Stochastic Modelling Solutions to Exercises on Time Series

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1 Stochastic Modelling Solutions to Exercises on Time Series Dr. Iqbal Owadally March 3, 2003 Solutions to Elementary Problems Q1. (i) (1 0.5B)X t = Z t. The characteristic equation 1 0.5z = 0 does not have a unit root and the time series is ARIMA(1, 0, 0). The root of the characteristic equation is 2. Since 2 > 1, the time series is stationary. It is also (trivially) invertible. (ii) (1 0.5B)X t = (1 1.3B+0.4B 2 )Z t = (1 0.8B)(1 0.5B)Z t. Equation 1 0.5z = 0 does not have a unit root and the time series is ARIMA(1, 0, 2). The root of the characteristic equation is 2. Since 2 > 1, the time series is stationary. The roots of equation (1 0.8z)(1 0.5z) = 0 are 1.25 and 2. Since 1.25 > 2 and 2 > 1, the time series is invertible. (iii) (1 1.5B + 0.6B 2 )X t = Z t. The characteristic equation does not have a unit root and the time series is ARIMA(2, 0, 0). The roots of the c.e. are (15 ± i 15)/12. Since (15 ± i 15)/12 = ( )/12 2 = 5/3 > 1, the time series is stationary. It is also (trivially) invertible. (iv) 0.2 causes only a translation of the mean level of the process and may be ignored here: let Y t = X t 0.2 and represent the time series as (1 B)(1 0.2B)Y t = (1 0.5B)Z t. The characteristic equation does have a unit root and the time series is ARIMA(1, 1, 1). The time series is non-stationary (being an I(1) process) but is invertible (the root of 1 0.5z = 0 being greater than 1 in magnitude). Q2. These processes are stationary: EX t and VarX t are constant and Cov [X t, X t k ] depends on k and not on t. (i) EX t + 0.5EX t 1 0.1EX t 2 = EZ t = 0 and hence EX t = 0 t and ρ k = ρ k = E [X t X t k ] /VarX t. Furthermore, Z t is independent of X t k for k 1, so that E [Z t X t k ] = EZ t EX t k = 0. Multiplying across X t + 0.5X t 1 0.1X t 2 = Z t by X t k for k 1 gives X t X t k + 0.5X t 1 X t k 0.1X t 2 X t k = Z t X t k. Solutions to past exam questions are adapted from examiners reports to the Exam Board of the Institute and Faculty of Actuaries and to the Exam Board, Faculty of Actuarial Science and Statistics, Cass Business School, City University. Contact details: Cass Building Room 5071, extension 8478, iqbal@city.ac.uk. 1

