The productivity cost of sovereign default: evidence from the European debt crisis


We calibrate the cost of sovereign defaults using a continuous time model, where government default decisions may trigger a change in the regime of a stochastic TFP process. We calibrate the model to a sample of European countries from 2009 to 2012. By comparing the estimated drift in default relative to that in no-default, we find that TFP falls in the range of 3.70–5.88 %. The model is consistent with observed falls in GDP growth rates and subsequent recoveries and illustrates why fiscal multipliers are small during sovereign debt crises.

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  1. 1.

    As Mendoza and Yue (2012) report, in almost every episode GDP fell below trend, external financing shut down, interest rates peaked, external debt built up and labour input fell dramatically, imposing large potential costs on each economy that experienced default.

  2. 2.

    They find a much stronger correlation between the stock market and future productivity during the Great Depression than in US postwar cyclical fluctuations.

  3. 3.

    This would be similar to assuming a benevolent government that tries to maximise the utility of a representative household with a separable utility function in private and public consumption.

  4. 4.

    We assume neutral agents. For the consequences of assuming agents concerned with the worst case scenario see Araujo (2015).

  5. 5.

    This is a reduced form of an enforcement mechanism or an optimal debt contract. A theoretical characterisation of the effect of enforcement on the interest rate can be found in Krasa et al. (2008). Optimal debt contracts are designed in Hvide and Leite (2010) and Mateos-Planas and Seccia (2014).

  6. 6.

    This productivity level is chosen in the same way as S&P and Fitch classify bonds as AAA or Moodys as Aaa. An obligor that has issued a prime quality bond is considered as having an extremely strong capability of meeting its financial commitments.

  7. 7.

    We obtained the data from Bloomberg.

  8. 8.

    We are using Germany as the risk-free option, but we do not use this country explicitly in our estimations.

  9. 9.

    We normalise both simulated series and data at the beginning of our estimation period for each country.

  10. 10.

    We use the average debt-to-GDP and taxes-to-GDP ratios for the countries in our sample as exogenous inputs for our calibration algorithm.

  11. 11.

    Danthine and Jin (2007) show that financial volatility is a multiple of macroeconomic volatility.

  12. 12.

    Let \({\bar{x}}=\log \left( \frac{y}{y_{d}}\right) \) (where \(y_{d}\) is the default threshold in terms of GDP) be a random variable. In this case, recovery is defined as \({\bar{x}}=0\). A recovery date can be defined as \(T(x)=\{T : {\bar{x}} \ge 0\}\) and, following Harrison (1985), the distribution of recovery dates can be written as

    $$\begin{aligned} P(T(x)>t)=\left[ \phi \left( \frac{x-\mu _y t}{\sigma _y \sqrt{t}}\right) -e^{\displaystyle \frac{2\mu _y x}{\sigma _y^2}}\phi \left( \frac{-x-\mu _y t}{\sigma _y \sqrt{t}}\right) \right] \end{aligned}$$

    where \(\phi \) is a \( N(0,1) \) distribution function.

  13. 13.

    The solid line represents the average.


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We thank T. J. Kehoe and several participants at the 2013 Economic Theory Meeting in Paris for helpful comments. This article has also benefited from helpful comments and suggestions by an anonymous referee.

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Corresponding author

Correspondence to Jorge Alonso-Ortiz.

Additional information

Jose Maria Da Rocha gratefully acknowledges financial support from Xunta de Galicia (Ref. GRC 2015/014 and ECOBAS).



This appendix presents the solution of the second-order differential Eqs. (2),  (3) and (4).

