Topic 4 Unit Roots. Gerald P. Dwyer. February Clemson University

Transcription

1 Topic 4 Unit Roots Gerald P. Dwyer Clemson University February 2016

2 Outline 1 Unit Roots Introduction Trend and Difference Stationary Autocorrelations of Series That Have Deterministic or Stochastic Trends Removing The Trend Why Care About Unit Roots? Dickey-Fuller tests Augmented Dickey-Fuller Tests Lag length selection ARMA model and Autoregressions Number of Crossings of Mean Summary

3 Classic time series decomposition Trend, cyclical and seasonal

4 Slutzky and others Stochastic processes look much like cycles y t = α 0 + β i ε t i Irregular cycle

5 Deterministic trend Deterministic trend only y t = α 0 + κt Deterministic trend with white-noise innovation y t = α 0 + κt + ε t, E ε t = 0, E ε 2 t = σ 2, E ε t ε s = 0 t = s Also can have y t = α 0 + α 1 y t 1 + α 2 y t 2 + κt + ε t, E ε t = 0, E ε 2 t = σ 2, E ε t ε s = 0 t = s

6 Deterministic trend only

7 Deterministic trend with stochastic innovations

8 Trend stationary A series is trend stationary if it can be written y t = κt + stationary component Examples of possibly stationary terms are ε t, E ε t = 0, E ε 2 t = σ 2, E ε t ε s = 0 t = s α 1 y t 1 + α 2 y t 2 β 1 ε t 1 + β 2 ε t 2

9 Stochastic trend A series has a stochastic trend if it has a component that can be written y t = y 0 + i=1 ε t i + stationary component with ε t white noise A series with a stochastic trend and no other aspect can be written y t = ε t, E ε t = 0, E ε 2 t = σ 2, E ε t ε s = 0 t = s Can see the relationship of the two representations from y t = y t 1 + ε t y t = y t 2 + ε t + ε t 1 y t = y t 3 + ε t + ε t 1 + ε t 2 y t = y t (k+1) + ε t + ε t 1 + ε t ε t k

10 Stochastic trend alone A series with a stochastic trend and no other aspect can be written Which results in y t = ε t, E ε t = 0, E ε 2 t = σ 2, E ε t ε s = 0 t = s t 1 y t = y 0 + i=0 ε t i The term t 1 i=0 ε t i is called a stochastic trend because the future of this apparent trend is not predictable

11 Stochastic trend and deterministic trend Deterministic trend is called that because deterministic means perfectly predictable from its own past and a linear trend is perfectly predictable y t = κt implies E t y t+100 = κ (t + 100) Stochastic trend is called that because this trend is not predictable y t = y 0 + t 1 i=0 ε t i implies E t y t+100 = y 0 + E t t i=0 t 1 ε t i = y 0 + E t ε t i = y t i=0

12 Stochastic trend

13 Random walk with drift Random walk with drift is y t = α 0 + y t 1 + ε t, E ε t = 0, E ε 2 t = σ 2, E ε t ε s = 0 t = s Stationary in terms of first differences Evolution over time y t = α 0 + ε t t 1 y t = y 0 + α 0 t + i=0 ε t i α 0 t is a deterministic trend t 1 i=0 ε t i is a stochastic trend Predictable increases E y t = α 0

14 Random Walk with Drift

15 Autocorrelation of stochastic trend Series with a stochastic trend is not stationary or strictly stationary Random walk is y t = y t 1 + ε t, E ε t = 0, E ε 2 t = σ 2, E ε t ε s = 0 t = s Expected value of random walk If condition expected value on y 0, that is if use E [y t y 0 ], then E [y t y 0 ] = y 0 + E [ ] t 1 ε t i y 0 = y 0 i=0 If not conditioned on y 0, then complicated problem if even well posed

