Growth and unemployment in an OLG economy with public pensions

Abstract

This paper develops an overlapping generations model including (1) a productive externality as an engine of endogenous growth and (2) wage setting by trade unions as the cause of unemployment. Within this framework, the paper considers growth and unemployment affected by public pensions under the following two types of pension system: the proportionate pension system where only the contributors, that is, the employed, receive pensions, and the lump-sum pension system where both the employed and the unemployed receive pensions. It is shown that public pensions create a trade-off between growth and employment in the former system, whereas they produce no trade-off in the latter.

Appendix

1.1 Appendix A: Derivation of Eqs. 10 and 11

From the savings functions Eqs. 3 and 4, the capital-market-clearing condition is:

$$ K_{t+1}=L_{t}\frac{w_{t}\big(1-\tau ^{w}-\theta ^{w}\big)-\phi (R)b_{\!t+1}}{1+\phi (R)R}+(N-L_{t})\frac{u_{t}-\phi (R)\eta b_{\!t+1}}{1+\phi (R)R}. $$

Multiplying both sides by 1 + ϕ(R)R, we have:

$$ \begin{array}{lll} (1+\phi (R)R)K_{t+1} &=&w_{t}L_{t}\big(1-\tau ^{w}-\theta ^{w}\big)+\big(N-L_{t}\big)u_{t} \\[4pt] &&-\,\big\{L_{t}+\big(N-L_{t}\big)\eta \big\}\phi (R)b_{\!t+1}\\[4pt] &=&\big\{\big(1-\tau ^{w}-\theta ^{w}\big)+\big(\theta ^{w}+\theta ^{f}\big)\big\}w_{t}L_{t} \\[4pt] &&-\,\phi (R)\big(\tau ^{w}+\tau ^{f}\big)w_{t+1}L_{t+1}\\[4pt] &=&\left\{\big(1-\tau ^{w}-\theta ^{w}\big)+\big(\theta ^{w}+\theta ^{f}\big)\right\}\frac{(1-\alpha )AK_{t}}{1+\tau ^{f}+\theta ^{f}}\\[4pt] &&-\,\phi (R)\big(\tau ^{w}+\tau ^{f}\big)\frac{ \big(1-\alpha \big)AK_{t+1}}{1+\tau ^{f}+\theta ^{f}}, \end{array} $$

where the equality in the second line is derived by using Eqs. 7 and 8 and the equality in the third line is derived by using Eq. 9. Rearranging the terms of the above equation, we obtain Eqs. 10 and 11.

1.2 Appendix B: Derivation of Eq. 12

The first-order condition for union wage setting Eq. 6 is rewritten as:

$$ \begin{array}{lll} (1-\alpha )(1-\tau ^{w}-\theta ^{w})w_{t}L_{t} &=&u_{t}L_{t}+\frac{(-1+\eta )b_{\!t+1}L_{t}}{R} \\[5pt] &=&\frac{(\theta ^{w}+\theta ^{f})w_{t}L_{t}}{N-L_{t}}L_{t}+\frac{(-1+\eta ) }{R}\\[5pt] &&\cdot \frac{(\tau ^{w}+\tau ^{f})w_{t+1}L_{t+1}}{L_{t}+(N-L_{t})\eta } L_{t} \\[5pt] &=&\frac{(\theta ^{w}+\theta ^{f})w_{t}L_{t}}{N-L_{t}}L_{t}+\frac{(-1+\eta ) }{R}\\[5pt] &&\cdot \frac{(\tau ^{w}+\tau ^{f})Gw_{t}L_{t}}{L_{t}+(N-L_{t})\eta }L_{t}, \end{array} $$

where the equality in the second line is derived by using Eqs. 7 and 8 and the equality in the third line is derived by using w _t + 1 L _t + 1 = Gw _t L _t . Dividing both sides by w _t L _t and rearranging the terms, we obtain Eq. 12.

1.3 Appendix C: Proof of u _t  > b _t + 1/R (Footnote 9)

From Eqs. 7 and 8, u _t  > b _t + 1/R in the case of η = 0 is rewritten as:

$$ \frac{(\theta ^{w}+\theta ^{f})w_{t}L_{t}}{N-L_{t}}>\frac{(\tau ^{w}+\tau ^{f})w_{t+1}L_{t+1}}{RL_{t}}=\frac{(\tau ^{w}+\tau ^{f})Gw_{t}L_{t}}{RL_{t}} . $$

Dividing both sides by w _t N and rearranging the terms, we have:

$$ (\theta ^{w}+\theta ^{f})l>\frac{(\tau ^{w}+\tau ^{f})G(1-l)}{R}. $$

We substitute the equilibrium employment rate Eq. 12 into the above inequality and rearrange the terms. Then, we find that u _t  > b _t + 1/R is rewritten as:

$$ (1-\alpha )(1-\tau ^{w}-\theta ^{w})+\frac{(\tau ^{w}+\tau ^{f})G}{R}>\frac{ (\tau ^{w}+\tau ^{f})G}{R}. $$

This inequality holds for any set of parameters.

