The Cost of Pollution on Longevity, Welfare and Economic Stability

Abstract

This paper presents an overlapping generations model where pollution, private and public health expenditures are all determinants of longevity. Public expenditures, financed through labour taxation, provide both public health and abatement. We study the role of these three components of longevity on welfare and economic stability. At the steady state, we show that an appropriate fiscal policy may enhance welfare. However, when pollution is heavily harmful for longevity, the economy might experience aggregate instability or endogenous cycles. Nonetheless, a fiscal policy, which raises the share of public spending devoted to health, may display stabilizing virtues and rule out cycles. This allows us to recommend the design of the public policy that may comply with the dynamic and welfare objectives.

Appendices

Appendix 1: Proof of Lemma 1

First of all, to prove Lemma 1, we can easily show that $\lim _{x_t\rightarrow 0} \theta _{t}= 0$. Let us note $\Gamma _t>0$ the left-hand side of inequality (7), i.e. $\Gamma _t \equiv \frac{1}{1+\theta _{t}}\frac{\alpha }{x_{t}}$. Then, $\lim _{x_t\rightarrow 0}\Gamma _t =+ \infty$. Therefore, $\Gamma _t<\frac{1-\gamma }{\gamma s_t}$ cannot hold for $x_t=0$, which implies that $x_t>0$ and $\Gamma _t=\frac{1-\gamma }{\gamma s_t}$.

Appendix 2: Proof of Proposition 1

Any steady state should satisfy $P> 0$. Under Assumption 2, $P=\psi (k)> 0$ for all $k>0$. In contrast, $P=\varphi (k) > 0$ requires:

$$\begin{aligned}&w(k)/k> 1/(1-\tau ) \end{aligned}$$

(31)

$$\begin{aligned}&\left( \alpha \frac{\gamma }{1-\gamma }+1\right) /(1-\tau ) > w(k)/k \end{aligned}$$

(32)

Since $w(k)/k=A(1-\sigma )k^{\sigma -1}$, we can easily show that $\lim _{k\rightarrow 0}w(k)/k=+\infty$, $\lim _{k\rightarrow +\infty }w(k)/k=0$ and w(k) / k is strictly decreasing. This shows the existence of $\overline{k}$ and $\underline{k}$, defined by (20) and (21) respectively, such that $\underline{k}<\overline{k}$. Any $k\in (\underline{k}, \overline{k})$ satisfies inequalities (31) and (32).

We show now the existence of a steady state that belongs to $(\underline{k}, \overline{k}]$. By direct inspection of (18), we deduce that $\varphi (\underline{k})=+\infty$ and $\varphi (\overline{k})=0$. Moreover,

$$\begin{aligned} \psi (\underline{k})<&\frac{1}{m}\epsilon _1 f(\underline{k})<+\infty \\ \psi (\overline{k})>&\frac{1}{m}f(\overline{k})[\epsilon _1-\epsilon _2 (1-\mu )\tau ]>0 \end{aligned}$$

By continuity, there is at least one solution $k^*\in (\underline{k}, \overline{k})$ solving $\varphi (k^*)=\psi (k^*)$.

To show uniqueness, let us note $\epsilon _\varphi (k) \equiv \varphi '(k)k/\varphi (k)$ and $\epsilon _\psi (k) \equiv \psi '(k)k/\psi (k)$. Using (18) and (19), we get:

$$\begin{aligned} \epsilon _\psi (k)= & {} \sigma >0 \end{aligned}$$

(33)

$$\begin{aligned} \epsilon _\varphi (k)= & {} \frac{1}{\beta }\frac{2(1-\tau )w(k) \sigma -[1+\alpha +(1-\alpha )\sigma ]k}{(1-\tau )w(k)-k} \nonumber \\&-\frac{1}{\beta }\frac{[1+\alpha \gamma /(1-\gamma )]k-(1-\tau )w(k) \sigma }{[1+\alpha \gamma /(1-\gamma )]k-(1-\tau )w(k)} \end{aligned}$$

(34)

Under Assumptions 1 and 2, the first term in (34) is lower than $1/\beta$, whereas the second one is larger than $1/\beta$. We conclude that $\epsilon _\varphi (k)<0$, which shows the uniqueness of the steady state.

