Lab 6

Jonathan Zhang 44311103 Zhi Yi Ye 40357105 Dustin Johnson 11338118

data <- read.csv("C:/Users/Jonathan/SkyDrive/Jonathan Zhang/UBC/Stat Courses/Stat 443/dataTempPG.csv")
data.ts <- ts(data$Annual)

You can also embed plots, for example:

plot(data.ts)

Not stationary, there is clearly a trend

par(mfrow = c(2, 1))
acf(data.ts)
pacf(data.ts)

residuals.ts <- ts(lm(data$Annual ~ data$Year)$residuals)

par(mfrow = c(3, 1))
plot(residuals.ts)
acf(residuals.ts)
pacf(residuals.ts)

We are deciding whether it is ARIMA(0,0,1) or ARIMA(0,0,9). Because the p and d parameters are 0, we are basically asking if our residuals follow an MA(1) or MA(9) process. We think it follows an MA(1) process because the acf cuts off after 1 (somewhat). MA(9) is definitely a wrong fit.

arima1 <- arima(residuals.ts, order = c(0, 0, 1), include.mean = F)
arima1

## 
## Call:
## arima(x = residuals.ts, order = c(0, 0, 1), include.mean = F)
## 
## Coefficients:
##         ma1
##       0.235
## s.e.  0.103
## 
## sigma^2 estimated as 0.972:  log likelihood = -126.5,  aic = 256.9

tsdiag(arima1)

confint(arima1)

##       2.5 % 97.5 %
## ma1 0.03294 0.4373

Model: X_t = 0.2351 Z_t

arima9 <- arima(residuals.ts, order = c(0, 0, 9), include.mean = F)
arima9

## 
## Call:
## arima(x = residuals.ts, order = c(0, 0, 9), include.mean = F)
## 
## Coefficients:
##          ma1    ma2    ma3     ma4    ma5    ma6     ma7     ma8     ma9
##       -0.037  0.063  0.003  -0.199  0.155  0.130  -0.088  -0.476  -0.552
## s.e.   0.103  0.131  0.097   0.132  0.115  0.121   0.112   0.118   0.109
## 
## sigma^2 estimated as 0.628:  log likelihood = -113,  aic = 246.1

tsdiag(arima9)

confint(arima9)

##        2.5 %   97.5 %
## ma1 -0.23802  0.16428
## ma2 -0.19455  0.32038
## ma3 -0.18722  0.19404
## ma4 -0.45809  0.05914
## ma5 -0.07092  0.38176
## ma6 -0.10748  0.36834
## ma7 -0.30849  0.13159
## ma8 -0.70649 -0.24453
## ma9 -0.76521 -0.33851

Looking at the confidence intervals, all our 9 parameters are insignificant, this model is probably incorrect! Now we check AIC

AIC(arima1)

## [1] 256.9

AIC(arima9)

## [1] 246.1

Our AIC results do not agree with the model we deemed as appropriate. However, this is not surprising because the ARIMA(0,0,9) model obviously overfits the model (more parameters) and AIC does not penalize the number of parameters that much. It does not agree with our intuition. We think a smaller model is better, however AIC may not agree, this is normal.