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Explicit formula for the optimal government debt ceiling

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Abstract

Motivated by the current debt crisis in the world, we develop a stochastic debt control model to study the optimal government debt ceiling, or equivalently the optimal ceiling for government debt. We consider a government that wants to control its debt by imposing an upper bound or ceiling on its debt-to-GDP ratio. We assume that debt generates a cost for the country, and this cost is an increasing and convex function of debt ratio. The government can intervene to reduce its debt ratio, but there is a cost generated by this reduction. The goal of the government is to find the optimal control that minimizes the expected total cost. We obtain an explicit solution for the government debt problem, that gives an explicit formula for optimal government debt ceiling. Moreover, we derive a rule for optimal debt policy in terms of the optimal government debt ceiling. In an extension of the model, we find that countries with a strong positive link between debt and economic growth should have a high optimal debt ceiling. This paper provides the first theoretical model for the optimal government debt ceiling.

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Notes

  1. For the sake of further clarification, we provide the definition of other terms related to government debt (but different from “debt ceiling”). “Optimal debt” is the level of debt that comes as a result of a welfare analysis that considers both the benefits and costs of increasing debt. On the one hand, government debt enhances the liquidity of households by providing an additional manner of smoothing consumption. On the other hand, government debt, paid by future taxation, may have adverse wealth distribution and incentive effects. See, for example, the study of optimal debt by Aiyagari and McGrattan (1998) and Barro (1999). “Credit ceiling” (or “credit limit”) is the maximum level of debt that a country is allowed to borrow. This level is imposed by lenders based on the characteristics of the country, such as its history of default. For a reference of this term, see Eaton and Gersovitz (1981). “Debt limit” is the level of debt at which a debt crisis occurs. In particular, at this level of debt the debt service payments are unsustainable (see, for example, Ostry et al. (2010) and Stein (2006) ). Thus, the “debt limit” is strictly larger than the “debt ceiling”. Another difference between these two terms is that a government can select its “debt ceiling”, but a government cannot select its “debt limit”. Summarizing, “optimal debt”, “credit ceiling”, “debt limit”, and “optimal debt ceiling” are different terms.

  2. See Lima et al. (2008) for an empirical work with a definition of debt ceiling different from ours.

  3. Recently Égert (2015) shows that the results of Reinhart and Rogoff (2010) are extremely sensitive to the time dimension and country coverage.

  4. We point out that in a paper that studies the link between fiscal and monetary policies, Sargent and Wallace (1981) consider \(r > g\), that is, \(\mu >0\).

  5. In contrast to the study of the exchange rate control by Cadenillas and Zapatero (1999), we have not found compelling evidence of the existence of fixed costs in our problem.

  6. For references on models that consider default, but do not study the debt ceiling, see Eaton and Gersovitz (1981), Aguiar and Giponath (2006), Arellano (2008), and Mendoza and Yue (2012).

  7. Fleming and Stein (2004), Stein (2006), and Stein (2012) present a classical stochastic control model for the ratio of debt to net worth (where net worth is defined as the difference between capital and debt), with the goal of providing a technique to predict debt crisis.

  8. This is known as the “Skorokhod problem” (see Skorokhod 1961) in the stochastic analysis literature.

  9. Romer (1986) develops a general equilibrium version of Baumol (1952) and Tobin (1956).

  10. See Panizza (2008) for a study regarding domestic and external public debt in developing countries.

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Acknowledgments

We are grateful to the Associate Editor and the Reviewers for their valuable comments. Preliminary versions of this paper have been presented at the Advanced Finance and Stochastics 2013 conference organized by the Steklov Mathematical Institute, EURO-INFORMS 26th European Conference on Operational Research Rome, INFORMS Annual Meeting 2013 Minnesota, INFORMS Applied Probability Society Conference 2013 Costa Rica, Second International Conference on Probability and Statistics organized by the Catholic University of Perú, Quantitative Finance Seminar of the Fields Institute for Research in Mathematical Sciences, Eight World Congress of the Bachelier Finance Society 2014 Brussels, SIAM Conference on Financial Mathematics and Engineering 2014 Chicago, Stern School of Business, National University of Singapore, University of Alberta, and Ajou University. We are grateful for comments from participants at these conferences and seminars. Existing errors are our sole responsibility. This research was supported by the Social Sciences and Humanities Research Council of Canada. The work of A. Cadenillas was also supported by the World Class University (WCU) program through the Korea Science and Engineering Foundation funded by the Ministry of Education, Science and Technology (R31-2009-000-20007).

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Correspondence to Abel Cadenillas.

Appendices

Appendix 1: Proof of Proposition 3.1

Proof

The value function V is nonnegative because J is nonnegative.

Consider \(x_{1} < x_{2}\) with corresponding controls \(Z^{(1)} \in \mathcal{A}(x_{1} )\) and \(Z^{(2)} \in \mathcal{A}(x_{2} )\). Since h is an strictly increasing function,

$$\begin{aligned} V(x_{1} ) \le J\left( x_{1}, Z^{(2)}\right) < J\left( x_{2}, Z^{(2)}\right) . \end{aligned}$$

Hence \( V(x_{1} ) \le V(x_{2} )\). Thus, V is increasing.

Consider \(x_{1} \le x_{2}\) with corresponding controls \(Z^{(1)} \in \mathcal{A}(x_{1} )\) and \(Z^{(2)} \in \mathcal{A}(x_{2} )\). Let \(\gamma \in [0,1]\). We define \(Z^{(3)}:=\gamma Z^{(1)} + (1-\gamma ) Z^{(2)}\) and \(x_{3}:=\gamma x_{1} + (1-\gamma ) x_{2}\). We denote by \(X^{(j)}\) the trajectory that starts at \(x_j\) and is determined by the control \(Z^{(j)}\), for \(j=1,2,3\). Since Eq. (1) is linear, we observe that for every \(t\ge 0\):

$$\begin{aligned} X^{(3)}_t=\gamma X^{(1)}_t + (1-\gamma ) X^{(2)}_t. \end{aligned}$$

