Post on 19-Feb-2022
Econometrics of money and financeLecture nine: multivariate modeling I
Zongxin Qian
School of Finance, Renmin University of China
November 4, 2014
Revisit the example from monetary economics
yt = a1yt−1 + a2Etyt+1 − a3(Rt − Etπt+1) + e1t (1)
πt = b1πt−1 + b2Etπt+1 + b3yt + e2t (2)
Rt = φπEtπt+1 + e3t (3)
I We estimated the Phillips curve as a single equation in thelast lecture
I We can also estimate the three-equation model together.
The monetary economics example in matrix form
I Recall that the three equation model can be reduced to:
yt = a1yt−1 + a2Etyt+1 − a3(φπ − 1)Etπt+1 + e1t − a1e3t(4)
πt = b1πt−1 + b2Etπt+1 + b3yt + e2t (5)
I In matrix form, this is
Yt = A0 + A1EtYt+1 + A2Yt−1 + Ut (6)
I Can you write out Ai s?
The monetary economics example in matrix form
I Recall that the three equation model can be reduced to:
yt = a1yt−1 + a2Etyt+1 − a3(φπ − 1)Etπt+1 + e1t − a1e3t(4)
πt = b1πt−1 + b2Etπt+1 + b3yt + e2t (5)
I In matrix form, this is
Yt = A0 + A1EtYt+1 + A2Yt−1 + Ut (6)
I Can you write out Ai s?
Solution of the Matrix form
I The matrix form of the model has a unique solution undercertain condition (Blachard-Khan condition).
I The solution has a vector autoregression (VAR) form.However, the solution method is at the master’s level.
I To proceed, I make the following simplifying assumptionEtYt+1 = Yt−1, Etπt+1 = πt−1.
I In addition, I modify the Taylor rule as Rt = φππt + e3t .
I Now it is easy to show that the solution is a SVARB0Yt = B1 + B2Yt−1 + ut . Can you tell the parametermatrices?
I If B0 is invertible, the SVAR can be written as a restrictedVAR Yt = B−10 B1 + B−10 B2Yt−1 + B−10 ut .
The unrestricted VAR
I For now, let’s forget about the restriction imposed by thestructural model.
I The unrestricted VAR can be estimated by either OLS orMLE, just as the single equation AR model.
I Note that the consequence of not imposing a valid restrictionis efficiency loss.
I However, the structural restriction might be wrong and leadto biased estimates and we don’t want to test the restrictionswith a wrong model!!!
I This tradeoff is at the center of various debates amongeconomists, particularly in the field of monetary economics.
The unrestricted VAR
I For now, let’s forget about the restriction imposed by thestructural model.
I The unrestricted VAR can be estimated by either OLS orMLE, just as the single equation AR model.
I Note that the consequence of not imposing a valid restrictionis efficiency loss.
I However, the structural restriction might be wrong and leadto biased estimates and we don’t want to test the restrictionswith a wrong model!!!
I This tradeoff is at the center of various debates amongeconomists, particularly in the field of monetary economics.
Estimation of the unrestricted VAR
I The density function of a k dimensional multivariate normaldistribution is (2π)−
k2 |Ω|− 1
2 exp[− 1
2 (X − µ)′Ω−1(X − µ)].
I Consider the VAR(1): Yt = BYt−1 + Ut , Ut ∼ (0, Ω).
I Set Y0 = 0, Y1 ∼ (0, Ω); Yt |Yt−1 ∼ (BYt−1, Ω) for t > 2.
I Following the same procedure of AR(1) estimation, we obtainl(B, Ω) = − kT
2 ln(2π) + T2 ln|Ω−1| −
12Σt=T
t=1 [(Yt −BYt−1)′Ω−1(Yt − BYt−1)]
I Maximization of the log-likelihood function with respect to Bis equivalent to minimizeΣt=Tt=1 [(Yt − BYt−1)′Ω−1(Yt − BYt−1)]
I This is a GLS estimator. But it will be equivalent to OLS if allequations in the VAR have a same set of regressors (Zellner,1962).
Estimation of the unrestricted VAR
I The density function of a k dimensional multivariate normaldistribution is (2π)−
k2 |Ω|− 1
2 exp[− 1
2 (X − µ)′Ω−1(X − µ)].
I Consider the VAR(1): Yt = BYt−1 + Ut , Ut ∼ (0, Ω).
I Set Y0 = 0, Y1 ∼ (0, Ω); Yt |Yt−1 ∼ (BYt−1, Ω) for t > 2.
