Class 4: Non-stationary Time Series

Transcription

1 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC Approach Homework Jacek Suda, BdF and PSE January 24, 2013

2 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC Approach Homework Outline Outline: 1 Unit Root (Augmented) Dickey-Fuller Tests Phillips-Perron Stationarity Tests Variance Ratio Test 2 Structural Breaks

3 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachDFHomework ADF UNIT ROOT

4 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachDFHomework ADF Autoregressive Unit Root Tests ARMA(p, q) process: φ(l)y t = θ(l)ε t, ε t WN φ(z) = 0, where φ(z) is characteristic equation. φ(l) = 1 φ 1 L φ 2 L 2... φ p L p φ(z) = 0 1 φ 1 z φ 2 z 2... φ p z p = 0 (1 1 z)(1 1 z) (1 1 z) = 0 λ 1 λ 2 λ p Consider hypothesis: H 0 : φ(z) = 0 has (at least) one root on unit circle. An alternative for H 0 is H 1 : φ(z) = 0 has all roots outside unit circle y t = φ(l) 1 θ(l)ε t y t = Ψ(L)ε t = u t I(0)

5 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachDFHomework ADF Dickey-Fuller: Case 1 Consider AR(1): Then y t = φy t + ε t, ε t WN φ(l) = 1 φl H 0 : φ(z) = 0 has unit root φ = 1 H 1 : φ(z) = 0 has roots outside unit circle φ < 1 Test statistics: ˆt φ=1 = ˆφ φ d SE( ˆ ˆφ) DF, where ˆφ comes from OLS on y t = ˆφy t 1 + ˆε t.

6 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachDFHomework ADF Dickey-Fuller distribution Dickey-Fuller distribution Does not have a closed form representation. Is not centered around 0. 5% critical value for 1-side test is 1.94 ( 1.65 for Normal) 1% critical value for 1-side test is 2.57 ( 2.32 for Normal) Note: 1.65 is the 9.45% quantile of the DF distribution. It has less power under H 0, but higher size adjusted power. Additionally, ˆφ is super-consistent; that is, ˆφ p φ at rate T instead of the usual rate T 1/2.

7 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachDFHomework ADF Dickey-Fuller: Case 2 Constant + AR(1): (y t µ) = φ(y t 1 µ) + ε t, ε t iid WN Under H 0 : φ = 1 y t = y 0 + t ε j I(1), y 0 = µ, j=1 i.e. ε, shock, never dies out - its effect will be forever present in the series. Alternatively, under H 1 : φ < 1 y t = c + φy t 1 + ε t I(0), c = µ(1 φ), with shocks dying out over time.

8 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachDFHomework ADF Dickey-Fuller distribution The t statistics is t µ φ=1 = ˆφ 1 ˆ SE( ˆφ) from OLS regression y t = ĉ + ˆφy t 1 + ˆε t, Dickey-Fuller shows that, under H 0 : φ = 1, it is with t µ φ=1 1 d DF µ 0 = Wµ (r)dw(r) ( 1 0 Wµ (r) 2 dr), 1/2 W µ (r) = W(r) 1 0 W(r)dr the de-meaned Wiener process, 1 0 Wµ (r) = 0.

9 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachDFHomework ADF Dickey-Fuller: Case 3 The test regression is y t c β t = φ(y t 1 c β (t 1)) + ε t and includes a constant and deterministic time trend to capture the deterministic trend under the alternative. H 0 : φ = 1: y t = c + β t + t ε j I(1) with drift, j=1 where c + β t denotes deterministic component, and t j=1 ε j the random walk component. H 1 : φ < 1: y t = c + β t + φy t 1 + ε t Trend stationary y t β t I(0)