Since h is a convex function,

$$\begin{aligned} \int _{0}^{\infty } e^{-{\uplambda }t} h \left( X^{(3)}_{t}\right) dt\le & {} \int _{0}^{\infty } e^{-{\uplambda }t} \left( \gamma h \left( X^{(1)}_{t}\right) + (1-\gamma )h\left( X^{(2)}_{t}\right) \right) dt\\= & {} \gamma \left( \int _{0}^{\infty } e^{-{\uplambda }t} h \left( X^{(1)}_{t}\right) dt\right) + (1-\gamma )\left( \int _{0}^{\infty } e^{-{\uplambda }t} h\left( X^{(2)}_{t}\right) dt \right) . \end{aligned}$$

As a consequence, we obtain

$$\begin{aligned} J\left( x_3;Z^{(3)}\right) \le \gamma J\left( x_1;Z^{(1)}\right) + (1-\gamma )J\left( x_2;Z^{(2)}\right) . \end{aligned}$$

Hence

$$\begin{aligned} V\left( \gamma x_{1} + (1-\gamma ) x_{2} \right)\le & {} J\left( \gamma x_{1} + (1-\gamma ) x_{2} , \gamma Z^{(1)} + (1-\gamma ) Z^{(2)} \right) \\= & {} J\left( x_3;Z^{(3)}\right) \\\le & {} \gamma J\left( x_{1} , Z^{(1)}\right) + (1-\gamma ) J\left( x_{2} , Z^{(2)}\right) . \end{aligned}$$

Consequently,

$$\begin{aligned} V\left( \gamma x_{1} + (1-\gamma ) x_{2} \right) \le \gamma V( x_{1} ) + (1-\gamma ) V( x_{2}), \end{aligned}$$

which shows that V is convex.

It remains to prove the last assertion. Since

$$\begin{aligned} \int _{0}^{\infty } e^{-{\uplambda }t}\beta dt \le E_{x}\left[ \int _{0}^{\infty } e^{-{\uplambda }t} \left\{ \alpha X_{t}^{2n} + \beta \right\} dt + \int _{0}^{\infty } e^{-{\uplambda }t} kdZ_t \right] , \end{aligned}$$

we have \( \frac{\beta }{{\uplambda }} \, \le \, J(x;Z)\), which yields \(\frac{\beta }{{\uplambda }}\le V(x)\). Hence \(\frac{\beta }{{\uplambda }}\le V(0+)\). On the other hand, by Example 2.1, condition (4) implies

$$\begin{aligned} V(x)\le J(x;0)=\frac{\alpha x^{2n} }{{\uplambda }-\sigma ^2 n(2n-1)-2\mu n} + \frac{\beta }{{\uplambda }}, \end{aligned}$$

from which we conclude that \(V(0+)\le \frac{\beta }{{\uplambda }}\), as required. This completes the proof. \(\square \)

Appendix 2: Proof of Lemma 4.1

We will use Lemmas 7.17.3 to prove Lemma 4.1.

We recall \(\tilde{\mu }:=\mu -\frac{1}{2}\sigma ^2\), by definition given in Eq. (13). Let us consider the constant

$$\begin{aligned} \bar{y}: =2n^2 \sigma ^2+2\tilde{\mu }n=\sigma ^2n(2n-1)+2\mu n. \end{aligned}$$

We observe that \({\uplambda }> \bar{y}\), because we are assuming condition (4).

Let us consider the function \(f:[-\frac{\tilde{\mu }^2}{2\sigma ^2} ,\infty )\rightarrow \mathbb R\) defined by

$$\begin{aligned} f(y) := \frac{\sqrt{\tilde{\mu }^2+2y\sigma ^2}}{\sigma ^2}- \frac{\tilde{\mu }}{\sigma ^2}. \end{aligned}$$

Lemma 7.1

  1. (i)

    f is strictly increasing on its domain \([-\frac{\tilde{\mu }^2}{2\sigma ^2} ,\infty )\),

  2. (ii)

    \(f(\bar{y})=2n\)    if    \(\tilde{\mu }+2n\sigma ^2 \ge 0\),

  3. (iii)

    \(f(0)>4n\)    if    \(\tilde{\mu }+2n\sigma ^2 < 0\),

  4. (iv)

    \(f(\mu )=1\)    if    \( \mu > 0\),

  5. (v)

    \(f(0)\ge 1\)    if    \(\mu \le 0\).

Proof

  1. (i)

    Taking the first derivative of f, we obtain

    $$\begin{aligned} f^{\prime }(y)=\frac{1}{\sqrt{\tilde{\mu }^2+2y\sigma ^2}}>0, \quad \text {for} \quad y\in \left( -\frac{\tilde{\mu }^2}{2\sigma ^2} ,\infty \right) . \end{aligned}$$

    This establishes the desired result.

  2. (ii)

    We have

    $$\begin{aligned} f(\bar{y})&=\frac{\sqrt{\tilde{\mu }^2+2\left( 2n^2 \sigma ^2+2\tilde{\mu }n\right) \sigma ^2}}{\sigma ^2}- \frac{\tilde{\mu }}{\sigma ^2} \\&=\frac{\sqrt{\tilde{\mu }^2+4n^2\sigma ^4+4\tilde{\mu }\sigma ^2 n}}{\sigma ^2}- \frac{\tilde{\mu }}{\sigma ^2} \\&=\frac{\sqrt{\left( \tilde{\mu }+2n\sigma ^2\right) ^2}}{\sigma ^2}- \frac{\tilde{\mu }}{\sigma ^2} \\&=\frac{\tilde{\mu }+2n\sigma ^2}{\sigma ^2}- \frac{\tilde{\mu }}{\sigma ^2} \\&=2n, \end{aligned}$$

    where we used \(\tilde{\mu }+2n\sigma ^2 \ge 0\) to get the fourth equality above.

