The transfer paradox in a pay-as-you-go pension system

Abstract

We examine how international transfers affect the welfare levels of a donor with a higher marginal propensity to save and a recipient with a lower marginal propensity to save when both countries adopt a pay-as-you-go (PAYG) pension system using a one-sector overlapping generations model. We demonstrate that in a dynamically efficient economy, except at the golden rule, when a per capita PAYG pension contribution of either a donor or a recipient increases marginally, the effect of the transfer on the donor’s welfare can be reduced, whereas whether the effect of the transfer on the recipient’s welfare is reduced is ambiguous. These results imply that the existence of a PAYG pension might hinder the effectiveness of the transfer on the donor’s welfare, and the adoption of a PAYG pension system is likely to cause a weak transfer paradox in which both a donor and a recipient immiserize. Our results also suggest that the introduction of a PAYG pension system, which is used as a domestic policy instrument for intergenerational income redistribution, reduces the donor’s incentive to make an international transfer to a recipient, which is a form of international income redistribution.

Appendices

Appendix A: The proof of Lemma 1

From Eq. 7 and r≡r _t = r _t+1, the world capital market equilibrium in the steady state is rewritten as follows:

$$ 2(1+n)k(r) =s^{D}(I^{D}, r) +s^{R}(I^{R}, r). $$

(A.1)

• (i) By the definition of a transfer from a donor to a recipient, $\frac {dI^{D}}{dT}=-1$ and $\frac {dI^{R}}{dT}=1$. By differentiating Eq. A.1 with respect to r and T, we obtain the following equation:

  $$ {\Gamma} dr =-{s^{D}_{I}}dT+{s^{R}_{I}}dT \Leftrightarrow \frac{dr}{dT} =-\frac{{s^{D}_{I}}-{s^{R}_{I}}}{\Gamma}. $$

  (A.2)

  Γ<0 holds in the steady state under the dynamically stable condition, Eq. 11. Thus, if ${s^{D}_{I}}>{s^{R}_{I}}$, $\frac {dr}{dT}>0$.

• (ii) By the definition of lifetime income I ⁱ, $\frac {dI^{i}}{dP^{i}}= -\frac {r-n}{1+r}<0$. By differentiating Eq. A.1 with respect to r and P ⁱ, we obtain the following equation:

  $$ {\Gamma} dr ={s^{i}_{I}}(-\frac{r-n}{1+r})dP^{i} \Leftrightarrow \frac{dr}{dP^{i}} =-\frac{(r-n){s^{i}_{I}}}{(1+r){\Gamma}}>0. $$

  (A.3)

Appendix B: The proof of Lemma 2

Differentiating $\frac {dr}{dT}$, which is given by Eq. A.2, with respect to P ⁱ, we obtain the following equation:

$$\begin{array}{@{}rcl@{}} \frac{d^{2}r}{dTdP^{D}} &=&\\ &&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!-\frac{[s_{II}^{D}(w^{\prime}\frac{dr}{dP^{D}}-\frac{r-n}{1+r}-\frac{(1+n)P^{D}}{(1+r)^{2}}\frac{dr}{dP^{D}}\!)\,+\,s_{Ir}^{D}\frac{dr}{dP^{D}}-s_{II}^{R}(w^{\prime}\frac{dr}{dP^{D}}-\frac{(1+n)P^{R}}{(1+r)^{2}}\frac{dr}{dP^{D}}\!)-s_{Ir}^{R}\frac{dr}{dP^{D}}]{\Gamma} -({s_{I}^{D}}-{s_{I}^{R}})\!\frac{d{\Gamma}} {dP^{D}}}{{\Gamma}^{2}}.\\ \end{array} $$

(A.4)

$$\begin{array}{@{}rcl@{}} &&\!\!\!\!\!\!\frac{d^{2}r}{dTdP^{R}}=\\ &&\!\!\!\!\!\!-\frac{[s_{II}^{D}(w^{\prime}\frac{dr}{dP^{R}}-\frac{(1+n)P^{D}}{(1+r)^{2}}\frac{dr}{dP^{R}}\!)\,+\,s_{Ir}^{D}\frac{dr}{dP^{R}}-s_{II}^{R}(w^{\prime}\frac{dr}{dP^{R}}-\frac{r-n}{1+r}-\frac{(1+n)P^{R}}{(1+r)^{2}}\frac{dr}{dP^{R}}\!)-s_{Ir}^{R}\frac{dr}{dP^{R}}]{\Gamma} -({s_{I}^{D}}-{s_{I}^{R}})\frac{d{\Gamma}} {dP^{R}}}{{\Gamma}^{2}}.\\ \end{array} $$

(A.5)