Appendix 3: Proof of Lemma 2

In the long run, the welfare W is defined by $W\equiv \pi (\theta )c^{1-\gamma }/(1-\gamma )$. Since consumption is given by $c=R(k)k/\pi (\theta )$, the welfare may be rewritten:

$$\begin{aligned} W= \pi (\theta )^\gamma (R(k)k)^{1-\gamma }/(1-\gamma ) \end{aligned}$$

(35)

Using (18), we have:

$$\begin{aligned} \theta = \frac{[(1-\tau )w(k)-k]^\alpha [\mu \tau w(k)]^{1-\alpha }}{P^\beta }=\frac{\left( \alpha \frac{\gamma }{1-\gamma }+1\right) k-(1-\tau )w(k)}{(1-\tau )w(k)-k}\equiv \theta (k) \end{aligned}$$

(36)

Substituting $\theta =\theta (k)$ into (35), the welfare becomes a function of k, namely $W\equiv W(k)$. Therefore, at the steady state $k^*$, the welfare is given by $W(k^*)$. Using Assumption 1, $R(k)k=\sigma Ak^\sigma$ is strictly increasing in k. We also know that $\pi (\theta )$ is increasing in $\theta$. Therefore, $W(k ^{*})$ is an increasing function of physical capital if $\theta '(k)>0$. Using (36), we get:

$$\begin{aligned} \frac{\theta '(k)k}{\theta (k)}=\frac{\left( \alpha \frac{\gamma }{1-\gamma }+1\right) k-(1-\tau )w(k)\sigma }{\left( \alpha \frac{\gamma }{1-\gamma }+1\right) k-(1-\tau )w(k)}-\frac{(1-\tau )w(k)\sigma -k}{(1-\tau )w(k)-k}>0 \end{aligned}$$

which concludes the proof.

Appendix 4: Proof of Proposition 2

Let us note $\epsilon _\varphi (\mu )\equiv \frac{\partial \varphi (k)}{\partial \mu }\frac{\mu }{k}$ and $\epsilon _\psi (\mu )\equiv \frac{\partial \psi (k)}{\partial \mu }\frac{\mu }{k}$. The steady state $k^*$ is given by $\varphi (k^*)=\psi (k^*)$. Differentiating this equation with respect to k and $\mu$, we get:

$$\begin{aligned} \frac{dk^*}{d\mu }\frac{\mu }{k^*}=\frac{\epsilon _\varphi (\mu )-\epsilon _\psi (\mu )}{\epsilon _\psi (k^*)-\epsilon _\varphi (k^*)} \end{aligned}$$

(37)

We have $\epsilon _\psi (k^*)>\epsilon _\varphi (k^*)$. Therefore, the sign of $\frac{dk^*}{d\mu }\frac{\mu }{k^*}$ is given by $\epsilon _\varphi (\mu )-\epsilon _\psi (\mu )$. Using (18) and (19), we have:

$$\begin{aligned} \epsilon _\varphi (\mu )= & {} \frac{1-\alpha }{\beta } >0 \end{aligned}$$

(38)

$$\begin{aligned} \epsilon _\psi (\mu )= & {} \frac{\epsilon _2 \mu \tau w(k^*)}{\epsilon _1 f(k^*)-\epsilon _2 (1-\mu )\tau w(k^*)} >0 \end{aligned}$$

(39)

Therefore, $\epsilon _\varphi (\mu )>\epsilon _\psi (\mu )$ if and only if:

$$\begin{aligned} (1-\alpha )[\epsilon _1-\epsilon _2 \tau (1-\sigma )]>\mu \epsilon _2\tau (1-\sigma )[\beta -(1-\alpha )] \end{aligned}$$

Since $\mu \in (0,1]$, this inequality is satisfied if $\beta \leqslant \widehat{\beta }$ or $\beta >\widehat{\beta }$ and $\mu <\widehat{\mu }$. On the contrary, when $\beta >\widehat{\beta }$ and $\mu \geqslant \widehat{\mu }$, we obtain $\epsilon _\varphi (\mu )\leqslant \epsilon _\psi (\mu )$.