10 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachDFHomework ADF Test statistics Test statistics where ˆφ is from OLS regression t β φ=1 = ˆφ 1 ŜE( ˆφ) y t = ĉ + ˆβ t + ˆφy t 1 + ε t. Both β and c are nuisance parameters. Under H 0 : φ = 1 t β φ=1 1 d DF µ 0 = Wβ (r)dw(r) ( 1 0 Wβ (r) 2 dr), 1/2 with ( W β (r) = W µ (r) 12 r 1 2 ) 1 0 ( s 1 ) W(s)ds, 2

11 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachDFHomework ADF Test statistics De-meaned and detrended Wiener process. The inclusion of a constant and trend in the test regression further shifts the distribution of t β φ=1to the left. 5% critical value for 1-side test is 3.41 ( 1.65 for Normal) 1% critical value for 1-side test is 3.96 ( 2.32 for Normal) 1.65 is the 77.52% quantile of the DF β distribution! Test DF µ has more power than DF β if β = 0.

12 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachDFHomework ADF AR(p): Hypothesis Basic AR(p) model φ(l)y t = ε t, ε t WN φ(l) = 1 φ 1 L... φ p L p Hypothesis H 0 : H 1 : φ(z) = 0 has one unit root φ(z) = (1 z)φ (z), φ (z) has no unit root. φ(z) = 0 has all roots outside unit circle.

13 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachDFHomework ADF Transformation Dickey-Fuller transformation of φ(l) y t = ρy t 1 + φ 1 y t 1 + φ 2 y t φ p 1 y t p 1 + ε t, where ρ = φ 1 + φ φ p p φ j = k=j+1 The hypothesis can be simply restated as φ k H 0 : H 1 : ρ = 1 unit root ρ < 1 I(0) in equation y t = ρy t 1 + u t, u t I(0) with u t containing lagged differences to capture serial correlation in u t.

14 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachDFHomework ADF Augmented Dickey-Fuller Test Augmented Dickey-Fuller Test (ADF Test) from OLS regression ˆt ρ=1 = ˆρ 1 ŜE( ˆρ) y t = ˆρy t 1 + ˆφ 1 y t ˆφ p 1 y t p 1 + ε t. The distribution of t-statistics is ˆt ρ=1 d DF ˆt µ ρ=1 ˆt β ρ=1 d DF µ d DF β.

15 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachStatistics Homework Phillips-Perron Unit-Root Test Phillips-Perron Unit-Root Test Model y t = ρy t 1 + u t, u t serially correlated residuals We do not specify how it is correlated, do not put any parametric approach. If φ is close to 1, ADF has terrible size. Phillips-Perron addresses this issue

16 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachStatistics Homework Phillips-Perron Unit-Root Test Phillips-Perron Unit-Root Test The PP tests correct for any serial correlation and heteroskedasticity in the errors u t of the test regression. It directly modifies the test statistics t ρ=0 : Z t = t ρ=0 = ( ˆσ 2 ˆλ 2 ) 1/2 t ρ=0 1 2 ˆρ ŜE( ˆρ) ( ) ( ˆλ2 ˆσ 2 ˆλ 2 ) T ŜE(ˆρ) ˆσ 2

17 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachStatistics Homework Phillips-Perron Unit-Root Test Phillips-Perron Unit-Root Test Terms ˆσ 2 and ˆλ 2 are consistent estimates of the variance parameters σ 2 = lim T T 1 λ 2 = lim T S T = T u t. t=1 T E(u 2 t ) t=1 T E(T 1 ST) 2 = long run variance t=1 Result: Under the null hypothesis that ρ = 0, the PP Z t statistic has the same asymptotic distributions as the ADF t-statistic.