When we evaluate Eqs. A.4 and A.5 at P ^D = P ^R=0, and noting that $\frac {r-n}{1+r} =-\frac {\Gamma }{{s_{I}^{i}}}\frac {dr}{dP^{i}}$ and $w^{\prime }=-k$, the following equations are obtained:

$$\begin{array}{@{}rcl@{}} \frac{d^{2}r}{dTdP^{D}}\left|{\vphantom{\frac{d^{2}r}{dTdP^{D}}}}{~}_{P^{D}=P^{R}=0}\right.\,=\, \frac{1}{{\Gamma}^{2}} \left[ \left\{ \underbrace{\frac{s_{II}^{D}{\Gamma}} {{s_{I}^{D}}}}{~}_{{(-)\ \text{or}\ 0} } \!+(s_{Ir}^{D}\,-\,s_{Ir}^{R})-(s_{II}^{D}\,-\,s_{II}^{R})k \right\} \underbrace{ \frac{(r\,-\,n){s_{I}^{D}}}{(1\!+r)}}_{{ (+)} } \,+\, \underbrace{ ({s_{I}^{D}}\,-\,{s_{I}^{R}}) \vphantom{ \frac{d{\Gamma}} {dP^{D}} } }{~}_{{ (+) } } \underbrace{ \frac{d{\Gamma}} {dP^{D}} }{~}_{{ (-) } } \right]. \end{array} $$

(A.6)

$$\begin{array}{@{}rcl@{}} \frac{d^{2}r}{dTdP^{R}}\left|{\vphantom{\frac{d^{2}r}{dTdP^{R}}}}{~}_{P^{D}=P^{R}=0}\right. &= \frac{1}{{\Gamma}^{2}} \left[ \left\{ \underbrace{ -\frac{s_{II}^{R}{\Gamma}} {{s_{I}^{R}}}}{~}_{{ (+)\ \text{or}\ 0 } } +(s_{Ir}^{D}-s_{Ir}^{R})-(s_{II}^{D}-s_{II}^{R})k \right\} \underbrace{ \frac{(r-n){s_{I}^{R}}}{(1+r)} }_{{ (+)} } + \underbrace{ ({s_{I}^{D}}-{s_{I}^{R}}) \vphantom{ \frac{d{\Gamma}} {dP^{R}} } }{~}_{{ (+) } } \underbrace{ \frac{d{\Gamma}} {dP^{R}} }{~}_{{ (-) } } \right].\\ \end{array} $$

(A.7)

If $s^{D}_{II}= s^{R}_{II}$ and $s^{D}_{Ir}= s^{R}_{Ir}$, $\frac {d^{2}r}{dTdP^{D}}\left |{\vphantom {\frac {d^{2}r}{dTdP^{R}}}}{~}_{P^{D}=P^{R}=0}\right .$ is negative from Eq. A.6, while only when $\frac {d{\Gamma }} {dP^{R}} < \frac {(r-n)s_{II}^{R}{\Gamma }} {(1+r)({s_{I}^{D}}-{s_{I}^{R}})}(<0)$, $\frac {d^{2}r}{dTdP^{R}}\left |{\vphantom {\frac {d^{2}r}{dTdP^{R}}}}{~}_{P^{D}=P^{R}=0}\right .$ is negative from Eq. A.7.

Appendix C: The derivation of $\frac {d{\Gamma }}{dP^{D}}$

From Eq. 11, the dynamic stability condition in the steady state can be rewritten as follows:

$$ {\Gamma} =2(1+n)k^{\prime} +\frac{(1+n)P^{D}}{(1+r)^{2}}{s_{I}^{D}} +\frac{(1+n)P^{R}}{(1+r)^{2}}{s_{I}^{R}} -{s_{r}^{D}} -{s_{r}^{R}} -{s_{I}^{D}}w^{\prime} -{s_{I}^{R}}w^{\prime}. $$

(A.8)

Differentiating Eq. A.8 with respect to P ^D, we obtain the following equation:

$$\begin{array}{@{}rcl@{}} \frac{d{\Gamma}} {dP^{D}} &=&2(1+n)k^{\prime\prime}\frac{dr}{dP^{D}} +\frac{1+n}{(1+r)^{3}}(1+r-2P^{D}\frac{dr}{dP^{D}}){s_{I}^{D}} \\ &&-\frac{(1+n)P^{D}}{(1+r)^{2}}\left[s_{II}^{D}\left\{k\frac{dr}{dP^{D}}+\frac{r-n}{1+r}+\frac{(1+n)P^{D}}{(1+r)^{2}}\frac{dr}{dP^{D}}\right\}-s_{Ir}^{D}\frac{dr}{dP^{D}}\right] \\ &&-2\frac{(1+n)P^{R}}{(1+r)^{3}}{s_{I}^{R}}\frac{dr}{dP^{D}} -\frac{(1+n)P^{R}}{(1+r)^{2}}\\&&\times\left[s_{II}^{R}\left\{k\frac{dr}{dP^{D}}+\frac{(1+n)P^{R}}{(1+r)^{2}}\frac{dr}{dP^{D}}\right\}-s_{Ir}^{R}\frac{dr}{dP^{D}}\right] \\ &&+\left[s_{rI}^{D}\left\{k\frac{dr}{dP^{D}}+\frac{r-n}{1+r}+\frac{(1+n)P^{D}}{(1+r)^{2}}\frac{dr}{dP^{D}}\right\}-s_{rr}^{D}\frac{dr}{dP^{D}}\right] \\ &&+\left[s_{rI}^{R}\left\{k\frac{dr}{dP^{D}}+\frac{(1+n)P^{R}}{(1+r)^{2}}\frac{dr}{dP^{D}}\right\}-s_{rr}^{R}\frac{dr}{dP^{D}}\right] \\ &&+{s_{I}^{D}}k^{\prime}\frac{dr}{dP^{D}}-k\left[s_{II}^{D}\left\{k\frac{dr}{dP^{D}}+\frac{r-n}{1+r}+\frac{(1+n)P^{D}}{(1+r)^{2}}\frac{dr}{dP^{D}}\right\}-s_{Ir}^{D}\frac{dr}{dP^{D}}\right] \\ &&+{s_{I}^{R}}k^{\prime}\frac{dr}{dP^{D}}-k\left[s_{II}^{R}\left\{k\frac{dr}{dP^{D}}+\frac{(1+n)P^{R}}{(1+r)^{2}}\frac{dr}{dP^{D}}\right\}-s_{Ir}^{R}\frac{dr}{dP^{D}}\right]. \end{array} $$

(A.9)

When marginally evaluating at P ^D=0 and P ^R=0, Eq. A.9 is arranged as follows:

$$\begin{array}{@{}rcl@{}} \frac{d{\Gamma}} {dP^{D}}\left|{\vphantom{\frac{d^{2}r}{dTdP^{R}}}}{~}_{P^{D}=P^{R}=0}\right. &=&2(1+n)k^{\prime\prime}\frac{dr}{dP^{D}} +\frac{1+n}{(1+r)^{2}}{s_{I}^{D}} +s_{rI}^{D}(k\frac{dr}{dP^{D}}+\frac{r-n}{1+r})\\&&-s_{rr}^{D}\frac{dr}{dP^{D}} +s_{rI}^{R}k\frac{dr}{dP^{D}}-s_{rr}^{R}\frac{dr}{dP^{D}} \\ &&+{s_{I}^{D}}k^{\prime}\frac{dr}{dP^{D}}-k\left\{s_{II}^{D}(k\frac{dr}{dP^{D}}+\frac{r-n}{1+r})-s_{Ir}^{D}\frac{dr}{dP^{D}}\right\} \\&&+{s_{I}^{R}}k^{\prime}\frac{dr}{dP^{D}}-k(s_{II}^{R}k\frac{dr}{dP^{D}}-s_{Ir}^{R}\frac{dr}{dP^{D}}). \end{array} $$

(A.10)

By arranging Eq. 10 with $\frac {dr}{dP^{D}} =-\frac {(r-n){s^{D}_{I}}}{(1+r){\Gamma }}>0$, we obtain the following equation:

$$\begin{array}{@{}rcl@{}} \frac{d{\Gamma}} {dP^{D}}\left|{\vphantom{\frac{d{\Gamma}} {dP^{D}}}}{~}_{P^{D}=P^{R}=0}\right. &=&\left[ 2(1+n)k^{\prime\prime}+\underbrace{({s_{I}^{D}}+{s_{I}^{R}})k^{\prime}}_{{(-)}}-\left\{ (s_{II}^{D}+s_{II}^{R})k^{2}-2(s^{D}_{Ir}+s^{R}_{Ir})k+ (s_{rr}^{D}+s_{rr}^{R}) \right\}\right.\\ &&\left.- \left\{ \frac{s_{rI}^{D}-s_{II}^{D}k}{{s_{I}^{D}}} +\frac{1+n}{(1+r)(r-n)} \right\} \underbrace{\Gamma}{~}_{{(-) } } \right] \times \underbrace{\frac{dr}{dP^{D}} }{~}_{{ (+) } }. \end{array} $$

(A.11)

The signs follow from $k^{\prime }=\frac {1}{f^{\prime \prime }}<0$, Γ<0, and $\frac {dr}{dP^{D}}>0$, from Lemma 1(ii). The sufficient condition under which $\frac {d{\Gamma }}{d P^{D}}\left |{\vphantom {\frac {d^{2}r}{dTdP^{R}}}}{~}_{P^{D}=P^{R}=0}\right .$ is negative is that the large bracket of the left-hand side of Eq. A.11 is negative. A similar procedure can be used to derive $\frac {d{\Gamma }}{dP^{R}}$.