Of course, using Lemma 2, we deduce the effects of $\mu$ on the stationary welfare $W^*$. We focus now on the effects induced by a change in $\mu$ on pollution $P^*$. Using (19) and (37), we have:

$$\begin{aligned} \frac{dP^*}{d\mu }\frac{\mu }{P^*}= & {} \epsilon _\psi (k^*) \frac{dk}{d\mu }\frac{\mu }{k^*}+\epsilon _\psi (\mu ) \\= & {} \frac{\epsilon _\psi (k^*)\epsilon _\varphi (\mu )-\epsilon _\varphi (k^*)\epsilon _\psi (\mu ) }{\epsilon _\psi (k^*)-\epsilon _\varphi (k^*)} \end{aligned}$$

Since $\epsilon _\psi (k^*)>0$, $\epsilon _\varphi (k^*)<0$, $\epsilon _\varphi (\mu )>0$ and $\epsilon _\psi (\mu ) >0$, we easily deduce that $\frac{dP^*}{d\mu }\frac{\mu }{P^*}>0$.

Appendix 5: Proof of Proposition 3

Using (18) and (19), there exists a unique value $m=m^*_\beta$ solving $\varphi (1)=\psi (1)$, with:

$$\begin{aligned} m^*_\beta \equiv [\epsilon _1 f(1)-\epsilon _2 (1-\mu )\tau w(1)]\frac{\left[ \alpha \frac{\gamma }{1-\gamma }+1-(1-\tau )w(1)\right] ^\frac{1}{\beta }}{[(1-\tau )w(1)-1]^\frac{\alpha +1}{\beta }[\mu \tau w(1)]^\frac{1-\alpha }{\beta }} \end{aligned}$$

(40)

Since $w(1)=(1-\sigma ) A$, we deduce that $m^*_\beta >0$ for $\underline{A}<A<\overline{A}$, whereas $m^*_\beta <1$ for A sufficiently close to $\overline{A}$. Finally, we clarify that $k^*=1$ belongs to $(\underline{k}, \overline{k})$ for $\underline{A}<A<\overline{A}$.

Appendix 6: Proof of Lemma 3

We differentiate the dynamic system (15)–(16) around the normalized steady state $k^*=1$. We get:

$$\begin{aligned} \frac{\textit{dP}_{t+1}}{P^*}= & {} (1-m^*_\beta )\frac{\textit{dP}_{t}}{P^*}+m^*_\beta \sigma \frac{\textit{dk}_{t}}{k^*} \\ \frac{\textit{dk}_{t+1}}{k^*}= & {} - \frac{\frac{\beta \theta \alpha \gamma /(1-\gamma )}{(1+\theta )(1+\theta +\alpha \gamma /(1-\gamma ))}}{1+\frac{\alpha }{(1-\tau )w-1}\frac{\theta \alpha \gamma /(1-\gamma )}{(1+\theta )(1+\theta +\alpha \gamma /(1-\gamma ))}} \frac{\textit{dP}_{t}}{P^*} \\&+ \sigma \frac{1+\frac{(1-\tau )w-(1-\alpha )}{(1-\tau )w-1}\frac{\theta \alpha \gamma /(1-\gamma )}{(1+\theta )(1+\theta +\alpha \gamma /(1-\gamma ))}}{1+\frac{\alpha }{(1-\tau )w-1}\frac{\theta \alpha \gamma /(1-\gamma )}{(1+\theta )(1+\theta +\alpha \gamma /(1-\gamma ))}}\frac{\textit{dk}_{t}}{k^*} \end{aligned}$$

where $\theta$ is given by (28). We deduce the trace $T(\beta )$ and the determinant $D(\beta )$ of the associated Jacobian matrix, given by (26) and (27) respectively.

Appendix 7: Proof of Lemma 4

The steady state is a saddle if either $P(-1)<0<P(1)$ or $P(1)<0<P(-1)$. Since $T(\beta )>0$ and $D(\beta )>0$, we have $P(-1)=1+T(\beta )+D(\beta )>0$. To determine the sign of $P(1)=1-T(\beta )+D(\beta )$, we note that $\Lambda _1\in (0,1)$ and $\Lambda _{2,\beta }>0$ [see (29) and (30)]. Using (26) and (27), we deduce that:

$$\begin{aligned} P(1)= & {} m^*_\beta (1-\Lambda _1)+m^*_\beta \Lambda _{2,\beta } >0 \end{aligned}$$

Since $P(1)>0$ and $P(-1)>0$, the steady state cannot be a saddle, but is either a source or a sink.

