Existence, multiplicity and policy prescriptions for debt sustainability in an OLG model with fiscal policy and debt

Abstract

We consider the overlapping generation model formulated in Dioikitopoulos (J Econ Dyn Control 93:260–276, 2018). Its innovative approach involves endogenous adaptations of the deficit/surplus to debt and income levels through an empirically estimated fiscal policy rule. We improve the analysis of the model in order to tackle the issue of debt sustainability. In detail, we derive the analytic expressions of stationary states of the model as well as necessary and sufficient existence conditions.Moreover, our mathematical analysis sheds new light on the role of fiscal parameters and policy prescriptions. As a result, low deficit levels can be associated with the presence of steady-state configurations, which is a prerequisite for sustainable economic patterns to be achieved.

Appendices

Appendix A: Proof of Proposition 1

Null variations in state variables occur at values of k and B that satisfy the following system of equations

$$\begin{aligned} \left\{ \begin{array}{l} (s(1-\alpha )+s(b-\gamma )-b)y(k) - k + (a(1-s)-R(k))B = 0\\ (R(k)-1-a)B + by(k) = 0 \end{array}\right. \end{aligned}$$

(A.18)

Step 1. Stationary states.

Consider the cases $a(1-s)-R(k)\ne 0$, $R(k)-1-a\ne 0$ and $y(k)\ne 0$. Values of B satisfying the first and the second equation in system (A.18) can be expressed in terms of k, respectively, as

$$\begin{aligned} \phi (k) = \dfrac{(s(1-\alpha )+s(b-\gamma )-b)y(k)-k}{R(k) -a(1-s)},\;\;\psi (k) = \dfrac{by(k)}{1+a-R(k)} \end{aligned}$$

Observing that $y(k) = kR(k)/\alpha$, the condition $\phi (k) = \psi (k)$ leads to

$$\begin{aligned} \dfrac{s(1-\alpha ) + s(b-\gamma )-b}{b} = \dfrac{\alpha }{bR(k)} + \dfrac{R(k)-a(1-s)}{1+a-R(k)} \end{aligned}$$

(A.19)

This equation can be rewritten in terms of the second-order polynomial Z in the variable $r = R(k)$ as

$$\begin{aligned} Z(r)&:=r^2(s(1-\alpha -\gamma ) + bs) +\nonumber \\&\quad -r(\alpha +s(1-\alpha -\gamma )(1+a)-b(1-s))+ \alpha (1+a) = 0 \end{aligned}$$

with discriminant

$$\begin{aligned} \Delta&=(\alpha +s(1-\alpha -\gamma )(1+a)-b(1-s))^2-4\alpha (1+a) (s(1-\alpha -\gamma ) + bs)\\&=\delta _1^2-\delta _2 \end{aligned}$$

and with roots

$$\begin{aligned} r_{\pm }&= \dfrac{1}{2(s(1-\alpha -\gamma ) + bs)}\Big (\alpha +s(1-\alpha -\gamma ) (1+a)-b(1-s)\pm \nonumber \\&\quad \pm \sqrt{(\alpha +s(1-\alpha -\gamma )(1+a)-b(1-s))^2-4\alpha (1+a) (s(1-\alpha -\gamma ) + bs)}\Big )\nonumber \\&=\dfrac{1}{2(s(1-\alpha -\gamma ) + bs)}\left( \delta _1 \pm \sqrt{\delta _1^2-\delta _2}\right) \end{aligned}$$

(A.20)

Hence, provided that inverses $R^{-1}(r_+)$ and $R^{-1}(r_-)$ exist, equation (A.19) is satisfied both at $k_+:=R^{-1}(r_+)$ and $k_-:=R^{-1}(r_-)$, where

$$\begin{aligned} k_{\pm } = \left( \dfrac{r_{\pm }}{\alpha A(\gamma A)^{\frac{\gamma }{1-\gamma }}} \right) ^{-\frac{1-\gamma }{1-\alpha -\gamma }} \end{aligned}$$

We then conclude that solutions of system (A.18), when existing, are the pairs $(k_-,B_-)$ and $(k_+,B_+)$, with $B_{+}:=\psi (k_{+})$ and $B_{-}:=\psi (k_{-})$, explicitly given by

$$\begin{aligned} B_{\pm } = \dfrac{bk_{\pm }r_{\pm }}{\alpha (1+a-r_{\pm })} \end{aligned}$$

(A.21)

To conclude Step 1, let us consider that $a(1-s)-R(k)= 0$ holds if and only if $k = k'$, where

$$\begin{aligned} k'=\left( \dfrac{a(1-s)}{\alpha A(\gamma A)^{\frac{\gamma }{1-\gamma }}} \right) ^{-\frac{1-\gamma }{1-\alpha -\gamma }} \end{aligned}$$

However, $k'$ solves the first equation of the system (A.18) whenever $a(1-s)(s(1-\alpha -\gamma )-b(1-s)) = \alpha$. Hence, the system (A.18) has no solution if $a(1-s)(s(1-\alpha -\gamma )-b(1-s))\ne \alpha$.