2 Taking expectation on both sides yields E [X t X t k ] + 0.5E [X t 1 X t k ] 0.1E [X t 2 X t k ] = 0 and dividing by VarX t on both sides gives ρ k + 0.5ρ k 1 0.1ρ k 2 = 0. When k = 1, ρ ρ 0 0.1ρ 1 = 0 which solves to ρ 1 = 0.5/0.9 = 5/9 (noting that ρ k = ρ k and ρ 0 = 1 by definition). When k = 2, ρ ρ 1 0.1ρ 0 = 0 yielding ρ 2 = ( 5/9) = 17/45. When k = 3, ρ ρ 2 0.1ρ 1 = 0 yielding ρ 3 = 0.1( 5/9) 0.5(17/45) = 11/45. (ii) {X t } in X t + 0.6X t 2 = Z t will clearly have zero autocorrelation at odd lags. ρ k + 0.6ρ k 2 = 0 for k 1. When k = 1, ρ ρ 1 = 0 and ρ 1 = 0. When k = 2, ρ ρ 0 = 0 and ρ 2 = 0.6. When k = 3, ρ ρ 1 = 0 and ρ 3 = 0. (iii) ρ k 1.1ρ k ρ k 2 = 0 for k 1. When k = 1, ρ 1 1.1ρ ρ 1 = 0 and ρ 1 = 1.1/1.18 = 55/59. When k = 2, ρ 2 1.1ρ ρ 0 = 0 and ρ 2 = 1.1(55/59) 0.18 = 1247/1475. When k = 3, ρ 3 1.1ρ ρ 1 = 0 and ρ 3 = 1.1(1247/1475) 0.18(55/59) = 5621/7375. (iv) ρ k + αρ k 1 + α 2 ρ k 2 + α 3 ρ k 3 = 0 for k 1. When k = 1, ρ 1 (1 + α 2 ) + α + α 3 ρ 2 = 0. When k = 2, ρ 1 (α + α 3 ) + α 2 + ρ 2 = 0. (Note in the above that ρ 0 = 1 and ρ k = ρ k ). Hence ρ 1 = α/(1 + α 2 ) and ρ 2 = 0. When k = 3, ρ 3 + αρ 2 + α 2 ρ 1 + α 3 ρ 0 = 0. Hence, ρ 3 = α 5 /(1 + α 2 ). Q3. (i) Y t = Z t βz t 1. VarY t = VarZ t + β 2 VarZ t 1 = σ 2 Z (1 + β2 ). Cov [Y t, Y t 1 ] = Cov [Z t βz t 1, Z t 1 βz t 2 ] = βσ 2 Z. Cov [Y t, Y t k ] = 0 for k 2. Since ρ k = γ k /γ 0, ρ 1 = β/(1 + β 2 ) and ρ k = 0 for k 2. (ii) Y t = Z t + 2.4Z t Z t 2. VarY t = σ 2 Z ( ) = 7.4σ 2 Z. Cov [Y t, Y t 1 ] = Cov [Z t + 2.4Z t Z t 2, Z t + 2.4Z t Z t 2 ] = ( )σ 2 Z = 7.4σ2 Z. Cov [Y t, Y t 2 ] = 0.8σ 2 Z. Cov [Y t, Y t k ] = 0 for k 3. Since ρ k = γ k /γ 0, ρ 1 = 4.32/7.4 = , ρ 2 = 0.8/7.4 = and ρ k = 0 for k 3. Q4. (i) The correlogram is a plot of autocorrelation function against lag, i.e. ρ k vs. k. An MA(q) process has a correlogram with a cutoff at lag q, whereas a stationary AR process has a correlogram that tapers off gradually (possibly with some damped oscillations) to zero as the lag increases. Comments. Plotting the correlogram is therefore a way of distinguishing MA from AR time series, although in practice estimation errors may mean that it is not so easy to discern a difference. 2

3 (ii) The process is assumed to be stationary and we can proceed to find the autocorrelation function by using the usual method for stationary time series. The time series is shifted by µ and EX t = µ t. Let Y t = X t µ. Then, EY t = 0 t. The autocovariance function is γ k = Cov [X t, X t k ] = Cov [Y t, Y t k ] = E[Y t Y t k ]. Recall that γ k = γ k. Also, EZ t = 0 and VarZ t = σ 2 Z t. Multiply Y t αy t 1 = Z t βz t 1 by Y t k on both sides and take expectations to obtain γ k αγ k 1 = E[Y t k Z t ] βe[y t k Z t 1 ]. Observe that Z t is statistically independent of Y t k, k 1. Therefore, when k = 0, γ 0 αγ 1 = E[Y t Z t ] βe[y t Z t 1 ], (1) when k = 1, γ 1 αγ 0 = βe[y t Z t 1 ], (2) when k > 1, γ k αγ k 1 = 0. (3) The non-zero cross-correlations involving Y t and Z t are easily found. Multiply Y t αy t 1 = Z t βz t 1 by Z t on both sides and take expectations to obtain E[Y t Z t ] = σz 2 t, noting that {Z t} is a sequence of independent and identically distributed random variables. Also, multiply Y t αy t 1 = Z t βz t 1 by Z t 1 on both sides and take expectations to obtain E[Y t Z t 1 ] αe[y t 1 Z t 1 ] = βσz 2. Hence, E[Y t Z t 1 ] = σz 2 (α β). Equation (1) may be simplified to γ 0 αγ 1 = σz 2 [1 αβ + β2 ] and equation (2) simplifies to γ 1 αγ 0 = βσz 2. Solving these two equations simultaneously yields: γ 0 = σ2 Z (1 2αβ + β2 ) 1 α 2, γ 1 = σ2 Z (α β)(1 αβ) 1 α 2. Finally, equation (3) solves to γ k = α k 1 γ 1, for k > 1. Since ρ k = γ k /γ 0 by definition, 1 k = 0 ρ k = (α β)(1 αβ)/(1 2αβ + β 2 ) k = 1 α k 1 ρ 1 k > 1. (iii) The correlogram of the ARMA(1, 1) time series has a kink at lag 1. At lag 1, both the MA and AR components affect the correlogram. From lag 2 onwards, it looks rather like the correlogram of an AR series and it decays exponentially. Q5. Comments. If the process were known (or assumed) to be stationary over time, then EX t = 0 and ρ k = 0.4 k, using the properties of a stationary AR(1) process. But the process has a finite history (with a given initial condition) and so it will be nonstationary for t <. Solution. (i) In terms of the backward shift operator, X t = Z t /(1 0.4B) = Z t + 0.4Z t Z t 2 + 3