1.4 Appendix D: Proof of Proposition 4

In the case of η = 0, the employment rate is given by:

$$ l_{t}=\left[ \frac{\theta ^{w}+\theta ^{f}}{(1-\alpha )(1-\tau ^{w}-\theta ^{w})+\frac{(\tau ^{w}+\tau ^{f})G}{R}}+1\right] ^{-1}. \label{a1} $$

(17)

• 1. Given that $\partial G/\partial \theta ^{w}=0$ (Proposition 1(1)), we immediately find that $\left. \partial l\right\vert _{\eta =0}/\partial \theta ^{w}<0$ holds. In order to determine the sign of $\left. \partial l\right\vert _{\eta =0}/\partial \theta ^{f}$, we focus on the first term within the square brackets, which is rewritten as:

  $$ \frac{\theta ^{w}+\theta ^{f}}{(1-\alpha )(1-\tau ^{w}-\theta ^{w})+\frac{ (\tau ^{w}+\tau ^{f})G}{R}}\!=\!\frac{1}{\underset{(a1)}{\underbrace{\frac{ (1-\alpha )(1-\tau ^{w}-\theta ^{w})}{\theta ^{w}+\theta ^{f}}}}+\underset{ (a2)}{\underbrace{\frac{(\tau ^{w}+\tau ^{f})G}{R(\theta ^{w}+\theta ^{f})}}} }. \label{a2} $$

  (18)

  The term (a1) is decreasing in θ ^f. The term (a2) is rewritten as:

  $$ (a2)=\frac{(\tau ^{w}+\tau ^{f})}{R}\times \frac{\frac{1}{1+\phi (R)R}\cdot \left\{ \frac{1-\tau ^{w}-\theta ^{w}}{\theta ^{w}+\theta ^{f}}+1\right\} (1-\alpha )A}{(1+\tau ^{f}+\theta ^{f})+\frac{\phi (R)}{1+\phi (R)R}(\tau ^{w}+\tau ^{f})(1-\alpha )A}, $$

  which shows that the term (a2) is decreasing in θ ^f. Thus, the left-hand side of Eq. 18 is increasing in θ ^f, which implies that $\left. \partial l\right\vert _{\eta =0}/\partial \theta ^{f}<0$.

• 2. Equation 17 implies that the sign of $\left. \partial l\right\vert _{\eta =0}/\partial \tau ^{f}$ is equivalent to the sign of $\partial (\tau ^{w}+\tau ^{f})G/\partial \tau ^{f}$. In order to determine the sign of $\partial (\tau ^{w}+\tau ^{f})G/\partial \tau ^{f}$, we rewrite (τ ^w + τ ^f)G as:

  $$ (\tau ^{w}+\tau ^{f})G=\frac{\frac{1}{1+\phi (R)R}\cdot \left\{ (1-\tau ^{w}-\theta ^{w})+(\theta ^{w}+\theta ^{f})\right\} (1-\alpha )A}{\frac{ 1+\tau ^{f}+\theta ^{f}}{\tau ^{w}+\tau ^{f}}+\frac{\phi (R)}{1+\phi (R)R} (1-\alpha )A}. $$

  This equation implies $\partial (\tau ^{w}+\tau ^{f})G/\partial \tau ^{f}>0$ because $\partial \{ (1+\tau ^{f}+\theta ^{f})/(\tau ^{w}+\tau ^{f})\} /\partial \tau ^{f}=\{\left( \tau ^{w}-(1+\theta ^{f})\right) \}/(\tau ^{w}+\tau ^{f})^{2}<0$ holds.

• 3. Equation 17 implies that the sign of $\left. \partial l\right\vert _{\eta =0}/\partial \tau ^{w}$ is equivalent to the sign of $\partial \{(1-\alpha )(1-\tau ^{w}-\theta ^{w})+(\tau ^{w}+\tau ^{f})G/R\}/\partial \tau ^{w}.$ We derive the sufficient condition for $\partial \{(1-\alpha )(1-\tau ^{w}-\theta ^{w})+(\tau ^{w}+\tau ^{f})G/R\}/\partial \tau ^{w}<0$.

