Public Debt Reduction in Advanced Countries and Its Impacts on Emerging Countries

Abstract

Financial crises associated with banking crises leave heavy fiscal legacies. For the USA e.g. an increase in the gross government debt to GDP ratio towards more than 100 % is predicted by 2012. Due to its unsustainability a significant reduction of public debt in the USA and in other advanced countries seems to be indispensable. However, as shown in this paper, the long run welfare effects of debt reduction in advanced countries at home as well as on emerging countries are not in accordance with debt reduction necessities. In particular, we show that domestic and foreign welfare decrease when the domestic country (USA) reduces public debt, given that this country has a negative external balance and a lower capital production share than the other country (China) and that dynamic inefficiency holds.

This chapter draws heavily on Bednar-Friedl and Farmer (Public debt, terms of trade and welfare: The role or capital production shares, external balances and dynamic (in)efficiency. International Economic Journal, 26(2), 317–349) as well as on Farmer and Schelnast (Public debt reduction in advanced countries and its impact on emerging countries. Paper presented at the Annual Conference of the International Atlantic Economic Society, Istanbul).

Mathematical Appendix

1.1 Proof of Proposition 15.1

It is easy to see that a solution of Eqs. 15.7, 15.8 and 15.9 is equivalent to solving the equations:

$$ k=F\left( {k;b,{b^{*}}} \right)\equiv \left\{ {\frac{{\left[ {\zeta \left( {1-\alpha } \right)\sigma +\left( {1-\zeta } \right)\left( {1-{\alpha^{*}}} \right)\sigma\eta (k)} \right]}}{{\left[ {\zeta +\left( {1-\zeta } \right)\left( {{{{{\alpha^{*}}}} \left/ {\alpha } \right.}} \right)\eta (k)} \right]}}} \right\}\frac{qk }{{\alpha {G^A}}}- $$

$$ -\left\{ {\frac{{\left[ {\zeta b\sigma +\left( {1-\zeta } \right){b^{*}}\sigma \left( {{k \left/ {{{k^{*}}}} \right.}} \right)\eta (k)} \right]}}{{\left[ {\zeta +\left( {1-\zeta } \right)\left( {{{{{\alpha^{*}}}} \left/ {\alpha } \right.}} \right)\eta (k)} \right]}}} \right\}\frac{q}{{{G^A}}}-\frac{{\left[ {\zeta b\left( {1-\sigma } \right)+\left( {1-\zeta } \right){b^{*}}\left( {1-\sigma } \right)\left( {{k \left/ {{{k^{*}}}} \right.}} \right)\eta (k)} \right]}}{{\left[ {\zeta +\left( {1-\zeta } \right)\left( {{{{{\alpha^{*}}}} \left/ {\alpha } \right.}} \right)\eta (k)} \right]}}, $$

$$ \eta (k)\equiv \frac{{q-\alpha {G^A}}}{{q-{\alpha^{*}}{G^A}}},k\in \left( {0,\tilde{k}} \right), $$

$$ \tilde{k}\equiv {{\left[ {\left( {\frac{\alpha }{{{\alpha^{*}}}}} \right)\left( {\frac{M}{{{G^A}}}} \right)} \right]}^{{\frac{1}{{1-\alpha }}}}}, $$

$$ e=\left[ {\frac{{\left( {1-\zeta } \right)}}{\zeta}\frac{k}{{{k^{*}}}}} \right]\left[ {\frac{\alpha }{{{\alpha^{*}}}}\frac{{\left( {q-\alpha {G^A}} \right)}}{{\left( {q-{\alpha^{*}}{G^A}} \right)}}} \right], $$

$$ {k^{*}}={{\left[ {\left( {\frac{{{\alpha^{*}}{M^{*}}}}{{\alpha M}}} \right)} \right]}^{{\frac{1}{{1-\alpha *}}}}}{k^{{\frac{{1-\alpha }}{{1-\alpha *}}}}}. $$

To prove proposition 15.1, we start with the special case of $b={b^{*}}=0$, where $F(k;0,0)$ = $[\zeta (1-\alpha )\sigma +(1-\zeta )(1-{\alpha^{*}}){\sigma^{*}}\eta (k)]/[\zeta +(1-\zeta )({\alpha^{*}}/\alpha )\eta (k)][qk/(\alpha {G^A})]$ holds. Note first that ${\lim_{{k\to 0}}}F(k)=0$ and ${\lim_{{k\to 0}}}{F_k}(k)=\infty$ since ${\lim_{{k\to 0}}}\eta (k)=1$.