4 The process is arbitrarily started at t = 0 with X 0 = x 0 and X j = 0 for j 1. Hence, Z 0 = x 0 and Z j = 0 for j 1. Therefore, X t = 0.4 j Z t j + x t. {Z t } is a sequence of zero-mean random variables. Hence, EX t = x t for t 0. (ii) Note that and that and also that X t+k EX t+k = X t EX t = 0.4 j Z t j t+k j Z t+k j = j= k 0.4 j+k Z t j Cov [X t, X t+k ] = E {(X t EX t )(X t+k EX t+k )}. For k 0, Cov [X t, X t+k ] = E 0.4 j Z t j j= k 0.4 j+k Z t j = σ2 Z 0.4 j 0.4 j+k (where σz 2 = VarZ t) because {Z t } is a sequence of zero-mean independent and identically distributed random variables. Hence, for k 0 and t 0. Cov [X t, X t+k ] = σ 2 Z0.4 k ( t )/( ) (4) (iii) The process is not stationary for t < as both EX t and Cov [X t, X t+k ] vary with t. (iv) The process is stationary in the limit since the magnitude of the root of the characteristic equation 1 0.4z = 0 is 2.5 and 2.5 > 1. As t, EX t = x t 0 and Cov [X t, X t+k ] σ 2 Z 0.4k /( ) and VarX t σ 2 Z /( ). In the limit, the autocorrelation coefficient ρ k = Cov [X t, X t k ] /VarX t = 0.4 k. (v) Suppose that the initial value x 0 at time 0 is itself a random variate that is normally distributed with zero mean and variance σ 2 Z /( ). This distribution is the same as that of X t t in the stationary AR(1) process { X t } given by X t = 0.4 X t 1 +Z t. One may therefore choose to put x 0 = X 0 = Z j0.4 j. Since X t = t 1 0.4j Z t j + x t, it follows that X t = 0.4 j Z t j t Z j 0.4 j = 0.4 j Z t j. In other words, an initial condition that is random and taken from N(0, σ 2 Z /( )) yields an AR(1) process that is stationary ab initio. 4

5 Q6. (i) EW t = EX t +EY t is constant over time. The autocovariance of the process {W t } is Cov [W t, W t k ] = Cov [X t + Y t, X t k + Y t k ] = Cov [X t, X t k ] + Cov [Y t, Y t k ] and depends on lag k and not on time t. (ii) X t = ln CPI t ln CPI t 1 and hence ln CPI t = t j= X j. Since X t is an AR(1) process, it is clear that ln CPI t is integrated once and is ARIMA(1, 1, 0). Alternatively, note that (1 B)(1 αb) ln CPI t = Z t + µ(1 α) and a unit root occurs. Solutions to Past Exam Questions Q1. (i) (a) γ 0 = Var(e t +β 1 e t 1 ) = (1+β 2 1 )σ2 e and γ 1 = Cov [e t + β 1 e t 1, e t 1 + β 1 e t 2 ] = β 1 σ 2 e, with γ k = 0 for k > 1. This gives ρ 0 = 1, ρ 1 = β 1 /(1 + β 2 1 ), ρ k = 0 otherwise. (b) Invertibility requires that β 1 < 1, so that the sum X t β 1 X t 1 +β 2 1 X t 2 + converges. µ and σ e are irrelevant. (ii) We need to solve (1 + β 2 1 )σ2 e = 14.5, β 1 σ 2 e = 5.0. Eliminating σ 2 e, we have 1 + β 2 1 = 2.9β 1, or β 1 = 1 2 (2.9 ± ( )) = 2.5 or 0.4. β 1 = 2.5 corresponds to σ 2 e = 2, whereas β 1 = 0.4 corresponds to σ 2 e = For invertibility, solve 1 + β 1 z = 0. In the first case, z = 0.4 (no good); in the second, z = 2.5 (OK). Q2. (i) (a) γ 1 = Cov [X t, X t 1 ] = Cov [α 1 X t 1 + α 2 X t 2 + e t, X t ] = α 1 γ 0 +α 2 γ 1 +0, since e t is independent of X t 1. (b) Similarly γ 2 = α 1 γ 1 + α 2 γ 0 and γ 0 = α 1 γ 1 + α 2 γ 2 + Cov [X t, e t ]. A further application of the same technique gives Cov [X t, e t ] = σe. 2 Thus, γ 1 = α ( ) 1 γ 0, γ 2 = α 2 + α2 1 γ 0. 1 α 2 1 α 2 (c) ρ k is found by the relation ρ k = γ k /γ 0. (ii) We have ˆα 1 = r 1 (1 ˆα 2 ) and ˆα 2 + ˆα 2 1 /(1 ˆα 2) = r 2, which are solved by ˆα 1 = r 1(1 r 2 ) 1 r 2 1, ˆα 2 = r 2 r1 2 1 r1 2. Q3. (i) γ 1 = 0.8γ 0 0.4γ and γ 2 = 0.8γ 1 0.4γ Hence, γ 1 = 4 7 γ 0 and γ 2 = 2 35 γ 0, implying that ρ 1 = 4 7 and ρ 2 = 2 φ 1 = ρ 1 = 4 7 and φ 2 = ρ 2 ρ 2 1 = ρ 2 1 (ii) The acf will reduce to zero as k increases; the pacf, however, will be equal to zero for all k >

