Optimal interregional redistribution and local budget rules with multidimensional heterogeneity

Abstract

In this study we analyse optimal interregional redistribution and local budget rules in a two-region, two-period federation model. The two regions differ in privately observable discount factors and in publicly observable durability of local public goods. We address the question of whether the contributor region of redistribution should face a weaker borrowing constraint than the recipient region. The answer to this question is yes for two cases: (1) the patient region is the recipient and has public goods durability no greater than that of the impatient region; (2) the impatient region is the recipient with smaller public goods durability and the regional difference in discount factors is small. Otherwise, the recipient region may face a debt floor rather than a debt limit. These differentiated budget rules solve the self-selection problem under asymmetric information and decentralized borrowing and spending decisions, internalize the positive intergenerational externality durable public goods entail, and hence are constrained efficient. Moreover, optimal interregional redistribution schemes feature that the region with an undistorted intertemporal allocation of local public goods is always the contributor.

Appendices

Appendix A: Proofs

Proof of Lemma 3.1

We complete the proof in 3 steps. To save on notation, region index R will be suppressed in the proof.

Step 1 By replacing the private consumptions in Equation (4) with public goods provision through application of budget constraints, we have the following:

$$\begin{aligned} \begin{aligned} u'\left( y_{1}+b+z-{\hat{G}}_{1}\right)&= g'\left( {\hat{G}}_{1}\right) +\theta \delta g'\left( \theta {\hat{G}}_{1}+{\hat{G}}_{2}\right) , \\ u'\left( y_{2}-b(1+r)-{\hat{G}}_{2}\right)&= g'\left( \theta {\hat{G}}_{1}+{\hat{G}}_{2}\right) . \end{aligned}\end{aligned}$$

(8)

To simplify notations in the following derivation, let us denote the utilities for period $t\in \{1,2\}$ by $u_{t}$ and $g_{t}$. We then differentiate both sides of the two equations in (8) with respect to b, and apply the Implicit Function Theorem and the Cramer’s Rule and find that

$$\begin{aligned} \begin{aligned} \phi _{b}&\ =\frac{u_{1}''(u_{2}''+g_{2}'')+(1+r)\delta \theta u_{2}''g_{2}''}{\Sigma }>0 \ \ \text {and} \ \ \\ \psi _{b}&\ =-\frac{(1+r)u_{2}''(u_{1}''+g_{1}''+\delta \theta ^{2}g_{2}'')+\theta u_{1}''g_{2}''}{\Sigma }<0, \end{aligned}\end{aligned}$$

(9)

where $\Sigma \equiv (u_{1}''+g_{1}'')(u_{2}''+g_{2}'')+\delta \theta ^{2} u_{2}''g_{2}''>0$. By the same token, we have

$$\begin{aligned} \phi _{\delta } =-\frac{\theta g_{2}'(u_{2}''+g_{2}'')}{\Sigma }>0 \ \ \text {and} \ \ \psi _{\delta } = \frac{\theta ^{2}g_{2}'g_{2}''}{\Sigma }<0, \end{aligned}$$

(10)

which immediately yields

$$\begin{aligned} \theta \phi _{\delta } + \psi _{\delta } = - \frac{\theta ^{2}g_{2}'u_{2}''}{\Sigma }>0. \end{aligned}$$

(11)

Step 2 Applying the Envelope Theorem to the value function (3), we obtain the slope of an indifference curve in the (b, z)-space as

$$\begin{aligned} \frac{\mathrm {d} z}{\mathrm {d} b}\Big |_{\mathrm {d} V=0} = - \frac{g'\left( {\hat{G}}_{1}\right) -(1+r-\theta )\delta g'\left( \theta {\hat{G}}_{1}+{\hat{G}}_{2}\right) }{g'\left( {\hat{G}}_{1}\right) +\theta \delta g'\left( \theta {\hat{G}}_{1}+{\hat{G}}_{2}\right) }. \end{aligned}$$

(12)

Differentiating both sides of Equation (12) with respect to b and rearranging the algebra, we have

$$\begin{aligned} \frac{\mathrm {d}^{2}z}{\mathrm {d} b^{2}}\Big |_{\mathrm {d} V=0} = \frac{(1+r)\delta [g_{1}'g_{2}''(\theta \phi _{b}+\psi _{b})-g_{2}'g_{1}''\phi _{b}]}{(g_{1}'+\theta \delta g_{2}')^{2}}>0 \end{aligned}$$

in which we have used (9) and

$$\begin{aligned} \theta \phi _{b}+\psi _{b} = - \frac{(1+r)u_{2}''g_{1}''+(1+r-\theta )u_{1}''u_{2}''}{\Sigma }<0. \end{aligned}$$

(13)

The proof of part (i) is thus completed by setting $\mathrm {d} z/\mathrm {d} b|_{\mathrm {d} V=0}=0$ in Equation (12).

Step 3 Differentiating both sides of Equation (12) with respect to $\delta$ and simplifying the algebra, we get

$$\begin{aligned} \frac{\mathrm {d} }{\mathrm {d} \delta } \left( \frac{\mathrm {d} z}{\mathrm {d} b}\Big |_{\mathrm {d} V = 0} \right) = \frac{Q}{(g_{1}'+\theta \delta g_{2}')^{2}}, \end{aligned}$$

in which

$$\begin{aligned} Q&\equiv \ g_{1}'\cdot [g_{1}''\cdot \phi _{\delta }+\theta g_{2}'+\theta \delta g_{2}''\cdot (\theta \phi _{\delta } + \psi _{\delta })]\\&\quad +\, (1+r)\{g_{1}'\cdot [g_{2}'+\delta g_{2}''\cdot (\theta \phi _{\delta } + \psi _{\delta })]-\delta g_{2}'\cdot g_{1}''\cdot \phi _{\delta }\}. \end{aligned}$$

Making use of (10) and (11), we have

$$\begin{aligned} g_{1}''\cdot \phi _{\delta }+\theta g_{2}'+\theta \delta g_{2}''\cdot (\theta \phi _{\delta } + \psi _{\delta }) =\frac{\theta u_{1}''\cdot (u_{2}''+g_{2}'')g_{2}'}{\Sigma }>0 \end{aligned}$$

and

$$\begin{aligned} g_{2}'+\delta g_{2}''\cdot (\theta \phi _{\delta } + \psi _{\delta }) = \frac{(u_{1}''+g_{1}'')(u_{2}''+g_{2}'')g_{2}'}{\Sigma }>0, \end{aligned}$$

so we must have $Q>0$ for any $\delta \in (0,1)$, as predicted in part (ii). $\square$

Proof of Proposition 3.1

We complete the proof in 6 steps.

Step 1 Using Lagrangian (5) and assuming the existence of an interior solution, we get the following first-order conditions (FOCs) for Region A:

$$\begin{aligned} \begin{aligned}&(1+\mu ^{A})[g'(G_{1}^{A})-(1+r-\theta )\delta ^A g'(\theta G_{1}^{A}+G_{2}^{A})]\\&\quad = \mu ^{B}[g'(G_{1}^{A})-(1+r-\theta )\delta ^{B}g'(\theta G_{1}^{A}+G_{2}^{A})]; \ \text {and} \\&(1+\mu ^{A})[g'(G_{1}^{A})+\theta \delta ^A g'(\theta G_{1}^{A}+G_{2}^{A})]\\&\quad = \mu ^{B}[g'(G_{1}^{A})+\theta \delta ^{B} g'(\theta G_{1}^{A}+G_{2}^{A})]+\lambda . \end{aligned}\end{aligned}$$

(14)

By symmetry we get the following FOCs for Region B:

$$\begin{aligned} \begin{aligned}&(1+\mu ^{B})[g'(G_{1}^{B})-(1+r-\theta )\delta ^B g'(\theta G_{1}^{B}+G_{2}^{B})]\\&\quad = \mu ^{A}[g'(G_{1}^{B})-(1+r-\theta )\delta ^{A}g'(\theta G_{1}^{B}+G_{2}^{B})]; \ \text {and} \\&(1+\mu ^{B})[g'(G_{1}^{B})+\theta \delta ^B g'(\theta G_{1}^{B}+G_{2}^{B})]\\&\quad = \mu ^{A}[g'(G_{1}^{B})+\theta \delta ^{A} g'(\theta G_{1}^{B}+G_{2}^{B})]+\lambda . \end{aligned}\end{aligned}$$

(15)

Let us suppose $\text {IC}_{B}$ is not binding such that $\mu ^B=0$. It then follows from Lemma 3.2 that $\text {IC}_{A}$ must be binding so that $\mu ^A >0$. Applying these assumptions to the FOCs (14) and (15) gives rise to

$$\begin{aligned} \begin{aligned} g'(G_{1}^{A}) - (1+r-\theta )\delta ^A g'(\theta G_{1}^{A}+G_{2}^{A})&\ =0,\\ g'(G_{1}^{A})+\theta \delta ^A g'(\theta G_{1}^{A}+G_{2}^{A})&\ = \frac{\lambda }{1+\mu ^{A}}; \end{aligned}\end{aligned}$$

(16)

and

$$\begin{aligned} \begin{aligned}&g'(G_{1}^{B}) -(1+r-\theta )\delta ^B g'(\theta G_{1}^{B}+G_{2}^{B}) \\&\quad = \mu ^{A}[g'(G_{1}^{B})-(1+r-\theta )\delta ^{A}g'(\theta G_{1}^{B}+G_{2}^{B})], \\&g'(G_{1}^{B}) +\theta \delta ^B g'(\theta G_{1}^{B}+G_{2}^{B}) = \mu ^{A}[g'(G_{1}^{B})+\theta \delta ^{A} g'(\theta G_{1}^{B}+G_{2}^{B})]+\lambda . \end{aligned}\end{aligned}$$

(17)

Step 2 Using (17) to solve for $g'(\theta G_{1}^{B}+G_{2}^{B})$ gives rise to

$$\begin{aligned} g'(\theta G_{1}^{B}+G_{2}^{B}) = \frac{\lambda }{(1+r)(\delta ^B - \mu ^{A} \delta ^A)}. \end{aligned}$$

(18)

