Working time regulation, unequal lifetimes and fairness

Abstract

We examine the redistributive impact of working time regulations in an economy with unequal lifetimes. We first compare the laissez-faire equilibrium with the ex post egalitarian optimum, where the realized lifetime well-being of the worst off (usually the short-lived) is maximized, and show that, unlike the laissez-faire, this social optimum involves an increasing working time age profile and equalizes the realized lifetime well-being of the short-lived and the long-lived. We then examine whether working time regulations can compensate the short-lived. It is shown that uniform working time regulations cannot improve the situation of the short-lived with respect to the laissez-faire, and can only reduce well-being inequalities at the cost of making the short-lived worse off. However, age-specific regulations involving lower working time for the young and higher working time for the old make the short-lived better off, even though such regulations may not fully eradicate well-being inequalities.

Appendix

1.1 Proof of Proposition 1

The FOCs of the individual’s problem are:

$$\begin{aligned} \pi u^{\prime }(c)= & {} \pi u^{\prime }(d) \\ wu^{\prime }(c)= & {} v^{\prime }(\ell _{y})+\rho _{y}-\varphi _{y} \\ wu^{\prime }(c)= & {} v^{\prime }(\ell _{o})+\frac{\rho _{o}}{\pi }-\frac{ \varphi _{o}}{\pi } \end{aligned}$$

as well as conditions $\rho _{y}\ge 0,1-\ell _{y}\ge 0$; $\rho _{o}\ge 0,1-\ell _{o}\ge 0$; $\varphi _{y}\ge 0,\ell _{y}\ge 0$ and $\varphi _{o}\ge 0,\ell _{o}\ge 0$ with complementary slackness.

We obviously have, from the first FOC, $c=d$. Regarding the working time profile, several cases can arise.

• If $0<\ell _{y},\ell _{o}<1$, we have $\rho _{y}=\rho _{o}=\varphi _{y}=\varphi _{o}=0.$ The FOCs become $wu^{\prime }(c)=v^{\prime }(\ell _{y})$ and $wu^{\prime }(c)=v^{\prime }(\ell _{o})$. Hence we have $0<\ell _{y}=\ell _{o}<1$.

• If $\ell _{y}=0$, we have $\rho _{y}=0$ and $\varphi _{y}\ge 0$. The second FOC becomes: $wu^{\prime }(c)=v^{\prime }(0)-\varphi _{y}\iff wu^{\prime }(c)=-\varphi _{y}$. Given that $u^{\prime }(c)>0$, that case is not possible.

• If $\ell _{o}=0$, we have $\rho _{o}=0$ and $\varphi _{o}\ge 0$. The third FOC becomes: $wu^{\prime }(c)=v^{\prime }(0)-\frac{\varphi _{o}}{\pi } \iff wu^{\prime }(c)=-\frac{\varphi _{o}}{\pi }$. Again that case is not possible.

• If $\ell _{y}=1$, we have $\rho _{y}\ge 0$ and $\varphi _{y}=0$. The second FOC becomes: $wu^{\prime }(c)=v^{\prime }(1)+\rho _{y}$.

  • If $0<\ell _{o}<1$, so that $\rho _{o}=\varphi _{o}=0$, we then have $wu^{\prime }(c)=v^{\prime }(1)+\rho _{y}$ and $wu^{\prime }(c)=v^{\prime }(\ell _{o})$, which implies $v^{\prime }(1)+\rho _{y}=v^{\prime }(\ell _{o})$. That case is not possible.

  • If $\ell _{o}=1$, so that $\rho _{o}\ge 0$ and $\varphi _{o}=0$, we then have: $wu^{\prime }(c)=v^{\prime }(1)+\rho _{y}$ and $wu^{\prime }(c)=v^{\prime }(1)+\rho _{o}$. That case is possible.

