On Equilibrium Determinacy in Overlapping Generations Models with Money

A straightforward calculation yields

	$\displaystyle D_{\xi}\Phi$	$\displaystyle=\begin{bmatrix}s_{w}\xi_{1}f^{\prime\prime}(\xi_{1})&1\\ 0&-f^{\prime}(\eta_{1})\end{bmatrix},$
	$\displaystyle D_{\eta}\Phi$	$\displaystyle=\begin{bmatrix}1-s_{R}f^{\prime\prime}(\eta_{1})&0\\ -\xi_{2}f^{\prime\prime}(\eta_{1})&1\end{bmatrix}.$

If $D_{\eta}\Phi$ is nonsingular, we may apply the implicit function theorem to obtain

	$\displaystyle D_{\xi}\phi$	$\displaystyle=-[D_{\eta}\Phi]^{-1}D_{\xi}\Phi$
		$\displaystyle=\frac{1}{1-s_{R}f^{\prime\prime}(\eta_{1})}\begin{bmatrix}-s_{w}\xi_{1}f^{\prime\prime}(\xi_{1})&-1\\ -s_{w}\xi_{1}\xi_{2}f^{\prime\prime}(\xi_{1})f^{\prime\prime}(\eta_{1})&(1-s_{R}f^{\prime\prime}(\eta_{1}))f^{\prime}(\eta_{1})-\xi_{2}f^{\prime\prime}(\eta_{1})\end{bmatrix}.$

Setting $\xi=\eta=(k,P)$ yields (2.4). ∎

NMSS conditions are utility maximization (2.1), profit maximization $R=f^{\prime}(k^{*})$ and $w=f(k^{*})-k^{*}f^{\prime}(k^{*})$, and market clearing condition $k^{*}=s$. Let us construct such a model with $f^{\prime\prime}(k^{*})=c$. First, given $u$, $v$, $k^{*}>0$, $R>0$, and $w>k^{*}$, set $s=k^{*}$ and take a discount factor $\beta>0$ such that $u^{\prime}(w-s)=\beta Rv^{\prime}(Rs)$. Next, define the output $Y\coloneqq w+Rk^{*}$. Since $w>0$, we can take a nonnegative, increasing, and concave function $f$ such that $f(k^{*})=Y$, $f^{\prime}(k^{*})=R$, and $f^{\prime\prime}(k^{*})=c$. For instance, for small enough $\epsilon>0$, we can let

$$f(k)=Y+R(k-k^{*})+\frac{c}{2}(k-k^{*})^{2}=w+Rk+\frac{c}{2}(k-k^{*})^{2}$$

for $k\in[k^{*}-\epsilon,k^{*}+\epsilon]$ and linearly extrapolate outside that range. (The condition $w>0$ ensures that this $f$ is nonnegative as long as $\epsilon>0$ is small enough.) With this $f$, we have $R=f^{\prime}(k^{*})$ and $w=Y-Rk^{*}=f(k^{*})-k^{*}f^{\prime}(k^{*})$, so the first-order conditions for profit maximization are satisfied. Therefore $(k^{*},0)$ is indeed NMSS. ∎

For simplicity, suppose $c=0$. (The proof for $c<0$ sufficiently close to 0 is similar.) Since $f^{\prime}(k^{*})=R$ and $f^{\prime\prime}(k^{*})=c=0$, the Jacobian (2.5) reduces to

$$D_{\xi}\phi=\begin{bmatrix}0&-1\\ 0&R\end{bmatrix},$$

whose eigenvalues are $0,R$. If $R<1$, the steady state $(k^{*},0)$ is stable, and therefore for any $(k_{0},P_{0})$ sufficiently close to $(k^{*},0)$, there exists an equilibrium path converging to the steady state.

If $R>1$, the steady state $(k^{*},0)$ is a saddle point. Therefore for any $k_{0}$ sufficiently close to $k^{*}$, there exists a unique equilibrium path converging to the steady state. On such a path, because $\left\{{P_{t}}\right\}$ is bounded and $R>1$, we have $\lim_{t\to\infty}R^{-t}P_{t}=0$. Since the transversality condition for asset pricing holds (see Hirano and Toda (2024, §2)), the asset price $P_{t}$ must equal its fundamental value, so $P_{t}=0$ for all $t$. ∎

