Work environment and moral hazard

Abstract

We consider a firm’s provision of safety and health measures (working conditions) in a hidden action agency problem in which effort and working conditions interact in multiplicatively separable (neutral) manner in the cash flow process. Under this common formulation, the firm under supplies working conditions and effort at its second best, regardless of the share of accident damages borne by the firm. At this optimum, increases in the damage share paid by the firm decrease the compensation to the agent so as to render working conditions and effort unchanged. Shifting the damage share then does not impact the firm’s or the agent’s welfare. We show that direct regulation of working conditions can improve total surplus, but that the regulation of the damage share is ineffectual. Under first order approximations, we also examine the effects of changes in the hazard level of the job and the efficiency of working conditions. Finally, we show that our results can be changed if the neutral interaction between effort and working conditions is violated.

Appendix

Proof of Lemma 1

Divide (P) by $e>0$

$$\begin{aligned} \beta x(\gamma +\sigma y)-(1-s)L(x)\phi (y)-c(e)/e+\alpha /e\ge 0. \end{aligned}$$

This condition is met if

$$\begin{aligned} \beta x(\gamma +\sigma y)-(1-s)L(x)\phi (y)-c(e)/e\ge 0. \end{aligned}$$

From (1), substitute $c^{\prime }=\beta x(\gamma +\sigma y)-(1-s)L(x)\phi (y),$ so that (P) is met if

$$\begin{aligned} c^{\prime }(e)-c(e)/e\ge 0. \end{aligned}$$

This expression is strictly positive because $c^{\prime },c^{\prime \prime }>0$ and $c(0)=0.$

The principal’s Lagrangian can now be written as

$$\begin{aligned} L=-\alpha +\big [(1-\beta )x(\gamma +\sigma y)-sL(x)\phi (y)\big ]e(r_{a})-g(y)+\lambda _{\alpha }\alpha +\lambda _{\beta }\beta +\lambda _{y}y, \end{aligned}$$

where $\lambda _{i}\ge 0$, $i=\alpha ,\beta ,y,$ are the multipliers for the non-negativity constraints on $\alpha ,\beta$ and $y,$ respectively. The first order condition for $\alpha$ is $-1+\lambda _{\alpha }=0.$ Thus, it follows that $\lambda _{\alpha }>0$ and $\alpha =0.$ $\square$

Proof of Proposition 1

Let $H_{TS}$ denote the Hessian of TS in $(e,y).$ We have that $\partial e^{*}/\partial b=$ $(1/|H_{TS}|)(L(x)g^{ \prime \prime }y+L(x)e(x\sigma +bL(x)))>0$ and $\partial y^{*}/\partial b=$ $(1/|H_{TS}|)(L(x)ec^{\prime \prime }+L(x)y(x\sigma +bL(x)))>0.$ Further, $\partial e^{*}/\partial a=-(1/|H_{TS}|)$ $(L(x)g^{\prime \prime })<0$ and $\partial y^{*}/\partial a=-(1/|H_{TS}|)L(x)(\sigma x+L(x)b)<0.$ Finally, $\partial e^{*}/\partial \gamma$ $=(1/|H_{TS}|)g^{\prime \prime }x>0, \partial y^{*}/\partial \gamma =(1/|H_{TS}|)x(\sigma x+bL(x))>0,$ $\partial e^{*}/\partial \sigma =(1/|H_{TS}|)[g^{\prime \prime }\cdot xy+ex(x\sigma +bL(x))],$ and $\partial y^{*}/\partial \sigma =(1/|H_{TS}|)[exc^{\prime \prime }+yx(\sigma x+bL(x))].$ $\square$

Proof of Proposition 2

Because $TS$ is strictly concave and differentiable,

$$\begin{aligned} \sum \limits _{i=e}^{y}\frac{\partial TS^{o}}{\partial i}(i^{*}-i^{o})>TS(e^{*},y^{*})-TS(e^{o},y^{o})>0. \end{aligned}$$

