Negligence as Failure to Take Some Cost-Justified Precaution

Abstract

The notion of negligence is usually defined in terms of shortfall from the due care. There is, however, another notion of negligence under which a party is adjudged to be negligent iff there exists a cost-justified precaution which could have been taken but was not taken. That is to say, a party is deemed to be negligent iff it can be shown that the party could have averted some harm by taking care which would have cost less than the loss due to harm. These two notions of negligence are logically independent of each other. This chapter investigates the efficiency of liability rules when negligence is defined in terms of existence of cost-justified untaken precautions rather than in terms of shortfall from the due care. It is shown that if negligence is defined in terms of existence of cost-justified untaken precautions, then there is no liability rule which is efficient.

Appendix: The Impossibility Theorem

Proposition 5.1.

Let negligence be defined in terms of cost-justified untaken precautions. If a liability rule $f: [0,1]^{2}\mapsto [0,1]^{2}$ is efficient for every application belonging to $\mathcal{A}^{u}$ , then it satisfies the condition of negligence liability.

Proof.

Let f be any liability rule violating condition NL. As condition NL is violated, we must have:

$$\displaystyle{[\exists p \in [0,1)][f(p,1)\neq (1,0)]\; \vee \; [\exists q \in [0,1)][f(1,q)\neq (0,1)].}$$

Suppose $[\exists q \in [0,1)][f(1,q)\neq (0,1)]$ holds.

Suppose for q ₀ ∈ [0, 1) we have:

$$\displaystyle{f(1,q_{0}) = (x_{q_{0}},y_{q_{0}}),y_{q_{0}}\neq 1.}$$

Let t be a positive number. As $y_{q_{0}} \in [0,1)$, we have $y_{q_{0}}t < t$. Choose a positive number r such that $y_{q_{0}}t < r < t$.

As $q_{0}\neq 1,(1 - q_{0})\neq 0$. Let $d_{0} = \frac{r} {1-q_{0}}$.

Let 0 < ε and 0 < c ₀.

Now let C and D be specified as follows:

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

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

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

It should be noted that the above specification of C, D, and L is consistent with (A1)–(A4). Furthermore, the specification of L is done in such a way that no inconsistency would arise even if q ₀ = 0.

As ε > 0 and $t > r = (1 - q_{0})d_{0}$, it follows that (c ₀, d ₀) is the unique TSC-minimizing configuration.

The following array gives C ^u(c, d) for (c, d) ∈ C × D:

$$\displaystyle\begin{array}{rcl} C^{u}(c,d)& & {}\\ \begin{array}{|c|c|c|c|}\hline &d = 0&d = q_{0}d_{0} & d = d_{0} \\\hline c = 0 & \{c_{0}\} & \{c_{0}\} & \{c_{0}\} \\\hline c = c_{0} & \varnothing & \varnothing & \varnothing \\\hline \end{array} & & {}\\ \end{array}$$

And the following array gives D ^u(c, d) for (c, d) ∈ C × D:

$$\displaystyle\begin{array}{rcl} D^{u}(c,d)& & {}\\ \begin{array}{|c|c|c|c|}\hline &d = 0&d = q_{0}d_{0} & d = d_{0} \\\hline c = 0 & \{d_{0}\} & \{d_{0}\} & \varnothing \\\hline c = c_{0} & \{d_{0}\} & \{d_{0}\} & \varnothing \\\hline \end{array} & & {}\\ \end{array}$$

Therefore, we obtain:

$$\displaystyle{(\forall (c,d) \in C \times D)[\hat{c}(c,d) = c_{0}\; \wedge \;\hat{ d}(c,d) = d_{0}].}$$

Now, expected costs of the injurer at $(c_{0 }, q_{0 } d_{0 } ) = EC_{2 } (c_{0 }, q_{0 } d_{0 } )$

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

This establishes that (c ₀, d ₀) is not a Nash equilibrium. Thus f is not efficient for the application under consideration. Consequently, it is not the case that f is efficient for every application belonging to $\mathcal{A}^{u}$. In case $[\exists p \in [0,1)][f(p,1)\neq (1,0)]$ holds, an analogous argument shows that it is not the case that f is an efficient liability rule for every application belonging to $\mathcal{A}^{u}$.

Thus it follows that if f is efficient for every application belonging to $\mathcal{A}^{u}$, then it must satisfy condition NL, establishing the proposition. □ 

Proposition 5.2.

Let negligence be defined in terms of cost-justified untaken precautions. If a liability rule $f: [0,1]^{2}\mapsto [0,1]^{2}$ is efficient for every application belonging to $\mathcal{A}^{u}$ , then it violates the condition of negligence liability.

Proof.

Let f be any liability rule satisfying condition NL.

Let f(1, 1) = (x ^∗, y ^∗).

First suppose that x ^∗ > 0.

Let t be a positive integer such that $\frac{1} {t} < x^{{\ast}}$.⁶

Choose positive numbers $c_{0},d_{0},\theta _{1},\theta _{2},L_{0},\epsilon _{1},\epsilon _{2},\epsilon _{3},\epsilon _{4},\epsilon _{5}$ such that:

• (i) ε ₃ < ε ₂

• (ii) ε ₄ < ε ₁

• (iii) ε ₁ < ε ₄ +ε ₅

• (iv) ε ₂ < ε ₃ +ε ₅

• (v) $\epsilon _{3} +\epsilon _{4} +\epsilon _{5} <\epsilon _{1} +\epsilon _{2}$

• (vi) ε ₁ −ε ₄ < θ ₁

• (vii) ε ₂ −ε ₃ < θ ₂

• (viii) L ₀ > t ε, where ε is any number greater than 2 [max $\{\epsilon _{1},\epsilon _{2},\epsilon _{3},\epsilon _{4},\epsilon _{5}\}]$.⁷

