Public debt and total factor productivity

Abstract

This paper explores the role of public debt and fiscal deficits on factor productivity in an economy with credit market frictions and heterogeneous firms. When credit market conditions are sufficiently weak, low interest rates permit the government to run Ponzi schemes so that permanent primary deficits can be sustained. For small enough deficit ratios, the model has two steady states of which one is an unstable bubble and the other one is stable. The stable equilibrium features higher levels of credit and capital, but also a lower interest rate, lower total factor productivity and output. The model is calibrated to the US economy to derive the maximum sustainable deficit ratio and to examine the dynamic responses to changes in debt policy. A reduction in the primary deficit triggers an expansion of credit and capital, but it also leads to a deterioration of total factor productivity since more low-productivity firms prefer to remain active at the lower equilibrium interest rate.

Appendix

Lemma 1

A competitive equilibrium is described by solutions in $(z_t,k_t,d_t)$ to the dynamic system defined by (16), (17) and (18).

Proof

Rewrite Eqs. (7), (12) and (13) in terms of $k_t=K_t/A_t$, $\omega _t=W_t/A_t$, $d_t=(q_tD_t)/A_t$:

$$\begin{aligned} \omega _t= & {} f'(k_tZ_t)Z_tk_t+(1-\delta )k_t +d_t, \end{aligned}$$

(26)

$$\begin{aligned} k_{t+1}= & {} \beta \omega _t {\textstyle \frac{\displaystyle R_{t+1}(1-G(z_{t+1}))}{\displaystyle (1+g)[R_{t+1}-\lambda (1-\delta )]}}\end{aligned}$$

(27)

$$\begin{aligned} {\textstyle \frac{\displaystyle \beta R_{t+1}}{\displaystyle 1+g}}\omega _t G(z_{t+1})= & {} \lambda (1-\delta )k_{t+1}+d_{t+1}. \end{aligned}$$

(28)

Substitution of (26) into (27) directly yields (17). Combine (27) and (28) to obtain

$$\begin{aligned} k_{t+1}[R_{t+1}-\lambda (1-\delta )]= & {} \beta {\textstyle \frac{\displaystyle \omega _t}{\displaystyle 1+g}}[1-G(z_{t+1}]R_{t+1}\\= & {} \beta {\textstyle \frac{\displaystyle \omega _{t}}{\displaystyle 1+g}}R_{t+1}-d_{t+1}-\lambda (1-\delta )k_{t+1}. \end{aligned}$$

Substitution of (26) yields (16). With $X_t/A_t=\xi _ty_t=\xi _tf(Z_tk_t)$, the government budget constraint (4) is

$$\begin{aligned} \xi _tf(Z_tk_t)={\textstyle \frac{\displaystyle (1+g)d_{t+1}}{\displaystyle R_{t+1}}}-d_t, \end{aligned}$$

which is Eq. (18). $\square$

Proposition 1

A no-bubble stationary equilibrium $(z^N,R^N)$ exists if either Assumption 2 holds or if $\lambda$ is sufficiently small. The interest rate satisfies $R^N=\frac{\lambda (1-\delta )}{G(z^N)}<(1+g)/\beta$. Under Assumption 1, a no-bubble stationary equilibrium is unique.

Proof

Let $\underline{z}\ge 0$ and $\overline{z}>0$ denote the upper and lower bounds of the support of G, with $\overline{z}=\infty$ if G has unbounded support. The right-hand side (RHS) of (23) is zero at $z=\underline{z}$ and positive at higher values of z. Hence a solution exists if $\lambda$ is sufficiently small. Moreover, under Assumption 2 the RHS tends to $(1+g)/(\beta (1-\delta ))>1$ for $z\rightarrow \overline{z}$, so that (23) has a solution for any $\lambda \le 1$. Because of $z/Z(z)< 1$, it follows that $\lambda <G(z^N){\textstyle \frac{\displaystyle 1+g}{\displaystyle \beta (1-\delta )}}$ and hence $R^N={\textstyle \frac{\displaystyle \lambda (1-\delta )}{\displaystyle G(z^N)}}<{\textstyle \frac{\displaystyle 1+g}{\displaystyle \beta }}$.

Under Assumption 1, the RHS is strictly monotonically increasing. Hence, there can be at most one solution to Eq. (23).$\square$

Proposition 2

Suppose that Assumption 1 holds. Then a bubble stationary equilibrium exists if and only if $R^N<1+g$. Moreover, $z^B>z^N$ holds so that TFP is higher in the bubble stationary equilibrium.