I Following the same procedure of AR(1) estimation, we obtainl(B, Ω) = − kT
2 ln(2π) + T2 ln|Ω−1| −
12Σt=T
t=1 [(Yt −BYt−1)′Ω−1(Yt − BYt−1)]
I Maximization of the log-likelihood function with respect to Bis equivalent to minimizeΣt=Tt=1 [(Yt − BYt−1)′Ω−1(Yt − BYt−1)]
I This is a GLS estimator. But it will be equivalent to OLS if allequations in the VAR have a same set of regressors (Zellner,1962).
Statistical identification v.s. economic identification
I Economists are more interested in economic identificationwhich means distinguishing between different economicstories.
I For this reason, structural models with restrictions fromtheoretical models are popular these days.
I However, as we have mentioned, a wrong restriction may biasthe estimation.
I Moreover, restriction imposed on a misspecified model doesnot generate reliable results even if the restriction itself isvalid.
I For this reason, statistical identification is important. Toavoid problems with complicated structural models, theliterature has offered various solutions.
Solution I: the LSE approach-test, test, test
I Start with a general dynamic reduced-form model.
I Test whether the residuals follow a normal i.i.d.
I Testing down and simplifying dynamics.
I Test for exogeneity and reduce dimensionality.
I Identify and estimate the structural model.
Solution I: the LSE approach-test, test, test
I Start with a general dynamic reduced-form model.
I Test whether the residuals follow a normal i.i.d.
I Testing down and simplifying dynamics.
I Test for exogeneity and reduce dimensionality.
I Identify and estimate the structural model.
Solution I: the LSE approach-test, test, test
I Start with a general dynamic reduced-form model.
I Test whether the residuals follow a normal i.i.d.
I Testing down and simplifying dynamics.
I Test for exogeneity and reduce dimensionality.
I Identify and estimate the structural model.
Solution I: the LSE approach-test, test, test
I Start with a general dynamic reduced-form model.
I Test whether the residuals follow a normal i.i.d.
I Testing down and simplifying dynamics.
I Test for exogeneity and reduce dimensionality.
I Identify and estimate the structural model.
Solution I: the LSE approach-test, test, test
I Start with a general dynamic reduced-form model.
I Test whether the residuals follow a normal i.i.d.
I Testing down and simplifying dynamics.
I Test for exogeneity and reduce dimensionality.
I Identify and estimate the structural model.
Solution I: the LSE approach-test, test, test
I Start with a general dynamic reduced-form model.
I Test whether the residuals follow a normal i.i.d.
I Testing down and simplifying dynamics.
I Test for exogeneity and reduce dimensionality.
I Identify and estimate the structural model.
Normality test
I The JB (Jarque-Bera) test test for the null hypothesis ofnormality:
I The JB test statistics is N6 (S
2 + (K−34 )2), S is the skewness,K is the kurtosis
I Under the null hypothesis of normality, the JB statistics followa χ2 distribution with 2 degrees of freedom
I Eviews provide the JB test for residuals after estimation.View/Residual Diagnostics/Histogram-Normality
Normality test
I The JB (Jarque-Bera) test test for the null hypothesis ofnormality:
I The JB test statistics is N6 (S
2 + (K−34 )2), S is the skewness,K is the kurtosis
I Under the null hypothesis of normality, the JB statistics followa χ2 distribution with 2 degrees of freedom
I Eviews provide the JB test for residuals after estimation.View/Residual Diagnostics/Histogram-Normality
Normality test in Eviews
Normality test for a multivariate model
I The JB (Jarque-Bera) test has multivariate versions.
I The complications is that the residuals can be correlated.That is, the residual vector u ∼ N(0,C ), where is notdiagonal.
I To perform the test we first orthogonalize the residuals so thatthe transformed residuals no longer correlate with each other.
I One can use either the Cholesky (Ltkepohl) or triangle(Doornik and Hansen) decomposition to achieve this.
I Eviews can do these things for us (see next two slides).
Normality test for a VAR model in Eviews
Normality test for a VAR model in Eviews
VAR Residual Normality TestsOrthogonalization: Residual Correlation (Doornik-Hansen)Null Hypothesis: residuals are multivariate normalDate: 11/26/13 Time: 20:03Sample: 1 993Included observations: 991
Component Skewness Chi-sq df Prob.
1 -0.286836 13.27030 1 0.00032 0.131291 2.865489 1 0.0905
Joint 16.13578 2 0.0003
Component Kurtosis Chi-sq df Prob.
