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Learning, robust monetary policy and the merit of precaution

  • Marine Charlotte André ORCID logo EMAIL logo and Meixing Dai


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


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


Using Etπt+1at and (26), we get:


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




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


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


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


Comparing (39) and (41) yields:




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




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:


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:


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


After simplification, we get:


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:


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:




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


We rewrite p1, after some tedious calculus, as


or alternatively simplify it as


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:


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


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:


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


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:


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) :


It follows from (42)–(43) that


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


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


A.3 The effects of learning

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


Deriving cπcg with respect to γ yields:


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]


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


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


Deriving H with respect to γ yields


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


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


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


It follows that


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:


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]


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:


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


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


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


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

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