Proof

Let $\hat{z}$ be the unique productivity value where $G(\hat{z})=\lambda (1-\delta )/(1+g)<1$. A stationary bubble equilibrium is a solution $z^B$ of (24) such that $z^B>\hat{z}$. Write $\phi (z)$ for the RHS of (24). Since $\phi$ is strictly decreasing and tends to zero for $z\rightarrow \overline{z}$, a bubble stationary equilibrium exists if and only if the LHS of (24) is smaller than $\phi (\hat{z})$ which is equivalent to

$$\begin{aligned} (1+g)\frac{1-\beta }{\beta }< \left[ \frac{Z(\hat{z})}{\hat{z}}-1\right] (g+\delta ). \end{aligned}$$

(29)

Now suppose that $R^N<1+g$. Because of

$$\begin{aligned} R^N=\frac{\lambda (1-\delta )}{G(z^N)}=1-\delta +\frac{z^N}{Z(z^N)}\left( \frac{1+g}{\beta }-1+\delta \right) <1+g, \end{aligned}$$

it follows that $z^N>\hat{z}$, and therefore

$$\begin{aligned} \frac{1+g}{\beta }-1+\delta <\frac{Z(z^N)}{z^N}(g+\delta )\le \frac{Z(\hat{z})}{\hat{z}}(g+\delta ), \end{aligned}$$

where the last inequality follows since Z(z) / z is non-increasing. This proves the existence of a bubble stationary equilibrium.

Conversely, suppose that

$$\begin{aligned} R^N=\frac{\lambda (1-\delta )}{G(z^N)}=1-\delta +\frac{z^N}{Z(z^N)}\left( \frac{1+g}{\beta }-1+\delta \right) \ge 1+g. \end{aligned}$$

This implies that $z^N\le \hat{z}$ and therefore

$$\begin{aligned} \frac{1+g}{\beta }-1+\delta \ge \frac{Z(z^N)}{z^N}(g+\delta )\ge \frac{Z(\hat{z})}{\hat{z}}(g+\delta ). \end{aligned}$$

This proves that no-bubble stationary equilibrium exists if $R^N\ge 1+g$.

Finally, it remains to show $z^B>z^N$ whenever a bubble stationary equilibrium exists. This holds iff

$$\begin{aligned} \frac{1-\beta }{\beta } [1+g-\lambda (1-\delta )] =\phi (z^B)<\phi (z^N). \end{aligned}$$

(30)

Because of

$$\begin{aligned} R^N=\frac{\lambda (1-\delta )}{G(z^N)}=1-\delta +\frac{z^N}{Z(z^N)}\left( \frac{1+g}{\beta }-1+\delta \right) \end{aligned}$$

one has

$$\begin{aligned} \phi (z^N)= & {} [1-G(z^N)] \left[ \frac{Z(z^N)}{z^N}-1\right] (g+\delta )\\= & {} [1-G(z^N)]\frac{g+\delta }{\beta (1-\delta )}\frac{G(z^N)(1+g)-\beta (1-\delta )\lambda }{\lambda -G(z^N)}. \end{aligned}$$

To show (30), it must be proven that

$$\begin{aligned} \frac{1-\beta }{\beta } [1+g-\lambda (1-\delta )] < \underbrace{[1-G]\frac{g+\delta }{\beta (1-\delta )}\frac{G(1+g)-\beta (1-\delta )\lambda }{\lambda -G}}_{\equiv \psi (G)} \end{aligned}$$

(31)

holds at $G=G(z^N)$. Because a bubble equilibrium exists, $R^N<1+g$ implies that $G(z^N)>\lambda (1-\delta )/(1+g)$, and (23) implies that $G(z^N)<\lambda$. It follows $\psi (\lambda (1-\delta )/(1+g))=(1/\beta -1 )[1+g-\lambda (1-\delta )]$ and $\psi (\lambda )=\infty$. Therefore (31) holds if $\psi '(G)>0$ for $G\in [\lambda (1-\delta )/(1+g),\lambda )$ which is easy to verify. This proves that $z^B>z^N$.$\square$

Proposition 3

Under Assumption 1, a credit expansion induced by an increase in parameter $\lambda$ raises $z^N$ and $z^B$, and therefore total factor productivity at any stationary equilibrium.

Proof

Consider first the no-bubble stationary equilibrium at $z^N$. Because the RHS in (23) strictly increases in z under Assumption 1, $z^N$ (and hence TFP which depends positively on Z(z)) increases if $\lambda$ increases. Second consider a bubble stationary equilibrium at $z^B$. Again Assumption 1 implies that the RHS of (24) decreases in z, and since the LHS decreases in $\lambda$, $z^B$ must strictly increase if $\lambda$ increases.$\square$

Proposition 4

Suppose that the production function is $f(Zk)=(Zk)^{\alpha }$. Under Assumption 1 and $R^N<1+g$, there is a maximum sustainable deficit $\overline{\xi }>0$ such that any economy with deficit-output ratio $\xi \in [0,\overline{\xi })$ has two stationary equilibria $(z_i,R_i)_{i=1,2}$ with $z^{N} \le z_1<z_2 \le z^{B}$ and $R^{N} \le R_1<R_2\le 1+g$.

