Real Exchange Rate and Public Debt in a Two-Advanced-Country OLG Model

Abstract

The recent euro crisis has caused concerns both with respect to public-debt sustainability and the stability of exchange rates of highly indebted countries. This chapter investigates these concerns in a two-good, two-country OLG model of the world economy with country-specific saving rates to mimic also Asian economies. We find that the concerns with respect to debt-sustainability are warranted since limits for national debt levels do exist. The concerns regarding exchange-rate stability are not warranted since unilateral debt expansion does not impact the real exchange rate at all or the impacts are independent of the external balance of the debt-expanding country.

This chapter draws heavily on Farmer (International Advances of Economic Research 17, 45–65, 2011) and for the mathematical aspects on Farmer (Public-debt sustainability, exchange rates, and country-specific saving rates. In A. Bayar, C. Erbil & D. Ozdemir (Eds.), Proceedings of Ecomod 2010: International Conference on Economic Modeling, Brussels, 2010).

Mathematical Appendix

1.1 Utility Maximizing Consumption and Saving Functions of Younger Households

In order to show how the equations of motion in world market equilibrium are derived, the optimal consumption and savings levels of households for $t=1,2,\ldots$are needed. We indicate roughly how optimal consumption and savings for households in Home are obtained. First, insert ${s_t}$ into the second budget constraint, while taking the international interest parity condition (14.8) into account. This implies:

$$ x_t^1+{e_t}y_t^1+{{{x_{t+1}^2}} \left/ {{(1+{i_{t+1 }})+{{{({e_{t+1 }}y_{t+1}^2)}} \left/ {{(1+{i_{t+1 }})}} \right.}}} \right.}={w_t}-{\tau_t}. $$

Second, maximize Eq. 14.7 subject to this intertemporal budget constraint and solve for optimal consumption quantities and optimal savings. An analogous procedure gives the optimal consumption quantities and optimal savings in Foreign.

$$ x_t^1=\Big[ {{\zeta \left/ {{\left( {1+\beta } \right)}} \right.}} \Big]\Big[ {w_t-{\tau_t}} \Big] $$

(14.25)

$$ x_t^{*,1 }=\Big[ {{\zeta \left/ {{\left( {1+{\beta^{*}}} \right)}} \right.}} \Big]\Big[ {w_t^{*}-\tau_t^{*}} \Big]{e_t} $$

(14.25*)

$$ y_t^1=\Big[ {{{{\left( {1-\zeta } \right)}} \left/ {{\left( {1+\beta } \right)}} \right.}} \Big]{{\left( {{e_t}} \right)}^{-1 }}\Big[ {w_t-{\tau_t}} \Big] $$

(14.26)

$$ y_t^{*,1 }={{{\left( {1-\zeta } \right)}} \left/ {{\left( {1+{\beta^{*}}} \right)\left( {w_t^{*}-\tau_t^{*}} \right)}} \right.} $$

(14.26*)

$$ x_0^2=\zeta \left( {1+{i_0}} \right){a_0}{G^L}\Big[ {k_0+b-{\Phi_0}} \Big],\;x_{t+1}^2={{{\beta \zeta }} \left/ {{\left( {1+\beta } \right)}} \right.}\left( {1+{i_{t+1 }}} \right)\left( {{w_t}-{\tau_t}} \right) $$

(14.27)

$$ \begin{array}{lllll} x_0^{*,2 }=\zeta {G^L}{a_0}\left[ {\left( {1+i_0^{*}} \right)\left( {k_0^{*}+{b^{*}}} \right)+\left( {1+{i_0}} \right){{{\left( {{e_0}} \right)}}^{-1 }}{\Phi_0}} \right]{e_0},\, \hfill \\ \,x_{t+1}^{*,2 } = {{{\left( {{\beta^{*}}\zeta } \right)}} \left/ {{\left( {1+{\beta^{*}}} \right)}} \right.}\left( {1+i_{t+1}^{*}} \right)\left( {w_t^{*}-\tau_t^{*}} \right){e_{t+1 }}\,\,\,\,\, \hfill \\ \end{array} $$

