The Stabilizing Virtues of Monetary Policy on Endogenous Bubble Fluctuations

Abstract

We explore the stabilizing role of monetary policy on the existence of endogenous fluctuations when the economy experiences a rational bubble. Considering an overlapping generations model, expectation-driven fluctuations are explained by a portfolio choice between three assets (capital, bonds and money), credit market imperfections and a collateral effect. They occur under a positive bubble on bonds. The key mechanism relies on the existence of gaps between the returns on assets due to financial distortions. Then, we study the stabilizing role of the monetary policy. Such a policy managed by a (standard) Taylor rule has no clear stabilizing virtues.

We would like to thank an anonymous referee for his comments. We also thank Stefano Bosi, Marion Davin, Takashi Kamihigashi, Carine Nourry and Alain Venditti for valuable suggestions. We also thank participants to the Conference in honor of Jean-Michel Grandmont held in Aix-Marseille University on June 2013, to the Conference 12th “Journées Louis-André Gérard-Varet”, to the Conference PET 2013 and to the 13th Annual SAET Conference. Any remaining errors are our own.

Appendix

1.1 Proof of Proposition 1

A steady state k is a solution of $h\left( k\right) =j\left( k\right)$, with:

$$\begin{aligned} h\left( k\right) \equiv \frac{u^{\prime }\left( c(k)\right) }{\beta v^{\prime }\left( d(k)\right) } \end{aligned}$$

(10.39a)

$$\begin{aligned} j\left( k\right) \equiv \frac{1}{1+i^{*}\gamma (k)} \end{aligned}$$

(10.39b)

where $\quad c\left( k\right) \equiv f(k)-k-d(k)$ and $d\left( k\right) \equiv$ $\displaystyle \frac{k[1-f^{\prime }(k)]}{i^{*}\eta _1(k)[1-\gamma (k)]}$.

We start by determining the admissible range of values for k. To ensure $d(k)>0$, we get at the steady state $f^{\prime }\left( k\right) <1$. Under Assumption 3, $f^{\prime }(k)$ is a decreasing function of k. Hence, $k>f^{\prime -1}\left( 1\right) =\underline{k}$.

Now, we want to determine the range of k such that $c(k)>0$. The decreasing returns on capital imply $f\left( \underline{k}\right) >\underline{k}f^{\prime }(\underline{k})$. Since $f^{\prime }(\underline{k})=1$, $c(\underline{k})=f(\underline{k})-\underline{k}>0$. In addition, as $d(k)>0$, we derive the following inequality:

$$\begin{aligned} \lim _{k\rightarrow +\infty }c\left( k\right)< & {} \lim _{k\rightarrow +\infty } f(k)-k=-\infty \end{aligned}$$

because $f^{\prime }(k)<1$ for k large enough. As a result, there exists one value $\overline{k}$ such that $\forall k<\overline{k}$, $c\left( \overline{k}\right) >0$. By construction, we have $\underline{k}< \overline{k}$, and therefore $(\underline{k}, \overline{k})$ is a nonempty subset.

To prove the existence of a stationary solution k, we use the continuity of $h\left( k\right)$ and $j\left( k\right)$. Using Eqs. (10.39a) and (10.39b), we determine the boundary values of $h\left( k\right)$ and $j\left( k\right)$:

$$\begin{aligned} \begin{array}{cc} \lim _{k\rightarrow \underline{k}}h\left( k\right) =\displaystyle \frac{u^{\prime }\left( c\left( \underline{k}\right) \right) }{\beta v^{\prime }(0)}=0^{+} &{} \lim _{k\rightarrow \overline{k}}h\left( k\right) =\displaystyle \frac{u^{\prime }\left( c\left( 0\right) \right) }{\beta v^{\prime }(d(\overline{k}))}=+\infty \\ \lim _{k\rightarrow \underline{k}}j\left( k\right) =\displaystyle \frac{1}{1+i(\underline{k})\gamma (\underline{k})}\in (0, 1] &{} \lim _{k\rightarrow \overline{k}}j\left( k\right) =\displaystyle \frac{1}{1+i(\overline{k})\gamma \left( \overline{k}\right) }\le 1 \end{array} \end{aligned}$$