Appendix D: The proof of Proposition 2

The proof procedure is similar to that of Proposition 1. By rearranging Eq. 20, we obtain the following equation:

$$ \frac{dV^{R}}{dT} =1 - \left[ k+\frac{(1+n)P^{R}}{(1+r)^{2}}-\frac{s^{R}}{1+r} \right] \frac{dr}{dT}. $$

(A.12)

By differentiating Eq. A.12 with respect to P ^R and P ^D, respectively, we obtain the second-order cross-partial derivatives. Evaluating them at P ^D = P ^R=0, we obtain the following equations:

$$\begin{array}{@{}rcl@{}} \frac{d^{2}V^{R}}{dTdP^{R}}\left|{\vphantom{\frac{dr}{dT}}}{~}_{P^{D}=P^{R}=0}\right. &=&\\ &&-{\frac{ \overbrace{ 1+n+(r-n){s_{I}^{R}}} ^{{(+)}} + \left[ s^{R}+ (1+r) \left\{ k{s_{I}^{R}}-{s_{r}^{R}}+(1+r)k^{\prime} \right\} \right] \overbrace{ \frac{dr}{dP^{R}}}^{{ (+) } } |_{P^{D}=P^{R}=0}}{(1+r)^{2}} \underbrace{ \frac{dr}{dT} \vphantom{ \biggl|_{P^{D}=P^{R}=0} } }{~}_{{ (+) } } \left|{\vphantom{\frac{dr}{dT}}}{~}_{P^{D}=P^{R}=0} \right.} \\ &&{- \overbrace{ \frac{(1+r)k-s^{R}}{1+r}}^{{ (+) } } \underbrace{ \frac{d^{2}r}{dTdP^{R}} \vphantom{ \biggl|_{P^{D}=P^{R}=0} } }{~}_{{ (-) } } \left|{\vphantom{\frac{dr}{dT}}}{~}_{P^{D}=P^{R}=0}\right. . } \end{array} $$

(A.13)

$$\begin{array}{@{}rcl@{}} \frac{d^{2}V^{R}}{dTdP^{D}}\left|{\vphantom{\frac{dr}{dT}}}{~}_{P^{D}=P^{R}=0}\right. &= &-\frac{s^{R}+(1+r)[k{s_{I}^{R}}-{s_{r}^{R}}+(1+r)k^{\prime}]}{(1+r)^{2}} { \underbrace{ \frac{dr}{dP^{D}}}{~} _{{ (+) } } \left|{\vphantom{\frac{dr}{dT}}}{~}_{P^{D}=P^{R}=0}\right. \underbrace{ \frac{dr}{dT} }{~}_{{ (+) } } \left|{\vphantom{\frac{dr}{dT}}}{~}_{P^{D}=P^{R}=0}\right.} \\ &&{ - \overbrace{ \frac{(1+r)k-s^{R}}{1+r}}^{{ (+) } } \underbrace{ \frac{d^{2}r}{dTdP^{D}} }{~}_{ { (-) } } \left|{\vphantom{\frac{dr}{dT}}}{~}_{P^{D}=P^{R}=0}\right.. } \end{array} $$

(A.14)

Note that $\frac {dr}{dT}>0$, $\frac {dr}{dP^{R}}>0$, and $\frac {d^{2}r}{dTdP^{i}}\left |{\vphantom {\frac {d^{2}r}{dTdP^{R}}}}{~}_{P^{D}=P^{R}=0}\right .<0$ from Lemmas 1 and 2, respectively. By the capital market equilibrium and the dynamically efficient condition, s ^R<(1 + n)k≤(1 + r)k. Thus, (1 + r)k−s ^R>0 always holds irrespective of the size of s ^R. Therefore, the second terms of Eqs. A.13 and A.14 are necessarily positive, while the sign of the first terms is not determined. For example, when ${s^{R}_{I}}\ge \frac {{s^{R}_{r}}-(1+r)k^{\prime }}{k}$, which is a similar condition to that used in Proposition 1, the first terms of Eqs. A.13 and A.14 are negative and therefore the signs of both Eqs. A.13 and A.14 depend on the relative sizes of the first term and the second term.