18 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachStatistics Homework Phillips-Perron Unit-Root Test Phillips-Perron Unit-Root Test The sample variance of the least squares residual û t is a consistent estimate of σ 2. The Newey-West long-run variance estimate of u t using û t is a consistent estimate of λ 2. m [ ˆλ 2 = ˆγ j ] ˆγ j m + 1 ˆγ 0 = 1 T ˆγ j = 1 T T t=1 T j=1 û 2 t t=j+11 û t û t j

19 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachUCHomework Model Unit MA root KPSS Test Stationarity Tests Nelson-Plosser found that most macro variables have unit roots. They failed to reject H 0 of the presence of unit root (we may reject it because of the low power of the test). H 0 in unit root test is that there is unit root and N-P fails to reject it. We want to reverse the problem and test if the series is stationary so that unit root would make that we reject stationary H 0. Kwiatkowski, Phillips, Schmidt and Shin (KPSS), 1992 JoE - non parametric approach; Leybourne and McCabe (1994, JEBS) - parametric approach.

20 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachUCHomework Model Unit MA root KPSS Test Unobserved Components Model Unobserved Components Model y t = µ t + ε t, where µ t = µ t 1 + u t u t iid(0, σu), 2 µ 0 = constant, ε t I(0) (i.e. φ(l)ε t = θ(l)η t ) µ t is unobserved component. it takes both cases of unit root and stationarity µ t = local mean + unobserved component: it changes every time. Even though we don t observe this shock we can still recover it.

21 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachUCHomework Model Unit MA root KPSS Test Stationarity Tests (KPSS) Hypotheses: H 0 : σ 2 u = 0 = y t I(0) H 1 : σ 2 u > 0 = y t I(1) Test is equivalent to testing for unit MA root in y t. How to estimate variance σ 2 = 0, in model that is already difficult to estimate? It is equivalent to testing for unit MA root in y t.

22 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachUCHomework Model Unit MA root KPSS Test Unit MA root So far we talked about AR unit root: σ 2 u > 0. There is alternative approach MA unit root to difference which corresponds to the general notion of overdifferencing. Unit MA root: y t = µ t + ε t apply (1 L) (1 L)y t = (1 L)µ t + (1 L)ε t y t = u t + ε t ε t 1 We never observe the shock (unless u t = 0). Under H 0 : σ 2 u = 0, u t = 0 so if there is a shock today then tomorrow will exactly offset it => no permanent shock to accumulate of the process. So unit MA root implies no permanent effect of shock.

23 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachUCHomework Model Unit MA root KPSS Test Granger representation If ε t iid, Granger representation theorem implies that y t = e t + θe t 1, where e t is unobservable forecast error. If cov(u t, ε t ) = 0 cov( y t, y t 1 ) = cov(u t + ε t ε t 1, u t 1 + ε t 1 ε t 2 ) = σ 2 ε and cov( y t, y t j ) = 0, j > 1 The same autocovariance structure as in MA(1) process.

24 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachUCHomework Model Unit MA root KPSS Test MA(1) representation y t = e t + θe t 1, For σ εu = 0 θ = (q + 2) + q 2 + 4q 2 q is the signal-to-noise ration., q = σ2 u σε 2. As q 0 we have unit MA root we have permanent shock but they are very very small. If σ 2 u = 0 then q = 0 = θ = 2 2 = 1 so Ψ (L) = 1 + θl has unit root.

25 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachUCHomework Model Unit MA root KPSS Test KPSS Test Testing Regress y t on MA(1) process and see if θ = 1. Not easy to do -> problem with power. KPSS proposes one-sided LM statistics for hypothes H 0 : σu 2 = 0 no random walk component, just constant H 1 : σu 2 > 0 LM statistics depends on process for y t

26 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachUCHomework Model Unit MA root KPSS Test KPSS: Case 1 Case 1: constat term only Test regression LM test: y t = µ t + ε t µ t = µ t 1 + u t, µ 0 = constant y t = α + ε t = ˆε t = y t ȳ ˆη µ = 1 T 2 T t=1 S 2 t Λ 2, where St 2 = t j=1 ˆεj is a partial sum over time of residuals, and Λ 2, spectral density at frequency 0 Sum up sample residual over time under H 0 they should not be a big number, they should cancel out. Otherwise, (under alternative) they should get larger and larger.