The differentiation leads to:

$$\begin{array}{lll} &&\frac{\partial \left[ (1-\alpha )\left(1-\tau ^{w}-\theta ^{w}\right)+\left(\tau ^{w}+\tau ^{f}\right)G/R\right]}{\partial \tau ^{w}} \\ &&{\kern1pc} =-(1-\alpha )+\frac{G}{R}+\frac{\left(\tau ^{w}+\tau ^{f}\right)}{R}\times \frac{\partial G}{\partial \tau ^{w}}, \label{a4} \end{array} $$

(19)

where $\partial G/\partial \tau ^{w}$ is given by:

$$ \frac{\partial G}{\partial \tau ^{w}}=(-1)G\frac{1+\frac{\phi (R)}{1+\phi (R)R}\cdot \frac{(1-\alpha )A}{1+\tau ^{f}+\theta ^{f}}\left(1+\tau ^{f}+\theta ^{f}\right)}{(1-\tau ^{w}+\theta ^{f})\left\{ 1+\frac{\phi (R)}{1+\phi (R)R}\left(\tau ^{w}+\tau ^{f}\right)\frac{(1-\alpha )A}{1+\tau ^{f}+\theta ^{f}}\right\} }. \label{a4b} $$

(20)

Equations 19 and 20 imply that $\left. \partial l\right\vert _{\eta =0}/\partial \tau ^{w}<0$ is equivalent to:

$$ \begin{array}{lll} &&\frac{1}{\alpha }\cdot \frac{1}{1+\phi (R)R}\left[ \left(1-\tau ^{w}+\theta ^{f}\right)-\frac{1+\frac{\phi (R)}{1+\phi (R)R}(1-\alpha )A}{\frac{1}{\tau ^{w}+\tau ^{f}}+\frac{\phi (R)}{1+\phi (R)R}\cdot \frac{(1-\alpha )A}{1+\tau ^{f}+\theta ^{f}}}\right] \\ &&{\kern1pc} <(1+\tau ^{f}+\theta ^{f})+\frac{\phi (R)}{1+\phi (R)R}\left(\tau ^{w}+\tau ^{f}\right)(1-\alpha )A. \label{a9} \end{array} $$

(21)

The left-hand side of Eq. 21, LHS, is decreasing in τ ^f whereas the right-hand side of Eq. 21, RHS, is increasing in τ ^f. Thus, Eq. 21 holds if $\left. \text{LHS}\right\vert _{\tau ^{f}=0}<\left. \text{RHS}\right\vert _{\tau ^{f}=0},$ that is, if:

$$\begin{array}{lll} &&\frac{1}{\alpha }\cdot \frac{1}{1+\phi (R)R}\left[ (1-\tau ^{w}+\theta ^{f})-\frac{1+\frac{\phi (R)}{1+\phi (R)R}(1-\alpha )A}{\frac{1}{\tau ^{w}}+ \frac{\phi (R)}{1+\phi (R)R}\cdot \frac{(1-\alpha )A}{1+\theta ^{f}}}\right] \\ &&{\kern1pc} <\left(1+\theta ^{f}\right)+\frac{\phi (R)}{1+\phi (R)R}\tau ^{w}(1-\alpha )A. \label{a10} \end{array} $$

(22)

Multiplying both sides of Eq. 22 by 1/τ ^w + {ϕ(R)/(1 + ϕ(R)R)}·{(1 − α)A/(1 + θ ^f)} and rearranging the terms, we obtain:

$$\begin{array}{lll} &&\left\{ \frac{1}{\tau ^{w}}+\frac{\phi (R)}{1+\phi (R)R}\cdot \frac{ (1-\alpha )A}{1+\theta ^{f}}\right\} \\ &&{\kern1pc} \times\left\{ \frac{1-\tau ^{w}+\theta ^{f}}{ \alpha (1+\phi (R)R)}-\left(1+\theta ^{f}\right)-\frac{\phi (R)}{1+\phi (R)R}\tau ^{w}(1-\alpha )A\right\} \\ &&{\kern1pc} <\frac{1}{\alpha (1+\phi (R)R)}\left\{ 1+\frac{\phi (R)}{1+\phi (R)R} (1-\alpha )A\right\} . \label{a12} \end{array} $$

(23)

Thus, $\left. \partial l\right\vert _{\eta =0}/\partial \tau ^{w}<0$ holds if Eq. 23 holds.