1 14.69512 1202.215 1 0.00002 6.284798 222.8960 1 0.0000
Joint 1425.111 2 0.0000
Component Jarque-Bera df Prob.
1 1215.485 2 0.00002 225.7615 2 0.0000
Joint 1441.247 4 0.0000
Serial correlation test
I The Breusch-Godfrey test test for the null hypothesis of noserial correlation:
I The original regression isyt = c + a1yt−1 + a2yt−2 + · · ·+ apyt−p + ut ,ut = φ1ut−1 + φ2ut−2 ++ · · ·+ φput−m + et , et is i. i. d.
I The orginal regression generates a series of residuals. Usingthose data, we can run an auxiliary regression as followsyt = d + b1yt−1 + b2yt−2 + · · ·+ bpyt−p + θ1ut−1 +θ2ut−2 + · · ·+ θput−m + wt
I The LM statistics is (T −m)R2, which follows a χ2
distribution with m degrees of freedom
I When the sample size is small the F test for the jointsignificance of uts is better than the χ2 test
Serial correlation test
I The Breusch-Godfrey test test for the null hypothesis of noserial correlation:
I The original regression isyt = c + a1yt−1 + a2yt−2 + · · ·+ apyt−p + ut ,ut = φ1ut−1 + φ2ut−2 ++ · · ·+ φput−m + et , et is i. i. d.
I The orginal regression generates a series of residuals. Usingthose data, we can run an auxiliary regression as followsyt = d + b1yt−1 + b2yt−2 + · · ·+ bpyt−p + θ1ut−1 +θ2ut−2 + · · ·+ θput−m + wt
I The LM statistics is (T −m)R2, which follows a χ2
distribution with m degrees of freedom
I When the sample size is small the F test for the jointsignificance of uts is better than the χ2 test
Serial correlation test in Eviews
Eviews provide the LM test for residuals after estimation.View/Residual Diagnostics/Serial Correlation LM Test
Breusch-Godfrey Serial Correlation LM Test:
F-statistic 3.654696 Prob. F(4,155) 0.0071Obs*R-squared 13.96215 Prob. Chi-Square(4) 0.0074
Serial correlation test (multivariate) in EviewsThere is also a multivariate version of the LM test. In Eviews, it isView/Residual Diagnostics/Serial Correlation LM Test
VAR Residual Serial Correlation LM TestsNull Hypothesis: no serial correlation at lag order hDate: 11/26/13 Time: 20:28Sample: 1 993Included observations: 991
Lags LM-Stat Prob
1 7.60051 0.10742 10.12839 0.03833 5.181823 0.26914 9.272124 0.05465 2.448518 0.65396 2.846312 0.58397 2.639085 0.61998 1.592152 0.81029 3.806246 0.4329
10 1.180014 0.881411 0.708716 0.950212 1.271123 0.8663
Probs from chi-square with 4 df.
Granger Causality/block exogeneity test
Consider a bivariate VAR of M1 growth and CPI inflation:(mt
πt
)=
(c1c2
)+
(b111 b112b121 b122
)(mt−1πt−1
)+
(b211 b212b221 b222
)(mt−2πt−2
)+
(v1tv2t
)
I Test for the null b112 = b212 = 0 is a test for whether inflationis a Granger cause of money growth.
I Test for the null b121 = b222 = 0 is a test for whether moneygrowth is a Granger cause of inflation.
I However, the name cause is misleading. The test cannot tellthe causal relationship. It is a test for forecasting power.
Granger Causality/block exogeneity test
Consider a bivariate VAR of M1 growth and CPI inflation:(mt
πt
)=
(c1c2
)+
(b111 b112b121 b122
)(mt−1πt−1
)+
(b211 b212b221 b222
)(mt−2πt−2
)+
(v1tv2t
)
I Test for the null b112 = b212 = 0 is a test for whether inflationis a Granger cause of money growth.
I Test for the null b121 = b222 = 0 is a test for whether moneygrowth is a Granger cause of inflation.
I However, the name cause is misleading. The test cannot tellthe causal relationship. It is a test for forecasting power.
Granger-Causality test in Eviews
Granger-Causality test in Eviews
VAR Granger Causality/Block Exogeneity Wald Tests
Date: 11/04/14 Time: 13:47
Sample: 1992Q1 2007Q1
Included observations: 58
Dependent variable: M1G
Excluded Chi-sq df Prob.
PI_SA 6.208926 2 0.0448
All 6.208926 2 0.0448
Dependent variable: PI_SA
Excluded Chi-sq df Prob.