Thus, $TS_{e}^{o}(e^{*}-e^{o})+TS_{y}^{o}(y^{*}-y^{o})>0.$ However, $TS_{e}^{o}>0$ while $TS_{y}^{o}=0,$ so that $e^{*}>e^{o}.$ Next note that $TS_{ey}=\sigma x-L(x)\phi ^{\prime }>0,$ by A.1. By $e^{*}>e^{o}$ and $TS_{ey}>0,$ it follows that

$$\begin{aligned} TS_{y}(e^{*},y^{*})-TS_{y}(e^{o},y^{*})>0. \end{aligned}$$

Because it is true that $TS_{y}(e^{*},y^{*})-TS_{y}(e^{o},y^{o})=0,$ it is, therefore, necessary that $y^{o}<y^{*},$ by $TS_{yy}<0.$ $\square$

Proof of Proposition 3

The second order conditions to the principal’s problem include

$$\begin{aligned} \pi _{yy}= & {} -sL(x)e\phi ^{\prime \prime }\!-\!g^{\prime \prime }\!+\!2e^{\prime }\big (\beta \sigma x\!-\!(1\!-\!s)L(x)\phi ^{\prime }\big )\big ((1\!-\!\beta )\sigma x\!-\!sL(x)\phi ^{\prime }\big ) \\&+(r_{f})\big (e^{\prime \prime }(\beta \sigma x-(1-s)L\phi ^{\prime }\big )^{2}-e^{\prime }\big ((1-s)L(x)\phi ^{\prime \prime }\big )<0 \end{aligned}$$

and

$$\begin{aligned} \pi _{\beta \beta }=-2e^{\prime }x^{2}(\gamma +\sigma y)^{2}+e^{\prime \prime }x^{2}(\gamma +\sigma y)^{2}r_{f}<0. \end{aligned}$$

The second order cross partial $\pi _{y\beta }$ is given by

$$\begin{aligned} \pi _{y\beta }= & {} \!-x\sigma e\!-\!x(\gamma \!+\!\sigma y)e^{\prime }\big [\sigma \beta x\!-\!(1\!-\!s)L(x)\phi ^{\prime }\big ]\!+\!\big [(1\!-\!\beta )x\sigma \!-\!sL(x)\phi ^{\prime }\big ]e^{\prime }\\&x(\gamma +\sigma y) +r_{f}e^{\prime }x\sigma +r_{f}e^{\prime \prime }x(\gamma +\sigma y)(\beta \sigma x-(1-s)L(x)\phi ^{\prime })]. \end{aligned}$$

Using $\pi _{\beta }=0,$ we have that $e=r_{f}e^{\prime },$ so that $\pi _{y\beta }$ can be rewritten as

$$\begin{aligned} \pi _{y\beta }\!=\!e^{\prime }x(\gamma \!+\!\sigma y)\big [(1\!-\!2\beta )x\sigma \,+\,(1\!-\!2s)L(x)\phi ^{\prime }\big ]+\,e^{\prime \prime }x(\gamma +\sigma y)r_{f}\big [\beta x\sigma -(1-s)\phi ^{\prime }L(x)\big ]. \end{aligned}$$

We will make use of

$$\begin{aligned} \frac{\partial }{\partial s}\pi _{y}\equiv & {} \pi _{ys}=-L(x)e\phi ^{\prime }-L(x)\phi e^{\prime }\big [\beta x\sigma -(1-s)\phi ^{\prime }L(x)\big ]\\&+\,\big [(1-\beta )x\sigma -sL(x)\phi ^{\prime }\big ]e^{\prime }L(x)\phi \\&+\,r_{f}e^{\prime }\phi ^{\prime }L(x)+r_{f}e^{\prime \prime }L(x)\phi \big [ \beta x\sigma -(1-s)\phi ^{\prime }L(x)\big ]. \end{aligned}$$

Employing $e=r_{f}e^{\prime },$ this can be rewritten as

$$\begin{aligned} \pi _{ys}=e^{\prime }L(x)\phi \big [(1-2\beta )x\sigma +(1-2s)L(x)\phi ^{\prime }\big ]+e^{\prime \prime }L(x)\phi r_{f}\big [\beta x\sigma -(1-s)L(x)\phi ^{\prime }\big ]. \end{aligned}$$

Moreover, we have

$$\begin{aligned} \frac{\partial }{\partial s}\pi _{\beta }\equiv \pi _{\beta s}=-2e^{\prime }L(x)\phi x(\gamma +\sigma y)+e^{\prime \prime }L\phi x(\gamma +\sigma y)r_{f}. \end{aligned}$$

