House prices and monetary policy

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

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:

subject to

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

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

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

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))e−∫tv(R(j)−π(j)+μ)djdv. From (6),

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

Therefore, the dynamic equation for individual consumption is

Appendix B. Aggregation

The per capita aggregate financial wealth is given by

Differentiating with respect to time yields

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

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

where k(t)=∫t∞(y(t)−τ(t))e−∫tv(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

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

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

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=q0∗≡Ψ(0)=−(1−α)/(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

Ψ(R)=Φ(R)⇔(R−P(R)−ρ)(R−P(R))−β(1−α)(ρ+μ)=(r−r+)(r−r−)=0,

where

r+≡ρ+ρ2+4β(1−α)(ρ+μ)2, r−≡ρ−ρ2+4β(1−α)(ρ+μ)2

and r=R–P(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∗={R≥0: R−P(R)=r+}. From the assumption that the monetary policy is globally active, R–P(R) is monotonically increasing in R, which means that R–P(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

trJ(R1∗)=r∗+αβ(ρ+μ)Φ′(R1∗)L′(R1∗)=2r∗−ρ+(1−P′(R1∗)L(R1∗)L′(R1∗)

and

detJ(R1∗)=r∗αβ(ρ+μ)(Φ′(R1∗)−Ψ′(R1∗)L′(R1∗))=(2r∗−ρ)(1−P′(R1∗))L(R1∗)(r∗−ρ)L′(R1∗)>0,

because

Φ′(R1∗)=q1∗(1−P′(R1∗)r∗−ρ+L′(R1∗)L(R1∗))>0,

yielding the eigenvalues λ₁=2r^*–ρ>0 and λ2=(1−P′(R1∗))L(R1∗)/L′(R1∗)>0 if the monetary policy is locally active, 1−P′(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

trJ(0)=−P(0)+αβ(ρ+μ)(Φ′(0)L′(0)+(Ψ(0)−Φ(0))L″(0)(L′(0))2)

and

detJ(0)=−P(0)αβ(ρ+μ)L′(0)(Φ′(0)−Ψ′(0)+(Ψ(0)−Φ(0))L″(0)L′(0)).

As

Ψ′(0)−Φ′(0)L′(0)=Ψ(0)−Φ(0)=(P(0)+r+)(P(0)+r−)αβ(ρ+μ)P(0)>0,

we have

trJ(0)=−(2P(0)+ρ)+(P(0)+r+)(P(0)+r−)P(0) L″(0)(L′(0))2 ,

detJ(0)=(P(0)+r+)(P(0)+r−)(1−L″(0)(L′(0))2).

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),

q˙=q(R−P(R))−(1−αα)L,L˙=L′(R)R˙=L(R−P(R)−ρ)−αβ(ρ+μ)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 ℝ+/(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.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.

dqdR|R˙=dqdR|ℰ1,−u=Φ′(R1∗)+r+r+−ρΨ′(R1∗)=−2r+−ρr+−ρq1∗(L′(R1∗)L(R1∗))<0.

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 ℰ1,−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∗−(q1∗−q0∗)/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(R−R1∗)}.

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

R˙|(A,B)=αβ(ρ+μ)(Φ(R)−Ψ(R))L′(R)=L(R)(R−P(R)−r+)(R−P(R)−r−)L′(R)(R−P(R))<0,

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

dqdR|(A,B)=0,

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, Φ−1(q0∗)),R˙=0 at point R=Φ−1(q0∗) and R˙>0 for R∈Φ−1(q0∗), RC, and, we have q˙>0 everywhere. As the slope of the vector field is given by

dqdR|(A,C)=L′(R)(R−P(R))(q0∗−Ψ(R))αβ(ρ+μ)(Φ(R)−q0∗) {<0,if R∈(0, Φ−1(q0∗)∞if R=Φ−1(q0∗)>0,if R∈(Φ−1(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 ℰ1,−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

dqdR|(B,C)−dqdR|ℰ1,−u=L′(R)(R−P(R))(q−Ψ(R))αβ(ρ+μ)(Φ(R)−q)+r+Ψ′(R1∗)r+−ρ<0.

The numerator is equivalent to

L′(R)(R−P(R))(q−Ψ(R))+(Φ(R)−q)(r+L′(R1∗)−L(R1∗)(1−P′(R1∗)))<L′(R)(R−P(R))(q−Ψ(R)) +(Φ(R)−q)((R−P(R))L′(R1∗)−L(R1∗)(1−P′(R1∗)))<L′(R1∗)(R−P(R))(Φ(R)−Ψ(R)) −(Φ(R)−q)L(R1∗)(1−P′(R1∗))<(Φ(R)−q)(L′(R1∗)(R−P(R))−L(R1∗)(1−P′(R1∗))) <(Φ(R)−q)(L′(R1∗)r+−L(R1∗)(1−P′(R1∗)))<0.

It is negative because R>R1∗ implies L′(R)<L′(R1∗),R–P(R)>r₊, q>Ψ(R), and because we assume L′(R1∗)r+<L(R1∗)(1−P′(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 (q–q₊)(q–q_–)=0, where

q±=(1+γϵ){P(0, 0)2±[(P(0, 0)2)2+1−ααϵ1+γ]1/2}.

As q_–<0<q₊, then

q0∗=q+=(1+γϵ)BP(0, 0)>0,

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

−P(0, q0∗)=ϵ1+γq0∗−P(0, 0)=BP(0, 0)−P(0, 0)=−P(0, 0)2+[(P(0, 0)2)2+1−ααϵ1+γ]1/2>0,

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)≡R–P(R, q), this condition is equivalent to (r–r₊)(r–r_–)=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 γR+ϵq=(1+γ)(P(0, 0)+r₊). Substituting in q=Ψ(R, q), we get the equation

L(R)=α1−α[(1+γϵ)(P(0, 0)+r+)−γϵR]r+.

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

λ1=2r+−ρ>0,λ2=ϵ1+γq1∗+γ1+γL(R1∗)L′(R1∗)>0

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∗):

λ1=2r+−ρ>0,λ2=γ1+γ−δL(R1∗)L′(R1∗),

where sign(λ₂)=sign(1+γ–δ).