Non-Stationary Time Series, Cointegration and Spurious Regression

Econometrics 2 Fall 25 Non-Stationary Time Series, Cointegration and Spurious Regression Heino Bohn Nielsen 1of32 Motivation: Regression with Non-Stationarity What happens to the properties of OLS if variables are non-stationary? Consider two presumably unrelated variables: CONS Danish private consumption in 1995 prices. BIRD Number of breeding cormorants (skarv) in Denmark. And consider a static regression model log(cons t )=β + β 1 log(bird t )+u t. We would expect (or hope) to get β b 1 and R 2. Applying OLS to yearly data 1982 21 gives the result: with R 2 =.688. log(cons t )=12.145 (8.9) +.95 (6.3) log(bird t)+u t, It looks like a reasonable model. But it is complete nonsense: spurious regression. 2of32

The variables are non-stationary. The residual, u t, is non-stationary and standard results for OLS do not hold. In general, regression models for non-stationary variables give spurious results. Only exception is if the model eliminates the stochastic trends to produce stationary residuals: Cointegration. For non-stationary variables we should always think in terms of cointegration. Only look at regression output if the variables cointegrate. 13.2 Consumption and breeding birds, logs Consumption Number of breeding birds 1 Regression residuals (cons on birds) 13.1 13. -1 12.9 1985 199 1995 2 1985 199 1995 2 3of32 Definitions and concepts: Outline (1) Combinations of non-stationary variables; Cointegration defined. (2) Economic equilibrium and error correction. Engle-Granger two-step cointegration analysis: (3) Static regression for cointegrated time series. (4) Residual based test for no-cointegration. (5) Models for the dynamic adjustment. Cointegration analysis based on dynamic models: (6) Estimation in the unrestricted ADL or ECM model. (7) PcGive test for no-cointegration. What if variables do not cointegrate? (8) Spurious regression revisited. 4of32

Cointegration Defined Let X t =(X 1t X 2t ) be two I(1) variables, i.e. they contain stochastic trends: X 1t = X t 1i + initial value + stationary process i=1 X 2t = X t 2i + initial value + stationary process. i=1 In general, a linear combination of X 1t and X 2t will also have a random walk. Define β =(1 β 2 ) and consider the linear combination: Z t = β X t = µ X 1 β 1t 2 = X X 1t β 2 X 2t 2t = X t X t 1i β 2 2i + initial value + stationary process. i=1 i=1 Important exception: There exist a β, sothatz t is stationary: We say that X 1t and X 2t co-integrate with cointegration vector β. 5of32 Remarks: (1) Cointegration occurs if the stochastic trends in X 1t and X 2t are the same so they cancel, P t i=1 1i = β 1 Pt i=1 2i. This is called a common trend. (2) You can think of an equation eliminating the random walks in X 1t and X 2t : If u t is I() (mean zero) then β =(1 X 1t = µ + β 2 X 2t + u t. ( ) β 2 ) is a cointegrating vector. (3) The cointegrating vector is only unique up to a constant factor. If β X t I(). Then so is β e X t = bβ X t for b 6=. We can choose a normalization µ µ 1 β = or β e β1 e =. β 2 1 This corresponds to different variables on the left hand side of ( ) (4) Cointegration is easily extended to more variables. The variables in X t = X 1t X 2t X pt cointegrate if Z t = β X t = X 1t β 2 X 2t... β p X pt I(). 6of32

Cointegration and Economic Equilibrium Consider a regression model for two I(1) variables, X 1t and X 2t,givenby X 1t = µ + β 2 X 2t + u t. ( ) The term, u t, is interpretable as the deviation from the relation in ( ). If X 1t and X 2t cointegrate, then the deviation u t = X 1t µ β 2 X 2t is a stationary process with mean zero. Shocks to X 1t and X 2t have permanent effects. X 1t and X 2t co-vary and u t I(). We can think of ( ) asdefining an equilibrium between X 1t and X 2t. If X 1t and X 2t do not cointegrate, then the deviation u t is I(1). There is no natural interpretation of ( ) as an equilibrium relation. 7of32 Empirical Example: Consumption and Income Time series for log consumption, C t,andlogincome,y t,arelikelytobei(1). Define a vector X t =(C t Y t ). Consumption and income are cointegrated with cointegration vector β =(1 1 ) if the (log-) consumption-income ratio, µ Z t = β Ct X t =(1 1 ) = C Y t Y t, t is a stationary process. The consumption-income ratio is an equilibrium relation. Consumption and income, logs Income Consumption. Income ratio, log(c t ) log(y t ) 6.25 -.5 6. 197 198 199 2 -.1 -.15 197 198 199 2 8of32

