Health Spending, Education and Endogenous Demographics in an OLG Model

Abstract

We present a model of endogenous aging with public expenditure on health and pensions financed by an income tax. We show that government policies on health and pensions might lift an economy from a low to a high income steady state. In particular, the impact of an increase in the income tax is non-monotonic and depends on the initial levels of income and longevity: it is positive at low levels, and negative at high levels. On the other hand, a change in the allocation of public spending from social security benefits to health expenditures, without varying the tax rate, always increases income, even though pension benefits might decrease, and the problem of population aging could worsen.

Appendices

Appendix 1: Dynamics of Income Per Capita

We present some general properties of y_t+1 with respect to y_t derived from Eq. (7.22).

• y_t+1 is strictly increasing and continuous in the interval (0, + ∞):

  $$\displaystyle \begin{aligned} \lim_{y_t \to \hat{y}^-} y_{t+1}=\lim_{y_t \to \hat{y}^+} y_{t+1}=\left\{\frac{\alpha\beta p(\hat{y})}{\gamma[\alpha+\tau(1-\alpha)(1-\lambda)]} \right\}^{\alpha} \end{aligned} $$

  (7.35)

• The function ∂y_t+1∕∂y_t is discontinuous at $\hat {y}$, in particular:

  $$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \lim_{y_t \to \hat{y}^-}\frac{\partial y_{t+1}}{\partial y_{t}}- \lim_{y_t \to \hat{y}^+}\frac{\partial y_{t+1}}{\partial y_{t}}= -Ap(\hat{y})^{\alpha}\\ &\displaystyle &\displaystyle \quad A\left\{\frac{\alpha \beta}{\gamma \delta [\alpha +\tau(1-\alpha)(1-\lambda)]}\right\}^{\alpha} \frac{\delta^2z(1-\alpha)}{1-\delta}<0\qquad \end{array} \end{aligned} $$

  (7.36)

  When we specify adult survival, from Eq. (7.28) we get that:

• ∂y_t+1∕∂y > 0 in the interval (0, + ∞), that is:

  $$\displaystyle \begin{aligned} \frac{\partial{y_{t+1}}}{\partial{y_t}}=\left\{ \begin{array}{ll} A \alpha \Psi (1+\epsilon) x^{\alpha} y_t^{\alpha(1+\epsilon)-1} &\ \ \text{if}\ 0 \leq y_{t}\leq \hat{y}, \\ {} A \Delta (x y_t^{\epsilon})^{\alpha}[z(1-\alpha)y_t-1]^{\alpha+\delta(1-\alpha)}\\ \begin{array}{l}\quad \left(\frac{\alpha\epsilon}{y_t}+\frac{[\alpha+\delta(1-\alpha)]z(1-\alpha)}{z(1-\alpha)y_t-1}\right)\end{array} &\text{ if}\ \hat{y}<y_t<\bar{y}, \\ {} A \Delta \bar{p}^{\alpha}[z(1-\alpha)y_t-1]^{\alpha+\delta(1-\alpha)-1}z(1-\alpha)\\ \begin{array}{l}\quad [\alpha+\delta(1-\alpha)]\end{array} &\text{ if}\ y_t>\bar{y} \end{array} \right. \end{aligned} $$

  (7.37)

• The function ∂y_t+1∕∂y_t is discontinuous at $\hat {y}$ and $\bar {y}$, in particular:

  $$\displaystyle \begin{aligned} \begin{cases} \lim_{y_t \to \hat{y}^-}\frac{\partial y_{t+1}}{\partial y_{t}}< \lim_{y_t \to \hat{y}^+}\frac{\partial y_{t+1}}{\partial y_{t}} &\\ \lim_{y_t \to \bar{y}^-}\frac{\partial y_{t+1}}{\partial y_{t}}> \lim_{y_t \to \bar{y}^+}\frac{\partial y_{t+1}}{\partial y_{t}} \end{cases} \end{aligned} $$

  (7.38)