Thus, $\lambda >0$ yields $\delta ^B > \mu ^{A} \delta ^A$. Rearranging the first equation of (17) shows that $\mu ^A <1$ and

$$\begin{aligned} \frac{g'(G_{1}^{B})}{\delta ^B g'(\theta G_{1}^{B}+G_{2}^{B})} = (1+r-\theta )\left( \frac{1}{1-\mu ^A}\right) \left( 1 - \mu ^{A}\cdot \frac{\delta ^A}{\delta ^B}\right) >1+r-\theta , \end{aligned}$$

(19)

in which we have used Assumption 2.1. Using (16), Equation (18), and Assumption 2.1, we have

$$\begin{aligned} \begin{aligned} \delta ^{B} g'(\theta G_{1}^{B}+G_{2}^{B})&= \frac{\lambda }{(1+r)[1 - \mu ^{A} (\delta ^A/\delta ^B)]} \\&\quad > \frac{\lambda }{(1+r)(1 + \mu ^{A})}=\delta ^{A} g'(\theta G_{1}^{A}+G_{2}^{A}). \end{aligned}\end{aligned}$$

(20)

By comparing the first equation of (16) to (19), and using (20), it follows that $g'(G_{1}^{B})>g'(G_{1}^{A})$, which immediately produces $G_{1}^{B}< G_{1}^{A}$ given the strict concavity of $g(\cdot )$. Applying this result and (20) to (4) reveals that $u'(c_{1}^{B})>u'(c_{1}^{A})$, which then gives $c_{1}^{B}<c_{1}^{A}$. Moreover, $\mu ^A >0$ implies that $(b^A, z^A)$ and $(b^B, z^B)$ lie on the same indifference U-shape curve of Region A, a property demonstrated by Lemma 3.1(i). By using the first equation of (16) we get that $(b^A, z^A)$ lies at the minimum of this curve, thereby yielding that $z^A<0<z^B$. In addition, using (19) and Lemma 3.1 shows that $b^B < b^A$.

Step 3 We now proceed to establish the critical condition of $\delta ^B/ \delta ^A >1+2\mu ^A$. We shall do this by means of contradiction. Using (20), we have that $\delta ^A + \mu ^A \delta ^A \ge \delta ^B - \mu ^A \delta ^A$ is equivalent to $\delta ^B/\delta ^A \le 1+ 2 \mu ^A$, under which we immediately arrive at $g'(\theta G_{1}^{A}+G_{2}^{A})\le g'(\theta G_{1}^{B}+G_{2}^{B})$. Therefore, if $\mu ^A < \delta ^B/\delta ^A \le 1+ 2 \mu ^A$ holds, we must have $\theta G_{1}^{A}+G_{2}^{A} \ge \theta G_{1}^{B}+G_{2}^{B}$ and $c_{2}^A \ge c_{2}^B$ by using (4). Consequently, this result combined with the result established in Step 2 reveals that

$$\begin{aligned}&V(b^B, z^B, \delta ^B) - V(b^A, z^A, \delta ^B) \\&\quad = \underset{-}{\underbrace{u(c_{1}^B) - u(c_{1}^A)}} \ + \ \underset{-}{\underbrace{g(G_{1}^B) - g(G_{1}^A)}}\\&\qquad +\, \delta ^B \underset{-/0}{\underbrace{[u(c_{2}^B) - u(c_{2}^A) + g(\theta G_{1}^B + G_{2}^{B}) - g(\theta G_{1}^A +G_{2}^{A})]}}, \end{aligned}$$

which means $\text {IC}_{B}$ is violated under this allocation. For Region A,

$$\begin{aligned}&V(b^A, z^A, \delta ^A) - V(b^B, z^B, \delta ^A) \\&\quad = \underset{+}{\underbrace{u(c_{1}^A) - u(c_{1}^B)}} \ + \ \underset{+}{\underbrace{g(G_{1}^A) - g(G_{1}^B)}}\\&\qquad +\, \delta ^A \underset{+/0}{\underbrace{[u(c_{2}^A) - u(c_{2}^B) + g(\theta G_{1}^A + G_{2}^{A}) - g(\theta G_{1}^B +G_{2}^{B})]}}, \end{aligned}$$

which means $\text {IC}_{A}$ is not binding under this allocation so that $\mu ^A =0$, which is an immediate contradiction. Therefore, we should instead have that $\delta ^B/ \delta ^A >1+2\mu ^A$, which combined with (20) and (4) shows that $\theta G_{1}^{A}+G_{2}^{A} < \theta G_{1}^{B}+G_{2}^{B}$ and $c_{2}^A < c_{2}^B$. The result of $G_{1}^{B}< G_{1}^{A}$ obtained in Step 2 implies that $G_{2}^{A} < G_{2}^{B}$, as predicted.

Step 4 We now proceed to the proof of part (ii). Applying $\mu ^A =0$ and $\mu ^B>0$ to (14) and (15) gives the following simplified FOCs:

$$\begin{aligned} \begin{aligned}&g'(G_{1}^{A}) -(1+r-\theta )\delta ^A g'(\theta G_{1}^{A}+G_{2}^{A}) \\&\quad = \mu ^{B}[g'(G_{1}^{A})-(1+r-\theta )\delta ^{B}g'(\theta G_{1}^{A}+G_{2}^{A})], \\&g'(G_{1}^{A}) +\theta \delta ^A g'(\theta G_{1}^{A}+G_{2}^{A}) = \mu ^{B}[g'(G_{1}^{A})+\theta \delta ^{B} g'(\theta G_{1}^{A}+G_{2}^{A})]+\lambda ; \end{aligned}\end{aligned}$$

(21)

and

$$\begin{aligned} \begin{aligned} g'(G_{1}^{B}) - (1+r-\theta )\delta ^B g'(\theta G_{1}^{B}+G_{2}^{B})&\ =0,\\ g'(G_{1}^{B})+\theta \delta ^B g'(\theta G_{1}^{B}+G_{2}^{B})&\ = \frac{\lambda }{1+\mu ^{B}}. \end{aligned}\end{aligned}$$

(22)

Solving for $g'(\theta G_{1}^{A}+G_{2}^{A})$ using (21) shows that

$$\begin{aligned} g'(\theta G_{1}^{A}+G_{2}^{A}) = \frac{\lambda }{(1+r)(\delta ^A - \mu ^{B} \delta ^B)}. \end{aligned}$$

(23)

Thus, $\lambda >0$ implies that $\mu ^B< \delta ^{A} / \delta ^B <1$ under Assumption 2.1. Rearranging the first equation of (21) reveals that

$$\begin{aligned} \frac{g'(G_{1}^{A})}{\delta ^A g'(\theta G_{1}^{A}+G_{2}^{A})} = (1+r-\theta )\left( \frac{1}{1-\mu ^B}\right) \left( 1 - \mu ^{B}\cdot \frac{\delta ^B}{\delta ^A}\right) <1+r-\theta \end{aligned}$$

(24)

under the condition $\mu ^B< \delta ^{A} / \delta ^B <1$. If we solve $\delta ^B g'(\theta G_{1}^{B}+G_{2}^{B})$ from (22) and compare it to (23), we arrive at the following:

$$\begin{aligned} \begin{aligned} \delta ^{A} g'(\theta G_{1}^{A}+G_{2}^{A})&= \frac{\lambda }{(1+r)[1 - \mu ^{B} (\delta ^B/\delta ^A)]}\\&> \frac{\lambda }{(1+r)(1 + \mu ^{B})}=\delta ^{B} g'(\theta G_{1}^{B}+G_{2}^{B}), \end{aligned}\end{aligned}$$

(25)

which then gives $\theta G_{1}^{A}+G_{2}^{A} < \theta G_{1}^{B}+G_{2}^{B}$ and $c_{2}^A < c_{2}^B$ by using Assumption 2.1, Condition (4), as well as the strict concavity of g and u.

Step 5 Moreover, applying $\mu ^B >0$, the first equation of (22), and inequality (24) to Lemma 3.1, we use the same reasoning in Step 2 to get $z^B<0 <z^A$ and $b^B < b^A$. Then, applying these results to the budget constraints gives

$$\begin{aligned} G_{2}^A -G_{2}^B = (1+r) \underset{-}{\underbrace{(b^B -b^A)}} + \underset{+}{\underbrace{c_{2}^B - c_{2}^A}} \end{aligned}$$

and

$$\begin{aligned} G_{1}^A - G_{1}^B = \underset{+}{\underbrace{b^A -b^B}} + \underset{+}{\underbrace{z^A- z^B}} + c_{1}^B - c_{1}^A. \end{aligned}$$

(26)

We now show $G_{2}^A <G_{2}^B$ by means of contradiction. If, instead, we let $G_{2}^A \ge G_{2}^B$, then we must have $G_{1}^A < G_{1}^B$. Applying this result to (26) immediately reveals $c_{1}^B < c_{1}^A$. By using (4) and Equation (25) we thus arrive at

$$\begin{aligned} 0&< u'(c_{1}^B) - u'(c_{1}^A) \\&= g'(G_{1}^B)- g'(G_{1}^A) + \theta \underset{-}{\underbrace{[\delta ^{B} g'(\theta G_{1}^{B}+G_{2}^{B})-\delta ^{A} g'(\theta G_{1}^{A}+G_{2}^{A})]}}, \end{aligned}$$

by which we must have $G_{1}^B <G_{1}^A$, an immediate contradiction. Analogously, we shall prove $G_{1}^A > G_{1}^B$ by means of contradiction. If, instead, we suppose that $G_{1}^A \le G_{1}^B$, then using (4) and Equation (25) again shows that

$$\begin{aligned}&u'(c_{1}^B) - u'(c_{1}^A) \\&\quad = \underset{-/0}{\underbrace{g'(G_{1}^B)- g'(G_{1}^A)}} + \theta \underset{-}{\underbrace{[\delta ^{B} g'(\theta G_{1}^{B}+G_{2}^{B})-\delta ^{A} g'(\theta G_{1}^{A}+G_{2}^{A})]}}, \end{aligned}$$

which yields $c_{1}^B > c_{1}^A$. Then by (25) we must have $V(b^B, z^B, \delta ^B)>V(b^A, z^A, \delta ^B)$, i.e., $\text {IC}_{B}$ is not binding under this allocation so that $\mu ^B =0$, which is an immediate contradiction.

