Problem Set 1 Solution Sketches Time Series Analysis Spring 2010

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1 Problem Set 1 Solution Sketches Time Series Analysis Spring Construct a martingale difference process that is not weakly stationary. Simplest e.g.: Let Y t be a sequence of independent, non-identically distributed r.v. s such that E(Y t ) = 0 for all t but var(y t ) varies with t. 2. Let ɛ t be an i.i.d. N(0, 1) process, and let X t be defined as X t = ɛ t if t is odd 2 2 (ɛ2 t 1 1) if t is even. Show that X t is white noise. (Hint: What are the moments of ɛ t? Calculate E(X t ), var(x t ), cov(x t, X t 1 ), cov(x t, X t 2 ), etc. for t even and odd.) Is X t a martingale difference sequence w.r.t. its own history? The solution involves straightforward (though somewhat tedious) calculation. What makes this example work is the fact that E(ɛ 3 t ) = 0 due to the normality assumption. Furthermore, if one is in an odd period, then next period s value is perfectly predictable, so clearly E(X t X t 1, X t 2,...) 0 for all t. 3. The usual factorization theorem for polynomials states that if a(z) = a 0 + a 1 z a n z n, a n 0, then one can write a(z) = a n (z z 1 )(z z 2 )... (z z n ) z, where z 1,..., z n are the n roots satisfying a(z i ) = 0, i = 1,..., n. Using this result, show that every polynomial of the form φ(z) = 1 φ 1 z φ 2 z 2... φ p z p, φ p 0, can be factored as φ(z) = (1 z1 1 z)(1 z2 1 z)... (1 zp 1 z), where z 1,..., z p are the p roots of the equation φ(z) = 0. 1

2 By the standard factorization theorem (the Fundamental Theorem of Algebra), φ(z) = φ p i=1 (z z i ) for all z. In particular, 1 = φ(0) = φ p i=1 ( z i). Hence, φ(z) = [ i=1 ( z i)] 1 i=1 (z z i) = i=1 (1 z 1z). (Obviously, none of the z i s is zero.) i 4. Consider the second order difference equation y t = 0.4y t y t 2 + ɛ t. Find an analytical expression for the impulse response function y t+j / ɛ t, j = 1, 2,.... (Hint: Use the partial fraction method shown in class.) Has the form Aλ j 1 + Bλj 2, where λ 1 and λ 2 are the inverse roots of the characteristic polynomial and A and B are obtained as shown in class. 5. This exercise develops a recursive method for calculating the inverse of autoregressive lag polynomials. Let φ(l) = 1 φ 1 L φ 2 L 2... φ p L p. Write the inverse as φ 1 (L) ψ(l) = ψ 0 + ψ 1 L + ψ 2 L 2 + ψ 3 L , where ψ 0, ψ 1, etc. are to be determined. The inverse must satisfy φ(l)ψ(l) 1, i.e. (1 φ 1 z φ 2 z 2... φ p z p )(ψ 0 + ψ 1 z + ψ 2 z 2 + ψ 3 z ) = 1 (1) must hold identically in z. a) Given φ 1,..., φ p and (1), derive a recursive formula for calculating ψ i, i = 0, 1, 2,.... This means that each ψ i must be expressed as a function of the φ s and ψ i 1, ψ i 2,..., ψ 0. Hint: Find a formula for ψ 0 first, then for ψ 1, then for ψ 2, etc. After a while you ll see the general pattern. Perform the multiplication on the lhs of (1), collect terms with the same power of z, and set the coefficients to zero. This yields the recursive formula ψ 0 = 1 and ψ n = n j=1 φ jψ n j. b) Write a computer code (e.g., in Matlab) that implements this formula. The routine should take as its input the numbers φ 1,..., φ p and a positive integer n. The output should be an 1 (n + 1) vector giving ψ 0, ψ 1,..., ψ n. 2