27 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachUCHomework Model Unit MA root KPSS Test KPSS: Case 1 Under H 0 : σu 2 = 0 where 1 d η µ V(r) 2 dr 0 V(r) = W(r) rw(1) = standard Brownian Bridge Reject at 5% if ˆη µ > To do it need to estimate Λ 2, spectral density at frequency 0. KPSS proposes against Bartlet kernel approach: depending how you choose the bandwidth you get different statistics depending on what bandwidth you choose -> big sample size distortions. Some people suggest using parametric approach to construct test statistics.

28 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachUCHomework Model Unit MA root KPSS Test KPSS: Case 2 Case 2: constat + trend y t = τ t + µ t + ε t, ε t I(0) µ t = µ t 1 + u t, u t iid(0, σu) 2 Test regression y t = α + τ t + ε t LM test: Reject H 0 at 5% if ˆη τ > ˆη µ = 1 T 2 T t=1 S 2 t Λ 2,

29 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachUCHomework Model Unit MA root KPSS Test Testing Now we have unit root and stationarity test: apply both. 1 Unit root test: you can t reject H 0; KPSS test: reject H 0. If put together, they imply that series has unit root. 2 If we can t reject both test: data give not enough observations. 3 Reject unit root, reject stationarity: both hypothesis are component hypothesis heteroskedasticity in series may make a big difference; if there is structural break it will affect inference. Power problem: if there is small random walk component (small variance σ 2 u), we can t reject unit root and can t reject stationarity. Economics: if the series is highly persistence we can t reject H 0 (unit root) highly persistent may be even without unit root but it also means we shouldn t treat/take data in levels. If we want to quantify how important the unit root is, we should use Variance Ratio Test.

30 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachIdea Homework Variance Example Result Idea Idea Cochrane, 1988 non-paramteric measure of economic importance of unit root Variance of random walk grows linearly with horizon (no unconditional variance) Variance of trend stationary process is finite (it may grow over short horizon but it will finally settle down) Test the behavior of variance.

31 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachIdea Homework Variance Example Result Variance Let V k = (1/k)var(Y t+k Y t kµ) V k the variance of k th period difference, µ is deterministic trend that does not affect the variance. If there is no random walk it should converge to zero as variance is converging to constant and k is growing. Rewrite it as V k = 1 k E [( y t+1 µ) + ( y t+2 µ) ( y t+k µ)] 2 k 1 ( ) k j = γ0 + 2 γj k j=1 = weighted average of auto-covariances γ j = cov( y t, y t+j ) Auto-covariances are important because everything about the covariance-stationary series is in auto-covariance generating function. If model reduced to covariance can do analysis non-parametrically don t have to specify the model when using sample auto-covariances.

32 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachIdea Homework Variance Example Result Variance ratio To make the test better suited to compare series: Variance ratio VR k = Vk, V 1 = γ0 V 1 Result lim V k = k k= (spectral density at frequency 0 for y) γ k = Λ 2

33 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachIdea Homework Variance Example Result Example: Random Walk Random walk Then Variance y t = y t 1 + ε t, ε t N(0, σ 2 ε) y t+k y t = y t = y 0 + t+k j=t+1 t t+k ε j, y t+k = y 0 + j=1 var(y t+1 y t ) = σ 2 ε, var(y t+k y t ) = kσ 2 ε Variance ratio: ε j j=1 V 1 = σ 2 ε; V k = 1 k var(y t+k y t ) = σ 2 ε k VR k = V k V 1 = σ2 ε = 1, k. σε 2 shock today has effect on series today and on series in the future. ε j

34 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachIdea Homework Variance Example Result Example: White Noise White noise y t = ε t, ε t N(0, σε) 2 Variance var(y t+1 y t ) = 2σε 2 var(y t+k y t ) = 2σε 2 Variance ratio VR k = 1 k 0 V k = 2σ2 ε k as k k So for different type of process the variance ratio behaves differently. In practice, we have to estimate VR k. Cochrane estimates VR k using Newey-West ˆΛ 2 NW, ˆγ 0, ˆγ j.

