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House prices and monetary policy

  • Paulo Brito , Giancarlo Marini and Alessandro Piergallini EMAIL logo


This paper analyzes global dynamics in an overlapping generations general equilibrium model with housing-wealth effects. It demonstrates that monetary policy cannot burst rational bubbles in the housing market. Under monetary policy rules of the Taylor-type, there exist global self-fulfilling paths of house prices along a heteroclinic orbit connecting multiple equilibria. From bifurcation analysis, the orbit features a boom (bust) in house prices when monetary policy is more (less) active. The paper also proves that booms or busts cannot be ruled out by interest-rate feedback rules responding to both inflation and house prices.

JEL: C61; C62; E31; E52

Corresponding author: Alessandro Piergallini, Department of Economics and Finance, Tor Vergata University, Via Columbia 2, 00133 Rome, Italy, Phone: +390672595431, Fax: +39062020500, e-mail:
aUECE (Research Unit on Complexity and Economics) is financially supported from national funds by FCT (Fundação para a Ciência e a Tecnologia), Portugal. This article is part of Strategic Project UID/ECO/00436/2013.


We are very grateful to an anonymous referee for many valuable comments and remarks. We also thank Paolo Canofari, Andrea Ferrero, Maurizio Fiaschetti, Ricardo Reis and participants to the 6th Annual Conference of the Portuguese Economic Journal, University of Porto, to the 54th Annual Conference of the Italian Economic Association, University of Bologna, and to the 2nd Macro Banking and Finance Workshop, University of Rome Tor Vergata, for very useful comments and suggestions. We gratefully acknowledge financial support from MIUR (Ministero dell’Istruzione, dell’Università e della Ricerca) and FCT (Fundação para a Ciência e a Tecnologia). The usual disclaimers apply.

Appendix A. Solution of the consumer’s problem

In the intertemporal optimization problem, the representative consumer born at time s chooses the optimal time path of total consumption, z̅(s, t), to maximize the lifetime utility function (2), given (6) and the constraints (3) and (4). Using the definition of total consumption, z̅(s, t)≡c̅(s, t)+R(t)m̅(s, t), and the optimal intratemporal condition (6), we can write

(49)logΛ(c¯(s,t),m¯(s,t))=logυ(t)+logz¯(s,t), (49)

where υ(t)Λ(Γ(R(t))Γ(R(t))+R(t),1Γ(R(t))+R(t)) is the same for all generations. Therefore, the intertemporal optimization problem can be formalized in the following terms:

(50)max{z¯(s,t),h¯(s,t)}0[α(logυ(t)+logz¯(s,t))+(1α)logh¯(s,t)]e(μ+ρ)tdt, (50)

subject to

(51)a¯˙(s,t)=(R(t)π(t)+μ)a¯(s,t)+y¯(s,t)τ¯(s,t)z¯(s,t)+[q˙(t)q(t)(R(t)π(t))]q(t)h¯(s,t), (51)

and given a̅(s, 0). The optimality conditions are

(52)z¯˙(s,t)=(R(t)π(t)ρ)z¯(s,t), (52)
(53)1ααz¯(s,t)=[(R(t)π(t))q˙(t)q(t)]q(t)h¯(s,t), (53)
(54)limta¯(s,t)e0t(R(j)π(j)+μ)dj=0. (54)

Therefore, the individual budget constraint (51) can be expressed as

(55)a¯˙(s,t)=(R(t)π(t)+μ)a¯(s,t)+y¯(s,t)τ¯(s,t)z¯(s,t)+[q˙(t)q(t)(R(t)π(t))]q(t)h¯(s,t)=(R(t)π(t)+μ)a¯(s,t)+y¯(s,t)z¯(s,t)1ααz¯(s,t)=(R(t)π(t)+μ)a¯(s,t)+y¯(s,t)1αz¯(s,t). (55)

Integrating forward (55), using the transversality condition (54) and (52), total consumption turns out to be a linear function of total wealth:

