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Fertility and labor supply of the old with pay-as-you-go pension and child allowances

  • Hung-Ju Chen and Koichi Miyazaki EMAIL logo

Abstract

We investigate the effects of pay-as-you-go pension and child allowances on fertility, labor supply of the old, and welfare. For this purpose, we analyze a small open overlapping-generations model in which fertility and an old agent’s labor supply (retirement time) are endogenized with pay-as-you-go pension and child allowances. We find that how the pay-as-you-go pension tax rate affects the fertility rate depends on whether an old agent retires. When an old agent fully retires, then the size of the interest rate and fertility rate determine the effect of the pay-as-you-go pension tax rate on the fertility rate. When an old agent works, the pay-as-you-go pension tax rate certainly reduces the fertility rate. In addition, how child allowances affect the fertility rate depends on whether an old agent works. If an old agent retires fully, then an increase in the child allowance tax rate increases the fertility rate. When an old agent works, this is not necessarily true, which suggests that an old agent’s labor status should be taken into account when we evaluate the effects of the social security system on economic variables. In addition, we examine the effect of the social security tax rates on welfare and provide numerical examples.

JEL Classification: J13; J26; H55; D91

Acknowledgement

We thank the Managing Editor, Tiago Cavalcanti, and an anonymous referee for their excellent comments and suggestions. Chen gratefully acknowledges financial support from the Program for Globalization Studies at the Institute for Advanced Studies in Humanities at National Taiwan University (grant number: NTU-ERP-103R890502). Miyazaki gratefully acknowledges financial support from the Japan Society for the Promotion of Science KAKENHI (grant number: 16K17086).

A Appendix

A.1 Proof of Proposition 1

Proof.

(i) Taking the derivative of n¯ with respect to τ, we obtain

n¯τ=ϕr¯w¯A+{ϕ(1τ)+η(1+β)}r¯w¯ϕw¯A2=r¯w¯A2[ϕζ(1+β+ϕ)r¯+ϕ2τw¯+{ϕ(1τ)+η(1+β)}ϕw¯]=r¯w¯A2ϕ[ζ(1+β+ϕ)r¯+ϕw¯+η(1+β)w¯].

The sign of n¯τ is greater (resp. smaller) than 0 if and only if η is greater (resp. smaller) than ηf=ζ(1+β+ϕ)r¯ϕw¯(1+β)w¯.

(ii) From Equation (13),

n¯η=(1+β)r¯w¯ζ(1+β+ϕ)r¯ϕτw¯>0.

   ⊡

A.2 Proof of Proposition 2

Proof.

Let

G(n):=(1τη)w¯ζ[ϕ1+β+γ+ϕ(1+θr¯)+η1η]+1nη(1τη)w¯ζ(1η)θ(1+β+ϕ)γr¯1+β+γ+ϕ.

Then, Equation (20) is written as nt+1 = G(nt). Note that when Equation (16) holds, η(1τη)w¯ζ(1η)θ(1+β+ϕ)γr¯1+β+γ+ϕ>0 for all τ ∈ [0, 1) and η ∈ [0, 1) satisfying τ + η < 1. Consider functions y = n and y = G(n). Both functions are continuous in n. The function y = n is strictly increasing in n, limn0n=0 and limn+n=+. The function y = G(n) is strictly decreasing in n, limn0G(n)=+, and limn+G(n)=(1τη)w¯ζ[ϕ1+β+γ+ϕ(1+θr¯)+η1η]<+. Thus, there is a unique stationary point, n¯ ∈ (0, + ∞).

To prove the local stability, first, we show that |G(n¯)|<1, where |G(n)| is the absolute value of G′(n). Since

G(n)=1n2η(1τη)w¯ζ(1η)θ(1+β+ϕ)γr¯1+β+γ+ϕ,1|G(n¯)|=11n¯2η(1τη)w¯ζ(1η)θ(1+β+ϕ)γr¯1+β+γ+ϕ=1(11n¯(1τη)w¯ζ[ϕ1+β+γ+ϕ(1+θr¯)+η1η])=1n¯(1τη)w¯ζ[ϕ1+β+γ+ϕ(1+θr¯)+η1η]>0,

where we use the fact that n¯=(1τη)w¯ζ[ϕ1+β+γ+ϕ(1+θr¯)+η1η]+1n¯η(1τη)w¯ζ(1η)θ(1+β+ϕ)γr¯1+β+γ+ϕ. Since G(n) is strictly decreasing and convex in n, |G(n)| is strictly decreasing in n. Since G′(n) is continuous in n and limn0G(n)=, ε ∈ (0, n¯) exists such that |G(n¯ε)|=1. Define N:=[n¯ε,G(n¯ε)]. Since |G(n)|1 for all n, G(n) ∈ N for all n. Hence, G is a mapping from N to N. On this set, we define

(27)L(n)={nG(n)if n<n¯G(n)nif nn¯.