How is the Equilibrium Sustained? There must be forces pulling X 1t or X 2t towards the equilibrium. Famous representation theorem: X 1t and X 2t cointegrate if and only if there exist an error correction model for either X 1t, X 2t or both. As an example, let Z t = X 1t β 2 X 2t be a stationary relation between I(1) variables. Then there exists a stationary ARMA model for Z t. Assume for simplicity an AR(2): This is equivalent to or Z t = θ 1 Z t 1 + θ 2 Z t 2 + t, θ(1) = 1 θ 1 θ 2 >. (X 1t β 2 X 2t ) = θ 1 (X 1t 1 β 2 X 2t 1 )+θ 2 (X 1t 2 β 2 X 2t 2 )+ t X 1t = β 2 X 2t + θ 1 X 1t 1 θ 1 β 2 X 2t 1 + θ 2 X 1t 2 θ 2 β 2 X 2t 2 + t, X 1t = β 2 X 2t + θ 2 β 2 X 2t 1 θ 2 X 1t 1 (1 θ 1 θ 2 ) {X 1t 1 β 2 X 2t 1 } + t. In this case we have a common-factor restriction. That is not necessarily true. 9of32 More on Error-Correction Cointegration is a system property. Both variables could error correct, e.g.: X 1t = δ 1 + Γ 11 X 1t 1 + Γ 12 X 2t 1 + α 1 (X 1t 1 β 2 X 2t 1 )+ 1t X 2t = δ 2 + Γ 21 X 1t 1 + Γ 22 X 2t 1 + α 2 (X 1t 1 β 2 X 2t 1 )+ 2t, We may write the model as the so-called vector error correction model, µ µ µ µ µ X1t δ1 Γ11 Γ = + 12 X1t 1 α1 + (X X 2t δ 2 Γ 22 X 2t 1 α 1t 1 β 2 X 2t 1 )+ 2 or simply Γ 21 X t = δ + Γ X t 1 + αβ X t 1 + t. µ 1t 2t, Note that β X t 1 = X 1t 1 β 2 X 2t 1 appears in both equations. For X 1t to error correct we need α 1 <. For X 2t to error correct we need α 2 >. 1 of 32

Consider a simple model for two cointegrated variables: µ µ X1t.2 = (X X 2t.1 1t 1 X 2t 1 )+ µ 1t. 2t (A) Two cointegrated variables (B) Deviation from equilibrium, β'x t =X 1t X 2t X 1t X 2t 2.5-5. -1-2.5 2 4 6 8 1 2 4 6 8 1 12.5 1. (C) Speed of adjustment of β'x t (D) Cross-plot X 1t X 2t 7.5 5. -5 2.5. -1 X 1-2.5 2 4 6 8 1-12.5-1. -7.5-5. -2.5. 11 of 32 OLS Regression with Cointegrated Series In the cointegration case there exists a β 2 so that the error term, u t,in X 1t = µ + β 2 X 2t + u t. ( ) is stationary. OLS applied to ( ) givesconsistent results, so that b β 2 β 2 for T. Note that consistency is obtained even if potential dynamic terms are neglected. This is because the stochastic trends in X 1t and X 2t dominate. We can even get consistent estimates in the reverse regression X 2t = δ + γ 1 X 1t + v t. Unfortunately, it turns out that b β 2 is not asymptotically normal in general. The normal inferential procedures do not apply to b β 2! We can use ( ) for estimation not for testing. 12 of 32