(56)z¯(s,t)=α(μ+ρ)(a¯(s,t)+k¯(s,t)), (56)

where k̅(s, t) is human wealth, defined as the present discounted value of after-tax labor income, k¯(s,t)t(y¯(s,t)τ¯(s,t))etv(R(j)π(j)+μ)djdv. From (6),

(57)z¯(s,t)=L(R(t))c¯(s,t), (57)

where L(R(t))≡1+R(t)/Γ(R(t)). Time-differentiating (57) yields

(58)z¯˙(s,t)=L(R(t))c¯(s,t)R˙(t)+L(R(t))c¯˙(s,t). (58)

Therefore, the dynamic equation for individual consumption is

(59)c¯˙(s,t)=(R(t)π(t)ρ)c¯(s,t)L(R(t))R˙(t)L(R(t))c¯(s,t). (59)

Appendix B. Aggregation

The per capita aggregate financial wealth is given by

(60)a(t)=βta¯(s,t)eβ(st)ds. (60)

Differentiating with respect to time yields

(61)a˙(t)=βa¯(t,t)βa(t)+βta¯˙(s,t)eβ(st)ds, (61)

where a̅(t, t) is equal to zero by assumption. Using (3) yields

(62)a˙(t)=βa(t)+μa(t)+(R(t)π(t))a(t)+y(t)τ(t)c(t)R(t)m(t)+[q˙(t)q(t)(R(t)π(t))]q(t)h(t)=(R(t)π(t)n)a(t)+y(t)τ(t)c(t)R(t)m(t)+[q˙(t)q(t)(R(t)π(t))]q(t)h(t). (62)

Using (11), the per capita aggregate consumption is given by

(63)c(t)=α(μ+ρ)L(R(t))(a(t)+k(t)), (63)

where k(t)=t(y(t)τ(t))etv(R(j)π(j)+μ)djdv is the per capita aggregate human wealth. Next differentiate with respect to time the definition of per capita aggregate consumption, to obtain

(64)c˙(t)=βc¯(t,t)βc(t)+βtc¯˙(s,t)eβ(st)ds. (64)

Note that c̅(t, t) denotes consumption of the newborn generation. Since a̅(t, t)=0, from (11) we have

(65)c¯(t,t)=α(μ+ρ)L(R(t))(k¯(t,t)). (65)

Using (12), (63) and (65) into (64) yields the time path of per capita aggregate consumption:

(66)c˙(t)=(R(t)π(t)ρ)c(t)L(R(t))R˙(t)L(R(t))c(t)αβ(ρ+μ)L(R(t))a(t). (66)

Appendix C. Proof of Proposition 1

Any equilibrium steady states, (R*, q*), are bounded points (R, q)∈(ℝ+, ℝ++) such that q˙=R˙=0 and the transversality condition holds if R*P(R*)>–μ. From equation (28), q˙=0 if and only if q=Ψ(R). From equation (29), R˙=0 if and only if R=0, because L′(0)=∞, or q=Φ(R). Therefore, there are two candidates for equilibrium steady states. First, we have R=0 and q=q0Ψ(0)=(1α)/(αP(0)) which constitute indeed an equilibrium point only if P(0)<0. In this case, the transversality condition is verified, because the condition –P(0)>–μ always holds. Second, from the condition q=Ψ(R)=Φ(R), there is a second candidate if the set {R≥0:Ψ(R)=Φ(R)>0} is non-empty. We have




and r=RP(R)>–μ from the transversality condition. Because β(1–α)(ρ+μ)>0, then r+>ρ>0>r. As L(R)>1, then the condition r>ρ must hold, which implies that the transversality condition is always met. Therefore, an equivalent condition for the existence of a second steady state is R1={R0:RP(R)=r+}. From the assumption that the monetary policy is globally active, RP(R) is monotonically increasing in R, which means that RP(R)∈[–P(0), +∞). Therefore, the steady state (R1,q1) exists if P(0)+r+≥0, where q1=Ψ(R1)=Φ(R1). As a result, there is multiplicity if 0>P(0)≥–r+, there is only one steady state (R1,q1) if P(0)>0, and there is only steady state (0,q0) if P(0)+r+<0. We set r*=r+.