A simple application of the implicit function theorem shows

	$\displaystyle s_{w}(w,R)$	$\displaystyle=\frac{u^{\prime\prime}(w-s)}{u^{\prime\prime}(w-s)+\beta R^{2}v^{\prime\prime}(Rs)}\in(0,1),$		(A.1a)
	$\displaystyle s_{R}(w,R)$	$\displaystyle=-\frac{\beta v^{\prime}(Rs)+\beta Rsv^{\prime\prime}(Rs)}{u^{\prime\prime}(w-s)+\beta R^{2}v^{\prime\prime}(Rs)}.$		(A.1b)

In MSS, the no-arbitrage condition (2.3) implies $f^{\prime}(k)=1$. Therefore letting $J$ be the Jacobian (2.4) and $c=f^{\prime\prime}<0$, we have

	$\displaystyle t$	$\displaystyle\coloneqq\operatorname{tr}J=1-\frac{(s_{w}k+P)c}{1-s_{R}c},$
	$\displaystyle d$	$\displaystyle\coloneqq\det J=-\frac{s_{w}kc}{1-s_{R}c}.$

Let the characteristic polynomial of $J$ be

$$p(x)\coloneqq\det(xI-J)=x^{2}-tx+d.$$

By assumption, $1-s_{R}c>0$. Since $s_{w}>0$ by (A.1a) and $c<0$, it follows that $p(0)=d>0$. Furthermore,

$$p(1)=1-t+d=\frac{Pc}{1-s_{R}c}<0.$$

Therefore the two roots $\lambda_{1},\lambda_{2}$ of $p$ satisfy $0<\lambda_{1}<1<\lambda_{2}$. Since the steady state is a saddle point, MSS is locally determinate.

Since $c<0$, the condition $1-s_{R}c>0$ trivially holds if $s_{R}\geq 0$. Noting that $u^{\prime\prime}<0$, $v^{\prime\prime}<0$, and $v^{\prime}(z)+zv^{\prime\prime}(z)=v^{\prime}(z)(1-\gamma_{v}(z))$, by (A.1b) a sufficient condition for $s_{R}\geq 0$ is $\gamma_{v}\leq 1$. ∎

We focus on the case $\rho\neq 1$, though the Cobb-Douglas case with $\rho=1$ can be analyzed by taking the limit $\rho\to 1$.

Define $g(k)=(\alpha k^{1-\rho}+1-\alpha)^{\frac{1}{1-\rho}}$. A straightforward calculation yields

	$\displaystyle g^{\prime}$	$\displaystyle=\alpha k^{-\rho}g^{\rho},$
	$\displaystyle g^{\prime\prime}$	$\displaystyle=-\rho\alpha(1-\alpha)k^{-\rho-1}g^{2\rho-1}.$

The equilibrium condition (2.2) at NMSS is

$$k=\beta w=\beta A(1-\alpha)g^{\rho}\iff A=\frac{1}{\beta(1-\alpha)}kg^{-\rho}.$$

(A.2)

Therefore $A>0$ is uniquely determined once we determine $\alpha,\rho$. At NMSS, we have

	$\displaystyle\lambda_{1}$	$\displaystyle=-\beta kf^{\prime\prime}(k)=\beta kA\rho\alpha(1-\alpha)k^{-\rho-1}g^{2\rho-1}$
		$\displaystyle=\rho\alpha k^{1-\rho}g^{\rho-1}=\rho\frac{\alpha k^{1-\rho}}{\alpha k^{1-\rho}+1-\alpha},$		(A.3a)
	$\displaystyle\lambda_{2}$	$\displaystyle=f^{\prime}(k)=A\alpha k^{-\rho}g^{\rho}+1-\delta$
		$\displaystyle=\frac{\alpha}{\beta(1-\alpha)}k^{1-\rho}+1-\delta.$		(A.3b)

Solving (A.3b) for $\alpha k^{1-\rho}$ and substituting into (A.3a), we obtain

$$\lambda_{1}=\rho\frac{\beta(\lambda_{2}-1+\delta)}{\beta(\lambda_{2}-1+\delta)+1}\iff\rho=\lambda_{1}\frac{\beta(\lambda_{2}-1+\delta)+1}{\beta(\lambda_{2}-1+\delta)}.$$

(A.4)

Therefore $\rho$ is uniquely determined as in (A.4). Because the right-hand side of (A.3b) is strictly increasing in $\alpha\in(0,1)$ and its range is $(1-\delta,\infty)$, there exists a unique $\alpha\in(0,1)$ satisfying (A.3b). ∎