To assess the existence of intersection points of $F(k)$ and the 45 ° line for $k\in (0,\tilde{k})$ in a ($k,F(k)$)-diagram (see Fig. 15.1), the cases of $\alpha <{\alpha^{*}}$ (case 1) and of $\alpha >{\alpha^{*}}$ (case 2) are to be distinguished.

In case 1 ${\lim_{{k\to \tilde{k}}}}\eta (k)=\infty$, hence ${\lim_{{k\to \tilde{k}}}}F(k)={\lim_{{k\to \tilde{k}}}}\{[\zeta (1-\alpha )\sigma ]/\eta (k)+$ $+(1-\zeta )(1-{\alpha^{*}}){\sigma^{*}}\}$/$[\zeta /\eta (k)+(1-\zeta )({\alpha^{*}}/\alpha )](M/{G^A}){{\tilde{k}}^{\alpha }}{k^{\alpha }}=(\alpha /{\alpha^{*}})(1-{\alpha^{*}}){\sigma^{*}}$ $(M/{G^A}){{\tilde{k}}^{\alpha }}=(1-{\alpha^{*}}){\sigma^{*}}\tilde{k}$. Since $F(k)$ is continuous on the interval ($0,\tilde{k}$), the slope of $F(k)$ at the origin is certainly larger than one (= slope of 45 ° line) and the value of $F(k)$ at $k=\tilde{k}$ (= lim $_{{k\to \tilde{k}}}$ $F(k)=(1-{\alpha^{*}}){\sigma^{*}}\tilde{k}$) is less than the value of the 45 ° line, an intermediate value theorem ensures the existence of at least one $k\in (0,\tilde{k})$ such that $k=F(k)$. The solution of this equation is unique, which can be seen by setting ${k^{{1-\alpha }}}=z$ and calculating the negative root of the polynomial $({\alpha^{*}}/\alpha ){{({G^A}/M)}^2}{z^2}-$ $\{[\zeta +(1-\zeta )({\alpha^{*}}/\alpha )]+[\zeta (1-\alpha )\sigma ({\alpha^{*}}/\alpha )+$ $(1-\zeta )(1-{\alpha^{*}}){\sigma^{*}}]\}$ $({G^A}/M)z+\zeta (1-\alpha )\sigma$ $+(1-\zeta )(1-{\alpha^{*}}){\sigma^{*}}=0.$ ¹¹ Moreover, it is immediate that $k=0$ is the second steady-state solution.

In case 2 ${\lim_{{k\to \tilde{k}}}}\eta (k)=-\infty$. Thus, there exists a $\hat{k}<\tilde{k}$ such that $\eta (\hat{k})=0\Leftrightarrow \hat{k}={{(M/{G^A})}^{{1/(1-\alpha )}}}$. As above for case 1, it is easy to see that lim $_{{k\to \tilde{k}}}$ $F(k)=(1-\alpha )\sigma \hat{k}<\hat{k}$. Substituting $\hat{k}$ for $\tilde{k}$ the argumentation above can be reiterated, which demonstrates the existence of a $k\in (0,\hat{k})$ such that $k=F(k)$. Clearly, this solution is also unique and there is also the second steady state $k=0$.

Next, consider the general $F(k;b,{b^{*}})$ function, which is governed also by the policy parameters $b\in (0,\bar{b})$ and ${b^{*}}\in (0,{{\bar{b}}^{*}})$. In contrast to the special case above, now ${\lim_{{k\to 0}}}F(k;b,{b^{*}})=-\infty$, while as above ${\lim_{{k\to 0}}}{F_k}(k;b,{b^{*}})=\infty$. As above, we have to distinguish case 1 ($\alpha <{\alpha^{*}}$) and case 2 ($\alpha >{\alpha^{*}}$) (for $\alpha ={\alpha^{*}}$ see Farmer 2010, 37).

In case 1 holds lim $_{{k\to \tilde{k}}}$ $F(k;b,{b^{*}})=(1-{\alpha^{*}}){\sigma^{*}}\tilde{k}-[1-{\sigma^{*}}(1-{\alpha^{*}})]{b^{*}}$ $\times {{(\alpha M/{\alpha^{*}}{M^{*}})}^{{{1 \left/ {{(1-{\alpha^{*}})}} \right.}}}}{{(\tilde{k})}^{{{{{(\alpha -{\alpha^{*}})}} \left/ {{(1-{\alpha^{*}})}} \right.}}}}$ $<\tilde{k}.$ Note that for special case 1 of $b={b^{*}}=0$there are exactly two steady-state solutions. Moreover, $F(k;b,{b^{*}})$ depends negatively on $b$ and ${b^{*}}$ with ${\lim_{{b\to \infty }}}F(k;b,{b^{*}})=-\infty$ and lim $_{{{b^{*}}\to \infty }}$ $F(k;b,{b^{*}})=-\infty$. Since $F(k;b,{b^{*}})$ depends continuously on $b$ and ${b^{*}}$, we can conclude that for $b\in (0,\bar{b})$ _. and ${b^{*}}\in (0,{{\bar{b}}^{*}})$ there exist exactly two non-trivial steady-state solutions ${k^L}=F({k^L};b,{b^{*}})$ and ${k^H}=F({k^H};b,{b^{*}})$ with $0<{k^L}<{k^H}<\tilde{k}$.