Step 6 Finally, we show that $\mu ^A, \mu ^B >0$ cannot occur, namely that $\text {IC}_{A}$ and $\text {IC}_{B}$ cannot be simultaneously binding. We shall prove this by means of contradiction. Let us suppose instead that both $\text {IC}_{A}$ and $\text {IC}_{B}$ are binding. Under the complementary slackness conditions, we then get the following:

$$\begin{aligned} \begin{aligned} 0&= V(b^B, z^B, \delta ^B) - V(b^A, z^A, \delta ^B)\\&= u(c_{1}^B) - u(c_{1}^A) + g(G_{1}^B) - g(G_{1}^A)\\&\quad +\, \delta ^B [u(c_{2}^B) - u(c_{2}^A) + g(\theta G_{1}^B + G_{2}^{B}) - g(\theta G_{1}^A +G_{2}^{A})], \end{aligned}\end{aligned}$$

(27)

and

$$\begin{aligned} \begin{aligned} 0&= V(b^A, z^A, \delta ^A) - V(b^B, z^B, \delta ^A)\\&= u(c_{1}^A) - u(c_{1}^B) + g(G_{1}^A) - g(G_{1}^B)\\&\quad +\, \delta ^A [u(c_{2}^A) - u(c_{2}^B) + g(\theta G_{1}^A + G_{2}^{A}) - g(\theta G_{1}^B +G_{2}^{B})]. \end{aligned}\end{aligned}$$

(28)

Applying (27) to (28) and simplifying the algebra, we have under Assumption 2.1 that

$$\begin{aligned} u(c_{2}^A) - u(c_{2}^B) = g(\theta G_{1}^B + G_{2}^{B}) - g(\theta G_{1}^A +G_{2}^{A}), \end{aligned}$$

which combined with the condition $u'(c_{2})=g'(\theta G_{1}+G_{2})$ given in (4) as well as the strict concavity of u and g implies that

$$\begin{aligned} c_{2}^A = c_{2}^B \ \ \text {and} \ \ \theta G_{1}^B + G_{2}^{B}= \theta G_{1}^A +G_{2}^{A}. \end{aligned}$$

(29)

It follows from rearranging the FOCs (14) and (15) that

$$\begin{aligned} \begin{aligned} (1+\mu ^A -\mu ^B) g'(G_{1}^A)&= [(1+\mu ^A)\delta ^A - \mu ^B \delta ^B](1+r-\theta )g'(\theta G_{1}^A +G_{2}^{A}),\\ \lambda&= [(1+\mu ^A)\delta ^A - \mu ^B \delta ^B](1+r)g'(\theta G_{1}^A +G_{2}^{A}); \end{aligned}\end{aligned}$$

(30)

and

$$\begin{aligned} \begin{aligned} (1+\mu ^B -\mu ^A) g'(G_{1}^B)&= [(1+\mu ^B)\delta ^B - \mu ^A \delta ^A](1+r-\theta )g'(\theta G_{1}^B +G_{2}^{B}),\\ \lambda&= [(1+\mu ^B)\delta ^B - \mu ^A \delta ^A](1+r)g'(\theta G_{1}^B +G_{2}^{B}). \end{aligned}\end{aligned}$$

(31)

Thus, applying $\lambda >0$, (29) and Assumption 2.1 to (30) and (31) yields the following critical condition:

$$\begin{aligned} \frac{\mu ^A}{1+\mu ^B}< \frac{\delta ^B}{\delta ^A} = \frac{1+2\mu ^A}{1+2\mu ^B}< \frac{1+\mu ^A}{\mu ^B} \ \ \text {with} \ \ \mu ^B< \mu ^A <1+\mu ^B. \end{aligned}$$

Under this critical condition and Assumption 2.1, we use (30) and (31) to obtain

$$\begin{aligned} \begin{aligned} \frac{g'(G_{1}^A)}{\delta ^A g'(\theta G_{1}^A +G_{2}^{A})}&= (1+r-\theta ) \left[ \frac{1+\mu ^A -\mu ^B (\delta ^B/\delta ^A)}{1+\mu ^A -\mu ^B} \right] < 1+r-\theta , \\ \frac{g'(G_{1}^B)}{\delta ^B g'(\theta G_{1}^B +G_{2}^{B})}&= (1+r-\theta ) \left[ \frac{1+\mu ^B -\mu ^A (\delta ^A/\delta ^B)}{1+\mu ^B -\mu ^A} \right] > 1+r-\theta , \end{aligned}\end{aligned}$$

(32)

revealing that the intertemporal allocations of both regions are distorted relative to the complete information welfare optimum. Making use of Equation (32) yields:

$$\begin{aligned} \frac{g'(G_{1}^A)}{g'(G_{1}^B)} = \underset{<1}{\underbrace{\frac{1+\mu ^B -\mu ^A}{1+\mu ^A -\mu ^B}}} \cdot \underset{=1}{\underbrace{\frac{(1+\mu ^A)\delta ^A - \mu ^B \delta ^B}{(1+\mu ^B)\delta ^B - \mu ^A \delta ^A}}}, \end{aligned}$$

in which we have used Assumption 2.1 and the critical condition given above. We then immediately get $G_{1}^A > G_{1}^B$, and we apply this result to (29) to get $G_{2}^A < G_{2}^B$. Consequently, $b^B <b^A$ follows from the fact that

$$\begin{aligned} \underset{-}{\underbrace{G_{2}^A -G_{2}^B}} \ = \ (1+r) (b^B -b^A) + \underset{0}{\underbrace{c_{2}^B -c_{2}^A}}. \end{aligned}$$

Moreover, applying $\mu ^A, \mu ^B>0$ shows that $(b^A, z^A)$ and $(b^B, z^B)$ lie on the same indifference curves of both regions. Nevertheless, since $(b^A, z^A)$ and $(b^B, z^B)$ are two different points, the single-crossing property established in Lemma 3.1(ii) is violated. $\square$

Proof of Lemma 4.1

We complete the proof in 3 steps.

Step 1 Using (6), the FOCs read as follows:

$$\begin{aligned} g'(G_{1}^{R}) - (1+r-\theta ^R)\delta ^R g'(\theta ^R G_{1}^{R}+G_{2}^{R})&=0,\\ g'(G_{1}^{R})+\theta ^R \delta ^R g'(\theta ^R G_{1}^{R}+G_{2}^{R})&= \lambda , \end{aligned}$$

by which we obtain

$$\begin{aligned} \begin{aligned} g'(G_{1}^R)&= \frac{\lambda (1+r-\theta ^R)}{1+r}, \\ g'(\theta ^R G_{1}^{R}+G_{2}^{R})&= \frac{\lambda }{(1+r)\delta ^R}. \end{aligned}\end{aligned}$$

(33)

ing Assumptions 2.1 and 4.1 to (33) shows that $\theta ^A G_{1}^{A}+G_{2}^{A}< \theta ^B G_{1}^{B}+G_{2}^{B}$ and $G_{1}^{A}<G_{1}^{B}$. By using (4) and (33) we then get $c_{2}^A <c_{2}^B$ and $c_{1}^A =c_{1}^B$, as anticipated.

Step 2 Note that

$$\begin{aligned} \theta ^A G_{1}^{A}+G_{2}^{A} - (\theta ^A G_{1}^{B}+G_{2}^{B})= & {} \theta ^A G_{1}^{A}+G_{2}^{A} - (\theta ^B G_{1}^{B}+G_{2}^{B}) - (\theta ^A-\theta ^B)G_{1}^B,\nonumber \\ \theta ^B G_{1}^{B}+G_{2}^{B} - (\theta ^B G_{1}^{A}+G_{2}^{A})= & {} \theta ^B G_{1}^{B}+G_{2}^{B} - (\theta ^A G_{1}^{A}+G_{2}^{A}) - (\theta ^B-\theta ^A)G_{1}^A.\nonumber \\ \end{aligned}$$

(34)

Applying $g(\cdot )=\ln (\cdot )$ to (33) gives

$$\begin{aligned} \begin{aligned} G_{1}^R&= \frac{1+r}{\lambda (1+r-\theta ^R)} \ \ \text {and} \\ \theta ^R G_{1}^{R}+G_{2}^{R}&= \frac{(1+r)\delta ^R}{\lambda }, \end{aligned}\end{aligned}$$

(35)

which by applying to (34) and simplifying the algebra gives rise to

$$\begin{aligned} \begin{aligned} \theta ^A G_{1}^{A}+G_{2}^{A} - (\theta ^A G_{1}^{B}+G_{2}^{B})&\le 0 \Leftrightarrow \ \delta ^B -\delta ^A \ge \frac{\theta ^B -\theta ^A}{1+r - \theta ^B}, \\ \theta ^B G_{1}^{B}+G_{2}^{B} - (\theta ^B G_{1}^{A}+G_{2}^{A})&\ge 0 \Leftrightarrow \ \delta ^B -\delta ^A \ge \frac{\theta ^B -\theta ^A}{1+r - \theta ^A}. \end{aligned}\end{aligned}$$

(36)

We note that from applying the results established in Step 1 to (3),

$$\begin{aligned} \begin{aligned}&V (b^A, z^A, \theta ^A, \delta ^A) - V(b^B, z^B, \theta ^A, \delta ^A)\\&\quad = \underset{-}{\underbrace{g(G_{1}^A)-g(G_{1}^B)}} + \delta ^A \underset{-}{\underbrace{[u(c_{2}^A) - u(c_{2}^B)]}}\\&\qquad +\, \delta ^A \cdot \underset{?}{\underbrace{[g(\theta ^A G_{1}^{A}+G_{2}^{A}) - g(\theta ^A G_{1}^{B}+G_{2}^{B})]}} \end{aligned}\end{aligned}$$

(37)

and

$$\begin{aligned} \begin{aligned}&V (b^B, z^B, \theta ^B, \delta ^B) - V(b^A, z^A, \theta ^B, \delta ^B) \\&\quad = \underset{+}{\underbrace{g(G_{1}^B)-g(G_{1}^A)}} + \delta ^B \underset{+}{\underbrace{[u(c_{2}^B) - u(c_{2}^A)]}}\\&\qquad +\, \delta ^B \cdot \underset{?}{\underbrace{[g(\theta ^B G_{1}^{B}+G_{2}^{B}) - g(\theta ^B G_{1}^{A}+G_{2}^{A})]}}, \end{aligned}\end{aligned}$$

(38)

applying (36) to (37) and (38) leads to the desired assertion (i) under Assumptions 2.1 and 4.1.

