Heterogeneity and monetary policy

Abstract

In a simple environment with heterogeneous agents, we demonstrate that the central bank delivers a higher inflation rate than when population is homogeneous. This tendency to choose a higher level of inflation than efficiency dictates is due to the efficiency-vs-equity trade-off that the central bank faces in this heterogeneous economy: up to a certain level, inflation decreases inequality. Optimal delegation involves appointing a central bank that puts a higher weight on the utility of the high-productivity workers than society does. This effect of delegation that improves the macroeconomic outcome disappears as homogeneity is restored. However, the inflation level under optimal delegation does not reach the efficiency level.

Appendix

1.1 Properties of the equilibrium: homogenous case

We have the following first-order conditions:

$$\begin{aligned} \frac{u_{ct}}{p_{t}}= & {} \frac{\beta \, R_t}{1+a_t} \cdot \frac{u_{c t+1}}{p_{t+1}} \,, \end{aligned}$$

(A.1)

$$\begin{aligned} \frac{u_{ct}}{u_{nt}}= & {} -\frac{R_t}{w} \,. \end{aligned}$$

(A.2)

Step 1: Stationarity of equilibrium. We first will establish that the maximum-welfare equilibrium is stationary. In this homogeneous environment, equilibrium aggregate and individual money holdings coincide, i.e. $M_t=M_t^a$, and there is no borrowing and lending in equilibrium; therefore we have

$$\begin{aligned} m_t=1, \qquad b_t=0. \end{aligned}$$

(A.3)

Then from the CIA and budget constraints (1)–(2) it follows that

$$\begin{aligned} p_t c_t= & {} m_{t-1}+a_t \,, \end{aligned}$$

(A.4)

$$\begin{aligned} p_t w n_t= & {} m_t(1+a_t) \,. \end{aligned}$$

(A.5)

Conditions (A.1), (A.4), (A.3) and the functional form of the utility function (9) imply

$$\begin{aligned} R_t = \frac{1+a_t}{\beta }\cdot \frac{p_{t+1}c_{t+1}}{p_tc_t} = \frac{1+a_t}{\beta }\cdot \frac{m_t+a_{t+1}}{m_{t-1}+a_t} = \frac{1+a_{t+1}}{\beta } \,. \end{aligned}$$

(A.6)

Conditions (A.2), (A.6) and the functional form of the utility function (9) imply

$$\begin{aligned} \frac{1-n_t}{\kappa \cdot c_t} = \frac{1+a_{t+1}}{\beta }\cdot \frac{1}{w} \,. \end{aligned}$$

(A.7)

By dividing (A.4) by (A.5) and using (A.3) we obtain

$$\begin{aligned} c_t =w n_t \,, \end{aligned}$$

(A.8)

which together with (A.7) yields

$$\begin{aligned} \frac{1-n_t}{\kappa n_t} = \frac{1+a_{t+1}}{\beta } \,. \end{aligned}$$

(A.9)

This means that employment $n_t$ is uniquely determined by $a_{t+1}$ alone. In other words, we can think of employment as a function of $a_{t+1}$, i.e. $n_t = n_t(a_{t+1})$. From this and (A.8) we have consumption as a function of $a_{t+1}$ too: $c_t = c_t(a_{t+1})$.

Then the CB’s problem in period t becomes a static problem of maximization of

$$\begin{aligned} u(c_t(a_{t+1}),n_t(a_{t+1})) \end{aligned}$$

by choosing optimal $a_{t+1}$. Therefore $a_{t+1}$ is stationary, i.e. $a_{t+1}=a$. This implies stationarity of employment, consumption and the rest of variables.

Step 2: Friedman rule. Condition (A.1) and stationarity yield

$$\begin{aligned} R_t = R = \frac{1+a}{\beta } \,. \end{aligned}$$

(A.10)

Let us now turn to the efficient allocation, which is found as a solution to the following problem: $\max u(c,n) \quad \text{ s.t } \quad c=wn \,.$ The first-order condition:

$$\begin{aligned} \frac{u_{c}}{u_{n}} = -\,\frac{1}{w} \,. \end{aligned}$$

(A.11)

Comparing (A.11) with the (due to stationarity of equilibrium) stationary version of (A.2), we conclude that in the maximum-welfare equilibrium,

• $R_t = R = 1$ (the Friedman rule holds);

• the money growth rate is constant and $a_t = a = \beta -1$ [this follows from (A.10)].