(14.27*)

$$ \begin{array}{lllllll} y_0^2=\left( {1-\zeta } \right){G^L}{a_0}\left( {1+{i_0}} \right){{\left( {{e_0}} \right)}^{-1 }}\left[ {k_0+b-{\Phi_0}} \right],\,\, \hfill \\ y_{t+1}^2={{{\beta \left( {1-\zeta } \right)}} \left/ {{\left( {1+\beta } \right)}} \right.}\left( {1+{i_{t+1 }}} \right)\left( {{w_t}-{\tau_t}} \right){{\left( {{e_{t+1 }}} \right)}^{-1 }} \hfill \\ \end{array} $$

(14.28)

$$ \begin{array}{lllll} y_0^{*,2 }=\left( {1-\zeta } \right){G^L}{a_0}\left[ {\left( {1+i_0^{*}} \right)\left( {k_0^{*}+{b^{*}}} \right)+{{{\left( {{e_0}} \right)}}^{-1 }}\left( {1+{i_0}} \right){\Phi_0}} \right],\,\, \hfill \\ y_{t+1}^{*,2 }={{{{\beta^{*}}\left( {1-\zeta } \right)}} \left/ {{\left( {1+{\beta^{*}}} \right)}} \right.}\left( {1+i_{t+1}^{*}} \right)\left( {w_t^{*}-\tau_t^{*}} \right) \hfill \\ \end{array} $$

(14.28*)

$$ s_t=\sigma\Big[ {w_t-{\tau_t}} \Big],\sigma \equiv {\beta \left/ {{\left( {1+\beta } \right)}} \right.} $$

(14.29)

$$ s_t^{*}={\sigma^{*}}\Big[ {w_t^{*}-\tau_t^{*}} \Big] $$

(14.29*)

1.2 Proof of Proposition 14.1

To prove proposition 14.1, let’s start with ${F_k}(k)=\alpha (1-\alpha )\xi ({M \left/ {{{G^A}}} \right.}){k^{{\alpha -1}}}+\alpha (1-\alpha )\vartheta ({M \left/ {{{G^A}}} \right.}){k^{{\alpha -2}}}-1$. It is immediate that (i) ${F_k}(k)$ is a continuous and strictly decreasing function, that (ii) ${\lim_{{k\to 0}}}{F_k}(k)=\infty$ and (iii) that ${\lim_{{k\to \infty }}}{F_k}(k)=0$. Hence, an Intermediate Value Theorem guaranties for each $\vartheta$ the existence of a $\kappa$ which solves ${F_k}\left( {\kappa, \vartheta } \right)=1$. Moreover, the solution is unique since ${F_{kk }}\left( {k,\vartheta } \right)<0$. Hence, $\kappa =K\left( \vartheta \right)$. Note also that $K\left( \vartheta \right)$ is a strictly increasing function because ${F_{kk }}\left( {k,\vartheta } \right)<0$ and ${F_{{k\vartheta }}}\left( {k,\vartheta } \right)>0$. To obtain ER2, combine ${F_k}(\kappa, \vartheta )=1$ and $F(\kappa, \vartheta, {b^{*}})=\kappa$. ${F_k}(\kappa, \vartheta )=1\Leftrightarrow \alpha (1-\alpha )({M \left/ {{{G^A}}} \right.})(\xi \kappa +\vartheta ){\kappa^{{\alpha -2}}}=1$. $F(\kappa, \vartheta, {b^{*}})=\kappa \Leftrightarrow \kappa ={\kappa^{{\alpha -2}}}({M \left/ {{{G^A}}} \right.})[\xi (1-\alpha ){\kappa^2}-\alpha \vartheta \kappa ]-[{{{(1-\sigma )}} \left/ {\sigma } \right.}]\vartheta -(1-\zeta ){\mu^{-1 }}{b^{*}}(1-{{{{\sigma^{*}}}} \left/ {\sigma } \right.})$. Eliminating ${\kappa^{{\alpha -2}}}$ from both equations gives: ${{[\alpha (1-\alpha )({M \left/ {{{G^A}}} \right.})(\xi \kappa +\vartheta )]}^{-1 }}=\{\kappa +[{{{(1-\sigma )}} \left/ {\sigma } \right.}]\vartheta +(1-\zeta ){\mu^{-1 }}{b^{*}}(v)\}/$ $({M \left/ {{{G^A}}} \right.})[\xi (1-\alpha ){\kappa^2}-\alpha \vartheta \kappa ]$. Rearranging this equation and solving it for $\kappa$ yields: ${\kappa^2}-2\eta \kappa -\nu =0$ with $\eta$ and $\nu$ as defined in text after lemma 14.1. From this quadratic polynomial immediately follows: ${\kappa_{+}}\equiv \eta +{{({\eta^2}+\nu )}^{0.5 }}$. Reinserting ${\kappa_{+}}$ into $\alpha (1-\alpha )({M \left/ {{{G^A}}} \right.})(\xi \kappa +\vartheta ){\kappa^{{\alpha -2}}}=1$ results in ER2. ER1 follows from visual inspection of Fig. 14.1.▪