• From Eq. (7.37) it is easy to note that in the range $0\leq y_t \leq \hat {y}$, $\partial ^2 y_{t+1}/\partial y_t^2<0$ if:

  $$\displaystyle \begin{aligned} \epsilon<\frac{1-\alpha}{\alpha} \end{aligned} $$

  (7.39)

  In the interval $\hat {y} \leq y_t \leq \bar {y}$:

  $$\displaystyle \begin{aligned} \frac{\partial^2 y_{t+1}}{\partial y^2_t}=\chi\left\{ [z(1-\alpha)y_t]^2(2-E-Q)+2z(1-\alpha)y_t(E-1)-E\right\}, \end{aligned} $$

  (7.40)

  where: $\chi =y_{t+1} \alpha \epsilon z(1-\alpha ) [\alpha +\delta (1-\alpha )]/y^2_t z(1-\alpha ) [z(1-\alpha )y_t-1]^2$, E = (1 − α𝜖)∕[α + δ(1 − α)] and Q = (1 − α)(1 − δ)∕α𝜖.

  It is easy to note that if E > 1, then Q > 1 and therefore E + Q > 2. In this case, therefore, $\partial ^2 y_{t+1}/\partial y^2_t<0$ if:

  $$\displaystyle \begin{aligned} T(y_t)=-[z(1-\alpha)y_t]^2(E+Q-2)+2z(1-\alpha)y_t(E-1)-E<0. \end{aligned} $$

  (7.41)

  This is a concave downward function and thus if the maximum value of this function is negative then this function is always negative. In particular, given the maximum point:

  $$\displaystyle \begin{aligned} y^{MAX}=\frac{E-1}{(E+Q-2)z(1-\alpha)}, \end{aligned} $$

  (7.42)

  we get that the maximum value of the function is always negative, that is:

  $$\displaystyle \begin{aligned} T_{y_t=y^{MAX}}=\frac{1-EQ}{E+Q-2}=\frac{1}{-\delta(1-\alpha)-\alpha(1+\epsilon)}<0. \end{aligned} $$

  (7.43)

  In short $\partial ^2 y_{t+1}/\partial y^2_t<0$ if E > 1, that is:

  $$\displaystyle \begin{aligned} \epsilon<\frac{(1-\alpha)(1-\delta)}{\alpha}, \end{aligned} $$

  (7.44)

  note that if this condition is satisfied then so is (7.39).

  Finally when $y_t>\bar {y}$, from Eq. (7.37) it is easy to note that $\partial ^2 y_{t+1}/\partial y^2_t<0$.

1.1 Equilibria

Assuming that the condition in Eq. (7.44) holds then:

• In the range $0\leq y_t \leq \hat {y}$ the economy can reach the steady state:

  $$\displaystyle \begin{aligned} y_L=(A\Psi x^\alpha)^{\frac{1}{1-\alpha(1+\epsilon)}} \end{aligned} $$

  (7.45)

  The necessary and sufficient condition for the existence of such equilibrium is that $y_L<\hat {y}$ (or $\left . y_{t+1}\right \vert { }_{y_{t}=\hat {y}}<\hat {y}$ ). This holds if $A \leq \tilde {A}$, where:

  $$\displaystyle \begin{aligned} \tilde{A}=\frac{\hat{y}^{1-(1+\epsilon)\alpha}}{x^{\alpha}\Psi}=\frac{\hat{y}^{1-\alpha\epsilon}}{\Delta x^{\alpha}[z(1-\alpha)\hat{y}-1]^{\alpha+\delta(1-\alpha)}} \end{aligned} $$

  (7.46)

  In summary there exists one stable equilibrium in the range $y_t \in (0, \hat {y}]$ if:

  $$\displaystyle \begin{aligned} A\leq\tilde{A} \end{aligned} $$

  (7.47)

• In the range $y_t \in (\hat {y}, \bar {y}]$ different scenarios can arise.

