Real transfers and the Friedman rule

Abstract

We find that the Friedman rule is not optimal with real government transfers and distortionary taxation. As transfers cannot be taxed, a positive nominal net interest rate is the indirect way to tax the additional income derived from transfers. This result holds for heterogeneous agents, standard homogeneous preferences, and constant returns to scale production functions. The presence of real transfers changes the standard optimal taxation result of uniform taxation. Higher transfers imply higher optimal inflation rates. We calibrate a model with transfers to the US economy and obtain optimal values for inflation substantially above the Friedman rule.

A Appendix

We provide an alternative proof to Proposition 2 of Sect. 4.

Proposition 2

When transfers follow path 2, the inflation tax is a more efficient instrument than the labor income tax.

Proof

Consider the simple economy of Sect. 4 with the path of transfers following path 2, $e_{i}=1,$$g=0$, and $\tau _{c}=0.$ Define $\varGamma _{t} \equiv \frac{1}{1-\tau _{t}}$ and $c_{t}\equiv f \left( \varGamma _{t} R_{t} \right)$ as the value of consumption that solves Eqs. (43) and (47). Define the instantaneous indirect utility as $V \left( \varGamma _{t}R_{t} \right) \equiv U(f\left( \varGamma _{t}R_{t}\right) , 1-f\left( \varGamma _{t}R_{t}\right) )$, using the fact that $c_{t}=l_{t}.$ Since U is strictly concave, the optimal allocation is stationary, which implies that $\varGamma _{t}$ and $R_{t}$ should be stationary too. It is trivial to see that V is decreasing in $\varGamma R$. Therefore, the optimal tax policy solves the problem $\min \nolimits _{\varGamma ,R}\varGamma R$ subject to the government budget constraint

$$\begin{aligned} \dfrac{\varGamma R-1}{\varGamma } f\left( \varGamma R\right) =z. \end{aligned}$$

(A.1)

Suppose that $0<\tau <1$, $\varGamma \equiv \frac{1}{1-\tau }>1$ and that $\varGamma$ and R satisfy (A.1). We can show that it is always possible to decrease $\varGamma \ge 1$ and increase R so that $\varGamma R$ decreases and the constraint (A.1) is still satisfied. As a result, the solution of the problem cannot involve $\varGamma >1$. First, a change in $\varGamma$ together with a change in R so that $\frac{\mathrm{d}\varGamma }{\mathrm{d}R} = -\frac{\varGamma }{R}$ maintains the value of $\varGamma R$ constant. Consider an increase in R, $\mathrm{d}R>0$. If $\varGamma$ changes by $\mathrm{d}\varGamma = -\frac{\varGamma }{R} \mathrm{d}R -\varepsilon$, for $\varepsilon >0$, then, as $\mathrm{d}\left( \varGamma R\right) = \varGamma \mathrm{d}R+R\mathrm{d}\varGamma$, this change in $\varGamma$ and R implies a change in $\varGamma R$ equal to $-R\varepsilon <0$. On the other hand, $\mathrm{d}\varGamma$ and $\mathrm{d}R$ change government revenues by $(\frac{1}{\varGamma ^{2}}f\left( \varGamma R\right) + \frac{\varGamma R-1}{\varGamma }f^{\prime }\left( \varGamma R\right) R)\mathrm{d}\varGamma +(f\left( \varGamma R\right) +\left( \varGamma R-1\right) f^{\prime }\left( \varGamma R\right) )\mathrm{d}R$. With $\mathrm{d}\varGamma =-\frac{\varGamma }{R}\mathrm{d}R-\varepsilon$, this change in government revenues is equal to

$$\begin{aligned} \left( 1-\dfrac{1}{\varGamma R}\right) f\left( \varGamma R\right) \mathrm{d}R-\left( \dfrac{1}{\varGamma ^{2}} f \left( \varGamma R\right) + \dfrac{\varGamma R-1}{\varGamma } f^{\prime } \left( \varGamma R\right) R\right) \varepsilon . \end{aligned}$$

(A.2)

As $\varGamma R>1$ and $f\left( \varGamma R\right) >0$, the coefficient on dR is strictly positive. Therefore, for any $\mathrm{d}R>0$, there is a sufficiently small $\varepsilon >0$ such that the expression in (A.2) is positive. This means that $\varGamma R$ decreases, but government revenues increase. We then have that a pair $\left( \varGamma ,R\right)$, with $\varGamma >1$, cannot be the solution to the Ramsey problem as there is a decrease in the labor tax rate and an increase in the nominal interest rate such that the distortion $\varGamma R$ decreases and government revenues do not decrease. It follows that the solution to the Ramsey problem requires $\varGamma ^{*}=1$. Moreover, setting $\varGamma ^{*}=1$ in (A.1) implies that $R^{*}$ is equal to the smallest value that satisfies $\left( R-1\right) f(R) = z$. As $\varGamma = \frac{1}{1-\tau }$ and $z>0$, $\tau =0$ and $R>1$.$\square$