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Licensed Unlicensed Requires Authentication Published by De Gruyter February 6, 2018

Learning, robust monetary policy and the merit of precaution

  • Marine Charlotte André ORCID logo EMAIL logo and Meixing Dai

Abstract

We study in a New Keynesian framework the consequences of adaptive learning for the design of robust monetary policy. Compared to rational expectations, the fact that private sector follows adaptive learning gives the central bank an additional intertemporal trade-off between optimal behavior in the present and in later periods thanks to its ability to manipulate future inflation expectations. We show that adaptive learning imposes a more restrictive constraint on monetary policy robustness to ensure the dynamic stability of the equilibrium than under rational expectations but strengthens the argument in favor of a more aggressive monetary policy when the central bank fears for model misspecifications.

JEL Classification: C62; D83; D84; E52; E58

Acknowledgement

We are grateful to Jean-Bernard Chatelain, Francesco De Palma, Amélie Barbier-Gauchard, Rodolphe Dos Santos Ferreira, Moïse Sidiropoulos, Marco Maria Sorge and an anonymous reviewer for helpful remarks and suggestions.

A Appendix

In subsections A.1 and A.2, we closely follow Molnár and Santoro (2014) to find the equilibrium solution under learning. In subsections A.3 and A.4, we develop original techniques to show the effects of learning and robustness on the equilibrium.

A.1 The equilibrium solution of inflation under learning

Substituting λ2,t+1=0, λ3,t and λ3,t+1 given by (24) andλ1,t+1=ακxt+1 from (20) into (21), we obtain

(32)πt=ακxt+β(1γ)Et*πt+1+[αγβ2κ+β(1γ)ακ]Etxt+1.

Using Etπt+1at and (26), we get:

(33)xt=κθκ2θα(πtβatet),
(34)xt+1=κθκ2θα(πt+1βat+1et+1).

Substituting xt and xt+1 given by (33)–(34) into (32) and arranging the terms yields:

(35)Etπt+1=A11πt+A12at+P1et,

with

(36)A11κ2θα+αθ+θαγβ2[1γ(1β)]β{(κ2θα)(1γ)+θα[1γ(1β)]},
(37)A12αβθ[1β(1γ)[1γ(1β)]]β{(κ2θα)(1γ)+θα[1γ(1β)]},
(38)P1αθβ{(κ2θα)(1γ)+θα[1γ(1β)]}.

According to the proposition 1 from Blanchard and Kahn (1980), the ALM solution for inflation takes the following form :

(39)πt=cπcgat+dπcget.

Advancing (39) one period and taking the expectation of the resulting equation while using (6) yield:

(40)Etπt+1=cπcg[(1γ)at+γπt].

Using (35) to eliminate Etπt+1 in (40) and arranging the terms, we get:

(41)πt=A12cπcg(1γ)cπcgγA11at+P1cπcgγA11et.

Comparing (39) and (41) yields:

(42)cπ,tcg=A12cπcg(1γ)cπcgγA11,

and

(43)dπ,tcg=P1cπcgγA11.

We gather equations (6), (7) and (35), while using (33) to substitute xt to obtain the system of three equations:

Etyt+1=Atyt+Ptet

where

yt[πtatbt],A[A11A120γ1γ0κθγακ2θγβκθακ2θ1γ],andP[P10κθγακ2θ].

The above system is subject to three boundary conditions: a0, b0, and lims|Etπt+s|<. The eigenvalues of At are 1 − γ and the two eigenvalues of A1:

(44)A1=[A11A12γ1γ].

We can show that, in Appendix A.2, A1 has an eigenvalue inside and one outside the unit circle.□

A.2 The single stable solution

Among infinite stochastic sequences satisfying equation (42), we focus on a non-explosive solution, i.e. the solution corresponding to the eigenvalue of A1 given by (44) inside the unit circle. The trace and determinant of A1 are both positive. Thus, for A1 to have two real eigenvalues (μ1, μ2), one inside and one outside the unit circle, it is sufficient to show that (1μ1)(1μ2)<0. This can be rewritten as:

(45)μ1+μ2>1+μ1μ2.