Finally, we mention that, in both cases $y(k)=0$ and $R(k)-1-a= 0$, the system (A.18) has no solution.

Step 2. Existence and positiveness of $k_{+}$ and $k_-$.

Values $k_{+}$ and $k_-$ exist and are positive if and only if $r_-$ and $r_{+}$ exist and are positive, respectively. Both $r_+$ and $r_-$ exist provided that $\Delta = \delta _1^2-\delta _2\ge 0$ or, equivalently, if either $\delta _1\le -\sqrt{\delta _2}$ or $\sqrt{\delta _2}\le \delta _1$. Moreover, by noting that $0\notin [r_-,r_+]$, being the value of the convex parabola Z at $r = 0$ positive (i.e., $Z(0) = \alpha (1+a)>0$), either $r_-,r_+<0$ or $r_-,r_+>0$ hold. Hence, since $r_-\le r_+$, the conditions $r_->0$ and $r_+>0$ reduce to the unique $r_->0$. Taking into account that $\delta _2>0$, it is easy to verify that $r_->0$ together with $\Delta \ge 0$ are equivalent to $\sqrt{\delta _2}\le \delta _1$.

Step 3. Positiveness of $B_{\pm }$

Provided that $k_{+}$ and $k_-$ exist and are positive, the values $B_+ = \phi (k_+)$ and $B_{_-} = \psi (k_-)$ exist as well. Moreover, since under the same assumption $r_+,r_->0$ also hold, from expressions (A.21) it follows that $B_+$ and $B_{-}$ are positive if and only if $r_{+}<1+a$ and $r_-<1+a$, respectively. By noting that $1+a\notin [r_-,r_+]$, being the value of the convex parabola Z at $r = 1+a$ positive (i.e., $Z(1+a) = b(1+as)(1+a)>0$), either $r_{-},r_+<1+a$ or $r_{-},r_+>1+a$. Hence, conditions $r_{-},r_+<1+a$ reduce to the unique $r_+<1+a$.

Let us show that $r_+<1+a$ if and only if $\delta _1<2(s(1-\alpha -\gamma )+bs)(1+a)$. Assume $r_+<1+a$. This relation can be rewritten, using (A.20), as

$$\begin{aligned} \sqrt{\delta _1^2-\delta _2}<2(s(1-\alpha -\gamma )+bs)(1+a)-\delta _1 \end{aligned}$$

which implies $0<2(s(1-\alpha -\gamma )+bs)(1+a)-\delta _1$. Conversely, assume $\delta _1<2(s(1-\alpha -\gamma )+bs)(1+a)$. This relation can be rewritten as

$$\begin{aligned} \dfrac{\delta _1}{2(s(1-\alpha -\gamma )+bs)}<1+a \end{aligned}$$

since $0<s(1-\alpha -\gamma )+bs$. Hence, it holds

$$\begin{aligned} r_-&=\dfrac{\delta _1}{2(s(1-\alpha -\gamma )+bs)} -\dfrac{\sqrt{\delta _1^2-\delta _2}}{2(s(1-\alpha -\gamma )+bs)}\nonumber \\&<1+a-\dfrac{\sqrt{\delta _1^2-\delta _2}}{2(s(1-\alpha -\gamma )+bs)}\nonumber \\&<1+a \end{aligned}$$

implying $r_+<1+a$, since $1+a\notin [r_-,r_+]$. By noting that $2(s(1-\alpha -\gamma )+bs)(1+a) = \delta _2/(2\alpha )$, this completes the proof of Proposition 1.

Appendix B: Proof of Corollary 1

Part (i).

Step 1. Values of a for which the stationary $E_+$ and $E_-$ exist.

