THE UNIVERSITY OF CHICAGO Graduate School of Business Business 41202, Spring Quarter 2003, Mr. Ruey S. Tsay

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1 THE UNIVERSITY OF CHICAGO Graduate School of Business Business 41202, Spring Quarter 2003, Mr. Ruey S. Tsay Homework Assignment #2 Solution April 25, 2003 Each HW problem is 10 points throughout this quarter. The final grade for HW is the average (times 10) over the total number of HW problems.. 1. Consider the quarterly U.S. unemployment rate from 1948.I to 2003.I, for 221 observations. The quarterly data are obtained by averaging over seasonally adjusted monthly data, which are in turn obtained from the Federal Reserve Bank at St Louis, website Build a time series model for the series and use the model to forecast the unemployment rate for the 2nd and 3rd quarters of In addition, compute the average period of business cycles if they exist? [Note that there are more than one model fits the data well. You only need an adequate model.] Explanatory Analysis of the Series The first thing one should do is an exploratory analysis of the series to try and figure out what the data looks like. Let X t be the series of quarterly unemployment rate, in percentages. Figure 1 shows that there seems to be some sort of cycle, since every so often you have peaks, although they do not seem to occur at regular intervals. This may be an indication of a business cycle or a combination of business cycles. X(t) (in percentages) Time Figure 1: Time series plot for quarterly US unemployment rate. Figure 2 shows the ACF and the PACF for X t. On the ACF we see that the lags are decaying somewhat faster than linearly, but not quite exponentially. This could be an indication that the series may not be stationary and, thus, a (regular) difference could be needed. The PACF shows 1

2 ACF of X(t) PACF of X(t) ACF Partial ACF Lag Lag Figure 2: Autocorrelation function and partial autocorrelation function for quarterly US unemployment rate. that lag 1 and 2 are significantly different from zero; the first lag is quite big actually, very close to one, this is also an indication that a difference might be needed. Simple Solution After considering a couple of models, a simple model that fits reasonably well the series is (1 φ 1 B φ 2 B 2 φ 4 B 4 )(1 B)X t = a t (1) where (1 B)X t means that we have used one regular difference on the series. The commands to produce and estimate such model in SCA are tsm m1. (1,2,,4)series(1) = noise. estim m1. hold resid(r1). The output from SCA is displayed bellow. SUMMARY FOR UNIVARIATE TIME SERIES MODEL M1 1 UMP RANDOM ORIGINAL (1-B ) 1 UMP AR 1 1 NONE UMP AR 1 2 NONE UMP AR 1 4 NONE EFFECTIVE NUMBER OF OBSERVATIONS R-SQUARE RESIDUAL STANDARD ERROR E+00 2

3 The command for S-Plus is m1 <- arima.mle(series,model=list(order=c(4,1,0),ar.opt=c(t,t,f,t))) where series is simply the name you gave to the series. Also note that the results are exactly the same using both packages (although S-Plus does not give you the t-stats by default). Thus, the estimated model is (1 0.70B B B 4 )(1 B)X t = a t. If we solve that characteristic polynomial, we obtain the following complex roots (with their respective conjugates): { i, i} resulting in two business cycles. The first one with an average length of quarters (or 9 months); and the second one with an average length of quarters (almost 3 years.) Using SCA the forecast for the 2nd and 3rd quarters of 2003 with model (1) are fore m1. nofs 2. hold fore(f1), stderr(e1) 2 FORECASTS, BEGINNING AT 221 TIME FORECAST STD. ERROR ACTUAL IF KNOWN The results are the same using S-Plus. The command is fore.m1 <- arima.forecast(series,m1$model,n=2) Notice that we said that the model in equation (1) fits reasonably well because actually, if we compute the ACF and PACF on the residuals, there are lags that are significantly different from zero (individually for both and jointly for the ACF). In particular Q(12) = 20.2, and the cut-off point of a χ 2 0.5,ν = 16.92, where ν := m p q = = 9 degrees of freedom. If we look at the actual ACF and PACF (not shown here) we see that lags 4, 8 and 12 are significant, implying some sort of seasonality (remember that the data is quarterly and thus, the periodicity is 4.) Thus, a more detailed analysis is required. Solution Taking into Account Seasonality This part is not required in the solution of the homework, but it is shown as an example on how to deal with such an effect. After doing some exploratory analysis, a simple and good model was found. In the sense that it has the least number of parameters that produce a good fit to the data, and shows uncorrelated residuals. The model is (1 φ 1 B)(1 B)X t = (1 θ 4 B 4 θ 8 B 8 )a t. (2) The interpretation of this models goes as follows: rate at which the unemployment rate grows at a given quarter is influenced by the previous quarter (AR part); there will also be an influence on on the rate of growth from a shock that occurred in the same quarter but one and two years before 3

