On (Non-)Neutrality of Public Debt in Growing Economies

Abstract

In this paper we analyze effects of public debt on the long-run allocation of resources in a basic endogenous growth model with infinitely lived households. The government levies an income tax and issues government bonds to finance unproductive public spending. We demonstrate that in the case of flexible wages and elastic labour supply the balanced growth rate is the higher the smaller the ratio of public debt to GDP for a given income tax rate. When wages are rigid public debt is neutral in the sense that it does not affect the allocation of resources along the balanced growth path. Finally, in both cases the economy is stable only if the government puts a sufficiently high weight on stabilizing public debt.

Appendix

1.1 Proof of Proposition 1

To prove this proposition we set $\dot{C}/C =\dot{ B}/B,$ which must hold on a BGP in the case of permanent public deficits, giving

$$\displaystyle{\phi \,\psi \,c^{-\beta /(1-\beta +\gamma )}b^{-1} =\rho -\beta \;\;\;\;\;\;(A.1)}$$

Substituting this relation in $\dot{b}/b$ gives,

$$\displaystyle{\dot{b}/b = c -\rho -\psi c^{-\beta /(1-\beta +\gamma )}((1-\tau )(1-\alpha ) +\rho \,\phi /(\rho -\beta ))}$$

From (A. 1) we know that b > 0 implies that ϕ and ρ −β have the same sign so that ϕ∕(ρ −β) > 0 holds. With this, it is easily seen that the following relations hold,

$$\displaystyle{\lim _{c\rightarrow 0}(\dot{b}/b) = -\infty,\;\;\;\lim _{c\rightarrow \infty }(\dot{b}/b) = +\infty,\;\;\;\partial (\dot{b}/b)/\partial c > 0.}$$

This proves the existence of a unique c ^⋆ which solves $\dot{b}/b = 0$.

In case of a balanced budget or a slight deficit we have b ^⋆ = 0 since public debt is constant while the capital stock grows over time or grows less than capital and, in addition, we also have ϕ = 0 (see the proof of Lemma 7). Using this, the equation $\dot{c}/c$ can be written as

$$\displaystyle{\dot{c}/c = \left (c - c^{-\alpha /((1-\alpha )+\gamma )}\psi (1-\tau )(1 - (1-\alpha ))-\rho \right ),}$$

It is easily seen that the following relations hold,

$$\displaystyle{\lim _{c\rightarrow 0}(\dot{c}/c) = -\infty,\;\;\;\lim _{c\rightarrow \infty }(\dot{c}/c) = +\infty,\;\;\;\partial (\dot{c}/c)/\partial c > 0.}$$

This proves the existence of a unique c ^⋆ which solves $\dot{c}/c = 0$. □ 

1.2 Proof of Proposition 2

Setting $\dot{C}/C =\dot{ B}/B$ implies

$$\displaystyle{-\phi \,\psi \,c^{-\beta /(1-\beta +\gamma )} -\beta \, b = -\rho \,b}$$

Substituting in −ϕ ψ c ^{−β∕(1−β+γ)} −β b by −ρ b in (17) gives,

$$\displaystyle{c -\rho -c^{-\alpha /((1-\alpha )+\gamma )}\,\psi \,(1-\tau )(1 - (1-\alpha )) -\rho \, b = 0}$$

From this we get by implicit differentiation:

$$\displaystyle{\frac{dc} {db} = \frac{\rho } {1 + (\alpha /((1-\alpha )+\gamma ))c^{-1-\alpha /((1-\alpha )+\gamma )}\psi \,(1-\tau )(1 - (1-\alpha ))} > 0}$$

Since a higher c implies a lower balanced growth rate the proposition is proven. □ 

1.3 Proof of Lemma 7

To prove that lemma we see from (10) that setting ϕ = 0 and β = (1 −τ) gives $\dot{B} = 0$ (balanced budget scenario) and setting ϕ = 0 and ρ < β < (1 −τ) gives $0 <\dot{ B}/B <\dot{ C}/C = g$ (slight deficit). Both of these debt/deficit policies imply b ^⋆ = 0 so that $\dot{b} = 0$ always holds. The balanced growth rate, then, is determined by the solution of $\dot{c} = 0$ with respect to c. Since b ^⋆ = 0 and ϕ = 0 hold both in the balanced budget scenario and for the slight deficit scenario, both scenarios imply the same c ^⋆ and, therefore, the same balanced growth rate. □ 

1.4 Proof of Proposition 3

To prove Proposition 3, we first note that the balanced budget rule (rule (i)) implies b ^⋆ = 0 since public debt is constant while the capital stock grows over time. Further, ϕ = 0 and β = (1 −τ)r hold in this case.

