The Non-Stochastic Steady State

Σχετικά έγγραφα
α β

1. For each of the following power series, find the interval of convergence and the radius of convergence:


Introduction of Numerical Analysis #03 TAGAMI, Daisuke (IMI, Kyushu University)

Solve the difference equation

Ψηφιακή Επεξεργασία Εικόνας

Degenerate Perturbation Theory

Partial Differential Equations in Biology The boundary element method. March 26, 2013

On Generating Relations of Some Triple. Hypergeometric Functions

FREE VIBRATION OF A SINGLE-DEGREE-OF-FREEDOM SYSTEM Revision B

SUPERPOSITION, MEASUREMENT, NORMALIZATION, EXPECTATION VALUES. Reading: QM course packet Ch 5 up to 5.6

Homework for 1/27 Due 2/5

Last Lecture. Biostatistics Statistical Inference Lecture 19 Likelihood Ratio Test. Example of Hypothesis Testing.

The Heisenberg Uncertainty Principle

Presentation of complex number in Cartesian and polar coordinate system

Bessel function for complex variable

IIT JEE (2013) (Trigonomtery 1) Solutions

Inertial Navigation Mechanization and Error Equations

MATH 38061/MATH48061/MATH68061: MULTIVARIATE STATISTICS Solutions to Problems on Matrix Algebra

4.6 Autoregressive Moving Average Model ARMA(1,1)

n r f ( n-r ) () x g () r () x (1.1) = Σ g() x = Σ n f < -n+ r> g () r -n + r dx r dx n + ( -n,m) dx -n n+1 1 -n -1 + ( -n,n+1)

Phys460.nb Solution for the t-dependent Schrodinger s equation How did we find the solution? (not required)

Matrices and Determinants

SCHOOL OF MATHEMATICAL SCIENCES G11LMA Linear Mathematics Examination Solutions

Solutions: Homework 3

HOMEWORK 4 = G. In order to plot the stress versus the stretch we define a normalized stretch:

Concrete Mathematics Exercises from 30 September 2016

6.1. Dirac Equation. Hamiltonian. Dirac Eq.

p n r

EE512: Error Control Coding

Diane Hu LDA for Audio Music April 12, 2010

Notes on the Open Economy

A study on generalized absolute summability factors for a triangular matrix

Math221: HW# 1 solutions

Second Order RLC Filters

Areas and Lengths in Polar Coordinates

Srednicki Chapter 55

Lecture 17: Minimum Variance Unbiased (MVUB) Estimators

1. Matrix Algebra and Linear Economic Models

The Simply Typed Lambda Calculus

Binet Type Formula For The Sequence of Tetranacci Numbers by Alternate Methods

derivation of the Laplacian from rectangular to spherical coordinates

w o = R 1 p. (1) R = p =. = 1

Chapter 6: Systems of Linear Differential. be continuous functions on the interval

Other Test Constructions: Likelihood Ratio & Bayes Tests

Areas and Lengths in Polar Coordinates

Lecture 2: Dirac notation and a review of linear algebra Read Sakurai chapter 1, Baym chatper 3

Outline. M/M/1 Queue (infinite buffer) M/M/1/N (finite buffer) Networks of M/M/1 Queues M/G/1 Priority Queue

Biorthogonal Wavelets and Filter Banks via PFFS. Multiresolution Analysis (MRA) subspaces V j, and wavelet subspaces W j. f X n f, τ n φ τ n φ.

Approximation of distance between locations on earth given by latitude and longitude

Exercises 10. Find a fundamental matrix of the given system of equations. Also find the fundamental matrix Φ(t) satisfying Φ(0) = I. 1.

EN40: Dynamics and Vibrations

Homework 3 Solutions

3.4 SUM AND DIFFERENCE FORMULAS. NOTE: cos(α+β) cos α + cos β cos(α-β) cos α -cos β

C.S. 430 Assignment 6, Sample Solutions

LAD Estimation for Time Series Models With Finite and Infinite Variance

Homework 4.1 Solutions Math 5110/6830

Example Sheet 3 Solutions

Jesse Maassen and Mark Lundstrom Purdue University November 25, 2013

INTEGRATION OF THE NORMAL DISTRIBUTION CURVE

Every set of first-order formulas is equivalent to an independent set

Numerical Analysis FMN011

On Certain Subclass of λ-bazilevič Functions of Type α + iµ

Partial Trace and Partial Transpose

) 2. δ δ. β β. β β β β. r k k. tll. m n Λ + +

CHAPTER 103 EVEN AND ODD FUNCTIONS AND HALF-RANGE FOURIER SERIES

Outline. Detection Theory. Background. Background (Cont.)