Proof

A steady-state equilibrium (R, z) is a solution to Eqs. (20) and (22). For any given z, the RHS of (20) is a quadratic function of R which intersects the LHS uniquely at some $R>1-\delta$; see Fig. 6. This follows because $\hbox {LHS}(R=1-\delta )>\hbox {RHS} (R=1-\delta )>0$ and $\hbox {RHS}(R=0)=0$, so that the RHS is quadratic in R and strictly increasing in $R>1-\delta$. Since the RHS is strictly decreasing in z, Eq. (20) defines an upward-sloping solution $R(z)\ge 1-\delta$ such that $R(z)\rightarrow (1+g)/\beta$ when $z\rightarrow \overline{z}$.

Fig. 6

Equation (20) defines R(z)

It will now be shown that R(z) intersects twice with the curve that defines implicit solutions (R, z) to the other equilibrium condition (22) for small enough values of the deficit ratio $\xi$. Because two solutions at $z=z^N$ and at $z=z^B$ exist for $\xi =0$, the implicit function theorem also guarantees two (locally unique) solutions for small values of $\xi$, provided that the two curves (20) and (22) have different slopes at $z^B$ and at $z^N$ for $\xi =0$. But this is easy to verify. At $z=z^B$ and $\xi =0$, (22) defines the flat curve $R=1+g$, whereas at $z=z^N$, Eq. (22) defines the downward-sloping $R=\lambda (1-\delta )/G(z)$; see Fig. 7. As shown above, (20) defines an upward-sloping curve R(z). Therefore, for any small enough $\xi >0$, the two equations have two solutions $(z_i,R_i)$, $i=1,2$ such that $z_1\rightarrow z^N$ and $z_2\rightarrow z^B$ when $\xi \rightarrow 0$.

Fig. 7

Intersections between Eqs. (20) and (22) define $z_1$ and $z_2$

It still remains to verify that public debt is nonnegative at either solution. This follows trivially by continuity for the solution $z_2$ because of $R^BG(z^B)>\lambda (1-\delta )$. For the other solution $(z_1,R_1)$, rewrite Eq. (22) as

$$\begin{aligned} \phi (z,R,\xi )\equiv [RG(z)-\lambda (1-\delta )][1+g-R]-{\textstyle \frac{\displaystyle \xi }{\displaystyle \alpha }} (1-G(z))R(R+\delta -1){\textstyle \frac{\displaystyle Z(z)}{\displaystyle z}}. \end{aligned}$$

The local derivatives at $(z,R,\xi )=(z^N,R^N,0)$ have signs

$$\begin{aligned} \phi _1= & {} G'(z^N)R^N(1+g-R^N)\\&+{\textstyle \frac{\displaystyle \xi }{\displaystyle \alpha }}R^N(R^N+\delta -1){\textstyle \frac{\displaystyle (z^N)^2G'(z^N)+\int _{z^N}z\ \mathrm{d}G(z)}{\displaystyle (z^N)^2}}>0, \end{aligned}$$

and $\phi _3<0$ which implies that ${\textstyle \frac{\displaystyle dz}{\displaystyle d\xi }}=-\phi _3/\phi _1>0$ at any given R. Hence, the curve (22) lies above the curve $R=\lambda (1-\delta )/G(z)$ in (z, R) space for $\xi >0$, which implies that $z_1>z^N$ and $R_1>R^N$, and hence $R_1G(z_1)>\lambda (1-\delta )$, so that public debt is positive.

Lastly, because both solutions $(z_i,R_i)$, $i=1,2$, satisfy the upward-sloping relation $R=R(z)$, the equilibrium with the higher interest rate has higher TFP. $\square$

Calibration

Parameters $\alpha$, $\delta$ and g are directly calibrated as mentioned in the text. Given the calibration targets for the interest rate R and for the debt-output ratio (qD) / Y, the deficit-output ratio follows from $\xi =((qD)/Y)\cdot (g-R)/R$. The calibration target for K / Y together with the normalization $Z=1$ yields the steady-state value of $k=K/A=(K/Y)^{1/(1-\alpha )}$. This together with $R=zf'(Zk)+1-\delta$ yields the steady-state value of $z=(K/Y)\cdot (R+\delta -1)/\alpha$, so that the Pareto shape parameter follows from $\varphi =1/(1-z)$. The calibration target for the firm-credit-output ratio $Credit/Y=\theta K/Y=\lambda (1-\delta )/R\cdot (K/Y)$ identifies parameter $\lambda =(Credit/Y)\cdot R/(1-\delta )/(K/Y)$. Then steady-state Eq. (22) can be solved for G(z):

$$\begin{aligned} G(z)={\textstyle \frac{\displaystyle {\textstyle \frac{\displaystyle \xi }{\displaystyle \alpha z}}(R+\delta -1)+\lambda (1-\delta ){\textstyle \frac{\displaystyle g-R}{\displaystyle R}}}{\displaystyle g-R+{\textstyle \frac{\displaystyle \xi }{\displaystyle \alpha z}}(R+\delta -1)}}. \end{aligned}$$

With $G(z)=1-(z_0/z)^{\varphi }$, this yields the Pareto scale parameter $z_0$. Finally, the discount factor follows uniquely from steady-state Eq. (20).