16 Autocorrelation of stochastic trend Random walk is y t = y t 1 + ε t, E ε t = 0, E ε 2 t = σ 2, E ε t ε s = 0 t = s Variance of random walk Var [y t ] = Var [ y 0 + ] t 1 ε t i i=0 If condition variance on y 0, that is if use Var [y t y 0 ], then Var [y t y 0 ] = tσ 2 Also lim Var [y t y 0 ] = t

17 Autocovariance of stochastic trend Random walk is y t = y t 1 + ε t, E ε t = 0, E ε 2 t = σ 2, E ε t ε s = 0 t = s Covariance if condition on y 0 and have s > 0 [ ] t 1 t s 1 Cov [y t, y t s y 0 ] = Cov y 0 + ε t i, y 0 + ε t i y 0 = (t s) σ 2 i=0 i=0

18 Autocorrelation of stochastic trend Random walk is y t = y t 1 + ε t, E ε t = 0, E ε 2 t = σ 2, E ε t ε s = 0 t = s Autocorrelation ρ = Cov [y t, y t s ] SD [y t ] SD [y t s ] = (t s) σ 2 tσ 2 (t s) σ 2 (t s) t s = = t t s t If t is large relative to s, then the autocorrelations will be close to one Do fall but rather slowly if t is large relative to s

19 Autocorrelation of trend stationary series A trend stationary series is not covariance stationary A trend stationary series y t = κt + ε t, E ε t = 0, E ε 2 t = σ 2, E ε t ε s = 0 t = s The expected value of y t is E y t = κt which implies E y 1 = κ, E y 2 = 2κ, etc. Suppose we have T observations and for this series we compute y = T t=1 y t T = T t=1 κt T = κt

20 Autocorrelation of trend stationary series A trend stationary series y t = κt + ε t, E ε t = 0, E ε 2 t = σ 2, E ε t ε s = 0 t = s The expected value of the variance for T observations is s 2 = T 1 T t=1 (y t κt) 2 = T 1 T (κt + ε t κt) 2 t=1 = κt 1 T (κt κt + ε t ) 2 t=1 E s 2 = κ 2 T t=1 (t t) 2 + σ 2 T

21 Autocorrelation of trend stationary series A trend stationary series y t = κt + ε t, E ε t = 0, E ε 2 t = σ 2, E ε t ε s = 0 t = s The series is nonstationary by construction Simpler to examine test statistics for examples than to do cumbersome and problematic algebra for autocorrelations

22 Removing the trend A deterministic trend can be removed by y t α 0 κt = stationary component A stochastic trend can be removed by first differencing y t = y t y t 1 = stationary component A stochastic and deterministic trend can be removed from a series by first differencing and subtracting the mean y t α 0 = y t y t 1 α 0 = stationary component Note that y t is covariance stationary for a random walk with drift or a series that consists of a random walk with drift plus a stationary component

23 Implication of unit roots Suppose two series y t and z t are not related but both have unit roots Run regression y t = a 0 + a 1 z t + e t Then more likely to find a statistically significant relationship than usual five percent of the time As likely to find a positive as a negative relationship This implies this problem would not matter so much with lots of series on data (e.g. many series on M2 for the United States in the postwar period; almost a contradiction in terms) Residuals likely to be highly serially correlated Granger called this a spurious regression problem If series have unit roots, then simple Ordinary Least Squares with usual test statistics is a misleading way to analyze data

24 Common tests for unit roots Dickey-Fuller unit root tests Augmented Dickey-Fuller unit root tests

25 The Dickey-Fuller tests Dickey-Fuller unit root tests Null hypothesis of most general test y t = α 0 + ε t, E ε t = 0, E ε 2 t = σ 2, E ε t ε s = 0 t = s Random walk with drift Has a trend as we saw above Can be rewritten as y t = α 0 + y t 1 + ε t Alternative hypothesis y t = α 0 + α 1 y t 1 + βt + ε t, ε t has properties above α 1 < 1 Also has a trend This is a suitable comparison for a series with an evident upward trend For example, GDP or population in the United States