Appendix 8: Proof of Proposition 4

Using Lemma 4, the steady state can either be a sink, for $D(\beta )<1$, or a source, for $D(\beta )>1$. A Hopf bifurcation generically occurs when $D(\beta )$ crosses 1. Using (29), we deduce that $0<\Lambda _1<1$. This implies that $0<D(0)<1$.

We note that $\Lambda _{2,\beta }$ is strictly increasing in $\beta$, with $\Lambda _{2,\beta }=0$ when $\beta =0$ and $\lim _{\beta \rightarrow +\infty } \Lambda _{2,\beta }=+\infty$. In particular, $\Lambda _{2,\beta }\leqslant \Lambda _1$ for all $\beta \leqslant \widetilde{\beta }$, with:

$$\begin{aligned} \widetilde{\beta } \equiv \frac{(1+\theta )\left( 1+\theta +\frac{\alpha \gamma }{1-\gamma } \right) }{\theta \frac{\alpha \gamma }{1-\gamma }}+\frac{(1-\tau )w-(1-\alpha )}{(1-\tau )w-1} \end{aligned}$$

We immediately deduce that $D(\beta )<1$ for all $\beta \leqslant \widetilde{\beta }$. Using (27), we also have:

$$\begin{aligned} D'(\beta )=m^*_\beta \frac{\Lambda _{2,\beta }}{\beta }+\frac{\partial m^*_\beta }{\partial \beta }(\Lambda _{2,\beta }-\Lambda _1) \end{aligned}$$

(41)

Differentiating (40), we obtain:

$$\begin{aligned} \frac{\partial m^*_\beta }{\partial \beta }=-\frac{m^*_\beta }{\beta ^2} \ln \left[ \frac{\frac{\alpha \gamma }{1-\gamma }+1-(1-\tau )(1-\sigma )A}{((1-\tau )(1-\sigma )A-1)^{\alpha +1}(\mu \tau (1-\sigma )A)^{1-\alpha } } \right] \end{aligned}$$

Because $\underline{A}<A<\overline{A}$ and A sufficiently close to $\overline{A}$, the term into brackets is strictly smaller than 1, which implies that $\partial m^*_\beta /\partial \beta >0$. Therefore, since $\beta >\widetilde{\beta }$ means that $\Lambda _{2,\beta }>\Lambda _1$, we have $D'(\beta )>0$ for all $\beta >\widetilde{\beta }$. Taking into account that $\lim _{\beta \rightarrow +\infty } D(\beta )=+\infty$, there is a unique $\beta ^H>\widetilde{\beta }$ such that $D(\beta )<1$ for all $\beta <\beta ^H$, $D(\beta ^H)=1$, and $D(\beta )>1$ for all $\beta >\beta ^H$.

Appendix 9: Proof of Proposition 5

To analyze the stabilizing role of $\mu$, we evaluate how the critical value $\beta ^H$ evolves according to an increase of $\mu$. We recall that $\beta ^H$ solves $D(\beta ^H)=1$. We deduce that:

$$\begin{aligned} \frac{d \beta ^H}{d \mu }=-\frac{\partial D(\beta ^H)/\partial \mu }{D'(\beta ^H)} \end{aligned}$$

(42)

As shown in the proof of Proposition 4, $D'(\beta ^H)>0$. We also have $\partial D(\beta ^H)/\partial \mu$ $=(\Lambda _{2,\beta }-\Lambda _1)\partial m^*_\beta /\partial \mu$, where $\Lambda _{2,\beta }>\Lambda _1$ when $\beta = \beta ^H$ (see the proof of Proposition 4). Using (40), we get:

$$\begin{aligned} \frac{\partial m^*_\beta }{\partial \mu }\frac{\mu }{m^*_\beta }=\frac{\epsilon _2 \tau \mu (1-\sigma )}{\epsilon _1-\epsilon _2 \tau (1-\mu )(1-\sigma )} -\frac{1-\alpha }{\beta ^H} \end{aligned}$$

We easily deduce that $\partial m^*_\beta /\partial \mu <0$ if $\beta ^H\leqslant \widehat{\beta }$ or $\beta ^H>\widehat{\beta }$ and $\mu <\widehat{\mu }$, where $\widehat{\beta }$ and $\widehat{\mu }$ are given by (23), while $\partial m^*_\beta /\partial \mu >0$ if $\beta ^H>\widehat{\beta }$ and $\mu >\widehat{\mu }$. We conclude the proof using (42).