1.2 Proof of Proposition 1

Step 1. From (3) and (9) we obtain

$$\begin{aligned} R_t = \frac{1+a_t}{\beta }\cdot \frac{p_{t+1}c_{i,t+1}}{p_t c_{it}} \,. \end{aligned}$$

(A.12)

Rearranging and summing over i yields

$$\begin{aligned} \sum _i \alpha _i p_t c_{it} R_t = \frac{1+a_t}{\beta } \sum _i \alpha _i p_{t+1} c_{i,t+1} \, \end{aligned}$$

which together with (5), (7) and (8) result in

$$\begin{aligned} R_t = \frac{1+a_{t+1}}{\beta } \,. \end{aligned}$$

(A.13)

Step 2. From (6) it follows that

$$\begin{aligned} \frac{m_{ht}}{m_{lt}} = \frac{w_h n_{ht}}{w_l n_{lt}} \,. \end{aligned}$$

(A.14)

Combining this with (7) yields

$$\begin{aligned} m_{it} = \frac{w_i n_{it}}{\sum _j \alpha _j w_j n_{jt}} \,. \end{aligned}$$

(A.15)

Step 3. From (4) and (9) we have $\frac{1-n_{it}}{\kappa c_{it}}=\frac{R_t}{w_i}$. This and the use of (A.13) result in

$$\begin{aligned} c_{it} = \frac{\beta }{\kappa }\cdot \frac{w_i(1-n_{it})}{1+a_{t+1}} \,. \end{aligned}$$

(A.16)

Note that it follows from (A.12) that $\frac{c_{h,t+1}}{c_{h,t}}=\frac{c_{l,t+1}}{c_{lt}}$ and thus $\frac{c_{h,t+1}}{c_{l,t+1}} = \frac{c_{ht}}{c_{lt}}$, i.e. this ratio stays constant over time. Let the common value of this ratio be denoted $\phi$:

$$\begin{aligned} \frac{c_{ht}}{c_{lt}} = \phi , \quad t = 0, 1, 2, \ldots \,. \end{aligned}$$

(A.17)

Then from this and (A.16) we obtain

$$\begin{aligned} \frac{1-n_{ht}}{1-n_{lt}} = \phi \cdot \frac{w_l}{w_h}, \quad t = 0, 1, 2, \ldots \,. \end{aligned}$$

(A.18)

Step 4. Combining the goods market clearing condition $\sum _j \alpha _j c_{jt} = \sum _j \alpha _j w_j n_{jt}$ [which follows from (5)–(8)] with (A.16) yields

$$\begin{aligned} \frac{\beta }{\kappa }\cdot \frac{1}{1+a_{t+1}}\cdot \sum _j \alpha _j w_j (1-n_{jt}) = \sum _j \alpha _j w_j n_{jt} \end{aligned}$$

(A.19)

For a given value of $\phi$, Eqs. (A.18) and (A.19) are a system of two equations with two unknowns, $n_{ht}$ and $n_{lt}$, the solution to which is two functions of $a_{t+1}$, i.e. $n_{it}=n_{it}(a_{t+1})$. Then consumption $c_{it}$ can be found as functions of $a_{t+1}$ as well, i.e. $c_{it}=c_{it}(a_{t+1})$.

Then the problem of CB reduces to a sequence of static problems whereby at each period t, the weighted utility

$$\begin{aligned} \sum _i \gamma _i u(c_{it}(a_{t+1}),n_{it}(a_{t+1})) \end{aligned}$$

is maximized by choosing $a_{t+1}$. Thus, $a_{t+1}$ is stationary: $a_{t+1}=a^*$. This implies stationarity of emplyment and consumption values: $n_{it}=n_{it}(a^*)=n_i$ and $c_{it}=c_{it}(a^*)=c_i$ The stationarity of $m_{it}$ follows from (A.15). Then stationarity of $p_t$ and $R_t$ come from (6) and (A.13) respectively. The stationarity of the term $b_{i,t}$ follows from (5). Thus we have established stationarity of the maximum-welfare equilibrium. It may seem that all these stationary values depend on $\phi$. However, we can pin down the value of $\phi$ by requiring that $b_{it}=0$ in a stationary equbibrium (otherwise one type of agents would owe a positive amount to the other type and never return the debt).