  3. (iii)

    First of all, note that \(\tilde{\mu }+2n\sigma ^2 < 0\) implies that \(\tilde{\mu }\) is negative. To be more precise, it implies \(\tilde{\mu }<-2n\sigma ^2\), or equivalently \(-2\frac{\tilde{\mu }}{\sigma ^2}>4n\). Hence,

    $$\begin{aligned} f(0)&=\frac{\sqrt{\tilde{\mu }^2}}{\sigma ^2}- \frac{\tilde{\mu }}{\sigma ^2}=\frac{|\tilde{\mu }|}{\sigma ^2}- \frac{\tilde{\mu }}{\sigma ^2}=-2\frac{\tilde{\mu }}{\sigma ^2} > 4n. \end{aligned}$$
  4. (iv)

    We have

    $$\begin{aligned} f(\mu )&=\frac{\sqrt{\tilde{\mu }^2+2\mu \sigma ^2}}{\sigma ^2}- \frac{\tilde{\mu }}{\sigma ^2} \\&=\frac{\sqrt{\left( \mu -\frac{1}{2}\sigma ^2\right) ^2+2\mu \sigma ^2}}{\sigma ^2}- \frac{\left( \mu -\frac{1}{2}\sigma ^2\right) }{\sigma ^2} \\&=\frac{\sqrt{\left( \mu +\frac{1}{2}\sigma ^2\right) ^2}}{\sigma ^2}- \frac{\left( \mu -\frac{1}{2}\sigma ^2\right) }{\sigma ^2} \\&=\frac{\left( \mu +\frac{1}{2}\sigma ^2\right) }{\sigma ^2}- \frac{\left( \mu -\frac{1}{2}\sigma ^2\right) }{\sigma ^2}=1, \end{aligned}$$

    where we used \(\mu >0\) to obtain the fourth equality above.

  5. (v)

    If \(\mu \le 0\), then \(\tilde{\mu }\le -\frac{1}{2}\sigma ^2 \le 0\). Thus,

    $$\begin{aligned} f(0)&=\frac{\sqrt{\tilde{\mu }^2}}{\sigma ^2}- \frac{\tilde{\mu }}{\sigma ^2}=\frac{|\tilde{\mu }|}{\sigma ^2}- \frac{\tilde{\mu }}{\sigma ^2}=-2\frac{\tilde{\mu }}{\sigma ^2}=1-2\frac{\mu }{\sigma ^2} \ge 1. \end{aligned}$$

    This completes the proof.

\(\square \)

Lemma 7.2

If

  1. (i)

       \(2{\uplambda }(\sigma ^2 n +\mu )^2 \ge ({\uplambda }+ \sigma ^2 n) ({\uplambda }\sigma ^2 + 2\mu \sigma ^2n+2\mu ^2-\sigma ^2 \mu )\),

then

  1. (ii)

       \({\uplambda }\bar{y} \ge {\uplambda }({\uplambda }-\mu )+\mu \bar{y}\).

Proof

We will work with the left-hand side and the right-hand side of (i) separately. Subtracting \((2{\uplambda }\mu ^2+2{\uplambda }\mu \sigma ^2 n+{\uplambda }\sigma ^4 n)\) from the left-hand side of (i), we have

$$\begin{aligned} 2{\uplambda }\left( \sigma ^2 n +\mu \right) ^2&- \left( 2{\uplambda }\mu ^2+2{\uplambda }\mu \sigma ^2 n+{\uplambda }\sigma ^4 n\right) \\&=2{\uplambda }\left( \sigma ^4 n^2+2\mu \sigma ^2 n + \mu ^2\right) -\left( 2{\uplambda }\mu ^2+2{\uplambda }\mu \sigma ^2 n+{\uplambda }\sigma ^4 n\right) \\&=2{\uplambda }\sigma ^4 n^2+4{\uplambda }\mu \sigma ^2 n + 2{\uplambda }\mu ^2 -2{\uplambda }\mu ^2-2{\uplambda }\mu \sigma ^2 n-{\uplambda }\sigma ^4 n \\&=2{\uplambda }\sigma ^4 n^2 +2{\uplambda }\mu \sigma ^2 n- {\uplambda }\sigma ^4 n\\&= \sigma ^2 {\uplambda }\left( 2\sigma ^2 n^2+2\mu n -\sigma ^2 n\right) \\&= \sigma ^2 {\uplambda }\left( \sigma ^2 n(2n-1)+2\mu n\right) = \sigma ^2 {\uplambda }\bar{y}. \end{aligned}$$

Subtracting the same quantity from the right-hand side of (i), we obtain

$$\begin{aligned}&\left( {\uplambda }+ \sigma ^2 n\right) \left( {\uplambda }\sigma ^2 + 2 \mu \sigma ^2n+2\mu ^2-\sigma ^2 \mu \right) - \left( 2{\uplambda }\mu ^2+2{\uplambda }\mu \sigma ^2 n+{\uplambda }\sigma ^4 n\right) \\&\quad ={\uplambda }^2 \sigma ^2 + 2 {\uplambda }\mu \sigma ^2n+2{\uplambda }\mu ^2-{\uplambda }\sigma ^2 \mu + {\uplambda }\sigma ^4 n \\&\qquad +\,2 \mu \sigma ^4 n^2 +2\mu ^2 \sigma ^2 n -\mu \sigma ^4 n - \left( 2{\uplambda }\mu ^2\right. \\&\qquad \left. +\,2{\uplambda }\mu \sigma ^2 n+{\uplambda }\sigma ^4 n\right) \\&\quad = {\uplambda }^2 \sigma ^2 -{\uplambda }\mu \sigma ^2+2\mu \sigma ^4 n^2 + 2\mu ^2\sigma ^2 n - \mu \sigma ^4 n \\&\quad = {\uplambda }\sigma ^2 ({\uplambda }- \mu ) +\sigma ^2 \mu \left( 2\sigma ^2 n^2+2\mu n-\sigma ^2 n \right) \\&\quad = {\uplambda }\sigma ^2 ({\uplambda }- \mu ) +\sigma ^2 \mu \left( \sigma ^2 n(2n-1)+2\mu n \right) \\&\quad = {\uplambda }\sigma ^2 ({\uplambda }-\mu ) +\sigma ^2 \mu \bar{y} \\&\quad = \sigma ^2 \left( {\uplambda }({\uplambda }-\mu )+\mu \bar{y} \right) . \end{aligned}$$