Appendix D. Proof of Proposition 2

For the steady-state equilibrium (R,q)=(R1,q1), we have the trace and determinant of the Jacobian given by






yielding the eigenvalues λ1=2r*ρ>0 and λ2=(1P(R1))L(R1)/L(R1)>0 if the monetary policy is locally active, 1P(R1)>0. Then the steady state (R1,q1) is always a source.

Evaluating the trace and the determinant for equilibrium (R,q)=(0,q0), with q0=Ψ(0), we get






we have



Recall that this steady state exists only if P(0)<0, and observe that L′(0)=∞ and L″(0)/(L′(0))2=–∞. Then

trJ(0)=detJ(0)=(,if P(0)+r+>0+,if P(0)+r+<0

which means that the eigenvalues of the Jacobian are infinite and the steady state is singular. If P(0)+r+>0, the steady state is multiple (case (b) in Proposition 2) and is a kind of non-regular saddle point, and, if P(0)+r+<0, the steady state is unique (case (c) in Proposition 2) and is a kind of non-regular source. In any case, the flow approaches or diverges from (0,q0) at an infinite speed and is non-differentiable locally.

In order to study local dynamics, we can use several methods, such as, first, finding a de-singularized projection and studying its local dynamics, and, second, studying global dynamics.

If we use the first method, the natural way to remove the singularity introduced by L′(R) at R=0, would be to recast the system in variables (L, q),


where L≥1, R=R(L) is increasing and R(1)=0, R′(1)=0. However, in this case there is an unique steady state (L(R1),q1). Therefore, this method does not solve our de-singularization problem. We use the second method in the proof of Proposition 3. There, we show that the singular steady state (0,q0), for the case P(0)+r+>0 (i.e. case (b) in Proposition 2) behaves as a generalized saddle point.

Appendix E. Proof of Proposition 3

As the system (28)–(29) does not have an explicit solution, we must employ qualitative methods in order to study global dynamics. One possible method is to find a first integral of system (28)–(29), that is, a Lyapunov function V(·) such that V(R, q)=constant. We could not find this function. Another method is to determine a trapping area for the heteroclinic orbit. The rationale is the following: as the steady state (R1,q1) is a source, the unstable manifold is the set +/(R1,q1); as the steady state (0,q0) is a saddle point, the stable manifold is, locally, composed by a single trajectory belonging to ℝ+; therefore there is an intersection of the unstable manifold of the first point and of the stable manifold of the second which is non-empty. In order to prove that it exists, and to characterize it, we consider a trapping area for the heteroclinic orbit. In order to prove this, we start by determining the slopes of the heteroclinic orbit in the neighborhoods of the two equilibria, we build a trapping area enclosing the heteroclinic, and demonstrate that all the trajectories starting inside the trapping area escape from it, with the exception of those starting at any point along the heteroclinic orbit.

Appendix E.1. Slopes of the eigenspaces associated to the two equilibria

The unstable eigenspace 1u is the tangent space to the unstable manifold associated to equilibrium (R1,q1), in the space where (R,q)W+2 lie,


The unstable eigenspace 1u is the linear space which is tangent to W1u and is spanned by the eigenvectors (V1, 1) and (V2, 1) which are associated to eigenvalues λ1 and λ2, respectively, where