  In particular, $\left .y_{t+1} \right \vert { }_{y_{t}=\bar {y}} > \bar {y}$ if $A\geq \bar {A}$ with:

  $$\displaystyle \begin{aligned} \bar{A}=\frac{\bar{y}}{\Delta \bar{p}^{\alpha}[z(1-\alpha)\bar{y}-1]^{\alpha+\delta(1-\alpha)}}. \end{aligned} $$

  (7.48)

  In order to reduce the number of feasible scenarios for the evolution of the economy we suppose that $\bar {A}>\tilde {A}$. This is true if the following sufficient condition holds:

  $$\displaystyle \begin{aligned} \tau<\tau^{MAX}=\bar{p}^{1/\epsilon}\frac{\delta z(1-\delta)^{1/(E-1)}}{\lambda}. \end{aligned} $$

  (7.49)

  Thus in the range $\hat {y}\leq y_t\leq \bar {y}$, if $A>\bar {A}$ there is no steady state; if $\tilde {A}<A<\bar {A}$ there is one stable equilibrium and finally when $A<\tilde {A}<\bar {A}$ there are two steady states if:

  $$\displaystyle \begin{aligned} T[z(1-\alpha)y_t-1]=y_t^{E}, \end{aligned} $$

  (7.50)

  where $T=(A \Delta x^{\alpha })^{\frac {1}{\alpha +\delta (1-\alpha )}}$. Note that in all other cases there is no steady state. Both the LHS and RHS of (7.50) increase with respect y_t, in particular:

  $$\displaystyle \begin{aligned} \begin{aligned} \frac{\partial LHS(y_t)}{\partial y_t}>0; \\ \frac{\partial^2 LHS(y_t)}{\partial y^2_t}=0 \\ \lim \limits_{y \rightarrow 0}LHS(y_t)=-T \\ \frac{\partial RHS(y_t)}{\partial y_t}>0; \\ \lim\limits_{y_t \rightarrow 0}RHS(y_t)=0; \end{aligned} \end{aligned} $$

  (7.51)

  It easy to note that the sign of $\partial ^2 RHS(y_t)/\partial y^2_t$ can be either negative if E < 1 or positive if E > 1.

  However, as specified above in Eq. (7.44) we assume E > 1. In this case, therefore, Eq. (7.50) can show no or two steady states. In particular by defining the maximum point of the function LHS(y_t) − RHS(y_t) as y^MAX there are two steady states if:

  $$\displaystyle \begin{aligned} \left. LHS(y_t)-RHS(y_t) \right\vert {}_{y_{t}=y^{MAX}}>0, \end{aligned} $$

  (7.52)

  and:

  $$\displaystyle \begin{aligned} y^{MAX}>\hat{y}. \end{aligned} $$

  (7.53)

  The maximum point of LHS(y_t) − RHS(y_t) > 0 is:

  $$\displaystyle \begin{aligned} y^{MAX}=\left[\frac{Tz(1-\alpha)}{E}\right]^{\frac{1}{E-1}} \end{aligned} $$

  (7.54)

  Some calculations show that (7.52) holds if $A>\hat {A}$ where:

  $$\displaystyle \begin{aligned} \hat{A}=\left[\left(\frac{E}{z(1-\alpha)}\right)^E\left(\frac{1}{E-1}\right)^{E-1}\right]^{\alpha+\delta(1-\alpha)}\frac{1}{\Delta x^{\alpha}}, \end{aligned} $$

  (7.55)

  and $y^{MAX}>\hat {y}$ if A > A₂ where:

  $$\displaystyle \begin{aligned} A_2=\left[\left(\frac{1}{\delta z(1-\alpha)}\right)^{E-1}\left(\frac{E}{z(1-\alpha)}\right)\right]^{\alpha+\delta(1-\alpha)}\frac{1}{\Delta x^{\alpha}} \end{aligned} $$

  (7.56)

  In summary, assuming that the parameters’ condition in Eq. (7.44) holds, then when $\hat {y}<y_t<\bar {y}$ the economy shows two steady states if $A>{max}\{\hat {A},A_2\}$.