35 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachIdea Homework Variance Example Result Implications 1 If VR k 1 VR k k random walk, all shocks are permanent 2 If VR k 0 < VR < 1 VR k 1 VR 0 1 k mean reverting I(1) process

36 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachIdea Homework Variance Example Result Implications 3 If VR k VR > 1 VR k VR k 4 If VR k 0 mean averting I(1) process possible serial correlation in differences VR k k trend stationary I(0) process

37 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachIdea Homework Variance Example Result Implications If k too large you get spurious mean reversion. In sample it always the case that so mean reversion has to appear. T 1 ˆγ 0 + ˆγ h = 0 h=1 Which k to use? : balance the two effects. Non parametric approach makes problem in small sample. At long horizon GDP has neither VR 0 not VR 1, it s between. With standard errors, though, you can t reject any of them.

38 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachStructural Homework Breaks Trend Test statistics Perron (1989) Criticism Structural Breaks Failure to reject unit root could reflect structural breaks, not unit root Issue: frequency of permanent shocks how often do you have permanent shock to the mean or drift Trend stationary = never TS with one structural break = once TS with n structural breaks = n times TS with T structural breaks = Unit root (difference stationary) Perron, 1989 Econometrica, The Great Crash, The Oil Price Shock, and the Unit Root Hypothesis Given a level a level break in 1929, the unit root can be rejected for 11 of 14 Nelson-Plosser (1982) series including at 10% real GNP and nominal stock prices. (annual data). Given the break in growth in 1973, Perron rejects unit root for postwar quarterly GDP.

39 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachStructural Homework Breaks Trend Test statistics Perron (1989) Criticism Model A: The Great Crash Model Model A: The Great Crash Model Trend t m 0 T B t Model of trend fluctuations of GDP. Trend is deterministic, when removed one gets the covariance stationary series. Trend t = µ 1 + β t + (µ 2 µ 1 )DU t + e t DU t = { 1 if t > TB 0

40 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachStructural Homework Breaks Trend Test statistics Perron (1989) Criticism Model B: The Oil Shock Model Model B: The Oil Shock Model Trend t Slope b 2 Slope b 1 0 m T B t Trend t = µ + β 1 t + (β 2 β 1 )DT t + e t DT t = { t TB if t > T B 0

41 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachStructural Homework Breaks Trend Test statistics Perron (1989) Criticism Model C: The Combo Model Model C: The Combo Model Trend t Slope b 2 Slope b 1 m 1 0 T B t Trend t = µ 1 + β 1 t + (µ 2 µ 1 )DU t + (β 2 β 1 )DT t + e t DU t = DT t = { 1 if t > TB 0 { t TB if t > T B 0

42 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachStructural Homework Breaks Trend Test statistics Perron (1989) Criticism Test statistics Maybe there is more structural breaks but only one big. Perron: do unit root test as always, OLS: p 1 y t = ρy t 1 + Trend t (λ) + φ k y t k + ε t k=1 λ = T B T denotes location of break date. Lagged difference to capture serial correlation. The t-statistics ˆt ρ=1 (λ) = ˆρ(λ) 1 ŜE(ˆρ(λ)) We have more nuisance parameters that affect distribution 1 ˆt ρ=1 (λ) A 0 W λ (r)dw λ ( (r) ) 1 1/2, 0 W λ (r)2 dr W λ is demeaned, detrended, dedummied Brownian motion. The distribution is shifted further left than ADF.