Step 3 Applying $g(\cdot ) \equiv (\cdot )^\alpha$, for a constant parameter $\alpha \in (0,1)$, to (33) gives

$$\begin{aligned} \begin{aligned} G_{1}^R&= \left[ \frac{\alpha (1+r)}{\lambda (1+r-\theta ^R)}\right] ^{1/(1-\alpha )} \ \ \text {and} \\ \theta ^R G_{1}^{R}+G_{2}^{R}&= \left[ \frac{\alpha (1+r)\delta ^R}{\lambda }\right] ^{1/(1-\alpha )}, \end{aligned}\end{aligned}$$

(39)

which applying to (34) and simplifying the algebra reveals that

$$\begin{aligned} \begin{aligned}&\theta ^A G_{1}^{A}+G_{2}^{A} - (\theta ^A G_{1}^{B}+G_{2}^{B}) \le 0 \\&\quad \Leftrightarrow \ (\delta ^B)^{1/(1-\alpha )} -(\delta ^A)^{1/(1-\alpha )} \ge (\theta ^B -\theta ^A)(1+r - \theta ^B)^{1/(\alpha -1)}, \\&\theta ^B G_{1}^{B}+G_{2}^{B} - (\theta ^B G_{1}^{A}+G_{2}^{A}) \ge 0 \\&\quad \Leftrightarrow \ (\delta ^B)^{1/(1-\alpha )} -(\delta ^A)^{1/(1-\alpha )} \ge (\theta ^B -\theta ^A)(1+r - \theta ^A)^{1/(\alpha -1)}. \end{aligned}\end{aligned}$$

(40)

Consequently, applying (40) to (37) and (38) leads to the desired assertion (ii) under Assumptions 2.1 and 4.1. Similarly, applying $g(\cdot ) \equiv -\beta ^{-1} e^{-\beta (\cdot )}$, for a constant parameter $\beta >0$, to (33) shows that

$$\begin{aligned} \begin{aligned} G_{1}^R&= \frac{1}{\beta } \ln \left[ \frac{1+r}{\lambda (1+r-\theta ^R)}\right] \ \ \text {and} \\ \theta ^R G_{1}^{R}+G_{2}^{R}&= \frac{1}{\beta } \ln \left[ \frac{(1+r)\delta ^R}{\lambda }\right] , \end{aligned}\end{aligned}$$

(41)

which applying to (34) and simplifying the algebra reveals that

$$\begin{aligned} \begin{aligned} \theta ^A G_{1}^{A}+G_{2}^{A} - (\theta ^A G_{1}^{B}+G_{2}^{B})&\le 0 \ \Leftrightarrow \ \frac{\delta ^B}{\delta ^A} \ge \left[ \frac{1+r}{\lambda (1+r-\theta ^B)}\right] ^{\theta ^B -\theta ^A}, \\ \theta ^B G_{1}^{B}+G_{2}^{B} - (\theta ^B G_{1}^{A}+G_{2}^{A})&\ge 0 \ \Leftrightarrow \ \frac{\delta ^B}{\delta ^A} \ge \left[ \frac{1+r}{\lambda (1+r-\theta ^A)}\right] ^{\theta ^B -\theta ^A}. \end{aligned}\end{aligned}$$

(42)

Therefore, applying (42) to (37) and (38) leads to the desired assertion (iii) under Assumptions 2.1 and 4.1. $\square$

Proof of Proposition 4.1

We complete the proof in 3 steps.

Step 1 Suppose u and g take the logarithmic utility functional form, then we get from (4) that $c_{2}^R = \theta ^R G_{1}^R + G_{2}^R$. Then, using (35) and $(b^B - b^A)(1+r)= G_{2}^A - G_{2}^B + c_{2}^A - c_{2}^B$ gives rise to

$$\begin{aligned} b^B -b^A = \frac{1}{\lambda } \left[ 2(\delta ^A-\delta ^B) +\frac{\theta ^B}{1+r-\theta ^B}-\frac{\theta ^A}{1+r-\theta ^A}\right] , \end{aligned}$$

which when rearranged gives the cross-region debt comparison result in part (i). Applying this result and Lemma 4.1 to Period-1 budget constraints indicates that $z^A -z^B=G_{1}^A-G_{1}^B + b^B - b^A = (2/\lambda )(\delta ^A-\delta ^B)<0$ under Assumption 2.1.

Step 2 Let us suppose u and g take the same power utility functional form, then it follows from applying (4) and (39) to $(b^B - b^A)(1+r)= G_{2}^A - G_{2}^B + c_{2}^A - c_{2}^B$ that

$$\begin{aligned}&(b^B -b^A)(1+r)\left[ \frac{\alpha (1+r)}{\lambda }\right] ^{1/(\alpha -1)} = 2 \left[ (\delta ^A)^{1/(1-\alpha )}-(\delta ^B)^{1/(1-\alpha )} \right] \\&\qquad +\, \theta ^B \left( \frac{1}{1+r-\theta ^B}\right) ^{1/(1-\alpha )}-\theta ^A \left( \frac{1}{1+r-\theta ^A}\right) ^{1/(1-\alpha )}, \end{aligned}$$

which when rearranged gives the cross-region debt comparison result in part (ii). Applying this result and Lemma 4.1 to Period-1 budget constraints shows that

$$\begin{aligned}&(z^A -z^B)(1+r)\left[ \frac{\alpha (1+r)}{\lambda }\right] ^{1/(\alpha -1)} = 2 \underset{-}{\underbrace{\left[ (\delta ^A)^{1/(1-\alpha )}-(\delta ^B)^{1/(1-\alpha )} \right] }}\\&\qquad +\, \underset{-}{\underbrace{\left( \frac{1}{1+r-\theta ^A}\right) ^{\alpha /(1-\alpha )} - \left( \frac{1}{1+r-\theta ^B}\right) ^{\alpha /(1-\alpha )}}} \end{aligned}$$

under Assumptions 2.1 and 4.1.

Step 3 If we suppose u and g take the same exponential utility functional form, it then follows from (41) that $1+r>\lambda$. Then, applying (4) and (41) to $(b^B - b^A)(1+r)= G_{2}^A - G_{2}^B + c_{2}^A - c_{2}^B$ gives

$$\begin{aligned} (b^B -b^A)(1+r)\beta&= \underset{-}{\underbrace{2 \ln \left( \frac{\delta ^A}{\delta ^B}\right) }} + \underset{+}{\underbrace{\left( \theta ^B - \theta ^A \right) \ln \left( \frac{1+r}{\lambda }\right) }} \\&\quad +\, \underset{+}{\underbrace{\theta ^B \ln \left( \frac{1}{1+r-\theta ^B} \right) - \theta ^A \ln \left( \frac{1}{1+r-\theta ^A} \right) }} \end{aligned}$$

under Assumptions 2.1 and 4.1. Then, rearranging the algebra reveals that

$$\begin{aligned} b^B < b^A \ \Leftrightarrow \ \frac{\delta ^B}{\delta ^A}>\sqrt{\left( \frac{1+r}{\lambda }\right) ^{\theta ^B - \theta ^A} \frac{(1+r-\theta ^A)^{\theta ^A}}{(1+r-\theta ^B)^{\theta ^B}}}, \end{aligned}$$

as predicted in part (iii). Applying the expression of $b^B -b^A$ given above, (41), and Lemma 4.1 to Period-1 budget constraints, we arrive at

$$\begin{aligned}&(z^A -z^B)(1+r)\beta \\&\quad = \underset{-}{\underbrace{2 \ln \left( \frac{\delta ^A}{\delta ^B}\right) }} + \underset{+}{\underbrace{\left( \theta ^B - \theta ^A \right) \ln \left( \frac{1+r}{\lambda }\right) }} \\&\qquad +\, \underset{?}{\underbrace{(1+r-\theta ^B) \ln (1+r-\theta ^B) - (1+r-\theta ^A) \ln (1+r-\theta ^A)}} \end{aligned}$$

under Assumptions 2.1 and 4.1. If we note that

$$\begin{aligned} \frac{\partial (1+r-\theta ) \ln (1+r-\theta )}{\partial \theta } \ge 0 \ \Leftrightarrow \ \theta \ge 1+r-e^{-1} \end{aligned}$$

and

$$\begin{aligned} 2 \ln \left( \frac{\delta ^A}{\delta ^B}\right) + \left( \theta ^B - \theta ^A \right) \ln \left( \frac{1+r}{\lambda }\right) \ge 0 \ \Leftrightarrow \ \frac{\delta ^B}{\delta ^A}\le \left( \frac{1+r}{\lambda }\right) ^{(\theta ^B - \theta ^A)/2}, \end{aligned}$$

the predicted assertion in part (iii) is thus confirmed. $\square$

Proof of Proposition 4.2

We complete the proof in 3 steps.