By substituting $\hat{k}$ for $\tilde{k}$ in case 2 we can reiterate the argumentation of case 1 and show that for $b\in (0,\bar{b})$ and ${b^{*}}\in (0,{{\bar{b}}^{*}})$ there also exist exactly two non-trivial steady-state solutions ${k^L}=F({k^L};b,{b^{*}})$ and ${k^H}=F({k^H};b,{b^{*}})$ with $0<{k^L}<{k^H}<\hat{k}$.

1.2 The Jacobian Matrix

The Jacobian matrix of the dynamic system 15.7, 15.8 and 15.9 $J(e,k,{k^{{*}}})$ reads as follows:

$$ J(e,k,{k^{*}})=\left[ {\begin{array}{lllllllll}{{{{\partial {e_{t+1 }}}} \left/ {{\partial {e_t}}} \right.}} & {{{{\partial {e_{t+1 }}}} \left/ {{\partial {k_t}}} \right.}} & {{{{\partial {e_{t+1 }}}} \left/ {{\partial k_t^{*}}} \right.}} \\ {{{{\partial k_{t+1 }}} \left/ {{\partial {e_t}}} \right.}} & {{{{\partial {k_{t+1 }}}} \left/ {{\partial {k_t}}} \right.}} & {{{{\partial {k_{t+1 }}}} \left/ {{\partial k_t^{*}}} \right.}} \\ {{{{\partial k_{t+1}^{*}}} \left/ {{\partial {e_t}}} \right.}} & {{{{\partial k_{t+1}^{*}}} \left/ {{\partial {k_t}}} \right.}} & {{{{\partial k_{t+1}^{*}}} \left/ {{\partial k_t^{*}}} \right.}} \\ \end{array}} \right]\equiv \left[ {\begin{array}{llllllllll}{{j_{11 }}} & {{j_{12 }}} & {{j_{13 }}} \\ {{j_{21 }}} & {{j_{22 }}} & {{j_{23 }}} \\ {{j_{31 }}} & {{j_{32 }}} & {{j_{33 }}} \\ \end{array}} \right], $$

(15.15)

with

$$ {j_{11 }}=1+\left( {1-\alpha } \right)\left\{ {{{{\left[ {\left( {1-\zeta } \right)H-\zeta \Phi } \right]}} \left/ {k} \right.}} \right\}+\left( {1-{\alpha^{*}}} \right)\left\{ {{{{\left[ {\zeta {H^{*}}-\left( {1-\zeta } \right){\Phi^{*}}} \right]}} \left/ {{{k^{*}}}} \right.}} \right\}, $$

$$ \begin{array}{llllll} {j_{12 }}=-\left[ {{{{\left( {1+i} \right)}} \left/ {{{G^A}}} \right.}} \right]\left\{ {\left[ {{{{\left( {1-{\alpha^{*}}} \right)}} \left/ {{{k^{*}}}} \right.}} \right]\left( {1-\zeta } \right)\left[ {1-\left( {1-\alpha } \right)\sigma \left( {1+{b \left/ {k} \right.}} \right)} \right]+} \right. \hfill \\ \qquad+ \left. {\left[ {{{{\left( {1-\alpha } \right)}} \left/ {k} \right.}} \right]e\left[ {\left( {1-\zeta } \right)+\zeta \left( {1-\alpha } \right)\sigma \left( {1+{b \left/ {k} \right.}} \right)} \right]} \right\}, \hfill \\ \end{array} $$

$$ {j_{13 }}= e\left[ {{{{\left( {1+i} \right)}} \left/ {{{G^A}}} \right.}} \right]\left\{ {\left[ {{{{\left( {1-{\alpha^{*}}} \right)}} \left/ {{{k^{*}}}} \right.}} \right]\left[ {\zeta +\left( {1-\zeta } \right)\left( {1-{\alpha^{*}}} \right){\sigma^{*}}\left( {1+{{{{b^{*}}}} \left/ {{{k^{*}}}} \right.}} \right)} \right]+} \right. \hfill \\+ \left. {\left[ {{{{\left( {1-\alpha } \right)}} \left/ {k} \right.}} \right]e\zeta \left[ {1-\left( {1-{\alpha^{*}}} \right){\sigma^{*}}\left( {1+{{{{{{{b^{*}}}} \left/ {k} \right.}}}^{*}}} \right)} \right]} \right\}, \hfill<!endaligned> $$