1.3 Proof of Claim 1

From (11), using the functional form of the utility function $u(c,n) = \ln c + \kappa \ln (1-n)$ assumed in (9) we obtain

$$\begin{aligned} c_i = \frac{\beta w_i}{\kappa } \cdot \frac{1-n_i}{1+a} \,. \end{aligned}$$

(A.20)

Using this to eliminate $c_i$ and (13) will result in a system of 3 equations with 3 unknowns: $p, n_h$, and $n_l$. Solving it yields

$$\begin{aligned} p= & {} \frac{(1+a)(\kappa +\beta +\kappa a)}{\beta \bar{w}} \,,\nonumber \\ n_i= & {} \beta \, \frac{(\kappa +\beta )w_i + \kappa \,(w_i - \bar{w})\,a}{w_i(\kappa +\beta )(\kappa +\beta +\kappa a)} \,, \end{aligned}$$

(A.21)

where

$$\begin{aligned} \bar{w}= {\scriptstyle \sum }_i \alpha _i w_i \,. \end{aligned}$$

(A.22)

The derivative of $n_i$ with respect to a is

$$\begin{aligned} n_i'(a) = -\, \frac{\kappa \beta \bar{w}}{(\kappa +\beta +\kappa a)^2 w_i} < 0. \end{aligned}$$

(A.23)

Similarly, using this and (A.20) we get

$$\begin{aligned} c_i'(a) = -\, \frac{\beta \, \mathcal {C}_i}{(\kappa +\beta )(\kappa +\beta +\kappa a)^2(1+a)^2} \,, \end{aligned}$$

where $\mathcal {C}_i = (w_i\kappa ^2+\bar{w}\beta \kappa )(a+1)^2 + 2\beta \kappa (w_i-\bar{w})(a+1) + \beta ^2(w_i-\bar{w}) \,.$ For $i=h$, clearly $\mathcal {C}_h > 0$ since $a+1 \ge \beta >0$ as $w_h > \bar{w}$. Thus, $c_h'(a)<0$.

In case of $i=l$, determining the sign of $\mathcal {C}_l$ is not straightforward. Since $\mathcal {C}_l$ is a quadratic function of a, for large enough values of a, its sign will be positive. Let us assess its value at the lower bound $a=\beta -1$. It is easy to show that $\mathcal {C}_l|_{a=\beta -1} = \beta ^2 \left\{ w_l \kappa ^2 + [\beta \bar{w}-2\alpha _h(w_h-w_l)]\kappa -\alpha _h(w_h-w_l) \right\} ,$ where $\bar{w}={\scriptstyle \sum }\alpha _i w_i$. The quadratic equation $w_l \kappa ^2 + [\beta \bar{w}-2\alpha _h(w_h-w_l)]\kappa -\alpha _h(w_h-w_l) = 0$ with respect to $\kappa$ has only one positive root $\tilde{\kappa }$. Since $c_l'(a)$ and $\mathcal {C}_l$ have opposite signs it is clear that $c_l'|_{a=\beta -1} > 0$ for $\kappa \in (0,\tilde{\kappa })$.

1.4 Proof of Claim 2

Step 1. Using the functional form (9) of the utility function as well as the relationship between equilibrium consumption and labor (A.20) we have $U_i'(a) = \frac{c_i'}{c_i} -\frac{\kappa \, n_i'}{n_i} = -\, (1+\kappa )\frac{n_i'}{1-n_i} - \frac{1}{1+a}.$ Then using (A.21), (A.22) and (A.23) we obtain

$$\begin{aligned} U_i'(a) = \frac{\beta (1+\kappa )(\kappa +\beta )\bar{w}}{(\kappa +\beta +\kappa a)[w_i(\kappa +\beta )+(w_i\kappa +\beta \bar{w})a]} - \frac{1}{1+a}. \end{aligned}$$

Step 2. We will find the value of $U_i'$ at $a=\beta -1$: $U_i'(\beta -1) = \frac{\kappa +1}{\beta }\cdot \frac{\bar{w}-w_i}{w_i(\kappa +1)+\bar{w}(\beta -1)} \,.$ Thus, $U_h'(\beta -1) <0 \,, \quad U_l'(\beta -1) >0 \,.$

Step 3.$U_i'$ can be written in this form: $U_i'(a) = \frac{A_2}{A_0+A_1a}\cdot \frac{1}{B_{i0}+B_{i1}a} - \frac{1}{1+a} \,,$ where

$$\begin{aligned} A_2= & {} \beta (1+\kappa )(\kappa +\beta )\bar{w}\,, \quad A_0 = \kappa +\beta \,, \quad A_1 = \kappa \,, \quad B_{i0} = (\kappa +\beta )w_i \,, \\ B_{i1}= & {} \kappa w_i+\beta \bar{w}\,. \end{aligned}$$