Since we are subtracting the same quantity from both sides of inequality (i), we get the following equivalent expression

$$\begin{aligned} \sigma ^2 {\uplambda }\bar{y} \ge \sigma ^2 \left( {\uplambda }({\uplambda }-\mu )+\mu \bar{y} \right) . \end{aligned}$$

Recalling that \(\sigma >0\), we see that inequality (ii) is satisfied. \(\square \)

Lemma 7.3

Let us consider the function \(\phi : [-\frac{\tilde{\mu }^2}{2\sigma ^2},\infty ) \rightarrow \mathbb R\) defined by

$$\begin{aligned} \phi ( p ) := 2 p-\tilde{\mu }-\sqrt{\tilde{\mu }^2+2p\sigma ^2}. \end{aligned}$$

Then,

  1. (i)

    \(\phi \) is strictly increasing on the interval \([\max \{\mu ,0\}, \infty )\),

  2. (ii)

    \(\phi (\mu )=0\)    if   \(\mu >0\),

  3. (iii)

    \(\phi (0)=0\)    if   \(\mu \le 0\).

Proof

Before proceeding with the specific proof of (i)–(iii), we observe that for every \(p\in \left( -\frac{\tilde{\mu }^2}{2\sigma ^2},\infty \right) \):

$$\begin{aligned} \phi ^{\prime }( p )= & {} 2-\frac{\sigma ^2}{\left( \tilde{\mu }^2+2p\sigma ^2\right) ^{\frac{1}{2}}} \\ \phi ^{\prime \prime }( p )= & {} \frac{\sigma ^4}{\left( \tilde{\mu }^2+2p\sigma ^2\right) ^{\frac{3}{2}}}>0. \end{aligned}$$

Since \(\phi ^{\prime \prime }>0\), we conclude that \(\phi ^{\prime }\) is strictly increasing on the open interval above. We notice that \(\tilde{\mu }^2+2\mu \sigma ^2= (\mu -\frac{1}{2}\sigma )^2 + 2\mu \sigma ^2 =(\mu +\frac{1}{2}\sigma )^2\). If \(\mu >0\) we obtain

$$\begin{aligned} \phi ^{\prime }(\mu )=2-\frac{\sigma ^2}{\left( \tilde{\mu }^2+2\mu \sigma ^2 \right) ^\frac{1}{2}}=2-\frac{\sigma ^2}{\mu +\frac{1}{2}\sigma ^2}=\frac{2\mu }{\left( \mu +\frac{1}{2}\sigma ^2\right) }>0. \end{aligned}$$

On the other hand, if \(\mu \le 0\) we obtain

$$\begin{aligned} \phi ^{\prime }(0)=2-\frac{\sigma ^2}{\sqrt{\tilde{\mu }^2}} =2-\frac{\sigma ^2}{|\tilde{\mu }|} = 2+\frac{\sigma ^2}{\left( \mu -\frac{1}{2}\sigma ^2\right) } =\frac{2\mu }{\left( \mu -\frac{1}{2}\sigma ^2\right) } \ge 0. \end{aligned}$$
  1. (i)

    To prove (i), we proceed by considering cases. First, we consider the case \(\mu >0\). Then, we need to show that \(\phi \) is strictly increasing on \([\mu , \infty )\). To this end, all we need to show is that \(\phi ^{\prime }( p )>0\) for all \(p>\mu \). Indeed, since \(\phi ^{\prime \prime }\) is strictly positive, we obtain for every \(p>\mu \):

    $$\begin{aligned} \phi ^{\prime }( p )>\phi ^{\prime }(\mu )>0. \end{aligned}$$

    This shows that if \(\mu >0\), then \(\phi \) is strictly increasing on the interval \([\mu ,\infty )=[\max (\mu ,0), \infty )\). Now we consider the case \(\mu \le 0\). We have to show that \(\phi \) is strictly increasing on \([0,\infty )\). Using again that \(\phi ^{\prime \prime }\) is strictly positive, we have for every \(p>0\):

    $$\begin{aligned} \phi ^{\prime }( p )>\phi ^{\prime }(0)\ge 0. \end{aligned}$$

    This shows that if \(\mu \le 0\), then \(\phi \) is strictly increasing on the interval \([0,\infty ) =[\max \{\mu ,0\},\infty )\). Hence in both cases \(\phi \) is strictly increasing on \([\max \{\mu ,0\},\infty )\).

  2. (ii)

    Let us show the second assertion. Since \(\mu >0\) and \(\tilde{\mu }^2+2\mu \sigma ^2=(\mu +\frac{1}{2}\sigma ^2)^2\), we have

    $$\begin{aligned} \phi (\mu )=2\mu -\tilde{\mu }-\sqrt{\left( \mu +\frac{1}{2}\sigma ^2\right) ^2} = 2\mu -\tilde{\mu }-\left( \mu +\frac{1}{2}\sigma ^2\right) =0. \end{aligned}$$
  3. (iii)

    Now we consider the third assertion. We note that \(\sqrt{\tilde{\mu }^2}=-\tilde{\mu }\) because \(\tilde{\mu }\le 0\) due to \(\mu \le 0\). Consequently,

    $$\begin{aligned} \phi (0)=-\tilde{\mu }-\sqrt{\tilde{\mu }^2}=0. \end{aligned}$$

    This completes the proof.

\(\square \)

Proof of Lemma  4.1 We observe that Eq. (15) can be written as \(\gamma _2=f({\uplambda })\).

  1. (i)

    It follows immediately from the definition of \(\zeta \) [given in Eq. (16)] and condition (4).

  2. (ii)

    We need to distinguish two cases. Consider first the case \(\mu >0\). From

    $$\begin{aligned} \bar{y}:=\sigma ^2n(2n-1)+2\mu n, \end{aligned}$$

    defined at the beginning of this appendix it follows obviously that \(\bar{y}>\mu \). This result and \({\uplambda }>\bar{y}\) (recall the beginning of this appendix) imply \({\uplambda }>\mu \). Let us consider now the case \(\mu \le 0\). From Sect. 2, we recall that \({\uplambda }\in (0,\infty )\). Under this condition, we have \({\uplambda }>0\ge \mu \). This proves that \({\uplambda }>\mu \) for every value of \(\mu \).