$$ {j_{21 }}={e^{-1 }}\left[ {\zeta \Phi -\left( {1-\zeta } \right)H} \right],{j_{22 }}=\left[ {{{{\left( {1+i} \right)}} \left/ {{{G^A}}} \right.}} \right]\left[ {1-\zeta +\zeta \sigma \left( {1-\alpha } \right)\left( {1+{b \left/ {k} \right.}} \right)} \right], $$

$$ {j_{23 }}=-\zeta e\left[ {{{{\left( {1+i} \right)}} \left/ {{{G^A}}} \right.}} \right]\left[ {1-\left( {1-{\alpha^{*}}} \right){\sigma^{*}}\left( {1+{{{{b^{*}}}} \left/ {{{k^{*}}}} \right.}} \right)} \right], $$

$$ {j_{31 }} ={{{\left[ {\zeta {H^{*}}-\left( {1-\zeta } \right){\Phi^{*}}} \right]}} \left/ {e} \right.},{j_{32 }} =-\left[ {{{{\left( {1-\zeta } \right)}} \left/ {e} \right.}} \right]\left[ {{{{\left( {1+i} \right)}} \left/ {{{G^A}}} \right.}} \right]\left[ {1-\left( {1-\alpha } \right)\sigma \left( {1+{b \left/ {k} \right.}} \right)} \right], $$

$$ {j_{33 }}=\left[ {{{{\left( {1+i} \right)}} \left/ {{{G^A}}} \right.}} \right]\left[ {\zeta +\left( {1-\zeta } \right)\left( {1-{\alpha^{*}}} \right){\sigma^{*}}\left( {1+{{{{b^{*}}}} \left/ {{{k^{*}}}} \right.}} \right)} \right],\ \mathrm{ whereby} $$

$$ H\equiv \left( {{M \left/ {{{G^A}}} \right.}} \right){k^{\alpha }}-k=\left[ {{{{\left( {1+i} \right)}} \left/ {{\left( {{G^A}\alpha } \right)}} \right.}-1} \right]k\;\mathrm{ and} $$

$$ \Phi \equiv k+b\left\{ {\sigma \left[ {{{{\left( {1+i} \right)}} \left/ {{{G^A}-1}} \right.}} \right]+1} \right\}-{\sigma_0}{k^{\alpha }}. $$

The elements of the Jacobian matrix (15.15) can be analogously obtained as in the appendix of Chap. 14.

1.3 Proof of Proposition 15.2

To prove the saddle-path stability of the steady state with larger capital intensities we need the eigenvalues of the Jacobian in the neighborhood of the steady-state solution. The calculation of the eigenvalues commences with the calculation of the determinant of the Jacobian (15.15). To calculate this determinant, denoted detJ, multiply the second row of Eq. 15.15 by $e{{{(1-\alpha )}} \left/ {k} \right.}$ and the third row by $-e{{{(1-{\alpha^{*}})}} \left/ {{{k^{*}}}} \right.}$ and add the results to the first row, and you will obtain:

$$ J=\left( {\begin{array}{*{20}{c}} 1 & 0 & 0 \\ {\frac{{\zeta \boldsymbol{\Phi} -\left( {1-\zeta } \right)H}}{e}} & {\frac{{q\left[ {1-\zeta +\zeta \sigma \left( {1-\alpha } \right)\left( {1+\frac{b}{k}} \right)} \right]}}{{{G^A}}}} & {\frac{{-\zeta qe\left[ {1-{\sigma^{*}}\left( {1-{\alpha^{*}}} \right)\left( {1+\frac{{{b^{*}}}}{{{k^{*}}}}} \right)} \right]}}{{{G^A}}}} \\ {\frac{{\zeta {H^{*}}-\left( {1-\zeta } \right){\boldsymbol{\Phi}^{*}}}}{e}} & {\frac{{\left( {\zeta -1} \right)q\,\left[ {1-\sigma \left( {1-\alpha } \right)\left( {1+\frac{b}{k}} \right)} \right]}}{{e{G^A}}}} & {\frac{{q\left[ {\zeta +\left( {1-\zeta } \right){\sigma^{*}}\left( {1-{\alpha^{*}}} \right)\left( {1+\frac{{{b^{*}}}}{{{k^{*}}}}} \right)} \right]}}{{{G^A}}}} \\ \end{array}} \right). $$

By applying the method of cross-diagonal multiplication we obtain det J= ${(1+i)^2}[\zeta (1-\alpha )\sigma ()+(1-\zeta )(1-{\alpha^{*}}){\sigma^{*}}(1+{b^{*}}/{k^{*}})]/{{({G^A})}^2}$ , acknowledging that $q=1+i$ .