Solution of Equation (3). For any \(A_{d}>0\), using \(x=\log \left( \frac{A}{A_{d}}\right) \), the debt price is the solution of the boundary-value problem that consists of solving the equation:

$$\begin{aligned} -rq(x)+{\hat{\mu }}_{nd}q^{\prime }(x)+\frac{\sigma _{nd}^{2}}{2}q^{\prime \prime }(x)=0 \end{aligned}$$

with boundary conditions \(q(0) =0\) and \(q^{\prime }(0) =k\), where \({\hat{\mu }}_{nd}=\mu _{nd}-\frac{1}{2}\sigma _{nd}^{2}<0\) and k is an arbitrary constant. We solve the boundary-value problem using Laplace transforms, \({\mathscr {L}}[q(x)]\). Laplace transforms are given by

$$\begin{aligned} {\mathscr {L}}[q'(x)]= & {} s {\mathscr {L}}[q(x)] - q(0),\\ {\mathscr {L}}[q''(x)]= & {} s^2 {\mathscr {L}}[q(x)] -sq(0)-q'(0). \end{aligned}$$

By applying Laplace transforms in equation (A)

$$\begin{aligned} \left( \frac{\sigma _{nd}^{2}}{2} s^2 +{\hat{\mu }}_{nd} s -r\right) {\mathscr {L}}[q(x)]- (s +{\hat{\mu }}_{nd}) q(0)- \frac{\sigma _{nd}^{2}}{2}q'(0)= 0 \end{aligned}$$

and the boundary condition \(g(0)=0\), we obtain:

$$\begin{aligned} {\mathscr {L}}[q(x)]=\frac{\sigma _{nd}^{2}}{2} \frac{k}{(s-z_{1})(s-z_{2})}, \end{aligned}$$

where \(z_{i} =\left( -{\hat{\mu }}_{nd}\pm \sqrt{{\hat{\mu }}_{nd}^{2}+2r\sigma _{nd}^{2}} \right) \left( \sigma _{nd}^{2}\right) ^{-1}\) , \(i=1,2\). We obtain the solution by solving the Laplace inverses given by:

$$\begin{aligned} q(x)={\mathscr {L}}^{-1}\left[ \frac{k\sigma _{nd}^{2}/2}{(s-z_1)(s-z_2)}\right]= & {} \frac{k\sigma _{nd}^{2}/2}{(z_1-z_2)}\left( e^{z_1x} - e^{z_2x}\right) \\= & {} \frac{1}{1+r}\left( \frac{e^{z_{2}x}-e^{z_{1}x}}{e^{z_{2}{\overline{x}} }-e^{z_{1}{\overline{x}}}}\right) \end{aligned}$$

and taking into account the second boundary condition, \(\lim _{x\rightarrow {\overline{x}}}q(x)=\frac{1}{1+r}\) where \( {\overline{x}}=\log \left( \frac{A^*}{A_{d}}\right) \).

Solution of Equation (2). Equation default regions are characterised by the non-homogeneous second-order differential equation

$$\begin{aligned} rW(x)-{\hat{\mu }}_{nd}W^{\prime }(x)-\frac{\sigma _{nd}^{2}}{2}W^{\prime \prime }(x)=\tau A_{d}e^{x}+[q(x)-1]b \end{aligned}$$

with boundary conditions \(W(0) =\displaystyle \frac{\tau A_{d}}{r-\mu _{d}+\sigma ^2_d/2}\) and \(W^{\prime }(0) =\displaystyle \frac{\tau A_{d}}{r-\mu _{d}+\sigma ^2_d/2}\). Taking the Laplace Transform of both sides of the differential equation, default regions are characterised by solving

$$\begin{aligned} \left( r-{\hat{\mu }}_{nd}s-\frac{\sigma _{nd}^{2}}{2}s^{2}\right) {\mathscr {L}}[W(x)]= & {} -\left( {\hat{\mu }}_{nd}+\frac{\sigma _{nd}^{2}}{2}s\right) W(0)-\frac{ \sigma _{nd}^{2}}{2}W^{\prime }(0)\nonumber \\&\ -\frac{b}{s}+\frac{\tau A_{d}}{s-1} +b{\mathscr {L}}[q(x)], \end{aligned}$$


$$\begin{aligned} {\mathscr {L}}[q(x)]=\frac{1}{(1+r)(e^{z_1{\overline{x}}}- e^{z_2{\overline{x}}})} \left[ \frac{1}{s-z_{1}}+\frac{1}{s-z_{2}}\right] . \end{aligned}$$