The effect of $s$ on $y$ is given by

$$\begin{aligned} \partial y^{o}/\partial s=(1/H)(-\pi _{ys}\pi _{\beta \beta }+\pi _{\beta s}\pi _{y\beta })=0. \end{aligned}$$

We will show that $[\pi _{ys}(-\pi _{\beta \beta })+\pi _{\beta s}\pi _{y\beta }]=0.$ Let us use $Z\equiv [(1-2\beta )x\sigma +(1-2s)L(x)\phi ^{\prime }]$ and $\partial r_{a}/\partial y=[\beta x\sigma -(1-s)L(x)\phi ^{\prime }]$ to rewrite this expression:

$$\begin{aligned}&\big [e^{\prime }L\phi Z+e^{\prime \prime }L\phi r_{f}(\partial r_{a}/\partial y)\big ]\big [2e^{\prime }x^{2}(\gamma +\sigma y)^{2}-e^{\prime \prime }x^{2}(\gamma +\sigma y)^{2}r_{f}\big ] \\&\quad +\,\big [-2e^{\prime }L\phi x(\gamma +\sigma y)+e^{\prime \prime }L\phi x(\gamma +\sigma y)r_{f}\big ] \\&\quad \times \,\big [e^{\prime }x(\gamma +\sigma y)Z +\,e^{\prime \prime }x(\gamma +\sigma y)r_{f}(\partial r_{a}/\partial y)\big ]. \end{aligned}$$

Multiplying terms

$$\begin{aligned}&\!\!2(e^{\prime })^{2}L\phi x^{2}(\gamma \!+\!\sigma y)^{2}Z\!-\!L\phi Zr_{f}x^{2}(\gamma \!+\!\sigma y)^{2}e^{\prime }e^{\prime \prime }\!+\!2L\phi r_{f}(\partial r_{a}/\partial y)x^{2}(\gamma \!+\!\sigma y)^{2}e^{\prime }e^{\prime \prime } \\&\quad -\,L\phi (r_{f})^{2}(\partial r_{a}/\partial y)x^{2}(\gamma +\sigma y)^{2}(e^{\prime \prime })^{2}-2(e^{\prime })^{2}L\phi x^{2}(\gamma +\sigma y)^{2}Z\\&\quad +\,L\phi Zr_{f}x^{2}(\gamma +\sigma y)^{2}e^{\prime }e^{\prime \prime } \\&\quad -\,2L\phi r_{f}(\partial r_{a}/\partial y)x^{2}(\gamma +\sigma y)^{2}e^{\prime }e^{\prime \prime }+L\phi (r_{f})^{2}x^{2}(\gamma +\sigma y)^{2}(\partial r_{a}/\partial y)(e^{\prime \prime })^{2} \\&\quad =0. \end{aligned}$$

Thus, $\partial y^{o}/\partial s=0.$

Next consider

$$\begin{aligned} \partial \beta ^{o}/\partial s=\frac{(-\pi _{yy}\pi _{\beta s} +\pi _{ys}\pi _{y\beta })}{\pi _{yy}\pi _{\beta \beta }-(\pi _{y\beta })^{2}}. \end{aligned}$$

From above we know that $\pi _{ys}=(\pi _{\beta s}\pi _{y\beta })/\pi _{\beta \beta }$, so that

$$\begin{aligned} \partial \beta ^{o}/\partial s= & {} \frac{\Big (-\pi _{yy}\pi _{\beta s} +\frac{\big (\pi _{\beta s}\pi _{y\beta }\big )}{\pi _{\beta \beta }}\pi _{y\beta }\Big )}{\pi _{yy}\pi _{\beta \beta }-(\pi _{y\beta })^{2}}=\frac{\frac{-\pi _{\beta s}}{\pi _{\beta \beta }}\big (\pi _{yy}\pi _{\beta \beta }-(\pi _{y\beta })^{2}\big )}{\pi _{yy}\pi _{\beta \beta }-(\pi _{y\beta })^{2}} \\= & {} \frac{-\pi _{\beta s}}{\pi _{\beta \beta }}=\frac{2e^{\prime }L(x)\phi x(\gamma +\sigma y)-e^{\prime \prime }L(x)\phi x(\gamma +\sigma y)r_{f}}{ -2e^{\prime }x^{2}(\gamma +\sigma y)^{2}+e^{\prime \prime }x^{2}(\gamma +\sigma y)^{2}r_{f}} \\= & {} \frac{-L(x)\phi }{x(\gamma +\sigma y)}. \end{aligned}$$