4 (seasonal MA part). In this model there is no stochastic cycle, we are already assessing from some periodicity via the seasonal moving average part. The specification and estimation, using SCA, gives tsm m2. model is (1)series(1)=(4,8)noise estim m2. hold resid(r2). method exact. SUMMARY FOR UNIVARIATE TIME SERIES MODEL M2 1 UMP RANDOM ORIGINAL (1-B ) 1 UMP MA 1 4 NONE UMP MA 1 8 NONE UMP AR 1 1 NONE EFFECTIVE NUMBER OF OBSERVATIONS R-SQUARE RESIDUAL STANDARD ERROR E+00 X(t) (in percentages) Time Figure 3: Forecast for the 2nd and 3rd quarters of 2003, for the US unemployment rate. The forecasts, shown in Figure 3, are fore m2. nofs 2. hold fore(f2), stderr(e2) 2 FORECASTS, BEGINNING AT 221 TIME FORECAST STD. ERROR ACTUAL IF KNOWN

5 They seem to be a bit smaller than the ones obtained by the model (1). In S-Plus do m2 <- arima.mle(series,model=list(list(order=c(1,1,0)),list(order=c(0,0,2),period=4)),max.iter=40,max.fcal=40) diag.m2 <- arima.diag(m2,type="l",lag.max=16) fore.m2 <- arima.forecast(series,model=m2$model,n=2) giving exactly the same results. Note that in the forecast we do not need to add the mean, since the differenced process has already mean zero and this has been specified into the model. Finally, Figure 4 shows the diagnostic analysis for the residuals in model (2). There seems not be no pattern or information left. PACF ACF ARIMA Model Diagnostics: unemp.ts Plot of Standardized Residuals ACF Plot of Residuals PACF Plot of Residuals P values of Ljung Box Chi Squared Statistics p value Lag ARIMA() Model with Mean 0 Figure 4: Diagnostics for the residuals of model (2). 2. Consider the monthly log returns of CRSP equal-weighted index from January 1962 to December of (a) Build an AR model for the series and check the fitted model. From inspection of the series (ACF and PACF), we observe that the mean is significantly different from zero (with a t-stat of 3.7); we can also see that the 1st and 2nd lags in the PACF are significant, indicating a possible AR(2) model. The 8th lag of the PACF is also significant, but it could be due to randomness only; plus it does not have a clear interpretation. NAME OF THE SERIES EW TIME PERIOD ANALYZED TO 492 MEAN OF THE (DIFFERENCED) SERIES STANDARD DEVIATION OF THE SERIES T-VALUE OF MEAN (AGAINST ZERO)

6 AUTOCORRELATIONS ST.E Q I IX+XXX XI XI XI I I XI X+XI I IX IX IXX PARTIAL AUTOCORRELATIONS ST.E I IX+XXX XXI I I I XI XI X+XI IX IX IX IXX Let rt ew denote the monthly log returns on the equally-weighted index. After evaluating a couple of different models, we came up with the following The estimated coefficients are tsm m3. model is (1)ew = c + noise. estim m3. hold resid(r3). (1 φ 1 B) t = φ 0 + a t (3) 6

7 SUMMARY FOR UNIVARIATE TIME SERIES MODEL M3 EW RANDOM ORIGINAL NONE 1 C CNST 1 0 NONE EW AR 1 1 NONE EFFECTIVE NUMBER OF OBSERVATIONS R-SQUARE RESIDUAL STANDARD ERROR E+01 We can see that the two coefficients are significant, but the fit is rather poor, since the R 2 is not even 5%. The residuals, however, do not show visible pattern, and Q(12) = 16.1 which, compared to χ ,11 = does not give enough evidence to reject the null hypothesis (of no serial correlation). (b) Build an MA model for the series and check the fitted model. Form the ACF of the original series, we observe that lag 1 is significant, pointing toward an MA(1). Again, although 8 is seems significant, it would create a very complicated model that may not necessarily be a better predictor. Thus, we can estimate the simple model with estimates tsm m4. model is ew = c + (1)noise. estim m4. hold resid(r4). method exact. t = φ 0 + (1 θ 1 B)a t (4) SUMMARY FOR UNIVARIATE TIME SERIES MODEL M4 EW RANDOM ORIGINAL NONE 1 C CNST 1 0 NONE EW MA 1 1 NONE EFFECTIVE NUMBER OF OBSERVATIONS R-SQUARE RESIDUAL STANDARD ERROR E+01 The fit close to the model in Eq. (3). Similarly, the residuals do not show enough evidence of serial correlation. In particular Q(12) = 13.6 which, compared to the same cut-off point as before, does not lead us to reject the null. 7