If $\rho=1$, the equilibrium condition (2.2) at NMSS is

$$k=s=\beta A(1-\alpha)k^{\alpha}\iff k_{f}\coloneqq[\beta A(1-\alpha)]^{\frac{1}{1-\alpha}},$$

which uniquely exists. At this $k_{f}$, (A.3) implies $\lambda_{1}=\alpha<1$ and $\lambda_{2}=\frac{\alpha}{\beta(1-\alpha)}+1-\delta$. Therefore NMSS is locally indeterminate if and only if $\lambda_{2}<1$, which is equivalent to $\beta\delta(c-1)>1$ because $c=1/\alpha$. If MSS exists, we have

$$1=f^{\prime}(k)=A\alpha k^{\alpha-1}+1-\delta\iff k_{b}\coloneqq(A\alpha/\delta)^{\frac{1}{1-\alpha}}.$$

The equilibrium condition (2.2) at $k=k_{b}$ implies

$$P=s-k=\beta A(1-\alpha)k^{\alpha}-k=(A\alpha/\delta)^{\frac{1}{1-\alpha}}\left(\beta\delta\frac{1-\alpha}{\alpha}-1\right).$$

Therefore a necessary and sufficient condition for the existence (and uniqueness) of MSS is $P>0$, or equivalently, $\beta\delta(c-1)>1$. ∎

The uniqueness of $k$ follows from the strict concavity of $f$.

If MSS exists, using (A.3b), we have

$$1=f^{\prime}(k)=A\alpha(g/k)^{\rho}+1-\delta\iff g/k=(A\alpha/\delta)^{-1/\rho}.$$

Define $x=\alpha k^{1-\rho}$. Then this condition is equivalent to

$$\frac{x+1-\alpha}{x/\alpha}=(A\alpha/\delta)^{\frac{\rho-1}{\rho}}\iff x=\frac{1-\alpha}{\alpha^{-\frac{1}{\rho}}(\delta/A)^{\frac{1-\rho}{\rho}}-1}=\frac{1-\alpha}{c-1},$$

(A.5)

so $c>1$ is necessary. Under this condition, the steady state capital is $k=(x/\alpha)^{\frac{1}{1-\rho}}=\left(\frac{1/\alpha-1}{c-1}\right)^{\frac{1}{1-\rho}}$.

For this $k$ to be MSS, it is necessary and sufficient that $P=\beta w-k>0$, where $w$ is the wage. Using (A.2), we obtain

$$\beta w/k=\beta A(1-\alpha)g^{\rho}/k=\beta A(1-\alpha)\alpha^{\frac{1}{1-\rho}}\psi(x)^{\frac{1}{1-\rho}},$$

(A.6)

where $\psi(x)\coloneqq(x+1-\alpha)^{\rho}/x$. Using (A.5), we obtain

		$\displaystyle x+1-\alpha=\frac{c(1-\alpha)}{c-1}=cx$
	$\displaystyle\iff$	$\displaystyle\psi(x)=\frac{(x+1-\alpha)^{\rho}}{x}=c^{\rho}x^{\rho-1}$
	$\displaystyle\iff$	$\displaystyle\psi(x)^{\frac{1}{1-\rho}}=c^{\frac{\rho}{1-\rho}}\frac{c-1}{1-\alpha}=\alpha^{-\frac{1}{1-\rho}}(\delta/A)\frac{c-1}{1-\alpha}.$

Therefore by (A.6), we obtain

	$\displaystyle P>0$	$\displaystyle\iff\beta A(1-\alpha)\alpha^{\frac{1}{1-\rho}}\alpha^{-\frac{1}{1-\rho}}(\delta/A)\frac{c-1}{1-\alpha}>1$
		$\displaystyle\iff\beta\delta(c-1)>1.\qed$

The existence and uniqueness of MSS are immediate from Lemmas A.1 and A.2. Since $\gamma_{v}(z)=1$, Theorem 2 implies that MSS is locally determinate. ∎

We use the same notation as in the proof of Lemma A.2. By (A.6), NMSS $x$ satisfies

$$\beta A(1-\alpha)\alpha^{\frac{1}{1-\rho}}\psi(x)^{\frac{1}{1-\rho}}\iff\psi(x)=\frac{1}{\alpha[\beta A(1-\alpha)]^{1-\rho}}.$$