The remaining task is to derive the sufficient condition for which Eq. 23 holds. Because the right-hand side of Eq. 23 is positive, Eq. 23 holds if the left-hand side of Eq. 23 is nonpositive; that is, if:

$$ \frac{1+\theta ^{f}}{\alpha }\left\{1-\alpha (1+\phi (R)R)\right\}\leq \tau ^{w}\left[ \frac{1}{\alpha }+\phi (R)(1-\alpha )A\right] . \label{a13} $$

(24)

Condition (24) holds if:

$$ \left\{ \begin{array}{c} \alpha (1+\phi (R)R)\geq 1,\text{ or} \\ \alpha (1+\phi (R)R)<1\text{ and }\frac{(1+\theta ^{f})\left\{1-\alpha (1+\phi (R)R)\right\}}{1+\alpha \phi (R)(1-\alpha )A}\leq \tau ^{w}\text{.} \end{array} \right. \label{a14} $$

(25)

Thus, Eq. 25 is the sufficient condition for $\left. \partial l\right\vert _{\eta =0}/\partial \tau ^{w}<0$.

1.5 Appendix E: Proof of Proposition 5

1.5.1 Derivation of the net replacement rate

Suppose that the net replacement rate is fixed at an exogenously given level for all t ≥ T. In the case of η = 1, the net replacement rate, $x_{t}\equiv b_{\!t}/(1-\tau ^{w}-\theta ^{w})w_{t},$ is rewritten as:

$$ x_{t}=\frac{\left(\tau ^{w}+\tau ^{f}\right)w_{t}}{\left(1-\tau ^{w}-\theta ^{w}\right)w_{t}}l_{t}= \frac{(\tau ^{w}+\tau ^{f})(1-\alpha )}{(\theta ^{w}+\theta ^{f})+(1-\alpha )(1-\tau ^{w}-\theta ^{w})}\equiv \bar{x}^{1}, \label{a15} $$

(26)

where the first equality is derived from the government budget constraint Eq. 7 and the second equality is derived from the equilibrium employment rate Eq. 12. Under the constraint of $x_{t}=\bar{x}^{1}$ for all t ≥ T, Eq. 26 leads to the following marginal relationship between τ ^w and τ ^f:

$$ \frac{d\tau ^{f}}{d\tau ^{w}}=(-1)\left(1+\bar{x}^{1}\right). \label{a16} $$

(27)

In the case of η = 0, the government budget constraint Eq. 7 is rewritten as $b_{\!t}=(\tau ^{w}+\tau ^{f})w_{t}l_{t}/l_{t-1}$. We substitute this into the definition of the net replacement rate to obtain:

$$ x_{t}=\left\{ \begin{array}{c} \frac{(\tau ^{w}+\tau ^{f})l_{T}}{(1-\tau ^{w}-\theta ^{w})l_{T-1}}\text{ if }t=T \\[8pt] \bar{x}^{0}\equiv \frac{(\tau ^{w}+\tau ^{f})}{(1-\tau ^{w}-\theta ^{w})} \text{ if }t\geq T+1. \end{array} \right. \label{a18} $$

(28)

Note that l _T  ≠ l _T − 1 because l _T − 1 is not affected by the changes in the contribution rates in period T. We hereafter focus on the net replacement rate in period t ≥ T + 1.

Under the constraint of $x_{t}=\bar{x}^{0}$ for all t ≥ T + 1, Eq. 28 leads to the following marginal relationship between τ ^w and τ ^f:

$$ \frac{d\tau ^{f}}{d\tau ^{w}}=(-1)\left(1+\bar{x}^{0}\right). \label{a19} $$

(29)

1.5.2 Growth effect for the case of η = 1

The differentiation of G with respect to τ ^w leads to:

$$ \begin{array}{lll} \left. \frac{dG}{d\tau ^{w}}\right\vert _{x=\bar{x}} &=&\frac{(1-\alpha )A}{ \left\{ \left(1+\phi (R)R\right)\left(1+\tau ^{f}+\theta ^{f}\right)+\phi (R)\left(\tau ^{w}+\tau ^{f}\right)(1-\alpha )A\right\} ^{2}} \\[6pt] &&\times \left[ (-1)\left\{ \left(1+\phi (R)R\right)\left(1+\tau ^{f}+\theta ^{f}\right)+\phi (R)\left(\tau ^{w}+\tau ^{f}\right)(1-\alpha )A\right\} \right. \\[6pt] &&{\kern10pt} +\,(-1)\left(1-\tau ^{w}+\theta ^{f}\right) \\[6pt] &&{\kern10pt} \times\left.\left\{ (1+\phi (R)R)d\tau ^{f}/d\tau ^{w} +\,\phi (R)\left( 1+d\tau ^{f}/d\tau ^{w}\right) (1-\alpha )A\right\} \right] . \end{array} $$

