Central Bank of Chile October 29-31, 2013 Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91. Bruce E.

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1 Forecasting Lecture 3 Structural Breaks Central Bank of Chile October 29-31, 2013 Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91 Bruce E. Hansen

2 Organization Detection of Breaks Estimating Breaks Forecasting after Breaks Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

3 Types of Breaks Breaks in Mean Breaks in Variance Breaks in Relationships Single Breaks Multiple Breaks Continuous Breaks Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

4 Example Simple AR(1) with mean and variance breaks y t = ρy t 1 + µ t + e t e t N(0, σ 2 t ) Ey t = var(y t ) = µ t 1 ρ σ 2 t 1 ρ 2 µ t and/or σ 2 t may be constant or may have a break at some point in the sample Sample size n Questions: Can you guess: Is there a structural break? If so, when? Is the shift in the mean or variance? How large do you guess? Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

5 Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

6 Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

7 Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

8 Terminology Sample Period: t = 1,..., n Breakdate: T 1 Date of change Breakdate fraction: τ 1 = T 1 /n Pre-Break Sample: t = 1,..., T 1 T 1 observations Post-Break Sample: t = T 1 + 1,..., n n T 1 observations Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

9 Structural Break Model Full structural break y t = β 1 x t + e t, t T 1 y t = β 2 x t + e t, t > T 1 or y t = β 1 x t1 (t T 1 ) + β 2 x t1 (t > T 1 ) + e t Partial structural break y t = β 0 z t + β 1 x t1 (t T 1 ) + β 2 x t1 (t > T 1 ) + e t Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

10 Variance Break Model y t = β x t + e t, var (e t ) = σ 2 1, t T 1 var (e t ) = σ 2 2, t > T 1 Breaks do not necessarily affect point forecasts Affects forecast variances, intervals, fan charts, densities Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

11 Detection of Breaks Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

12 Testing for a Break Classic Test (Chow) Assume T 1 is known Test H 0 : β 1 = β 2 Use classic linear hypothesis test (F, Wald, LM, LR) Least-Squares y t = β 0 z t + β 1 x t1 (t T 1 ) + β 2 x t1 (t > T 1 ) + ê t Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

13 Full Break Model Y 1 = X 1 β 1 + e 1 Y 2 = X 1 β 2 + e 2 β 1 = (X 1X 1 ) 1 (X 1Y 1 ) β 2 = (X 2X 2 ) 1 (X 2Y 2 ) ê 1 = Y 1 X 1 β1 ê 2 = Y 2 X 2 β2 SSE (T 1 ) = ê 1ê 1 + ê 2ê 2 ˆσ 2 (T 1 ) = 1 (ê n m 1 ê 1 + ê 2ê ) 2 Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

14 F Test Statistic F test F (T 1 ) = (SSE SSE (T 1)) /k SSE (T 1 )(n m) where k = dim(β 1 ), m =all parameters, (full sample estimate) SSE = ẽ ẽ σ 2 = 1 (ẽ ẽ ) n k ẽ = Y X β F test assumes homoskedasticity, better to use Wald test Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

15 Wald Test Statistic W (T 1 ) = n ( β1 β ) ( ) n n 1 2 V 1 + V 2 ( β1 T 1 n T β ) 2 1 where V 1 and V 2 are standard asymptotic variance estimators for β 1 and β 2 (on the split samples: V 1 = Q 1 1 Ω 1 Q 1 1 V 2 = Q 1 2 Ω 2 Q 1 2 Q 1 = 1 T 1 X 1X 1 Q 2 = 1 n T 1 X 2X 2 Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

16 HAC variance options For iid e t Ω1 = σ 2 Q1 For homoskedastic (within regiome ˆσ 2 1 = ˆσ 2 2 = Ω 2 = σ 2 Q 2 1 (ê ) T 1 k 1 ê 1 1 (ê ) n T 1 k 2 ê 2 For serially uncorrelated but possibly heteroskedastic Ω 1 = Ω 2 = T 1 1 T 1 k x t x tê t 2 t=1 n 1 n T 1 k x t x tê t 2 t=t 1 +1 Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

17 For serially correlated (e.g. h > 1) Ω 1 = Ω 2 = T 1 1 T 1 k x t x tê t 2 t=1 + 1 T 1 k h 1 j=0 n 1 n T 1 k x t x tê t 2 t=t n T 1 k T 1 j ( xt x t+jê t ê t+j + x t+j x ) tê t+j ê t t=1 h 1 j=0 n j t=t 1 +1 ( xt x t+jê t ê t+j + x t+j x tê t+j ê t ) Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