Super-Consistency For stationary series, the variance of b β 1 declines at a rate of T 1. For cointegrated I(1) series, the variance of b β 1 declines at a faster rate of T 2. Intuition: If b β 1 = β 1 then u t is stationary. If b β 1 6= β 1 thentheerrorisi(1)andwill have a large variance. The information on the parameter grows very fast. 3 Stationary, T=5 3 Stationary, T=1 3 Stationary, T=5 2 2 2 1 1 1.5 1. 1.5.5 1. 1.5.5 1. 1.5 3 Non-Stationary, T=5 3 Non-Stationary, T=1 3 Non-Stationary, T=5 2 2 2 1 1 1.5 1. 1.5.5 1. 1.5.5 1. 1.5 13 of 32 Test for No-Cointegration, Known β 1 Suppose that X 1t and X 2t are I(1). Also assume that β =(1 β 2 ) is known. The series cointegrate if is stationary. Z t = X 1t β 2 X 2t ThiscanbetestedusinganADFunitroottest,e.g.thetestforH : π =in kx Z t = δ + Z t i + πz t 1 + η t. i=1 The usual DF critical values apply to t π=. Note, that the null, H : π =, is a unit root, i.e. no cointegration. 14 of 32

Test for No-Cointegration, Estimated β 1 Engle-Granger (1987) two-step procedure. If β =(1 β 2 ) is unknown, it can be (super-consistently) estimated in X 1t = µ + β 2 X 2t + u t. ( ) bβ is a cointegration vector if bu t = X 1t bµ b β 2 X 2t is stationary. ThiscanbetestedusingaDFunitroottest,e.g.thetestforH : π =in Remarks: bu t = kx bu t i + πbu t 1 + η t. i=1 (1) The residual bu t has mean zero. No deterministic terms in DF regression. (2) The critical value for t π= still depends on the deterministic regressors in ( ). (3) The fact that b β 1 is estimated also changes the critical values. OLS minimizes the variance of bu t. Look as stationary as possible. Critical value depends on the number of regressors. 15 of 32.6.5 Estimated parameters in cointegrating regression (with constant in the regression (***)) 7 6 5 4 3 2 1 DF(constant).4 N(,1).3.2.1-8 -7-6 -5-4 -3-2 -1 1 2 3 4 5 16 of 32

Critical values for the Dickey-Fuller test for no-cointegration are given by: Case 1: A constant term in ( ). Number of estimated Significance level parameters 1% 5% 1% 3.43 2.86 2.57 1 3.9 3.34 3.4 2 4.29 3.74 3.45 3 4.64 4.1 3.81 4 4.96 4.42 4.13 Case 2: A constant and a trend in ( ). Number of estimated Significance level parameters 1% 5% 1% 3.96 3.41 3.13 1 4.32 3.78 3.5 2 4.66 4.12 3.84 3 4.97 4.43 4.15 4 5.25 4.72 4.43 17 of 32 Dynamic Modelling Given the estimated cointegrating vector we can define the error correction term ecm t = bu t = X 1t bµ β b 2 X 2t, which is, per definition, a stationary stochastic variable. Since b β 2 converges to β 2 very fast we can treat it as a fixed regressor and formulate an error correction model conditional on ecm t 1,i.e. X 1t = δ + λ 1 X 1t 1 + κ X 2t + κ 1 X 2t 1 α ecm t 1 + t, where α> is consistent with error-correction. Given cointegration, all terms are stationary, and normal inference applies to δ, λ, κ, κ 1,andα. 18 of 32