By using Eqs. 26 and 27, we can rewrite the above equation as follows:

$$\begin{array}{lll} \left. \frac{dG}{d\tau ^{w}}\right\vert _{x=\bar{x}} &=&\frac{(-1)(1-\alpha )A}{\left\{ \left(1+\phi (R)R\right)\left(1+\tau ^{f}+\theta ^{f}\right)+\phi (R)\left(\tau ^{w}+\tau ^{f}\right)(1-\alpha )A\right\} ^{2}} \\[6pt] &&\times \left[ (1+\phi (R)R)\underset{(a3)}{\underbrace{\left\{ \left(1+\tau ^{f}+\theta ^{f}\right)-\left(1-\tau ^{w}+\theta ^{f}\right)\left(1+\bar{x}^{1}\right)\right\} }}\right. \\[3pt] &&\left. +\,\phi (R)(1-\alpha )A\underset{(a4)}{\underbrace{\left\{ \left(\tau ^{w}+\tau ^{f}\right)-\left(1-\tau ^{w}+\theta ^{f}\right)\bar{x}^{1}\right\} }}\right] . \label{a17} \end{array} $$

(30)

The terms (a3) and (a4) in Eq. 30 are rewritten as $(a3)=(a4)=\alpha (\theta ^{w}+\theta ^{f})\bar{x}^{1}/(1-\alpha ).$ Therefore, Eq. 30 is reduced to:

$$ \left. \frac{dG}{d\tau ^{w}}\right\vert _{x=\bar{x}}\!=\!\frac{(-1)(1-\alpha )A\big\{(1+\phi (R)R)+\phi (R)(1-\alpha )A\big\}\frac{\alpha }{1-\alpha }\left(\theta ^{w}+\theta ^{f}\right)\bar{x}^{1}}{\left\{ \left(1+\phi (R)R\right)\left(1+\tau ^{f}+\theta ^{f}\right)+\phi (R)\left(\tau ^{w}+\tau ^{f}\right)\left(1-\alpha \right)A\right\} ^{2}}\!<\!0. $$

1.5.3 Growth effect for the case of η = 0

Given the constraint Eq. 29, the differentiation of G with respect to τ ^w leads to:

$$ \begin{array}{lll} \left. \frac{dG}{d\tau ^{w}}\right\vert _{x=\bar{x}} &=&\frac{(-1)(1-\alpha )A}{\left\{ \left(1+\phi (R)R\right)\left(1+\tau ^{f}+\theta ^{f}\right)+\phi (R)\left(\tau ^{w}+\tau ^{f}\right)\left(1-\alpha \right)A\right\} ^{2}} \\ &&\times \left[ \left(1+\phi (R)R\right)\underset{(a5)}{\underbrace{\left\{ \left(1+\tau ^{f}+\theta ^{f}\right)-\left(1-\tau ^{w}+\theta ^{f}\right)\left(1+\bar{x}^{0}\right)\right\} }}\right. \\ &&\left. +\,\phi (R)(1-\alpha )A\underset{(a6)}{\underbrace{\left\{ \left(\tau ^{w}+\tau ^{f}\right)-\left(1-\tau ^{w}+\theta ^{f}\right)\bar{x}^{0}\right\} }}\right] . \label{a25} \end{array} $$

(31)

By using Eq. 28, we can rewrite the terms (a5) and (a6) in Eq. 31 as $(a5)=(a6)=(-1)(\theta ^{w}+\theta ^{f})\bar{x}^{0}$. Therefore, Eq. 30 is rewritten as:

$$ \left. \frac{dG}{d\tau ^{w}}\right\vert _{x=\bar{x}}=\frac{\left(1-\alpha\right )A\left(\theta ^{w}+\theta ^{f}\right)\bar{x}^{0}\big\{\left(1+\phi (R)R\right)+\phi (R)(1-\alpha )A\big\} }{\left\{ \left(1+\phi (R)R\right)\left(1+\tau ^{f}+\theta ^{f}\right)+\phi (R)\left(\tau ^{w}+\tau ^{f}\right)\left(1-\alpha \right)A\right\} ^{2}}>0. \label{a20} $$

(32)

1.6 Appendix F: Proof of Proposition 6

1.6.1 Employment effect for the case of η = 1

Equation 12 implies that $\left. l\right\vert _{\eta =1}$ is decreasing in τ ^w and is independent of τ ^f. Thus, we obtain $\left. \left. dl\right\vert _{\eta =1}/d\tau ^{w}\right\vert _{x=\bar{x}}<0$.