We have $\lim _{k\rightarrow \underline{k}}h\left( k\right) <\lim _{k\rightarrow \underline{k}}j\left( k\right)$ and $\lim _{k\rightarrow \overline{k}}h\left( k\right) >\lim _{k\rightarrow \overline{k}}j\left( k\right)$. Therefore, there exists at least one value $k^{*} \in \left( \underline{k},\overline{k}\right)$ such that $h\left( k^{*}\right) =j\left( k^{*}\right)$.    $\blacksquare$

1.2 Proof of Example in Section 10.4.2

Let $\underline{\sigma } \equiv 1-\frac{1}{2+\nu (\eta _1)}$, $\bar{\alpha }\equiv \frac{1}{2+\nu (\eta _1)}$, $\underline{A}\equiv \frac{1+\nu (\eta _1)}{\alpha +\nu (\eta _1)}$ and $\bar{A} \equiv \frac{1/\alpha }{1+\nu (\eta _1)}$. For $a\in (0,1)$, $c>1$, $b\in (ac,c)$ and $\epsilon >0$, Assumption 2 is satisfied at the normalized steady state. Assumption 3 requires $A<1/\alpha$ and $\sigma >1-\alpha$. For $A>\underline{A}$, Assumption 5 is satisfied. Moreover, the bubble is positive at the normalized steady state when $A<\overline{A}$. As a consequence, the set $(\underline{A},\bar{A})$ must be non-empty. This is true for $\alpha <\bar{\alpha }$ and $\sigma >\underline{\sigma }$.

1.3 Proofs of Proposition 3, Corollaries 1 and 2

$$\begin{aligned} \text { Recall that } \psi= & {} f^{\prime }(1),\,\, \nu (\eta _1)= \eta _1i^{*}\left( 1-\gamma \right)>0,\,\, c^{*}= f(1) -1- \frac{1-\psi }{\nu (\eta _1)}>0, \\ a_\phi= & {} \phi /(1-\phi ) \in (-\infty , -1[\cup [0,+\infty ) \text { and let } \varepsilon _{dk}=\frac{\psi }{1-\psi } \frac{1-\alpha }{\sigma }+ \eta _2>0,\text { and } \nonumber \\ \eta _1^{\theta }= & {} \frac{1-\psi }{\psi }\frac{1}{i^{*}(1-\gamma )}\nonumber .\end{aligned}$$

(10.40)

First of all, we suppose for the rest of the proof that $\eta _1<\eta _1^{\theta }$, because $\forall$ $\eta _1<\eta _1^{\theta }$, $\theta >0$ (see condtion (10.35)).

From Lemma 2, the expressions of $1-T+D$, $1+T+D$ and $1-D$ can be written as follows:

$$\begin{aligned} 1-T+D= & {} \varepsilon _{dk}\frac{1-\psi \left[ 1+\nu (\eta _1)\right] }{\xi _1(a_\phi )}\frac{\varepsilon _v-\varepsilon _v^{s}}{\varepsilon _v-\bar{\varepsilon }_v}\end{aligned}$$

(10.41)

$$\begin{aligned} 1+T+D= & {} \frac{\xi _3(a_\phi )}{\xi _1(a_\phi )}\frac{\varepsilon _v-\varepsilon _v^{f}}{\varepsilon _v-\bar{\varepsilon }_v}\end{aligned}$$

(10.42)

$$\begin{aligned} 1-D= & {} \frac{\varepsilon _v-\varepsilon _v^{h}}{\varepsilon _v-\bar{\varepsilon }_v} \end{aligned}$$

(10.43)

$$\begin{aligned} \text { where } \varepsilon _v^{s}= & {} \frac{1-\psi }{\nu (\eta _1)}\frac{\nu (\eta _1)+\varepsilon _{dk}}{\varepsilon _{dk}} \frac{\varepsilon _u^{s}-\varepsilon _u}{c^{*}},\,\,\,\, \varepsilon _v^{f}=\frac{\xi _4(a_\phi )}{\xi _3(a_\phi )}\equiv \varepsilon _v^{f}(a_\phi ),\nonumber \\ \varepsilon _v^{h}= & {} \frac{\xi _5(a_\phi )}{\xi _1(a_\phi )}\equiv \varepsilon _v^{h}(a_\phi )\text { and } \bar{\varepsilon }_v=-\frac{\xi _2(a_\phi )}{\xi _1(a_\phi )}\equiv \bar{\varepsilon }_v(a_\phi ) \end{aligned}$$

