ECON 4160, Autumn term 2017 Lecture 9

Transcription

1 ECON 4160, Autumn term 2017 Lecture 9 Structural VAR (SVAR) Ragnar Nymoen Department of Economics 26 Oct / 14

2 Parameter of interest: impulse responses I Consider again a partial market equilibrium example, as in Lecture 8, and with first order dynamics: Q t + b 12 P t = b 10 + φ 11 Q t 1 + φ 12 P t 1 + ɛ 1t, (1) b 12 > 0, φ 11 = 0, φ 12 = 0 (demand) b 21 Q 1t + P t = b 20 + φ 21 Q t 1 + φ 22 P t 1 + ɛ 2t, (2) b 12 < 0, φ 21 = 0, φ 22 = 0 (supply) with symmetric, but not diagonal covariance matrix Ω (hence ω 12 = 0 in general). This SEM is not identified on the O/R conditions. Hence there are no consistent estimators for any of the structural coeffi cients. 2 / 14

3 Parameter of interest: impulse responses II But assume that we do not mind that, because our interest is how the two endogenous variables respond dynamically to random shocks (also called impulses). Those response functions can be found from the reduced form,the VAR(1): ( Qt P t ) = ( ϕ10 ϕ 20 and are partial derivatives: ) ( ϕ11 ϕ + 12 ϕ 21 ϕ 22 ) ( Q1t 1 P 2t 1 ) ( ε1t + ε 2t ) (3) Q t+h and P t+h, h = 0, 1, 2,..., j = 1, 2. of the solutions for Q t and P t. 3 / 14

4 Parameter of interest: impulse responses III Since only reduced form coeffi cients are used in the expressions for Q t+h, the responses are consistently estimated by OLS. But in another important sense the VAR impulse responses are not identified: We cannot identify the shock (impulses) as demand-shocks or supply shocks, because ε 1t and ε 2t are composite errors: ɛ 1t ε 1t = b 12ɛ 2t 1 b 12 b 21 1 b 12 b 21 (4) ε 2t = b 21ɛ 1t ɛ 2t + 1 b 12 b 21 1 b 12 b 21 (5) 4 / 14

5 Parameter of interest: impulse responses IV Hence, we cannot say whether for example Q t+h is due to an impulse to demand (ɛ 1t changes) or to supply (ɛ 2t changes), without making further assumptions. 5 / 14

6 Cholesky decomposition I The most popular way of identifying impulse responses, is known by the technical name Cholesky decompostion. In our example, applying a Cholesky decompostion amounts to imposing a restriction: b 21 = 0 (lower triangularization), or b 12 = 0 (upper triangularization). Setting b 21 = 0, implies for the VAR error-terms: ε 1t = b21 =0 ɛ 1t b 12 ɛ 2t (6) ε 2t = b21 =0 ɛ 2t (7) 6 / 14

7 Cholesky decomposition II and we can see that two of the VAR impulse responses Q t+h ε 2t and P t+h ε 2t, h = 0, 1,... are the dynamic responses with respect to a supply-shock. They are identified. Now consider the two remaining responses: Q t+h and P t+h, h = 0, 1,... in the light of (6). For Q t (the instantaneous response) we get: Q t ε 2t = 1 b 12 7 / 14

8 Cholesky decomposition III where ε 2t = ω 12 /ω2 2. Hence, if we impose a second restriction: ω 12 = 0 (8) the responses with respect to changes in ε 1t are identified as responses to a demand shock. 8 / 14

9 Formulating a SVAR as a recursive model I From Lecture 8, we have that the two restrictions on the SEM: b 21 = 0 and ω 12 = 0 (9) defines a recursive model, which we can write as Q t = φ 0 + φ 1 Q t 1 + β 0 P t + β 1 P t 1 + ɛ 1t (10) P t = ϕ 20 b 20 + ϕ 21 φ 21 Q t 1 + ϕ 22 φ 22 P t 1 + ε 2t ɛ 2t (11) where we have chosen parameter symbols to remind ourselves that the first equation is interpreted as the conditional model equation for Q t given P t and the second is the marginal model equation for P t. 9 / 14

10 Formulating a SVAR as a recursive model II The conditional + marginal interpretation means that Cov(ɛ 1t, ε 2t ) = ω 12 = 0, and so provides us with a practical way of implementing the required ω 12 = 0 condition. OLS on (10) gives for ˆβ 0,OLS : plim( ˆβ 0,OLS ) = Cov(ε 1t,ε 2t ) Var(ε 2t ) σ 12 σ 2 2 DIY: Use (4) and (5) to show that for a recursive model (remember (9)) we have: σ 12 = b 12 ω 2 2 (12) σ 2 2 = ω 2 2 (13) 10 / 14

11 Formulating a SVAR as a recursive model III Therefore showing identification. plim( ˆβ 0,OLS ) = b / 14

12 Example using the program I Formulate and estimate the VAR(1) for the fish-data with stormy, mixed and hol as exogenous variables Let the program calculate 12 impulse responses, choosing first the Standard deviation radio button and then Orthogonalized radio button. Formulate (10) and (11) as SEM in the program and estimate by OLS on each equation ( forcing ω 12 = 0). Let the program calculate 12 impulse responses, choose the Standard deviation radio button. Compare with VAR impulse responses. 12 / 14

13 Recursive model and SVAR I Using the Cholesky decompostion to formulate a SVAR is the same as choosing a recursive model. And a recursive model is a SVAR. A recursive model can also be interpreted as consisting of conditional and marginal model equations. Notice that the SVAR (recursive) identifying assumptions only involve the contemporaneous coeffi cient matrix (and the Ω matrix). Not the autoregressive matrix of the SEM These can be used to test overidentifying restrictions Or indeed to formulate other exact identification schemes, which we do not go into here 13 / 14

14 Recursive model and SVAR II In summary: for unclear reasons, SVARs are sometimes presented as more different from SEMs than they are. 14 / 14