Malpractice liability, technology choice and negative defensive medicine

Abstract

We extend the theoretical literature on the impact of malpractice liability by allowing for two treatment technologies, a safe and a risky one. The safe technology bears no failure risk, but leads to patient-specific disutility since it cannot completely solve the health problems. By contrast, the risky technology (for instance a surgery) may entirely cure patients, but fail with some probability depending on the hospital’s care level. Tight malpractice liability increases care levels if the risky technology is chosen at all, but also leads to excessively high incentives for avoiding the liability exposure by adopting the safe technology. We refer to this distortion toward the safe technology as negative defensive medicine. Taking the problem of negative defensive medicine seriously, the second best optimal liability needs to balance between the over-incentive for the safe technology in case of tough liability and the incentive to adopt little care for the risky technology in case of weak liability. In a model with errors in court, we find that gross negligence where hospitals are held liable only for very low care levels outperforms standard negligence, even though standard negligence would implement the first best efficient care level.

Appendix

Proof of the first best benchmark

part (ii). $\frac{\partial \widetilde{\theta }}{{\rm d}\alpha }$ can be written as

$$ \frac{\partial \widetilde{\theta }}{{\rm d}\alpha }=\frac{\frac{\partial x}{ \partial \alpha }\left( 1+\alpha \right) \left[ p^{\prime }\left( x\right) \left( 1+\alpha \right) +1\right] -x}{\left( 1+\alpha \right) ^{2}}<0 $$

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since $\left[ p^{\prime }\left( x\right) \left( 1+\alpha \right) +1\right] =0$ as this is the first order condition for the optimal care level expressed by Eq. 3. □

Proof of Proposition 2

Part (i). Applying the implicit function theorem to the hospital’s first order condition

$$ A\equiv \left[ p^{\prime }\left( x\right) d\left( x,\widehat{x},q\right) +p\left( x\right) \frac{\partial d\left( x,\widehat{x},q\right) }{\partial x}\right] +p^{\prime }\left( x\right) \alpha +1=0 $$

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yields

$$ \frac{{\rm d}x^{\ast }}{{\rm d}\widehat{x}}=-\frac{p^{\prime }\left( x\right) \frac{ \partial {\rm d}(\cdot )}{\partial \widehat{x}}+p\left( x\right) \frac{\partial ^{2}{\rm d}(\cdot )}{\partial x\partial \widehat{x}}}{\frac{\partial ^{2}A\left( \cdot \right) }{\partial x^{2}}}>0 $$

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since $p^{\prime }\left( x\right) <0$, $\frac{\partial {\rm d}(\cdot )} {\partial \widehat{x}}=\frac{1}{2q}>0$, $\frac{\partial ^{2}{\rm d}(\cdot )}{\partial x\partial \widehat{x}}=0$, and because $\frac{\partial ^{2}A}{\partial x^{2}}>0$ is required for an interior solution for the hospital’s care level. Part (ii). Applying again the implicit function theorem, and making use of $\frac{\partial {\rm d}(\cdot )}{\partial q}=-\frac{\left( \widehat{x}-x\right) }{2q^{2}}$ and $\frac{\partial ^{2}{\rm d}(\cdot )}{\partial x\partial q}=\frac{1}{2q^{2}}$, we get

$$ \frac{\partial x^{\ast }}{\partial q}=-\frac{-p^{\prime }\left( x\right) \frac{\left( \widehat{x}-x\right) }{2q^{2}}+p\left( x\right) \frac{1}{2q^{2}} }{\frac{\partial ^{2}A\left( \cdot \right) } {\partial x^{2}}}=\frac{ p^{\prime }\left( x\right) \left( \widehat{x}-x\right) -p\left( x\right) }{ 2q^{2}\cdot \frac{\partial ^{2}A\left( \cdot \right) }{\partial x^{2}}}, $$

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so that the sign depends on $Z\equiv \left( \widehat{x}-x\right) p^{\prime }(x)+p(x)$. If $x<\widehat{x}$, then Z < 0 due to $p^{\prime }\left( x\right) <0$, and hence $\frac{\partial x^{\ast }(\cdot )}{\partial q}<0$. For $x\geq \widehat{x}$ we get $\frac{\partial x^{\ast }(\cdot )}{\partial q}<0$ if and only if $-\frac{p(x)}{p^{\prime }(x)}<\left( x-\widehat{x}\right)$. Part (iii). We prove by contradiction. Define x ^*₁ (x ^*₂ ) as the cost minimizing care levels for $\widehat{x}_{1}$ ($\widehat{x}_{2}$), and assume $\widehat{x}_{2}<\widehat{x}_{1}$. Now suppose that, for $\widehat{x}_{2}$, the hospital does not choose x ^*₂ but x ^*₁ . Then,