In order to calculate the eigenvalues of $J(e,k,{k^{*}})$, we first consider the case of $b={b^{*}}=0$. Since $J(e,k,{k^{*}})$ is a square matrix, we use the theorem which asserts that the product of the eigenvalues equals its determinant and the sum of its eigenvalues is equal to its trace (see for example Azariadis 1993, 124). Denoting the eigenvalues of $J(e,k,{k^{*}})$ by ${\lambda_i},i=1,2,3$ and the trace by $\mathrm{ tr}J$, the theorem reads as follows:

$$ {\lambda_1}{\lambda_2}{\lambda_3}=\det J, $$

(15.16)

$$ {\lambda_1}+{\lambda_2}+{\lambda_3}=\mathrm{ tr}J. $$

(15.17)

Next, from extensive numerical calculations for all admissible parameter values we guess that ${\lambda_2}=\alpha$ and ${\lambda_3}={\alpha^{*}}$. Inserting these guess values into Eq. 15.16 and solving for ${\lambda_1}$, we obtain:

$$ {\lambda_1}=\left\{ {{{{{{{\left( {1+i} \right)}}^2}}} \left/ {{\left[ {{{{\left( {{G^A}} \right)}}^2}\alpha{\alpha^{*}}} \right]}} \right.}} \right\}\left[ {\zeta \left( {1-\alpha } \right)\sigma +\left( {1-\zeta } \right)\left( {1-{\alpha^{*}}} \right){\sigma^{*}}} \right]. $$

(15.18)

To verify these guess solutions for the eigenvalues we show that by setting ${\lambda_1}=\mathrm{ tr}J-\alpha -{\alpha^{*}}$ (using Eq. 15.17) equal to ${\lambda_1}$ in Eq. 15.18 we obtain an identity. Acknowledging the Jacobian (15.15) under $b={b^{*}}=0$, it is immediate that $\mathrm{tr}J=1+(1-\alpha ){{{[(1-\zeta )H-\zeta \varPhi ]}} \left/ {k} \right.}+(1-{\alpha^{*}}){{{[\zeta {H^{*}}-(1-\zeta ){\varPhi^{*}}]}} \left/ {{{k^{*}}}} \right.}+[{(1+i) \left/ {{{G^A}}} \right.}]\times$ $\times [1+\zeta \sigma \left( {1-\alpha } \right)+(1-\zeta ){\sigma^{*}}(1-{\alpha^{*}})]$. Using the determinant of $J(e,k,{k^{*}})$, the trace of $J(e,k,{k^{*}})$ can be equivalently written as follows:

$$ \mathrm{tr}J=1+(q/{G^A})+({G^A}/q)\det\,J+\varDelta, $$

(15.19)

whereby $\Delta \equiv (1-\alpha )[(1-\zeta )H-\zeta \Phi ]/k + (1-{\alpha^{*}})[\zeta {H^{*}}-\left( {1-\zeta } \right){\Phi^{*}}]/{k^{*}}.$ To proceed note that for $b={b^{*}}=0$,${\Phi \left/ {{k=1-{{{(1-\alpha )\sigma (1+i)}} \left/ {{(\alpha {G^A})}} \right.}}} \right.}$and ${{{{\Phi^{*}}}} \left/ {{{k^{*}}=1-{{{(1-{\alpha^{*}}){\sigma^{*}}(1+i)}} \left/ {{({\alpha^{*}}{G^A})}} \right.}}} \right.}$. Moreover, ${H \left/ {k= } \right.}{{{(1+i-\alpha {G^A})}} \left/ {{(\alpha {G^A})}} \right.}$ and ${{{{H^{*}}}} \left/ {{{k^{*}}=}} \right.}{{{(1+i-{\alpha^{*}}{G^A})}} \left/ {{({\alpha^{*}}{G^A})}} \right.}$ hold true. Inserting these expressions into the definition of $\Delta$ in Eq. 15.19 and the assumption that $\zeta =1-\zeta$ without loss of generality lead to the following equivalent expression for $\varDelta$:

$$ \varDelta =\zeta \{(1+i)\{(1-\alpha ){\alpha^{*}}[1+(1-\alpha )\sigma ]+(1-{\alpha^{*}})\alpha [1+(1-{\alpha^{*}}){\sigma^{*}}]\}- $$

$$ -2{G^A}\alpha {\alpha^{*}}(2-\alpha -{\alpha^{*}})\}/(\alpha{\alpha^{*}}{G^A}). $$