26 Dickey-Fuller test with trend Why not just run an OLS regression such as y t = α 0 + α 1 y t 1 + βt + e t and test whether α 1 = 1, using ( α 1 1) /SD( α 1 ) where SD( α 1 ) is the usual OLS standard error? Could also test β = 0 at the same time but everyone focuses on α 1 The asymptotic distribution of α 1 /SD( α 1 ) is not N(0,1)

27 Dickey-Fuller test with trend Properties of OLS regression estimating y t = α 0 + α 1 y t 1 + βt + e t α 1 estimated by OLS is a consistent estimator of α 1 In fact, it is super consistent: The estimator converges to α 1 at the rate T instead of the usual rate T 2 1 Faster convergence and therefore called a super-consistent estimator But ( α 1 1) /SD( α 1 ) a N(0,1) The distribution of α 1 /SD( α 1 ) is the Dickey-Fuller distribution Dickey and Fuller tabulated this distribution

28 Dickey-Fuller test with trend Use statistic τ τ for the test whether the coefficient α 1 is one in y t = α 0 + α 1 y t 1 + βt + e t Run what is casually called the Dickey-Fuller regression y t = α 0 + ( α 1 1) y t 1 + βt + e t Just subtract y t 1 from both sides of the first regression to get this one Test whether α 1 1 α τ = 0 Now testing whether a coefficient is zero The t-statistic for this test is the usual one from a regression program and can be compared with the Dickey-Fuller distribution

29 Use tabulated distribution of test statistics Use statistic τ τ for the test whether the coefficient α τ = 0 in y t = α 0 + α τ y t 1 + βt + e t The t-statistic can be read from the regression output and compared with the Dickey-Fuller distribution of τ τ From Wayne Fuller, Introduction to Statistical Time Series, Second edition

30 Tabulated distribution for ττ

31 Distributions Can use p-values from programs Distributions allow for heteroskedasticity and serial correlation of error term Computers much faster than they were in the 1970s

32 Summary of Dickey-Fuller test with trend Use statistic τ τ for the test whether the coefficient α 1 is one in y t = α 0 + α 1 y t 1 + βt + ε t Run a Dickey-Fuller regression y t = α 0 + α τ y t 1 + βt + e t Test whether α τ = 0 using usual t-statistic from OLS program Compare t-statistic with the Dickey-Fuller distribution

33 Dickey-Fuller test without trend and with nonzero mean Null hypothesis is Alternative hypothesis y t = y t 1 + ε t Neither representation has a trend y t = α 0 + α 1 y t 1 + ε t, α 1 < 1

34 Dickey-Fuller test without trend and with nonzero mean Null hypothesis is y t = y t 1 + ε t Alternative hypothesis y t = α 0 + α 1 y t 1 + ε t, α 1 < 1 Subtract y t 1 from both sides of the regression and get y t = α 0 + (α 1 1) y t 1 + ε t y t = α 0 + α µ y t 1 + ε t Subtracting y t 1 from both sides of the first regression is a linear transformation that does not change the residual and therefore does not affect statistical properties Test whether α 1 1 α µ = 0 Again just testing whether a coefficient is zero The t-statistic for this test is the usual one from a regression program and can be compared with the Dickey-Fuller distribution of τ µ

35 Tabulated distribution for τµ

36 Dickey-Fuller test with zero mean Null hypothesis is Alternative hypothesis y t = y t 1 + ε t y t = α 1 y t 1 + ε t, α 1 < 1 Neither representation has a trend and the autoregression has mean zero

37 Dickey-Fuller test with zero mean Null hypothesis is Alternative hypothesis y t = y t 1 + ε t y t = α 1 y t 1 + ε t, α 1 < 1 Subtract y t 1 from both sides of the regression and get y t = (α 1 1) y t 1 + ε t y t = αy t 1 + ε t Test whether α 1 1 α = 0 Again just testing whether a coefficient is zero The t-statistic for this test is the usual one from a regression program and can be compared with the Dickey-Fuller distribution of τ

