Stat 153 Time Series. Problem Set 4

Transcription

1 Stat Time Series Problem Set Problem I generated 000 observations from the MA() model with parameter 0.7 using the following R function and then fitted the ARMA(, ) model to the data. order = c(, 0, )), which gave the following output: > ma. <-arima.sim(n = 000, list(ma = 0.7)) > arima(x = ma., order = c(, 0, )) Call: arima(x = ma., order = c(, 0, )) Coefficients: ar ma ma intercept s.e sigma^ estimated as 0.98: log likelihood = -., aic = 8. Because the data have been simulated from an MA() model with parameter 0.7, I expected the estimated coefficients of ar and ma to be close to zero and the estimated coefficient of ma to be close to 0.7. However, the fitted model does not appear to be close to the model that generated the data because of parameter redundancy. The fitted model can be written as ( 0.78B)X t = ( 0.08B 0.B )W t ( 0.78B)X t = ( + 0.9B)( 0.7B)W t We see that we can almost cancel ( 0.78B) with ( 0.7B). If we could do this we would have a model that is almost AR() with parameter λ = 0.7. Thus, we conclude that although our model may not, at first glance, look close to the AR() model generating the data, in reality it is very close. Problem (a) There are many approaches here. If we look at contiguous subseries we can see that the mean for the subseries increases monotonically. We can also examine the and observe that it appears to decay roughly linearly, suggestive of a trend. In addition, you are welcome to use kpss.test(), PP.test(), adf.test(), etc. Make sure you know the null hypothesis of these tests and how to interpret the result. Again, these tests sometimes might have different outcomes (To be safe, I usually try at least two). If errors are discovered, please kindly report them to yalunsu@berkeley.edu

2 (b) > yt <- jj/lag(jj,-) > plot(yt); acf(yt); pacf(yt) Series yt Series yt yt Partial Time 0 Figure : Plot of y t, Sample, and P (c) Similar to (a), there are many approaches here. Instead of using built-in functions, we can estimate a separate auto-covariance function for disjoint, contiguous subseries and see if they are similar. The figure on the next page is a bit difficult to interpret, but it seems that different windows have markedly different sample s. Thus, we conclude the series is probably not stationary. Again, we will be open minded in grading this problem as long as you provide a valid explanation. my_acovf <- function(h, ts){ /length(ts) * sum((ts - mean(ts)) * (lag(ts, h) - mean(ts))) } my_acorf <- function(h, ts){ ts <- as.ts(ts) my_acovf(h,ts) / my_acovf(0, ts) } my_acorf_vec <- function(ts){ sapply(:(length(ts)-), my_acorf, ts = ts) } windowed_acf <- simplifyarray(tapply(jj, rep(:, each = 7), my_acorf_vec)) matplot(t(windowed_acf), type = "b")

3 t(windowed_acf) Figure : Problem > eurogdp <- read.csv(file=" > eurogdp <- as.numeric(head(t(unname(as.vector(eurogdp[,])))[-(:)], -)) Series eurogdp Series diff(eurogdp, ) Series diff(eurogdp, ) Series diff(eurogdp, ) Figure :

4 (a) Differencing at order seems the most appropriate because there is no real indication of a trend in the for this order of differencing. (b) > arima...0 <- arima(eurogdp, order= c(,,0)) > arima...0 Call: arima(x = eurogdp, order = c(,, 0)) Coefficients: ar ar s.e sigma^ estimated as 0: log likelihood = -80.9, aic = 97.7 (c) Since the best linear forecast is a linear operator, we then have Y t t+ = P(X t+ + Y t Y,, Y t ) = Y t + P(X t+ Y,, Y t ) () = Y t + X t t+ The last equality holds because X t+ is a function of X,, X t, which can, using the definition of X t be expressed as a function of Y,, Y t. (d) Y t t+ = Y t + X t t+ = Y t + λ X t + λ X t = Y t ( + λ ) + Y t (λ λ ) + λ Y t () Again, this calculation can be easily done in R. You are welcome to use predict(arima...0, ), but I used forecast package here. > forecast(arima...0,, 9) Point Forecast Lo 9 Hi Problem This is the same data we used for Problem Set. Certainly, you may find some very complex transformations/models with the lowest AIC or BIC. Nonetheless, simpler models are nice for various reasons. Also, AIC & BIC have their limitations. You will not lose any point if you have showed sufficient reasoning to support your final model provided that it is not too complicated. ARIMA models produced by auto.arima will not be accepted in this homework setting. There are many discussions on auto.arima arguing its performance, but it is a very good tool to play around with the data at the first glance.