The Jacobian of the dynamic system (17)–(18) is given by

$$\displaystyle{J = \left [\begin{array}{cc} a_{11} & - c(1-\tau )\,\alpha \,\psi \,c^{-\alpha /(1-\alpha +\gamma )} \\ 0 & - g\\ \end{array} \right ],}$$

with c and b evaluated at the rest point {c ^⋆, 0} and a ₁₁ given by

$$\displaystyle{a_{11} = c + (1-\tau )(1-\alpha )\,\psi \,(\alpha /(1 -\alpha +\gamma ))\,c^{-\alpha /(1-\alpha +\gamma )}.}$$

The eigenvalues are a ₁₁ > 0, and − g < 0 so that the BGP is saddle point stable.

For slight deficits (rule (ii)) we also have b ^⋆ = 0 and ϕ = 0. The reaction coefficient now is ρ < β < (1 −τ)r and the Jacobian J is obtained as,

$$\displaystyle{J = \left [\begin{array}{cc} a_{11} & -\beta \, c \\ 0 & \rho -\beta \\ \end{array} \right ],}$$

with c and b evaluated at the rest point {c ^⋆, 0}, a ₁₁ as above and where we used that $\dot{C}/C =\dot{ K}/K$ holds on the BGP. Since ρ < β holds for rule (ii) one eigenvalue is negative and one is positive implying saddle point stability, which proves the first part of the proposition.

In the case of permanent deficits (rule (iii)), the Jacobian matrix evaluated at the rest point of (17)–(18). The Jacobian is given by

$$\displaystyle{J = \left [\begin{array}{cc} a_{11} & -\beta \, c \\ a_{21} & \phi \,\psi \,c^{-\alpha /(1-\alpha +\gamma )}b^{-1} -\beta \, b\\ \end{array} \right ],}$$

with c and b evaluated at the rest point {c ^⋆, b ^⋆} and a ₁₁ and a ₂₁ given by

$$\displaystyle\begin{array}{rcl} a_{11}& =& c\left (1 + (\alpha /(1 -\alpha +\gamma ))\psi \,c^{-1-\alpha /(1-\alpha +\gamma )}(\phi +(1-\tau )(1-\alpha )\right ) {}\\ a_{21}& =& b\left (1 + (\alpha /(1 -\alpha +\gamma ))\psi \,c^{-1-\alpha /(1-\alpha +\gamma )}(\phi (1 + b^{-1}) + (1-\tau )(1-\alpha ))\right ) {}\\ \end{array}$$

The determinant of the Jacobian matrix can be computed as

$$\displaystyle\begin{array}{rcl} \det J& =& (\rho -\beta )\left (c + (\alpha /(1 -\alpha +\gamma ))\psi \,c^{-\alpha /(1-\alpha +\gamma )}(1-\tau )(1-\alpha )\right ) + {}\\ & & \rho \,(\alpha /(1 -\alpha +\gamma ))\phi \,\psi \,c^{-\alpha /(1-\alpha +\gamma )} {}\\ \end{array}$$

From $\dot{C}/C =\dot{ B}/B,$ which must hold on a BGP, we have ϕ ψ c ^{−α∕(1−α+γ)} b ⁻¹ = ρ −β. Using this we can rewrite the determinant as follows,

$$\displaystyle\begin{array}{rcl} \det J& =& (\rho -\beta )\left (c + (\alpha /(1 -\alpha +\gamma ))\psi \,c^{-\alpha /(1-\alpha +\gamma )}(1-\tau )(1-\alpha )\right ) + {}\\ & & \rho \,(\alpha /(1 -\alpha +\gamma ))(\rho -\beta )\,b {}\\ \end{array}$$

For β > ρ the determinant is negative since b > 0 holds.

For β < ρ the determinant is positive. To show that the BGP is unstable we have to compute the trace of the Jacobian, tr J, which is given by,

$$\displaystyle\begin{array}{rcl} tr\:J& =& c -\beta \, b + (\alpha /(1 -\alpha +\gamma ))\psi \,c^{-\alpha /(1-\alpha +\gamma )}(\phi +(1-\tau )(1-\alpha )) {}\\ & & +\phi \,\psi b^{-1}c^{-\alpha /(1-\alpha +\gamma )}. {}\\ \end{array}$$

To see that tr J is positive we first note that a positive value of b implies ϕ > 0 for β < ρ. Further, from $\dot{c}/c = 0$ we get c −β b = ρ + c ^{−α∕(1−α+γ)}(ϕ + (1 −τ)(1 −α)) > 0, so that the trace of the Jacobian is positive, too. Since the trace and the determinant are both positive, the BGP is unstable for β < ρ.