\(H(s)={\mathscr {L}}[W(x)]\) satisfies,

$$\begin{aligned} H(s)=\frac{P_{1}+P_{2}s+P_{3}s^{2}+P_{4}s^{3}+P_{5}s^{4}+P_{6}s^{5}}{ s(s-1)(s-z_{1})^{2}(s-z_{2})^{2}} \end{aligned}$$

where the vector \({\mathbf {P}}\) is given by

$$\begin{aligned} {\mathbf {P}}= & {} \left[ \begin{array}{ccccc} 0 &{} -1 &{}0 &{}0 &{} 0\\ -1 &{} 1+z_1+z_2 &{} 0&{} 0&{}0 \\ 1+z_1+z_2 &{} -z_1z_2+z_1+z_2 &{} -1 &{} 1 &{} 0\\ -(z_1z_2+z_1+z_2) &{} z_1z_2 &{} 1+z_1+z_2 &{} -(z_1+z_2) &{} -(z_2-z_1)\\ z_1z_2 &{} 0 &{} -(z_1z_2+z_1+z_2) &{} z_1z_2 &{} z_2-z_1\\ 0 &{} 0 &{} z_1z_2 &{} 0 &{} 0 \end{array}\right] \nonumber \\&\ \left[ \begin{array}{c} {\hat{\mu }}_{nd}W(0)+\frac{\sigma ^2_1W'(0)}{2}\\ \frac{\sigma ^2_1W'(0)}{2}\\ b\\ \tau A_d\\ \frac{(1+r)b}{\left( e^{z_2 {\overline{x}}}-e^{z_1 {\overline{x}}}\right) } \end{array}\right] \end{aligned}$$

Expanding H(s) in partial fractions

$$\begin{aligned} H(s)=\frac{C_{1}}{s}+\frac{C_{2}}{s-1}+\frac{C_{3}}{(s-z_{1})}+\frac{C_{4}}{ (s-z_{2})}+\frac{C_{5}}{(s-z_{1})^{2}}+\frac{C_{6}}{(s-z_{2})^{2}}, \end{aligned}$$

Applying the Laplace inverses given by:

$$\begin{aligned} W(x) = {\mathscr {L}}^{-1}[H(s)]{=} C_1-C_2*e^x+C_3 x e^{z_1x}+C_4 x e^{z_2 x}+C_5 x^2 e^{2z_1x}+C_6 x^2 e^{2z_2 x} \end{aligned}$$

we can find the solution of W(x) by solving a system of linear equations which can be written in matrix notation as:

$$\begin{aligned} \left[ \begin{array}{cccccc} C_{1}&C_{2}&C_{3}&C_{4}&C_{5}&C_{6} \end{array} \right] ^{T}=\mathbf {\Lambda } ^{-1}{\mathbf {P}} \end{aligned}$$

and \(\mathbf {\Lambda }\) is equal to

Given \(\mu _{nd}\), \(\sigma ^2_{nd}\) \(\mu _{d}\), \(\sigma ^2_{d}\) and r, b and \(\tau \), \(A_d\) is obtained by solving \(W(x)={\mathscr {L}}^{-1}[H(s)]\) at \(x=0\), i.e.

$$\begin{aligned} W(0)=\left. W(x)\right| _{x=0}=C_0(A_d) -C_1(A_d) =\displaystyle \frac{\tau A_d}{r-\mu _{d}+\sigma ^2_d/2}. \end{aligned}$$

