Learning, robust monetary policy and the merit of precaution

Marine Charlotte André; Meixing Dai

doi:10.1515/bejm-2016-0236

Published by De Gruyter February 6, 2018

Learning, robust monetary policy and the merit of precaution

Marine Charlotte André and Meixing Dai

From the journal The B.E. Journal of Macroeconomics

https://doi.org/10.1515/bejm-2016-0236

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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.

Keywords: adaptive learning; model uncertainty; optimal monetary policy; robust control

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, λ_3,t and λ3,t+1 given by (24) andλ1,t+1=−ακxt+1 from (20) into (21), we obtain

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

Using Etπt+1≡at and (26), we get:

(33)xt=κθκ2θ−α(πt−βat−et),

(34)xt+1=κθκ2θ−α(πt+1−βat+1−et+1).

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

(35)Et∗πt+1=A11πt+A12at+P1et,

with

(36)A11≡κ2θ−α+αθ+θαγβ2[1−γ(1−β)]β{(κ2θ−α)(1−γ)+θα[1−γ(1−β)]},

(37)A12≡−αβθ[1−β(1−γ)[1−γ(1−β)]]β{(κ2θ−α)(1−γ)+θα[1−γ(1−β)]},

(38)P1≡−αθβ{(κ2θ−α)(1−γ)+θα[1−γ(1−β)]}.

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)Et∗πt+1=cπcg[(1−γ)at+γπt].

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

(41)πt=A12−cπcg(1−γ)cπcgγ−A11at+P1cπcgγ−A11et.

Comparing (39) and (41) yields:

(42)cπ,tcg=A12−cπcg(1−γ)cπcgγ−A11,

and

(43)dπ,tcg=P1cπcgγ−A11.

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

Et∗yt+1=Atyt+Ptet

where

yt≡[πtatbt],A≡[A11A120γ1−γ0−κθγα−κ2θγβκθα−κ2θ1−γ],andP≡[P10κθγα−κ2θ].

The above system is subject to three boundary conditions: a₀, b₀, and lims→∞|Et∗πt+s|<∞. The eigenvalues of A_t are 1 − γ and the two eigenvalues of A₁:

(44)A1=[A11A12γ1−γ].

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 (1−μ1)(1−μ2)<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:

κ2θ−α+αθ+θαγβ2[1−γ(1−β)]β{(κ2θ−α)(1−γ)+θα[1−γ(1−β)]}+1−γ>1+(1−γ)κ2θ−α+αθ+θαγβ2[1−γ(1−β)]β{(κ2θ−α)(1−γ)+θα[1−γ(1−β)]}+γαβθ[1−β(1−γ)[1−γ(1−β)]]β{(κ2θ−α)(1−γ)+θα[1−γ(1−β)]}.

After simplification, we get:

(κ2θ−α)[1−β(1−γ)]+αθ(1−β){1−β[1−γ(1−β)]}>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)2−A11cπcg−A12+cπcg(1−γ)=0 and substituting A₁₁ and A₁₂ by their expressions, we obtain:

(47)p2(cπcg)2+p1cπcg+p0=0

where

p0=αβθ{1−β(1−γ)[1−γ(1−β)]}>0,p2=γβ{(κ2θ−α)(1−γ)+αθ[1−γ(1−β)]},p1=α+β(1−γ){(κ2θ−α)(1−γ)+αθ[1−γ(1−β)]}−{θ(κ2+α)+αβ2γθ[1−γ(1−β)]}.

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

(48)cπcg=−p0+p2(cπcg)2p1≡f(cπcg).

We rewrite p₁, after some tedious calculus, as

(49)p1=−(κ2θ−α)[1−β(1−γ)]−αθ(1−β){1−β[1−γ(1−β)])}−p0−p2,

or alternatively simplify it as

(50)p1=−βp2−θ(α+κ2)−αθαβp0.