1.3 The Jacobian Matrix and Its Derivation

The Jacobian matrix of the dynamical system 14.15, 14.16, and 14.17 $J(e,k,{k^{{*}}})$ reads as follows:

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

(14.30)

with

$$ {j_{11 }}=1+\left( {1-\alpha } \right)\left( {{H \left/ {k} \right.}} \right),\,\,{j_{12 }}=-{{{\left[ {\left( {1-\alpha } \right)\left( {1+i} \right)e} \right]}} \left/ {{\left( {{G^A}k} \right)}} \right.}, $$

$$ {j_{13 }}={{{\left[ {\left( {1-\alpha } \right)\left( {1+i} \right)e} \right]}} \left/ {{\left( {{G^A}{k^{*}}} \right)}} \right.}, $$

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

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

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

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

$$ H\equiv \left( {{M \left/ {{{G^A}}} \right.}} \right){k^{\alpha }}-k=\Big[ {{{{\left( {1+i} \right)}} \left/ {{\left( {{G^A}\alpha } \right)}} \right.}-1} \Big]k\mathrm{ and}\kern0.5em $$

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

To show roughly how we obtained the elements of the Jacobian matrix (14.30), we now describe the main steps taken in the derivation of ${{{\partial {e_{t+1 }}}} \left/ {{\partial {e_t}=1+(1-\alpha )({H \left/ {k} \right.})}} \right.}$. First, take the total differential of Eq. 14.15 with respect to all variables: $d{e_{t+1 }}={{({{{{M^{*}}}} \left/ {M} \right.})}^{-1 }}{k^{{\alpha -1}}}{{({k^{*}})}^{{1-\alpha }}}d{e_t}-e({M \left/ {{{M^{*}}}} \right.})(1-\alpha ){k^{{\alpha -2}}}{{({k^{*}})}^{{1-\alpha }}}d{k_{t+1 }}+e({M \left/ {{{M^{*}}}} \right.})(1-\alpha ){k^{{\alpha -1}}}{{({k^{*}})}^{{-\alpha }}}dk_{t+1}^{*}$. Second, solve the left-hand sides of Eqs. 14.16 and 14.17 simultaneously with respect to the total differentials of ${k_{t+1 }}$ and $k_{t+1}^{*}$. Third, form the partial differentials $\partial {{{{k_{t+1 }}}} \left/ {{\partial {e_t}}} \right.}$ and ${{{\partial k_{t+1}^{*}}} \left/ {{\partial {e_t}}} \right.}$ while taking the results of the second step into account. Fourth, evaluate the total differential of the first step at a steady-state solution and consider the infinitesimal changes of ${k_{t+1 }}$ and $k_{t+1}^{*}$ only with respect to ${e_t}$: ${{{\partial {e_{t+1 }}}} \left/ {{\partial {e_t}=1+}} \right.}e(1-\alpha )[{{({k^{*}})}^{-1 }}\partial k_{t+1}^{*}/\partial {e_t}-{k^{-1 }}\partial {k_{t+1 }}/\partial {e_t}]$. The last step is to insert the partial differentials evaluated at a steady-state solution from step three into the above equation.