1.6.2 Employment effect for the case of η = 0

The equilibrium employment rate is given by:

$$ \begin{array}{lll} \left. l_{t}\right\vert _{\eta =0} &=&\left[ \frac{\theta ^{w}+\theta ^{f}}{ (1-\alpha )(1-\tau ^{w}-\theta ^{w})+\frac{(\tau ^{w}+\tau ^{f})G}{R}}+1 \right] ^{-1} \\ &=&\left[ \frac{\theta ^{w}+\theta ^{f}}{(1-\tau ^{w}-\theta ^{w})\left\{ (1-\alpha )+\bar{x}^{0}G/R\right\} }+1\right] ^{-1}, \end{array} $$

where the second equality is derived by using Eq. 28. This equation implies that the sign of $\left. \left. dl\right\vert _{\eta =0}/d\tau ^{w}\right\vert _{x=\bar{x}}<0$ is equivalent to the sign of $d(1\!-\!\tau ^{w}\!-\!\theta ^{w})\big\{ (1\!-\!\alpha )\!+\!\bar{x}^{0}G/R\big\} /d\tau ^{w}\big\vert _{x=\bar{x}^{0}}$.

Given the constraint Eq. 29, the differentiation of $(1-\tau ^{w}-\theta ^{w})\left\{ (1-\alpha )+\bar{x}^{0}G/R\right\}$ with respect to τ ^w leads to:

$$ \begin{array}{lll} &&{\kern-6pt} \left. \frac{d(1-\tau ^{w}-\theta ^{w})\left\{ (1-\alpha )+xG/R\right\} }{ d\tau ^{w}}\right\vert _{x=\bar{x}^{0}} \\ &&{\kern3pt} =(-1)\left\{ (1-\alpha )+\bar{x}^{0}\frac{G}{R}\right\} +(1-\tau ^{w}-\theta ^{w})\frac{\bar{x}^{0}}{R} \\ &&{\kern16pt}\times \frac{(1-\alpha )A(\theta ^{w}+\theta ^{f})\bar{x}^{0}\{(1+\phi (R)R)+\phi (R)(1-\alpha )A\}}{\left\{ (1+\phi (R)R)(1+\tau ^{f}+\theta ^{f})+\phi (R)(\tau ^{w}+\tau ^{f})(1-\alpha )A\right\} ^{2}} \\ &&{\kern3pt} =(-1)(1-\alpha )-\bar{x}^{0}\frac{G}{R} \\ &&{\kern16pt} \times \left[ 1-\tfrac{(\theta ^{w}+\theta ^{f})(\tau ^{w}+\tau ^{f})\{(1+\phi (R)R)+\phi (R)(1-\alpha )A\}}{\left\{ (1+\phi (R)R)(1+\tau ^{f}+\theta ^{f})+\phi (R)(\tau ^{w}+\tau ^{f})(1-\alpha )A\right\} (1-\tau ^{w}+\theta ^{f})}\right] \\ &&{\kern3pt} =(-1)(1-\alpha )-\bar{x}^{0}\frac{G}{R} \\ &&{\kern16pt} \times \tfrac{\left\{ (\theta ^{w}+\theta ^{f})(1-\tau ^{w}+\theta ^{f})+(1+\tau ^{f}+\theta ^{f})(1-\tau ^{w}-\theta ^{w})\right\} (1+\phi (R)R)+(\tau ^{w}+\tau ^{f})(1-\tau ^{w}-\theta ^{w})\phi (R)(1-\alpha )A\}}{ \left\{ (1+\phi (R)R)(1+\tau ^{f}+\theta ^{f})+\phi (R)(\tau ^{w}+\tau ^{f})(1-\alpha )A\right\} (1-\tau ^{w}+\theta ^{f})} \\ &&{\kern3pt} <0, \end{array} $$

where the first equality is derived by the substitution of Eq. 32 and the second equality is derived by using Eqs. 10 and 11.