The conditions imposed on θ to ensure that p2>0, i.e. θ>αα(1+γβ1−γ)+κ2, and that p1<0, i.e. θ>ακ2+α(1+γ2β31−β(1−γ)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(1−β)+κ2, which is more restrictive than the condition (12), i.e. θ>αα+κ2. Thus, the stability condition is θ>αα(1−β)+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 (κ2θ−α)[1−β(1−γ)]+αθ(1−β){1−β[1−γ(1−β)])}>0. This implies that:

(51)θ>αα(1−β)[1−γβ21−β(1−γ)]+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=−p1−p12−4p2p02p2.

The other possible solution cπcg=−p1+p12−4p2p02p2 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(β−cπcg)+γβ(1−γ){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(1−β)+κ2θ−αθα+κ2θ−αp2+θαβθα+κ2θ−αp2, p0=−θα(1−β)+κ2θ−αθαβp0+θα+κ2θ−αθαβp0 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(κ2θ−α)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) :

A11≡θ(α+κ2)−αβ[θ(α+κ2)−α]=1β,

A12≡−αθ(1−β)θ(α+κ2)−α,

P1≡−αθβ[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

A11≡θ(α+κ2+αβ3)−ααβ2θ,A12≡−1β,P1≡−1β2.

Substituting the latter into (36)–(38) leads to p2=θαβ2>0, p1=α−κ2θ−αθ−θαβ3<0 and p0=αβθ>0, and hence

cπcg=θ(α+κ2)−α+θαβ3−[θ(α+κ2)−α+θαβ3]2−4θ2α2β32θαβ2,dπcg=αθθ(α+κ2)−α+β2θα(β−cπcg).

A.3 The effects of learning

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

∂p0∂γ=αβ2θ[(2−β)(1−γ)+γβ]>0,∂p1∂γ=−β∂p2∂γ−θ(α+κ2)−αθαβ∂p0∂γ<0,∂p2∂γ=β{(κ2θ−α)(1−2γ)+θα[1−2γ(1−β)]},=−1β∂p1∂γ−θ(α+κ2)−αθαβ∂p0∂γ.

Deriving cπcg with respect to γ yields:

∂cπcg∂γ=[−∂p1∂γ−1p12−4p2p0(p1∂p1∂γ−2p0∂p2∂γ−2p2∂p0∂γ)]p2−(−p1−p12−4p2p0)∂p2∂γ2p22,

which can be rewritten, using ∂p2∂γ=−1β∂p1∂γ−1βθ(α+κ2)−αθαβ∂p0∂γ and the definition of p₀, p₁ and p₂, p1=−βp2−θ(α+κ2)−αθαβp0 and cπcg=−p1−p12−4p2p02p2 and after fastidious arrangements of terms, as:^[19]

∂cπcg∂γ=1−θ(α+κ2)−αθαβcπcgβp2p12−4p2p0(p0∂p1∂γ−p1∂p0∂γ).

Using cπcg<θαβθ(α+κ2)−α, we obtain: 1−θαβθ(α+κ2)−αcπcg>1−θαβθ(α+κ2)−αθ(α+κ2)−αθαβ=0. To determine the sign of H≡p0∂p1∂γ−p1∂p0∂γ, 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{α−θ[κ2−αβ(1−β2)]}<0

if κ2−αβ(1−β2)<0; otherwise, we must impose:

θ>ακ2−αβ(1−β2)>ακ2.

Deriving H with respect to γ yields

∂H∂γ=p0∂2p1∂2γ−p1∂2p0∂2γ=2αθβ3(1−β){[1−γ(1−γβ)]{θ[κ2+α(1−β)]−α}+β3γ2αθ}>0.

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

∂cπcg∂γ<0.