6 Q4. (i) Identification: Determination of the parameters p, d and q in the ARIMA(p, d, q) model. Estimation: Determination of the p AR parameters α 1, α 2, α 3,..., α p and the q MA parameters β 1, β 2, β 3,..., β q in the stationary ARMA(p, q) model for the dth differences of the observed series. Diagnosis: Testing the goodness of fit of the proposed model. (ii) The optimal value of d is the smallest value that produces a stationary series. Three criteria used are: (a) sample acf should tend rapidly to zero (b) sample variance should be mimised (c) series should exhibit a constant mean (iii) Parismony is using the lowest values for the parameters p, d and q which adequately model the observed series, i.e. additional parameters are only added if they significantly improve the fit of the model. (iv) Q 10 = k=1 r2 k = = 9.07 Under null hypothesis of independence of residuals, Q 10 has a χ 2 distribution with 10 1 = 9 degrees of freedom. From tables, 95th percentile of χ 2 9 is Hence, accept H 0. (v) Let M = number of tps = 48. Let N = number of residuals = 78. Under null hypothesis of independence of residuals, ( ) M app 2 16N 29 N (N 2), 3 90 Then, M N ( , 1219 ) 90. ( ) P (M 48) = P Z (1219/90) = P (Z 0.59) = 1 Φ(0.59) = accept H 0 at 5% level. (Exam Board, Faculty of Actuarial Science and Statistics, Cass Business School, City University) Q5. (i) Consumer prices do tend to exhibit regular seasonal variation, though not a great deal these days. And, since prices tend to go up rather more than they come down, it is probably worth including a trend term in any model. It is certainly possible to test whether the trend term is equal to zero. (ii) (a) X n+1 x n = α(x n x n 1 ) + e n+1 + βe n. (b) The parameters are α, β and σe. 2 The trend removal process would have accounted for any µ parameter. (iii) ˆx n (1) = E [X n+1 x n,..., x 1 ] = x n + α(x n x n 1 ) + E [e n+1 + βe n x n,..., x 1 ]. Now e n+1 has mean 0 and is conventionally supposed independent of everything that happens before n. On the other hand, e n can be deduced from past data, e.g. e n = x n x n 1 α(x n 1 x n 2 ) βe n 1, which may be iterated back to get e n in terms of the known x and the known e 0. Thus, ˆx n (1) = x n + α(x n x n 1 ) + βe n. Similarly, ˆx n (2) = E [X n+2 F n ] = E [X n+1 + α(x n+1 x n ) + e n+2 + βe n+1 F n ] = (1 + α)ˆx n (1) αx n. 6

7 We see that X n+1 ˆx n (1) = e n+1, so that the prediction variance is just Var(e n+1 ) = σ 2 e. (iv) Since e n = x n ˆx n 1 (1), we have ˆx n (1) = x n + α(x n x n 1 ) + β(x n ˆx n 1 (1)). If we set α = 0 and β = ξ, the equation is identical to the updating equation for exponential smoothing. (v) An ARIMA(p, d, q) model is I(d); in this case, x is I(1). A stationary (I(0)) model has an equilibrium distribution: the distribution of the forecast of X n+k would converge to equilibrium for large k. An I(1) process is the partial sum of an I(0) process, so would have increasing variance, even if the mean happened to be stable. (vi) Two series {x} and {y} are cointegrated if both are I(1) but there are some constants a and b such that {ax + by} is stationary. Two processes are likely to be cointegrated if one drives the other, or if both are driven by the same underlying process. In the given instance the suggestion is certainly worth investigating. 7