• If $\ell _{o}=1$, we have $\rho _{o}\ge 0$ and $\varphi _{o}=0$. The third FOC becomes: $wu^{\prime }(c)=v^{\prime }(1)+\frac{\rho _{o}}{\pi }$. If $0<\ell _{y}<1$, so that $\rho _{y}=\varphi _{y}=0$, we then have $wu^{\prime }(c)=v^{\prime }(\ell _{y})$ and $wu^{\prime }(c)=v^{\prime }(1)+ \frac{\rho _{o}}{\pi }\iff v^{\prime }(1)+\frac{\rho _{o}}{\pi }=v^{\prime }(\ell _{y})$. That case is not possible.

In sum, only two cases can arise: either $0<\ell _{y}=\ell _{o}<1$ or $\ell _{y}=\ell _{o}=1$. Given that consumption is smoothed, we have $c=\ell w+\pi \ell w-\pi c$, implying $c=d=w\ell$. Hence whether one case arises or the other depends on whether $wu^{\prime }(w)\gtrless v^{\prime }(1)$. If $wu^{\prime }(w)<v^{\prime }(1)$, the laissez-faire equilibrium is interior and we have $0<\ell _{y}=\ell _{o}<1$. On the contrary, when $wu^{\prime }(w)\ge v^{\prime }(1)$, we have the corner solution $\ell _{y}=\ell _{o}=1$.

1.2 Analytical example: the threshold for w

In order to discuss whether the long-lived is better off than the short-lived at the laissez-faire, let us take the following functional forms for $u\left( \cdot \right)$ and $v\left( \cdot \right)$:

$$\begin{aligned} u(c)=\frac{\left( w\ell \right) ^{1-\sigma }}{1-\sigma }-\alpha \quad \text { and }\quad v\left( \ell \right) =\frac{\beta \ell ^{2}}{2}. \end{aligned}$$

The condition for optimal working time $wu^{\prime }(w\ell )=v^{\prime }(\ell )$ and the egalitarian condition $u(w\ell )=v(\ell )$ can be rewritten as, respectively:

$$\begin{aligned} w\left( w\ell \right) ^{-\sigma }=\beta \ell \iff \ell =\left( \frac{ w^{1-\sigma }}{\beta }\right) ^{\frac{1}{1+\sigma }}\quad \text { and }\quad \frac{\left( w\ell \right) ^{1-\sigma }}{1-\sigma }-\alpha =\frac{\beta \ell ^{2}}{2}. \end{aligned}$$

Hence substituting for the first equation in the second, we obtain:

$$\begin{aligned} U^{LL}\gtrless U^{SL}\iff w\gtrless \left[ \frac{\alpha }{\beta ^{\frac{ \sigma -1}{1+\sigma }}\left[ \frac{1}{1-\sigma }-\frac{1}{2}\right] }\right] ^{\frac{1+\sigma }{2(1-\sigma )}}. \end{aligned}$$

The RHS consists of the threshold for the wage beyond which the long-lived is better off than the short-lived. The threshold depends on $\alpha$, $\beta$ and $\sigma$. The higher the wage is, the more likely it is that the LHS exceeds the RHS, implying that $U^{LL}>U^{SL}$.

1.3 Proof of Proposition 2

The FOCs of the social planner’s problem are:

$$\begin{aligned} \pi u^{\prime }(c)= & {} \mu u^{\prime }(d) \\ wu^{\prime }(c)= & {} v^{\prime }(\ell _{y})+\kappa _{y}-\gamma _{y} \\ \pi wu^{\prime }(c)= & {} \mu v^{\prime }(\ell _{o})+\kappa _{o}-\gamma _{o} \end{aligned}$$

as well as the conditions $\mu \ge 0,u(d)-v(\ell _{o})\ge 0$; $\kappa _{y}\ge 0,1-\ell _{y}\ge 0$;$\kappa _{o}\ge 0,1-\ell _{o}\ge 0$; $\gamma _{y}\ge 0,\ell _{y}\ge 0$ and $\gamma _{o}\ge 0,\ell _{o}\ge 0$ with complementary slackness.