8 (c) Compute 1- and 2-step ahead forecasts of the AR and MA models built in the previous questions. Table 1 shows the forecasts and their standard errors (in parenthesis). As expected, the MA forecasts go back to the mean after one step, while the AR forecasts tend to the mean in a slower fashion. Table 1: 1-step to 5-step ahead forecasts for t T T T T T AR(1) MA(1) (1) (5.7081) (5.6896) (2) (5.8299) (5.8396) (3) (5.8352) (5.8396) (4) (5.8354) (5.8396) (5) (5.8354) (5.8396) (with standard errors). (d) Compare the fitted AR and MA fitted models. The MA seems to give a slightly better fit. It has a higher R 2 and the residuals seem to behave better. The 1-step ahead forecast of the MA(1) model is low because of a large shock in the last period of the data. The standard errors for the predictions are huge, making the forecasts basically not significant. This could be an indication of no arbitrage. Note that for this problem, the results obtained in SCA and S-Plus may vary a little bit, in particular with respect of the forecasts. 3. The file m-geln2602.dat contains monthly log returns, in percentage, of GE stock from 1926 to 2002 for 924 observations. (a) Find the sample mean and sample standard deviation of the series. Is the sample mean different from zero at the 5% significance level? This is simple to do in SCA. Once you compute the ACF, the first part of the output gives the result. In S-Plus, simply use the command t.test(x), where x is the name of your variable. As shown in part (c), the sample mean of the series is The t-statistic is 3.6, well above the cut-off point (1.96) leading, thus, to reject the null that the mean is equal to zero at the 5% significance level. (b) Compute the PACF of the series to confirm that lags 1 and 3 are significant at the 5% level. Indeed, the plot shows that the 3rd lag is significant. We can also see this from the estimate itself and the standard error. PARTIAL AUTOCORRELATIONS ST.E

9 I IXX XI X+XI IXX IX XI I IXX I I I IXX (c) Fit an AR(3) model to the series. What is the fitted model? What is the expectation of the series implied by the model? After removing the non-significant coefficients, we obtained the model (via re-estimation) with estimates tsm m5. model is (3)ge = c + noise. estim m5. hold resid(r5). (1 φ 3 B 3 )r vw t = φ 0 + a t, (5) SUMMARY FOR UNIVARIATE TIME SERIES MODEL M5 GE RANDOM ORIGINAL NONE 1 C CNST 1 0 NONE GE AR 1 3 NONE EFFECTIVE NUMBER OF OBSERVATIONS R-SQUARE RESIDUAL STANDARD ERROR E+01 From (5), it is easy to see that the expectation of the series implied by the model is which is very close to our original estimate. E(r vw t ) = φ 0 1 φ 3 = = ; (d) Compute the ACF of the series to confirm that lag 1 and lag 3 are significant at the 5% level. Similarly we have that the plot shows a significant 3rd lag. We can also see this from the estimate itself and the standard error, as well as from the Q-statistic which is quite large. 9

10 AUTOCORRELATIONS ST.E Q I IXX XI X+XI IX IX XI I IXX IX I XI IXX (e) Fit a MA(3) model to the series. What is the fitted model? What is the expectation of the series implied by the model? The model is r vw t = φ 0 + (1 θ 1 B θ 3 B 3 )a t (6) with estimates tsm m6. model is ge = c + (1,3)noise. estim m6. hold resid(r5). method exact. SUMMARY FOR UNIVARIATE TIME SERIES MODEL M6 GE RANDOM ORIGINAL NONE 1 C CNST 1 0 NONE GE MA 1 1 NONE GE MA 1 3 NONE EFFECTIVE NUMBER OF OBSERVATIONS R-SQUARE RESIDUAL STANDARD ERROR E+01 In this case, equation (6) implies that E(r vw t ) = φ 0 = ; which is even closer to the original estimate of the mean. 10

11 4. Revisit the file d-hwp3dx9302.dat of HW #1. (a) Compute the first 100 lags of ACF of the absolute daily simple returns of Hewlett-Packard stock. Is there evidence of long-range dependence? Why? Note that long memory means ACF decays slowly. Figure 5 shows the ACF of the absolute daily simple returns of Hewlett-Packard stock. Disregarding lag 0 (which is obviously 1), we see that there is a permanent effect in the autocorrelations and that is not decaying with time. This gives evidence of long memory. ACF for HP absolute daily returns ACF Lag Figure 5: ACF for absolute daily simple returns of HP stock (b) Compute the first 100 lags of ACF of the squared daily log returns of value-weighted index. Is there any evidence of long-range dependence? Why? Similarly, Figure 6 displays the ACF of the squared daily log returns of value-weighted index. It also shows a slow decay, implying some sort of long-range dependence. ACF for squared simple daily returns of VW index ACF Lag Figure 6: ACF for squared daily log returns of value-weighted index 11