Problem-1: Consider random walk with drift plus a linear time trend: y t = c + y t 1 + δ t + ϵ t, (1) where {ϵ t } is white noise with E[ϵ 2 t ] = σ 2 >, and y is a non-stochastic initial value. (a) Show that y t = ( c + δ ) t + δ 2 2 t2 + y + t ϵ j. (2) j=1 (b) Compute E[y t ]. (c) Compute V ar[y t ] E[(y t E[y t ]) 2 ]. (d) Compute Cov[y t, y t h ] E[(y t E[y t ])(y t h E[y t h ])] for h 1. (e) Is {y t } covariance stationary? Explain why or why not. (f) Suppose that we simulate the process (1) with y = and ϵ t N(,.5). We consider four cases for (c, δ): Case A. (c, δ) = (.4,.1). Case B. (c, δ) = ( 1.1,.1). Case C. (c, δ) = (.4,.1). Case D. (c, δ) = (1.1,.1). Figure 1 plots simulated sample paths with sample size n = 1. Panels 1-4 of Figure 1 match Cases A-D but possibly with a different order. Answer with a brief reason which panel matches which case. 1
Figure 1: Random Walk with Drift plus Linear Time Trend 8 6 4 2 2 4 6 8 1 1. Which case? 15 1 5-5 -1 2 4 6 8 1 2. Which case? -1-2 -3-4 -5-6 2 4 6 8 1 3. Which case? 15 1 5-5 -1 2 4 6 8 1 4. Which case? Case A: (c, δ) = (.4,.1). Case B: (c, δ) = ( 1.1,.1). Case C: (c, δ) = (.4,.1). Case D: (c, δ) = (1.1,.1). 2
Problem-2: Consider testing for the unit root hypothesis of a time series {y t }. Suppose that we run the Dickey-Fuller test with Approach #2, under which the regression model is written as y t = α + ϕy t 1 + u t = x tθ + u t (3) with x t = [1, y t 1 ] and θ = [α, ϕ]. (a) Write the expression of the ordinary least squares (OLS) estimator ˆθ n, using x t and y t. (b) Write the expression of the test statistic S n. (c) State the 5% critical value c n when sample size is n = 1. (Instruction: Your answer does not have to be exactly the same as what we learned in class. You get credits if your answer is reasonably close to the true critical value.) (d) Do you reject or accept H : ϕ = 1 at the 5% level when S n < c n? (e) Recall from class that Approach #2 is valid if a true DGP is y t = ϕ y t 1 + ϵ t with ϕ 1 and ϵ t (, σ 2 ). Hereafter we analyze the consequence of having serially correlated errors instead of IID. Assume that the true DGP is given by y t = ϕ y t 1 + ϵ t, ϵ t = ν t + β ν t 1, ν t N(, 1), ϕ {1,.95,.9}, β {,.2,.4}. (4) (Remark: When β =, this DGP reduces to the IID case elaborated in class.) Show 3
that E[ϵ t ] =, γ() E[ϵ 2 t ] = 1 + β, 2 γ(1) E[ϵ t ϵ t 1 ] = β, γ(h) E[ϵ t ϵ t h ] = for any h 2. (f) We simulate J = 1 Monte Carlo samples with sample size n {1, 5, 1} from DGP (4). In Table 1, we report rejection frequencies of Approach #2 of the Dickey-Fuller test with nominal size 5%. Discuss the empirical size and power of Approach #2. (g) To achieve correct size for any β, Phillips (1987) and Phillips and Perron (1988) proposed a modified version of the Dickey-Fuller test. 1 Keep the regression model (3) and modify the test statistic S n as follows. S n = S n B n, where S n is the usual test statistic used in Approach #2, and the extra term B n is defined as follows. B n = ˆΣ 1 n (2, 2) ˆγ n (1), where û t = y t x tˆθ n, ˆγ n (1) = 1 n n û t û t 1, t=2 1 P. C. B. Phillips (1987). Time Series Regression with a Unit Root. Econometrica, 55, 277-31; P. C. B. Phillips and P. Perron (1988). Testing for a Unit Root in Time Series Regression. Biometrika, 75, 335-346. 4
and 1 ˆΣ n (2, 2) is the (2, 2)-element of ˆΣ 1 n = [ 1 n 2 1 n x t x t]. t=1 (Remark: It can be shown that 1 ˆΣ n (2, 2) > ; ˆΣ 1 n (2, 2) = O p (1) under H : ϕ = 1; ˆγ n (1) p γ(1) = β.) We use the same critical value c n as in Approach #2 (i.e. we reject H : ϕ = 1 at the 5% level if and only if S n < c n ). Provide an intuitive explanation on why the Phillips-Perron test statistic S n is supposed to achieve correct size for any β. (h) In Table 2 we report rejection frequencies of Approach #2 and the Phillips-Perron test. (Remark: The rejection frequencies of the former are all the same as those in Table 1.) Compare the two tests in terms of empirical size and power. 5