Then

$$\begin{aligned} U_i'(a) = \frac{-A_1B_{i1}a^2+[A_2-A_0B_{i1}-A_1B_{i0}]a+(A_2-A_0B_{i0})}{(A_0+A_1a)(B_{i0}+B_{i1}a)(1+a)} = \frac{\mathcal {N}_i(a)}{\mathcal {M}_i(a)} \,. \end{aligned}$$

Since $\mathcal {M}_i>0$, the sign of $U_i'$ is the same as that of $\mathcal {N}_i(a)$. Note that $\mathcal {N}_i(a)$ is a parabola that reaches its maximum at

$$\begin{aligned} \tilde{a}_i = \frac{A_2-A_0B_{i1}-A_1B_{i0}}{2A_1B_{i1}} = \frac{\kappa +\beta }{2}\cdot \frac{\beta \bar{w}-2w_i}{\kappa w_i+\beta \bar{w}} \,. \end{aligned}$$

Let us show that $\tilde{a}_h < \beta -1$. To simplify notation, let us use $\xi$ to denote $\bar{w}/w_h$. Then we need to show that $\frac{\kappa +\beta }{2}\cdot \frac{\beta \xi -2}{\beta \xi +\kappa } < \beta -1 \,,$ which holds if and only if $(\kappa +\beta )(\beta \xi -2) < (2\beta -2)(\kappa +\beta \xi ) \,.$ The latter is equivalent to

$$\begin{aligned} \xi < 1+ \frac{\kappa +\beta }{\kappa +2-\beta } \,, \end{aligned}$$

which obviously holds as $\xi =\bar{w}/w_h<1$. Therefore the function $\mathcal {N}_h(a)$ is decreasing for $a \ge \beta -1$. And since $U_h'(\beta -1)<0$ and thus $\mathcal {N}_h(\beta -1)<0$, it is clear that $\mathcal {N}_h(a)<0$ for $a \ge \beta -1$, and thus

$$\begin{aligned} U_h'(a)<0 \quad a \ge \beta -1. \end{aligned}$$

Let us turn to $U_l'(a)$. Since $\mathcal {N}_l(a)$ is a parabola which eventually becomes negative for large values of a, and $\mathcal {N}_l(\beta -1)>0$, the equation $\mathcal {N}_l(\beta -1)=0$ has a unique solution for $a \ge \beta -1$. Let us denote it by $a^*$. Since both $U_i'(a)<0$ for $a>a^*$, it is reasonable to assume that $a \in [\underline{a},\overline{a}] = [\beta -1,a^*] \,.$ Then indeed $U_h'(a)<0 \,, \,\, U_l'(a)>0 \,, \,\, \text{ for } \text{ all }\,\, a \ge \beta -1 \,.$

1.5 Proof of Proposition 2

We will demonstrate that $\mathcal {C}_h/\mathcal {C}_l$ is a decreasing function of a. From (15) and (16) it follows that

$$\begin{aligned} \frac{\mathcal {C}_h}{\mathcal {C}_l} = \frac{w_h(1-n_h)}{w_l(1-n_l)} \cdot \frac{\alpha _h}{\alpha _l} = \frac{a\beta \bar{w}+ w_h(a\kappa +\beta +\kappa )}{a\beta \bar{w}+ w_l(a\kappa +\beta +\kappa )} \cdot \frac{\alpha _h}{\alpha _l} \end{aligned}$$

As $w_h>w_l$, the derivative of this ratio with respect to a:

$$\begin{aligned} \frac{\partial }{\partial a}\left( \frac{\mathcal {C}_h}{\mathcal {C}_l}\right) = -\, \frac{\beta (\beta +\kappa )\bar{w}(w_h-w_l)}{(a\beta \bar{w}+ w_l(a\kappa +\beta +\kappa ))^2} \cdot \frac{\alpha _h}{\alpha _l} < 0. \end{aligned}$$

1.6 Proof of Claim 3

The solution to the problem is characterized by the following equations:

$$\begin{aligned} \frac{u_c(i)}{u_n(i)}&= -\, \frac{1}{w_i} \,, \end{aligned}$$

(A.24)

$$\begin{aligned} \frac{u_c(h)}{u_c(l)}&= \frac{\delta _l}{\delta _h}\cdot \frac{\alpha _h}{\alpha _l} \,. \end{aligned}$$

(A.25)

These three equations together with the goods market clearing condition ${\scriptstyle \sum }\alpha _i c_i = {\scriptstyle \sum }\alpha _i w_i n_i$ yield the following solution:

$$\begin{aligned} n_i = \frac{\alpha _i w_i (\kappa +1) - \delta _i\kappa {\scriptstyle \sum }\alpha _i w_i}{\alpha _i w_i (\kappa +1)} = \frac{\alpha _i w_i (\kappa +1) - \delta _i\kappa \,\bar{w}}{\alpha _i w_i (\kappa +1)} \,, \qquad \quad c_i = \frac{w_i(1-n_i)}{\kappa } \,, \end{aligned}$$