(15.20)

On the other hand, $\mathrm{tr}\,J-\alpha -{\alpha^{*}}={\lambda_1}={{{\det\,J}} \left/ {{(\alpha\,{\alpha^{*}})}} \right.}$ can be equivalently written as follows:

$$ \alpha\,{\alpha^{*}}[1+{(1+i) \left/ {{{G^A}}} \right.}]+\alpha\,{\alpha^{*}}[{{{{G^A}}} \left/ {(1+i) } \right.}]\det\,J+\alpha\,{\alpha^{*}}\varDelta -\alpha\,{\alpha^{*}}(\alpha +{\alpha^{*}})=\det\,J. $$

(15.21)

Inserting Eq. 15.20 and $\det\,J$ into the left-hand side of Eq. 15.21 we get the result after some simple but tedious calculations that the equality in Eq. 15.21 is equivalent to the following equality:

$$ [({{{(1+i)\zeta )}} \left/ {{{G^A}}} \right.}][\alpha (1+{\sigma^{*}})+{\alpha^{*}}(1+\sigma )-\alpha {\alpha^{*}}(\sigma +{\sigma^{*}})]-\alpha {\alpha^{*}}=\det J. $$

(15.22)

Setting Eq. 15.8 equal to 15.9 and acknowledging $\zeta =1-\zeta$ we get:

$$ \frac{{\left( {1+i-{\alpha^{*}}{G^A}} \right)}}{{\left( {1+i-\alpha {G^A}} \right)}}=\frac{{\left[ {\left( {1-{\alpha^{*}}} \right){\sigma^{*}}(1+i)-{\alpha^{*}}{G^A}} \right]}}{{\left( {\alpha {G^A}-\left( {1-\alpha } \right)\sigma (1+i)} \right)}}. $$

(15.23)

Cross multiplication in Eq. 15.23 and collecting terms yield exactly the equality in Eq. 15.22. Since $0<\alpha <1$ and $0<{\alpha^{*}}<1$ by assumption, ${\lambda_1}>1,{\lambda_2}<1,{\lambda_3}<1$ as claimed.

Next, we consider the more general case of $b>0,\;{b^{*}}=0.$ We guess that ${\lambda_3}={\alpha^{*}.}$Insertion into Eqs. 15.16 and 15.17 yields ${\lambda_2}=({1 \left/ {2} \right.})(\mathrm{ tr}J-{\alpha^{*}})$ $-({1 \left/ {2} \right.}){{[{{(\mathrm{ tr}J-{\alpha^{*}})}^2}-4\det J/{\alpha^{*}}]}^{0.5 }}$, ${\lambda_1}=({1 \left/ {2} \right.})(\mathrm{ tr}J-{\alpha^{*}})+({1 \left/ {2} \right.})[{{(\mathrm{ tr}J-{\alpha^{*}})}^2}-$ $-4\det J/{\alpha^{*}}{]^{0.5 }}$. Since ${\alpha^{*}}<1$, ${\alpha^{*}}{{(\mathrm{ tr}J-{\alpha^{*}})}^2}>4\det J$ and $1+{\alpha^{*}}-\mathrm{ tr}\ J+$ $+{{{\det J}} \left/ {{{\alpha^{*}}}} \right.}<0$ ¹² at $({e^H},{k^H},{k^{*,H }})$, ${\lambda_2}<1,{\lambda_1}>1$. For the general case of $b>0,$ ${b^{*}}>0$Descartes Rule of Sign (Sydsaeter et al. 2010, 8) has to be applied on the polynomial: ${\lambda^3}+(3-\mathrm{ tr}J){\lambda^2}+(3+\mathrm{ mi}J-2\mathrm{ tr}J)\lambda +1$ $-\mathrm{ tr}\,\,J+\mathrm{ mi}\ J-\det J=0,$where $\mathrm{mi}\,J={(1+i) \left/ {{{G^A}+(1+{{{{G^A}}} \left/ {{(1+i))\det J}} \right.}}} \right.}+({(1+i) \left/ {{{G^A}}} \right.})\{(1-\alpha )(1-{\alpha^{*}})\times$ $(1-\zeta )\sigma [{(H \left/ {k} \right.})(1+{b \left/ {{k)+{{{({H^{*}}}} \left/ {{{k^{*}}}} \right.})(1+{{{{b^{*}}}} \left/ {{{k^{*}})}} \right.}}} \right.}]-[(1-\alpha )\zeta {{{(\Phi }} \left/ {k} \right.})+(1-{\alpha^{*}})(1-\zeta )$ ${{{({\Phi^{*}}}} \left/ {{{k^{*}}}} \right.})]\}$ denotes the principal minor of $J(e,k,{k^{*}})$ . Since $({e^H},{k^H},{k^{*,H }})$ $3-\mathrm{ tr}J<0,$ $3+\mathrm{ mi}J-2<0,$ $1-\mathrm{ tr}J+\mathrm{ mi}J-\det J<0$,¹³ ${\lambda_1}>1,{\lambda_2}<1,{\lambda_3}<1$ follow.