(10.44)

$$\begin{aligned} \text {with }\varepsilon _u^{s}&= c^{*}\frac{\nu (\eta _1)}{1-\psi }\frac{1}{\nu (\eta _1)+\varepsilon _{dk}}\frac{\nu (\eta _1)}{1+i^{*}\gamma },\end{aligned}$$

(10.45)

$$\begin{aligned} \xi _1(a_\phi )&=-\psi \nu (\eta _1)\left( 1-\small {\frac{1-\alpha }{\sigma }}\right) \frac{1+i^{*}}{i^{*}} a_\phi +\left\{ 1-\psi \left[ 1+\nu (\eta _1)\right] \right\} \varepsilon _{dk}\nonumber \\&\quad -\frac{\psi }{1-\psi }\nu (\eta _1)\left( 1-\frac{1-\alpha }{\sigma }\right) ,\nonumber \\ \end{aligned}$$

(10.46)

$$\begin{aligned} \xi _2(a_\phi )&= \left( 1-\psi \right) \frac{1+i^{*}}{i^{*}}\left\{ -\frac{\varepsilon _u}{c^{*}}+\frac{\nu (\eta _1)}{1+i^{*}\gamma }\left[ 1+i^{*}\gamma \frac{\psi }{1-\psi }\left( 1-\frac{1-\alpha }{\sigma }\right) \right] \right. \nonumber \\&\left. -\varepsilon _{dk}\frac{i^{*}\gamma }{1+i^{*}\gamma }\right\} a_\phi -\psi \left[ 1+\nu (\eta _1)\right] \frac{\varepsilon _u}{c^{*}}+\nu (\eta _1)^{2}\frac{\psi }{1+i^{*}\gamma }+\nu (\eta _1)\psi \nonumber \\&\left( 1-\frac{1-\alpha }{\sigma }+\frac{1}{1+i^{*}\gamma }\right) -\left( 1-\psi \right) \varepsilon _{dk},\nonumber \\ \end{aligned}$$

(10.47)

$$\begin{aligned} \xi _3(a_\phi )&= -\psi \nu (\eta _1)\left( 1-\frac{1-\alpha }{\sigma }\right) \frac{1+i^{*}}{i^{*}} a_\phi +\left\{ 1-\psi \left[ 1+\nu (\eta _1)\right] \right\} \varepsilon _{dk}\nonumber \\&\quad -2\frac{\psi }{1-\psi }\nu (\eta _1) \left( 1-\frac{1-\alpha }{\sigma }\right) , \end{aligned}$$

(10.48)

$$\begin{aligned} \xi _4(a_\phi )&= 2 (1-\psi )\frac{1+i^{*}}{i^{*}}\left\{ \left( 1+\psi \frac{1-\alpha }{\sigma }\right) \frac{\varepsilon _u}{c^{*}}-\frac{i^{*}\gamma \nu (\eta _1)}{1+i^{*}\gamma }\left[ 1+ \frac{\psi }{1-\psi }\left( 1-\frac{1-\alpha }{\sigma }\right) \right] \right. \nonumber \\&\quad \left. +\frac{\varepsilon _{dk}i^{*}\gamma }{1+i^{*}\gamma }\right\} a_\phi +\left\{ 2\psi \left[ 1+\nu (\eta _1)+\frac{1-\alpha }{\sigma }\right] +\right. \nonumber \\&\quad \left. \frac{1-\psi }{\nu (\eta _1)}\left[ \nu (\eta _1)+\varepsilon _{dk}\right] \left\{ 1-\psi \left[ 1+\nu (\eta _1)\right] \right\} \right\} \frac{\varepsilon _u}{c^{*}}-\frac{\nu (\eta _1)^{2}\psi }{1+i^{*}\gamma }\nonumber \\&\quad -2\nu (\eta _1)\psi \left( 1-\frac{1-\alpha }{\sigma }\right) -\nu (\eta _1)\frac{1+\psi }{1+i^{*}\gamma }+2(1-\psi )\varepsilon _{dk},\end{aligned}$$