• 1. Let us suppose that the egalitarian constraint is satisfied, so that $\mu \ge 0$ and$u(d)-v(\ell _{o})=0$.

  • (a) In the interior case ($0<\ell _{o},\ell _{y}<1$), we have $\kappa _{y}=\kappa _{o}=\gamma _{y}=\gamma _{o}=0$, so that: $v^{\prime }(\ell _{y})=\frac{\mu }{\pi }v^{\prime }(\ell _{o})$. We thus have either an increasing or a decreasing consumption profile, depending on whether $\frac{ \mu }{\pi }\lessgtr 1$.

  • (b) If now $\kappa _{y}\ge 0$ so that $\ell _{y}=1$, and $\kappa _{o}=0$ so that $\ell _{o}<1$. In that situation, we have $\gamma _{y}=0$ (since $\ell _{y}=1$). We then have: $wu^{\prime }(c)=v^{\prime }(1)+\kappa _{y}$ and $wu^{\prime }(c)=\frac{\mu }{\pi }v^{\prime }(\ell _{o})-\gamma _{o}\implies v^{\prime }(1)+\kappa _{y}=\frac{\mu }{\pi }v^{\prime }(\ell _{o})-\frac{\gamma _{o}}{\pi }$.

    • If $\gamma _{o}=0$, i.e. $\ell _{o}>0$, two cases can arise depending on whether $\frac{\mu }{\pi }\gtrless 1$. (1) If $\frac{\mu }{\pi }<1$, we have a contradiction since the above equality would imply $v^{\prime }(\ell _{o})>v^{\prime }(1)+\kappa _{y}$ and thus $\ell _{o}>1$. That corner case $\left( \ell _{y}=1,\ell _{o}<1\right)$ is not possible in affluent economies. (2) When $\frac{\mu }{\pi }>1$, $\left( \ell _{y}=1,\ell _{o}<1\right)$ is possible.

    • If $\gamma _{o}\ge 0$, $\ell _{o}=0$, we have $v^{\prime }(1)+\kappa _{y}=\frac{\mu }{\pi }v^{\prime }(0)-\gamma _{o}$ which is possible only if $\mu /\pi >1$ (i.e. poor economy). But the case $\left( \ell _{y}=1,\ell _{o}=0\right)$ is not possible when $\frac{\mu }{\pi }<1$ (affluent economy).

  • (c) If $\kappa _{y}=0$ so that $\ell _{y}<1$, and $\kappa _{o}\ge 0$ and $\ell _{o}=1$. We then have that $\gamma _{o}=0$ and $wu^{\prime }(c)=\frac{ \mu }{\pi }v^{\prime }(1)+\frac{\kappa }{\pi }=v^{\prime }(\ell _{y})-\gamma _{y}$.

    • If $\gamma _{y}=0$ implying $\ell _{y}>0$, when $\frac{\mu }{\pi }>1$, we have a contradiction. That case $\left( \ell _{y}<1,\ell _{o}=1\right)$ is thus not possible. But when $\frac{\mu }{\pi }<1$ (affluent economies) that case $\left( \ell _{y}<1,\ell _{o}=1\right)$ is possible.

    • If $\gamma _{y}\ge 0$ implying $\ell _{y}=0$, we have $\frac{\mu }{ \pi }v^{\prime }(1)+\frac{\kappa _{o}}{\pi }=v^{\prime }(0)-\gamma _{y}$, which is impossible.

  • (d) Consider now the double corner case, where $\ell _{o}=\ell _{y}=1$, so that $\kappa _{o}\ge 0$ , $\kappa _{y}\ge 0$ and $\gamma _{y}=\gamma _{o}=0$. We then have $wu^{\prime }(c)=\frac{\mu }{\pi }v^{\prime }(1)+ \frac{\kappa _{o}}{\pi }=v^{\prime }(1)+\kappa _{y}$. When $\frac{\mu }{\pi } >1$, we have $\kappa _{y}>\frac{\kappa _{o}}{\pi }$. When $\frac{\mu }{\pi } <1$, we have the opposite. That case is possible.