Step 1 Differentiating Lagrangian (6) with respect to $b^A$ and $z^A$ and assuming the existence of an interior solution, we give the FOCs for Region A as follows:

$$\begin{aligned} \begin{aligned}&(1+\mu ^{A})[g'(G_{1}^{A})-(1+r-\theta ^A)\delta ^A g'(\theta ^A G_{1}^{A}+G_{2}^{A})]\\&\quad = \mu ^{B}[g'(G_{1}^{A})-(1+r-\theta ^B)\delta ^{B}g'(\theta ^B G_{1}^{A}+G_{2}^{A})]; \ \text {and} \\&(1+\mu ^{A})[g'(G_{1}^{A})+\theta ^A \delta ^A g'(\theta ^A G_{1}^{A}+G_{2}^{A})]\\&\quad = \mu ^{B}[g'(G_{1}^{A})+\theta ^B \delta ^{B} g'(\theta ^B G_{1}^{A}+G_{2}^{A})]+\lambda . \end{aligned}\end{aligned}$$

(43)

By symmetry we get the following FOCs for Region B:

$$\begin{aligned} \begin{aligned}&(1+\mu ^{B})[g'(G_{1}^{B})-(1+r-\theta ^B)\delta ^B g'(\theta ^B G_{1}^{B}+G_{2}^{B})]\\&\quad = \mu ^{A}[g'(G_{1}^{B})-(1+r-\theta ^A)\delta ^{A}g'(\theta ^A G_{1}^{B}+G_{2}^{B})]; \ \text {and} \\&(1+\mu ^{B})[g'(G_{1}^{B})+\theta ^B \delta ^B g'(\theta ^B G_{1}^{B}+G_{2}^{B})]\\&\quad = \mu ^{A}[g'(G_{1}^{B})+\theta ^A \delta ^{A} g'(\theta ^A G_{1}^{B}+G_{2}^{B})]+\lambda . \end{aligned}\end{aligned}$$

(44)

Let us suppose $\text {IC}_{B}$ is not binding so that $\mu ^B=0$. Then, it follows from Lemma 4.1 that $\text {IC}_{A}$ tends to be binding so that $\mu ^A >0$. Applying these assumptions to the FOCs (43) and (44) gives rise to

$$\begin{aligned} \begin{aligned} g'(G_{1}^{A}) - (1+r-\theta ^A)\delta ^A g'(\theta ^A G_{1}^{A}+G_{2}^{A})&\ =0,\\ g'(G_{1}^{A})+\theta ^A \delta ^A g'(\theta ^A G_{1}^{A}+G_{2}^{A})&\ = \frac{\lambda }{1+\mu ^{A}}; \end{aligned}\end{aligned}$$

(45)

and

$$\begin{aligned} \begin{aligned}&g'(G_{1}^{B}) -(1+r-\theta ^B)\delta ^B g'(\theta ^B G_{1}^{B}+G_{2}^{B})\\&\quad = \mu ^{A}[g'(G_{1}^{B})-(1+r-\theta ^A)\delta ^{A}g'(\theta ^A G_{1}^{B}+G_{2}^{B})], \\&g'(G_{1}^{B}) +\theta ^B \delta ^B g'(\theta ^B G_{1}^{B}+G_{2}^{B}) = \mu ^{A}[g'(G_{1}^{B})+\theta ^A \delta ^{A} g'(\theta ^A G_{1}^{B}+G_{2}^{B})]+\lambda . \end{aligned}\end{aligned}$$

(46)

Step 2 In (46), by plugging the expression of $g'(G_{1}^{B})+\theta ^B \delta ^B g'(\theta ^B G_{1}^{B}+G_{2}^{B})$ from the second equation in the first equation and simplifying the algebra, we have

$$\begin{aligned} \frac{\lambda }{1+r} = \delta ^B g'(\theta ^B G_{1}^{B}+G_{2}^{B}) - \mu ^{A} \delta ^{A} g'(\theta ^A G_{1}^{B}+G_{2}^{B})>0, \end{aligned}$$

(47)

in which we have used the fact that $\lambda >0$. Then, applying (47) to the first equation of (46) reveals that

$$\begin{aligned} (1-\mu ^A)g'(G_{1}^{B}) > (1+r-\theta ^B)\delta ^B g'(\theta ^B G_{1}^{B}+G_{2}^{B})\left( \frac{\theta ^A - \theta ^B}{1+r-\theta ^B} \right) . \end{aligned}$$

(48)

Considering the case of $\mu ^A >1$, we rearrange (48) to get

$$\begin{aligned} \frac{g'(G_{1}^{B})}{\delta ^B g'(\theta ^B G_{1}^{B}+G_{2}^{B})} < (1+r-\theta ^B)\underset{+}{\underbrace{\left( \frac{1}{\mu ^A -1} \right) \left( \frac{\theta ^B - \theta ^A}{1+r-\theta ^B} \right) }} \end{aligned}$$

under Assumption 4.1. It is thus easy to verify the desired assertion:

$$\begin{aligned} \frac{g'(G_{1}^{B})}{\delta ^B g'(\theta ^B G_{1}^{B}+G_{2}^{B})} < 1+r-\theta ^B \ \ \text {for} \ \ 1 + \frac{\theta ^B - \theta ^A}{1+r-\theta ^B} \le \mu ^A. \end{aligned}$$

(49)

It follows from (45) that

$$\begin{aligned} \delta ^A g'(\theta ^A G_{1}^{A}+G_{2}^{A}) = \frac{\lambda }{(1+r)(1+\mu ^{A})}. \end{aligned}$$

(50)

Applying $\lambda >0$, Assumption 4.1 and the strict concavity of g to (47) gives rise to

$$\begin{aligned} \frac{\delta ^B}{\mu ^A \delta ^A}> \frac{g'(\theta ^A G_{1}^{B}+G_{2}^{B})}{g'(\theta ^B G_{1}^{B}+G_{2}^{B})} >1 \ \ \text {and} \ \ \frac{\lambda }{1+r} < g'(\theta ^B G_{1}^{B}+G_{2}^{B}) (\delta ^B- \mu ^{A} \delta ^{A}), \end{aligned}$$

and from which we obtain

$$\begin{aligned} g'(\theta ^B G_{1}^{B}+G_{2}^{B}) > \frac{\lambda }{(1+r)(\delta ^B- \mu ^{A} \delta ^{A})}. \end{aligned}$$

(51)

As a consequence, comparing (51) to (50) reveals that

$$\begin{aligned} g'(\theta ^B G_{1}^{B}+G_{2}^{B}) >g'(\theta ^A G_{1}^{A}+G_{2}^{A}) \ \ \text {for} \ \ \frac{\delta ^B}{\delta ^A} \le 1+2\mu ^A, \end{aligned}$$

which combined with (4) shows that $c_{2}^B <c_{2}^A$, as predicted. Applying (45), (49), and $\mu ^A>0$ to Lemma 3.1 demonstrates that $(b^A, z^A)$ lies at the minimum of, while $(b^B, z^B)$ lies on the increasing part of, a U-shaped indifference curve of Region A. We, therefore, must have $z^A<0<z^B$ and $b^A <b^B$, as desired in Proposition 4.2(i). Applying these results to Period-1 budget constraints yields

$$\begin{aligned} G_{1}^{A} - G_{1}^{B} = \underset{-}{\underbrace{b^A- b^B}} + \underset{-}{\underbrace{z^A -z^B}} + c_{1}^B -c_{1}^A. \end{aligned}$$

(52)

We now prove $G_{1}^{A} < G_{1}^{B}$ by means of contradiction. If, instead, we let $G_{1}^{A} \ge G_{1}^{B}$, then using (52) gives $c_{1}^B >c_{1}^A$. Moreover, using (50) and (51) reveals that

$$\begin{aligned} \theta ^B \delta ^B g'(\theta ^B G_{1}^{B}+G_{2}^{B})&> \frac{\lambda \theta ^B}{(1+r)[1- \mu ^{A} (\delta ^{A}/\delta ^B)]}\\&> \frac{\lambda \theta ^A}{(1+r)(1+\mu ^{A})} =\theta ^A \delta ^A g'(\theta ^A G_{1}^{A}+G_{2}^{A}) \end{aligned}$$

under Assumptions 2.1 and 4.1. As such, we use (4) and get the following:

$$\begin{aligned} u'(c_{1}^B) - u'(c_{1}^A)&= \underset{+/0}{\underbrace{g'(G_{1}^B) - g'(G_{1}^A)}} \\&\quad + \underset{+}{\underbrace{\theta ^B \delta ^B g'(\theta ^B G_{1}^{B}+G_{2}^{B})-\theta ^A \delta ^A g'(\theta ^A G_{1}^{A}+G_{2}^{A})}}, \end{aligned}$$

which yields $c_{1}^B < c_{1}^A$, an immediate contradiction. By using Assumption 4.1, $G_{1}^{A} < G_{1}^{B}$, and $\theta ^B G_{1}^{B}+G_{2}^{B}<\theta ^A G_{1}^{A}+G_{2}^{A}$, we accordingly get $G_{2}^{B}< G_{2}^{A}$.

Step 3 We now proceed to the proof of Proposition 4.2(ii). Applying $\mu ^A=0$ and $\mu ^B>0$ to (43) and (44) gives the following FOCs:

$$\begin{aligned} \begin{aligned} g'(G_{1}^{B}) - (1+r-\theta ^B)\delta ^B g'(\theta ^B G_{1}^{B}+G_{2}^{B})&=0,\\ g'(G_{1}^{B})+\theta ^B \delta ^B g'(\theta ^B G_{1}^{B}+G_{2}^{B})&= \frac{\lambda }{1+\mu ^{B}}; \end{aligned}\end{aligned}$$

(53)

and

$$\begin{aligned} \begin{aligned}&g'(G_{1}^{A}) -(1+r-\theta ^A)\delta ^A g'(\theta ^A G_{1}^{A}+G_{2}^{A})\\&\quad = \mu ^{B}[g'(G_{1}^{A})-(1+r-\theta ^B)\delta ^{B}g'(\theta ^B G_{1}^{A}+G_{2}^{A})], \\&g'(G_{1}^{A}) +\theta ^A \delta ^A g'(\theta ^A G_{1}^{A}+G_{2}^{A}) = \mu ^{B}[g'(G_{1}^{A})+\theta ^B \delta ^{B} g'(\theta ^B G_{1}^{A}+G_{2}^{A})]+\lambda . \end{aligned}\end{aligned}$$

(54)

We then get the following in light of (54) and $\lambda >0$:

$$\begin{aligned} \frac{\lambda }{1+r} = \delta ^A g'(\theta ^A G_{1}^{A}+G_{2}^{A}) - \mu ^{B} \delta ^{B} g'(\theta ^B G_{1}^{A}+G_{2}^{A})>0, \end{aligned}$$

which by applying Assumption 4.1 to (54) shows that

$$\begin{aligned} (1-\mu ^B)g'(G_{1}^{A})> (1+r-\theta ^A)\delta ^A g'(\theta ^A G_{1}^{A}+G_{2}^{A})\left( \frac{\theta ^B - \theta ^A}{1+r-\theta ^A} \right) >0. \end{aligned}$$

We must then have $\mu ^B <1$ and

$$\begin{aligned} \frac{g'(G_{1}^{A})}{\delta ^A g'(\theta ^A G_{1}^{A}+G_{2}^{A})} > (1+r-\theta ^A) \left( \frac{1}{1-\mu ^B}\right) \left( \frac{\theta ^B - \theta ^A}{1+r-\theta ^A} \right) , \end{aligned}$$

which enables us to arrive at the expected conclusion:

$$\begin{aligned} \frac{g'(G_{1}^{A})}{\delta ^A g'(\theta ^A G_{1}^{A}+G_{2}^{A})} > 1+r-\theta ^A \ \ \text {for} \ \ \mu ^B \ge \frac{1+r-\theta ^B}{1+r-\theta ^A}. \end{aligned}$$

(55)

We now apply the first equation of (53), (55), and $\mu ^B>0$ to Lemma 3.1, which leads us to $z^B<0<z^A$ and $b^A< b^B$. $\square$

Proof of Proposition 4.3

Note that the FOCs are still given by (33). Using Assumption 4.2 and (33), we have $G_{1}^{A}>G_{1}^{B}$. Similarly, using Assumption 2.1 and (33), we have $\theta ^{A}G_{1}^{A}+G_{2}^{A}<\theta ^{B}G_{1}^{B}+G_{2}^{B}$. Then, when combined with Assumption 4.2 these two results reveal that $G_{2}^{A}<G_{2}^{B}$. We can also get that $c_{2}^{A}<c_{2}^{B}$ and $c_{1}^{A}=c_{1}^{B}$ by using (4). We therefore obtain $b^{B}<b^{A}$ by using $G_{2}^{A}- G_{2}^{B} = (1+r)(b^{B}-b^{A}) + c_{2}^{B}-c_{2}^{A}$.