Knowing that μ1+μ2 is equal to the trace of A1 and μ1μ2 equal to its determinant, we rewrite (45) as:

κ2θα+αθ+θαγβ2[1γ(1β)]β{(κ2θα)(1γ)+θα[1γ(1β)]}+1γ>1+(1γ)κ2θα+αθ+θαγβ2[1γ(1β)]β{(κ2θα)(1γ)+θα[1γ(1β)]}+γαβθ[1β(1γ)[1γ(1β)]]β{(κ2θα)(1γ)+θα[1γ(1β)]}.

After simplification, we get:

(κ2θα)[1β(1γ)]+αθ(1β){1β[1γ(1β)]}>0,

which is verified given that β]0,1[ and γ[0,1] and condition (27).

There exists a unique solution to the model, whose ALM takes the following form:

(46)πt=cπcgat+dπcget.

To have a converging (and non-explosive) inflation, we must have cπcg[0,1]. Rewriting (42) as γ(cπcg)2A11cπcgA12+cπcg(1γ)=0 and substituting A11 and A12 by their expressions, we obtain:

(47)p2(cπcg)2+p1cπcg+p0=0

where

p0=αβθ{1β(1γ)[1γ(1β)]}>0,p2=γβ{(κ2θα)(1γ)+αθ[1γ(1β)]},p1=α+β(1γ){(κ2θα)(1γ)+αθ[1γ(1β)]}{θ(κ2+α)+αβ2γθ[1γ(1β)]}.

To characterize the two solutions of cπcg, we rewrite (47) as:

(48)cπcg=p0+p2(cπcg)2p1f(cπcg).

We rewrite p1, after some tedious calculus, as

(49)p1=(κ2θα)[1β(1γ)]αθ(1β){1β[1γ(1β)])}p0p2,

or alternatively simplify it as

(50)p1=βp2θ(α+κ2)αθαβp0.

The conditions imposed on θ to ensure that p2>0, i.e. θ>αα(1+γβ1γ)+κ2, and that p1<0, i.e. θ>ακ2+α(1+γ2β31β(1γ)2), are less restrictive than condition (27), i.e. θ>ακ2 that is imposed to ensure that current output gap decreases with a rise in expected inflation or a positive cost-push shocks.

Under RE, to ensure the dynamic stability of the equilibrium, we must have according to (9) that αθβθ(α+κ2)α<1, or equivalently θ>αα(1β)+κ2, which is more restrictive than the condition (12), i.e. θ>αα+κ2. Thus, the stability condition is  θ>αα(1β)+κ2.

For θ>ακ2, we always have p2>0, and p1<0. This implies that f(cπcg):[0,1][0,1], with f(0)=p0+p2p1>0 and 0<f(1)=p0+p2p1<1 and f(cπcg)=2p2p1cπcg>0. Hence, the Brower theorem and the fact that f(cπcg) is strictly monotonously increasing in the interval cπcg[0,1] imply that there is a unique solution in this interval. The other possible solution is greater than unit and is excluded because it leads to an explosive evolution of inflation.

To ensure that p1>p0+p2 and hence the existence of a stable solution, we must have (κ2θα)[1β(1γ)]+αθ(1β){1β[1γ(1β)])}>0. This implies that:

(51)θ>αα(1β)[1γβ21β(1γ)]+κ2.

The stability condition given by (51) is too loose compared to condition (27), i.e. θ>ακ2. As a result, the stability condition is θ>ακ2 instead of (51).

The stable solution of cπcg is given by

(52)cπcg=p1p124p2p02p2.

The other possible solution cπcg=p1+p124p2p02p2 is greater than unit and is excluded to avoid an explosive evolution of inflation. Substituting A11 and P1 into (43) and rearranging the terms leads to:

(53)dπcg=αθθ(α+κ2)α+θαγ2β2(βcπcg)+γβ(1γ){θαβ[θ(α+κ2)α]cπcg}.