43 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachStructural Homework Breaks Trend Test statistics Perron (1989) Criticism Perron (1989)

44 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachStructural Homework Breaks Trend Test statistics Perron (1989) Criticism Perron (1989)

45 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachStructural Homework Breaks Trend Test statistics Perron (1989) Criticism Perron (1989)

46 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachStructural Homework Breaks Trend Test statistics Perron (1989) Criticism Criticism Data mining: How Perron know the structural break was in 1929? He looked into data. λ must be chosen independently of the data for the correct size of the test (or else there is bias against unit root, Zivot and Andrews, 1992 JBES) size: if H 0 true how often you reject it power: if H 1 is true (H 0 false) how often do you reject Small size and large power is optimal. We normally fix size (e.g.5% size and make test as powerful as possible. t 5% = 3.8 critical value λ chosen after looking at data: choosing λ so that it generates the largest t-statistics test distributions is ever ore shifted so actual size might be bigger. Actual size might be 30% even though was set to 5%.

47 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachStructural Homework Breaks Trend Test statistics Perron (1989) Criticism Criticism: Zivot and Andrews, 1992 Need a model of break date selection procedure (Zivot and Andrews, 1992) ˆλ INF = break dates that produces the largest value of ˆt ρ=1 (λ) over all λs in the sample. ˆt ρ=1 (ˆλ INF ) = inf λ Λ {ˆt ρ=1 (λ)} Zivot and Andrews find that for 8 of 14 Nelson-Plosser series (including GNP) ˆλ INF = 1929 are stationary and the t-statistics is distributed as 1 ˆt ρ=1 (ˆλ INF ) A 0 inf W λ (r)d W λ (r) ( λ Λ ) 1 1/2 0 W λ (r)2 dr. It shifts distribution further left of Perron s.

48 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachDetrending HomeworkTrend/Cycle Example Example: AR(1) Estimation State-Space Form Detrending Need stationary series: Y t = X t β + ε t Granger and Newbold (1974, JoE, Spurious Regressions in Econometrics ) If y t and X t are independent random walk (β = 0), ˆβ OLS non-zero random variable, and ˆt β=0 is large: spurious regression phenomenon. Taking difference instead of levels (so we get stationary series) will bring larger standard errors => cannot reject hypothesis. Detrending still allows to analyze levels. Sometimes we are interested in trend alone.

49 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachDetrending HomeworkTrend/Cycle Example Example: AR(1) Estimation State-Space Form Trend/Cycle Observable series y t y t = τ t + c t τ t is trend, and c t is transitory component, (I(0)). τ t = µ + τ t 1 + η t If trend contains stochastic component, random walk, then if we apply HP we get spurious cycle. We have two unobserved components and if we can model the cycle we can try to use unobserved component estimation.

50 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachDetrending HomeworkTrend/Cycle Example Example: AR(1) Estimation State-Space Form Unobserved Components Approach Watson (1986, JME), Clark (1987, QJE), Morley, Nelson, Zivot (2003, ReStat) Approach: parametric model for c t Model ( Structural") y t = τ t + c t τ t = µ + τ t 1 + η t, η t iidn(0, ση) 2 φ(l)c t = ε t, ε t iidn(0, σ 2 ε), cov(η t, ε t ) = σ εη

51 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachDetrending HomeworkTrend/Cycle Example Example: AR(1) Estimation State-Space Form Problem: Identification We have 1 observable series and 2 unobservable components. To get 2 unobservable components, we need some identification assumptions. Identification: If c t = ε t or c t = φc t 1 + ε t, then σ εη is not identified from the data. There can be infinitely many values of σ εη that would produce the same autocovariance generating function for the first series. However, that does not mean that all values of σ εη are equal. If it is set to zero, it imposes restriction on autocovariance generating function of 1 st differences.

