(c^{{\ast}},d^{{\ast}}). {}\end{array}$$

(4.14)

$$\displaystyle\begin{array}{rcl} & & (1 - x^{{\ast}}) > 0 \wedge \mbox{ (4.14)} \rightarrow c^{{\prime}} < c^{{\ast}},\text{ which contradicts the hypothesis that }c^{{\prime}} > c^{{\ast}}.{}\end{array}$$

(4.15)

$$\displaystyle\begin{array}{rcl} & & (1 - x^{{\ast}}) = 0 \wedge \mbox{ (4.14)} \rightarrow 0 < 0,\text{ a contradiction. }{}\end{array}$$

(4.16)

(4.15) and (4.16) establish that (4.14) cannot hold. Therefore it follows that:

$$\displaystyle\begin{array}{rcl} c^{{\prime}} > c^{{\ast}}\rightarrow \mbox{ (4.12)}\text{ cannot hold. }& &{}\end{array}$$

(4.17)

(4.13) and (4.17) establish that (4.12) cannot hold.

By an analogous argument one can show that $(\exists d^{{\prime}}\in D)[d^{{\prime}} + y[p(c^{{\ast}}),q(d^{{\prime}})]$ $L(c^{{\ast}},d^{{\prime}}) < d^{{\ast}} + y^{{\ast}}L^{{\ast}}]$ cannot hold.

This establishes the theorem. □