  3. (iii)

    To prove (iii) of Lemma 4.1, we apply Lemma 7.1. We need to distinguish two cases. Consider first the case \(\tilde{\mu }+2n\sigma ^2\ge 0\). Then, Lemma 7.1 (ii) implies \(f(\bar{y})=2n\). Since f is strictly increasing, we obtain \(f({\uplambda })>f(\bar{y})\). Hence \(\gamma _2>2n\). Now we consider the case \(\tilde{\mu }+2n\sigma ^2<0\). Then, Lemma 7.1 (iii) yields \(f(0)>4n\). Again, since f is strictly increasing, we conclude \(\gamma _2=f({\uplambda })>f(0)>4n\). This completes the proof of this assertion.

  4. (iv)

    Aiming at a contradiction, suppose \({\uplambda }\le \sigma ^{2}n(\gamma _2-1)+\gamma _2 \mu \). Then

    $$\begin{aligned} \left( \sigma ^2n+\mu \right) \gamma _2\ge {\uplambda }+\sigma ^2 n>0, \end{aligned}$$

    which implies \(\sigma ^2n+\mu >0\), because \(\gamma _2>1\). Based on this observation, we can express \({\uplambda }\le \sigma ^{2}n(\gamma _2-1)+\gamma _2 \mu \) equivalently as

    $$\begin{aligned} \gamma _2 \ge \frac{{\uplambda }+\sigma ^2n}{\sigma ^2n+\mu }. \end{aligned}$$

Using the definition of \(\gamma _2\), we have the following equivalent inequalities:

$$\begin{aligned} \frac{\sqrt{\tilde{\mu }^2+2{\uplambda }\sigma ^2}}{\sigma ^2}- \frac{\tilde{\mu }}{\sigma ^2} = \gamma _2&\ge \frac{{\uplambda }+\sigma ^2n}{\sigma ^2n+\mu }, \nonumber \\ \sqrt{\tilde{\mu }^2+2{\uplambda }\sigma ^2}&\ge \frac{\sigma ^2\left( {\uplambda }+\sigma ^2n\right) }{\sigma ^2n+\mu } +\tilde{\mu }. \end{aligned}$$
(28)

At this point, we note that

$$\begin{aligned} \frac{\sigma ^2\left( {\uplambda }+\sigma ^2n\right) }{\sigma ^2n+\mu } +\tilde{\mu }= & {} \frac{\sigma ^2\left( {\uplambda }+\sigma ^2n\right) + \left( \mu -\frac{1}{2}\sigma ^2\right) \left( \sigma ^2n+\mu \right) }{\sigma ^2n+\mu }\\= & {} \frac{{\uplambda }\sigma ^2+\frac{1}{2}\sigma ^4n+\mu ^2+\mu \sigma ^2\left( n-\frac{1}{2}\right) }{\sigma ^2n+\mu }>0, \end{aligned}$$

where we used \(\tilde{\mu }:=\mu -\frac{1}{2}\sigma ^2\). The above inequality together with (28) imply

$$\begin{aligned} \tilde{\mu }^2+2{\uplambda }\sigma ^2\ge & {} \left( \frac{\sigma ^2\left( {\uplambda }+\sigma ^2n\right) }{\sigma ^2n+\mu } +\tilde{\mu }\right) ^2,\\ 2{\uplambda }\sigma ^2\ge & {} \left( \frac{\sigma ^2\left( {\uplambda }+\sigma ^2n\right) }{\sigma ^2n+\mu } +\tilde{\mu }\right) ^2-\tilde{\mu }^2,\\ 2{\uplambda }\ge & {} \sigma ^2\frac{\left( {\uplambda }+\sigma ^2n\right) ^2}{\left( \sigma ^2n +\mu \right) ^2} +2\frac{\left( {\uplambda }+\sigma ^2n\right) }{\left( \mu + \sigma ^2n\right) }\tilde{\mu },\\ 2{\uplambda }\left( \sigma ^2 n +\mu \right) ^2\ge & {} \left( {\uplambda }+ \sigma ^2 n\right) \left( \sigma ^2 \left( {\uplambda }+\sigma ^2n\right) + 2\left( \mu + \sigma ^2n\right) \tilde{\mu }\right) \\= & {} \left( {\uplambda }+ \sigma ^2 n\right) \left( \sigma ^2 \left( {\uplambda }+\sigma ^2n\right) + 2\left( \mu + \sigma ^2n\right) \left( \mu -\frac{1}{2}\sigma ^2\right) \right) \\= & {} \left( {\uplambda }+ \sigma ^2 n\right) \left( \sigma ^2 {\uplambda }+ \sigma ^4 n + 2 \left[ \mu ^2 - \frac{\mu }{2} \sigma ^2 + \mu \sigma ^2 n - \frac{1}{2} \sigma ^4 n \right] \right) \\= & {} \left( {\uplambda }+ \sigma ^2 n\right) \left( {\uplambda }\sigma ^2 + 2\mu \sigma ^2n+2\mu ^2-\mu \sigma ^2 \right) . \end{aligned}$$

Hence, we obtain

$$\begin{aligned} 2{\uplambda }\left( \sigma ^2 n +\mu \right) ^2 \ge \left( {\uplambda }+ \sigma ^2 n\right) \left( {\uplambda }\sigma ^2 + 2 \mu \sigma ^2n+2\mu ^2-\mu \sigma ^2 \right) . \end{aligned}$$

By virtue of Lemma 7.2, we have

$$\begin{aligned} {\uplambda }\bar{y} \ge {\uplambda }({\uplambda }-\mu )+\mu \bar{y}. \end{aligned}$$

Grouping terms and factoring \(({\uplambda }-\mu )\), we get the equivalent forms

$$\begin{aligned} \bar{y}({\uplambda }-\mu ) \ge {\uplambda }({\uplambda }-\mu ),\quad \left( \bar{y}-{\uplambda }\right) ({\uplambda }-\mu ) \ge 0. \end{aligned}$$

Since \({\uplambda }>\mu \), it follows that \(\bar{y} \ge {\uplambda }\). Thus, we obtain a contradiction. Therefore, we must have \( {\uplambda }> \sigma ^{2}n(\gamma _2-1)+\gamma _2 \mu \). This completes the proof of inequality (iv).