[we made use of (A.22)] with the following property: $c_h = \frac{\delta _h}{\delta _l}\cdot \frac{\alpha _l}{\alpha _h}\, c_l \,.$

Compare the F.O.C.s (10) and (11) of a competitive equilibrium with (A.24) and (A.25). For a competitive equilibrium allocation to be Pareto efficient, it should satisfy the following conditions:

$$\begin{aligned} \frac{u_n(h)}{u_n(l)}&= -\, \frac{1-\delta _h}{\delta _h}\cdot \frac{\alpha _h w_h}{\alpha _l w_l} \,, \end{aligned}$$

(A.26)

$$\begin{aligned} \frac{1+a}{\beta }&= 1 \,. \end{aligned}$$

(A.27)

Suppose it satisfies (A.27). Since the function $\delta \rightarrow (1-\delta )/\delta$ takes on all positive values when $\delta \in (0,1)$ there exists $\delta _h\in (0,1)$ such that (A.26) is satisfied.

1.7 Proof of Proposition 4

Step 1. For a fixed level of money growth a, the social planner first chooses T that maximizes $\alpha _h u\big (c_h-T,n_h\big )+\alpha _l u\big (c_l+T\cdot \frac{\alpha _h}{\alpha _l},n_l\big )$, which leads to the condition

$$\begin{aligned} u_c\big (c_h-T,n_h\big ) = u_c\Big (c_l+T\cdot \frac{\alpha _h}{\alpha _l},n_l\Big ) \,. \end{aligned}$$

(A.28)

Let T(a) be the optimal choice of lump-sum transfer T. And let

$$\begin{aligned} \mathcal {V}_{\alpha } (a) = \alpha _h\cdot u\Big (c_h(a)-T(a),n_h(a)\Big ) + \alpha _l\cdot u\Big (c_l(a)+T\cdot \frac{\alpha _h}{\alpha _l},n_l(a)\Big ) \,. \end{aligned}$$

Combining (A.28), (9), and (16) yields

$$\begin{aligned} \mathcal {V}_{\alpha } (a) = \ln \left( \sum _i \alpha _i w_i n_i\right) + \kappa \sum _i \alpha _i \ln (1-n_i) \,, \end{aligned}$$

(A.29)

where $n_i$ is as in (15).

Step 2. Let us show that

$$\begin{aligned} \mathcal {V}'_{\alpha } (a) < V'_{\alpha }(a) \,, \end{aligned}$$

(A.30)

where $V'_{\alpha }(a)$ as in (22). From (A.29) and (22) we have

$$\begin{aligned} \mathcal {V}'_{\alpha } (a) - V'_{\alpha }(a) = \left[ \ln \left( \sum _i \alpha _i c_i\right) \right] ' - \left[ \sum _i \alpha _i \ln c_i \right] ' = \sum _i \alpha _i \frac{c_i'}{c_i} - \frac{\sum _i \alpha _i c_i'}{\sum _i \alpha _i c_i} \,. \end{aligned}$$

(A.31)

Given that $c_i>0$, it can be shown that the last expression in (A.31) is negative if and only if

$$\begin{aligned} (c_h-c_l)\cdot (c_l'\cdot c_h - c_h'\cdot c_l) > 0 \,. \end{aligned}$$

(A.32)

It is easy to show that consumption of the high type is larger than that of the low type; from (16) and (15) it follows that

$$\begin{aligned} c_h-c_l = \frac{\beta (w_h-w_l)}{(\kappa +\beta )(1+a)} > 0 \,. \end{aligned}$$

Thus, (A.32) holds if and only if

$$\begin{aligned} c_l'\cdot c_h - c_h'\cdot c_l > 0 \,. \end{aligned}$$

(A.33)

Using (16) we get

$$\begin{aligned} c_l'\cdot c_h - c_h'\cdot c_l = \frac{\beta ^2 w_h w_l}{\kappa ^2(1+a)^2} \left[ -(1-n_h)n_l' + (1-n_l)n_h' \right] \,. \end{aligned}$$

(A.34)

From (15) and (A.23) we obtain

$$\begin{aligned} -(1-n_h)n_l' + (1-n_l)n_h' = \frac{\kappa \beta \bar{w}}{(\kappa +\beta +\kappa a)^3(\kappa +\beta )w_h w_l} \big [\mathcal {B}_h-\mathcal {B}_l\big ] \,, \end{aligned}$$