We now show that f(cπcg) defined in (48) is bounded, i.e. f(cπcg):[0;αβθθ(α+κ2)α]]0;αβθθ(α+κ2)α[. Knowing that f(0)>0 and substituting cπcg by αβθθ(α+κ2)α into the function f(cπcg), we find

(54)f(αβθθ(α+κ2)α)=p0+p2[αβθθ(α+κ2)α]2p1=αβθθ(α+κ2)α{θ(α+κ2)ααβθp0+αβθθ(α+κ2)αp2}p1.

Using p2=θα(1β)+κ2θαθα+κ2θαp2+θαβθα+κ2θαp2, p0=θα(1β)+κ2θαθαβp0+θα+κ2θαθαβp0 and the definition of p0, p1, and p2 given above, we substitute p1 using (50), we obtain:

f(αβθθ(α+κ2)α)=p0+p2[αβθθ(α+κ2)α]2p1=αβθθ(α+κ2)α{θ(α+κ2)ααβθp0+αβθθ(α+κ2)αp2}β(κ2θα)θ(α+κ2)αp2+θ(α+κ2)αθαβp0+θαβθ(α+κ2)αp2<αβθθ(α+κ2)α.

Given that f(cπcg)=2p2p1cπcg>0 for cπcg[0,1], f(cπcg) is strictly increasing in the interval [0;αβθθ(α+κ2)α]. This property and the fact that f(cπcg):[0;αβθθ(α+κ2)α]]0;αβθθ(α+κ2)α[ imply that there is a unique solution for cπcg so that 0<cπcg<αβθθ(α+κ2)α.

The case whereγ= 0. We obtain by substituting γ = 0 into (36)–(38) :

A11θ(α+κ2)αβ[θ(α+κ2)α]=1β,
A12αθ(1β)θ(α+κ2)α,
P1αθβ[θ(α+κ2)α].

It follows from (42)–(43) that

cπcg=αβθθ(α+κ2)α,dπcg=αθθ(α+κ2)α.

The case whereγ= 1. Inserting γ = 1 into (36)–(38) yields

A11θ(α+κ2+αβ3)ααβ2θ,A121β,P11β2.

Substituting the latter into (36)–(38) leads to p2=θαβ2>0, p1=ακ2θαθθαβ3<0 and p0=αβθ>0, and hence

cπcg=θ(α+κ2)α+θαβ3[θ(α+κ2)α+θαβ3]24θ2α2β32θαβ2,dπcg=αθθ(α+κ2)α+β2θα(βcπcg).

A.3 The effects of learning

Deriving p0, p1 and p2 with respect to γ and using (50), we get:

p0γ=αβ2θ[(2β)(1γ)+γβ]>0,p1γ=βp2γθ(α+κ2)αθαβp0γ<0,p2γ=β{(κ2θα)(12γ)+θα[12γ(1β)]},=1βp1γθ(α+κ2)αθαβp0γ.

Deriving cπcg with respect to γ yields:

cπcgγ=[p1γ1p124p2p0(p1p1γ2p0p2γ2p2p0γ)]p2(p1p124p2p0)p2γ2p22,

which can be rewritten, using p2γ=1βp1γ1βθ(α+κ2)αθαβp0γ and the definition of p0, p1 and p2, p1=βp2θ(α+κ2)αθαβp0 and cπcg=p1p124p2p02p2 and after fastidious arrangements of terms, as:[19]

cπcgγ=1θ(α+κ2)αθαβcπcgβp2p124p2p0(p0p1γp1p0γ).

Using cπcg<θαβθ(α+κ2)α, we obtain: 1θαβθ(α+κ2)αcπcg>1θαβθ(α+κ2)αθ(α+κ2)αθαβ=0. To determine the sign of Hp0p1γp1p0γ, we first check its value for γ = 1 and then its derivative with respect to γ.

It is easy to check that for γ = 1, we have

H=αβ3θ{αθ[κ2αβ(1β2)]}<0

if κ2αβ(1β2)<0; otherwise, we must impose:

θ>ακ2αβ(1β2)>ακ2.