  1. (v)

    We will apply Lemma 7.3 to prove this inequality. Accordingly, we distinguish two cases. Consider first the case \(\mu >0\). Since \(\phi (\mu )=0\) and \({\uplambda }>\mu \), we conclude that \(\phi ({\uplambda })>\phi (\mu )=0\), because \(\phi \) is strictly increasing on \([\mu , \infty )\). Now consider the case \(\mu \le 0\). From \(\phi (0)=0\), and \({\uplambda }>0\), it follows \(\phi ({\uplambda })>\phi (0)=0\), because \(\phi \) is strictly increasing on \([0,\infty )\). Consequently, \(\phi ({\uplambda })=2 {\uplambda }-\tilde{\mu }-\sqrt{\tilde{\mu }^2+2{\uplambda }\sigma ^2}>0\) for every \(\mu \), as was to be shown.

\(\square \)

Appendix 3: Proof of Lemma 4.2

Proof

We recall that for every \(x \in (0,\infty )\):

$$\begin{aligned} L(x):=\frac{1}{2} \sigma ^2x^2v^{\prime \prime }(x)+\mu x v^{\prime }(x)-{\uplambda }v(x)+ h(x). \end{aligned}$$

Using Eq. (25), we have

$$\begin{aligned} L(x)=\mu x k -{\uplambda }(k x +D)+\alpha x^{2n}+\beta \quad \text {for}\;\; x \in [b,\infty ). \end{aligned}$$

We observe that L is (infinitely) differentiable on \((b,\infty )\). Taking the first derivative, we obtain for every \(x\in (b,\infty )\):

$$\begin{aligned} L^{\prime }(x)=(\mu -{\uplambda })k+2\alpha n x^{2n-1}. \end{aligned}$$

Hence

$$\begin{aligned} L^{\prime }(b+) = (\mu -{\uplambda })k + 2 \alpha n b^{2n-1}. \end{aligned}$$

By continuity of \(v^{\prime }\), condition (19), we have

$$\begin{aligned} k=v^{\prime }(b)= 2\alpha \zeta n b^{2n-1} + B\gamma _2 b^{\gamma _2-1}. \end{aligned}$$

From Eqs. (23) and (16), we have

$$\begin{aligned} L^{\prime }(b+)&= (\mu -{\uplambda })\left\{ 2\alpha \zeta n b^{2n-1} + B\gamma _2 b^{\gamma _2-1}\right\} + 2 \alpha n b^{2n-1}\\&= (\mu -{\uplambda })\left\{ 2\alpha \zeta n b^{2n-1} - 2\alpha \zeta n \frac{(2n-1)}{\gamma _2(\gamma _2-1)}b^{2n-\gamma _2}\gamma _2 b^{\gamma _2-1}\right\} + 2 \alpha n b^{2n-1}\\&= (\mu -{\uplambda })\left\{ 2\alpha \zeta n b^{2n-1} -2\alpha \zeta n\frac{(2n-1)}{(\gamma _2-1)} b^{2n-1}\right\} + 2 \alpha n b^{2n-1} \\&= 2\alpha n b^{2n-1}\left\{ (\mu -{\uplambda })\zeta \left( 1-\frac{2n-1}{\gamma _2-1}\right) +1\right\} \\&= 2\alpha n b^{2n-1}\left\{ ({\uplambda }-\mu )\zeta \left( \frac{2n-\gamma _2}{\gamma _2-1} \right) +1\right\} \\&=2\alpha n b^{2n-1} \left\{ \frac{({\uplambda }-\mu )}{\left( {\uplambda }-\sigma ^2 n(2n-1)-2\mu n\right) } \frac{\left( 2n-\gamma _2\right) }{\left( \gamma _2-1\right) } +1\right\} \\&= 2\alpha n b^{2n-1} \left\{ \frac{({\uplambda }-\mu )\left( 2n-\gamma _2\right) +\left( \gamma _2-1\right) \left( {\uplambda }-\sigma ^2 n(2n-1)-2\mu n\right) }{\left( {\uplambda }-\sigma ^2 n(2n-1)-2\mu n\right) (\gamma _2-1)}\right\} . \end{aligned}$$

We notice that the numerator of the term in curly brackets above can be written in a convenient manner; namely

$$\begin{aligned}&({\uplambda }-\mu )(2n -\gamma _2)+(\gamma _2-1) \left( {\uplambda }-\sigma ^2 n(2n-1)-2\mu n \right) \\&\quad = 2{\uplambda }n -{\uplambda }\gamma _2 -2 \mu n +\mu \gamma _2 + {\uplambda }\gamma _2 \\&\qquad -\, \sigma ^2n(2n-1)\gamma _2 -2\mu n \gamma _2 - {\uplambda }+ \sigma ^2n(2n-1)+2 \mu n\\&\quad ={\uplambda }(2n-1)-\mu \gamma _2(2n-1)-\sigma ^2n(\gamma _2-1)(2n-1)\\&\quad = (2n-1)\left( {\uplambda }-\sigma ^2 n (\gamma _2-1)-\mu \gamma _2 \right) . \end{aligned}$$

Therefore,

$$\begin{aligned} L^{\prime }(b+)&=\, 2\alpha n b^{2n-1} \frac{(2n-1)\left( {\uplambda }-\sigma ^2 n (\gamma _2-1)-\mu \gamma _2 \right) }{\left( {\uplambda }-\sigma ^2 n(2n-1)-2\mu n\right) (\gamma _2-1)}\\&= \frac{ 2\alpha n (2n-1) b^{2n-1} \zeta \left( {\uplambda }-n\sigma ^2(\gamma _2-1)-\mu \gamma _2 \right) }{(\gamma _2-1)} >0, \end{aligned}$$

because \(\zeta >0\), \(\gamma _2>1\) and \({\uplambda }-n\sigma ^2(\gamma _2-1)-\mu \gamma _2>0\), by Lemma 4.1. \(\square \)

Appendix 4: Solution of the model with \(g(X) = g_{0} + g_{1}X\) and \(r(X) =r_{0} + r_{1}X\)

The debt dynamics is given by

$$\begin{aligned} X_{t} = x+\int _{0}^{t} (\rho X_s + \mu )X_s ds +\int _{0}^{t}\sigma X_{s} dW_{s} -Z_t, \end{aligned}$$
(27)

where \(\rho =r_{1}-g_{1}\in (-\infty , +\infty )\), \(\mu =r_{0}-g_{0}\in (-\infty , +\infty )\) and \(\sigma \in (0,\infty )\) are constants.

