Testing for Indeterminacy: An Application to U.S. Monetary Policy. Technical Appendix

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1 Testing for Indeterminacy: An Application to U.S. Monetary Policy Technical Appendix Thomas A. Lubik Department of Economics Johns Hopkins University Frank Schorfheide Department of Economics University of Pennsylvania June 2003 Not intended for Publication Mergenthaler Hall, 3400 N. Charles Street, Baltimore, MD 228. Tel.: 40) Fax: 40) McNeil Building, 378 Locust Walk, Philadelphia, PA 904. Tel.: 25) Fax: 25)

2 Solving the New Keynesian Model The simplified version of the DSGE model presented in Sections 5 of the paper can be expressed in terms of ξ t = ξt x, ξt π as ξ t = ψ ξ t + 0 ɛ t + ψ η t. ) } 0 {{ } } {{ } } {{ } } {{ } Γ 0 Γ Ψ Π Premultiply the system by Γ 0 = / 0 / 2) to obtain ξ t = + ψ ) Γ + Ψ ξ t 3) ɛ t + + ψ ) Π η t. The eigenvalues of Γ are a solution to the equation + /) λ 0 = det ψ /) 4) / / λ which can be rewritten as 0 = λ 2 λ The solution of this quadratic equation is λ, λ 2 = + + ) l + + ) + + ψ). 5) ) l 2 ψ). 6) The eigenvectors have to satisfy the relationship + / ψ /) x = l x + l 2 x. 7) / /

3 2 Thus, x solves /)x + / = l + l2, 8) which implies that x = l + l 2 ). 9) Thus, we obtain the following Jordan decomposition of Γ Γ = JΛJ, 0) where J = Λ = J = l + l 2 ) l l 2 ) ) l l 2 0 2) 0 l + l 2 2l 2 + l + l 2 l + l 2. 3) Let w t = J ξ t. We can now write the LRE system in terms of the transformed variables w t = Λw t + J Ψ ɛ t + J Π η t. 4) Determinacy If both λ and λ 2 are unstable then the unique) solution is η t = Π Ψ ɛ t 5) = + ψ ) ɛ t / / ψ) = + / / + ψ / + / ψ = ɛ t. + ψ ɛ t

4 3 The law of motion for output, inflation, and interest rates is given by x t ψ ɛ π t = R,t ɛ + ψ g,t. 6) ψ ψ R t ɛ z,t Indeterminacy In order to solve the system under indeterminacy we have to calculate J Ψ = + l + l 2. 7) 2l 2 l + l 2 The second) row of this vector corresponding to the unstable eigenvalue is Ψ J x = 2l 2 λ 2 ) since + / + / = 2l and λ 2 = l + l 2. Moreover, J Π = + l + l 2 2l 2 l + l 2, 8) + ψ ) 9) and the second) row of this matrix corresponding to the unstable eigenvalue is Ψ J x = = 2 / / + l l 2 2l 2 ψ /) + / l + l 2 λ 2. 2l 2 λ 2 ψ Thus, the stability condition can be expressed as 20) ɛ R,t + ɛ g,t + λ 2 )ɛ z,t λ 2 η y t + λ 2 ψ η π t = 0. 2) Note that in the paper we are using Π J x = λ 2 λ 2 ψ) and Ψ J x = λ 2 ),

5 4 which corresponds to a slightly different normalization of the matrix of eigenvectors J, than the one used in above derivations. The singular value decomposition of Π J x yields U. = D = d = λ 2 ) 2 + λ 2 ψ) 2 V. = λ 2 λ 2 ψ)/d V.2 = λ 2 ψ) λ 2 /d. Finding the Boundary of the Determinacy Region To analyze the standard New Keynesian model without restrictions on the autocorrelation parameters ρ R, ρ g, and ρ z it is useful to write it in terms of ξ x t, ξ π t, and R t : ξ x t ξ π t = + + ρ R)ψ 2 + ρ R)ψ ρ R 0 ξ x t ξ π t 22) R t ρ R )ψ 2 ρ R )ψ ρ R Γ θ) R t + Ψ ɛ R,t g t + Πη t z t The stability properties of the system depend on the eigenvalues of Γ θ). At the boundary of the determinacy region it has to be the case that det Γ θ) I 3 3 ) = 0 23) since the matrix Γ θ) has at least one unit eigenvalue. Tedious but straightforward algebra shows that 23) implies that ψ = ψ 2 ) 24)

6 5 which is the formula that we used to define the function gθ) in Section 4. Note that these calculations do not formally prove that 24) is the boundary of the determinacy region as they do not reveal the size of the other eigenvalues of Γ θ). To obtain a formal proof techniques as in Bullard and Mitra 2002) or Lubik and Marzo 200) have to be applied. Habit Formation Suppose that the period utility function is of the form ν t = C t/c γ t / / 25) where C γ t is the habit formation reference consumption level. The resulting firstorder condition is of the form λ t = Ct Ct Now define We then have u t = Ct C γ t C γ t which leads to the Euler equation In the steady state ) / Ct+ ) / γie t C γ t ) / and u t = u t γie t u t+.. 26) λ t = u t C t, 27) u t = IE t u t R t /π t+. 28) C t C t+ u = C γ) /), u = γ)u, and λ = u /C Log-linearization gives us ũ t C t = IE t C t+ IE t C t+ + R t IE t π t+ ), 29)

7 6 and the two definitional equations γ)ũ t = ũ t γie t ũ t+ 30) ũ t = ) C t γ ) C t. 3) Thus, γ)ũ t = ) γ C t γ ) IE t C t+ + γ 2 ) C t 32) ) C t. Combining 29) and 32) these three equations leads to + + γ2 + γ γ = γ ) γ x t + ) + x t + γ + γ) ) γ γ γ )IE t+2 x t+2 R t IE t π t+ ) + g t, IE t x t+ 33) where C t is replace by x t and g t captures the residual effect of the discrepancy between output and consumption on the Euler equation.