1.4 Comparative Steady-State Effects of Debt Changes

To derive the steady-state effects of a marginal change of public debt in Home, we differentiate Eqs. 15.7, 15.8 and 15.9 totally with respect to $e,k,{k^{*}}$ and $b$. After inserting the result of Eq. 15.7 into differentiated Eqs. 15.8 and 15.9 this yields:

$$ \left[ {\begin{array}{ccccccc}{{\Phi^{*}}} & {e\frac{{\left( {1-\alpha } \right){k^{*}}}}{{\left( {1-{\alpha^{*}}} \right)k}}\frac{{\partial {\Phi^{*}}}}{{\partial {k^{*}}}}+\frac{{\partial \Phi }}{{\partial k}}} \\ {\zeta {H^{*}}} & {\,\frac{{e\zeta \left( {1-\alpha } \right){k^{*}}}}{{\left( {1-{\alpha^{*}}} \right)k}}\frac{{\partial {H^{*}}}}{{\partial {k^{*}}}}-\left( {1-\zeta } \right)\frac{{\partial H}}{{\partial k}}} \\ \end{array}} \right]\left[ {\begin{array}{ccccccc} {de} \\ {dk} \\ \end{array}} \right]=\left[ {\begin{array}{ccccccc} {\frac{{-\partial \Phi }}{{\partial b}}} \\ 0 \\ \end{array}} \right]db $$

(15.24)

Solving Eq. 15.24 by using Cramer’s rule for ${de \left/ {db } \right.}$ and ${dk \left/ {db } \right.}$, we obtain Eq. 15.10 in the main text.

1.5 Proof of Proposition 15.3

Noting the definition of the slopes of the AA- and CC-curve in the main text, it is not difficult to see that $de/d{k_{|CC }}=[q(\alpha -{\alpha^{*}})-{G^A}\alpha {\alpha^{*}}(\alpha -{\alpha^{*}})]$ ${{{[(1-\zeta )\left( {{{{\partial H}} \left/ {{\partial k}} \right.}} \right){k^{*}}]}} \left/ {{[\zeta {{{({H^{*}})}}^2}{G^A}(1-{\alpha^{*}})\alpha {\alpha^{*}}]}} \right.}$ Acknowledging that ${G^A}=q\Leftrightarrow$ $\partial H/\partial k=0$, it follows that $de/d{k_{|CC }}=0$, while for ${G^A}<q\Leftrightarrow \partial H/\partial k>0$ and hence ${de \left/ {{d{k_{|CC }}}} \right.}>rless 0\Leftrightarrow \alpha >rless {\alpha^{*}}$ and for ${G^A}>q\Leftrightarrow \partial H/\partial k<0$, ${de \left/ {{d{k_{|CC }}}} \right.}\lessgtr 0\Leftrightarrow$ $\alpha >rless {\alpha^{*}}$.

The slope of the AA-curve is given by:

$$ {{\frac{de }{dk}}_{{\left| {AA} \right.}}}=-{{({\Phi^{*}})}^{-1 }}\left\{ {\frac{{(1-\zeta )}}{\zeta}\frac{{{\alpha^{*}}}}{\alpha}\frac{{(1-\alpha )}}{{(1-{\alpha^{*}})}}\frac{{(q-\alpha {G^A})}}{{(q-{\alpha^{*}}{G^A})}}\left[ {1-(1-{\alpha^{*}}){\sigma^{*}}{(q \left/ {{{G^A}}} \right.})(1+{{{{b^{*}}}} \left/ {{{k^{*}}}} \right.})} \right]} \right. $$

$$ \left. {+\left[ {1-(1-\alpha )\sigma {(q \left/ {{{G^A}}} \right.})(1+{b \left/ {k} \right.})} \right]} \right\}. $$

Since by assumption $(1-\alpha )\sigma q/{G^A}(1+{b \left/ {k) } \right.}<1$ and $(1-{\alpha^{*}}){\sigma^{*}}q/$ ${G^A}(1+{{{{b^{*}}}} \left/ {{{k^{*}})}} \right.}<1$ hold, the term in curled brackets is certainly larger than zero. Thus, ${de \left/ {{d{k_{{\left| {AA} \right.}}}}} \right.}>rless 0\Leftrightarrow {\Phi^{*}}\lessgtr 0\Leftrightarrow \Phi >rless 0$.