52 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachDetrending HomeworkTrend/Cycle Example Example: AR(1) Estimation State-Space Form Example: AR(1) Example: AR(1) y t = τ t + c t τ t = µ + τ t 1 + η t c t = φc t 1 + ε t Structural model: 5 parameters: µ, ση, 2 σε, 2 φ, σ εη. How many parameters can be identified from data? Reduced-Form First-difference equation y t = τ t + c t (1 L)y t = (1 L)τ t + (1 L)c t y t = µ + η t + (1 L)(1 φl) 1 ε t

53 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachDetrending HomeworkTrend/Cycle Example Example: AR(1) Estimation State-Space Form Example: AR(1) Multiply both sides by (1 φl): (1 φl) y t = (1 φl)µ + (1 φl)η t + (1 L)ε t = c + η t + φη t 1 + ε t ε t 1, c = (1 φ)µ. They are unobserved but we have a sum of two iid series η t + ε t + ( φ)η t 1 + ( 1)ε t 1 The sum of two white noise processes = white noise: same moments as MA(1). So this model is observationally equivalent to y t = c + φ y t 1 + e t + θe t 1 ARMA(1,1) = 4 parameters: c, φ, θ, σ 2 e, that s how many we can estimate. We have 5 parameters but only 4 observed. So far estimates assumes one of parameters fixed.

54 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachDetrending HomeworkTrend/Cycle Example Example: AR(1) Estimation State-Space Form Estimation Assume σ εη = 0 (Watson, Harvey, Clark). => shocks that drive transitory movements are not correlated with those that drive long-run behavior. With this assumption the model can be estimated: 1 Find match (functional) of observed/estimated parameters with the ones from structural model, or 2 Cast the model in a state space form and estimate via Kalman Filter:

55 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachDetrending HomeworkTrend/Cycle Example Example: AR(1) Estimation State-Space Form State-Space Form Observation equation State equation [ τt c t ] = y t = [ 1 1 ] [ τ t c t ] y t = Hβ t [ µ 0 ] [ φ ] [ τt 1 c t 1 ] [ ] ηt +, ε t β t = ˆµ + Fβ t 1 + e t, e t N(0, Q), [ ] σ 2 Q = η 0 0 σε 2

56 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachDetrending HomeworkTrend/Cycle Example Example: AR(1) Estimation State-Space Form Kalman Filter: Results Kalman Filter does not care about how we came up with state form. KF: τ t t and c t t, τ t T, c t T. We say τ t and c t are uncorrelated with each other, by assumption. corr(η t t, ε t t ) = 1 even though we assume corr(η t, ε t ) = 0. In classical approach corr(x t, ˆε t ) = 0 by construction, even though true relationship is corr(x t, ε t ) 0. Estimates of correlation rather than sample correlation of estimates. Identification: If we estimate the model without assuming σ εη Gauss will not converge as there is many numbers of σ εη for which likelihood doesn t decrease.

57 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachDetrending HomeworkTrend/Cycle Example Example: AR(1) Estimation State-Space Form Morley, Nelson and Zivot (2003) RW + AR(2) makes model identified. Why? AR(1) cycles is not observationally different from RW. AR(2) has this feature that cannot be proxied by RW. Morley, Nelson and Zivot (2003): σ εη identified for c t ARMA(p, q), with p q + 2.

58 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachDetrending HomeworkTrend/Cycle Example Example: AR(1) Estimation State-Space Form Example: AR(2) Model: y t = τ t + c t τ t = µ + τ t 1 + η t c t = φ 1 c t 1 + φ 2 c t 2 + ε t 6 parameters: µ, φ 1, φ 2, σ 2 η, σ 2 ε, σ εη. Pre-multiplying both sides with (1 L): y t = (1 L)τ t + (1 L)c t = µ + η t + (1 L)(1 φ 1 L φ 2 L 2 ) 1 ε t (1 φ 1 L φ 2 L 2 ) y t = (1 φ 1 φ 2 )µ + η t φ 1 η t 1 φ 2 η t 2 + ε t ε t 1 The model is observationally equivalent to ARMA(2,2) model: y t ARMA(2, 2) with 6 parameters: c, φ 1, φ 2, θ 1, θ 2, σ 2 e.