18 Classic Theory Under H 0, if the number of observations pre- and post-break are large, then F (T 1 ) d χ 2 k k under homoskedasticity, and in general W (T 1 ) d χ 2 k We can reject H 0 in favor of H 1 if the test exceeds the critical value Thus find a break if the test rejects Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

19 Modern Approach Break dates are unknown Sup tests (Andrews, 1993) SupF = sup F (T 1 ) T 1 SupW = sup W (T 1 ) T 1 The sup is taken over all break dates T 1 in the region [t 1, t 2 ] where t 1 >> 1 and t 2 << n The region [t 1, t 2 ] are candidate breakdates. If the proposed break is too near the beginning or end of sample, the estimates and tests will be misleading Recommended rule t 1 = [.15n], t 2 = [.85n] Numerically, calculate SupW using a loop Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

20 Example US GDP Quarterly data 1960:2011 k = 7 Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

21 Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

22 Evidence for Structural Break? SupW=27 Is this significant? Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

23 Theorem (Andrews) Under H 0, if the regressors x t are strictly stationary, then SupF, SupW, etc, converge to non-standard asymptotic distributions which depend on k (the number of parameters tested for constancy π 1 = t 1 /n π 2 = t 2 /n Only depend on π 1 and π 2 through λ = π 2 (1 π 1 )/(π 1 (1 π 2 )) Critical values in Andrews (2003, Econometrica, pp ) p-value approximation function in Hansen (1997 JBES, pp 60-67) Critical values much larger than chi-square Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

24 Evidence for Structural Break? SupW=27 k = 7 1% asymptotic critical value = Asymptotic p-value=0.008 Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

25 Non-Constancy in Marginal or Conditional? The model is y t = β 0 z t + β 1 x t1 (t T 1 ) + β 2 x t1 (t > T 1 ) + e t The goal is to check for non-constancy in the conditional relationship (in the coeffi cients β) while being agnostic about the marginal (the distribution of the regressors x t ) Andrews assume that x t are strictly stationary, which excludes structural change in the regressors In Hansen (2000, JoE) I show that this assumption is binding If x t has a structural break in its mean or variance, the asymptotic distribution of the SupW test changes This can distort inference (a large test may be due to instability in x t, not regression instability) There is a simple solution: Fixed Regressor Bootstrap Requires h = 1 (no serial correlation) Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

26 Fixed Regressor Bootstrap Similar to a bootstrap, a method to simulate the asymptotic null distribution Fix (z t, x t, ê t ), t = 1,..., n Let yt be iid N(0, êt 2 ), t = 1,..., n Estimate the regression y t = β 0 z t + β 1 x t1 (t T 1 ) + β 2 x t1 (t > T 1 ) + êt Form the Wald, SupW statistics on this simulated data W (T 1 ) = n ( β 1 (T 1) β ) ( 2 (T 1) V 1 (T 1 ) n + V n 2 (T 1 ) T 1 n T 1 ( β 1 (T 1) β ) 2 (T 1) SupW = sup T 1 W (T 1 ) ) 1 Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

27 Repeat this B 1000 times. Let SupWb denote the b th value Fixed Regressor bootstrap p-value p = 1 B N 1 (SupWb SupW ) b=1 Fixed Regressor bootstrap critical values are quantiles of empirical distribution of SupW b Important restriction: Requires serially uncorrelated errors (h = 1) Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

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29 Evidence for Structural Break? SupW=27 Asymptotic p-value=0.008 Fixed regressor bootstrap p-value=0.106 Bootstrap eliminates significance! Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

30 Recommendation In small samples, the SupW test highly over-rejects The Fixed Regressor Bootstrap (h = 1) greatly reduces this problem Furthermore, it makes the test robust to structural change in the marginal distribution For h > 1, tests not well investigated Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

31 Testing for Breaks in the Variance y t = β x t + e t, var (e t ) = σ 2 1, t T 1 var (e t ) = σ 2 2, t > T 1 Since var (e t ) = Eet 2, this is the same as a test for a break in a regression of et 2 on a constant Estimate constant-parameter model y t = β x t + ê t Obtain squared residuals ê 2 t Apply Andrews SupW test to a regression of êt 2 k = 1 critical values on a constant Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