Outline of an Engle-Granger Analysis (1) Test individual variables, e.g. X 1t and X 2t, for unit roots. (2) Run the static cointegrating regression X 1t = µ + β 2 X 2t + u t. Note that the t ratios cannot be used for inference. (3) Test for no-cointegration by testing for a unit root in the residuals, bu t. (4) If cointegration is not rejected estimate a dynamic (ECM) model like X 1t = δ + λ 1 X 1t 1 + κ X 2t + κ 1 X 2t 1 αbu t 1 + t. All terms are stationary. Remaining inference is standard. 19 of 32 Empirical Example: Danish Interest Rates Consider two Danish interest rates: r t : Money market interest rate b t : Bond Yield for the period t = 1972 : 1 23 : 2. Test for unit roots in r t and b t (5% critical value is 2.89): c r t =.638118 (1.35) c b t =.116558 (.658).12629 ( 1.39) +.395115 (4.67) r t 1.23433 ( 2.7) b t 1.128941 (.99) r t 4.826987 ( 1.8) b t 1 r t 1 We cannot reject unit roots. Test if s t = r t b t is I(1) (5% crit. value is 2.89): c s t =.848594 ( 3.71) +.2766 (2.56) s t 3.379449 ( 5.35) It is easily rejected that b t and r t are not cointegrating. s t 1. 2 of 32

.2 Bond Yield and money market interest rate Bond yield Money market interest rate.5 Interest rate spread..1 -.5 197 198 199 2 197 198 199 2.5 Residuals from r t =α+ β b t + ε t 1.5 1.25 1. Impulse responses from b t to r t 1.19928..75.5.866611 -.5.25. 198 199 2 5 1 21 of 32 Instead of assuming β 1 =1we could estimate the coefficient Modelling IMM by OLS (using PR312.in7) The estimation sample is: 1974 (3) to 23 (2) Coefficient Std.Error t-value t-prob Part.R^2 Constant -.46856.5545 -.845.4.62 IBZ.845524.4495 18.8..7563 sigma.224339 RSS.573738644 R^2.756314 F(1,114) = 353.8 [.]** log-likelihood 276.885 DW.82 no. of observations 116 no. of parameters 2 mean(imm).919727 var(imm).22967 22 of 32

We could test for a unit root in the residuals (5% crit. value is 3.34): b t =.2321 (2.95) Again we reject no-cointegration. b t 3.499443 ( 6.77) b t 1. Finally we could estimate the error correction models based on the spread: c r t =.77426 ( 3.23) c b t =.18162 ( 2.11) +1.17725 (4.55) +.43897 (4.16) b t.46456 (5.22) b t 1.673997 ( 2.1) (r t 1 b t 1 ) r t.638286 (2.22) Note that the short-rate, r t, error corrects, while the bond-yield, b t,doesnot. (r t 1 b t 1 ) 23 of 32 Estimation of β In the ADL/ECM The estimator of β 2 from a static regression is super-consistent...but (1) b β 2 is often biased (due to ignored dynamics). (2) Hypotheses on β 2 cannot be tested. An alternative estimator is based on an unrestricted ADL model, e.g. X 1t = δ + θ 1 X 1t 1 + θ 2 X 1t 2 + φ X 2t + φ 1 X 2t 1 + φ 2 X 2t 2 + t, where t is IID. This is equivalent to an error correction model: X 1t = δ + λ 1 X 1t 1 + κ X 2t + κ 1 X 2t 1 + γ 1 X 1t 1 + γ 2 X 2t 1 + t. An estimate of β 2 can be found from the long-run solutions: bβ 2 = bγ 2 bγ 1 = b φ + b φ 1 + b φ 2 1 b θ 1 b θ 2. The main advantage is that the analysis is undertaken in a well-specified model. The approach is optimal if only X 1t error corrects. Inference on b β 2 is possible. 24 of 32

Testing for No-Cointegration Due to representation theorem, the null hypothesis of no-cointegration corresponds to the null of no-error-correction. Several tests have been designed in this spirit. The most convenient is the so-called PcGive test for no-cointegration. Consider the unrestricted ADL or ECM: X 1t = δ + λ 1 X 1t 1 + κ X 2t + κ 1 X 2t 1 + γ 1 X 1t 1 + γ 2 X 2t 1 + t. (#) Test the hypothesis H : γ 1 = against the cointegration alternative, γ 1 <. This is basically a unit root test (not a N(, 1)). The distribution of the t ratio, t γ1 = = bγ 1 SE(bγ 1 ), depends on the deterministic terms and the number of regressors in (#). 25 of 32 Critical values for the PcGive test for no-cointegration are given by: Case 1: A constant term in (#). Number of variables Significance level in X t 1% 5% 1% 2 3.79 3.21 2.91 3 4.9 3.51 3.19 4 4.36 3.76 3.44 5 4.59 3.99 3.66 Case 2: A constant and a trend in (#). Number of variables Significance level in X t 1% 5% 1% 2 4.25 3.69 3.39 3 4.5 3.93 3.62 4 4.72 4.14 3.83 5 4.93 4.34 4.3 26 of 32