1.4 Proof of Lemma 14.2

To calculate the eigenvalues (and the eigenvectors) of the Jacobian $J\equiv J(e,k,{k^{*}})$ we use the characteristic equation of the Jacobian (14.30): $(J-{\lambda_i}I){v_i}=0$, whereby $I$ denotes the identity matrix, ${\lambda_i},$ $i=1,2,3$ denotes the eigenvalues (characteristic values), ${{(\upsilon_i^e,\upsilon_i^k,\upsilon_i^{*})}^T},$ $i=1,2,3$ is the transpose of the eigenvector (characteristic vector) associated with eigenvalue $i=1,2,3$, and the characteristic equation in expanded form reads as follows:

$$ \left( {J-{\lambda_i}I} \right){v_i}=0\Leftrightarrow \left( {\begin{array}{cccc}{{j_{11 }}-{\lambda_i}} & {{j_{12 }}} & {{j_{13 }}} \\ {{j_{21 }}} & {{j_{22 }}-{\lambda_i}} & {{j_{23 }}} \\ {{j_{31 }}} & {{j_{32 }}} & {{j_{33 }}-{\lambda_i}} \\ \end{array}} \right)\left( {\begin{array}{cccc}{\upsilon_i^e} \\ {\upsilon_i^k} \\ {\upsilon_i^{*}} \\ \end{array}} \right)=\left( {\begin{array}{ccc} 0 \\ 0 \\ 0 \\ \end{array}} \right). $$

Multiplying the second row of this matrix equation by $-\mu$, adding it to the last row, and multiplying the first row by $k/(1-\alpha )$ yields the following equivalent equation:

$$ \left( {\begin{array}{ccccc}{\frac{k}{{\left( {1-\alpha } \right)}}\left( {1-{\lambda_i}} \right)+H} & {-\frac{{\left( {1+i} \right)e}}{{{G^A}}}} & {\frac{{\left( {1+i} \right)e{\mu^{-1 }}}}{{{G^A}}}} \\ {{j_{21 }}} & {{j_{22 }}-{\lambda_i}} & {{j_{23 }}} \\ {\frac{{\mu H}}{e}} & {\mu \left[ {{\lambda_i}-\frac{{\left( {1+i} \right)}}{{{G^A}}}} \right]} & {\frac{{\left( {1+i} \right)}}{{{G^A}}}-{\lambda_i}} \\ \end{array}} \right)\left( {\begin{array}{ccccccc}{\upsilon_i^e} \\ {\upsilon_i^k} \\ {\upsilon_i^{*}} \\ \end{array}} \right)=\left( {\begin{array}{ccccccc} 0 \\ 0 \\ 0 \\ \end{array}} \right). $$

By multiplying the first row of the revised matrix equation by ${{{-\mu }} \left/ {e} \right.}$ times and adding it to the last row we get:

$$ \left( {\begin{array}{ccccccc} {\frac{k}{{\left( {1-\alpha } \right)}}\left( {1-{\lambda_i}} \right)+H} & {-\frac{{\left( {1+i} \right)e}}{{{G^A}}}} & {\frac{{\left( {1+i} \right)e{\mu^{-1 }}}}{{{G^A}}}} \\ {{j_{21 }}} & {{j_{22 }}-{\lambda_i}} & {{j_{23 }}} \\ {-\frac{{k\mu \left( {1-{\lambda_i}} \right)}}{{e\left( {1-\alpha } \right)}}} & {\mu {\lambda_i}} & {-{\lambda_i}} \\ \end{array}} \right)\left( {\begin{array}{ccccccc}{\upsilon_i^e} \\ {\upsilon_i^k} \\ {\upsilon_i^{*}} \\ \end{array}} \right)=\left( {\begin{array}{ccccccccc} 0 \\ 0 \\ 0 \\ \end{array}} \right). $$

Finally subtracting the last row ${{{{\mu^{-1 }}(1+i)e}} \left/ {{({\lambda_i}{G^A})}} \right.}$ times from the first row leads to:

$$ \left( {\begin{array}{ccccccc}{\frac{k}{{\left( {1-\alpha } \right)}}\left( {1-{\lambda_i}} \right)\left( {1-\frac{1+i }{{{\lambda_i}}}} \right)+H} & 0 & 0 \\ {{j_{21 }}} & {{j_{22 }}-{\lambda_i}} & {{j_{23 }}} \\ {-\frac{{k\mu \left( {1-{\lambda_i}} \right)}}{{e\left( {1-\alpha } \right)}}} & {\mu {\lambda_i}} & {-{\lambda_i}} \\ \end{array}} \right)\left( {\begin{array}{cccccccc}{\upsilon_i^e} \\ {\upsilon_i^k} \\ {\upsilon_i^{*}} \\ \end{array}} \right)=\left( {\begin{array}{cccccc}0 \\ 0 \\ 0 \\ \end{array}} \right). $$