Finally, we note that on the BGP we have from (14) the relation ρ = (1 −τ)r − g. Thus, Proposition 3 is proven. □ 

1.5 Proof of Proposition 4

To prove that proposition we set $\dot{x}/x = 0$ and solve that equation with respect to c giving

$$\displaystyle{c =\theta -\theta \,\frac{(1-\tau )L^{d}/L +\kappa (L - L^{d})/L} {(1-\tau )\left (1 -\gamma (1-\alpha )\right )} +\beta \, b + (\phi -\tau )(L^{d})^{\alpha } + (L^{d})^{\alpha } -\delta +\kappa (L - L^{d})\,x,}$$

with L ^d = α ^1∕(1−α) x ^1∕(α−1). Inserting that c in $\dot{c}$ leads to the following equation that we denote by f:

$$\displaystyle\begin{array}{rcl} f& =& (1-\alpha )(1-\tau )\,\alpha ^{\alpha /(1-\alpha )}x^{\alpha /(\alpha -1)} -\left ( \frac{\theta (1-\tau )-\theta \kappa } {(1-\tau )(1 -\gamma (1-\alpha ))}\right )\alpha x^{1/(\alpha -1)}/L - (\rho +\delta ) + {}\\ & & \theta (1 -\kappa /((1-\tau )(1 -\gamma (1-\alpha )))). {}\\ \end{array}$$

A solution f = 0 with respect to x gives a rest point of (31)–(33). As regards f we have

$$\displaystyle{\lim _{x\rightarrow 0}f = -\infty,\;\lim _{x\rightarrow \infty }f = -(\rho +\delta ) +\theta \left (1 - \frac{\kappa } {(1-\tau )(1 -\gamma (1-\alpha ))}\right ).}$$

The first derivative of f is given by

$$\displaystyle\begin{array}{rcl} \frac{\partial f} {\partial x}& =& -\left ( \frac{\alpha } {1-\alpha }\right )x^{-1+\alpha /(\alpha -1)}\alpha ^{\alpha /(1-\alpha )}(1-\alpha )(1-\tau ) + {}\\ & & \left ( \frac{1} {1-\alpha }\right )x^{-1+1/(\alpha -1)}\left ( \frac{\theta (1-\tau )-\theta \kappa } {(1-\tau )(1 -\gamma (1-\alpha ))}\right )\alpha /L. {}\\ \end{array}$$

The second derivative of f is

$$\displaystyle\begin{array}{rcl} \frac{\partial ^{2}f} {\partial x^{2}}& =& \frac{\alpha } {(1-\alpha )^{2}}\,x^{-2+\alpha /(\alpha -1)}\alpha ^{\alpha /(1-\alpha )}(1-\alpha )(1-\tau ) + {}\\ & & (-1)\, \frac{(2-\alpha )} {(1-\alpha )^{2}}\,x^{-2+1/(\alpha -1)}\left ( \frac{\theta (1-\tau )-\theta \kappa } {(1-\tau )(1 -\gamma (1-\alpha ))}\right )\alpha /L. {}\\ \end{array}$$

Setting ∂ f∕∂ x = 0 gives

$$\displaystyle{x = x_{m} = \frac{\alpha } {\alpha \,L\,\alpha ^{\alpha /(1-\alpha )}(1-\alpha )(1-\tau )}\left ( \frac{\theta (1-\tau )-\theta \kappa } {(1-\tau )(1 -\gamma (1-\alpha ))}\right ).}$$

Inserting x _m in ∂ ² f∕∂ x ² shows that the sign of the resulting expression is equivalent to

$$\displaystyle{-\alpha \,\alpha ^{\alpha /(1-\alpha )}(1-\alpha )(1-\tau ) < 0.}$$

This demonstrates that the function f reaches a maximum for x = x _m and it has a unique turning point given by

$$\displaystyle{x = x_{w} = x_{m} \cdot \left (1 + (1-\alpha )\right ).}$$

Thus, the function f is concave-convex, starts at −∞, reaches a maximum at x = x _m , has a turning point at x = x _w and converges to − (ρ +δ) +θ(1 −κ∕((1 −τ)(1 −γ(1 −α)))) for x → ∞. This implies that for − (ρ +δ) +θ(1 −κ∕((1 −τ)(1 −γ(1 −α)))) > 0 there exists a unique rest point of (31)–(33) and for − (ρ +δ) +θ(1 −κ∕((1 −τ)(1 −γ(1 −α)))) < 0 there exist two rest points for Eqs. (31)–(33) or no rest point if f does not intersect the horizontal axis. □ 