In general, V2>0 and the sign of V1 is ambiguous. Observe that the the slope of 1u,dqdR|1u, is the opposite to the slope of isocline q˙=0, locally at the steady state (R1,q1). Let us call 1,+u(1,u) the eigenspace related to the dominant (non-dominant) eigenvalue. The following conditions can be proved: (1) if r+>(1P(R1))L(R1)/L(R1), then 2r+ρ>(1P(R1))L(R1)/L(R1), which is equivalent to λ1>λ2, and therefore 1,+u={(R,q)W:(qq1)=V1(RR1)} and 1,u={(R,q)W:(qq1)=V2(RR1)}. In this case, V1<0 and V2>0 and the slope associated to the dominant eigenvalue is negative and the slope associated to the non-dominant eigenvalue is positive; (2) if λ1<λ2, which is equivalent to 2r+ρ<(1P(R1))L(R1)/L(R1), then r+<(1P(R1))L(R1)/L(R1), and 1,+u={(R,q)W:(qq1)=V2(RR1)} and 1,u={(R,q)W:(qq1)=V1(RR1)}. In this case, V1>V2>0 and the slope associated to the both eigenvalues are both positive, but the one associated with 1,u is steeper.

The stable manifold associated to steady state (0,q0) is defined as


However, we saw that the projection of the steady state (0,q0) in the space 𝒲 is singular. This means that the solution approaches the singular steady state asymptotically with an infinite speed. In order to characterize the dynamics in the space 𝒲 in the neighborhood of (0,q0), we have to take a different approach: observe that, as R′(1)=0, then a naïve calculation for the slope of the stable manifold in the neighborhood of the singular equilibria could be dq/dR=(r+ρ)/(αβ(ρ+μ)R′(1))=∞.

Instead, observe that along the singular surface R=0 we have R˙=0 and q˙=P(0)(Ψ(0)q). Then, from any point along this surface where qq0=Ψ(0), an unstable trajectory unfolds. This means that any trajectory coming from R>0 will be deflected away from the equilibrium point on hitting the surface R=0, with the exception of the one which converges to the equilibrium point (0,q0). The (global) direction of the vector field generated by equations (28)–(29) is given by

(67)dqdR|(q˙,R˙)=L(R)(RP(R))(qΨ(R))αβ(ρ+μ)(Φ(R)q). (67)

In order to determine the slope of the trajectory which converges to the singular steady state, we determine the slope of the vector field hitting the surface R=0 using equation (67). We have

dqdR|R=0=L(0)P(0)(qΨ(0))αβ(ρ+μ)(Φ(0)q)={,if qq00,if q=q0

because L′(0)=∞. Therefore, the stable manifold associated to the singular steady state (0,q0) is horizontal in the space 𝒲.

The heteroclinic orbit, Ω=W0sW1u, is tangent to a horizontal line in the neighborhood of the equilibrium point (0,q0) and is positively slopped in the neighborhood of (R1,q1), because it is tangent to 1,u.

Appendix E.2. Trapping area

Next we consider the case in which λ1<λ2, which is depicted in Figure 1. Recall that Ω is tangent to a line dq/dR=0 in the neighborhood of point (0,q0). Observe that, in the neighborhood of point (R1,q1), the slope of the eigenspace associated to the non-dominant eigenvalue (λ1) and of the isocline R˙=0 are both positive, but the former is less steep that the later because (see Figure 7)

Figure 7: Proof of Proposition 3.
Figure 7:

Proof of Proposition 3.


As the heteroclinic Ω is tangent to that eigenspace in the heighborhood of that equilibrium point, it will lie between the isocline R˙=0 and 1,u and will never cross this line.

This allows to consider the trapping area whose sides are given by the segment of the isocline q˙=0 between the two equilibria (recall that the two equilibria lay along this isocline), by a line passing through the steady state (R1,q1), whose slope is given by the eigenvector which is associated to the non-dominant eigenvalue and by the horizontal segment such that q=q0, between the q-axis and the previous eigenvector-line.

Formally, the trapping area [A, B, C] is defined by the vertices A(0,q0),B(R1,q1), and C(RC,q0)=(R1(q1q0)/V1,q0) and the sides (A,B)={(R,q)(0,R1)×(q1,q0):q˙=0},(A,C)={(R,q):R(0,R1),q=q0}, and (B,C)={(R,q)(R1,RC)×(q1,q0):q=q1+V1(RR1)}.