CHAPTER 48 APPLICATIONS OF MATRICES AND DETERMINANTS

Solutions to Exercise Sheet 5

SOLUTIONS TO MATH38181 EXTREME VALUES AND FINANCIAL RISK EXAM

Απόκριση σε Μοναδιαία Ωστική Δύναμη (Unit Impulse) Απόκριση σε Δυνάμεις Αυθαίρετα Μεταβαλλόμενες με το Χρόνο. Απόστολος Σ.

CHAPTER 25 SOLVING EQUATIONS BY ITERATIVE METHODS

MATH423 String Theory Solutions 4. = 0 τ = f(s). (1) dτ ds = dxµ dτ f (s) (2) dτ 2 [f (s)] 2 + dxµ. dτ f (s) (3)

Tridiagonal matrices. Gérard MEURANT. October, 2008

Statistical Inference I Locally most powerful tests

DESIGN OF MACHINERY SOLUTION MANUAL h in h 4 0.

The Neutrix Product of the Distributions r. x λ

Μηχανική Μάθηση Hypothesis Testing

Chapter 6: Systems of Linear Differential. be continuous functions on the interval

Lecture 3: Asymptotic Normality of M-estimators

forms This gives Remark 1. How to remember the above formulas: Substituting these into the equation we obtain with

B.A. (PROGRAMME) 1 YEAR

= λ 1 1 e. = λ 1 =12. has the properties e 1. e 3,V(Y

ΕΛΛΗΝΙΚΗ ΔΗΜΟΚΡΑΤΙΑ ΠΑΝΕΠΙΣΤΗΜΙΟ ΚΡΗΤΗΣ. Ψηφιακή Οικονομία. Διάλεξη 7η: Consumer Behavior Mαρίνα Μπιτσάκη Τμήμα Επιστήμης Υπολογιστών

2. THEORY OF EQUATIONS. PREVIOUS EAMCET Bits.

Problem Set 9 Solutions. θ + 1. θ 2 + cotθ ( ) sinθ e iφ is an eigenfunction of the ˆ L 2 operator. / θ 2. φ 2. sin 2 θ φ 2. ( ) = e iφ. = e iφ cosθ.

ST5224: Advanced Statistical Theory II

Συστήματα Διαχείρισης Βάσεων Δεδομένων

Finite Field Problems: Solutions

Lecture 34 Bootstrap confidence intervals

Derivation of Optical-Bloch Equations

Gauss Radau formulae for Jacobi and Laguerre weight functions

CRASH COURSE IN PRECALCULUS

F19MC2 Solutions 9 Complex Analysis

2 Composition. Invertible Mappings

Solution Series 9. i=1 x i and i=1 x i.

Reminders: linear functions

Econ 2110: Fall 2008 Suggested Solutions to Problem Set 8 questions or comments to Dan Fetter 1

Lecture 26: Circular domains

Supplemental Material: Scaling Up Sparse Support Vector Machines by Simultaneous Feature and Sample Reduction

Transcript:

July 25, 24 Techical Appedix: The Dyamic System I this appedix I preset the full derivatio of the approximated dyamic system, its properties, ad the procedure for computig secod momets. I rely heavily o Burside (1997) ad follow his otatio. 1 A The No-Stochastic Steady State I the steady state two relatios determie the vacacy rate ad the uemploymet rate: (i) All labor force variables grow at the same gross rate G L. Start from the uemploymet dyamics relatio: 4U = δn + 4L M Dividig throughout by N ad usig the steady state relatioship: This becomes: 4L L = 4U U = GL 1 Which ca also be writte as: (G L 1) U N = δ +(GL 1)(1 + U N ) M N (G L 1) + δ = m = Qv Isertig the matchig fuctio: ³ u σ ³ v 1 σ (G L 1) + δ = (1) (ii) Vacacy creatio is give by the combiatio of the relevat F.O.C.: φv +(1 φ)qv Θ (φ +(1 φ)q)( N ) γ G X β = Q [1 (1 δ)g X β] π (2) 1 I would like to express my gratitude to Craig Burside for his kid advice ad for the provisio of his programs ad extesive otes. 1

