esercitazione SVAR 2 - economia.unipv.iteconomia.unipv.it/pagp/pagine_personali/dean/esercitazione...

Structural-VAR II

Dean Fantazzini

Dipartimento di Economia Politica e Metodi Quantitativi

University of Pavia

Overview of the Lecture

1st Structural VARs with Combinations of I(1) and I(0) Data: The

Blanchard and Quah (1989) model

Advanced Time Series Econometrics

Dean Fantazzini 10 − 15 July 2006 2




2nd Identifying the SVAR Using Long-Run Restrictions


Dean Fantazzini 10 − 15 July 2006 2-a





3rd Estimating the SVAR in the Presence of Long-Run Restrictions


Dean Fantazzini 10 − 15 July 2006 2-b






4rd An empirical example with the BQ model (I)


Dean Fantazzini 10 − 15 July 2006 2-c







5th An empirical example: Impulse responses with the BQ model (II)


Dean Fantazzini 10 − 15 July 2006 2-d







5th An empirical example: Impulse responses with the BQ model (II)

6th A review empirical example: Cholesky factorization as structural

factorization


Dean Fantazzini 10 − 15 July 2006 2-e

Structural VARs with Combinations of I(1) and I(0) Data:The Blanchard and Quah (1989) model

Consider two observed series y1t and y2t such that y1t is I(1) and y2t is

I(0). This is the case, for example, in the famous analysis in Blanchard and

Quah (1989):

y1 = log of real GDP

y2 = unemployment rate

Define

yt = (∆y1t, y2t)′ =⇒ yt ∼ I(0)

Suppose yt has the structural representations

Byt = γ0 + Γ1yt−1 + εt

yt = µ + Θ(L)εt

Θ(L) = Ψ(L)B−1

E[εtε′

t] = D (= diagonal)




The reduced form representations are:

yt = B−1

γ0 + B−1

Γ1yt−1 + B−1

εt = a0 + A1yt−1 + ut

= µ + Ψ(L)ut,

Ψ(L) = (I2 − A1L)−1,

E[utu′

t] = B−1

E[εtε′

t]B−1′

= B−1

DB−1′

= Ω

Note: BQ loosely interpret ε1t as a permanent (supply) shock since it is the

innovation to the I(1) real output series yt and interpret ε2t as a transitory

(demand) shock since it is the innovation to the I(0) unemployment series.

The structural IRFs are given by

∂∆y1t+s

∂ε1t

= θ(s)11 ,

∂∆y1t+s

∂ε2t

= θ(s)12

∂y2t+s

∂ε1t

= θ(s)21 ,

∂y2t+s

∂ε2t

= θ(s)22




−→ Since y1t (output) is I(1), the long-run impacts on the level of y1 of

shocks to ε1 and ε1 are

lims→∞

∂y1t+s

∂ε1t

= θ11(1) =

∞∑

s=0

θ(s)11

lims→∞

∂y1t+s

∂ε2t

= θ12(1) =

∞∑

s=0

θ(s)12

−→ Since y2t (unemployment) is I(0), the long-run impacts on the level of

y2 of shocks to ε1 and ε2 are zero:

lims→∞

∂y2t+s

∂ε1t

= lims→∞

θ(s)2j = 0

For y2,

θ21(1) =

∞∑

s=0

θ(s)21

θ22(1) =

∞∑

s=0

θ(s)22

represent the cumulative impact of shocks to ε1 and ε2 on the level of y2.



Identifying the SVAR Using Long-Run Restrictions

BQ (1989) achieve identification of the SVAR/SMA by assuming

• Transitory (demand) shocks (shocks to ε2) have no long-run impact on

the level of output or unemployment.

• They allow permanent (supply) shocks (shocks to ε1) to have a

long-run impact on the level of output but not on the level of

unemployment.

The restriction that shocks to ε2 have no long-run impact on the level of y1

implies that

θ12(1) =∞

∑

s=0

θ(s)12 = 0.




The restriction that shocks to ε1 and ε2 have no long-run effect on the

level of y2 is

lims→∞

∂y2t+s

∂εjt

= lims→∞

θ(s)2j = 0

which follows automatically since y2 ∼ I(0).

BQ (1989) assumptions imply that the long-run impact matrix Θ(1) is

lower triangular

Θ(1) =

θ11(1) 0

θ21(1) θ22(1)

Claim: The lower triangularity of Θ(1) can be used to identify B.




To see why, consider the long-run covariance matrix of yt defined from the

Wold representation

Λ = Ψ(1)ΩΨ(1)′ = (I2 − A1)−1Ω(I2 − A1)−1′

Since

Ω = B−1

DB−1′

Θ(1) = Ψ(1)B−1 = (I2 − A1)−1B

−1,

Λ may be re-expressed as,

Λ = (I2 − A1)−1B

−1DB

−1′(I2 − A1)−1′ = Θ(1)DΘ(1)′

In order to identify B, BQ (1989) make the additional assumption

D = I2

so that the structural shocks ε1t and ε2t have unit variances. Then

Λ = Θ(1)Θ(1)′.