$$ \begin{aligned} &C\left( x_{1}^{\ast },\widehat{x}_{2},q\right) -C\left( x_{1}^{\ast }, \widehat{x}_{1},q\right)\\ &\quad =\left[ p\left( x_{1}^{\ast }\right) d\left( x_{1}^{\ast },\widehat{x} _{2},q\right) +p\left( x_{1}^{\ast }\right) \alpha +x_{1}^{\ast }\right] - \left[ p\left( x_{1}^{\ast }\right) d\left( x_{1}^{\ast },\widehat{x} _{1},q\right) +p\left( x_{1}^{\ast }\right) \alpha +x_{1}^{\ast }\right] \\ &\quad =p\left( x_{1}^{\ast }\right) \left[ d\left( x_{1}^{\ast },\widehat{x} _{2},q\right) -d\left( x_{1}^{\ast },\widehat{x}_{1},q\right) \right] <0 \\ \end{aligned} $$

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as $\frac{\partial d\left( x,\widehat{x},q\right) }{\partial \widehat{x}}>0$. Furthermore, $C\left( x_{2}^{\ast },\widehat{x}_{2},q\right) -C\left( x_{1}^{\ast },\widehat{x}_{2},q\right) <0$ by definition of the optimality of x ^*₂ . Combining the two inequalities yields $C\left( x_{2}^{\ast },\widehat{x}_{2},q\right) -C\left( x_{1}^{\ast },\widehat{x}_{1},q\right) <0$, and thus $\frac{\partial C\left( x^{\ast },\widehat{x},q\right) }{\partial \widehat{x}}>0$. Part (iv). The hospital chooses the risky technology if

$$ p\left( x^{\ast }\right) d\left( x^{\ast },\widehat{x},q\right) +p\left( x^{\ast }\right) \alpha +x^{\ast }\leq \alpha \theta, $$

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which yields a threshold of

$$ \widetilde{\theta }^{N}=\frac{p\left( x^{\ast }\right) d\left( x^{\ast }, \widehat{x},q\right) +p\left( x^{\ast }\right) \alpha +x^{\ast }}{\alpha }. $$

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Note that we need to compare to the optimal threshold under the assumption that $x^{\ast }\left( \widehat{x},q\right)$ is chosen as this is the relevant case given that a negligence rule is in place. Thus, the relevant second best optimal threshold is $\widetilde{\theta }^{S}=\frac{p\left( x^{\ast }\right) \left( 1+\alpha \right) +x^{\ast }}{1+\alpha }$, and not $\widetilde{\theta }^{f}$ (using $\widetilde{\theta }^{f}$ would only reinforce the result, though). Comparing these two thresholds gives

$$ \begin{aligned} \Updelta \widetilde{\theta }^{N} \equiv &\widetilde{\theta }^{N}-\widetilde{ \theta }^{S}= \frac{p\left( x^{\ast }\right) d\left( x^{\ast },\widehat{x} ,q\right) +p\left( x^{\ast }\right) \alpha +x^{\ast }}{\alpha }-\frac{ p\left( x^{\ast }\right) \left( 1+\alpha \right) +x^{\ast }}{1+\alpha}\\ =&\frac{\left[ x^{\ast }+d\left( x^{\ast },\widehat{x},q\right) p\left( x^{\ast }\right) \right] \left( 1+\alpha \right) }{\alpha \left( 1+\alpha \right)}>0. \end{aligned} $$

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□

Proof of Proposition 3

For $\widehat{x}=0$, we have $x^{\ast }\left( \widehat{x},q\right) =0$ as $d\left( x^{\ast },\widehat{x},q\right) =0\quad\forall x$, and since $\frac{\partial C\left( \cdot \right) }{\partial x}>0\quad\forall x$ if ${\rm d}\left( \cdot \right) =0$. Furthermore, we know from part (i) of Proposition 2 that $\frac{\partial x^{\ast }}{\partial \widehat{x}}>0$ as long as ${\rm d}\left( \cdot \right) \in \left[ 0,1\right]$. And if ${\rm d}\left( \cdot \right) =1$, then we are back to strict liability where we know from Proposition 1 that $x^{\ast }=x^{f}$. □