This completes the proof. $\square$

Proof of Proposition 4

Note that, given a general $\phi ,$ $e(r_{a})=r_{a},e(r_{a})=e^{\prime }(r_{a})r_{f}$, $e^{\prime }=1,$ so that $r_{a}=r_{f}.$ It immediately follows that $s\gtreqqless \frac{1}{2}$ implies $\beta \lesseqqgtr \frac{1}{2}.$ Now let $\phi =a-by.$ For this case, at a point where the first order conditions are met,

$$\begin{aligned} \pi _{yy}= & {} 2\big [(1-\beta )\sigma x+sLb)(\beta \sigma x+(1-s)Lb\big ]-g^{\prime \prime }, \\ \pi _{\beta \beta }= & {} -2x^{2}(\gamma +\sigma y)^{2}, \\ \pi _{y\beta }= & {} x(\gamma +\sigma y)\big [(1-2\beta )\sigma x-(1-2s)Lb\big ]. \end{aligned}$$

Whence, $\pi _{y\beta }\gtreqqless 0$ as $s\gtreqqless \frac{1}{2}.$ Using the notation of the proof of Proposition 3,

$$\begin{aligned} \pi _{ya}= & {} -(1-s)L\big [(1-\beta )\sigma x+sLb)-sL (\beta \sigma x+(1-s)Lb\big ]<0, \\ \pi _{\beta a}= & {} x(\gamma +\sigma y)(1-2s)L. \end{aligned}$$

We have

$$\begin{aligned} \partial y^{o}/\partial a=(1/H)(-\pi _{ya}\pi _{\beta \beta } +\pi _{\beta a}\pi _{y\beta }). \end{aligned}$$

With $|H|$ $>0,$ the $sign$ $\partial y^{o}/\partial a=$ $sign$ $(-\pi _{ya}\pi _{\beta \beta }+\pi _{\beta a}\pi _{y\beta }).$ Clearly, $-\pi _{ya}\pi _{\beta \beta }<0,$ $\pi _{\beta a}\gtreqqless 0$ as $s\lesseqqgtr \frac{1}{2},$ and $\pi _{y\beta }\gtreqqless 0$ as $s\gtreqqless \frac{1}{2} .$ Thus, $\pi _{\beta a}\pi _{y\beta }<0$ and $\partial y^{o}/\partial a<0.$

Next note that

$$\begin{aligned} \partial \beta ^{o}/\partial a=(1/H)(-\pi _{\beta a} \pi _{yy}+\pi _{ya}\pi _{y\beta }). \end{aligned}$$

If $s\ge 1/2,$ then $\pi _{\beta a}<0$ and $-\pi _{\beta a}\pi _{yy}\le 0.$ Further, $s\ge 1/2$ implies $\pi _{y\beta }\ge 0,$ with $\pi _{ya}\pi _{y\beta }\le 0.$ Whence, $s\ge 1/2$ implies $\partial \beta ^{o}/\partial a\le 0.$ The converse argument for $s<1/2$ implies $\partial \beta ^{o}/\partial a>0.$

Next consider a change in $b.$ We have

$$\begin{aligned} \partial y^{o}/\partial b=(1/H)(-\pi _{yb}\pi _{\beta \beta } +\pi _{\beta b}\pi _{y\beta }). \end{aligned}$$

Moreover,

$$\begin{aligned} \pi _{yb}= & {} sLr_{a}\!+\Ly\!+\![\beta \sigma x\!+\!(1\!-\!s)bL]sLy\!+\!r_{f}(1\!-\!s)L, \\ \pi _{\beta b}= & {} x(\gamma +\sigma y)yL(2s-1). \end{aligned}$$