For a function \(v:(0,\infty ) \rightarrow {\mathbb R}\) in \(C^2((0,\infty ))\), the HJB equation is given by

$$\begin{aligned} \min \left\{ {\widetilde{\mathcal{L}}} v(x) + h(x), k-{v}^{\prime }(x) \right\} =0. \end{aligned}$$
(29)

where

$$\begin{aligned} \widetilde{\mathcal{L}} v(x) := \frac{1}{2} {\sigma }^{2} x^{2} v^{\prime \prime }(x) + \rho x^{2} v^{\prime }(x) + \mu x v^{\prime }(x) - {\uplambda }v(x). \end{aligned}$$

A solution v of the HJB equation defines the regions \(\mathcal {C} = \mathcal {C}^{v}\) and \({\varSigma }={\varSigma }^{v}\) by

$$\begin{aligned} \mathcal {C}&= {\mathcal {C} ^{v}} := \left\{ x \in (0,\infty ): \widetilde{\mathcal{L}} v(x) + h(x) = 0 \quad \text{ and }\quad k-{v}^{\prime }(x) > 0 \right\} \end{aligned}$$
(30)
$$\begin{aligned} {{\varSigma }}&= {{\varSigma }^{v}} := \left\{ x \in (0,\infty ): \widetilde{\mathcal{L}} v(x) + h(x) \ge 0 \quad \text{ and } \quad k-{v}^{\prime }(x) = 0 \right\} . \end{aligned}$$
(31)

The HJB equation in the continuation region \(\mathcal C=(0,b)\) is

$$\begin{aligned} \frac{1}{2} {\sigma }^{2} x^{2} v^{\prime \prime }(x) + \rho x^{2} v^{\prime }(x) + \mu x v^{\prime }(x) - {\uplambda }v(x) = - \alpha x^{2n}-\beta , \end{aligned}$$

and in the intervention region \({\varSigma }= [b,\infty )\) is

$$\begin{aligned} v(x) = v(b) + k (x-b). \end{aligned}$$

Consequently, we have the differential equation

$$\begin{aligned} \left\{ \begin{array}{ll} \frac{1}{2} {\sigma }^{2} x^{2} v^{\prime \prime }(x) + \rho x^{2} v^{\prime }(x) + \mu x v^{\prime }(x) - {\uplambda }v(x) = - \alpha x^{2n}-\beta , &{} \quad \text{ if }\;x < b\\ v^{\prime }(x)=k &{} \quad \text{ if }\;x \ge b. \end{array} \right. \end{aligned}$$
(32)

If the debt ratio follows the dynamics (27), the value function for Problem 2.1 is the function \(V_{1}:(0,\infty ) \rightarrow \mathbb R\) that satisfies the differential equation (32), and the following conditions:

$$\begin{aligned} V_{1}(0+)=\frac{\beta }{{\uplambda }}, \quad V_{1}(b_{1}+) = V_{1}(b_{1}-) , \quad V_{1}^{\prime }(b_{1}+) = V_{1}^{\prime }(b_{1}-), \quad V_{1}^{\prime \prime }(b_{1}+) = V_{1}^{\prime \prime }(b_{1}-). \end{aligned}$$
(33)

The solution of the above differential equation depends on whether \(\rho >0\) or \(\rho <0\). From now on, without loss of generality, we assume \(\mu \ne 0\).

Consider first \(\rho >0\). Then the value function is given by

$$\begin{aligned} V_{1}(x) = \left\{ \begin{array}{ll} \frac{\beta }{{\uplambda }} + \sum _{j=2n}^{\infty }\zeta _j x^{j} + A_{1} H_{1}(x) + B_{1} G_{1}(x) &{} \quad \text{ if }\;x < b_{1}\\ k x + D_{1} &{}\quad \text{ if }\;x \ge b_{1}. \end{array} \right. \end{aligned}$$
(34)

Here

$$\begin{aligned} H_{1}(x)&:= x^{\widetilde{\gamma }}\exp {\left\{ -2\rho /\sigma ^2 x\right\} }H\left( \widetilde{\gamma }+ 2\mu /\sigma ^{2}, \gamma + 1, 2\rho /\sigma ^2x\right) ,\end{aligned}$$
(35)
$$\begin{aligned} G_{1}(x)&:= x^{\widetilde{\gamma }}\exp {\left\{ -2\rho /\sigma ^2 x\right\} }\,L\left( -\widetilde{\gamma }- 2\mu /\sigma ^{2},\gamma , 2\rho /\sigma ^2 x\right) , \end{aligned}$$
(36)

where H and L stands for the confluent hipergeometric function and the generalized Laguerre polynomial, respectively. In addition,

$$\begin{aligned}&\displaystyle \gamma = \sqrt{1 + 4\frac{\mu ^{2}}{\sigma ^{4}} + 8 \frac{{\uplambda }}{\sigma ^{2}} - 4 \frac{\mu }{\sigma ^{2}} }, \end{aligned}$$
(37)
$$\begin{aligned}&\displaystyle \widetilde{\gamma } = \frac{1}{2}\left( 1 + \gamma - 2\frac{\mu }{\sigma ^{2}} \right) , \end{aligned}$$
(38)
$$\begin{aligned}&\displaystyle \zeta _{2n} = \frac{\alpha }{{\uplambda }-n(2n-1)\sigma ^2-2n\mu }, \end{aligned}$$
(39)
$$\begin{aligned}&\displaystyle \zeta _{2n+m} = \frac{\left( {\begin{array}{c}2n-1+m\\ 2n-1\end{array}}\right) \alpha \rho ^m m!}{\prod \nolimits _{j=0}^{m}\left( {\uplambda }-(2n+j)(2n+j-1)\sigma ^2/2-(2n+j)\mu \right) }, \quad \forall m\in \{1,2,3,\ldots \}. \nonumber \\ \end{aligned}$$
(40)

The four constants \(A_{1}\), \(B_{1}\), \(b_{1}\), and \(D_{1}\) are determined by the system of four equations (33).