Profits are give by: ½ ¾ φv +(1 φ)qv 1 α π =[1 η] (1 α) Θ( ) γ+1 N 1 ηpλ (3) Notig that: This becomes: π = Thewageshareisgiveby: s = η (1 α)+θ 1 λ = G X β [1 (1 δ)g X β] π φv +(1 φ)qv [1 η] (1 α) Θ( N 1+ηP G X β [1 (1 δ)g X β] ) γ+1 h 1 α 1 io (1 α) φv +(1 φ)qv ( ) γ+1 + P N G X β [1 (1 δ)g X β] π (4) (5) B The Log-Liear Approximatio For each variable Y deote b Y t = Y t Y Y l Y t l Y where Y is the steady state value. B.1 The Itratemporal Coditio This coditio is give by: Defie: Θ (φ +(1 φ)q t.t+1 )( φv t +(1 φ)q t,t+1 V t N t ) γ = Q t,t+1 λ t (6) eq t,t+1 = φ +(1 φ)q t,t+1 The: Θ eq t,t+1 ( e Q t,t+1 V t N t ) γ = Q t,t+1 λ t (7) 2

Approximatig the itratemporal coditio, we get: b e Q t,t+1 + γ( b e Qt,t+1 + bv t b t )= b Q t,t+1 + b λ t (8) Usig the defiitio of Q ad u =1 : Approximatig: Defie: The: σ vt Q t,t+1 = (9) 1 t bq t,t+1 = σ bv t + 1 b t b e Q t,t+1 = Ω (1 φ)q Q φ +(1 φ) Q b t,t+1 (1 φ)q φ +(1 φ) Q (1) b e Q t,t+1 = Ω bq t,t+1 Isertig ito the approximated coditio: ΩQ b t,t+1 + γ(ωq b t,t+1 + bv t b t ) = Q b t,t+1 + λ b t bq t,t+1 (Ω + γω 1) + γbv t = γb t + λ b t σ bv t + 1 b t (Ω + γω 1) + γbv t = γb t + λ b t [ σ(ω + γω 1) + γ] bv t = γ + σ (Ω + γω 1) b t + λ 1 b t (11) Whe φ =the Ω =1ad the expressio becomes: (1 σ)γbv t =(γ + σγ 1 )b t + λ b t Whe φ =1the Ω =ad the expressio becomes: 3

B.2 The Flow Equatio [σ + γ] bv t = γ σ 1 Approximatig the dyamic equatio for employmet: b t + b λ t t+1 G L t+1 =(1 δ t,t+1 ) t + Q t,t+1 v t (12) We get (assumig G bl t+1 =): Usig (1): We get: G L b t+1 =(1 δ)b t δ b δ t,t+1 + Qv( bq t,t+1 + bv t ) (13) G L b t+1 =(1 δ)b t δ b δ t,t+1 + Qv G L b t+1 = Qv(1 σ)bv t δ b δ t,t+1 B.3 The Itertemporal Coditio This coditio is give by: σ bv t + 1 b t + bv t (14) (1 δ) Qvσ b t (15) 1 λ t G X t+1 " ³ ³ (1 α)+θ = E t β t+1 1 (1 α) γ+1 ( φv t+1+(1 φ)q t+1,t+2 v t+1 t+1 ) (1 η) ηp t+1,t+2 λ t+1 (16) Deote: +E t (1 δ t+1,t+2 )β t+1 λ t+1 π t+1 = "Ã ³ +Θ 1 (1 α) (1 α) ( v t+1 e Q t+1,t+2 t+1 ) γ+1! (1 η) ηp t+1,t+2 λ t+1 G X t+1 F t+1/n t+1 F t /N t 4