Proof of Proposition 4

Suppose that, with negligence, the regulator sets $\widehat{x}^{N}$, i.e. the due care level that implements the first best care level x ^f. As the care level is then the same with strict liability and with negligence, the welfare ranking depends only on the incentive to adopt the risky technology. Furthermore, we know from Proposition 1 and Proposition 2, part (iv), respectively, that this incentive is too small under both liability rules. Hence, the lower $\widetilde{\theta }$, the better. Comparing the hospital’s costs with negligence and with strict liability for $\widehat{x}^{N}$ gives

$$ \begin{aligned} \Updelta C &\equiv C^{N}\left( x^{f},\widehat{x}^{N}\right) -C^{\rm SL}\\ &=p\left( x^{f}\right) d\left( x^{f},\widehat{x},q\right) +p\left( x^{f}\right) \alpha +x^{f}-\left( p\left( x^{f}\right) \left( 1+\alpha \right) +x^{f}\right)\\ &=-p\left( x^{f}\right) \left[ 1-d\left( x^{f},\widehat{x},q\right) \right] <0 \\ \end{aligned} $$

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which proves that $\widetilde{\theta }^{N}\left( \widehat{x}^{N}\right) <\widetilde{\theta }^{\rm SL}$ and thus $SC\left( \widehat{x}^{N}\right) <SC^{\rm SL}$. Finally, it follows by definition of optimality that social costs with the second best optimal due care level $\widehat{x}^{S}$ are weakly below those with $\widehat{x}^{N}$, hence

$$ C^{N}\left( x^{\ast },\widehat{x}^{S}\right) \leq C^{N}\left( x^{f},\widehat{ x}^{N}\right) <C^{\rm SL}.$$

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□

Proof of Proposition 5

Social costs are

$$ SC=\int\limits_{0}^{\widetilde{\theta }(\widehat{x})}\left( 1+\alpha \right) \theta f(\theta ){\rm d}\theta +\int\limits_{\widetilde{\theta }(\widehat{x})}^{1}\left[ \left( 1+\alpha \right) p\left( x\left( \widehat{x}\right) \right) +x\left( \widehat{x}\right) \right] f(\theta ){\rm d}\theta . $$

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All patients with $\theta \leq \widetilde{\theta }(\widehat{x})$ are treated with the safe technology, and social costs are then $\left( 1+\alpha \right) \theta$ for patient type θ. All other patients are treated with the risky technology, and social costs for each patient are then given by the term in angled brackets. Taking the derivative with respect to the due care level $\widehat{x}$ yields

$$ \begin{aligned} \frac{\partial \left( SC\right) }{\partial \widehat{x}}&=A+B+C=\frac{ \partial \widetilde{\theta }(\widehat{x})}{\partial \widehat{x}} \left( 1+\alpha \right) \widetilde{\theta }(\widehat{x})f(\widetilde{\theta }( \widehat{x})) -\frac{\partial \widetilde{\theta }(\widehat{x})}{\partial \widehat{x}}\left( 1+\alpha \right) p\left( x\left( \widehat{x}\right) \right) f(\widetilde{\theta }(\widehat{x})) \\ &\quad +\int\limits_{\widetilde{\theta }(\widehat{x})}^{1}\left[ \frac{\partial x\left( \widehat{x}\right) }{\partial \widehat{x}}\left( \left( 1+\alpha \right) p^{\prime }+1\right) \right] f(\theta ){\rm d}\theta . \end{aligned} $$

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We now prove by inspection. The first part (A) is positive due to $\frac{\partial \widetilde{\theta }(\widehat{x})}{\partial \widehat{x}}>0$ and expresses that social costs are increasing in $\widehat{x}$ as more patients are then treated with the safe technology. Again due to $\frac{\partial \widetilde{\theta }(\widehat{x})}{\partial \widehat{x}}>0$, the second part (B) is negative and captures the benefit from the fact that fewer patients are treated with the risky technology. Finally, the third part (C) expresses the marginal social costs for patients treated with the risky technology, and will hence be zero at x = x ^f.

We already know from Proposition 2, part (iv), that, at x = x ^f, too many patients are treated with the risky technology which is identical to saying that A >  −B. Since C = 0 at x ^f, this implies that $\frac{\partial \left( SC\right) }{\partial \widehat{x}}>0$ at $\widehat{x}^{N}$ (recall that $x^{\ast }\left( \widehat{x}^{N}\right) =x^{f}$ by definition of $\widehat{x}^{N}$). Hence, $\widehat{x}$ must be reduced. This reduces $\widetilde{\theta }(\widehat{x})$, and hence the first part (A). Furthermore, note that the third part is then negative (C < 0) since $-p^{\prime }\left( \widehat{x}^{S}\right) >-p^{\prime }\left(\widehat{x}^{N}\right)$ for $\widehat{x}^{S}<\widehat{x}^{N}$ due to $p^{\prime \prime }(x)>0$. □