Solution of Equation (4). To solve firm value if the government has not defaulted \(V_{nd}(A)\), we rewrite the switching problem through the following change of variable g \(x=\log \left( \frac{A}{A_{d}}\right) \)

$$\begin{aligned} \left[ r-e^{-\left( 1 -\frac{2\mu _{nd}}{\sigma _{nd}^{2}}\right) x }\right] V_{nd}(x_{t})={\hat{\mu }}_{nd}V_{nd}^{\prime }(x_{t})+\frac{\sigma _{nd}^{2}}{2} V_{nd}^{\prime \prime }(x_{t})+e^{-\left( 1 -\frac{2\mu _{nd}}{\sigma ^{2}}\right) x } A_{d}^{\beta _{d}}e^{\beta _{d}x} \end{aligned}$$

where boundary conditions are given by \(V_{nd}(0) =A_{d}e^{\beta _{d} }\) and \(V_{nd}^{\prime }(0) =\beta _{0}A_{d}e^{\beta _{d} }\), and the probability of defaulting is \(e^{-\left( 1 -\frac{2\mu _{nd}}{\sigma ^{2}}\right) x }\). We solve Eq. (A) with a power series expansion. The basic idea is similar to that in the method of undetermined coefficients: We assume that the solutions of a given differential equation have power series expansions, and then we attempt to determine the coefficients so as to satisfy the differential equation. Equation (A) can be rewritten as

$$\begin{aligned} \left[ r-e^{\left( -a_{0}x\right) }\right] V-a_{1}V^{\prime }-a_{2}V^{\prime \prime }=a_3e^{bx}. \end{aligned}$$

We use the notation \(V=V_{nd}(0)\). Consider a Taylor expansion

$$\begin{aligned} V(x)=V+V^{\prime }x+\sum _{k=2}^{n}\frac{1}{k!}V^{^{\prime }(n)}x^{n}. \end{aligned}$$

Differentiating Eq. (A) n times yields a linear system

where \(\lambda _{n,j}=(-1)^{n+j+1}\left( \begin{array}{c} n! \\ j! \end{array} \right) \) are the Pascal’s triangle numbers (in absolute value). Given this recurrence relationship, the successive coefficients can be evaluated one by one by writing the recurrence relationship first for \(n = 0\), then for \(n = 1\), and so on. Therefore, the solution is merely a function of the boundary conditions \(V_{0}\) and \(V_{0}^{\prime }\), i.e.

$$\begin{aligned}&\left[ \begin{array}{c} V^{\prime \prime } \\ V^{(3)} \\ V^{(4)} \\ V^{(5)} \\ \ldots \\ V^{(n+2)} \end{array} \right] \nonumber \\&\quad =\left[ \begin{array}{cccccc} -a_{2} &{} 0 &{} \ldots &{} 0 &{} 0 &{} 0 \\ -a_{1} &{} -a_{2} &{} \ldots &{} 0 &{} 0 &{} 0 \\ (r-1) &{} -a_{1} &{} \ldots &{} 0 &{} 0 &{} 0 \\ 3a_{0} &{} (r-1) &{} \ldots &{} 0 &{} 0 &{} 0 \\ \ldots &{} \ldots &{} \ldots &{} \ldots &{} \ldots &{} \ldots \\ a_{0}^{n-1} &{} \ldots &{} na_{0} &{} (r-1) &{} -a_{1} &{} -a_{2} \\ &{} &{} &{} &{} &{} \end{array} \right] ^{-1}\nonumber \\&\qquad \left( a_3 \left[ \begin{array}{c} 1 \\ b \\ b^{2} \\ b^{3} \\ \ldots \\ b^{n-1} \end{array} \right] -\left[ \begin{array}{cc} (r-1) &{} -a_{1} \\ a_{0} &{} (r-1) \\ -a_{0}^{2} &{} 2a_{0} \\ a_{0}^{3} &{} -3a_{0}^{2} \\ \ldots &{} \ldots \\ a_{0}^{n+1} &{} (n+1)a_{0}^{n} \end{array} \right] \left[ \begin{array}{c} V_{0} \\ V_{0}^{\prime } \end{array} \right] \right) . \end{aligned}$$

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Alonso-Ortiz, J., Colla, E. & Da-Rocha, JM. The productivity cost of sovereign default: evidence from the European debt crisis. Econ Theory 64, 611–633 (2017).

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  • Default
  • Sovereign debt
  • Financial markets
  • Productivity

JEL Classification

  • E30
  • E44
  • G15