Since G(n) and n are continuous in n, L(n) is also continuous in n. From the above argument, L(n) < 0 for all nN{n¯} and L(n¯) = 0. Consider any point in [n¯ε, n¯). In this case, L(G(n)) = G(G(n)) − G(n), because G(n) > n¯. Thus,

L(G(n))L(n)=G(G(n))G(n)n+G(n)=G(n)n[G(n)G(G(n))]>G(n)n|G(n)|(G(n)n)0,

where we use the fact that G is strictly decreasing and strictly convex in n, and |G(n)|1 for all nN. Analogously, for any n ∈ (n¯, G(n¯ε)],

L(G(n))L(n)=G(n)G(G(n))G(n)+n>|G(G(n))|[nG(n)]G(n)+n0,

where |G(G(n)))|1 because G(n) ∈ N. It is clear that L(G(n¯)) − L(n¯) = 0, because n¯ = G(n¯). From Lemma 6.2 in Stokey, Rucas, and Prescott (1989), the function, L, is a Liapounov function, and n¯ is glacially stable on N, that is, n¯ is locally stable.    ⊡

A.3 Proof of Proposition 3

Proof.

(i) By differentiating both sides in Equation (21) with respect to τ, we obtain

(28)n¯τ=ϕn¯2(1η)(1+θr¯)+η{(1+β+ϕ)θγr¯}n¯+ηn¯2(1+β+γ+ϕ)ζn¯2(1+β+γ+ϕ)(1η)+(1τη){(1+β+ϕ)θγr¯}ηw¯w¯<0.

(ii) Taking the derivative of l¯o in Equation (22) with respect to τ, we obtain

l¯oτ=1(1η)θw¯[γr¯c¯yτ+w¯(θ+n¯)+τw¯n¯τ]

Since η < 1, the sign of l¯oτ is determined by the sign of γr¯c¯yτ+w¯(θ+n¯)+τw¯n¯τ. Since

c¯yτ=11+β+γ+ϕ(1+θr¯)w¯

and w¯ > 0, the sign of γr¯c¯yτ+w¯(θ+n¯)+τw¯n¯τ is equivalent to the sign of

γr¯1+β+γ+ϕ(1+θr¯)+θ+n¯+τn¯τ.

Rearranging this equation, we obtain

θ(γ1+β+γ+ϕ+1)γr¯1+β+γ+ϕ+n¯+τn¯τ=θ(1+β+ϕ)γr¯1+β+γ+ϕ+n¯+τn¯τ.

Since l¯o < 1, from Equation (22),

τ1τηn¯<θ(1+β+ϕ)γr¯1+β+γ+ϕ.

Thus,

θ(1+β+ϕ)γr¯1+β+γ+ϕ+n¯+τn¯τ>1η1τηn¯+τn¯τ.

From Equation (28), 1η1τηn¯+τn¯τ is

n¯(1η1τηϕn¯(1η)(1+θr¯)+η{(1+β+ϕ)θγr¯}+ηn¯(1+β+γ+ϕ)ζn¯2(1+β+γ+ϕ)(1η)+(1τη){(1+β+ϕ)θγr¯}ηw¯τw¯)=n¯h(n¯)(1τη)[ζn¯2(1+β+γ+ϕ)(1η)+(1τη){(1+β+ϕ)θγr¯}ηw¯],

where

h(n):=ζ(1+β+γ+ϕ)(1η)2n2[ϕ(1η)(1+θr¯)+η(1+β+γ+ϕ)](1τη)τw¯n+(1τη)2{(1+β+ϕ)θγr¯}ηw¯.

From Equation (21), equilibrium n¯ is a positive solution to

(29)ζn2(1τη)w¯[ϕ1+β+γ+ϕ(1+θr¯)+η1η]nη(1τη)w¯1ηθ(1+β+ϕ)γr¯1+β+γ+ϕ=0.

Note that this equation has one negative solution and one positive solution when θ>γr¯1+β+ϕ. Plugging n¯ in h(n) and applying the fact that n¯ is a solution to Equation (29), we obtain

h(n¯)=(1+β+γ+ϕ)(1η)2{[1τη1ηη+ϕ1τη1+β+γ+ϕ(1+θr¯)]w¯n¯+η(1τη)w¯1ηθ(1+β+ϕ)γr¯1+β+γ+ϕ}[ϕ(1η)(1+θr¯)+η(1+β+γ+ϕ)](1τη)τw¯n¯+(1τη)2{(1+β+ϕ)θγr¯}ηw¯=(1τη)2[(1+β+γ+ϕ)η+ϕ(1η)(1+θr¯)]w¯n¯+(1τη)τη{(1+β+ϕ)θγr¯}w¯>0

for all τ ∈ [0, 1 − η). Thus,

l¯oτ=1(1η)θ[θ(1+β+ϕ)γr¯1+β+γ+ϕ+n¯+τn¯τ]>1(1η)θ[1η1τηn¯+τn¯τ]>0

for all τ ∈ [0, 1 − η).    ⊡

A.4 Proof of Corollary 1

Proof.