38 Augmented Dickey-Fuller tests Augmented Dickey-Fuller tests includes additional lags Illustrate without trend because the principle is the same for the alternative assumptions

39 Augmented Dickey-Fuller test with nonzero mean and no trend Null hypothesis of at least one unit root y t = Alternative hypothesis y t = α 0 + k α i y t i + ε t, i=1 k i=1 α i y t i + ε t, k α i = 1 i=1 k i=1 α i < 1 Neither representation has a trend and the autoregression has a nonzero mean

40 Augmented Dickey-Fuller test with nonzero mean and no trend Alternative hypothesis y t = α 0 + k i=1 α i y t i + ε t, k i=1 α i < 1 Subtract y t 1 from both sides of the regression and get y t = α 0 + (α 1 1) y t 1 + k i=2 α i y t i + ε t Don t care whether α 1 1 = 0 Want joint deviation of 1 from α 1 and all other α i s because the condition for a unit root is k i=1 α i 1 = 0 Can do this if rewrite in terms of lagged changes

41 Augmented Dickey-Fuller test with nonzero mean and no trend Null hypothesis of at least one unit root with second-order autoregression rewritten in terms of first difference on left y t = α 0 + (α 1 1) y t 1 + α 2 y t 2 + ε t

42 Augmented Dickey-Fuller test with nonzero mean and no trend Null hypothesis of at least one unit root with second-order autoregression rewritten in terms of first difference on left y t = α 0 + (α 1 1) y t 1 + α 2 y t 2 + ε t Subtract α 2 y t 1 from the term in parentheses and add it to the right-hand side of the equation y t = α 0 + (α 1 + α 2 1) y t 1 α 2 y t 1 + α 2 y t 2 + ε t

43 Augmented Dickey-Fuller test with nonzero mean and no trend Have y t = α 0 + (α 1 + α 2 1) y t 1 α 2 y t 1 + α 2 y t 2 + ε t

44 Augmented Dickey-Fuller test with nonzero mean and no trend Have y t = α 0 + (α 1 + α 2 1) y t 1 α 2 y t 1 + α 2 y t 2 + ε t which can be written y t = α 0 + (α 1 + α 2 1) y t 1 α 2 (y t 1 y t 2 ) + ε t = α 0 + (α 1 + α 2 1) y t 1 α 2 y t 1 + ε t

45 Augmented Dickey-Fuller test with nonzero mean and no trend Have y t = α 0 + (α 1 + α 2 1) y t 1 α 2 y t 1 + α 2 y t 2 + ε t which can be written y t = α 0 + (α 1 + α 2 1) y t 1 α 2 (y t 1 y t 2 ) + ε t = α 0 + (α 1 + α 2 1) y t 1 α 2 y t 1 + ε t Under the null hypothesis of a unit root, the estimated coefficient of y t 1 is zero

46 Augmented Dickey-Fuller test with nonzero mean and no trend Have y t = α 0 + (α 1 + α 2 1) y t 1 α 2 y t 1 + α 2 y t 2 + ε t which can be written y t = α 0 + (α 1 + α 2 1) y t 1 α 2 (y t 1 y t 2 ) + ε t = α 0 + (α 1 + α 2 1) y t 1 α 2 y t 1 + ε t Under the null hypothesis of a unit root, the estimated coefficient of y t 1 is zero Same distribution of test statistic as with only one lag and just use table as before

47 Augmented Dickey-Fuller test with nonzero mean and no trend What if three lags? Can rewrite as y t = α 0 + α 1 y t 1 + α 2 y t 2 + α 2 y t 3 + ε t y t = α 0 + (α 1 + α 2 + α 3 1) y t 1 α 2 y t 1 α 3 y t 2 + ε t and so on DO three lagged levels yourself In general, estimate y t = α 0 + α µ y t 1 + k i=1 α i y t i + ε t Test whether α µ = 0 to see whether there is at least one unit root