Since \(\lim _{x\rightarrow 0+}H_{1}(x)=+\infty \), the first condition in (33) implies \(A_{1}=0\). Hence the value function becomes

$$\begin{aligned} V_{1}(x) = \left\{ \begin{array}{ll} \frac{\beta }{{\uplambda }} + \sum _{j=2n}^{\infty }\zeta _j\, x^{j} + B_{1}\, G_{1}(x) &{} \quad \text{ if }\;x < b_{1}\\ k x + D_{1} &{}\quad \text{ if }\;x \ge b_{1}. \end{array} \right. \end{aligned}$$
(41)

In view of the last three conditions in (33), the three parameters \(B_{1}\), \(b_{1}\), \(D_{1}\) are determined by the following system of equations:

$$\begin{aligned} V_{1}(b_{1}+) = k b_{1} + D_{1}, \quad V_{1}^{\prime }(b_{1}+) = k, \quad V_{1}^{\prime \prime }(b_{1}+)&= 0. \end{aligned}$$
(42)

In particular, using the above equations, the optimal debt ceiling \(b_{1}\) is determined by the following equation (see below for numerical solutions):

$$\begin{aligned} G_{1}^{\prime }(b_{1})\sum _{j=2n}^{\infty }j(j-1)\zeta _j b_{1}^{j-2} + \left( k - \sum _{j=2n}^{\infty }j\zeta _j b_{1}^{j-1}\right) G_{1}^{\prime \prime }(b_{1})=0. \end{aligned}$$
(43)

Similarly, if \(\rho < 0\) the value function is given by

$$\begin{aligned} V_{2}(x) = \left\{ \begin{array}{ll} \frac{\beta }{{\uplambda }} + \sum _{j=2n}^{\infty }\zeta _j x^{j} + A_{2} H_{2}(x) + B_{2} G_{2}(x) &{} \quad \text{ if }\; x < b_{2}\\ k x + D_{2} &{}\quad \text{ if } \;x \ge b_{2}. \end{array} \right. \end{aligned}$$
(44)

Here

$$\begin{aligned} H_{2}(x)&:= x^{\widetilde{\gamma }} H\left( \widetilde{\gamma }, \gamma + 1, -2\rho /\sigma ^2 x\right) \end{aligned}$$
(45)
$$\begin{aligned} G_{2}(x)&:= x^{\widetilde{\gamma }} L\left( -\widetilde{\gamma }, \gamma , -2\rho /\sigma ^2 x\right) , \end{aligned}$$
(46)

where H and L stands for the confluent hipergeometric function and the generalized Laguerre polynomial, respectively. The parameters \(\gamma \), \(\widetilde{\gamma }\), \(\zeta _{2n}\), and \(\{\zeta _{2n+m}\}_{m=1}^{\infty }\) are given in Eqs. (37)–(40).

Since \(\lim _{x\rightarrow 0+}H_{2}(x)=+\infty \), the first condition in (33) implies \(A_{2}=0\). Hence the value function becomes

$$\begin{aligned} V_{2}(x) = \left\{ \begin{array}{ll} \frac{\beta }{{\uplambda }} + \sum _{j=2n}^{\infty }\zeta _j x^{j} + B_{2} G_{2}(x) &{} \quad \text{ if }\;x < b_{2}\\ k x + D_{2} &{}\quad \text{ if } \;x \ge b_{2}. \end{array} \right. \end{aligned}$$
(47)

Form the last three conditions in (33), the three parameters \(B_{2}\), \(b_{2}\), and \(D_{2}\) are determined by the following system of three equations:

$$\begin{aligned} V_{2}(b_{2}+) = k b_{2} + D_{2}, \quad V_{2}^{\prime }(b_{2}+) = k, \quad V_{2}^{\prime \prime }(b_{2}+) = 0. \end{aligned}$$
(48)

In particular, using the above equations, the optimal debt ceiling \(b_{2}\) is determined by the following equation (see below for numerical solutions):

$$\begin{aligned} G_{2}^{\prime }(b_{2})\sum _{j=2n}^{\infty }j(j-1)\zeta _j b_{2}^{j-2} + \left( k - \sum _{j=2n}^{\infty }j\zeta _j b_{2}^{j-1}\right) G_{2}^{\prime \prime }(b_{2})=0. \end{aligned}$$
(49)

We present some numerical computations to analyze the effects of the parameters on the optimal debt ceiling. The basic parameters that we use are given in Table 5.

Table 5 Basic parameter values

In Table 6 row \(b(\rho =0)\) shows the values for the optimal debt ceiling for \(\rho =0\), which are calculated using the explicit formula given in (26). In row \(b(\rho =-0.05)\) we present the values for the optimal debt ceiling that corresponds to \(\rho =-0.05\), which are calculated numerically from (49). Similarly, row \(b(\rho =0.05)\) shows the values for the optimal debt ceiling corresponding to \(\rho =0.05\), which are obtained numerically using (43). Thus, for instance, the optimal debt ceiling 0.745428 is calculated using \(\rho =-0.05\), \(\alpha =0.5\), and the other parameters are specified in Table 5.

Table 6 Effect of \(\alpha \), \(\mu \) and \(\sigma \) on the optimal debt ceiling b

As with the debt ratio dynamics (1), we observe that the optimal debt ceiling b is a decreasing function of \(\alpha \), an increasing function of \(\mu =g-r\) (hence an increasing function of economic growth g and a decreasing function of the interest rate r), and an increasing function of volatility \(\sigma \). Therefore, the qualitative effects of the economic parameters \(\alpha \), g, r and \(\sigma \) on the optimal debt ceiling b are the same in model (1) and in model (27).

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Cadenillas, A., Huamán-Aguilar, R. Explicit formula for the optimal government debt ceiling. Ann Oper Res 247, 415–449 (2016). https://doi.org/10.1007/s10479-015-2052-9

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