M1G 1.681850 2 0.4313
All 1.681850 2 0.4313
Solution II: minimal structural VARs
I Since one big problem for structural models is thatassumptions imposed are too strong, a simple way out is torelax the restrictions.
I Structural VARs which are popular in current studies imposeonly minimal structural assumptions, such as the timing of thestructural shocks.
I Still face a problem of omitted variables? Try a structuralFAVAR model which is a VAR augmented by factor analysis.
Solution II: minimal structural VARs
I Since one big problem for structural models is thatassumptions imposed are too strong, a simple way out is torelax the restrictions.
I Structural VARs which are popular in current studies imposeonly minimal structural assumptions, such as the timing of thestructural shocks.
I Still face a problem of omitted variables?
Try a structuralFAVAR model which is a VAR augmented by factor analysis.
Solution II: minimal structural VARs
I Since one big problem for structural models is thatassumptions imposed are too strong, a simple way out is torelax the restrictions.
I Structural VARs which are popular in current studies imposeonly minimal structural assumptions, such as the timing of thestructural shocks.
I Still face a problem of omitted variables? Try a structuralFAVAR model which is a VAR augmented by factor analysis.
The triangular decomposition
I For any positive definitive symmetric matrix Ω, there is aunique decomposition Ω = ADA′.
I A =
1 0 0 . . . 0a21 1 0 . . . 0a31 a32 1 . . . 0
......
... . . ....
an1 an2 an3 . . . 1
and
D =
d11 0 0 . . . 00 d22 0 . . . 00 0 d33 . . . 0...
...... . . .
...0 0 0 . . . dnn
.
The triangular decomposition: details
I Introduce a matrix E1 ≡
1 0 0 . . . 0
−Ω21Ω−111 1 0 . . . 0−Ω31Ω−111 0 1 . . . 0
......
... . . ....
−Ωn1Ω−111 0 0 . . . 1
,
I It can be shown that
E1ΩE ′1 = H ≡
h11 0 0 . . . 00 h22 h23 . . . h2n0 h32 h33 . . . h3n...
...... . . .
...0 hn2 hn3 . . . hnn
.
The triangular decomposition: details continued
I Introduce another matrix E2 ≡
1 0 0 . . . 00 1 0 . . . 00 −h32h−122 1 . . . 0...
...... . . .
...0 −hn2h−122 0 . . . 1
,
I It can be shown that
E2HE′2 = K ≡
k11 0 0 . . . 00 k22 0 . . . 00 0 k33 . . . k3n...
...... . . .
...0 0 kn3 . . . knn
.
The triangular decomposition: details continued
I Carrying on similar procedures, we can find a series of Ei s,such that En−1 . . . E2E1ΩE ′1E
′2 . . . E ′n−1 = D
I So A = (En−1 . . . E2E1)−1
The Cholesky decomposition
I Ω = PP ′, where P ≡ AD1/2
I diag(P) =
√d11 0 0 . . . 00
√d22 0 . . . 0
0 0√d33 . . . 0
......
... . . ....
0 0 0 . . .√dnn
The recursive SVAR
I Consider the following reduced-form VAR
Yt = A1Yt−1 + . . . + ApYt−p + Ut , (7)
where Ut ∼ (0, ΣU) and ΣU is a non-diagonal matrix.
I It can be written as a recursive SVAR
Yt = A1Yt−1 + . . . + ApYt−p +W εt , (8)
where εt ∼ (0, Σε) and Σε = DD ′ is a diagonal matrix.
I W ≡ PD−1 and D = diag(P).
I Alternatively, one can define W ≡ P and Σε = I .
The recursive SVAR
I Consider the following reduced-form VAR
Yt = A1Yt−1 + . . . + ApYt−p + Ut , (7)
where Ut ∼ (0, ΣU) and ΣU is a non-diagonal matrix.
I It can be written as a recursive SVAR
Yt = A1Yt−1 + . . . + ApYt−p +W εt , (8)
where εt ∼ (0, Σε) and Σε = DD ′ is a diagonal matrix.
I W ≡ PD−1 and D = diag(P).
I Alternatively, one can define W ≡ P and Σε = I .
The recursive SVAR
I Consider the following reduced-form VAR
Yt = A1Yt−1 + . . . + ApYt−p + Ut , (7)
where Ut ∼ (0, ΣU) and ΣU is a non-diagonal matrix.
I It can be written as a recursive SVAR
Yt = A1Yt−1 + . . . + ApYt−p +W εt , (8)
where εt ∼ (0, Σε) and Σε = DD ′ is a diagonal matrix.