(i) If τ ≥ (1 − η)2, then the left-hand side (LHS) of Equation (24) is less than or equal to 0, while the RHS is positive. Thus, n¯η<0.

(ii) For any given η ∈ [0, 1), if τ moves to 0, then the LHS of Equation (24) moves to

1+(1+β+ϕ)θγr¯n~(1+β+γ+ϕ)>1,

where n~ is n¯ at τ = 0. If Equation (25) is satisfied, because θ ∈ (0, 1]

1>ϕ1+β+γ+ϕ(1+θr¯).

Since the LHS of Equation (24) is continuous in τ, for sufficiently small τ ∈ [0, 1), the inequality in Equation (24) is reversed, which implies n¯η>0.

(iii) Differentiating l¯o with respect to η, we obtain

l¯oη=1(1η)2θw¯[γr¯c¯y+τw¯(θ+n¯)γr¯(1η)1τηc¯y+(1η)τw¯n¯η]=1(1η)2θw¯[γr¯τ1τηc¯y+τw¯(θ+n¯)+(1η)τw¯n¯η],

where we use the fact that c¯yη=c¯y1τη in the first equation. From Equation (22),

γr¯c¯y=τw¯(θ+n¯)(1η)θw¯l¯o.

Then,

γr¯τ1τηc¯y+τw¯(θ+n¯)=τ(1η)θw¯(1l¯o)+τw¯n¯(1η)1τη>0.

Thus,

l¯oη=τ(1η)θ[θ(1l¯o)+n¯1τη+n¯η]>0

if n¯η>0.    ⊡

A.5 Proof of Proposition 5

Proof.

Define g(η):=η(1+β+ϕ)(τw¯ζr¯)+ϕ(1τ)τw¯. If τζr¯w¯, then g(η) > 0 for all η ∈ [0, 1 − τ). If τ<ζr¯w¯, then g(η) is a strictly decreasing function in η. Note that g(0)=ϕ(1τ)τw¯>0. At η = 1 − τ,

g(1τ)=(1τ)[τw¯(1+β+2ϕ)(1+β+ϕ)ζr¯].

If τ(1+β+ϕ)ζr¯(1+β+2ϕ)w¯, g(1 − τ) ≥ 0. This implies that for all η ∈ [0, 1 − τ), Wη(τ,η)>0. If τ<(1+β+ϕ)ζr¯(1+β+2ϕ)w¯, then g(1 − τ) < 0, which implies that for η sufficiently close to 1 − τ, Wη(τ,η)<0. This completes the proof.    ⊡

A.6 Proof of Proposition 6

Proof.

First, at (τ, 0), c¯y, l¯o and n¯ are

c¯y=1τ1+β+γ+ϕ(1+θr¯)w¯,l¯o=1θw¯[γr¯c¯y+τw¯(θ+n¯)],n¯=ϕζc¯y.

Taking the derivative of W with respect to τ and evaluating it at (τ, 0), we obtain

Wτ(τ,0)=1+βc¯yc¯yτ+ϕn¯n¯τ+γl¯olo¯τ=1+βc¯yc¯yτ+ϕn¯{ϕζ(1+β+γ+ϕ)(1+θr¯)w¯=ϕζc¯yτ}+γl¯o1θw¯[γr¯c¯yτ+w¯(θ+n¯)+τw¯n¯τ]=c¯yτ[1+βc¯y+ϕn¯ϕζ+γl¯oγr¯θw¯+γl¯oτθϕζ]+γlo¯θ+n¯θ=c¯yτ=c¯y1τ[1+β+ϕc¯y+γθw¯γr¯c¯y+τw¯(θ+n¯)γr¯θw¯+γθw¯γr¯c¯y+τw¯(θ+n¯)τθϕζ]+γθw¯γr¯c¯y+τw¯(θ+n¯)θ+n¯θ=11τ[1+β+ϕ+γ2r¯c¯yγr¯c¯y+τw¯(θ+n¯)+γτϕw¯c¯yζ[γr¯c¯y+τw¯(θ+n¯)]]+γw¯(θ+n¯)γr¯c¯y+τw¯(θ+n¯)=[ζ{γr¯c¯y+τw¯(θ+n¯)}(1+β+ϕ)ζγ2r¯c¯yγτϕw¯c¯y+(1τ)ζγw¯(θ+n¯)]ζ(1τ)[γr¯c¯y+τw¯(θ+n¯)]