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

∂dπcg∂γ=κ2θ−ακθ∂dxcg∂γ=−κ2θ−ασκθ∂drcg∂γ=−αβθ[Φ−γ[αβγθ+(1−γ)Θ]∂cπcg∂γ][Θ+θαγ2β2(β−cπcg)+γβ(1−γ)(θαβ−Θcπcg)]2,

where Θ≡θ(α+κ2)−α and Φ≡2αβγθ(β−cπcg)+(1−2γ)(αβθ−Θcπcg). Using the fact that β is very close to one and hence 2β−1>0, and the fact that cπcg<θαβθ(α+κ2)−α, which implies that β−cπcg>0 and θαβ−Θcπcg>0, we find that

Φ=βθαγ(2β−1)(β−cπcg)+βγ(θκ2−α)cπcg+β(1−γ){θαβ−[θ(α+κ2)−α]cπcg}>0.

It follows that

∂dπcg∂γ<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:

∂p0∂θ=αβ{1−β(1−γ)[1−γ(1−β)]}=p0θ>0,∂p1∂θ=−κ2[1−β(1−γ)2]−α(1−β){1−β[1−γ(1−β)])}−∂p0∂θ−∂p2∂θ<0,∂p2∂θ=γβ{κ2(1−γ)+α[1−γ(1−β)]}>0.

Deriving cπcg given by (52) with respect to θ, and using ∂p2∂θ=−1β∂p1∂θ−(α+κ2)αβ2∂p0∂θ, the definition of p₀, p₁ and p₂, p1=−βp2−θ(α+κ2)−αθαβp0 and cπcg=−p1−p12−4p2p02p2 yield:^[20]

∂cπcg∂θ=12p22(I∂p0∂θ+J∂p1∂θ),

where I=2p2p12−4p2p0{p2+(α+κ2)αβ2p0+(α+κ2)αβ2p1cπcg} and J=2p2βp12−4p2p0[(1−θ(α+κ2)−αθαβcπcg)p0]. Using these definitions and the expressions of ∂p0∂θ and ∂p1∂θ derived in the above, we obtain:

∂cπcg∂θ=(−∂p1∂θ−12p1∂p1∂θ−4p0∂p2∂θ−4p2∂p0∂θp12−4p2p0)p2−(−p1−p12−4p2p0)∂p2∂θ2p22=(−∂p1∂θ−12p1∂p1∂θ−4p0∂p2∂θ−4p2∂p0∂θp12−4p2p0)2p2−1p2cπcg∂p2∂θ.

Using p1=−βp2−θ(α+κ2)−αθαβp0, the definition of p₀, p₁ and p₂, we can show that p0∂p2∂θ−p2∂p0∂θ>0 and hence

∂cπcg∂θ=−12p222p2p12−4p2p0{[1−θ(α+κ2)−αθαβcπcg](p0∂p2∂θ−p2∂p0∂θ)+1θβcπcgp2∂p0∂θ}<0.

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

∂dπcg∂θ=−α2[1−βγ(1−γ)cπcg]+αθ{θαγ2β2+γβ(1−γ)[θ(α+κ2)−α]}∂cπcg∂θ{θ(α+κ2)−α+θαγ2β2(β−cπcg)+γβ(1−γ){θαβ−[θ(α+κ2)−α]cπcg}}2<0.

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

∂cxcg∂θ=−1σ∂crcg∂θ=κα(κ2θ−α)2(β−cπcg)+κθκ2θ−α∂cπcg∂θ,

∂dxcg∂θ=−1σ∂drcg∂θ=ακ(κ2θ−α)2(1−dπcg)+κθκ2θ−α∂dπcg∂θ.

To ensure that ∂cxcg∂θ=−1σ∂crcg∂θ>0, and ∂dxcg∂θ=−1σ∂drcg∂θ > 0, we must have ∂cπcg∂θ>−α(β−cπcg)θ(κ2θ−α) and ∂dπcg∂θ>−α(1−dπcg)θ(κ2θ−α), 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

Learning, robust monetary policy and the merit of precaution

Abstract

Acknowledgement

A Appendix

A.1 The equilibrium solution of inflation under learning

A.2 The single stable solution

A.3 The effects of learning

A.4 Effects of robustness

References

Article note

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