(14.31)

Equation 14.31 can be solved if and only if the determinant of the matrix in Eq. 14.31 vanishes, i.e. if either

$$ \left| {\begin{array}{llllll} {{j_{22 }}-{\lambda_i}} & {{j_{23 }}} \\\mu & {-1} \\\end{array}} \right|=0 $$

(14.32)

or

$$ \frac{k}{{\left( {1-\alpha } \right)}}\left( {1-{\lambda_i}} \right)\left( {1-\frac{1+i }{{{\lambda_i}}}} \right)+H=0. $$

(14.33)

There are thus two cases to be distinguished: case 1 in which Eq. 14.32 holds and case 2 for which Eq. 14.33 is true. Let us consider both cases in turn.

Case 1: Using the definition of ${j_{22 }}$ and ${j_{23 }}$, Eq. 14.32 straightforwardly leads to:

$$ {\lambda_3}={j_{22 }}+\mu {j_{23 }}=\frac{{\left( {1+i} \right)\left( {1-\alpha } \right)}}{{{G^A}}}\left( {\zeta \sigma \left( {1+\frac{b}{k}} \right)+\left( {1-\zeta } \right){\sigma^{*}}\left( {1+\frac{{{b^{*}}}}{{{k^{*}}}}} \right)} \right). $$

To determine its corresponding eigenvector, we make use of Eq. 14.31. Because $(k/(1-\alpha ))(1-{\lambda_i})(1-(1+i)/{\lambda_i})+H\ne 0$, it follows that $\upsilon_3^e=0$, and thus Eq. 14.31 turns into:

$$ \left( {\begin{array}{ccccccc}{{j_{22 }}-{\lambda_3}} & {{j_{23 }}} \\ {\mu {\lambda_3}} & {-{\lambda_3}} \\ \end{array}} \right)\left( {\begin{array}{ccccccc}{\upsilon {}_3^k} \\ {\upsilon_3^{*}} \\ \end{array}} \right)=\left( {\begin{array}{ccccccc} 0 \\ 0 \\ \end{array}} \right), $$

with the solution claimed in lemma 14.2. This can be seen as follows: the first row leads to $\upsilon_3^{*}=\mu \upsilon_3^k$, and the second row as a result of the value of ${\lambda_3}$ can then be solved identically, i.e. we can choose $\upsilon_3^k=1$.

Case 2: In this case we know that $(k/(1-\alpha ))(1-{\lambda_i})(1-(1+i)/{\lambda_i})+H=0$. Since $H/k=(1+i)/\alpha -1$, it follows that ${{{\left( {1+i} \right)}} \left/ {\alpha } \right.}-1>1$ and ${\lambda_2}=\alpha$ as claimed in lemma 14.2.

The eigenvector associated with the second eigenvalue can be found as follows: $(k/(1-\alpha ))(1-{\lambda_i})(1-(1+i)/{\lambda_i})+H=0$ implies that $\upsilon_2^e$ can be chosen freely, so, for instance, we can take $\upsilon_2^e={e \left/ {k} \right.}$. Therefore Eq. 14.31 reduces to:

$$ \left( {\begin{array}{cccccccc}{{j_{22 }}-{\lambda_i}} & {{j_{23 }}} \\ {\mu \alpha } & {-\alpha } \\ \end{array}} \right)\left( {\begin{array}{cccccccc} {\upsilon {}_2^k} \\ {\upsilon_2^{*}} \\ \end{array}} \right)=\left( {\begin{array}{cccccccc} {-{j_{21 }}\frac{e}{k}} \\ {-\mu } \\ \end{array}} \right). $$

The second row of this matrix equation yields $\upsilon_2^{*}=\mu (\upsilon_2^k+{\alpha^{-1 }})$. The first row equals:

$$ \left( {{j_{22 }}-\alpha } \right)\upsilon_2^k+{j_{23 }}\mu \left( {\upsilon_2^k+{\alpha^{-1 }}} \right)=-{j_{21 }}\left( {{e \left/ {k} \right.}} \right). $$