$|{de \left/ {{d{k_{|AA }}}} \right.}|>|{de \left/ {{d{k_{|CC }}}} \right.}|$ is true in cases (ii) and (iii), since ${de \left/ {{d{k_{{\left| {AA} \right.}}}}} \right.}={{{\{e{{{[(1-\alpha ){k^{*}}]}} \left/ {{[(1-{\alpha^{*}})k]({{{\partial {\Phi^{*}}}} \left/ {{\partial {k^{*}}}} \right.}}} \right.})+{{{\partial \Phi }} \left/ {{\partial k\}}} \right.}}} \left/ {{(-{\Phi^{*}})}} \right.} >$ ${{{\{(1-\zeta ){{{(\partial H}} \left/ {{\partial k)-e\zeta {{{(1-\alpha )}} \left/ {{(1-{\alpha^{*}}){{{({k^{*}}}} \left/ {k} \right.}){{{(\partial {H^{*}}}} \left/ {{\partial {k^{*}})}} \right.}}} \right.}\}}} \right.}}} \left/ {{(\zeta {H^{*}})}} \right.} = ({de \left/ {{d{k_{|CC }}}} \right.})$.

1.6 Derivation of Domestic Welfare Effect of Debt Change

To start with, write the steady-state indirect utility function of the younger household as follows: $V\equiv U({x^1}(w-\tau ),{y^1}(w-\tau, e),{x^2}(w-\tau, q),{y^2}(w-\tau, q,e)),$ $q=1+i$. Total differentiation and acknowledging the FOCs for direct utility maximization (see the main text) yield: $dV={{{\partial U}} \left/ {{\partial {x^1}\{[}} \right.}{{{\partial {x^1}}} \left/ {\partial } \right.}(w-\tau )+{{{\partial {y^1}}} \left/ {\partial } \right.}(w-\tau )+$ ${{{{q^{-1 }}\partial {x^2}}} \left/ {\partial } \right.}(w-\tau )+{{{({e \left/ {q) } \right.}\partial {y^2}}} \left/ {\partial } \right.}(w-\tau )]d(w-\tau )+{q^{-1 }}[{{{\partial {x^2}}} \left/ {{\partial q}} \right.}+e{{{\partial {y^2}}} \left/ {{\partial q]dq}} \right.}$ $+e[{{{\partial {y^1}}} \left/ {{\partial e}} \right.}+{q^{-1 }}{{{\partial {y^2}}} \left/ {{\partial e]de}} \right.}$. Noting the optimal consumption functions from appendix to chapter 14 we obtain:

$$ {{{\partial {x^1}}} \left/ {\partial } \right.}(w-\tau )=\zeta /(1+\beta ),\;{{{\partial {y^1}}} \left/ {\partial } \right.}(w-\tau )=(1-\zeta )/(1+\beta ){e^{-1 }}, $$

$$ {{{\partial {x^2}}} \left/ {\partial } \right.}(w-\tau )=[\beta \zeta /(1+\beta )]q,\;{{{\partial {y^2}}} \left/ {\partial } \right.}(w-\tau )={{{[\beta (1-\zeta )q]}} \left/ {{[(1+\beta )e}} \right.}], $$

$$ {{{\partial {x^2}}} \left/ {\partial } \right.}q=[\beta \zeta /(1+\beta )](w-\tau ),\;{{{\partial {y^2}}} \left/ {\partial } \right.}q=[\beta \zeta /(1+\beta )](w-\tau ){e^{-1 }}, $$

$$ {{{\partial {y^1}}} \left/ {\partial } \right.}e=-(1-\zeta )(w-\tau )/(1+\beta ){e^{-2 }},\;{{{\partial {y^2}}} \left/ {\partial } \right.}e=-{{{[\beta (1-\zeta )q(w-\tau )]}} \left/ {{[(1+\beta ){e^2}}} \right.}]. $$

Inserting these relations into the total indirect utility differential, you will obtain: $dV={{{\partial U}} \left/ {{\partial {x^1}[d(w-\tau )}} \right.}+$ $(s/q)dq-(1-\zeta )(w-\tau ){e^{-1 }}de]$ with $s=\sigma (w-\tau )$. From Eqs. 13.3 and 13.10 follows: $d(w-\tau )=$ $[\partial {{{(w-\tau )}} \left/ {{\partial k]dk}} \right.}+(q-{G^A})db$. Inserting this result into the indirect utility differential yields the total differential of the indirect utility function in the main text.