I W ≡ PD−1 and D = diag(P).
I Alternatively, one can define W ≡ P and Σε = I .
The recursive SVAR
I Elements in Ut cannot be taken as independent shocks sinceΣU is non-diagonal.
I That is why we would like to calculate εt , of which theelements are called structural shocks.
I Defining W ≡ P and Σε = I suggests that all structuralshocks have the same variance. This is quite a strongassumption.
I Defining W ≡ PD−1 and D = diag(P) allows the variancesof structural shocks to be different.
The recursive SVAR
I Elements in Ut cannot be taken as independent shocks sinceΣU is non-diagonal.
I That is why we would like to calculate εt , of which theelements are called structural shocks.
I Defining W ≡ P and Σε = I suggests that all structuralshocks have the same variance. This is quite a strongassumption.
I Defining W ≡ PD−1 and D = diag(P) allows the variancesof structural shocks to be different.
The recursive SVAR
I Elements in Ut cannot be taken as independent shocks sinceΣU is non-diagonal.
I That is why we would like to calculate εt , of which theelements are called structural shocks.
I Defining W ≡ P and Σε = I suggests that all structuralshocks have the same variance. This is quite a strongassumption.
I Defining W ≡ PD−1 and D = diag(P) allows the variancesof structural shocks to be different.
The recursive SVAR
I W is a lower triangular matrix, so
W εt =
W11ε1t
W21ε1t +W22ε2tW31ε1t +W32ε2t +W33ε3t
...Wn1ε1t +Wn2ε2t +Wn3ε3t + . . . +Wnnεnt
I Therefore, the ordering of endogenous variables in a VAR
matters a lot for the interpretation of its recursive SVARversion.
I The first VAR variable is only contemporaneously affected bystructural shock to itself.
I The second VAR variable is contemporaneously affected bystructural shocks to itself and to the first variable...
The recursive SVAR
I W is a lower triangular matrix, so
W εt =
W11ε1t
W21ε1t +W22ε2tW31ε1t +W32ε2t +W33ε3t
...Wn1ε1t +Wn2ε2t +Wn3ε3t + . . . +Wnnεnt
I Therefore, the ordering of endogenous variables in a VAR
matters a lot for the interpretation of its recursive SVARversion.
I The first VAR variable is only contemporaneously affected bystructural shock to itself.
I The second VAR variable is contemporaneously affected bystructural shocks to itself and to the first variable...
Impulse response analysis
I Recall that an AR model can be rewritten as a MA(∞) model.
I Similarly, a VAR model can also be written asYt = µ + ut + ψ1ut−1 + ψ2ut−2 + . . ..
I ∂Yt+h
∂ut= ψh, where the ij ’s element of ψh is the IRF of the jth
shock on variable i in period t + h.
I An important application of the SVAR is IRF analysis.
The recursive SVAR: an example
I Consider a VAR with three variables (πt , yt ,Mt), where Ihave replaced the usual monetary policy variable in athree-equation SVAR, policy interest rate, by the growth rateof aggregate money supply, M2. This is because moneysupply is more important as a policy tool in the Chineseexample we are going to look at.
I The ordering I use means that monetary policy actions haveonly lagged impact on inflation and output.
I Moreover, shocks to output have only lagged impact oninflation.
The recursive SVAR: an example
I Consider a VAR with three variables (πt , yt ,Mt), where Ihave replaced the usual monetary policy variable in athree-equation SVAR, policy interest rate, by the growth rateof aggregate money supply, M2. This is because moneysupply is more important as a policy tool in the Chineseexample we are going to look at.
I The ordering I use means that monetary policy actions haveonly lagged impact on inflation and output.
I Moreover, shocks to output have only lagged impact oninflation.
Impulse responses in the exampleHere we use the ordering (πt , yt ,Mt). We use VAR(2)and samplefrom 2008m10 to 2013m11.