1.6 Proof of Proposition 5

We know that a BGP is given for a value of x such that the function f in Proposition 4 equals zero. Looking at f it is immediately seen that this function does neither depend on the ratio of public debt to capital, b, nor on the parameters β and ϕ. □ 

1.7 Proof of Proposition 6

We compute the Jacobian matrix evaluated at the rest point of (31)–(33). For the balanced budget (rule (i)) and for the slight deficit (rule (ii)), the Jacobian is given by

$$\displaystyle{J = \left [\begin{array}{ccc} c &c\left (\partial (\dot{C}/C)/\partial x - \partial (\dot{K}/K)/\partial x\right ) & -\beta c \\ x&x\left (\partial (\dot{w}/w)/\partial x - \partial (\dot{K}/K)/\partial x\right )& -\beta x \\ 0 & 0 & a_{33} \end{array} \right ],}$$

where c and x evaluated at the BGP and where we have used that b = 0 holds on the BGP with a balanced government budget or a slight deficit. In case of a balanced budget we have $a_{33} = -\dot{K}/K$ and for the slight deficit a ₃₃ = ρ −β < 0, since the slight deficit is obtained for 0 < ρ < β < (1 −τ)r. The eigenvalues of that matrix are given by $-\dot{K}/K = -g < 0$ (balanced budget) and by ρ −β < 0 (slight deficit) and by the eigenvalues of the matrix J ₁ which is

$$\displaystyle{J_{1} = \left [\begin{array}{ccc} c &c\left (\partial (\dot{C}/C)/\partial x - \partial (\dot{K}/K)/\partial x\right )\\ x &x\left (\partial (\dot{w}/w)/\partial x - \partial (\dot{K}/K)/\partial x \right ) \end{array} \right ]}$$

The determinant of that matrix is obtained as

$$\displaystyle{\det J_{1} = cx\left (\partial (\dot{w}/w)/\partial x - \partial (\dot{C}/C)/\partial x\right ) = (-1)\,c\,x\,(\partial f/\partial x),}$$

with f from the proof of Proposition 5. With a unique rest point of (31)–(33), f has a positive derivative at f = 0 implying that the determinant is negative so that J ₁ has one negative and one positive eigenvalue.

In case of permanent deficits (rule (iii)) we have

$$\displaystyle{J_{2} = \left [\begin{array}{ccc} c &c\left (\partial (\dot{C}/C)/\partial x - \partial (\dot{K}/K)/\partial x\right ) & -\beta c \\ x&x\left (\partial (\dot{w}/w)/\partial x - \partial (\dot{K}/K)/\partial x\right )& -\beta x \\ b &b\left (\partial (\dot{B}/B)/\partial x - \partial (\dot{K}/K)/\partial x\right )& -\beta b + (\rho -\beta ) \end{array} \right ],}$$

The determinant is given by

$$\displaystyle{\det J_{2} = (-1)\,c\,x\,(\rho -\beta )\,(\partial f/\partial x)}$$

and the trace, tr J ₂, is

$$\displaystyle{\mbox{ tr }\:J_{2} = c+x\left (\partial (\dot{w}/w)/\partial x - \partial (\dot{K}/K)/\partial x\right )+(-1)(\beta b-(\rho -\beta )) = C_{1}-(\beta -\rho ),\;\;\;(A.1)}$$

with C ₁ containing terms that are independent of β and independent of c and b that are determined by β on the BGP.

Setting β = 0 we can explicitly compute the eigenvalues ev _i , i = 1, 2, 3 as

$$\displaystyle{ev_{1} =\rho,\;ev_{2,3} = (1/2)\,\left (\mbox{ tr}\:J_{1} \pm \sqrt{\left (\mbox{ tr} \:J_{1 } \right ) ^{2 } - 4\det J_{1}}\right ),}$$

with J ₁ as for the balanced budget that has one positive and one negative eigenvalues. The determinant of the Jacobian J ₂ is negative in this case. This shows that two eigenvalues of J ₂ are positive and one is negative. For reasons of continuity this also holds in an ε environment around β = 0.

If β is sufficiently large, i.e. at least larger than ρ, the determinant of the Jacobian J ₂ is positive, since (∂ f∕∂ x) > 0. Because of detJ ₂ = ev ₁ ⋅ ev ₂ ⋅ ev ₃ we know that in this case there is no negative eigenvalue or two negative eigenvalues (or eigenvalues with negative real parts in case of complex conjugate eigenvalues). Further, we know that we have for the trace, tr J ₂ = ev ₁ + ev ₂ + ev ₃. From (A. 1) we see that tr J ₂ monotonically declines with β so that it becomes negative for a sufficiently large β implying that there is at least one negative eigenvalue. But, because of detJ ₂ > 0 there must be two negative eigenvalues in that case. □