Values of a for which the condition $\sqrt{\delta _2}\le \delta _1$ is satisfied are those such that $\delta _1>0$ and $\delta _2\le \delta _1^2$. Moreover, the equation $\delta _2=\delta _1^2$ can be rewritten as

$$\begin{aligned} X(1+a)&=s^2(1-\alpha -\gamma )^2(1+a)^2+\\&\quad -2s(1+a)(\alpha (1+2b-\alpha -\gamma )\\&\quad +b(1-s) (1-\alpha -\gamma ))+(\alpha -b(1-s))^2 = 0 \end{aligned}$$

where X is a second-order polynomial in the variable $1+a$. Polynomial X has strictly positive discriminant

$$\begin{aligned} \Delta _a&:=4s^2\Big (\left( \alpha (1+2b-\alpha -\gamma ) +b(1-s)(1-\alpha -\gamma )\right) ^2\\&\quad -(1-\alpha -\gamma )^2(\alpha -b(1-s))^2\Big )\\&=16s^2\alpha b\Big (\alpha b + (1-s)(1-\alpha -\gamma )^2 +(\alpha +b(1-s))(1-\alpha -\gamma )\Big ) \end{aligned}$$

and roots given by $a_{+}+1$ and $a_-+1$, where $a_+$ and $a_-$ are reported in the statement of Corollary 1. Values $a_{+}$ and $a_-$ are both acceptable solutions for the equation $\sqrt{\delta _2} = \delta _1$ expressed in terms of a. Indeed, it can be easily verified that $\delta _1>0$ holds both at $a = a_-$ and $a = a_+$. In addition, the function $\delta _1-\sqrt{\delta _2}$ is convex w.r.t. a, being

$$\begin{aligned} \dfrac{d^2}{da^2}(\delta _1-\sqrt{\delta _2}) = \dfrac{\sqrt{\alpha (s (1-\alpha - \gamma )+bs)}}{2(1+a)^{3/2}}>0 \end{aligned}$$

Hence, relation $\delta _1-\sqrt{\delta _2}\ge 0$, which ensures the existence of stationary states $E_+$ and $E_-$, holds if and only if $a\notin (a_-,a_+)$.

Step 2. Values of a for which stationary states $E_+$ and $E_-$ exist and are componentwise positive.

In order to find values of a for which relation $\delta _1<\delta _2/(2\alpha )$ holds, let us rewrite it in terms of the parameter a as

$$\begin{aligned} a>\dfrac{\alpha -b(1-s)}{s(1+2b-\alpha -\gamma )}-1:=a_0 \end{aligned}$$

Through algebraic manipulations, it can be verified that $a_0<a_+$. Indeed, by multiplying both terms $a_0$ and $a_+$ by the positive term $1+2b-\alpha -\gamma$, relation $a_0<a_+$ is equivalent to

$$\begin{aligned} \alpha -b(1-s)&<\alpha \left( \dfrac{1+2b-\alpha -\gamma }{1-\alpha -\gamma }\right) ^2 +b(1-s)\dfrac{1+2b-\alpha -\gamma }{1-\alpha -\gamma }\\&\quad +\dfrac{1+2b-\alpha -\gamma }{(1-\alpha -\gamma )^2}\sqrt{\Delta _a} \end{aligned}$$

A further step forward leads to

$$\begin{aligned} \alpha \left( 1-\left( \dfrac{1+2b-\alpha -\gamma }{1-\alpha -\gamma }\right) ^2\right)&-b(1-s)\left( 1+\dfrac{1+2b-\alpha -\gamma }{1-\alpha -\gamma }\right) \\&\quad <\dfrac{1+2b-\alpha -\gamma }{(1-\alpha -\gamma )^2}\sqrt{\Delta _a} \end{aligned}$$

which is always satisfied since $1<\frac{1+2b-\alpha -\gamma }{1-\alpha -\gamma }$ and $\Delta _a\ge 0$.

To conclude, we proved that $\sqrt{\delta _2}\le \delta _1$ is equivalent to $a\notin (a_-,a_+)$ and $\delta _1<\delta _2/(2\alpha )$ is equivalent to $a_0< a$. Hence, considering that $a_0<a_+$, the two relations hold together if and only if $a_+\le a$. This completes the proof of Part (i) of Corollary 1.

Part (ii).

Step 1. Values of b for which the stationary $E_+$ and $E_-$ exist. Taking into account that $\delta _2> 0$, values of b for which the condition $\sqrt{\delta _2}\le \delta _1$ is satisfied are those such that $0<\delta _1$ and $\delta _2\le \delta _1^2$ hold. In particular, the relation $0<\delta _1$, expressed in terms of parameter b, can be rewritten as

$$\begin{aligned} b<b_0:=\dfrac{\alpha +s(1-\alpha -\gamma )(1+a)}{1-s} \end{aligned}$$

while relation $\delta _2\le \delta _1^2$ can be rearranged in the following equivalent form

$$\begin{aligned} Z(b)&= (1-s)^2b^2 - (2\alpha (1-s) + 2s(1-\alpha -\gamma )(1-s)(1+a)+4\alpha s(1+a))\\&\quad +(\alpha -s(1-\alpha -\gamma )(1+a))^2 \ge 0 \end{aligned}$$

where Z is a second-order polynomial in b with strictly positive discriminant and with roots

$$\begin{aligned} b_{\pm }&= \dfrac{\alpha }{1-s}+\dfrac{s(1-\alpha -\gamma )(1+a)}{1-s} +\dfrac{2\alpha s(1+a)}{(1-s)^2}\pm \\&\quad \pm \dfrac{2}{(1-s)^2}\sqrt{\alpha s(1+a)(1+as)(\alpha +(1-s)(1-\alpha -\gamma ))} \end{aligned}$$