3 6. Using a computer, simulate and plot the impulse response function ( y t+j / ɛ t, j = 0, 1, 2,..., 50) for the following processes: i) Y t = 0.6Y t Y t 2 + ɛ t ii) Y t = 0.25Y t Y t 2 + ɛ t iii) Y t = 1.3Y t 1 0.2Y t 2 0.1Y t 3 + ɛ t iv) Y t = Y t 1 0.2Y t 2 0.1Y t 3 + ɛ t v) Y t = 0.7Y t 1 + ɛ t + 0.6ɛ t 4 In each case calculate the roots of the corresponding AR characteristic polynomial (you can use the roots command in Matlab). Comment on the relationship between the size of the roots and the patterns you observe. You can use the code you just wrote for problem 5 part b). Alternatively, you can write another Matlab code following this algorithm: Create a 1 54 vector ɛ with ɛ(4) = 1 and ɛ(t) = 0 for t 4. Create a 1 54 vector y so that y(1) = y(2) = y(3) = 0 and y(t) is generated by the appropriate difference equation for t = 4,..., 54. E.g., in case i) y(t) = 0.6 y(t 1) 0.58 y(t 2) + ɛ(t), t = 4,..., 54. You can use a for cycle to do this. In Matlab, simply put plot(y) to create a graph of the impulse response function. i) All roots inside unit circle effects of a shock die out over time; reversion to the mean. Oscillation due to complex roots (see p in book for explanation). ii) One root inside unit circle effects of a shock accumulate over time, process explodes. Initial oscillation due to negative root. iii) One unit root, no constant (or drift ). Shocks have permanent effects; process settles down to a new level. iv) Technically, the impulse response function is the same as in last case the shock has a permanent effect. However, we do not see the process settle down to a fixed level as the interaction between the unit root and the constant generates a linear trend. (We will see later how this happens.) 3

4 v) Root of AR polynomial outside unit circle mean reversion. However, the shock hits more than once, due to the MA part. 7. Total annual rainfall in Austin was recorded over the last 100 years. The estimated first order autocorrelations of the data are ˆρ 1 = 0.45, ˆρ 2 = 0.14, ˆρ 3 = 0.15 ˆρ 4 = 0.19, ˆρ 5 = 0.05, ˆρ 6 = Can you propose a simple model model for the data? Many of you had problems here. First, the statistical significance of the estimated autocorrelations needs to be taken into account. The approx. 95% confidence band is ±2/ 100 = Only the first order autocorrelation is clearly outside the band; the fourth is borderline. The fifth and sixth order autocorrelations are insignificant. Higher order autocorrelations are not reported, but one can speculate that they too are likely insignificant. This suggests an MA(1) model (due to the sharp dropoff in the ACF after lag 1), though there is marginal evidence for MA(4). Estimating the parameters wasn t required but in case of an MA(1) model one can simply solve θ/(1 + θ 2 ) = 0.45 and take root less than one in absolute value. (To propose an AR(p) process one would generally want to see a more gradual decline in the ACF.) 8. Consider the process defined by X t = 0.6X t 1 + ɛ t 0.9ɛ t ɛ t 2, where ɛ t is white noise. It is suggested that the autocorrelation function ρ h of this process is zero for h 2. Can this be so? This is an overparameterized MA(1) process. Write in lag polynomial notation and factor the MA polynomial on the rhs. The term (1 0.6L) cancels out. 9. Let Y t = β 1 Y t 1 + ɛ t and ɛ t = β 2 ɛ t 1 + u t, where β 1 0, β 2 < 1 and u t is an i.i.d. process with mean zero and variance σ 2. a) Show that Y t follows an AR(2) process. (Hint: This is easiest if you use lag operator notation.) In particular, show that Y t = α 1 Y t 1 + α 2 Y t 2 + u t and find α 1 and α 2 in terms of β 1 and β 2. Given α 1 and α 2, is it possible to back out β 1? 4

5 Write (1 β 1 L)Y t = ɛ t and (1 β 2 L)ɛ t = u t. Multiplying both sides of the first equation by (1 β 2 L) gives (1 β 2 L)(1 β 1 L)Y t = u t, an AR(2) process. Clearly, α 1 = β 1 + β 2 and α 2 = β 1 β 2. Note that the role of β 1 and β 2 in these equations is completely symmetric, hence they cannot generally be distinguished (exceptions exist, e.g. when β 1 1). b) Is Y t is covariance stationary? From part a), the process is covariance stationary iff β 1 < 1 (in addition to β 2 < 1). c) Consider regressing Y t on Y t 1 using OLS. Will the regression coefficient be consistent for β 1? A fairly informal argument suffices. Assume stationarity. In the equation Y t = β 1 Y t 1 + ɛ t, Y t 1 and ɛ t are correlated, hence OLS is not consistent for β 1. To see that there is correlation, note that Y t 1 = (1 β 2 L) 1 (1 β 1 L) 1 u t 1 and ɛ t = (1 β 2 L) 1 u t. (Indeed, we now know that by regressing Y t on Y t 1, one consistently estimates the coefficient from the linear projection of Y t on Y t 1. This projection coefficient cannot be β 1, since ɛ t = Y t β 1 Y t 1 is correlated with Y t 1.) d) Consider regressing Y t on Y t 1 and Y t 2. Will the regression coefficients be consistent for α 1 and α 2? Again, a fairly informal argument suffices. Yes, because u t is uncorrelated with Y t 1 and Y t 2. This can be seen from the representation Y t = (1 β 2 L) 1 (1 β 1 L) 1 u t. 5