(10.49)

$$\begin{aligned} \xi _5(a_\phi )&=\left( 1-\psi \right) \frac{1+i^{*}}{i^{*}}\left\{ \left( 1-\psi \frac{1-\alpha }{\sigma }\right) \frac{\varepsilon _u}{c^{*}}-\frac{\nu (\eta _1)}{1+i^{*}\gamma }\right. \nonumber \\&\left. \quad \left[ 1+\frac{i^{*}\gamma \psi }{1-\psi }\left( 1-\frac{1-\alpha }{\sigma }\right) \right] +\frac{\varepsilon _{dk}i^{*}\gamma }{1+i^{*}\gamma }\right\} a_\phi +\psi \left[ 1+\nu (\eta _1)-\frac{1-\alpha }{\sigma }\right] \frac{\varepsilon _u}{c^{*}}-\frac{\nu (\eta _1)^{2}\psi }{1+i^{*}\gamma }\nonumber \\&\quad -\nu (\eta _1)\psi \left( 1-\frac{1-\alpha }{\sigma }+\frac{1}{1+i^{*}\gamma }\right) +(1-\psi )\varepsilon _{dk}\nonumber \\ \end{aligned}$$

(10.50)

We aim to determine the range of parameter values ($a_\phi$ and $\varepsilon _v$) for which local indeterminacy conditions (i)-(iii) are satisfied. To do this, we must analyze the functions $\varepsilon _v^{f}$, $\varepsilon _v^{h}$, $\bar{\varepsilon }_v$ and $\xi _i$ with $i=\lbrace 1,2,3,4,5 \rbrace$, then draw $\varepsilon _v^{f}$, $\varepsilon _v^{h}$ and $\bar{\varepsilon }_v$ in the plane $(a_\phi , \varepsilon _v)$.

We observe that $\xi _i$ with $i=\lbrace 1,2,3,4,5 \rbrace$ are linear functions of $a_\phi$, i.e. $\xi _i^{a} a_\phi +~\xi _i^{b}$. Note that $\xi _1^{a}<0$ and $\xi _3^{a}<0$. Furthermore, there exist $\eta _1^{\xi _1^{b}}>0$, $\eta _1^{\xi _2^{a}}>0$, $\eta _1^{\xi _2^{b}}>0$, $\eta _1^{\xi _3^{b}}>0$, $\eta _1^{\xi _4^{a}}>0$, $\eta _1^{\xi _4^{b}}>0$, $\eta _1^{\xi _5^{a}}>0$ and $\eta _1^{\xi _5^{b}}>0$ such that $\forall$ $\eta _1<min \lbrace \eta _1^{\xi _1^{b}}, \eta _1^{\xi _2^{a}}, \eta _1^{\xi _2^{b}}, \eta _1^{\xi _3^{b}}, \eta _1^{\xi _4^{a}}, \eta _1^{\xi _4^{b}}, \eta _1^{\xi _5^{a}}, \eta _1^{\xi _5^{b}} \rbrace \equiv \tilde{\eta }_1$, one has $\xi _1^{b}>0$, $\xi _2^{a}<0$, $\xi _2^{b}<0$, $\xi _3^{b}>0$, $\xi _4^{a}>0$, $\xi _4^{b}>0$, $\xi _5^{a}>0$ and $\xi _5^{b}>0$. Therefore, we deduce that $\xi _1(a_\phi )\ge 0$ when $a_\phi \le \frac{\xi _1^{b}}{\xi _1^{a}}$ and $\xi _1(a_\phi )<0$ otherwise. $\xi _2(a_\phi )\ge 0$ when $a_\phi \le \frac{\xi _2^{b}}{\xi _2^{a}}<0$ and $\xi _2(a_\phi )< 0$ otherwise. $\xi _3(a_\phi )\ge 0$ when $a_\phi \le \frac{\xi _3^{b}}{\xi _3^{a}}$ and $\xi _3(a_\phi )<0$ otherwise. $\xi _4(a_\phi )\le 0$ when $a_\phi \le \frac{\xi _4^{b}}{\xi _4^{a}}<0$ and $\xi _4(a_\phi )>0$ otherwise. $\xi _5(a_\phi )\le 0$ when $a_\phi \le \frac{\xi _5^{b}}{\xi _5^{a}}<0$ and $\xi _5(a_\phi )>0$ otherwise.