(A.35)

where $\mathcal {B}_i = w_i(\kappa +\beta )(\kappa +\beta +\kappa a) - \beta (\kappa +\beta )w_i - \beta \kappa (w_i-\bar{w})a$. Simplifications yield

$$\begin{aligned} \mathcal {B}_h - \mathcal {B}_l = \kappa (w_h-w_l)(\kappa +\beta +\kappa a) > 0 \,, \end{aligned}$$

which together with (A.34) and (A.35) proves that (A.33) holds and therefore $\mathcal {V}'_{\alpha } (a) - V'_{\alpha }(a) < 0$.

Recall that $a_{\alpha }$ solves $V'_{\alpha }(a)=0$. Therefore $\mathcal {V}'_{\alpha } (a_{\alpha })<0$. Thus, the social planner wants to choose a monetary growth level at a lower level than $a_{\alpha }$. We have shown that the optimal money growth level $a^*<a_{\alpha }$.

Step 3. Let us show $a^*>\beta -1$. It suffices to prove that $\mathcal {V}'_{\alpha } (\beta -1) >0$ so that the social planner would want to move the money growth rate up from the eficient level $\beta -1$. Differentiating (A.29) and using (A.23) yields after some simplification

$$\begin{aligned} \mathcal {V}'_{\alpha } = \frac{\kappa }{\kappa +\beta +\kappa a} \left[ -1 + \beta (\kappa +\beta )\bar{w}\sum _i \frac{\alpha _i}{(\kappa +\beta +\kappa a)w_i + \beta \bar{w}a} \right] \,. \end{aligned}$$

(A.36)

Then

$$\begin{aligned} \mathcal {V}'_{\alpha }(\beta -1) = \frac{\kappa }{\beta (\kappa +1)} \left[ -1 + (\kappa +\beta )\bar{w}\frac{(\kappa +1)\bar{\bar{w}}+\bar{w}(\beta -1)}{\prod _i[(\kappa +1)w_i + (\beta -1)\bar{w}]} \right] \,, \end{aligned}$$

(A.37)

where $\bar{\bar{w}}=\alpha _h w_l + \alpha _l w_h$. Further simplification results in

$$\begin{aligned} \mathcal {V}'_{\alpha }(\beta -1)= & {} \frac{\kappa }{\beta } \cdot \frac{(\kappa +1)(-w_h w_l+\bar{\bar{w}}\bar{w})}{\prod _i[(\kappa +1)w_i + (\beta -1)\bar{w}]} \,. \end{aligned}$$

Note that clearly $(\kappa +1)w_h + (\beta -1)\bar{w}> 0$, and $(\kappa +1)w_l + (\beta -1)\bar{w}> 0$ due to the parameter restriction $\frac{w_h}{w_l} < \frac{\kappa +\beta +\alpha _h(1-\beta )}{(1-\beta )\alpha _h}$ (see Appendix A.10). Therefore

$$\begin{aligned} \frac{\kappa }{\beta } \cdot \frac{(\kappa +1)}{\prod _i[(\kappa +1)w_i + (\beta -1)\bar{w}]} > 0 \,. \end{aligned}$$

and $\mathcal {V}'_{\alpha }(\beta -1)>0$ if and only if $-w_h w_l+\bar{\bar{w}}\bar{w}>0$. One can easily show that

$$\begin{aligned} -w_h w_l+\bar{\bar{w}}\bar{w} = \alpha _h \alpha _l (w_h-w_l)^2 > 0 \,. \end{aligned}$$

Therefore $\mathcal {V}'_{\alpha }(\beta -1)>0$. From here we conclude that $a^* > \beta -1$.

Step 4. From Steps 2 and 3 we conclude that $\beta -1<a^*<a_{\alpha }$. Since $a=\beta -1$ and $a=a_{\alpha }$ are delivered by CBs with preferences $\gamma _h=\bar{\gamma }_h$ and $\gamma _h=\alpha _h$ respectively, we conclude that the optimal money growth rate $a^*$ is delivered by a CB with the weights $\hat{\gamma }$ such that $\alpha _h< \hat{\gamma }_h < \bar{\gamma }_h$. Putting all this together, we have:

$$\begin{aligned} a^*=a_{\hat{\gamma }}\,, \qquad \beta -1< a_{\hat{\gamma }}< a_{\alpha }\,, \qquad \alpha _h< \hat{\gamma }_h < \bar{\gamma }_h \,. \end{aligned}$$