48 More than one unit root What if can t reject null hypothesis of one unit root in y t = α 0 + α 1 y t 1 + α 2 y t 3 + α 2 y t 3 + ε t There could be another Run augmented Dickey-Fuller test on second differences 2 y t = α 0 + α µ y t 1 + k i=1 α i 2 y t i + ε t Test whether α µ = 0 to see whether there is at least one unit root in this equation If so, then there are at least two unit roots in the data Practically speaking, not that common to find two unit roots and I at least have not seen an inability to reject three unit roots

49 More than one unit root Alternative strategy in the textbook is to test whether there are, for example, at least two unit roots first If reject null hypothesis, then test whether there is at least one unit root What is the difference? My suggested strategy will find too few unit roots sometimes Text s suggested strategy will find too many unit roots sometimes Which are you most worried about? Null hypothesis is at least N unit roots I am inclined to think this rejects the null hypothesis too seldom when the alternative is correct So I am more concerned about finding too many unit roots rather than too few

50 How many lags? Estimate a well behaved difference equation under the null hypothesis of a unit root Test for lag length y t = α 0 + k i=1 α i y t i + ε t

51 How many lags? Estimate a well behaved difference equation under the alternative hypothesis of not a unit root y t = α 0 + k i=1 α i y t i + ε t This is the same as estimating and testing with Test for lag length k 1 y t = α 0 + α y t 1 + i=1 α i y t i + ε t

52 Which is better? I If the series has a unit root, it makes little difference whether or not you include the lagged level If the series does not have a unit root, then the mis-specified first difference equation is likely to include many lags to reduce the estimated residuals to low autocorrelation y t = α 0 + First differenced, this becomes y t = k i=1 α i y t i + ε t k α i y t i + ε t i=1 α (L) y t = (1 L) ε t

53 Which is better? II Dividing by (1 γ L) with γ close to one yields roughly serially uncorrelated residuals (1 γ L) 1 α (L) y t = (1 L) (1 γ L) ε t where the error term is closer to serially uncorrelated as γ approaches one and (1 γ L) 1 = 1 + γ L +γ 2 L 2 +γ 3 L indicate many lags with large coefficients are included in the equation

54 Determination of lag length by successive regressions Sequential regression strategy to determine lag length Use t-ratio on last estimated coefficient in regression initially longer than plausible 1 Select lag length k greater than plausible 2 Estimate regression of lag length k 3 If last coefficient statistically significant using t-ratio, stop; otherwise reduce lag length (k) by one and go to 2 Will include too many lags if use typical 5 percent significance level for each coefficient

55 AIC, BIC and Hannan-Quinn for determining lag length Normal distribution of innovations with k parameters Akaike information criterion AIC (k) = ln σ 2 k + 2k T (Schwarz) Bayesian information criterion text calls it SBC BIC (k) = ln σ 2 k + k ln T T Hannan-Quinn HQ = ln σ 2 k + 2k ln (ln T ) T

56 General definitions of AIC, BIC and Hannan-Quinn Akaike information criterion AIC (k) = 2 ln L k + 2k (Schwarz) Bayesian information criterion Hannan-Quinn BIC (k) = 2 ln L k + k ln T HQ(k) = 2 ln L T + 2k T ln(ln T )

57 Which criterion is best? Successive reduction in lag length, AIC and BIC could pick different models depending on which criterion used AIC and BIC most commonly used in Economics in recent years Which is better? BIC is consistent AIC overestimates lag length BIC tends to have more variability than AIC in number of lags across samples It s a matter of preference to some degree which to use as far as I can tell I prefer too many lags to too few, so have tended to use t-ratios to reduce lag length AIC common choice given current knowledge In my experience, lag lengths typically are the same or close Very important practical point: Keep T fixed across estimates in all estimates