Stability

It can be verified numerically that the no-bubble steady-state equilibrium is locally stable, whereas bubble equilibria are unstable saddle points. To this end, consider again Eqs. (16)–(18) which define the dynamics in the three variables $(k_t,z_t,d_t)$. Substitution of the squared bracket expression in (16) into (17) shows that $d_{t+1}$ depends only on $k_{t+1}$ and $z_{t+1}$, but is otherwise independent of $(k_t,z_t,d_t)$. This implies that along any perfect foresight solution $d_t$ is linked to $(k_t,z_t)$ as follows:

$$\begin{aligned} d_t={\textstyle \frac{\displaystyle k_t [G(z_t)R(k_t,z_t)-\lambda (1-\delta )]}{\displaystyle 1-G(z_t)}}\equiv D(k_t,z_t). \end{aligned}$$

Therefore, local stability properties can be analyzed for the two-dimensional dynamic system in the variables $(z_t,k_t)$ defined by (16) and (18) with $d_t=D(k_t,z_t)$.¹⁶ Linearization of those two equations at a steady-state (k, z) with $d=D(k,z)$, $R=R(k,z)$, $Z=Z(z)$, $f'=f'(Zk)$, $f''=f''(Zk)$ yields

$$\begin{aligned}&\Big [1-\frac{1}{R}\frac{\partial D}{\partial k}-\frac{d}{R^2}\frac{\partial R}{\partial k}\Big ] dk_{t+1} +\Big [ \frac{1}{R}\frac{\partial D}{\partial z}-\frac{d}{R^2}\frac{\partial R}{\partial z}\Big ]dz_{t+1} \nonumber \\&\quad =\frac{\beta }{1+g}\Big [ (f'+Zkf'')Z+1-\delta +\frac{\partial D}{\partial k}\Big ] dk_t\nonumber \\&\qquad +\,\frac{\beta }{1+g}\Big [ (f'+Zkf'')Z'k+\frac{\partial D}{\partial z}\Big ] dz_t, \end{aligned}$$

(32)

$$\begin{aligned}&\quad \Big [ (\xi f+d)\frac{1}{1+g}\frac{\partial R}{\partial k}-\frac{\partial D}{\partial k}\Big ] dk_{t+1} +\Big [ (\xi f+d)\frac{1}{1+g}\frac{\partial R}{\partial z}-\frac{\partial D}{\partial z}\Big ]dz_{t+1} \nonumber \\&\qquad = -\frac{R}{1+g}\Big [\frac{\partial D}{\partial k}+\xi f'Z\Big ]dk_t -\frac{R}{1+g}\Big [\frac{\partial D}{\partial z}+\xi f'kZ'\Big ]dz_t. \end{aligned}$$

(33)

From the definitions of R and D, the partial derivatives in these expressions are:

$$\begin{aligned} \frac{\partial R}{\partial k}&=z\cdot f''\cdot Z,\\ \frac{\partial R}{\partial z}&=f'+z\cdot f'' \cdot k\cdot Z',\\ \frac{\partial D}{\partial k}&=\frac{G(z)R-\lambda (1-\delta )+kG(z)\frac{\partial R}{\partial k}}{1-G(z)},\\ \frac{\partial G}{\partial z}&=k\frac{[G'(z)R+G(z)\frac{\partial R}{\partial z}](1-G(z))+G'(z)[G(z)R-\lambda (1-\delta )]}{(1-G(z))^2}. \end{aligned}$$

The linearized system (32) and (33) can be solved as $(k_{t+1},z_{t+1})'=J\cdot (k_t,z_t)'$ with $2\times 2$-matrix J. At the calibrated steady-state equilibrium $(k,z,d)=(4.089, 0.625, 0.763)$, the eigenvalues are 0.928 and 0.991; hence, the steady state is locally stable. Moreover, one can confirm that the no-bubble steady states located on the solid branches in Fig. 1 are stable, whereas the bubble steady states on the dashed branches are saddle points. For example, with zero deficit, the no-bubble steady state at $d=0$ has eigenvalues 0.930 and 0.984, whereas the bubble steady state at $d=4.03$ has eigenvalues 0.920 and 1.017. Further, at the maximum sustainable deficit level $\xi =0.837\,\%$, there is a bifurcation where the larger eigenvalue equals one.