The numerator of Wτ(τ,0) is

c¯y[ζγr¯(1+β+ϕ)τw¯ϕ(1+β+ϕ)ζγ2r¯γτϕw¯+(1τ)γw¯ϕ]ζτw¯θ(1+β+ϕ)+(1τ)ζγw¯θ=1τ1+β+γ+ϕ(1+θr¯)w¯[ϕw¯(1+β+2γ+ϕ)τζγr¯(1+β+γ+ϕ)+ϕw¯γ]+ζw¯θ[τ(1+β+γ+ϕ)+γ]=w¯V(τ),

where

V(τ):=ϕ(1+β+2γ+ϕ)1+β+γ+ϕ(1+θr¯)w¯τ2{ϕ(1+β+3γ+ϕ)1+β+γ+ϕ(1+θr¯)w¯+ζ[θ(1+β+ϕ)γr¯]}τζγr¯+ϕγ1+β+γ+ϕ(1+θr¯)w¯.

Notice that V(1)=ζθ(1+β+ϕ)<0 and

V(0)=ζγr¯+ϕγ1+β+γ+ϕ(1+θr¯)w¯.

V(0) < 0 if and only if

r¯ζ+ϕ1+β+γ+ϕ(1+θr¯)w¯<0θ<θ^=r¯[(1+β+ϕ+γ)ζr¯ϕw¯1].

Hence, if θ < θ^, from the shape of V, Wτ(τ,0)<0 for all τ ∈ [0, 1).

Suppose θ > θ^. Then, a unique τ~ ∈ (0, 1) exists such that V(τ~) = 0 and V(τ) > 0 if and only if τ < τ~. Thus, Wτ(τ,0)>0 for all τ < τ~, as long as Equation (16) is satisfied.    ⊡

A.7 Proof of Lemma 2

Proof.

When τ = 0,

c¯y=1η1+β+γ+ϕ(1+θr¯)w¯,l¯o=1(1η)θw¯γr¯c¯y=γr¯θw¯(1+β+γ+ϕ)(1+θr¯)w¯.

Recall that n¯ is a positive solution to Equation (29), that is,

(30)ζn2[η+ϕ(1η)1+β+γ+ϕ(1+θr¯)]w¯nηw¯[θ(1+β+ϕ)γr¯]1+β+γ+ϕ=0.

Taking the derivative of W with respect to η and evaluating it at (0, η), we obtain

Wη(0,η)=1+βc¯yc¯yη=c¯y1η+ϕn¯n¯η+γl¯olo¯η=0.

From Equation (23), the second term is

(31)n¯η(0,η)=ϕ1+β+γ+ϕ(1+θr¯)+1+θ(1+β+ϕ)γr¯n¯(1+β+γ+ϕ)ζ+θ(1+β+ϕ)γr¯n¯2(1+β+γ+ϕ)w¯=ϕ(1+θr¯)n¯2+n¯2(1+β+γ+ϕ)+[θ(1+β+ϕ)γr¯]n¯ζn¯2(1+β+γ+ϕ)+[θ(1+β+ϕ)γr¯]w¯.

By applying Equation (30) to the denominator of Equation (31), we obtain

n¯η(0,η)=ϕ(1+θr¯)n¯2+n¯2(1+β+γ+ϕ)+[θ(1+β+ϕ)γr¯]n¯[η+ϕ(1η)1+β+γ+ϕ(1+θr¯)]w¯n¯(1+β+γ+ϕ)+ηw¯[θ(1+β+ϕ)γr¯]1+β+γ+ϕ(1+β+γ+ϕ)+[θ(1+β+ϕ)γr¯]w¯.

Then,

Wη(0,η)=1+β1η+ϕ2(1+θr¯)n¯+ϕn¯(1+β+γ+ϕ)+[θ(1+β+ϕ)γr¯]ϕ[η+ϕ(1η)1+β+γ+ϕ(1+θr¯)]w¯n¯(1+β+γ+ϕ)+(1+ηw¯)[θ(1+β+ϕ)γr¯]w¯.

Since the denominators of both terms are positive, the sign of Wη is determined by the sign of

(1+β)[η+ϕ(1η)1+β+γ+ϕ(1+θr¯)]w¯n¯(1+β+γ+ϕ)(1+β)(1+ηw¯)[θ(1+β+ϕ)γr¯](1η)ϕ2(1+θr¯)n¯w¯+(1η)ϕn¯(1+β+γ+ϕ)w¯+(1η)[θ(1+β+ϕ)γr¯]ϕw¯=(1+β+γ+ϕ)w¯n¯[(1+β+ϕ)η+ϕ]ϕ(1+θr¯)w¯n¯(1+β+ϕ)(1η)+[θ(1+β+ϕ)γr¯][(1+β+ϕ)η+ϕ]w¯[θ(1+β+ϕ)γr¯](1+β).

This is less than 0 if ηϕ1+β+ϕ.    ⊡

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Published Online: 2017-8-19

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