We analyze now $\varepsilon _v^{f}$, $\varepsilon _v^{h}$, $\varepsilon _v^{s}$ and $\bar{\varepsilon }_v$. Suppose that $\eta _1<min\lbrace \eta _1^{\theta },\tilde{\eta }_1\rbrace$.

First, $\varepsilon _v^{s}$ does not depend on $a_\phi$. Second, the different critical and bifurcation values ($\bar{\varepsilon }_v$, $\varepsilon _v^{f}$, $\varepsilon _v^{h}$) are homographic functions of $a_\phi$. $\varepsilon _v^{f}$ has a vertical asymptote at $a_\phi =\frac{\xi _3^{b}}{\xi _3^{a}}\equiv \tilde{a}_\phi >0$. $\bar{\varepsilon }_v$ and $\varepsilon _v^{h}$ have the same vertical asymptote at $a_\phi =\frac{\xi _1^{b}}{\xi _1^{a}}>\tilde{a}_\phi$. The first derivatives of $\varepsilon _v^{f}$, $\varepsilon _v^{h}$ and $\bar{\varepsilon }_v$ with respect to $a_{\phi }$ are given by:

$$ \frac{\partial \varepsilon _v^{f}}{\partial a_{\phi }}=\frac{\xi _4^{a}\xi _3^{b}-\xi _4^{b}\xi _3^{a}}{\xi _3(a_\phi )^{2}}>0,\frac{\partial \varepsilon _v^{h}}{\partial a_{\phi }}=\frac{\xi _5^{a}\xi _1^{b}-\xi _5^{b}\xi _1^{a}}{\xi _1(a_\phi )^{2}}>0 \text { and } \frac{\partial \bar{\varepsilon }_v}{\partial a_{\phi }}=-\frac{\xi _2^{a}\xi _1^{b}-\xi _2^{b}\xi _1^{a}}{\xi _1(a_\phi )^{2}}>0.$$

It would be useful to locate the different bifurcation and critical values ($\varepsilon _v^{f}$, $\varepsilon _v^{h}$, $\varepsilon _v^{s}$ and $\bar{\varepsilon }_v)$ when $a_\phi =0$. We can note that $\varepsilon _v^{f}>0$, $\varepsilon _v^{h}>0$ and $\bar{\varepsilon }_v>0$ when $a_\phi =0$. Moreover, $\varepsilon _v^{s}>0$ if and only if $\varepsilon _u\in (\tilde{\varepsilon }_u, \varepsilon _u^{s})$, where $\varepsilon _u>\tilde{\varepsilon }_u$ is required for the second-order conditions, with:

$$\begin{aligned} \tilde{\varepsilon }_u\equiv & {} c^{*}\frac{\nu (\eta _1)^{2}}{\eta _2\left( 1-\psi \right) \left( 1+i^{*}\gamma \right) ^{2}} \end{aligned}$$

(10.51)

The set ($\tilde{\varepsilon }_u, \varepsilon _u^{s}$) is nonempty if and only if

$$\begin{aligned} \eta _2(1-\psi )i^{*}\gamma> & {} \nu (\eta _1)\left( 1-\psi \right) +\psi \frac{1-\alpha }{\sigma } \end{aligned}$$

(10.52)

As $\nu (\eta _1)=i^{*}\eta _1\left( 1-\gamma \right)$, the condition (10.52) holds if

$$\begin{aligned} \eta _1 < \frac{1}{i^{*}\left( 1-\gamma \right) } \frac{\eta _2(1-\psi )i^{*}\gamma -\psi \left( 1-\alpha \right) /\sigma }{1-\psi }\equiv \overline{\overline{\eta }}_1\text { and }\eta _2 > \frac{\psi }{1-\psi }\frac{\left( 1-\alpha \right) /\sigma }{i^{*}\gamma }\equiv \bar{\eta }_2 \end{aligned}$$

(10.53)

Therefore, $\forall$ $\eta _1<\overline{\overline{\eta }}_1$ and $\eta _2>\bar{\eta }_2$, we have $\tilde{\varepsilon }_u<\varepsilon _u^{s}$. We suppose now that $\eta _1<\min \lbrace \eta _1^{\theta }, \tilde{\eta }_1, \overline{\overline{\eta }}_1\rbrace$ and $\eta _2>\bar{\eta }_2$. Note that using the function $\gamma (k)$ given by Eq. (10.7), $\eta _2>\bar{\eta }_2$ is equivalent to:

$$\begin{aligned} \epsilon >\frac{1+c}{c-1}\left( \frac{\psi }{1-\psi }\frac{1-\alpha }{\sigma i \gamma }-1\right) \equiv \bar{\epsilon } \end{aligned}$$

(10.54)

where $\gamma =1-(a+b)/(1+c)$.