Using the same arguments as above, $-\pi _{yb}\pi _{\beta \beta }>0$ and $s\gtreqqless \frac{1}{2}$ implies $\pi _{\beta b}\pi _{y\beta }\ge 0.$ Thus, $\partial y^{o}/\partial b>0.$ Likewise,

$$\begin{aligned} \partial \beta ^{o}/\partial b=(1/H)(-\pi _{\beta b} \pi _{yy}+\pi _{yb}\pi _{y\beta }). \end{aligned}$$

We have $-\pi _{\beta b}\pi _{yy}$ $\ge 0$ as $s\ge \frac{1}{2}.$ Moreover, $s\ge \frac{1}{2}$ implies that $\pi _{yb}\pi _{y\beta }\ge 0.$ Thus, $\partial \beta ^{o}/\partial b\ge 0.$ For $s<\frac{1}{2},$ the converse argument holds.

A change in $\gamma$ will be considered next. We have

$$\begin{aligned} \pi _{y\gamma }= & {} [(1-\beta )\sigma x+sLb]\beta x+[\beta \sigma x+(1-s)Lb](1-\beta )x>0, \\ \pi _{\beta \gamma }= & {} x^{2}(\gamma +\sigma y)(1-2\beta ), \end{aligned}$$

The sign of $\partial y^{o}/\partial \gamma$ is that of $-\pi _{y\gamma }\pi _{\beta \beta }+\pi _{y\beta }\pi _{\beta \gamma }.$ From above, $-\pi _{y\gamma }\pi _{\beta \beta }>0.$ Further, $\pi _{y\beta },\pi _{\beta \gamma }\gtreqqless 0$ as $s\gtreqqless \frac{1}{2},$ so that $\pi _{y\gamma }\pi _{\beta \gamma }>0$. Whence, $\partial y^{o}/\partial \gamma >0.$ The sign of $\partial \beta ^{o}/\partial \gamma$ is that of $-\pi _{yy}\pi _{\beta \gamma }+\pi _{y\gamma }\pi _{\beta y}.$ We have that $\pi _{\beta \gamma },\pi _{\beta y}\gtreqqless 0$ as $s\gtreqqless \frac{1}{2}$ and $-\pi _{yy}\pi _{\beta \gamma }\gtreqqless 0$ as $s\gtreqqless \frac{1}{2}.$ Thus, $\partial \beta ^{o}/\partial \gamma \gtreqqless 0$ as $s\gtreqqless \frac{1}{2}.$

Finally, we examine a change in $\sigma .$ For this case we have that

$$\begin{aligned} \pi _{y\sigma }= & {} r_{a}x+xy\{\beta [(1-\beta )\sigma x+sLb]+(1-\beta )[\beta \sigma x+(1-s)Lb]\}>0, \\ \pi _{\beta \sigma }= & {} x^{2}(\gamma +\sigma y)(1-2\beta ). \end{aligned}$$

The sign of $\partial y^{o}/\partial \sigma$ is that of $-\pi _{y\sigma }\pi _{\beta \beta }+\pi _{y\beta }\pi _{\beta \sigma }.$ We have that $-\pi _{y\sigma }\pi _{\beta \beta }>0$ and $\pi _{y\beta },\pi _{\beta \sigma }\gtreqqless 0$ as $s\gtreqqless \frac{1}{2}.$ It follows that $\partial y^{o}/\partial \sigma >0.$ The sign of $\partial \beta ^{o}/\partial \sigma$ is that of $-\pi _{\beta \sigma }\pi _{yy}+\pi _{y\beta }\pi _{y\sigma }.$ We have $-\pi _{\beta \sigma }\pi _{yy}\gtreqqless 0$ as $s\gtreqqless \frac{ 1}{2}$ and $\pi _{y\beta }\pi _{y\sigma }\gtreqqless 0$ as $s\gtreqqless \frac{1}{2},$ so that $\partial \beta ^{o}/\partial \sigma \gtreqqless 0$ as $s\gtreqqless \frac{1}{2}.$ $\square$