Hence, $Z(b)\ge 0$ holds whenever $b\le b_-$ or $b_+\le b$. However, only the values of b such that $b\le b_-$ are acceptable for the relation $\sqrt{\delta _2}\le \delta _1$ to be satisfied. Indeed, let us show that the occurrence of $b_+\le b$ implies $\delta _1< 0$:

$$\begin{aligned} \delta _1&= \alpha +s(1-\alpha -\gamma )(1+a)-b(1-s)\\&\le \alpha +s(1-\alpha -\gamma )(1+a)-b_+(1-s)\\&= -\dfrac{2\alpha s(1+a)}{1-s}-\dfrac{2}{(1-s)}\sqrt{...}< 0 \end{aligned}$$

Differently, relation $b_-<b_0$ always holds. To show this, let us rearrange that relation as

$$\begin{aligned} \dfrac{2\alpha s(1+a)}{(1-s)^2}-\dfrac{2}{(1-s)^2} \sqrt{\alpha s(1+a)(1+as)(\alpha +(1-s)(1-\alpha -\gamma ))}<0 \end{aligned}$$

or, equivalently, as

$$\begin{aligned} \alpha s(1+a)<\sqrt{\alpha s(1+a)(1+as)(\alpha +(1-s)(1-\alpha -\gamma ))} \end{aligned}$$

Since both terms in this last relation are positive, we obtain equivalent relation by rising both terms to square. Then, we get

$$\begin{aligned} \alpha ^2s^2(1+a)^2<\alpha s(1+a)(1+as)(\alpha +(1-s)(1-\alpha -\gamma )) \end{aligned}$$

that leads to

$$\begin{aligned} -\alpha <(1+as)(1-\alpha -\gamma ) \end{aligned}$$

which is always satisfied. To conclude, we proved that relation $b<b_-$ is necessary and sufficient condition for stationary states $E_-$ and $E_+$ to exist. The threshold $b_-$ is denoted by $b_{\max }$ in the statement of the corollary.

Step 2. Values of b for which the stationary $E_+$ and $E_-$ exist and are componentwise positive.

In order to find values of b for which the condition $\delta _1<\delta _2/(2\alpha )$ holds, let us rewrite it in terms of the parameter b as

$$\begin{aligned} b>b_{\min }:=\dfrac{\alpha -s(1-\alpha -\gamma )(1+a)}{1-s+2 s(1+a)} \end{aligned}$$

To conclude, we proved that $\sqrt{\delta _2}\le \delta _1$ is equivalent to $b\le b_-$ and $\delta _1<\delta _2/(2\alpha )$ is equivalent to $b>b_{\min }$. Hence, the two relations hold together if and only if $b_{\min }<b\le b_-$ and this completes the proof of Part (ii) of Corollary 1.

Appendix C: Comparison of thresholds from Corollary 1 and Proposition 1 in Dioikitopoulos (2018)

The existence condition provided in Dioikitopoulos (2018) is $b<s(1-\alpha -\gamma )/(1-s):=b_{\max }^D$ and $a<\alpha /(1-s)(s(1-\alpha -\gamma )-(1-s)b)$. Note that the threshold $b_{\max }^D$ is constant with respect to variations in a. The condition can be rewritten as

$$\begin{aligned} b\in [b_{\min }^D,b_{\max }^D] \end{aligned}$$

(C.22)

where

$$\begin{aligned} b_{\min }^D =\max \left\{ 0,b_{\max }^D-\dfrac{\alpha }{a(1-s)^2}\right\} \end{aligned}$$

The existence condition (C.22) is satisfied at small values of a, provided that capital sensitivity b satisfies $b<b_{\max }^D$, since $b_{\min }\longrightarrow 0$ as $a\longrightarrow 0$. Moreover, for any $a>0$ the interval $[b_{\min }^D,b_{\max }^D]$ is nonempty, being the relation $b_{\min }^D<b_{\max }^D$ always satisfied. On the one hand, this implies that, if $b>b_{\max }^D$, no $a>0$ exists for which existence condition is fulfilled. On the other hand, for any $a>0$, there always exist $b\in [b_{\min }^D,b_{\max }^D]$ for which condition (C.22) is fulfilled. We refer the reader to Fig. 4 for an exemplary graphical comparison between different existence regions in the parameter space.