32 GDP Example: Break in Variance? Apply test to squared OLS residuals Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

33 Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

34 Break in Variance? SupW=15.96 k = 1 1% asymptotic critical value =12.16 Asymptotic p-value=0.002 Fixed regressor bootstrap p-value=0.000 Strong rejection of constancy in variance Great moderation Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

35 End-of-Sample Breaks Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

36 End-of-Sample Breaks The SupW tests are powerful against structural changes which occur in the interior of the sample T 1 [.15n,.85n] Have low power against breaks at the end of the sample Yet for forecasting, this is a critical period Classic Chow test allows for breaks at end of sample But requires finite sample normality New end-of-sample instability test by Andrews (Econometrica, 2003) Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

37 End-of-Sample Test Write model as Y 1 = X 1 β 1 + e 1 Y 2 = X 1 β 2t + e 2 where Y 1 is n 1, Y 2 is m 1 and X has k regressors m is known but small Test is for non-constancy in β 2t Let β be full sample (n + m) LS estimate, ê = Y X β full-sample residuals Partition ê = (ê 1, ê 2 ) Test depends on Σ = E (e 2 e 2 ) First take case Σ = I m σ 2 Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

38 Ifm d If m < d S = ê 2X 2 ( X 2 X 2 ) 1 X 2 ê 2 P = ê 2X 2 X 2ê 2 If m is large we could use a chi-square approximation But when m is small we cannot Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

39 Andrews suggests a subsampling-type p-value p = n m+1 1 n m + 1 j=1 1 (S S j ) S j = ê j ( ) X j X 1 j X j X j ê j X j = {x t : t = j,..., j + m 1} Y j = {y t : t = j,..., j + m 1} ê j = Y j X j β(j) and β (j) is least-squares using all observations except for t = j,..., j + [m/2] Similar for P test You can reject end-of-sample stability if p is small (less than 0.05) Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

40 Weighted Tests Andrews suggested improved power by exploiting correlation in e 2 where S = ê 2 ˆΣ 1 X 2 ( X 2 ˆΣ 1 X 2 ) 1 X 2 ˆΣ 1 ê 2 ˆΣ = 1 n+1 ( ) ( ) n + 1 Y j X j β Y j X j β j=1 The subsample calculations are the same as before except that S j = ê j ˆΣ 1 X j ( X j ˆΣ 1 X j ) 1 X j ˆΣ 1 ê j Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

41 Example: End-of-Sample Instability in GDP Forecasting? m = 12 (last 3 years) S statistics (p-values) Unweighted Weighted h = h = h = h = h = h = h = h = Evidence does not suggest end-of-sample instability Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

42 Breakdate Estimation Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

43 Breakdate Estimation The model is a regression y t = β 0 z t + β 1 x t1 (t T 1 ) + β 2 x t1 (t > T 1 ) + e t Thus a natural estimator is least squares The SSE function is S(β, T 1 ) = 1 n n ( yt β 0 z t β 1 x t1 (t T 1 ) β 2 x t1 (t > T 1 ) ) 2 t=1 ( β, T 1 ) = argmin S(β, T 1 ) The function is quadratic in β, nonlinear in T 1 Convenient solution is concentration Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

44 Least-Squares Algorithm where ( β, T 1 ) = argmin β,t 1 S(β, T 1 ) = argmin min S(β, T 1 ) T β 1 = argmin S(T 1 ) T 1 S(T 1 ) = min β S(β, T 1 ) = 1 n and ê t (T 1 ) are the OLS residuals from with T 1 fixed. n ê t (T 1 ) 2 t=1 y t = β 0 z t + β 1 x t1 (t T 1 ) + β 2 x t1 (t > T 1 ) + ê t (T 1 ) Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

45 Least-Squares Estimator T 1 = argmin T 1 S(T 1 ) For each T 1 [t 1, t 2 ], estimate structural break regression, calculate residuals and SSE S(T 1 ) Find T 1 which minimizes S(T 1 ) Even if n is large, this is typically a quick calculation. Plots of S(T 1 ) against T 1 are useful The sharper the peak, then better T 1 is identified Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

46 Example: Breakdate Estimation in GDP Plot SSE as function of breakdate Break Date Estimate is lowest point of graph Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