59 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachDetrending HomeworkTrend/Cycle Example Example: AR(1) Estimation State-Space Form Results We can map parameters of ARMA(2,2) to our structural model or estimate KF with. [ ] σ 2 Q = η σ εη σ εη For US real GDP, setting σ εη = 0 can be rejected: ρ εη = 0.9. τ t is volatile Structural model with ARMA(3) has 7 structural parameters but is observationally equivalent to reduced-form version ARMA(3,3) with 8 parameters: overidentification. σ 2 ε Not such a big problem; ρ εη < 0 still holds.

60 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachQuestion Homework 1 Homework 1 Written Exercise For an AR(1) with φ = 0.5, calculate and plot the true ACF and PACF. and the IRF for j = 0, 1,..., 5. Do the same for an AR(2) with φ 1 = 1.25 and φ 2 = Eviews question

61 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachQuestion Homework 1 Homework: Question 1 1 Autocorrelation Function, ACF: ρ j γ j γ 0 corr(y t, Y t j ) 2 Partial Autocorrelation Function, PACF: Regression coefficient (for the population) φ kk in k-th order autoregression: Y t = c + φ 1k Y t 1 + φ 2k Y t φ kk Y t k + ε t 3 Impulse Response Function, IRF : X t+j ε t = ψ j

62 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachQuestion Homework 1 Homework: Question 1 Question 1: For an AR(1) with φ = 0.5, calculate and plot the true ACF and PACF and the IRF for j = 0, 1,..., 5. For AR(1), Y t = φy t 1 + ε t : 1 ACF : ρ j = φρ j 1 = ρ j = φ j. 2 PACF: φ 11 = φ and φ kk = 0 for k > 1. 3 IRF : ψ j = φ j. AR(1): φ = 0.5 j ACF(j) PACF(j) IRF(j)

63 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachQuestion Homework 1 Homework: Question 1 For AR(2), Y t = φ 1 Y t 1 + φ 2 Y t 2 + ε t : 1 ACF: Using definition, ρ 1 = ρ 1, and Yule-Walker equation we get: ρ 0 = 1, ρ 1 = φ1 1 φ 2, and ρ j = φ 1 ρ j 1 + φ 2 ρ j 2 for j > 1. 2 PACF: the vector φ (k) can be calculated from φ (k) 1 γ 0 γ 1 γ k 1 φ (k) 2. = γ 1 γ 0 γ k φ (k) γ k 1 γ k 2 γ 0 k Using ρ j = γj γ 0 and [ ] φ (2) 1 φ (2) 2 = [ ] 1 [ ] γ0 γ 1 γ1 = γ 1 γ 0 γ 2 1 [ ρ1 ρ 1ρ 2 1 ρ 2 1 ρ 2 ρ ρ 2 1 γ 1 γ 2. γ k ], φ (1) 1 = γ1 γ 0, φ (2) 2 = ρ2 ρ2 1 1 ρ 2 1 and φ (k) k = 0 for k > 2.

64 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachQuestion Homework 1 Homework: Question 1 For AR(2), Y t = φ 1 Y t 1 + φ 2 Y t 2 + ε t : 3 IRF: According to Yule-Walker equation for AR(2) ψ j = φ 1 ψ j 1 + φ 2 ψ j 2 or recursively solving Y t Y t = (φ φ 3 1φ 2 + 3φ 1 φ 2 2)Y t 5 + (φ 4 1φ 2 + 3φ 2 1φ φ 3 2)Y t 6 + (φ φ 2 1φ 2 + φ 2 2)ε t 4 + (φ φ 1 φ 2 )ε t 3 + (φ φ 2 )ε t 2 + φ 1 ε t 1 + ε t

65 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachQuestion Homework 1 Homework: Question 1 Question 1: For an AR(2) with φ 1 = 1.25 and φ 2 = 0.75, calculate and plot the true ACF and PACF and the IRF for j = 0, 1,..., 5. AR(2): φ 1 = 1.25, φ 2 = 0.75 j ACF PACF IRF

66 DF Phillips-Perron Stationarity Tests Variance Ratio Structural Break UC ApproachQuestion Homework 1 Homework: Question 1 Question 1: IRF ACF PACF