1.8 On the sign of $\psi '(a_0)$

Let $W_h(a)$ be defined as the left-hand side of (27), i.e.

$$\begin{aligned} W_h(a) = u(c_h(a)-\psi (a),n_h(a)) \,. \end{aligned}$$

(A.38)

Since the right-hand side of (27) is constant, $W_h'(a)=0$. In other words,

$$\begin{aligned} W_h' = u_c(c_h(a)-\psi (a),n_h(a))\cdot [c_h'-\psi '] + u_n(c_h(a)-\psi (a),n_h(a))\cdot n_h' = 0 \,. \end{aligned}$$

(A.39)

Then

$$\begin{aligned} \psi ' = \frac{u_c(c_h(a)-\psi (a),n_h(a))\cdot c_h'+u_n(c_h(a)-\psi (a),n_h(a))\cdot n_h'}{u_c(c_h(a)-\psi (a),n_h(a))} \,. \end{aligned}$$

(A.40)

Recall that $U_i(a)$ in (18) is defined as $U_i(a)=u(c_i(a),n_i(a))$ and thus

$$\begin{aligned} U_i' = u_c(c_i(a),n_i(a))\cdot c_i' + u_n(c_i(a),n_i(a))\cdot n_i' \,. \end{aligned}$$

(A.41)

Since $\psi (a_0)=0$, from (A.40) and (A.41) we have

$$\begin{aligned} \psi '(a_0)= & {} \frac{u_c(c_h(a_0),n_h(a_0))\cdot c_h'(a_0)+u_n(c_h(a_0),n_h(a_0))\cdot n_h'(a_0)}{u_c(c_h(a_0),n_h(a_0))} \nonumber \\= & {} \frac{U_h'(a_0)}{u_c(c_h(a_0),n_h(a_0))} \,. \end{aligned}$$

(A.42)

Since $U_h'(a_0)<0$ (Claim 2) and $u_c>0$, we have $\psi '(a_0)<0$.

1.9 Proof of Claim 6

It is sufficient to show that $W_l'(a_0)<0$.

Step 1. Similarly to derivation of (A.39) and by using (A.41) evaluated at $a_0$, we obtain

$$\begin{aligned} W_l'(a_0) = U_l'(a_0) + u_c(c_l(a_0),n_l(a_0))\cdot \frac{\alpha _h}{\alpha _l}\cdot \psi '(a_0) \,. \end{aligned}$$

(A.43)

From (A.43) and (A.42) we obtain

$$\begin{aligned} W_l'(a_0) = U_l'(a_0) + \frac{u_c(c_l(a_0),n_l(a_0))}{u_c(c_h(a_0),n_h(a_0))}\cdot \frac{\alpha _h}{\alpha _l}\cdot U_h'(a_0) = U_l'(a_0) + \frac{\alpha _h c_h(a_0)}{\alpha _l c_l(a_0)}\cdot U_h'(a_0) \,. \end{aligned}$$

or

$$\begin{aligned} \alpha _l c_l(a_0)W_l'(a_0) = \sum _i \alpha _i c_i(a_0)U_i'(a_0) \,. \end{aligned}$$

Thus $W_l'(a_0)<0$ if and only if $\sum _i \alpha _l c_i(a_0)U_i'(a_0) <0$.

Step 2. Let us determine the sign of $\sum _i \alpha _i c_i(a_0)U_i'(a_0) <0$. From (A.41) and (16), using (9) we have

$$\begin{aligned} 1-n_i = \frac{\kappa (1+a)}{\beta w_i}\cdot c_i \,. \end{aligned}$$

(A.44)

Thus

$$\begin{aligned} n_i' = -\,\frac{\kappa }{\beta w_i} \cdot [c_i + (1+a)c_i'] \,. \end{aligned}$$

(A.45)

From (A.41), (A.44) and (A.45) it follows that

$$\begin{aligned} c_iU_i' = \frac{\kappa c_i+(1+a)(1+\kappa )c_i'}{(1+a)} \,, \end{aligned}$$

(A.46)

Differentiating (A.20) yields

$$\begin{aligned} c_i' = \frac{\beta w_i}{\kappa }\cdot \frac{-n_i'(1+a)-(1-n_i)}{(1+a)^2} \end{aligned}$$

(A.47)