So: λ t G X t+1 Approximatig this coditio we get: = E t β t+1 π t+1 + E t (1 δ t+1,t+2 )β t+1 λ t+1 λ G (b λ X t bg X t+1) =E t βπ( β b t+1 + bπ t+1 )+(1 δ)βλ( β b t+1 + λ b t+1 ) δβλ b o δ t+1,t+2 Approximatig π t+1 : (17) πbπ t+1 = ηpλ( P b t+1,t+2 + λ b t+1 ) (18) (1 α) +Θ 1 (1 η)( v eq )γ+1 (1 + γ)(bv t+1 + ΩQ b t+1,t+2 b t+1 ) The first term o the RHS may be derived from: Usig (1) oe period forward: Isertig ito (18): πbπ t+1 = ηpλ +Θ 1 P t+1,t+2 = Q t+1,t+2v t+1 U t+1 = Q t+1,t+2v t+1 1 t+1 bp t+1,t+2 = b Q t+1,t+2 + bv t+1 + bp t+1,t+2 = 1 b t+1 (19) 1 (1 σ)b t+1 +(1 σ)bv t+1 (2) 1 (1 σ)b t+1 +(1 σ)bv t+1 + λ b t+1 (1 α) (1 η)( v eq )γ+1 (1 + γ)(bv t+1 + Ω Thus isertig ito (17): 5 σ bv t+1 + 1 b t+1 b t+1 )

λ G (b λ X t bg X t+1) = E t [βπ +(1 δ)βλ] β b t+1 (21) +(1 δ)βλλ b t+1 δβλ b δ t+1,t+2 h ηpλ (1 σ)b 1 t+1 +(1 σ)bv t+1 + b i λ t+1 ³ +β +Θ 1 (1 α) (1 η)( v Q e )γ+1 (1 + γ) bvt+1 + Ω σ bv t+1 + bt+1 1 b t+1 Re-arragig: = " βηpλ 1 (1 σ)+βθ 1 (1 α) (1 η)( v eq )γ+1 (1 + γ) 1+Ωσ E t b t+1 1 +[(1 δ)βλ βηpλ] λ b t+1 λ λ G b X t (22) " (1 α) βηpλ(1 σ) βθ(1 η) 1 ( v eq )γ+1 (1 + γ)[1 σω] E t bv t+1 λ G X E t b G X t+1 [βπ +(1 δ)βλ] E t b βt+1 +δβλe t b δt+1,t+2 C The System i Matrix Form The dyamic system is solved by the regular methods of differece equatios. I briefly outlie here the mai steps; for a extesive discussio, albeit i a differet cotext, see Kig, Plosser ad Rebelo (1988) ad Burside (1997). The geeral problem is give by: xt M cc u t = M cs + M λ ce z t (23) t MssE xt+1 t + Mss λ 1 xt = M t+1 λ t sce t u t+1 + Mscu 1 t + MseE t z t+1 + Msez 1 t. (24) 6

where x,λ are the edogeous state ad co-state variables, u are the cotrol variables, ad z is a vector of exogeous variables. Combiig we get or (Mss MscM cc 1 xt+1 M cs )E t +(Mss λ 1 M 1 t+1 scmcc 1 M cs ) (Mse + MscM cc 1 M ce )E t z t+1 +(Mse 1 + MscM 1 cc 1 M ce )z t M sse xt+1 t + M ss 1 xt = M λ t+1 λ t see t z t+1 + M sez 1 t xt λ t = Solvig for the edogeous variables: xt+1 E t = ( λ M t+1 ss) 1 1 xt M ss +( λ M t ss) 1 M se E t z t+1 +( M ss) 1 1 M se z t xt = W + RE λ t z t+1 + Qz t. (25) t I this paper the followig defiitios apply (LHS variables use otatio above while RHS variables are the actual variables here): x t = t λ t = λ t u t = v t z t = bg X t bβ t b δt,t+1 The other variables i the model - u, s, Q ad P - shall be cotaied i the vector f t ad are expressible as a liear combiatio of the cotrol, state ad exogeous variables: Here: f t = F c u t + F x x t + F z z t f t = bu t bs t bq t,t+1 bp t,t+1 \ Q t,t+1 V t 7

So: f t = F v bv t + F b t + F z z t The relevat equatios are: bu t = 1 b t (26) bq t,t+1 = σ bv t + 1 b t (27) Note too that hirig is give by Q t,t+1 V t ad that i log deviatios from steady state this is: Q\ t,t+1 V t = bq t,t+1 + bv t = σ bv t + 1 b t [1 σ] bv t σ 1 b t + bv t Ad give: bp t,t+1 = 1 (1 σ)b t +(1 σ)bv t (28) s t = η (1 α)+θ 1 1 α ( φv t +(1 φ)q t,t+1 v t ) γ+1 + P t,t+1 λ t à t = η (1 α)+θ 1 1 α! γ+1 vt Qt,t+1 e + P t,t+1λ t t We have: Isertig: bs t = η s Θ 1 (1 α) ( v eq )γ+1 (1 + γ)(bv t + ΩQ b t,t+1 b t ) + η s Pλ ³ bpt,t+1 + b λ t 8