47 Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

48 Break Date Estimate T 1 (minimum of SSE (T 1 ) = 1980:4 (82nd observation) Minimum not well defined Consistent with weak break or no break Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

49 Example: Breakdate Estimation for GDP Variance Plot SSE as function of breakdate Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

50 Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

51 Break Date Estimate for Variacne ˆT 1 (minimum of SSE (T 1 ) = 1983:4 (93rd observation) Well defined minimum Sharp V shape Consistent with strong single break Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

52 Distribution Theory and Confidence Intervals Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

53 Distribution of Break-Date Estimator Bai (Review of Economics and Statistics, 1997) Define Q = E (x t x t ) Ω = E ( x t x t ) e2 t δ = β 2 β 1 Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

54 Theorem [If δ 0, and the distribution of (x t, e t ) does not change at T 1, then] ( δ Qδ ) 2 ( ) T 1 T 1 δ Ωδ [ d ζ = argmax s W (s) s 2 where W (s) is a double-sided Brownian motion. The distribution of ζ for x 0 is x G (x) = 1 + ( 2π exp x ) x + 5 ( ) ( x Φ + 3ex Φ 3 ) x 2 and G (x) = 1 G ( x). If the errors are iid, then Ω 1 = Q 1 σ 2 1 and δ Q 1 δ σ 2 1 ( ) T 1 T 1 d ζ ] Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

55 Critical Values (Bai Method) Critical values for ζ can be solved by inverting G (x) : Coverage c 80% % % 11.0 Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

56 Confidence Intervals for Break Date (Bai Method) Point Estimate T 1 Theorem: T 1 T 1 + Confidence interval is then δ Ωδ ( δ Qδ ) 2 ζ T 1 ± ˆδ ˆΩˆδ (ˆδ ˆQ ˆδ) 2 c where ˆδ = β 2 β 1 ˆQ = 1 n n x t x t t=1 ˆΩ = 1 n k n x t x tê t t=1 n k h 1 j=0 T 1 j ( xt x t+jê t ê t+j + x t+j x ) tê t+j ê t t=1 Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

57 Confidence Intervals under Homoskedasticity T 1 ± n ˆσ 2 ( β2 β 1 ) (X X ) ( β2 β 1 )c Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

58 Example Break for GDP Forecast Point Estimate: 1980:4 Bai 90% Interval: 1979:2-1982:2 Break for GDP Variance Point Estimate: 1983:3 Bai 90% Interval: 1983:2-1983:4 Very tight Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

59 Slope Estimators Estimate slopes from regression with estmate ˆT 1 y t = β 0 z t + β 1 x t1 (t T 1 ) + β 2 x t1 (t > T 1 ) + ê t (T 1 ) In the case of full structural change, this is the same as estimation on each sub-sample. Asymptotic Theory: The sub-sample slope estimates are consistent for the true slopes If there is a structural break, their asymptotic distributions are conventional You can treat the structural break as if known Compute standard erros using conventional HAC formula Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

60 Example: Variance Estimates Pre 1983: ˆσ 2 1 = 14.8 (2.3) Post 193: ˆσ 2 2 = 4.9 (1.0) Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

61 Multiple Breaks Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

62 Multiple Structural Breaks Breaks T 1 < T 2 y t = β 0 z t + β 1 x t1 (t T 1 ) + β 2 x t1 (T 1 < t T 2 ) +β 3 x t1 (t > T 2 ) + e t Testing/estimation: Two approaches Joint testing/estimation Sequential Major contributors: Jushan Bai, Pierre Perron Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

63 Joint Methods Testing Test the null of constant parameters against the alternative of two (unknown) breaks Given T 1, T 2, construct Wald test for non-constancy Take the largest test over T 1 < T 2 Asymptotic distribution a generalization of Andrews Estimation The sum-of-squared errors is a function of (T 1, T 2 ) The LS estimates ( ˆT 1, ˆT 2 ) jointly minimize the SSE Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

64 Joint Methods - Computation For 2 breaks, these tests/estimates require O(n 2 ) regressions cumbersome but quite feasible For 3 breaks, naive estimation requires O(n 3 ) regressions, not feasible Bai-Perron developed effi cient computer code which solves the problem of order O(n 2 ) for arbitrary number of breaks Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