We now proceed to the characterization of the optimal interregional redistribution policy. Firstly, we have by definition and the above results that $z^{A}-z^{B}= G_{1}^{A}-G_{1}^{B}+b^{B}-b^{A}$. We then consider the three utility functional forms. If $u=g=\ln$, then we get from Step 1 of the proof of Proposition 4.1 that $z^A -z^B= (2/\lambda )(\delta ^A-\delta ^B)<0$ under Assumption 2.1, as predicted in part (i). For part (ii), we note from Step 2 of the proof of Proposition 4.1 that

$$\begin{aligned}&(z^A -z^B)(1+r)\left[ \frac{\alpha (1+r)}{\lambda }\right] ^{1/(\alpha -1)} = 2 \underset{-}{\underbrace{\left[ (\delta ^A)^{1/(1-\alpha )}-(\delta ^B)^{1/(1-\alpha )} \right] }}\\&\quad +\, \underset{+}{\underbrace{\left( \frac{1}{1+r-\theta ^A}\right) ^{\alpha /(1-\alpha )} - \left( \frac{1}{1+r-\theta ^B}\right) ^{\alpha /(1-\alpha )}}} \end{aligned}$$

under Assumptions 2.1 and 4.2. For part (iii), we apply Assumptions 2.1 and 4.2, in view of Step 3 of the proof of Proposition 4.1, and then we have

$$\begin{aligned}&(z^A -z^B)(1+r)\beta \\&\quad = \underset{-}{\underbrace{2 \ln \left( \frac{\delta ^A}{\delta ^B}\right) }} + \underset{-}{\underbrace{\left( \theta ^B - \theta ^A \right) \ln \left( \frac{1+r}{\lambda }\right) }} \\&\qquad +\, \underset{-/0}{\underbrace{(1+r-\theta ^B) \ln (1+r-\theta ^B) - (1+r-\theta ^A) \ln (1+r-\theta ^A)}} \end{aligned}$$

whenever $(1+r-\theta )\ln (1+r-\theta )$ is nondecreasing in $\theta$. $\square$

Proof of Proposition 4.4

Firstly, applying Assumption 4.2 to (48) yields

$$\begin{aligned} (1-\mu ^A)g'(G_{1}^{B}) > \underset{+}{\underbrace{(1+r-\theta ^B)\delta ^B g'(\theta ^B G_{1}^{B}+G_{2}^{B})\left( \frac{\theta ^A - \theta ^B}{1+r-\theta ^B} \right) }}. \end{aligned}$$

We thus have $\mu ^A <1$ and

$$\begin{aligned} \frac{g'(G_{1}^{B})}{\delta ^B g'(\theta ^B G_{1}^{B}+G_{2}^{B})} > (1+r-\theta ^B) \left( \frac{1}{1-\mu ^A} \right) \left( \frac{\theta ^A - \theta ^B}{1+r-\theta ^B}\right) \ge 1+r-\theta ^B, \end{aligned}$$

in which the second inequality follows from letting

$$\begin{aligned} \left( \frac{1}{1-\mu ^A} \right) \left( \frac{\theta ^A - \theta ^B}{1+r-\theta ^B}\right) \ge 1. \end{aligned}$$

In so doing, we get the desired interregional redistribution and local borrowing policies. The proof of part (i) is thus complete.

For part (ii), using (54) and Assumption 4.2 yields

$$\begin{aligned} (1-\mu ^B)g'(G_{1}^{A}) > (1+r-\theta ^A)\delta ^A g'(\theta ^A G_{1}^{A}+G_{2}^{A})\underset{-}{\underbrace{\left( \frac{\theta ^B - \theta ^A}{1+r-\theta ^A} \right) }}. \end{aligned}$$

Letting $\mu ^B >1$, we thus obtain

$$\begin{aligned} \frac{g'(G_{1}^{A})}{\delta ^A g'(\theta ^A G_{1}^{A}+G_{2}^{A})} < (1+r-\theta ^A) \left( \frac{1}{1-\mu ^B}\right) \left( \frac{\theta ^B - \theta ^A}{1+r-\theta ^A} \right) \le 1+r-\theta ^A \end{aligned}$$

for

$$\begin{aligned} \underset{+}{\underbrace{\left( \frac{1}{1-\mu ^B}\right) \left( \frac{\theta ^B - \theta ^A}{1+r-\theta ^A} \right) }} \le 1, \end{aligned}$$

which is expected. The remaining details of the proof are omitted to save space. $\square$

Proof of Proposition 5.1

We complete the proof in 2 steps.

Step 1 Applying the Envelope Theorem to value function (3), we have for $R \in \{A, B\}$ that

$$\begin{aligned} \begin{aligned} V_{b} (b^R, z^R, \delta ^R)&= g'(G_{1}^R) - (1+r-\theta )\delta ^R g'(\theta G_{1}^R+G_{2}^R),\\ V_{bb} (b^R, z^R, \delta ^R)&= g_{1}''\cdot \phi _{b}- (1+r-\theta )\delta ^R g_{2}''\cdot (\theta \phi _{b}+ \psi _{b})<0, \end{aligned}\end{aligned}$$

(56)

in which we have used (9), (13), and the strict concavity of g. We now prove Proposition 5.1(i). This is equivalent to showing that by imposing the upper bound denoted by ${\bar{b}}\equiv b^{B*}$ for regions of any type, setting the interregional redistribution scheme denoted by $(z^{A*}, z^{B*})$ implements the asymmetric information optimum when spending and borrowing decisions are decentralized at the regional level.

We first consider the case of $\mu ^A >\mu ^B =0$. Let us suppose that Region A receives the federal transfer $z^{B*}$ from the center, then its maximization problem becomes: $\max _{b^A} V(b^A, z^{B*}, \delta ^A)$, subject to the constraint $b^A \le b^{B*}$. Using (56) and evaluating $V_{b}(b^A, z^{B*}, \delta ^A)$ at $b^A = b^{B*}$ yields:

$$\begin{aligned} V_{b}(b^{B*}, z^{B*}, \delta ^A)&= g'(G_{1}^{B*}) - (1+r-\theta )\delta ^A g'(\theta G_{1}^{B*}+G_{2}^{B*})\\&> \ g'(G_{1}^{B*}) - (1+r-\theta )\delta ^B g'(\theta G_{1}^{B*}+G_{2}^{B*})>0 \end{aligned}$$

in which we have used Assumption 2.1 and Proposition 3.1(i). Consequently, using $V_{bb}<0$ leads to $0< V_{b}(b^{B*}, z^{B*}, \delta ^A) \le V_{b}(b^A, z^{B*}, \delta ^A)$ for any $b^A \le b^{B*}$. Taking the federal transfer $z^{B*}$ as given, Region A must choose local debt $b^A = b^{B*}$ accordingly; hence, allocation $(b^{B*}, z^{B*})$ is realized. Since Region A is indifferent between $(b^{B*}, z^{B*})$ and $(b^{A*}, z^{A*})$ under $\mu ^A>0$, it will report its type truthfully instead of mimicking Region B.

For Region B, in view of Proposition 3.1(i) we have $V_{b}(b^{B*}, z^{B*}, \delta ^B)>0$, thus using $V_{bb}<0$ yields $0< V_{b}(b^{B*}, z^{B*}, \delta ^B) \le V_{b}(b^B, z^{B*}, \delta ^B)$ for any $b^B \le b^{B*}$. This implies that $(b^{B*}, z^{B*})$ is also realized by Region B. Since we have $z^{A*}<0<z^{B*}$ according to Proposition 3.1(i) and that $V_{z}=g_{1}'+\theta g_{2}'>0$ holds true for any type, we must have $V(b^B, z^{B*}, \delta ^B)> V(b^B, z^{A*}, \delta ^B)$ for any feasible $b^B \le b^{B*}$. Therefore, Region B has no incentive to misreport its type as A. In summary, the incentive compatibility constraints for both regions are fulfilled under this local budget rule.

Step 2 We now prove Proposition 5.1(ii). This is equivalent to showing that, by imposing the lower bound denoted by ${\underline{b}}\equiv b^{A*}$ for regions of any type, setting the interregional redistribution scheme denoted by $(z^{A*}, z^{B*})$ implements the asymmetric information optimum when spending and borrowing decisions are decentralized at the regional level.