We can show that $\varepsilon _v^{s}<1$. Since $\varepsilon _{dk}=\frac{\psi }{1-\psi }\frac{1-\alpha }{\sigma }+\eta _2$ and the condition (10.52) is satisfied, we have $\nu (\eta _1)<\left( 1+i^{*}\gamma \right) \varepsilon _{dk}$. Therefore, $\varepsilon _v^{s}= \frac{\nu (\eta _1)}{1+i^{*}\gamma }\frac{1}{\varepsilon _{dk}}-\frac{\varepsilon _u}{c^{*}} \frac{\nu (\eta _1)+\varepsilon _{dk}}{\varepsilon _{dk}}\frac{1-\psi }{\nu (\eta _1)}<~1$.

Furthermore, $\varepsilon _v^{h}<\bar{\varepsilon }_v$ and $\varepsilon _v^{h}>1$ for a sufficiently small $\eta _1$. Indeed $\varepsilon _v^{h}>1$ is equivalent to $\psi Q(\nu (\eta _1))>0$, where $Q(\nu (\eta _1))$ is a quadratic polynomial defined on $\mathbb {R}_{+}$ such that:

$$\begin{aligned} Q(\nu (\eta _1))=&-\frac{\nu (\eta _1)^{2}}{1+i^{*}\gamma } +\nu (\eta _1)\left[ \frac{\varepsilon _u}{c^{*}}-\left( 1-\frac{1-\alpha }{\sigma }+\frac{1}{1+i\gamma }\right) \right. \\&+ \left. \varepsilon _{dk}+\frac{1-(1-\alpha )/\sigma }{1-\psi }\right] +\left( 1-\frac{1-\alpha }{\sigma }\right) \frac{\varepsilon _u}{c^{*}} \end{aligned}$$

$Q(\nu (\eta _1))$ is a concave function with $Q(\nu (0))>0$. As a consequence, there is a threshold $\hat{\eta }_1>0$ such that $\forall \eta _1<\hat{\eta }_1$, $Q(\nu (\eta _1))>0$.

Concerning $\varepsilon _v^{f}$, we can show that $\bar{\varepsilon }_v<\varepsilon _v^{f}$ for $\eta _1$ small enough. Note that $\varepsilon _v^{f}>\xi _4^{b}/\xi _1^{b}$. $\xi _4^{b}/\xi _1^{b}>\bar{\varepsilon }_v$ is satisfied if $-\nu (\eta _1) \left[ \psi \left( 1-\frac{1-\alpha }{\sigma }\right) +\frac{1}{1+i^{*}\gamma }\right] +\left( 1-\psi \right) \varepsilon _{dk}>0$. Therefore, there exists $\underline{\eta }_1>0$ such that $\forall$ $\eta _1<\underline{\eta }_1$, this inequality is satisfied. Hence, $\forall$ $\eta _1<\underline{\eta }_1$, $\bar{\varepsilon }_v<\varepsilon _v^{f}$.

Therefore, $\forall$ $\eta _1<\min \lbrace \eta _1^{\theta }, \tilde{\eta }_1, \overline{\overline{\eta }}_1,\hat{\eta }_1,\underline{\eta }_1 \rbrace$, one has $\varepsilon _v^{s}<1<\varepsilon _v^{h}<\bar{\varepsilon }_v<\varepsilon _v^{f}$ when $a_\phi =0.$ Moreover, we can show that there exists $\eta _1^{\prime }>0$ such that $\forall$ $\eta _1<\eta _1^{\prime }$, $\varepsilon _v^{s}<\varepsilon _v^{h}<\bar{\varepsilon }_v<\varepsilon _v^{f}$ when $a_\phi =-1$. For the rest of the proof, we assume that $\eta _1<\min \lbrace \eta _1^{\theta }, \tilde{\eta }_1, \overline{\overline{\eta }}_1,\hat{\eta }_1,\underline{\eta }_1, \eta _1^{\prime }\rbrace$.