  • (e) Consider now the double corner case, where $\ell _{o}=\ell _{y}=0$, so that $\kappa _{o}=\kappa _{y}=0$ and $\gamma _{y},\gamma _{o}>0$. This implies $c=d=0.$ But the egalitarian constrain would then become $u(0)=v(0)$ , which is impossible, since the LHS is negative and the RHS is zero. The case $\left( \ell _{y}=0,\ell _{o}=0\right)$ is not possible here.

  • (f) Consider now the case where $\ell _{y}=0$ so that $\kappa _{y}=0$ and $\gamma _{y}\ge 0$. We have: $wu^{\prime }(c)=\frac{\mu }{\pi }v^{\prime }(\ell _{o})+\frac{\kappa _{o}}{\pi }-\frac{\gamma _{o}}{\pi }=v^{\prime }(0)-\gamma _{y}$. If $0<\ell _{o}<1$, $\kappa _{o}=\gamma _{o}=0$, so that we have $\frac{\mu }{\pi }v^{\prime }(\ell _{o})=-\gamma _{y}$. This is not possible.

  • (g) Consider now the case where $\ell _{o}=0$ so that $\kappa _{o}=0$ and $\gamma _{o}\ge 0$. We have: $wu^{\prime }(c)=-\frac{\gamma _{o}}{\pi } =v^{\prime }(\ell _{y})+\kappa _{y}-\gamma _{y}$. If $0<\ell _{y}<1$, $\kappa _{y}=\gamma _{y}=0$, implying $-\frac{\gamma _{o}}{\pi }=v^{\prime }(\ell _{y}).$ which is not possible either.

• 2. Suppose now that the egalitarian constraint does not hold so that $\mu =0$. The FOC for d is now strictly negative and equal to $-\pi u^{\prime }(c)$ which implies that $d=0$. However, under the egalitarian constraint, we need to have $u(0)-v(\ell _o)>0$ which is not possible since $u(0)<0$. Hence, this case is not possible.

1.4 Proof of Proposition 4

The FOCs of the social planner’s problem are:

$$\begin{aligned} \left[ wu^{\prime }(w\bar{\ell })-v^{\prime }(\bar{\ell })\right] \left( 1+\mu \right) =\kappa -\gamma \end{aligned}$$

as well as the conditions $\mu \ge 0,u\left( w\bar{\ell }\right) -v\left( \bar{\ell }\right) \ge 0$; $\gamma \ge 0,\bar{\ell }\ge 0$ and $\kappa \ge 0,1-\bar{\ell }\ge 0$ with complementary slackness.

• 1. Suppose the egalitarian constraint holds so that $\mu \ge 0$ and $u\left( w\bar{\ell }\right) =v\left( \bar{\ell }\right)$. The FOC becomes $\left[ u^{\prime }(c)w-v^{\prime }(\bar{\ell })\right] =\frac{\kappa -\gamma }{1+\mu }$. For any interior value of $\bar{\ell }$ (i.e. $\gamma =\kappa =0$), we would have $u^{\prime }(w\bar{\ell })w=v^{\prime }(\bar{\ell })$ which cannot hold at the same time as the egalitarian constraint ($u(w\bar{\ell } )=v(\bar{\ell })$). The same reasoning applies when $\bar{\ell }=0$ (i.e. $\gamma \ge 0$ and $\kappa =0$) and when $\bar{\ell }=1$ (i.e. $\gamma =0$ and $\kappa \ge 0$). Hence, the egalitarian constraint can never be binding.