A straightforward calculation yields

$$\psi^{\prime}(x)=\frac{(x+1-\alpha)^{\rho-1}}{x^{2}}((\rho-1)x-(1-\alpha)).$$

(A.7)

If $\rho<1$, then $\psi^{\prime}(x)<0$, $\psi(0)=\infty$, and $\psi(\infty)=0$, so a unique NMSS exists. If $\rho>1$, then $\psi$ achieves a unique minimum at $m\coloneqq\frac{1-\alpha}{\rho-1}$ and $\psi(0)=\psi(\infty)=\infty$. Therefore NMSS exists if and only if

$$\rho^{\rho}\left(\frac{1-\alpha}{\rho-1}\right)^{\rho-1}=\psi(m)\leq\psi(x)=\frac{1}{\alpha[\beta A(1-\alpha)]^{1-\rho}},$$

which is equivalent to the desired condition. Furthermore, since $\psi^{\prime}(x)\gtrless 0$ according as $x\lessgtr m$, we obtain the claim. ∎

We use the same notation as in the proof of Lemma A.2. Let $x_{b},x_{f}$ and $k_{b},k_{f}$ be the $x$ and $k$ in the (necessarily unique) MSS and NMSS, where $x=\alpha k^{1-\rho}$. Then $\rho<1$, (A.6), the monotonicity of $\psi$, and the strict concavity of $f$ imply that MSS exists if and only if

		$\displaystyle\beta A(1-\alpha)\alpha^{\frac{1}{1-\rho}}\psi(x_{b})^{\frac{1}{1-\rho}}>1=\beta A(1-\alpha)\alpha^{\frac{1}{1-\rho}}\psi(x_{f})^{\frac{1}{1-\rho}}$		(A.8)
	$\displaystyle\iff$	$\displaystyle\psi(x_{b})>\psi(x_{f})\iff x_{b}<x_{f}$
	$\displaystyle\iff$	$\displaystyle k_{b}<k_{f}\iff 1=f^{\prime}(k_{b})>f^{\prime}(k_{f})=\lambda_{2}.\qed$

The case $\rho=1$ follows from Lemma A.1. If $\rho<1$, then (A.3a) implies $\lambda_{1}\in(0,1)$ at NMSS. The conclusion holds by Lemmas A.2 and A.4. ∎

(i) Obvious from Lemma A.3.

(ii) Solving for $\alpha A^{1-\rho}$, the existence conditions for MSS in Proposition 3.2 and for NMSS in Lemma A.3 are, respectively,

	$\displaystyle\alpha A^{1-\rho}$	$\displaystyle<\delta(1/\beta+\delta)^{-\rho},$		(A.9a)
	$\displaystyle\alpha A^{1-\rho}$	$\displaystyle\leq\rho^{-\rho}[\beta(\rho-1)]^{\rho-1}.$		(A.9b)

Using calculus, it is straightforward to show that the maximum of the right-hand side of (A.9a) over $\delta>0$ is achieved at $\delta=\frac{1}{\beta(\rho-1)}$, in which case the maximum value is equal to the right-hand side of (A.9b). Therefore whenever MSS exists, there exist exactly two NMSS.

Suppose MSS $x_{b}$ exists and denote the two NMSS by $x_{f}<x_{f}^{\prime}$. Then $\rho>1$ and (A.8) imply $\psi(x_{b})<\psi(x_{f})=\psi(x_{f}^{\prime})$. Since $\psi$ is strictly quasi-concave, it must be $x_{f}<x_{b}<x_{f}^{\prime}$. Since $x=\alpha k^{1-\rho}$ and $\rho>1$, we have $k_{f}>k_{b}>k_{f}^{\prime}$. Since $f$ is strictly concave, we have

$$0<\lambda_{2}=f^{\prime}(k_{f})<f^{\prime}(k_{b})=1<f^{\prime}(k_{f}^{\prime})=\lambda_{2}^{\prime}.$$

Again, the strict quasi-concavity of $\psi$ implies $x_{f}<m<x_{f}^{\prime}$, where $m=\frac{1-\alpha}{\rho-1}$. Therefore using (A.3a), we obtain

$$0<\lambda_{1}=\rho\frac{x_{f}}{x_{f}+1-\alpha}<\rho\frac{m}{m+1-\alpha}=1<\rho\frac{x_{f}^{\prime}}{x_{f}^{\prime}+1-\alpha}=\lambda_{1}^{\prime}.$$

Note this last inequality is independent of whether $x_{b}$ exists or not. ∎