Recall that $a_\phi$ is defined on $(-\infty , -1) \cup [0, +\infty )$. At this stage of the proof, we can state by analyzing $1-T+D$, $1+T+D$ and $1-D$ given by Eqs. (10.41)–(10.43) that if $a_\phi \in (-\infty , -1)$, local indeterminacy occurs when $\varepsilon _v<min \lbrace \bar{\varepsilon }_v, \varepsilon _v^{f}, \varepsilon _v^{h}, \varepsilon _v^{s}\rbrace$ or when $\varepsilon _v> max \lbrace \bar{\varepsilon }_v, \varepsilon _v^{f}, \varepsilon _v^{h}, \varepsilon _v^{s}\rbrace .$ From the previous results, we deduce that if $a_\phi \in [0,\tilde{a}_\phi )$, local indeterminacy occurs when $\varepsilon _v<\varepsilon _v^{s}$ or when $\varepsilon _v>\varepsilon _v^{f}$. Finally, if $a_\phi >\tilde{a}_\phi$, local indeterminacy occurs when $\varepsilon <\varepsilon _v^{s}$. For the case $a_\phi \in (-\infty , -1)$, we should determine the location of $\varepsilon _v^{f}$, $\varepsilon _v^{h}$, $\varepsilon _v^{s}$ and $\bar{\varepsilon }_v$ in the plane $(a_\phi , \varepsilon _v).$

The functions $\varepsilon _v^{f}$, $\varepsilon _v^{h}$ and $\bar{\varepsilon }_v$ are continuous and monotone increasing on $a_\phi \in (-\infty , -1)$. We can show that the graph of these functions cross the horizonntal axis on $a_\phi \in (-\infty , -1)$. Let introduce the different following points $a_\phi ^{\xi _2}$, $a_\phi ^{\xi _4}$ and $a_\phi ^{\xi _5}$, which corresponds to the points at which $\bar{\varepsilon }_v$, $\varepsilon _v^{f}$ and $\varepsilon _v^{h}$ cross the horizontal axis. $a_{\phi }^{\xi _2}$ is defined by $\bar{\varepsilon }_v=0$ such that $a_\phi ^{\xi _2}=-\frac{\xi _2^{b}}{\xi _2^{a}}<0$. $a_\phi ^{\xi _4}$ is defined by $\varepsilon _v^{f}=0$ such that $a_\phi ^{\xi _4}=-\frac{\xi _4^{b}}{\xi _4^{a}}<0$. $a_\phi ^{\xi _5}$ is defined by $\varepsilon _v^{h}=0$ such that $a_\phi ^{\xi _5}=-\frac{\xi _5^{b}}{\xi _5^{a}}<0$.

After some algebra, we can show that since $\psi <1/(1+i^{*}\gamma )$, there exists $\eta _1^{a}>0$ such that $\forall$ $\eta _1<\eta _1^{a}$, $a_\phi ^{\xi _5}<a_\phi ^{\xi _2}$. Furthermore, either $a_\phi ^{\xi _2}<a_\phi ^{\xi _4}$ $\forall$ $\eta _1>0$ or there exists $\eta _1^{b}$ such that $\forall$ $\eta _1<\eta _1^{b}>0$, $a_\phi ^{\xi _2}<a_\phi ^{\xi _4}$. Hence, if $\eta _1<min \lbrace \eta _1^{a}, \eta _1^{b}, \eta _1^{\theta }, \tilde{\eta }_1, \overline{\overline{\eta }}_1,\hat{\eta }_1,\underline{\eta }_1, \eta _1^{\prime } \rbrace$, one has $a_\phi ^{\xi _5}<a_\phi ^{\xi _2}< a_\phi ^{\xi _4}<0$.

Let $\eta _1^{\prime \prime }=min \lbrace \eta _1^{a}, \eta _1^{b}, \eta _1^{\theta }, \tilde{\eta }_1, \overline{\overline{\eta }}_1,\hat{\eta }_1,\underline{\eta }_1, \eta _1^{\prime } \rbrace$. For the rest of the proof, we suppose that $\eta _1<\eta _1^{\prime \prime }$.