Now consider the following application belonging to $\mathcal{A}^{u}$:

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

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

$$\displaystyle\begin{array}{rcl} L(c,d)& & {}\\ \begin{array}{|c|c|c|c|}\hline & d = 0 & d = d_{0} & d = d_{0} +\epsilon _{2} \\\hline c = 0 &c_{0} + d_{0} +\theta _{1} +\theta _{2} + L_{0} & c_{0} +\theta _{1} + L_{0} & c_{0} +\theta _{1} + L_{0} -\epsilon _{3} \\\hline c = c_{0} & d_{0} +\theta _{2} + L_{0} & L_{0} & L_{0} -\epsilon _{3} \\\hline c = c_{0} +\epsilon _{1} & d_{0} +\theta _{2} + L_{0} -\epsilon _{4} & L_{0} -\epsilon _{4} & L_{0} -\epsilon _{3} -\epsilon _{4} -\epsilon _{5} \\\hline \end{array} & & {}\\ \end{array}$$

(c ₀, d ₀) is the unique TSC-minimizing configuration.

We obtain C ^u(c, d), (c, d) ∈ C × D, as given in the following array:

$$\displaystyle\begin{array}{rcl} C^{u}(c,d)& & {}\\ \begin{array}{|c|c|c|c|}\hline & d = 0 & d = d_{0} & d = d_{0} +\epsilon _{2} \\\hline c = 0 &\{c_{0},c_{0} +\epsilon _{1}\} & \{c_{0},c_{0} +\epsilon _{1}\} & \{c_{0},c_{0} +\epsilon _{1}\} \\\hline c = c_{0} & \varnothing & \varnothing & \{c_{0} +\epsilon _{1}\} \\\hline c = c_{0} +\epsilon _{1} & \varnothing & \varnothing & \varnothing \\\hline \end{array} & & {}\\ \end{array}$$

and D ^u(c, d), (c, d) ∈ C × D, as given in the following array:

$$\displaystyle\begin{array}{rcl} D^{u}(c,d)& & {}\\ \begin{array}{|c|c|c|c|}\hline & d = 0 &d = d_{0} & d = d_{0} +\epsilon _{2} \\\hline c = 0 &\{d_{0},d_{0} +\epsilon _{2}\} & \varnothing & \varnothing \\\hline c = c_{0} & \{d_{0},d_{0} +\epsilon _{2}\} & \varnothing & \varnothing \\\hline c = c_{0} +\epsilon _{1} & \{d_{0},d_{0} +\epsilon _{2}\} & \{d_{0} +\epsilon _{2}\} & \varnothing \\\hline \end{array} & & {}\\ \end{array}$$

Therefore, we obtain $\hat{c}(c,d)$, (c, d) ∈ C × D, as given in the following array:

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

And $\hat{d}(c,d)$, (c, d) ∈ C × D, as given in the following array:

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

Now, expected costs of the victim at $(c_{0},d_{0}) = EC_{1}(c_{0},d_{0})$

$$\displaystyle\begin{array}{rcl} & & =\ c_{0} + x^{{\ast}}L(c_{ 0},d_{0}) {}\\ & & =\ c_{0} + x^{{\ast}}L_{ 0} {}\\ & & EC_{1}(c_{0} +\epsilon _{1},d_{0}) {}\\ & & =\ c_{0} +\epsilon _{1} + x(1, \frac{d_{0}} {d_{0} +\epsilon _{2}})L(c_{0} +\epsilon _{1},d_{0}) {}\\ & & =\ c_{0} +\epsilon _{1} {}\\ & & \text{as }x(1, \frac{d_{0}} {d_{0} +\epsilon _{2}}) = 0\text{ by condition NL.} {}\\ \end{array}$$

$$\displaystyle\begin{array}{rcl} & & EC_{1}(c_{0},d_{0}) - EC_{1}(c_{0} +\epsilon _{1},d_{0}) {}\\ & & =\ c_{0} + x^{{\ast}}L_{ 0} - c_{0} -\epsilon _{1} {}\\ & & =\ x^{{\ast}}L_{ 0} -\epsilon _{1} {}\\ \end{array}$$

Now,

$$\displaystyle\begin{array}{rcl} & & x^{{\ast}} > \frac{1} {t} \wedge L_{0} > t\epsilon \wedge t > 0\wedge \epsilon > 0 \rightarrow x^{{\ast}}L_{ 0} >\epsilon {}\\ & & \rightarrow x^{{\ast}}L_{ 0} >\epsilon _{1} {}\\ & & \rightarrow x^{{\ast}}L_{ 0} -\epsilon _{1} > 0. {}\\ \end{array}$$

Therefore it follows that (c ₀, d ₀) is not a Nash equilibrium. Thus f is not efficient for the application under consideration. Consequently, it is not the case that f is an efficient liability rule for every application belonging to $\mathcal{A}^{u}$. By an analogous argument it can be shown that if y ^∗ > 0, then it is not the case that f is an efficient rule for every application belonging to $\mathcal{A}^{u}$. As we must have $(x^{{\ast}} > 0 \vee y^{{\ast}} > 0)$, it follows that it is not the case that f is efficient for every application belonging to $\mathcal{A}^{u}$.

This establishes that if f is efficient for every application belonging to $\mathcal{A}^{u}$, then it must violate condition NL. □ 

Theorem 5.1.

There is no liability rule which is efficient for every application belonging to $\mathcal{A}^{u}$ .

Proof.

Follows immediately from Propositions 5.1 and 5.2. □