Proof of Lemma 2

From (12), we have that

$$\begin{aligned} \beta ^{\prime }(y)=\frac{x(\gamma +\sigma y)e^{\prime }(r_{a})((1-2\beta )\sigma x+(1-2s)L\phi ^{\prime })+x(\gamma +\sigma y)e^{\prime \prime }(r_{a})(\beta \sigma x-(1-s)L\phi ^{\prime })r_{f}}{2x^{2}(\gamma +\sigma y)^{2}e^{\prime }(r_{a})-x^{2}(\gamma +\sigma y)^{2}r_{f}e^{\prime \prime }(r_{a})}. \end{aligned}$$

From (13), $\partial e/\partial y=e^{\prime }(r_{a})[\beta ^{\prime }(y)(\gamma +\sigma y)x+\beta \sigma x-(1-s)L(x)\phi ^{\prime }].$ Substituting for $\beta ^{\prime }(y),$

$$\begin{aligned} \partial e/\partial y= & {} \frac{x(\gamma \!+\!\sigma y)e^{\prime }(r_{a})\{x(\gamma \!+\!\sigma y)e^{\prime }(r_{a})[(1\!-\!2\beta )x\!+\!(1\!-\!2s)L\phi ^{\prime }]\!+\!x(\gamma \!+\!\sigma y)e^{\prime \prime }(r_{a})[\beta \sigma x\!-\!(1\!-|!s)L\phi ^{\prime }]r_{f}\}}{2x^{2}(\gamma \!+\!\sigma y)^{2}e^{\prime }(r_{a})\!-\!x^{2}(\gamma \!+\!\sigma y)^{2}r_{f}e^{\prime \prime }(r_{a})} \\&+\,e^{\prime }(r_{a})[\beta \sigma x-(1-s)L\phi ^{\prime }]. \end{aligned}$$

Placing this expression over a common denominator and simplifying we obtain

$$\begin{aligned} \partial e/\partial y=\frac{e^{\prime }(r_{a})^{2}(\sigma x-L\phi ^{\prime }) }{2e^{\prime }(r_{a})-r_{f}e^{\prime \prime }(r_{a})}. \end{aligned}$$

The assumption that the second order condition $\pi _{\beta \beta }<0$ holds implies that $2e^{\prime }(r_{a})-r_{f}e^{\prime \prime }(r_{a})>0,$ and A.1 implies that $(\sigma x-L\phi ^{\prime })>0.$ Thus, $\frac{\partial e}{ \partial y}>0.$ $\square$

Proof of Proposition 5

By strict concavity, $\sum \nolimits _{i=e}^{y} \frac{\partial TS^{o}}{\partial i}(\hat{\imath }-i^{o})>TS(\hat{e},\hat{y} )-TS(e^{o},$ $y^{o})>0.$ Thus, $TS_{e}^{o}(\hat{e}-e^{o})-TS_{y}^{o}(\hat{y} -y^{o})>0.$ We know that $TS_{y}^{o}=0$ while $TS_{e}^{o}>0.$ It follows that $\hat{e}>e^{o}.$ Next note that $TS_{y}(e^{o},y^{o})=0,$ while $TS_{y}( \hat{e},y^{o})>0,$ by $TS_{ye}>0.$ Given that $TS_{y}(\hat{e},\hat{y})<0,$ it must be that $\hat{y}>y^{o},$ because $TS_{yy}<0.$ $\square$

Non-neutral interactions between effort and working conditions The second and first best solutions are given by

$$\begin{aligned} \beta ^{o}= & {} \frac{aLs[2bL(1-s)z^{2}-w]+x[w+bL(1-s)z^{2}]}{ x[w+bL(1-2s)z^{2}]}, \quad y^{o}=\frac{z^{2}(aL-x)(w+bLz^{2})}{ [w+z^{2}bL(1-2s)]^{2}}, \\ e^{o}= & {} \frac{z(aL-x)[-w^{2}(1-z)+2z^{4}b^{2}L^{2}(1-s)s-z^{2}bwL(1-2s-z)]}{ [w+z^{2}bL(1-2s)]^{2}},\\ y^{*}= & {} \frac{(aL-x)(w+bLz^{2})}{ (z^{2}L^{2}b^{2}+2bwL-1)},{\text { and }} \; e^{*} =\frac{z(aL-x)(bwL-1)}{(z^{2}L^{2}b^{2}+2bwL-1)}. \end{aligned}$$