Since the functions $\varepsilon _v^{f}$, $\varepsilon _v^{h}$ and $\bar{\varepsilon }_v$ are continuous and monotone increasing on $(-\infty , -1) \cup [0, \tilde{a}_\phi )$, we can now locate $\varepsilon _v^{f}$, $\varepsilon _v^{h}$ and $\bar{\varepsilon }_v$ in the plane $(a_\phi , \varepsilon _v).$

Let $\hat{a}_\phi \equiv a_\phi ^{\xi _4}$. Since $\psi <1/(1+i^{*}\gamma ),$ $a_\phi ^{\xi ^{5}}<a_\phi ^{\xi _2}< \hat{a}_\phi$. Using the expressions of $1-T+D$, $1+T+D$ and $1-D$ given by Eqs. (10.41)–(10.43), we state that if $a_\phi \in (-\infty , \hat{a}_\phi )$, local indeterminacy occurs when $\varepsilon _v>max \lbrace \varepsilon _v^{f}, \varepsilon _v^{h}, \varepsilon _v^{s} \rbrace .$ If $a_\phi \in [\hat{a}_\phi , -1)\cup [0,\tilde{a}_\phi )$, local indeterminacy occurs when $\varepsilon _v<min\lbrace \varepsilon _v^{f}, \varepsilon _v^{s} \rbrace$ or when $\varepsilon _v>max \lbrace \varepsilon _v^{f}, \varepsilon _v^{h}, \varepsilon _v^{s} \rbrace .$

We have shown that $\varepsilon _v^{s}$ can be positive. In such a case, it would be useful to determine when $\varepsilon _v^{f}$, $\varepsilon _v^{h}$ and $\bar{\varepsilon }_v$ cross $\varepsilon _v^{s}$. Suppose that $\varepsilon _v^{s}>0$. $\varepsilon _v^{h}=\varepsilon _v^{s}$ when $a_\phi = - \left( \varepsilon _v^{s}\xi _1^{b}-\xi _5^{b}\right) /(\varepsilon _v^{s}\xi _1^{a}-\xi _5^{a})\equiv a_\phi ^{1} <0$, $\bar{\varepsilon }_v=\varepsilon _v^{s}$ when $a_\phi = -\left( \varepsilon _v^{s}\xi _1^{b}+\xi _2^{b}\right) /(\varepsilon _v^{s}\xi _1^{a}+\xi _2^{a})\equiv a_\phi ^{2}<0$, and $\varepsilon _v^{f}=\varepsilon _v^{s}$ when $a_\phi = -\left( \varepsilon _v^{s}\xi _3^{b}-\xi _4^{b}\right) /(\varepsilon _v^{s}\xi _3^{a}-\xi _4^{a})\equiv a_\phi ^{3}<0$.

Because $\psi <1/(1+i^{*}\gamma )$, we can show after some algebra that either $a_\phi ^{1}<a_\phi ^{2}$ $\forall$ $\eta _1>0$ or there exists $\eta _1^{e}>0$ such that $\forall$ $\eta _1<\eta _1^{e}$, $a_\phi ^{1}<a_\phi ^{2}$. It is difficult to determine the location of $a_\phi ^{3}$. Nevertheless, if $a_\phi ^{1}<a_\phi ^{3}<a_\phi ^{2}<0$ or if $a_\phi ^{3}<a_\phi ^{1}<a_\phi ^{2}<0$, we get $1-T+D<0,$ $1+T+D<0$ and $D>1$ for some values of $\varepsilon _v$. Since this is not feasible, we can eliminate these configurations. Therefore, if $\eta _1<min \lbrace \eta _1^{e}, \eta _1^{\prime \prime } \rbrace$, one has $a_\phi ^{1}<a_\phi ^{2}<a_\phi ^{3}<0$.

Let $\bar{\eta }_1=min \lbrace \eta _1^{e}, \eta _1^{\prime \prime } \rbrace$. All conditions on $\eta _1$ required in this proof are satisfied when $\eta _1<\bar{\eta }_1$. We can now derive the dynamic properties of the model. The properties of local dynamics are depicted by Fig. 10.1.¹⁹ Grey areas in Fig. 10.1 correspond to the different regions in which the steady state is a sink, in other words to the indeterminacy regions.

We deduce Proposition 3 and Corollary 1 from Fig. 10.1. Since $a_\phi =\phi /(1-\phi )$ is increasing with $\phi$, we can derive Corollary 2 from Proposition 3 and Fig. 10.1.    $\blacksquare$