Then combining (A.47), (A.23), and (A.20) results in

$$\begin{aligned} \kappa c_i + (1+a)(1+\kappa ) c_i' = -\, \beta \left\{ \frac{\bar{w}\beta (1+a)+\kappa (1+a)w_i+\beta (w_i-\bar{w})}{(\kappa +\beta ) [\kappa (1+a)+\beta ](1+a)} - \frac{(1+\kappa )\beta \bar{w}}{[\kappa (1+a)+\beta ]^2} \right\} \,. \end{aligned}$$

(A.48)

Then from (A.46) and (A.48), after some simplification, we obtain

$$\begin{aligned} \sum _i \alpha _i c_i U_i'(a) = -\, \frac{\beta \bar{w}}{1+a}\cdot \frac{\kappa (1+a-\beta )}{[\kappa (1+a)+\beta ]^2} \,. \end{aligned}$$

(A.49)

Since $a_0>\beta -1$ by assumption, we have $\sum _i \alpha _i c_i U_i'(a_0)<0$. Therefore, it follows from the conclusion of Step 1 that $W_l'(a_0)<0$.

1.10 Conditions on parameters

We impose some conditions on model parameters to ensure that the first-order-condition approach works. More precisely, we will make sure that $n_i \in (0,1)$ in both competitive equilibrium and social optimum. It is shown below that the following conditions are sufficient:

$$\begin{aligned} \beta -1\le & {} a < \max \left( \kappa +\beta , \frac{1-\beta }{\kappa } \right) \,, \end{aligned}$$

(A.50)

$$\begin{aligned} \frac{w_h}{w_l}< & {} \min \left( \frac{1+\alpha _h\kappa }{\alpha _h\kappa },\frac{\kappa +\beta +\alpha _h(1-\beta )}{\alpha _h(1-\beta )} \right) \,. \end{aligned}$$

(A.51)

From (15) it follows that in competitive equilibrium, $n_i>0$ if and only if

$$\begin{aligned} (\kappa +\beta )w_i + \kappa \,(w_i - {\scriptstyle \sum }_i \alpha _i w_i)a > 0 \,. \end{aligned}$$

This clearly holds for $i=h$ since $w_h > {\scriptstyle \sum }_i \alpha _i w_i$. For case $i=l$ the inequality above is equivalent to

$$\begin{aligned} a < \frac{\kappa +\beta }{\kappa (\bar{w}/w_l-1)} \,, \end{aligned}$$

(A.52)

where, as before, $\bar{w}={\scriptstyle \sum }_i \alpha _i w_i$ [see (A.22)].

We also require $n_i<1$. This, using (15), can be shown to be equivalent to

$$\begin{aligned} \kappa +\beta - (\kappa +\beta \bar{w}/w_i)(1-\beta ) > 0 \,. \end{aligned}$$

Again, the above inequality holds for $i=h$ as $w_h>\bar{w}$. For case $i=l$ the inequality above is equivalent to this one:

$$\begin{aligned} \frac{w_h}{w_l} < \frac{\kappa +\beta +\alpha _h(1-\beta )}{\alpha _h(1-\beta )} \,. \end{aligned}$$

(A.53)

Let us turn to the social optimum. From (19) it is clear that $n_i^s<1$. We also require that $n_i^s>0$. Since $\bar{w}<w_h$, from (19) it follows that $n^s_h>0$. By requiring $n_l^s>0$ we derive the following:

$$\begin{aligned} \frac{w_h}{w_l} < \frac{\alpha _h\kappa +1}{\alpha _h\kappa } \,. \end{aligned}$$

(A.54)

Combining (A.53) and (A.54) we obtain

$$\begin{aligned} \frac{w_h}{w_l} < \min \left( \frac{\alpha _h\kappa +1}{\alpha _h\kappa }, \frac{\kappa +\beta +\alpha _h(1-\beta )}{\alpha _h(1-\beta )} \right) \,. \end{aligned}$$

(A.55)

Since $\bar{w}={\scriptstyle \sum }_i \alpha _i w_i$ and $\bar{w}/w_l-1 = \alpha _h(w_h/w_l-1)$, we obtain from (A.55) that

$$\begin{aligned} \frac{\kappa +\beta }{\kappa }\cdot \frac{1}{\bar{w}/w_l-1} > \frac{\kappa +\beta }{\kappa } \max \left( \kappa , \frac{1-\beta }{\kappa +\beta } \right) = \max \left( \kappa +\beta , \frac{1-\beta }{\kappa } \right) \,. \end{aligned}$$

From this and (A.52) we conclude that the following is a sufficient condition for the competitive allocation to have $n_i>0$:    $\beta -1 \le a < \max \left( \kappa +\beta , \frac{1-\beta }{\kappa } \right) \,.$