• 2. Suppose that the egalitarian constraint does not hold, so that we have $\mu =0$ and $u\left( w\bar{\ell }\right) >v\left( \bar{\ell }\right)$. The FOC becomes: $\left[ u^{\prime }(w\bar{\ell })w-v^{\prime }(\bar{\ell }) \right] =\kappa -\gamma$. If $\bar{\ell }$ is interior, we have $\kappa =\gamma =0$ and the solution is given by $u^{\prime }(w\bar{\ell } )w=v^{\prime }(\bar{\ell })$ as in the laissez-faire equilibrium. If $\bar{ \ell }=0$, we have $\kappa =0$ and $\gamma \ge 0$, so that $u^{\prime }(c)w=-\gamma$ which is not possible since $u^{\prime }(c)>0$. If $\bar{\ell }=1$, we have $\kappa \ge 0$ and $\gamma =0$, so that $\left[ u^{\prime }(w)w-v^{\prime }(1)\right] =\kappa$. That case is possible. Thus we have either $0<\bar{\ell }<1$ or $\bar{\ell }=1$.

Inequalities in realized lifetime well-being are: $U^{LL}-U^{SL}=u(w\bar{\ell })-v(\bar{\ell })$. To reduce $U^{LL}-U^{SL}$, the government must choose $\tilde{\ell }\ne \bar{\ell }$. We would then have: $u(w\tilde{\ell })-v(\tilde{ \ell })<u(w\bar{\ell })-v(\bar{\ell })$. Thus, there would be a reduction of $U^{LL}-U^{SL}$ with respect to the optimal regulation (and also with respect to the laissez-faire). But this reduction in inequalities would be achieved at the cost of making the short-lived worst-off than at the laissez-faire since his utility would now be $u(w\tilde{\ell })-v(\tilde{\ell })$ instead of $u(w\bar{\ell })-v(\bar{\ell })$.

1.5 Proof of Proposition 5

The FOCs of the social planner’s problem are:

$$\begin{aligned} u^{\prime }(c)\frac{w}{1+\pi }(1+\mu )= & {} v^{\prime }(\bar{\ell }_{y})+\kappa _{y}-\gamma _{y} \\ u^{\prime }(c)\frac{w}{1+\pi }(1+\mu )= & {} \frac{\mu }{\pi }v^{\prime }(\bar{ \ell }_{o})+\frac{\kappa _{o}}{\pi }-\frac{\gamma _{o}}{\pi } \end{aligned}$$

as well as the conditions $\mu \ge 0,u\left( w\frac{\bar{\ell }_{y}+\pi \bar{ \ell }_{o}}{1+\pi }\right) -v(\bar{\ell }_{o})\ge 0$; $\gamma _{y}\ge 0,\bar{ \ell }_{y}\ge 0$; $\kappa _{y}\ge 0,1-\bar{\ell }_{y}\ge 0$; $\gamma _{o}\ge 0,\bar{\ell }_{o}\ge 0$ and $\kappa _{o}\ge 0,1-\bar{\ell }_{o}\ge 0$ with complementary slackness.

• 1. Take the case where the egalitarian constraint is satisfied. We have $\mu \ge 0$ and $u\left( w\frac{\bar{\ell }_{y}+\pi \bar{\ell }_{o}}{1+\pi } \right) -v(\bar{\ell }_{o})=0.$

  • (a) If $0<\bar{\ell }_{o},\bar{\ell }_{y}<1$, we have $\kappa _{y}=\kappa _{o}=\gamma _{y}=\gamma _{o}=v^{\prime }(\bar{\ell }_{y})$ and $u^{\prime }(c) \frac{w}{1+\pi }(1+\mu )=\frac{\mu }{\pi }v^{\prime }(\bar{\ell }_{o})$. When $\frac{\mu }{\pi }<1$, we have thus $v^{\prime }(\bar{\ell }_{o})>v^{\prime }( \bar{\ell }_{y})$, implying $\bar{\ell }_{o}>\bar{\ell }_{y}$. When $\frac{\mu }{\pi }>1$, the opposite holds.