bp t,t+1 = 1 (1 σ)b t +(1 σ)bv t bλ t = γ + σ (Ω + γω 1) b t +[ σ(ω + γω 1) + γ] bv t 1 bq t,t+1 = σ bv t + 1 b t We get: bs t = η s Θ (1 α) 1 ( v eq )γ+1 (1 + γ)(bv t + Ω σ bv t + 1 b t b t ) + η s Pλ 1 (1 σ)b t +(1 σ)bv t γ + σ (Ω + γω 1) b t +[ σ(ω + γω 1) + γ] bv t 1 " η = s Θ (1 α) 1 ( v eq )γ+1 (1 + γ)(1 σω)+ η s Pλ(1 σ)+η Pλ[ σ(ω + γω 1) + γ] bv t s " ³ η + Θ 1 (1 α) ( v eq s )γ+1 (1 + γ)(1 + σω ) 1 + η Pλ (1 σ) η Pλ b γ + σ t (Ω + γω 1) s 1 s 1 Thematricesaregiveby: 9

F v = F = F e = " ³ η Θ 1 (1 α) ( v Q e s )γ+1 (1 + γ)(1 σω) + η s Pλ(1 σ)+ η Pλ[ σ(ω + γω 1) + γ] s σ (1 σ) [1 σ] " ³ 1 η Θ 1 (1 α) ( v eq s )γ+1 (1 + γ)(1 + σω ) 1 + η Pλ (1 σ) η Pλ γ + σ (Ω + γω 1) s 1 s 1 σ 1 (1 σ) 1 σ 1 1

M cc = [ σ(ω + γω 1) + γ] M cs = γ + σ (Ω + γω 1) 1 1 M ce = [] G L M ss = M 1 ss = M sc = M 1 sc = M se = ³ +βθ βηpλ (1 σ) 1 1 (1 α) (1 η)( v Q e (1 + γ) 1+Ωσ 1 (1 δ) Qvσ 1 M 1 se = [] " ³ βθ(1 η) λ G X βηpλ(1 σ) 1 (1 α) Qv(1 σ) δ λ [βπ +(1 δ)βλ] δβλ G X )γ+1 ( v eq )γ+1 (1 + γ)[1 σω] [(1 δ)βλ βηpλ] The implied matrices are: M ss = (Mss MscM cc 1 M cs ) M ss 1 = (Mss 1 MscM 1 cc 1 M cs ) M se = (Mse + MscM cc 1 M ce ) M se 1 = (Mse 1 + MscM 1 cc 1 M ce ) W = ( M ss) 1 M ss 1 R = ( M ss) 1 M se Q = ( M ss) 1 M se 1 11

D The Dyamic System D.1 Derivatio of the Solutio Thesystem(25)isgiveby: bt+1 bλ t+1 bt = W bλ t bg X t+1 +R bβ t+1 +Q b δt+1,t+2 bg X t bβ t b δt,t+1 (29) Defie W = P ΛP 1,whereΛ is a diagoal matrix with the eigevalues of W o its diagoal, ad P is a matrix whose colums are liearly idepedet eigevectors of W. The Λ is costructed with the eigevalues i icreasig order of modulus ad is decomposed ito Λ1 Λ = Λ 2 with all elemets of Λ 1 less tha 1, ad all values of Λ 2 greater tha 1. As a result the equatio for b t+1 should be solved backward, while the equatio for λ b t+1 should be solved forward. Partitio the matrices W, R, Q, P ad P 1 as follows: W11 W W = 12 Rb Qb R = Q = W 21 W 22 R bλ Q bλ ad P = P11 P 12 P 21 P 22 P P 1 11 P = 12 P 21 P 22. Sice W = P ΛP 1 we have W11 W 12 P11 Λ = 1 P 11 + P 12 Λ 2 P 21 P 11 Λ 1 P 12 + P 12 Λ 2 P 22 W 21 W 22 P 21 Λ 1 P 11 + P 22 Λ 2 P 21 P 21 Λ 1 P 12 + P 22 Λ 2 P 22. 12