-.2
.0
.2
.4
.6
1 2 3 4 5 6 7 8 9 10
Response of INF_SA to INF_SA
-.2
.0
.2
.4
.6
1 2 3 4 5 6 7 8 9 10
Response of INF_SA to YGAP
-.2
.0
.2
.4
.6
1 2 3 4 5 6 7 8 9 10
Response of INF_SA to M2_SA
-0.5
0.0
0.5
1.0
1 2 3 4 5 6 7 8 9 10
Response of YGAP to INF_SA
-0.5
0.0
0.5
1.0
1 2 3 4 5 6 7 8 9 10
Response of YGAP to YGAP
-0.5
0.0
0.5
1.0
1 2 3 4 5 6 7 8 9 10
Response of YGAP to M2_SA
-0.4
0.0
0.4
0.8
1.2
1 2 3 4 5 6 7 8 9 10
Response of M2_SA to INF_SA
-0.4
0.0
0.4
0.8
1.2
1 2 3 4 5 6 7 8 9 10
Response of M2_SA to YGAP
-0.4
0.0
0.4
0.8
1.2
1 2 3 4 5 6 7 8 9 10
Response of M2_SA to M2_SA
Response to Cholesky One S.D. Innovations ?2 S.E.
Impulse responses in the exampleNow we change the ordering to (yt , πt ,Mt).
-0.5
0.0
0.5
1.0
1 2 3 4 5 6 7 8 9 10
Response of YGAP to YGAP
-0.5
0.0
0.5
1.0
1 2 3 4 5 6 7 8 9 10
Response of YGAP to INF_SA
-0.5
0.0
0.5
1.0
1 2 3 4 5 6 7 8 9 10
Response of YGAP to M2_SA
-.2
.0
.2
.4
.6
1 2 3 4 5 6 7 8 9 10
Response of INF_SA to YGAP
-.2
.0
.2
.4
.6
1 2 3 4 5 6 7 8 9 10
Response of INF_SA to INF_SA
-.2
.0
.2
.4
.6
1 2 3 4 5 6 7 8 9 10
Response of INF_SA to M2_SA
-0.4
0.0
0.4
0.8
1.2
1 2 3 4 5 6 7 8 9 10
Response of M2_SA to YGAP
-0.4
0.0
0.4
0.8
1.2
1 2 3 4 5 6 7 8 9 10
Response of M2_SA to INF_SA
-0.4
0.0
0.4
0.8
1.2
1 2 3 4 5 6 7 8 9 10
Response of M2_SA to M2_SA
Response to Cholesky One S.D. Innovations ?2 S.E.
Estimate a VAR in Eviews
Specify your VAR in Eviews
Generate IRFs with your VAR in Eviews
Choose the Cholesky ordering in Eviews
The recursive SVAR: the impact of normalization
I The two specifications W = P and W = PD−1 can generatedifferent results.
I Eviews use W = P. You have to be aware of this when usingEviews.
I In the next slide, we preset an example where the twospecifications generate different results.
I We estimate a bivariate VAR(12) for exogenous tax changeand log output of the US. The sample period is1950q1-2007q4.
I Raw data is taken fromhttps://www.aeaweb.org/articles.php?doi=10.1257/aer.100.3.763
IRF of the bivariate VAR, W = P
-.012
-.008
-.004
.000
.004
.008
.012
.016
2 4 6 8 10 12 14 16 18 20
Response of LGDP to LGDP
-.012
-.008
-.004
.000
.004
.008
.012
.016
2 4 6 8 10 12 14 16 18 20
Response of LGDP to EXOR
-.1
.0
.1
.2
2 4 6 8 10 12 14 16 18 20
Response of EXOR to LGDP
-.1
.0
.1
.2
2 4 6 8 10 12 14 16 18 20
Response of EXOR to EXOR
Response to Cholesky One S.D. Innovations ?2 S.E.
IRF of the bivariate VAR, W = PD−1
This graph is generated with matlab programing.
3 6 9 12 15 18
0
20
40
60
80
100
Impulse response of exogenous
3 6 9 12 15 18
-1.2
-1
-0.8
-0.6
-0.4
-0.2
0
Impulse response of gdp
Application of recursive SVAR
I As mentioned before, economic theories usually imply morecomplicated restrictions than just the ordering of variables in arecursive VAR.
I A common practice to test theories is to match the impulseresponses of the structural model to those of a recursive VAR.
I Of course, this approach has limitations.
1. Not all theoretical models have a reduced-form VARrepresentation.
2. Estimation of VAR(∞) requires truncation which can causebias.
3. Some research topics might not have a clear time ordering ofvariables.
Application of recursive SVAR
I As mentioned before, economic theories usually imply morecomplicated restrictions than just the ordering of variables in arecursive VAR.
I A common practice to test theories is to match the impulseresponses of the structural model to those of a recursive VAR.
I Of course, this approach has limitations.
1. Not all theoretical models have a reduced-form VARrepresentation.
2. Estimation of VAR(∞) requires truncation which can causebias.
3. Some research topics might not have a clear time ordering ofvariables.