After substituting for the third eigenvalue, the result is:

$$ \upsilon_2^k=\frac{{1-{{{{\lambda_3}}} \left/ {k} \right.}+\zeta \left( {{b \left/ {k} \right.}} \right)\Big[ {1-\sigma +\left( {1+i} \right)\left( {{\sigma \left/ {{\left( {\alpha {G^A}} \right)}} \right.}} \right)} \Big]}}{{{\lambda_3}-\alpha }}. $$

1.5 Proof of Proposition 14.2

From lemma 14.2 we know that ${\lambda_2}=\alpha$ and by assumption $\alpha <1$ holds. Second, ${\lambda_1}={(1+i) \left/ {{({G^A}\alpha )}} \right.}$ $>1$ follows from ${k^L}<{k^H}<\bar{k}$, since $k<\bar{k}\Leftrightarrow$ $1<({M \left/ {{{G^A}}} \right.}){k^{{\alpha -1}}}$ $={(1+i) \left/ {{({G^A}\alpha )}} \right.}$. ${\lambda_3}<1\Leftrightarrow$ ${{{[(1+{i^H})(1-\alpha )]}} \left/ {{{G^A}}} \right.}\{\zeta \sigma (1+{{{b \left/ {k} \right.}}^H})+(1-\zeta ){\sigma^{*}}[1+{{{{b^{*}}}} \left/ {{{{{({k^{*}})}}^H}}} \right.}]\}$ $+(1-\zeta ){\sigma^{*}}[1+{{{{b^{*}}}} \left/ {{{{{({k^{*}})}}^H}}} \right.}]\}$. This follows from the facts that: (i) ${{{[(1+{i^H})(1-\alpha )]}} \left/ {{{G^A}}} \right.}\{\zeta \sigma (1+{{{b \left/ {k} \right.}}^H})+(1-\zeta ){\sigma^{*}}(1+{{{{b^{*}}}} \left/ {{{{{({k^{*}})}}^H}}} \right.})\}<1$ is equal to the derivative of the function $F(k)$ with respect to k evaluated at the higher steady state, (ii) function $F(k)$ is strictly concave and (iii) its graph cuts the 45° line at the higher steady state from above (see Fig. 14.1a in the main text). On the other hand, for ${\lambda_3}>1$ we have ${{{[(1+{i^L})(1-\alpha )]}} \left/ {{{G^A}}} \right.}\{\zeta \sigma (1+{{{b \left/ {k} \right.}}^L})+(1-\zeta ){\sigma^{*}}$ $[1+{{{{b^{*}}}} \left/ {{{{{({k^{*}})}}^L}}} \right.}]\}>1$, which is implied by the fact that the graph of $F(k)$ cuts the 45° line at the lower steady state from below.

1.6 Proof of Proposition 14.4

To start with the proof of proposition 14.4, we approximate Eqs. 14.15, 14.16, and 14.17 in a small neighborhood of $(e,{k^H},{k^{*,H }})$ as follows:

$$ \left[ {\begin{array}{ccccccc}{{e_{t+1 }}} \\ {{k_{t+1 }}} \\ {k_{t+1}^{*}} \\ \end{array}} \right]=\left[ {J\left( {e,{k^H},{k^{*,H }}} \right)} \right]\left[ {\begin{array}{ccccccc} {{e_t}} \\ {{k_t}} \\ {k_t^{*}} \\ \end{array}} \right]+\left[ {I-J\left( {e,{k^H},{k^{*,H }}} \right)} \right]\left[ {\begin{array}{ccccccc} e \\ {{k^H}} \\ {{k^{*,H }}} \\ \end{array}} \right]. $$

(14.34)

The general solution of the first-order linear difference equation system (14.34) takes the following form:

$$ {e_t}=e+{\kappa_2}\upsilon_2^e{{\left( {{\lambda_2}} \right)}^t}+{\kappa_3}\upsilon_3^e{{\left( {{\lambda_3}} \right)}^t}, $$

$$ k_t={k }+{\kappa_2}\upsilon_2^k{{\left( {{\lambda_2}} \right)}^t}+{\kappa_3}\upsilon_3^k{{\left( {{\lambda_3}} \right)}^t}, $$