Solvig the first equatio backward: Eb t+1 = Λ 1 b t +(P 11 R b + P 12 R bλ )E t z t+1 +(P 11 Q b + P 12 Q bλ )z t Solvig the other equatio forward: E tλt+1 b = Λ 2λt b +(P 21 R b + P 22 R bλ )E t z t+1 +(P 21 Q b + P 22 Q bλ )z t bλ t = Λ 1 2 E tλt+1 b Λ 1 2 (P 21 R b + P 22 R bλ )E t z t+1 Λ 1 2 (P 21 Q b + P 22 Q bλ )z t X = (P 21 R b + P 22 R bλ )E t z t+1+j +(P 21 Q b + P 22 Q bλ )E t z t+j. j= Λ (j+1) 2 Recall the AR(1) represetatio for z t : z t+1 = Πz t + Σ t+1 (3) The E t z t+j = Π j z t. Goig back to the solutio for b λ t give above bλ t = = = X j= X j= = Ψz t. " X j= Λ (j+1) 2 (P 21 R b + P 22 R bλ )E t z t+1+j +(P 21 Q b + P 22 Q bλ )E t z t+j Λ (j+1) 2 (Φ E t z t+1+j + Φ 1 E t z t+j ) Λ (j+1) 2 (Φ Π + Φ 1 )Π j z t A explicit formula for the rows of Ψ is obtaied by exploitig the diagoality of Λ 2. Defiig Λ 2i as the ith diagoal elemet of Λ 2,adΦ ji as the ith row of Φ j, j =, 1, it follows that the ith row of Ψ, deotedψ i,isgiveby " X Ψ i = Λ (j+1) 2i (Φ i Π + Φ 1i )Π j j= = Λ 1 2i (Φ iπ + Φ 1i ) X j= Λ j 2i Πj = Λ 1 2i (Φ iπ + Φ 1i )(I e Λ 1 2i Π) 1 13

The, the solutio for b t+1 is just b t+1 = (P 11 Λ 1 P11 1 )b t +(P 11 Λ 1 P 12 + P 12 Λ 2 P 22 )(P 22 ) 1 Ψz t + R b Πz t + Q b z t = (P 11 Λ 1 P11 1 )b t + (P 11 Λ 1 P 12 + P 12 Λ 2 P 22 )(P 22 ) 1 Ψ + R b Π + Q b zt = Υ bb b t + Υ bz z t (31) The solutio for b λ t is, ad the solutio for the cotrol is: bv t = Mcc 1 M cs = Υ bvb b t + Υ bvz z t. bλ t = (P 22 ) 1 P 21 b t +(P 22 ) 1 Ψz t = Υ bλb b t + Υ bλz z t (32) I (P 22 ) 1 P 21 b t + D.2 No Stochastic Dyamics M 1 cc M cs (P 22 ) 1 Ψ + M 1 cc M ce z t I order to characterize the properties of the dyamic path of b i the o-stochastic case, set all exogeous variables equal to their steady state values i.e.: o bg X t+1 o t= bβt o t= = for all t (34) bδt ad so the dyamics are give by: t= b t+1 = Υ bb b t (35) All other variables of iterest ca be represeted as liear fuctios of b t alog the uique path: bu t = 1 b t (36) bv t = Υ bvb b t (37) The liear coefficiets i these equatios are the elasticities of the origial variables with respect to deviatios of the employmet rate from its statioary value (as the hatted variables represet log deviatios). Thus they quatify the co-movemet of the differet variables alog the uique dyamic path. 14 (33)

D.3 Secod Momets The secod momets were computed usig the followig relatios: Defie vectors of state variables h t ad their iovatios t : bt h t = z t = t Σt Thus: (38) where h t+1 = Gh t + t+1 (39) G = 1 Υ bb Π. (4) The co-state ad cotrol variables are fuctios of this state vector: bλ t = Υ bλb Υ bλz ht (41) bv t = Υ bvb Υ bvz ht Therefore it is possible to represet all the variables, except the state ad exogeous variables, i the form: bλ t bv t = Hh t (42) f t The secod momets are therefore: E(h t h t) = Γ (43) E(h t h t i) = E (G i h t i + t + + G i 1 t i+1 )h t i = G i Γ E h t bλ t bv t f t = E(h t h th ) (44) = Γ H. We use the coefficiet matrix Π ad the variace-covariace matrix of Σ, estimated by the reduced-form VAR, ad the solutio of the model ( 1 ad the Υs)giveitermsof the model s parameters to compute Γ,G ad H. 15