5 (a) & (b) > df <-read.csv(file=" skip=) > df.ts <- ts(df$temp,start = 80) > acf(df.ts) > acf(diff(df.ts)) > pacf(diff(df.ts)) > fit <- Arima(df.ts,order = c(,,)) Series: df.ts ARIMA(,,) Coefficients: ar ma s.e sigma^ estimated as 0.00: log likelihood=9. AIC=-7. AICc=-7.9 BIC=-.8 > mean(diff(df.ts)) [] > sd(diff(df.ts)) [] 0. > tsdiag(fit) Series df.ts Series diff(df.ts) Series diff(df.ts) Partial Figure : of the undifferenced data and the first differences, and P of the first differences. We began by looking at the of the original data sequence, which seems to decay very slowly. In particular, the process is probably not an ARMA process. The and P of the first differences look much more plausible. If the first differences were an AR or an MA process, we

6 would expect either the or the P to cut off after a finite lag. Since this doesn t seem to happen, we will propose an ARMA model; the simplest candidate is ARIMA(,, ), so let s start with that. Using Arima function, we estimated the AR coefficient to be 0. (with a standard error of 0.0) and the MA coefficient to be 0.7 (with a standard error of 0.0). The mean (of the differenced sequence, so it corresponds to the drift of the original sequence) was estimated at with a standard error of 0.. (c) & (d) > tsdiag(fit) > shapiro.test(fit$residuals) Shapiro-Wilk normality test data: fit$residuals W = 0.997, p-value = 0. > fit <- Arima(df.ts, order=c(,,)) > fit Series: df.ts ARIMA(,,) Coefficients: ar ma ma ma s.e sigma^ estimated as 0.00: log likelihood=. AIC=-7.0 AICc=-7.8 BIC=-9. > shapiro.test(fit$residuals) Shapiro-Wilk normality test data: fit$residuals W = 0.997, p-value = 0.77 Some diagnostic plots can be seen in Figure. The correlations of the residuals were not significant, although some of were fairly close. The Shapiro Wilk test returned a fairly reasonable p-value and the QQ-plot suggests the the residuals are approximately normal. Nevertheless, we could attempt to reduce the almost-significant correlations at lags and by introducing some more MA terms. In fact, doing so reduces the AIC of the fitted model. After trying MA degrees of through, we found that an ARIMA(,, ) model had the lowest AIC (-7.0). It s Shapiro- Wilk p-value was 0.77 and the other diagnostic plots produced by tsdiag seemed reasonable (not included), but its BIC was actually higher than that of the ARIMA(,, ) model. In fact, the

7 ARIMA(,, ) model had the lowest BIC of all the alternatives we tried. Since simpler models are nice for various reasons, we decided to stick with ARIMA(,, ) for the predictions (i.e. ARMA(, ) after we difference the sequence). Standardized Residuals Time of Residuals p values for Ljung Box statistic p value lag Figure : Diagnostic plots for the ARMA(, ) model on the differenced data. Normal Q Q Plot Sample Quantiles Theoretical Quantiles Figure : Q-Q plots for the ARMA(, ) model on the differenced data. 7

8 (e) In this part, we are performing a hold-back forecasting. Thus, I redo the fitting using the data up to 00. The predictions are shown in Figure 7. > fit.holdback <- Arima(window(df.ts, end=00),order = c(,,)) > pred.matrix <- matrix(nrow =, ncol=) > pred.matrix[,] <- temp.pred$lower > pred.matrix[,] <- temp.pred$upper > pred.matrix[,] <- 0:0 > plot(window(df.ts, end=00),type="o", xlim=c(80, 08), ylim=c(-0.,.0), + ylab="global Temperature") > points(window(df.ts, start=0), col="green", type="o") > points(temp.pred$mean, col= "red", type="o") > lines(pred.matrix[,c(,)], col= "blue", lty=) > lines(pred.matrix[,c(,)], col= "blue", lty=) Global Temperature Time Figure 7: Predictions from the ARIMA(,, ) model. Otherwise, you cannot really forecast the data from 0 to 0 since the temperature data is given up to 0. 8