(14.35)

$$ k_t^{*}={k^{*}}+{\kappa_2}\upsilon_2^{*}{{\left( {{\lambda_2}} \right)}^t}+{\kappa_3}\upsilon_3^{*}{{\left( {{\lambda_3}} \right)}^t}. $$

Here ${\kappa_i},i=2,3$ denote constants determined by the initial conditions for capital intensities in Home and Foreign, while ${\upsilon_i}={{(\upsilon_i^e,\upsilon_i^k,\upsilon_i^{*})}^T},i=2,3$ is the eigenvector associated with the eigenvalues within the unit circle ${\lambda_i},i=2,3.$ Note that the eigenvector associated with the eigenvalue larger than unity is excluded from Eq. 14.35 by setting ${\kappa_1}=0$. However, this exclusion implies that the equilibrium dynamics must not start from any feasible combination $({e_0},{k_0},k_0^{*})$ in the neighborhood of $(e,{k^H},{k^{*,H }})$, and that the initial combination of dynamic variables has to be located on the stable sub-manifold in the $({e_t},{k_t},k_t^{*})$ -space. If $({e_0},{k_0},k_0^{*})$ belongs to this stable sub-manifold, the economy converges towards $(e,{k^H},{k^{*,H }})$, otherwise the system dynamics strays in finite time. From the proof of lemma 14.2 we know that the eigenvectors associated with the eigenvalues of the Jacobian within the unit circle ${\upsilon_i},$ $i=2,3$ read as follows:

$$ {\upsilon_2}={{(-{e \left/ {k} \right.},{1 \left/ {{\alpha +\gamma, \gamma \mu }} \right.})}^T},{\upsilon_3}={{(0,1,\mu )}^T},\gamma \equiv {{{[\zeta b(1-\sigma +\sigma {\lambda_1})]}} \left/ {{[k(\alpha -{\lambda_3})]}} \right.}. $$

Next, these eigenvectors have to be inserted into the second and third equation of (14.35) and solved simultaneously for ${\kappa_2}{{({\lambda_2})}^t}$ and ${\kappa_3}{{({\lambda_3})}^t}$. The results are as follows:

$$ {\kappa_2}{\lambda_2}^t={{{[{k_t}-{k^H}-(k_t^{*}-{k^{*,H }})]}} \left/ {{(\upsilon_2^k-\upsilon_3^k)=}} \right.}\alpha [{k_t}-{k^H}-{\mu^{-1 }}(k_t^{*}-{k^{*,H }})], $$

$$ {\kappa_3}{\lambda_3}^t=\alpha {\mu^{-1 }}[\upsilon_2^k(k_t^{*}-{k^{*,H }})-\upsilon_2^{*}({k_t}-{k^H})]=\alpha [({1 \left/ {\alpha } \right.}+\gamma ){\mu^{-1 }}(k_t^{*}-{k^{*,H }})-\gamma ({k_t}-{k^H})]. $$

The next step is to consider the second and the third equation of (14.35) for $t+1$ and $t$, and then to subtract the latter from the former. We get the following results: ${k_{t+1 }}-{k_t}=({1 \left/ {{\alpha +\gamma }} \right.})({\lambda_2}-1){\kappa_2}{\lambda_2}^t+({\lambda_3}-1){\kappa_3}{\lambda_3}^t$, $k_{t+1}^{*}-k_t^{*}=\gamma ({\lambda_2}-1){\kappa_2}{\lambda_2}^t$ $+({\lambda_3}-1){\kappa_3}{\lambda_3}^t$. The last step is to insert into these equations the equations for ${\kappa_2}{\lambda_2}^t$ and ${\kappa_3}{\lambda_3}^t$ from above, and to collect terms. As a consequence, Eqs. 14.23 and 14.24 are obtained. Finally, inserting the equations for ${\kappa_2}{\lambda_2}^t$ and ${\kappa_3}{\lambda_3}^t$ into the first equation of (14.35) and remembering that $\upsilon_2^e=-{e \left/ {k} \right.},\upsilon_3^e=0$ holds, we obtain Eq. 14.22.