65 Sequential Method If the truth is two breaks, but you estimate a one-break model, the SSE will (asymptotically) have local minima at both breakdates Thus the LS breakdate estimator will consistently estimate one of the two breaks, e.g. ˆT 1 for T 1 Given an estimated break, you can split the sample and test for breaks in each subsample You can then find ˆT 2 for T 2 Refinement estimator: Split the entire sample at T 2 Now re-estimate the first break ˆT 1 The refined estimators are asymptotically effi cient Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

66 Forecasting Focuses on Final Breakdate If you only want to find the last break First test for structural change on the full sample If it rejects, split the sample Test for structural change on the second half If it rejects, split again... Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

67 Forecasting After Breaks Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

68 Forecasting After Breaks There is no good theory about how to forecast in the presence of breaks There is a multitude of comflicting recommendations One important contribution: Pesaran and Timmermann (JoE, 2007) They show that in a regression with a single break, the optimal window for estimation includes all of the observations after the break, and some of the observations before the break By including more observations you decrease variance at the cost of some bias They provide empirical rules for selecting sample sizes Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

69 Recommentation The simulations in Persaran-Timmermann suggest that there little gain for the complicated procedures The simple rule Split the sample at the estimated break seems to work as well as anything else My recommendation Test for structural breaks using the Andrews or Bai/Perron tests If there is evidence of a break, estimate its date using Bai s least-squares estimator Calculate a confidence interval to assess accuracy Split the sample at the break, use the post-break period for estimation Use economic judgment to enhance statistical findings Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

70 Examples Revisited Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

71 Examples from Beginning of Class Simple AR(1) with mean and variance breaks y t = ρy t 1 + µ t (1 ρ) + e t e t N(0, σ 2 t (1 ρ 2 )) µ t and/or σ 2 t may be constant or may have a break at some point in the sample Sample size n Questions: Can you guess the timing and type of structural break? Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

72 Model A Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

73 Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

74 Results - Regression SupW = 0.01 (fixed regressor bootstrap p-value) Breakdate Estimate = 62 Bai Interval = [55, 69] Estimates y t = 0.03 (.60) (.67) y t 1 + e t, t 62 y t = 0.69 (.99) (.53) y t 1 + e t, t > 62 Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

75 Results - Variance SupW for Variance = 0.57 (fixed regressor bootstrap p-value) Breakdate Estimate = 78 Bai Interval = [9, 100] Estimates ˆσ 2 = 0.37 (.60), t 78 ˆσ 2 = 0.19 (.24), t > 78 Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

76 Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

77 Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

78 DGP (Model A) T 1 = 60 µ 1 = 0.2 µ 2 = 0.4 σ 2 1 = σ2 2 = 0.36 ρ = 0.8 Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

79 Model B Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

80 Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

81 Results - Regression SupW = 0.07 (fixed regressor bootstrap p-value) Breakdate Estimate = 37 Bai Interval = [20, 54] Estimates y t = 0.53 (1.12) (.40) y t 1 + e t, t 37 y t = 0.10 (.70) (.40) y t 1 + e t, t > 37 Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

82 Results - Variance SupW for Variance = 0.06 (fixed regressor bootstrap p-value) Breakdate Estimate = 15 Bai Interval = [0, 69] Estimates ˆσ 2 = 0.17 (.17), t 15 ˆσ 2 = 0.40 (.55), t > 15 Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

83 Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

84 Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

85 DGP (Model B) T 1 = 40 µ 1 = 0.5 µ 2 = 0.2 σ 2 1 = σ2 2 = 0.36 ρ = 0.8 Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

86 Model C Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

87 Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

88 Results SupW = 0.11 (fixed regressor bootstrap p-value) Regression Breakdate Estimate = 84 Bai Interval = [78, 90] Estimates y t = 0.27 (1.02) (.70) y t 1 + e t, t 37 y t = 1.53 (2.44) (1.47) y t 1 + e t, t > 37 Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

89 Results - Variance SupW for Variance = 0.13 (fixed regressor bootstrap p-value) Breakdate Estimate = 69 Bai Interval = [65, 73] Estimates ˆσ 2 = 0.43 (.51), t 69 ˆσ 2 = 1.77 (3.06), t > 69 Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

90 Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

91 Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91

92 DGP (Model C) T 1 = 70 µ 1 = µ 2 = 0.2 σ 2 1 = 0.36 σ 2 2 = 1.44 ρ = 0.8 Bruce Hansen (University of Wisconsin) Structural Breaks October 29-31, / 91