We first consider truthful implementation in Region B. Let us suppose that Region B receives the federal transfer $z^{A*}$ from the center, then its maximization problem becomes: $\max _{b^B} V(b^B, z^{A*}, \delta ^B)$, subject to the constraint $b^B \ge b^{A*}$. Evaluating $V_{b}(b^B, z^{A*}, \delta ^B)$ at $b^B = b^{A*}$ gives rise to

$$\begin{aligned} V_{b}(b^{A*}, z^{A*}, \delta ^B)&= g'(G_{1}^{A*}) - (1+r-\theta )\delta ^B g'(\theta G_{1}^{A*}+G_{2}^{A*})\\&< g'(G_{1}^{A*}) - (1+r-\theta )\delta ^A g'(\theta G_{1}^{A*}+G_{2}^{A*})<0, \end{aligned}$$

in which we have used Assumption 2.1 and Proposition 3.1(ii). Thus using $V_{bb}<0$ yields $V_{b}(b^{B}, z^{A*}, \delta ^B) \le V_{b}(b^{A*}, z^{A*}, \delta ^B)<0$ for any $b^B \ge b^{A*}$. If we take the federal transfer $z^{A*}$ as given, Region B must choose local debt $b^B = b^{A*}$, and hence allocation $(b^{A*}, z^{A*})$ is realized. Noting that Region B is indifferent between $(b^{B*}, z^{B*})$ and $(b^{A*}, z^{A*})$ under $\mu ^B>0$, it will report its type truthfully.

For Region A, given $V_{b}(b^{A*}, z^{A*}, \delta ^A)<0$ according to Proposition 3.1(ii), using $V_{bb}<0$ leads to $V_{b}(b^A, z^{A*}, \delta ^A)\le V_{b}(b^{A*}, z^{A*}, \delta ^A)<0$ for any $b^A \ge b^{A*}$. Therefore, $(b^{A*}, z^{A*})$ is also realized by Region A. Since we know that $z^{B*}<0<z^{A*}$ according to Proposition 3.1(ii) and that $V_{z}=g_{1}'+\theta g_{2}'>0$ always holds true, we must have $V(b^A, z^{A*}, \delta ^A)> V(b^A, z^{B*}, \delta ^A)$ for any feasible $b^A \ge b^{A*}$. In other words, Region A has no incentive to misreport its type as B. In summary, the incentive compatibility constraints for both regions are guaranteed under this local budget rule. $\square$

Proof of Proposition 5.2

We complete the proof in 3 steps.

Step 1 We firstly consider the case of $\mu ^A >\mu ^B =0$ corresponding to Proposition 4.2(i). Let us suppose that Region A receives the federal transfer $z^{B*}$ from the center, then its maximization problem becomes: $\max _{b^A} V(b^A, z^{B*}, \theta ^A, \delta ^A)$, subject to the constraint $b^A \ge b^{B*}$. In light of (56) and evaluation of $V_{b}(b^A, z^{B*}, \theta ^A, \delta ^A)$ at $b^A = b^{B*}$ we have:

$$\begin{aligned} V_{b}(b^{B*}, z^{B*}, \theta ^A, \delta ^A)&= g'(G_{1}^{B*}) - (1+r-\theta ^A)\delta ^A g'(\theta ^{A} G_{1}^{B*}+G_{2}^{B*})\\&\le \ g'(G_{1}^{B*}) - (1+r-\theta ^B)\delta ^B g'(\theta ^{B} G_{1}^{B*}+G_{2}^{B*})<0 \end{aligned}$$

under Proposition 4.2(i) and the following condition:

$$\begin{aligned} \underset{>1}{\underbrace{\frac{\delta ^B}{\delta ^A}}} \ \le \ \underset{>1}{\underbrace{\frac{(1+r-\theta ^A) g'(\theta ^{A} G_{1}^{B*}+G_{2}^{B*})}{(1+r-\theta ^B) g'(\theta ^{B} G_{1}^{B*}+G_{2}^{B*})}}}, \end{aligned}$$

in which we have used Assumptions 2.1 and 4.1. Given $V_{bb}<0$, we thus have $V_{b}(b^A, z^{B*}, \theta ^A, \delta ^A) \le V_{b}(b^{B*}, z^{B*}, \theta ^A, \delta ^A) <0$ for any $b^A \ge b^{B*}$. Taking the federal transfer $z^{B*}$ as given, Region A must choose local debt $b^A = b^{B*}$, and hence allocation $(b^{B*}, z^{B*})$ is realized. Since Region A is indifferent between $(b^{B*}, z^{B*})$ and $(b^{A*}, z^{A*})$ under $\mu ^A>0$, it has no incentive to mimic Region B.

For Region B, in view of Proposition 4.2(i) we have $V_{b}(b^{B*}, z^{B*}, \theta ^B, \delta ^B)<0$. Thus using $V_{bb}<0$ yields $V_{b}(b^B, z^{B*}, \theta ^B, \delta ^B) \le V_{b}(b^{B*}, z^{B*}, \theta ^{B}, \delta ^B) <0$ for any $b^B \ge b^{B*}$. Therefore, $(b^{B*}, z^{B*})$ is also realized by Region B. Given $z^{A*}<0<z^{B*}$ according to Proposition 4.2(i) and that $V_{z}=g_{1}'+\theta g_{2}'>0$ always holds true, we must have $V(b^B, z^{B*}, \theta ^B, \delta ^B)> V(b^B, z^{A*},\theta ^B, \delta ^B)$ for any feasible $b^B \ge b^{B*}$. As a result, Region B has no incentive to mimic Region A. In summary, the incentive compatibility constraints for both regions are guaranteed under this budget rule.

Step 2 The proof of Proposition 5.2(i-b) is similar to that of Proposition 5.2(i-a), so we only show the main difference. For Region B, we now use (56) and evaluate $V_{b}(b^B, z^{A*}, \theta ^B, \delta ^B)$ at $b^B = b^{A*}$, which gives rise to

$$\begin{aligned} V_{b}(b^{A*}, z^{A*}, \theta ^B, \delta ^B)&= g'(G_{1}^{A*}) - (1+r-\theta ^B)\delta ^B g'(\theta ^{B} G_{1}^{A*}+G_{2}^{A*})\\&\ge \ g'(G_{1}^{A*}) - (1+r-\theta ^A)\delta ^A g'(\theta ^{A} G_{1}^{A*}+G_{2}^{A*})>0 \end{aligned}$$

under Proposition 4.2(ii) and the following condition:

$$\begin{aligned} \underset{>1}{\underbrace{\frac{\delta ^B}{\delta ^A}}} \ \le \ \underset{>1}{\underbrace{\frac{(1+r-\theta ^A) g'(\theta ^{A} G_{1}^{A*}+G_{2}^{A*})}{(1+r-\theta ^B) g'(\theta ^{B} G_{1}^{A*}+G_{2}^{A*})}}}, \end{aligned}$$

in which we have used Assumptions 2.1 and 4.1.

Step 3 For Proposition 5.2(ii-a), note that for Region A,

$$\begin{aligned} V_{b}(b^{B*}, z^{B*}, \theta ^A, \delta ^A)&= g'(G_{1}^{B*}) - (1+r-\theta ^A)\delta ^A g'(\theta ^{A} G_{1}^{B*}+G_{2}^{B*})\\&> g'(G_{1}^{B*}) - (1+r-\theta ^B)\delta ^B g'(\theta ^{B} G_{1}^{B*}+G_{2}^{B*})>0, \end{aligned}$$

in which we have used Assumptions 2.1 and 4.2, and Proposition 4.4(i). Considering this region’s maximization problem: $\max _{b^A} V(b^A, z^{B*}, \theta ^A, \delta ^A)$ subject to the constraint $b^A \le b^{B*}$. By using $V_{bb}<0$, we thus have $V_{b}(b^A, z^{B*}, \theta ^A, \delta ^A) \ge V_{b}(b^{B*}, z^{B*}, \theta ^A, \delta ^A) >0$ for any $b^A \le b^{B*}$. Taking the federal transfer $z^{B*}$ as given, Region A must choose local debt $b^A = b^{B*}$ accordingly, and so allocation $(b^{B*}, z^{B*})$ is realized. Since Region A is indifferent between $(b^{B*}, z^{B*})$ and $(b^{A*}, z^{A*})$ under $\mu ^A>0$, it has no incentive to mimic Region B. For Region B, in view of Proposition 4.4(i) we have $V_{b}(b^{B*}, z^{B*}, \theta ^B, \delta ^B)>0$. Thus using $V_{bb}<0$ yields $V_{b}(b^B, z^{B*}, \theta ^B, \delta ^B) \ge V_{b}(b^{B*}, z^{B*}, \theta ^{B}, \delta ^B) >0$ for any $b^B \le b^{B*}$. Therefore, $(b^{B*}, z^{B*})$ is also realized by Region B. Given $z^{A*}<0<z^{B*}$ according to Proposition 4.4(i) and that $V_{z}=g_{1}'+\theta g_{2}'>0$ always holds true, we must have $V(b^B, z^{B*}, \theta ^B, \delta ^B)> V(b^B, z^{A*},\theta ^B, \delta ^B)$ for any feasible $b^B \le b^{B*}$. Put differently, Region B has no incentive to mimic Region A. Consequently, the incentive compatibility constraints for both regions are fulfilled under this budget rule.

The proof of Proposition 5.2(ii-b) is analogous to that of Proposition 5.2(ii-a), and hence one just needs to note that for Region B, we now have

$$\begin{aligned} V_{b}(b^{A*}, z^{A*}, \theta ^B, \delta ^B)&=g'(G_{1}^{A*}) - (1+r-\theta ^B)\delta ^B g'(\theta ^{B} G_{1}^{A*}+G_{2}^{A*})\\&< g'(G_{1}^{A*}) - (1+r-\theta ^A)\delta ^A g'(\theta ^{A} G_{1}^{A*}+G_{2}^{A*})<0, \end{aligned}$$

in which we have used Assumptions 2.1 and 4.2, and Proposition 4.4(ii). The remaining details are omitted to save space. $\square$

Appendix B: Discussion on multidimensional screening

We now consider the case in which both the degree of patience and the degree of intergenerational spillovers are privately observable by each region. As is well known, it is analytically intractable to obtain interesting results in the presence of multidimensional private information (Rochet and Choné 1998; Armstrong and Rochet 1999). For analytical tractability, we thus impose the following assumption.

Assumption 6.1

Let $\theta ^R \equiv \varphi (\delta ^R)$ for $R\in \{A,B\}$ and $\varphi (\cdot )$ be a continuously differentiable function satisfying:

• (a) $\varphi$ is publicly observable;

• (b) $\varphi '(\cdot )>0$;

  and (c) $\delta ^R \varphi '(\delta ^R)/\varphi (\delta ^R) \ge \max \left\{ g_{1}''/u_{1}'', [1+(g_{2}''/u_{2}'')][- \varphi (\delta ^R)G_{1} g_{2}''/g_{2}'] \right\}$.