  • (b) If $\bar{\ell }_{y}=0$, we have $\kappa _{y}=0$ and $\gamma _{y}\ge 0$ . We have $u^{\prime }(c)\frac{w}{1+\pi }(1+\mu )=v^{\prime }(0)-\gamma _{y} \frac{\mu }{\pi }v^{\prime }(\bar{\ell }_{o})+\frac{\kappa _{o}}{\pi }-\frac{ \gamma _{o}}{\pi }$. If $\bar{\ell }_{o}=0$, we have a contradiction: the egalitarian constraint cannot hold as $u(0)\ne v(0)$. If $0<\bar{\ell } _{o}<1$, we have $\kappa _{o}=0$ and $\gamma _{o}=0$, so that $-\gamma _{y}= \frac{\mu }{\pi }v^{\prime }(\bar{\ell }_{o})$ which is not possible. If $\bar{\ell }_{o}=1$, we have $\kappa _{o}\ge 0$ and $\gamma _{o}=0$, so that we have $-\gamma _{y}=\frac{\mu }{\pi }v^{\prime }(1)+\frac{\kappa _{o}}{\pi }$. That case is not possible either.

  • (c) Using the same reasoning as in (b) we can show that $\bar{\ell }_{o}=0$ is never possible.

  • (d) If $\bar{\ell }_{y}=1$ and $\bar{\ell }_{o}<1$, we have $\gamma _y=\gamma _o=0$, $\kappa _{y}\ge 0$ and $\kappa _{o}=0$. FOCs become $u^{\prime }(c)\frac{w}{1+\pi }=\frac{\kappa _{y}+v^{\prime }(1)}{1+\mu }= \frac{\frac{\mu }{\pi }v^{\prime }(\bar{\ell }_{o})}{1+\mu }$. If $\frac{\mu }{\pi }<1$, a contradiction is reached. If $\frac{\mu }{\pi }>1$, that case is possible.

  • (e) If $\bar{\ell }_{y}<1$ and $\bar{\ell }_{o}=1$, we have $\gamma _y=\gamma _o=0$, $\kappa _{y}=0,$ and $\kappa _{o}\ge 0.$ FOCs become $u^{\prime }(c)\frac{w}{1+\pi }=\frac{v^{\prime }(\bar{\ell }_{y})}{1+\mu }= \frac{\kappa _{o}+\mu v^{\prime }(1)}{\pi (1+\mu )}$. If $\frac{\mu }{\pi } >1$, that case is not possible. If $\frac{\mu }{\pi }<1$, that case is possible.

  • (f) If $\bar{\ell }_{y}=\bar{\ell }_{o}=1$, we have $\gamma _y=\gamma _o=0$, $\kappa _{y},\kappa _{o}\ge 0.$ FOCs become: $u^{\prime }(c)(1+\mu )\frac{w}{ 1+\pi }=\kappa _{y}+v^{\prime }(1)=\frac{\kappa _{o}}{\pi }+\frac{\mu }{\pi } v^{\prime }(1)$. That case is possible.

• 2. Take now the case where the egalitarian constraint does not hold. We have $\mu =0$ and $u\left( w\frac{\bar{\ell }_{y}+\pi \bar{\ell }_{o}}{1+\pi } \right) -v(\bar{\ell }_{o})>0$.

  In that case, the FOCs become: $u^{\prime }(c)\frac{w}{1+\pi }-v^{\prime }( \bar{\ell }_{y})=\kappa _{y}-\gamma _{y}$ and $u^{\prime }(c)\frac{w\pi }{ 1+\pi }=\kappa _{o}-\gamma _{o}$. Hence $u^{\prime }(c)\frac{w}{1+\pi }=v^{\prime }(\bar{\ell }_{y})+\kappa _{y}-\gamma _{y}=\frac{\kappa _{o}}{\pi } -\frac{\gamma _{o}}{\pi }$.