Outline of a (One-Step) Cointegration Analysis (1) Test individual variables, e.g. X 1t and X 2t, for unit roots. (2) Estimate an ADL model X 1t = δ + λ 1 X 1t 1 + κ X 2t + κ 1 X 2t 1 + γ 1 X 1t 1 + γ 2 X 2t 1 + t. (3) Test for no-cointegration with t γ1 =. If cointegration is found, the cointegrating relation is the long-run solution. (4) Derive the long-run solution X 1t = bµ + b β 2 X 2t. Inference on β 2 is standard (under some conditions). 27 of 32 Empirical Example: Interest Rates Revisited Estimation based on a ADL model. The significant terms are: Modelling IMM by OLS (using PR312.in7) The estimation sample is: 1973 (4) to 23 (2) Coefficient Std.Error t-value t-prob Part.R^2 IMM_1.615152.799 7.78..3447 Constant -.25456.4573 -.548.585.26 IBZ 1.19928.2347 5.11..1851 IBZ_1 -.865763.2648-3.27.1.851 sigma.182398 RSS.382594939 R^2.841437 F(3,115) = 23.4 [.]** log-likelihood 39.674 DW 2.16 no. of observations 119 no. of parameters 4 mean(imm).92754 var(imm).22764 28 of 32

The long-run solution is given in PcGive as Solved static long-run equation for IMM Coefficient Std.Error t-value t-prob Constant -.65791.1184 -.55.584 IBZ.866611.9491 9.13. Long-run sigma =.473948 Here the t valuescanbeusedfortesting!β 2 is not significantly different from unity. The test for no-cointegration is given by (critical value 3.69): PcGive Unit-root t-test: -4.8661 The impulse responses X 1t / X 2t, can be graphed. X 1t / X 2t 1,... and the cumulated P X 1t / X 2t i 29 of 32 Spurious Regression Revisited Recall that cointegration is a special case where all stochastic trends cancel. From an empirical point of view this an exception. What happens if the variables do not cointegrate? Assume that X 1t and X 2t are two totally unrelated I(1) variables. Then we would like the static regression to reveal that β 2 =and R 2 =. X 1t = µ + β 2 X 2t + u t, ($) This turns out not to be the case! The standard regression output will indicate a relation between X 1t and X 2t. This is called a spurious regression or nonsense regression result. With non-stationary data we always have to think in terms of cointegration. 3 of 32

Simulation: Stationary Case Consider first two independent IID variables: µ µµ X 1t = 1t 1t where N X 2t = 2t 2t for T =5, 1, 5. Here, we get standard results for the regression model X 1t = µ + β 2 X 2t + u t. µ 1, 1, 7.5 5. IID Case, estimates 5 1 5 Note the convergence to β 2 = as T diverges..4.2 IID Case, t-ratios N(,1) 5 1 5 Looks like a N(,1) for all T. Standard testing. 2.5 -.5 -.25..25.5-4 -2 2 4 31 of 32 Simulation: I(1) Spurious Regression Now consider two independent random walks µ µµ X 1t = X 1t 1 + 1t 1t where N X 2t = X 2t 1 + 2t 2t for T =5, 1, 5. µ 1, 1, Under the null hypothesis, β 2 =, the residual is I(1). The condition for consistency is not fulfilled..75.5 I(1) case, estimates 5 1 5 Looks unbiased, but NO convergence..75.5 I(1) case, t-ratios 5 1 5 The distribution gets increasingly dispersed..25.25 Note the scale as compared to a N(,1) -3-2 -1 1 2 3-75 -5-25 25 5 75 32 of 32