58 Relationship of autoregressions to ARMA models How is the AR(k) related to an ARMA(p,q)? y t = α 0 + k i=1 α i y t i + ε t

59 Relationship of autoregressions to ARMA models How is the AR(k) related to an ARMA(p,q)? y t = α 0 + k i=1 α i y t i + ε t Invert MA part and approximate length of finite-order ARMA(p,q) by AR(k)

60 Relationship of autoregressions to ARMA models How is the AR(k) related to an ARMA(p,q)? y t = α 0 + k i=1 α i y t i + ε t Invert MA part and approximate length of finite-order ARMA(p,q) by AR(k) Start with ARMA model with constant term suppressed α (L) y t = β (L) ε t

61 Relationship of autoregressions to ARMA models How is the AR(k) related to an ARMA(p,q)? y t = α 0 + k i=1 α i y t i + ε t Invert MA part and approximate length of finite-order ARMA(p,q) by AR(k) Start with ARMA model with constant term suppressed α (L) y t = β (L) ε t Then ARMA(p,q) can be written β (L) 1 α (L) y t = ε t

62 Relationship of autoregressions to ARMA models How is the AR(k) related to an ARMA(p,q)? y t = α 0 + k i=1 α i y t i + ε t Invert MA part and approximate length of finite-order ARMA(p,q) by AR(k) Start with ARMA model with constant term suppressed α (L) y t = β (L) ε t Then ARMA(p,q) can be written β (L) 1 α (L) y t = ε t In general, this equation has an infinite number of lags

63 Relationship of autoregressions to ARMA models Equation β (L) 1 α (L) y t = ε t

64 Relationship of autoregressions to ARMA models Equation β (L) 1 α (L) y t = ε t Can approximate the infinite number of lags in this equation by a finite number of lags which increases with the sample size

65 Relationship of autoregressions to ARMA models Equation β (L) 1 α (L) y t = ε t Can approximate the infinite number of lags in this equation by a finite number of lags which increases with the sample size If the number of lags approaches infinity as the number of observations approaches infinity and the ratio of the number of lags to the number of observations goes to zero, then can estimate the coefficients of lagged variables consistently lim k = and T lim k T T = 0

66 Relationship of autoregressions to ARMA models Equation β (L) 1 α (L) y t = ε t Can approximate the infinite number of lags in this equation by a finite number of lags which increases with the sample size If the number of lags approaches infinity as the number of observations approaches infinity and the ratio of the number of lags to the number of observations goes to zero, then can estimate the coefficients of lagged variables consistently lim k = and T lim k T T = 0 For example k = T 1/3 (Said and Dickey 1984)

67 French real exchange rate (log with zero mean)

68 Unit root and graph The number of crossings of the mean is an indicator of whether there is much evidence of a unit root A stationary autoregression reverts to the mean from any deviation from the mean A random walk has no tendency to revert to the sample mean Can build a test based on this Burridge and Guerre (1997, Econometric Theory) Generalization by García and Sansó (2006, Econometric Theory) Not more powerful than Dickey-Fuller test Logic is informative for an impression of the likely result of a test for a unit root

69 Summary I Trend stationary versus difference stationary Series has a unit root First autocorrelation close to one Slow decay of autocorrelations if a series has a unit root Series has a deterministic trend First autocorrelation close to one Slow decay of autocorrelations if a series has a deterministic trend Use unit root tests to distinguish between stationary series and those with a unit root Augmented Dickey-Fuller tests are fairly standard Often footnote material Lag length selection of some importance AIC and successive t-tests probably are most common way to pick lag lengths Why not estimate ARMA model to test for unit root? Autoregressions can approximate any ARMA model Number of crossings of mean is related to unit-root behavior Tests based on this perform similarly to augmented Dickey-Fuller tests