Since Θ(1) is lower triangular, Λ can be obtained uniquely using the

Choleski factorization; that is, Θ(1) can be computed as the lower

triangular Choleski factor of Λ. The Choleski factorization of Λ is

Λ = PP′ = Θ(1)Θ(1)′

=⇒ Θ(1) = P

Given that Θ(1) = P can be computed directly from Λ, B can then be

computed using

P = Θ(1) = Ψ(1)B−1 = (I2 − A1)−1B

−1

=⇒ B = [(I2 − A1)P]−1

and the SVAR model is exactly identified.



Estimating the SVAR in the Presence of Long-Run Restrictions

The estimation of B and Θ(L) using the BQ identification scheme can be

accomplished with the following steps:

• Estimate the reduced form VAR by OLS equation by equation:

yt = a0 + A1yt−1 + ut

Ω =1

T

T∑

t=1

utu′

t

• Compute a parametric estimate of the long-run covariance matrix:

Λ = (I2 − A1)−1Ω(I2 − A1)−1′

• Compute the Choleski factorization of Λ:

Λ = PP′



Estimating the SVAR in the Presence of Long-Run Restrictions

• Define the estimate of Θ(1) as the lower triangular Choleski factor of

Λ,

Θ(1) = P

• Estimate B using

B =[

(I2 − A1)Θ(1)]

−1

• Estimate Θk using

Θk = ΨkB−1 = A

k

1B−1

From the estimated Θk matrices the structural IRF may then be

computed. Also, estimates of the structural shocks ε1t and ε2t may be

extracted using εt = But.



An empirical example with the BQ model (I)

The following program replicates the long-run restriction model by

Blanchard-Quah (1989) (the data are not the same as those in the original

paper so that the results do not match). The ∆Y (output growth) and U

(unemployment) series are already demeaned following the procedure by

Blanchard-Quah (1989).

−→ The program estimates the factorization matrix by two different

methods which should all yield the same result.




’ Blanchard-Quah long-run restriction (11/5/99)

’ verify estimates using two methods

’change path to program path

%path = @runpath

cd %path

’ create workfile

wfcreate blanquah q 1948:1 1987:4

’ fetch data from database (already demeaned)

fetch(d=data svar) dy u

’ estimate VAR (no constant)

var var1.ls 1 8 dy u @




’-------------------------------------------------------------------

’ method 1: text form long-run restrictions

’-------------------------------------------------------------------

var1.cleartext(svar) var1.append(svar) @lr1(@u1)=0

freeze(tab1) var1.svar(rtype=text,conv=1e-4)

show tab1

’ store estimated A and B matrices from method 1

matrix mata1 = var1.@svaramat

’ A should be identity

matrix matb1 = var1.@svarbmat

’-------------------------------------------------------------------

’ method 2: long-run restrictions in short-run form

’ *not recommended; only for checking*

’-------------------------------------------------------------------




’ get unit (cumulated) long-run response

var1.impulse(imp=u)

matrix clr = var1.@lrrsp

var1.cleartext(svar)

var1.append(svar) @e1 = c(1)*@u1 + c(2)*@u2

var1.append(svar) @e2 = -clr(1,1)*c(1)/clr(1,2)*@u1 + c(4)*@u2


show tab2



’ A should be identity





You should get these results:

Structural VAR Estimates

Sample (adjusted): 1950Q2 1987Q4

Convergence achieved after 9 iterations

Structural VAR is just-identified

Model: Ae = Bu where E[uu’]=I

Restriction Type: long-run text form

Long-run response pattern:

0 C(2)

C(1) C(3)

Coefficient Std. Error z-Statistic Prob.

C(1) 4.043213 0.232661 17.37815 0.0000

C(2) 0.005186 0.000298 17.37815 0.0000

C(3) 0.008335 0.329032 0.025333 0.9798

Log likelihood 495.5765

Estimated A matrix:

1.000000 0.000000

0.000000 1.000000

Estimated B matrix:

0.009296 0.000746

-0.208221 0.219819




-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1 2 3 4 5 6 7 8 9 10

Response of DY to DY

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1 2 3 4 5 6 7 8 9 10

Response of DY to U

-50

-40

-30

-20

-10

0

10

1 2 3 4 5 6 7 8 9 10

Response of U to DY

-50

-40

-30

-20

-10

0

10

1 2 3 4 5 6 7 8 9 10

Response of U to U

Response to Nonfactorized One Unit Innovations




Structural VAR Estimates

Sample (adjusted): 1950Q2 1987Q4

Included observations: 151 after adjustments

Convergence achieved after 8 iterations

Structural VAR is just-identified


Restriction Type: short-run text form

@E1 = C(1)*@U1 + C(2)*@U2

@E2 = -CLR(1,1)*C(1)/CLR(1,2)*@U1 + C(4)*@U2

where

@e1 represents DY residuals

@e2 represents U residuals


C(1) 0.009296 0.000535 17.37815 0.0000

C(2) 0.000746 0.000758 0.984581 0.3248

C(4) 0.219819 0.021145 10.39558 0.0000

Log likelihood 495.5765

Estimated A matrix:

1.000000 0.000000

0.000000 1.000000

Estimated B matrix:

0.009296 0.000746

-0.208221 0.219819



An empirical example: Impulse responses with the BQ model (II)

The following program replicates the long-run restriction model by

Blanchard-Quah (1989) (the data are not the same as those in the original

paper so that the results do not match). The ∆Y (output growth) and U

(unemployment) series are already demeaned following the procedure by

Blanchard-Quah (1989).

−→ To plot the accumulated response and the level response in one graph,

we first store the impulse responses in separate matrices, extract and place

the relevant columns into a third matrix, and use the line view of that

matrix.




’ replicate impulse response graph

’ from long-run restriction (Blanchard-Quah 1989)


%path = @runpath

cd %path

’ create workfile

wfcreate blanquah q 1948:1 1987:4

’ fetch data from database (already demeaned)

fetch(d=data svar) dy u

’ change to percentage

dy = 100*dy

’ estimate VAR (no constant)

var var1.ls 1 8 dy u @




’ impose long-run restrictions

’ @u1 = aggregate demand shock

’ @u2 = aggregate supply shock


var1.append(svar) @lr1(@u1)=0

’ estimate factorization


’ set response horizon

!hrz = 40

’ replicate Figures 3-4 (p.662)

’ accumulate to get level response of output

freeze(fig3) var1.impulse(!hrz,imp=struct,se=a,a,matbys=rsp acc) dy @ 1

freeze(fig4) var1.impulse(!hrz,imp=struct,se=a,a) dy @ 2




freeze(fig34) fig3 fig4

show fig34

’ replicate Figures 5-6

freeze(fig5) var1.impulse(!hrz,imp=struct,se=a,matbys=rsp lvl) u @ 1

freeze(fig6) var1.impulse(!hrz,imp=struct,se=a) u @ 2

freeze(fig56) fig5 fig6

show fig56

’ replicate Figure 1

matrix(!hrz,2) mtmp

’ get accumulated output response

vector v = @columnextract(rsp acc,1)

colplace(mtmp,v,1)

’ get level unemployment response




v = @columnextract(rsp lvl,2)

colplace(mtmp,v,2)

freeze(fig1) mtmp.line

fig1.draw(line,left) 0

fig1.elem(1) legend(Output response to demand)

fig1.elem(2) legend(Unemployment response to demand)

show fig1

’ replicate Figure 2

’ get accumulated output response

vector v = @columnextract(rsp acc,3)

colplace(mtmp,v,1)

’ get level unemployment response

v = @columnextract(rsp lvl,4)




colplace(mtmp,v,2)

freeze(fig2) mtmp.line

fig2.draw(line,left) 0

fig2.elem(1) legend(Output response to supply)

fig2.elem(2) legend(Unemployment response to supply)

show fig2




-0.8

-0.4

0.0

0.4

0.8

1.2

1.6

5 10 15 20 25 30 35 40

Output response to demandUnemployment response to demand




-0.2

0.0

0.2

0.4

0.6

0.8

1.0

5 10 15 20 25 30 35 40

Output response to supplyUnemployment response to supply




-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

5 10 15 20 25 30 35 40

Accumulated Response of DY to StructuralOne S.D. Shock1

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

5 10 15 20 25 30 35 40

Accumulated Response of DY to StructuralOne S.D. Shock2




-.6

-.5

-.4

-.3

-.2

-.1

.0

.1

5 10 15 20 25 30 35 40

Response of U to StructuralOne S.D. Shock1

-.2

-.1

.0

.1

.2

.3

5 10 15 20 25 30 35 40

Response of U to StructuralOne S.D. Shock2



A review empirical example: Cholesky factorization asstructural factorization

−→ The following program illustrates the two forms of imposing the

identifying restrictions in structural decompositions.

For illustration purposes and to check that the restrictions are correctly

imposed, we impose restrictions that replicate the Cholesky factorization.

As you can see the results resulting from the three methods are the same.