Assumption 6.1(a) states that there is a publicly observable functional relationship between the degree of patience and the degree of intergenerational spillovers. Although the function $\varphi$ connecting the two parameters is public knowledge, $\theta ^R$ is still each region’s private information because $\delta ^R$ is privately observable. Assumption 6.1(b) is in line with Assumptions 2.1 and 4.1. It assumes that the region with a greater discount factor would entail a higher degree of intergenerational spillovers. Assumption 6.1(c) states that there is a lower bound for the elasticity of $\theta ^R$ with respect to $\delta ^R$, which is a technical requirement for establishing the following lemma.

Lemma 6.1

Suppose Assumptions 2.1 and 6.1 hold, then the single-crossing property is satisfied in the sense that

$$\begin{aligned} \frac{\mathrm {d} }{\mathrm {d} \delta ^R} \left( \frac{\mathrm {d} z}{\mathrm {d} b}\Big |_{\mathrm {d} V = 0} \right) <0 \ \text {for all} \ \delta ^R. \end{aligned}$$

Proof

To simplify notations, we will suppress the region index, R, throughout the proof. Under Assumption 6.1, the FOCs (8) are rewritten as follows:

$$\begin{aligned} \begin{aligned} u'\left( y_{1}+b+z-{\hat{G}}_{1}\right)&=g'\left( {\hat{G}}_{1}\right) + \varphi (\delta ) \delta g'\left( \varphi (\delta ) {\hat{G}}_{1}+{\hat{G}}_{2}\right) , \\ u'\left( y_{2}-b(1+r)-{\hat{G}}_{2}\right)&= g'\left( \varphi (\delta ) {\hat{G}}_{1}+{\hat{G}}_{2}\right) , \end{aligned}\end{aligned}$$

(57)

in which ${\hat{G}}_{1} \equiv \Phi (b,z,\delta )$ and ${\hat{G}}_{2}\equiv \Psi (b,z,\delta )$. Differentiating both sides of the two equations in (57) with respect to $\delta$, we then apply the Implicit Function Theorem and the Cramer’s Rule and find that

$$\begin{aligned} \begin{aligned} \Phi _{\delta }&= - \frac{[\varphi (\delta )+\delta \varphi '(\delta )]g_{2}'(u_{2}''+g_{2}'')+\delta \varphi (\delta )\varphi '(\delta )\Phi \cdot u_{2}''g_{2}''}{\Gamma } \ \ \text {and} \ \ \\ \Psi _{\delta }&=\frac{\varphi (\delta )[\varphi (\delta )+\delta \varphi '(\delta )]g_{2}'g_{2}'' -\varphi '(\delta )\Phi \cdot (u_{1}''+g_{1}'')g_{2}'' }{\Gamma }, \end{aligned}\end{aligned}$$

(58)

where $\Gamma \equiv (u_{1}''+g_{1}'')(u_{2}''+g_{2}'')+\delta \varphi (\delta )^{2} u_{2}''g_{2}''>0$. Additionally, Equation (12) can be rewritten as follows:

$$\begin{aligned} \frac{\mathrm {d} z}{\mathrm {d} b}\Big |_{\mathrm {d} V=0} = - \frac{g'\left( {\hat{G}}_{1}\right) -[1+r-\varphi (\delta )]\delta g'\left( \varphi (\delta ) {\hat{G}}_{1}+{\hat{G}}_{2}\right) }{g'\left( {\hat{G}}_{1}\right) +\varphi (\delta ) \delta g'\left( \varphi (\delta ) {\hat{G}}_{1}+{\hat{G}}_{2}\right) }. \end{aligned}$$

Then by rearranging the algebra, we arrive at

$$\begin{aligned} \begin{aligned} \frac{\Xi }{\Gamma }&\equiv \frac{\mathrm {d}}{\mathrm {d} \delta }\left( \frac{\mathrm {d} z}{\mathrm {d} b}\Big |_{\mathrm {d} V=0} \right) \cdot \frac{\left[ g_{1}' + \varphi (\delta ) \delta g_{2}'\right] ^{2}}{1+r} \\&= (g_{1}'-\delta \Phi _{\delta } g_{1}'')g_{2}'- \delta ^{2}\varphi '(\delta )(g_{2}')^2 + \delta g_{1}'g_{2}'' [\varphi '(\delta ) \Phi +\varphi (\delta ) \Phi _{\delta } + \Psi _{\delta } ]. \end{aligned}\end{aligned}$$

(59)

Using (58) and simplifying the algebra, we get

$$\begin{aligned} \begin{aligned}&\Gamma [\varphi '(\delta ) \Phi +\varphi (\delta ) \Phi _{\delta } + \Psi _{\delta } ] \\&\quad = \left\{ (u_{1}''+g_{1}'')\varphi '(\delta ) \Phi - \varphi (\delta )[\varphi (\delta )+\delta \varphi '(\delta )]g_{2}'\right\} u_{2}''>0. \end{aligned}\end{aligned}$$

(60)

Then, applying (60), (58) and Assumption 6.1 to (59) and simplifying the algebra, we have

$$\begin{aligned} \begin{aligned} 0 > \Xi&= \underset{-}{\underbrace{\delta ^{2}\varphi (\delta )\varphi '(\delta )u_{2}''g_{2}''g_{2}'(\Phi g_{1}''-g_{1}')- \delta ^{3} \varphi (\delta )^{2}\varphi '(\delta )(g_{2}')^{2} u_{2}''g_{2}''}} \\&\quad + \, \underset{-}{\underbrace{(u_{1}''+g_{1}'')g_{1}'}} \cdot \underset{+/0}{\underbrace{[(u_{2}''+g_{2}'') g_{2}'+\delta \varphi '(\delta )\Phi u_{2}''g_{2}'' ]}} \\&\quad +\, \underset{-}{\underbrace{\delta (u_{2}''+g_{2}'') (g_{2}')^{2}}} \cdot \underset{+/0}{\underbrace{[\varphi (\delta ) g_{1}'' - \delta \varphi '(\delta ) u_{1}'']}} \end{aligned}\end{aligned}$$

(61)

due to the fact that

$$\begin{aligned} \varphi (\delta ) g_{1}'' \ge \delta \varphi '(\delta ) u_{1}'' \ \Leftrightarrow \ \frac{\delta \varphi '(\delta )}{\varphi (\delta )} \ge \frac{g_{1}''}{u_{1}''} \end{aligned}$$

and

$$\begin{aligned} (u_{2}''+g_{2}'') g_{2}'+\delta \varphi '(\delta )\Phi u_{2}''g_{2}''\ge 0 \ \Leftrightarrow \ \frac{\delta \varphi '(\delta )}{\varphi (\delta )} \ge - \frac{(u_{2}''+g_{2}'') g_{2}'}{\varphi (\delta ) \Phi u_{2}''g_{2}''} \end{aligned}$$

under Assumption 6.1. Therefore, the desired assertion follows from applying (61) to (59). $\square$

In contrast to Lemma 3.1(ii), Lemma 6.1 yields that, in every point in the (b, z)-space, the indifference curve of H-region (or patient region) has a smaller slope than the indifference curve of L-region (or impatient region). We can then apply Lemma 6.1 and the reasoning used in the proof of Proposition 4.2 to establish the corresponding redistribution and local debt policies.

Similarly, when the public goods of the impatient region are more durable we can impose the following assumption:

Assumption 6.2

Let $\theta ^R \equiv \varphi (\delta ^R)$ for $R\in \{A,B\}$ and $\varphi (\cdot )$ be a continuously differentiable function satisfying: (a) $\varphi$ is publicly observable; (b) $\varphi '(\cdot )<0$; and (c)

$$\begin{aligned} \frac{\delta ^R \varphi (\delta ^R)g_{2}'}{\delta ^R \varphi (\delta ^R)g_{2}'-(u_{1}''+g_{1}'')\Phi } \le - \frac{\delta ^R \varphi '(\delta ^R)}{\varphi (\delta ^R)} \le 1. \end{aligned}$$

Assumption 6.2(b) yields that the impatient region’s public goods turn out to be more durable and is in line with Assumptions 2.1 and 4.2. Assumption 6.2(c) is a technical requirement for establishing the following lemma, which means that the absolute value of the elasticity of the degree of durability with respect to the degree of patience is no greater than 1 and no smaller than a positive threshold.

Lemma 6.2

Suppose Assumptions 2.1 and 6.2 hold, then the single-crossing property is satisfied in the sense that

$$\begin{aligned} \frac{\mathrm {d} }{\mathrm {d} \delta ^R} \left( \frac{\mathrm {d} z}{\mathrm {d} b}\Big |_{\mathrm {d} V = 0} \right) >0 \ \text {for all} \ \delta ^R. \end{aligned}$$

Proof

The proof is analogous to that of Lemma 6.1, one just needs to note that, under Assumption 6.2, we have $\Phi _{\delta } >0$, $\Gamma [\varphi '(\delta ) \Phi +\varphi (\delta ) \Phi _{\delta } + \Psi _{\delta } ]\le 0$, and hence

$$\begin{aligned}&\frac{\mathrm {d}}{\mathrm {d} \delta }\left( \frac{\mathrm {d} z}{\mathrm {d} b}\Big |_{\mathrm {d} V=0} \right) \cdot \frac{\left[ g_{1}' + \varphi (\delta ) \delta g_{2}'\right] ^{2}}{1+r} \\&\quad = \underset{+}{\underbrace{(g_{1}'-\delta \Phi _{\delta } g_{1}'')g_{2}'}} - \underset{-}{\underbrace{\delta ^{2}\varphi '(\delta )(g_{2}')^2}} + \underset{+/0}{\underbrace{\delta g_{1}'g_{2}'' [\varphi '(\delta ) \Phi +\varphi (\delta ) \Phi _{\delta } + \Psi _{\delta } ]}}. \end{aligned}$$

Thus, the desired assertion follows immediately. $\square$

We therefore conclude that, as shown in Lemma 3.1(ii), Lemma 6.2 yields that, in every point in the (b, z)-space, the indifference curves of the patient region have a greater slope than those of the impatient region. Using Lemma 6.2, one can characterize the corresponding asymmetric information optimum, as shown in Proposition 4.4. Due to paper length limitations, those details will be omitted.