  From Eqs. (7.46), (7.55) and (7.56) we get that $\hat {A}>A_2$ and $\tilde {A}>A_2$ if:

  $$\displaystyle \begin{aligned} E<\frac{1}{1-\delta} \Rightarrow \epsilon>\frac{(1-\alpha)(1-\delta)-\delta}{(1-\delta)\alpha} \end{aligned} $$

  (7.57)

  We assume that $\tilde {A}>\hat {A}$, that is:

  $$\displaystyle \begin{aligned} \left(\frac{E-1}{\delta E}\right)^{E}>\frac{(1-\delta)(E-1)}{\delta}. \end{aligned} $$

  (7.58)

• When $y_t \geq \bar {y}$ there is one stable equilibrium if $A>\bar {A}$. If $A<\bar {A}$ there are no or two steady states.

Appendix 2: Policy Effect

1.1 Income

From Eq. (7.26) we can observe that $\partial \bar {y}/\partial {\tau }<0$ and thus through this channel positively affects the evolution of income. From Eq. (7.28), in the range $0\leq y_t\leq \bar {y}$, ∂y_t+1∕∂τ > 0 if ∂( Ψx^α)∕∂τ > 0 and ∂( Δx^α)∕∂τ > 0. In particular, given η = α + τ(1 − λ)(1 − α) and x = [λτ(1 − α)]^𝜖, from Eqs. (7.23) and (7.24) both ∂( Ψx^α)∕∂τ > 0 and ∂( Δx^α)∕∂τ > 0 if ∂(τ^𝜖∕η)∕∂τ > 0 which holds if:

$$\displaystyle \begin{aligned} \tau<\hat{\tau}=\frac{\alpha\epsilon}{(1-\alpha)(1-\lambda)(1-\epsilon)}. \end{aligned} $$

(7.59)

Thus, ∂y_t+1∕∂y_t > 0 if $\tau <\hat {\tau }$ and ∂y_t+1∕∂y_t < 0if $\tau >\hat {\tau }$. This implies that both ∂y_L∕∂τ > 0 and ∂y_I∕∂τ > 0 if $\tau <\hat {\tau }$. See also Eq. (7.32) to back this result with respect to the equilibrium y_L. When $y_t\geq \bar {y}$ y_t+1 is always decreasing in τ.

With regard to the impact of the wage income tax on $\hat {A}$ and $\tilde {A}$ from Eqs. (7.29) and (7.30) we can observe that both $\partial \hat {A}/\partial \tau >0$ and $\partial \tilde {A}/\partial \tau >0$ if $\tau >\hat {\tau }$. From Eq. (7.31) we see that the impact of a higher wage income tax on $\bar {A}$ is ambiguous. However, with our set of parameters, the relationship between $\bar {A}$ and τ is always negative in the range 0 < τ ≤ τ^MAX.

1.2 Fertility

With respect to the impact of a higher wage income tax on fertility from Eq. (7.20) we derive that ∂n_t∕∂τ < 0 if:

$$\displaystyle \begin{aligned} -(1+\gamma)-\alpha\beta \frac{p(y_t)}{\eta}\left[1+(1-\tau)\frac{\partial(p(y_t)/\eta)/\partial \tau}{p(y_t)/\eta}\right]<0. \end{aligned} $$

(7.60)

From Eq. (7.25), note that when $y_t>\bar {y}$, this is always satisfied because ∂(p(y_t)∕η)∕∂τ = 0.

When $y_t<\bar {y}$ the term inside the brackets is given by:

$$\displaystyle \begin{aligned} 1+(1-\tau)\frac{[\alpha \epsilon-(1-\lambda)(1-\alpha)(1-\epsilon)\tau]}{\tau[\alpha+\tau(1-\lambda)(1-\alpha)]} \end{aligned} $$

(7.61)

which is positive since

$$\displaystyle \begin{aligned} \tau^2(1-\lambda)(1-\alpha)(2-\epsilon)+\tau(1-\epsilon)[\alpha-(1-\lambda)(1-\alpha)]+\alpha\epsilon>0\quad \end{aligned} $$

(7.62)

if the sufficient condition λ > (1 − 2α)∕(1 − α) holds. Thus under this parametric restriction, we can conclude that, a higher wage income tax always decreases fertility.