’ replicate cholesky factorization using SVAR


%path = @runpath

cd %path

’ create workfile

wfcreate cholsvar q 1948:1 1979:3

’ fetch data from database

fetch(d=data svar) rgnp rinv m1

’ take log of each series

series lgnp = log(rgnp)

series linv = log(rinv)

series lm1 = log(m1)

’ estimate unrestricted VAR




var var1.ls 1 4 lgnp linv lm1 @ c

’-------------------------------------------------------------------

’ method 1: short-run restrictions in text form

’-------------------------------------------------------------------


var1.append(svar) @e1 = c(1)*@u1

var1.append(svar) @e2 = -c(2)*@e1 + c(3)*@u2

var1.append(svar) @e3 = -c(4)*@e1 - c(5)*@e2 + c(6)*@u3


show tab1







’ compute factorization matrix

matrix fact1 = @inverse(mata1)*matb1

’-------------------------------------------------------------------

’ method 2: short-run restrictions in pattern matrix

’-------------------------------------------------------------------

’ create pattern matrices

matrix(3,3) pata

’ fill matrix in row major order

pata.fill(by=r) 1,0,0, na,1,0, na,na,1

matrix(3,3) patb

patb.fill(by=r) na,0,0, 0,na,0, 0,0,na

freeze(tab2) var1.svar(rtype=patsr,conv=1e-5,namea=pata,nameb=patb)

show tab2







’ compute factorization matrix

matrix fact2 = @inverse(mata2)*matb2

’-------------------------------------------------------------------

’ method 3: built-in cholesky

’-------------------------------------------------------------------

’ do impulse response with default Cholesky ordering

var1.impulse(10)

’ store cholesky factor

matrix fact3 = var1.@impfact





Restriction Type: short-run text form

@E1 = C(1)*@U1

@E2 = -C(2)*@E1 + C(3)*@U2

@E3 = -C(4)*@E1 - C(5)*@E2 + C(6)*@U3 where

@e1 represents LGNP residuals

@e2 represents LINV residuals

@e3 represents LM1 residuals


C(2) -1.267693 0.174224 -7.276223 0.0000

C(4) -0.163608 0.052334 -3.126221 0.0018

C(5) 0.024427 0.022646 1.078662 0.2807

C(1) 0.010597 0.000676 15.68439 0.0000

C(3) 0.020475 0.001305 15.68439 0.0000

C(6) 0.005142 0.000328 15.68439 0.0000

Estimated A matrix:

1.000000 0.000000 0.000000

-1.267693 1.000000 0.000000

-0.163608 0.024427 1.000000

Estimated B matrix:

0.010597 0.000000 0.000000

0.000000 0.020475 0.000000

0.000000 0.000000 0.005142




Model: Ae = Bu where E[uu’]=I Restriction Type: short-run pattern matrix

A =

1 0 0

C(1) 1 0

C(2) C(3) 1

B =

C(4) 0 0

0 C(5) 0

0 0 C(6)


C(1) -1.267693 0.174224 -7.276223 0.0000

C(2) -0.163608 0.052334 -3.126221 0.0018

C(3) 0.024427 0.022646 1.078662 0.2807

C(4) 0.010597 0.000676 15.68439 0.0000

C(5) 0.020475 0.001305 15.68439 0.0000

C(6) 0.005142 0.000328 15.68439 0.0000

Estimated A matrix:

1.000000 0.000000 0.000000

-1.267693 1.000000 0.000000

-0.163608 0.024427 1.000000

Estimated B matrix:

0.010597 0.000000 0.000000

0.000000 0.020475 0.000000

0.000000 0.000000 0.005142




-.010

-.005

.000

.005

.010

.015

1 2 3 4 5 6 7 8 9 10

Response of LGNP to LGNP

-.010

-.005

.000

.005

.010

.015

1 2 3 4 5 6 7 8 9 10

Response of LGNP to LINV

-.010

-.005

.000

.005

.010

.015

1 2 3 4 5 6 7 8 9 10

Response of LGNP to LM1

-.01

.00

.01

.02

.03

1 2 3 4 5 6 7 8 9 10

Response of LINV to LGNP

-.01

.00

.01

.02

.03

1 2 3 4 5 6 7 8 9 10

Response of LINV to LINV

-.01

.00

.01

.02

.03

1 2 3 4 5 6 7 8 9 10

Response of LINV to LM1

-.004

.000

.004

.008

.012

1 2 3 4 5 6 7 8 9 10

Response of LM1 to LGNP

-.004

.000

.004

.008

.012

1 2 3 4 5 6 7 8 9 10

Response of LM1 to LINV

-.004

.000

.004

.008

.012

1 2 3 4 5 6 7 8 9 10

Response of LM1 to LM1

Response to Cholesky One S.D. Innovations




−→ The matrices FACT1, FACT2 and FACT3 are all equal to:

0.010597 0.000000 0.000000

0.013433 0.020475 0.000000

0.001406 -0.000500 0.005142



esercitazione SVAR 2 - economia.unipv.iteconomia.unipv.it/pagp/pagine_personali/dean/esercitazione...

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