Deriving H with respect to γ yields

Hγ=p02p12γp12p02γ=2αθβ3(1β){[1γ(1γβ)]{θ[κ2+α(1β)]α}+β3γ2αθ}>0.

Consequently, given that H < 0 for γ = 1 and Hγ>0 for γ[0,1], we conclude that

cπcgγ<0.

Deriving dπcg given by (53) with respect to γ yields:

dπcgγ=κ2θακθdxcgγ=κ2θασκθdrcgγ=αβθ[Φγ[αβγθ+(1γ)Θ]cπcgγ][Θ+θαγ2β2(βcπcg)+γβ(1γ)(θαβΘcπcg)]2,

where Θθ(α+κ2)α and Φ2αβγθ(βcπcg)+(12γ)(αβθΘcπcg). Using the fact that β is very close to one and hence 2β1>0, and the fact that cπcg<θαβθ(α+κ2)α, which implies that βcπcg>0 and θαβΘcπcg>0, we find that

Φ=βθαγ(2β1)(βcπcg)+βγ(θκ2α)cπcg+β(1γ){θαβ[θ(α+κ2)α]cπcg}>0.

It follows that

dπcgγ<0.

Using the definition of cxcg, dxcg, crcg and drcg, it is straightforward to show the sign of their partial derivative with respect to γ.

A.4 Effects of robustness

Deriving p0, p1 and p2 with respect to θ and using (50), we get:

p0θ=αβ{1β(1γ)[1γ(1β)]}=p0θ>0,p1θ=κ2[1β(1γ)2]α(1β){1β[1γ(1β)])}p0θp2θ<0,p2θ=γβ{κ2(1γ)+α[1γ(1β)]}>0.

Deriving cπcg given by (52) with respect to θ, and using p2θ=1βp1θ(α+κ2)αβ2p0θ, the definition of p0, p1 and p2, p1=βp2θ(α+κ2)αθαβp0 and cπcg=p1p124p2p02p2 yield:[20]

cπcgθ=12p22(Ip0θ+Jp1θ),

where I=2p2p124p2p0{p2+(α+κ2)αβ2p0+(α+κ2)αβ2p1cπcg} and J=2p2βp124p2p0[(1θ(α+κ2)αθαβcπcg)p0]. Using these definitions and the expressions of p0θ and p1θ derived in the above, we obtain:

cπcgθ=(p1θ12p1p1θ4p0p2θ4p2p0θp124p2p0)p2(p1p124p2p0)p2θ2p22=(p1θ12p1p1θ4p0p2θ4p2p0θp124p2p0)2p21p2cπcgp2θ.

Using p1=βp2θ(α+κ2)αθαβp0, the definition of p0, p1 and p2, we can show that p0p2θp2p0θ>0 and hence

cπcgθ=12p222p2p124p2p0{[1θ(α+κ2)αθαβcπcg](p0p2θp2p0θ)+1θβcπcgp2p0θ}<0.

Deriving dπcg given by (53) with respect to θ yields

dπcgθ=α2[1βγ(1γ)cπcg]+αθ{θαγ2β2+γβ(1γ)[θ(α+κ2)α]}cπcgθ{θ(α+κ2)α+θαγ2β2(βcπcg)+γβ(1γ){θαβ[θ(α+κ2)α]cπcg}}2<0.

Deriving cxcg, dxcg, crcg and drcg with respect to θ leads to

cxcgθ=1σcrcgθ=κα(κ2θα)2(βcπcg)+κθκ2θαcπcgθ,
dxcgθ=1σdrcgθ=ακ(κ2θα)2(1dπcg)+κθκ2θαdπcgθ.

To ensure that cxcgθ=1σcrcgθ>0, and dxcgθ=1σdrcgθ > 0, we must have cπcgθ>α(βcπcg)θ(κ2θα) and dπcgθ>α(1dπcg)θ(κ2θα), respectively. For standard parameters values, these conditions are checked.

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Article note

A technical appendix is available upon request to the author.


Published Online: 2018-02-06

©2018 Walter de Gruyter GmbH, Berlin/Boston

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