  • (a) Take the case where both $\bar{\ell }_{y}$ and $\bar{\ell }_{o}$ are interior. We then have $v^{\prime }(\bar{\ell }_{y})=0$ leading to $\bar{\ell } _{y}=0$. A contradiction.

  • (b) If $\bar{\ell }_{y}=0$, we have $\kappa _{y}=0$ and $\gamma _{y}\ge 0$ . This leads to $-\gamma _{y}=\frac{\kappa _{o}}{\pi }-\frac{\gamma _{o}}{ \pi }$. The case where $\bar{\ell }_{y}=\bar{\ell }_{o}=0$ is not possible as it would not satisfy the egalitarian constraint ($u\left( 0\right) -v(0)<0$ in that case). If $0<\bar{\ell }_{o}<1$, we have $\kappa _{o}=\gamma _{o}=0$. This leads to $-\gamma _{y}=0$. That case is impossible. If $\bar{\ell } _{o}=1$, we have $\kappa _{o}\ge 0$ and $\gamma _{o}=0$. This leads to $-\gamma _{y}=\frac{\kappa _{o}}{\pi }$. That case is impossible.

  • (c) If $\bar{\ell }_{y}=1$, we have $\kappa _{y}\ge 0$ and $\gamma _{y}=0$ . This leads to $v^{\prime }(1)+\kappa _{y}=\frac{\kappa _{o}}{\pi }-\frac{ \gamma _{o}}{\pi }$. If $\bar{\ell }_{o}=0$, we have $\kappa _{o}=0$ and $\gamma _{o}\ge 0$. This leads to $v^{\prime }(1)+\kappa _{y}=-\frac{\gamma _{o}}{\pi }$. That case is impossible. If $0<\bar{\ell }_{o}<1$, we have $\kappa _{o}=\gamma _{o}=0$. This leads to $v^{\prime }(1)+\kappa _{y}=0$. That case is impossible. If $\bar{\ell }_{o}=1$, we have $\kappa _{o}\ge 0$ and $\gamma _{o}=0$. This leads to $v^{\prime }(1)+\kappa _{y}=\frac{\kappa _{o}}{\pi }$. That case is possible.

  • (d) If $\bar{\ell }_{o}=0$, we have $\kappa _{o}=0$ and $\gamma _{o}\ge 0$ . We have $v^{\prime }(\bar{\ell }_{y})+\kappa _{y}-\gamma _{y}=-\frac{\gamma _{o}}{\pi }$. From (b), we know that $\bar{\ell }_{y}=\bar{\ell }_{o}=0$ is not possible. If $0<\bar{\ell }_{y}<1$, we have $\kappa _{y}=\gamma _{y}=0$. We have $v^{\prime }(\bar{\ell }_{y})=-\frac{\gamma _{o}}{\pi }$. That case is impossible. If $\bar{\ell }_{y}=1$, we have $\kappa _{y}\ge 0$ and $\gamma _{y}=0$. We have $v^{\prime }(1)+\kappa _{y}=-\frac{\gamma _{o}}{\pi }$. That case is impossible.

  • (e) If $\bar{\ell }_{o}=1$, we have $\kappa _{o}\ge 0$ and $\gamma _{o}=0$ . We have $v^{\prime }(\bar{\ell }_{y})+\kappa _{y}-\gamma _{y}=\frac{\kappa _{o}}{\pi }.$ From (b), we know that $\bar{\ell }_{y}=0$ is impossible. If $0< \bar{\ell }_{y}<1$, we have $\kappa _{y}=\gamma _{y}=0$. We have $v^{\prime }( \bar{\ell }_{y})=\frac{\kappa _{o}}{\pi }$. That case is possible. If $\bar{ \ell }_{y}=1$, we have $\kappa _{y}\ge 0$ and $\gamma _{y}=0$. We have $v^{\prime }(1)+\kappa _{y}=\frac{\kappa _{o}}{\pi }$. That case is possible.