Next we have to demonstrate that the (global) direction of the vector field, given by equation (67), which is generated by equations (28)–(29), allows us to prove that all the trajectories hitting the three boundaries of the trapping exit the trapping area.

At side (A, B) we have q˙=0 and R˙<0, because


given the fact that –r<RP(R)<r+ if R[0,R1). Then, as the slope of the vector field in the interval is


then the vector field points globally out of the trapping area within (A, B) with a horizontal slope.

At side (A, C), the vector field corresponds to a horizontal line between point A and point C, which is in the intersection of a horizontal line passing through the q-axis and the direction defined by the eigenvector associated to the dominant eigenvalue at point B. These two lines meet at point C. Along (A, C), we have R˙<0 for R(0,Φ1(q0)),R˙=0 at point R=Φ1(q0) and R˙>0 for RΦ1(q0),RC, and, we have q˙>0 everywhere. As the slope of the vector field is given by

dqdR|(A,C)=L(R)(RP(R))(q0Ψ(R))αβ(ρ+μ)(Φ(R)q0){<0,if R(0,Φ1(q0)if R=Φ1(q0)>0,if R(Φ1(q0),RC)

then the vector field points globally out of the trapping area, at all points located at (A, C).

At side (B, C), which is a segment of the eigenspace 1,u between points (R1,q1) and (RC,q0), the vector field has local time-variations given by R˙>0 and q˙>0. As this side has a slope given by V1, then the trajectories exit the trapping area if the slope of the vector field is less steep than V1, that is, if and only if


The numerator is equivalent to


It is negative because R>R1 implies L(R)<L(R1),RP(R)>r+, q>Ψ(R), and because we assume L(R1)r+<L(R1)(1P(R1) from λ1<λ2. Then, all the trajectories reaching segment (B, C) will exit the trapping area.

Therefore, there is an unique trajectory starting from point B, (R1,q1), that does not hits the boundaries of the trapping area, and therefore converges to point A, (0,q0). This is the heteroclinic trajectory Ω, and it happens to cross the isocline R˙=0.

Appendix F. Proof of Proposition 4

We use the same methods as for the proof of Propositions 1, 2 and 3.

First, the steady-state conditions are q=Ψ(R, q) and R=0 or q=Φ(R, q), and a steady state is an equilibrium if R*P(R*, q*)+μ>0. A steady state exists and is an equilibrium if there is a q*>0 such that q*=Ψ(0, q*) and –P(0, q*)>–μ. Using the Taylor rule (21), the equilibrium condition is equivalent to (qq+)(qq)=0, where


As q<0<q+, then


which holds for any P(0, 0). As


then the transversality condition holds without further conditions.

An interior steady state is determined from the non-negative values of (R, q) such that q=Ψ(R, q)=Φ(R, q)>0. If we define r(R, q)≡RP(R, q), this condition is equivalent to (rr+)(rr)=0. As a necessary condition for a positive q is r(R, q)>ρ, then r(R, q)=r+>–μ, which means that the transversality condition is automatically verified. This is equivalent to γRq=(1+γ)(P(0, 0)+r+). Substituting in q=Ψ(R, q), we get the equation


After some algebra, we can prove that this equation has a non-negative solution for R if and only if r++P(0, 0)≥BP(0, 0)>0.

The local dynamics are determined in the same way as in the proofs of Propositions 2 and 3. But, in this case, the eigenvalues for the steady state (R1,q1) are


for the values of the parameters such that there are two steady states.

Appendix G. Proof of Proposition 5

It is easy to see that the steady state conditions are exactly the same as in the case of the conventional Taylor rule. The only thing that may change is related to the local and global dynamics of the model.

Applying the same methods as for the proof of Proposition 2, we find the eigenvalues for the steady state (R1,q1):


where sign(λ2)=sign(1+γδ).


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Published Online: 2016-3-1
Published in Print: 2016-6-1

©2016 by De Gruyter

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