Table 1: Rejection Frequencies of Approach #2 n = 1 β =.5.189.468 β =.2.15.61.191 β =.4.6.23.86 n = 5 β =.5.993 1. β =.2.16.95 1. β =.4.5.722 1. n = 1 β =.5 1. 1. β =.2.13 1. 1. β =.4.4 1. 1. DGP is y t = ϕ y t 1 + ϵ t, ϵ t = ν t + β ν t 1, ν t N(, 1), ϕ {1,.95,.9}, and β {,.2,.4}. We draw J = 1 Monte Carlo samples with sample size n {1, 5, 1} from this DGP. We run the Dickey-Fuller test with Approach #2 and report rejection frequencies with respect to the nominal size 5%. 6
Table 2: Rejection Frequencies of Approach #2 and Phillips-Perron Test n = 1 #2 PP #2 PP #2 PP β =.5.58.189.212.468.497 β =.2.15.5.61.181.191.445 β =.4.6.48.23.173.86.432 n = 5 #2 PP #2 PP #2 PP β =.5.52.993.991 1. 1. β =.2.16.5.95.991 1. 1. β =.4.5.49.722.991 1. 1. n = 1 #2 PP #2 PP #2 PP β =.5.52 1. 1. 1. 1. β =.2.13.51 1. 1. 1. 1. β =.4.4.51 1. 1. 1. 1. DGP is y t = ϕ y t 1 + ϵ t, ϵ t = ν t + β ν t 1, ν t N(, 1), ϕ {1,.95,.9}, and β {,.2,.4}. We draw J = 1 Monte Carlo samples with sample size n {1, 5, 1} from this DGP. We run the Dickey-Fuller test with Approach #2 and the Phillips-Perron (PP) test, and report rejection frequencies with respect to the nominal size 5%. 7
Problem-3: Consider two time series {y t } and {x t }. Suppose that a true DGP is given by y t = β x t + ϵ yt, (5) x t = d + x t 1 + ϵ xt, (6) where β, d, and ϵ t = [ϵ yt, ϵ xt ] follows a joint white noise with E[ϵ t ϵ t] = σ2 ϵy. σϵx 2 (a) Show that t x t = d t + x + ϵ xj, (7) j=1 where x is understood to be a non-stochastic initial value. (Remark: Since {x t } follows a random walk with drift, it is trivially nonstationary.) (b) Eq. (5) can be rewritten as y t = y + d t + and hence t ϵ yj + η yt (8) j=1 y t = d + y t 1 + ϵ yt + η yt, (9) where η yt = η yt η y,t 1 and ν t = [ϵ yt, η yt ] follows a joint white noise with E[ν t ν t] = E[(ϵ yt) 2 ]. (1) E[ηyt] 2 8
Properly characterize each of ( y, d, ϵ yt, η yt, E[(ϵ yt) 2 ], E[η 2 yt] ) in terms of (d, β, x, ϵ xt, ϵ yt, σ 2 ϵy, σ 2 ϵx). (Remark: Since {y t } follows a random walk with drift and noise, it is trivially nonstationary.) (c) Eqs. (5) and (6) can be rewritten as a structural VAR(1): n 11 n 12 y t = c 1 + m 11 m 12 y t 1 + ϵ yt, (11) n 21 n 22 x t c 2 m 21 m 22 x t 1 ϵ xt or compactly Nz t = c + Mz t 1 + ϵ t. Properly characterize each of (n 11, n 12, n 21, n 22, c 1, c 2, m 11, m 12, m 21, m 22 ) in terms of (d, β ). (d) Eqs. (5) and (6) can be rewritten as a vector error correction (VEC) form: y t = a 1 + α y (y t 1 β x t 1 ) + e yt. (12) x t a 2 α x e xt Properly characterize each of (a 1, a 2, α y, α x, e yt, e xt ) in terms of (d, β, ϵ yt, ϵ xt ). (e) Based on the speed-of-adjustment parameters (α y, α x ), explain how {y t } and {x t } correct a deviation from the equilibrium level y t 1 = β x t 1. (Instruction: Just a few lines of concise explanation should be enough.) (f) Hereafter we consider a specific example with β =.5, d =.5, ϵ t N(, I 2 ), and 9
sample size n = 1. Figure 2 plots a simulated path of {y t } and {x t } (Panel 1) and the equilibrium error ϵ yt = y t β x t = y t.5x t (Panel 2). Answer with a brief reason which of the two lines in Panel 1 depicts {y t }. (g) Suppose that we test whether {y t } and {x t } are cointegrated with a prescribed value β =.5. We apply Approach #1 of the Dickey-Fuller unit root test with respect to ϵ yt = y t.5x t plotted in Panel 2 of Figure 2. The OLS estimator from the regression model ϵ yt = ϕϵ yt 1 +ν t is ˆϕ n =.92, and the test statistic is S n = n( ˆϕ n 1) = 19.2. Which conclusion does the test reach cointegration or non-cointegration between {y t } and {x t }? (h) Suppose alternatively that we test whether {y t } and {x t } are cointegrated without any prescribed value of β. We run a linear regression model y t = α + βx t + u t in order to compute OLS estimators (ˆα n, ˆβ n ) and residuals û t = y t ˆα n ˆβx t. Then we apply Approach #1 of the Dickey-Fuller unit root test with the adjusted critical value for {û t }. The OLS estimator from the regression model û t = ϕû t 1 + ν t is ˆϕ n =.99, and the test statistic is S n = n( ˆϕ n 1) = 19.9. Which conclusion does the test reach cointegration or non-cointegration between {y t } and {x t }? 1
Figure 2: Simulated Bivariate Vector Error Correction Processes 5 4 3 2 1-1 2 4 6 8 1 1. Which is {y t }? Which is {x t }? 3 2 1-1 -2-3 2 4 6 8 1 2. {y t β x t } 11