Assignment_2_S07_Solution

Assignment_2_S07_Sol - Assignment#2 Solution Q1(a Since E Z t = 0 Var Z t = E Z t = s 2 we have 2 E X t = E(a bZ t 1 cZ t 2 = a k = Cov X t X t k =

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Assignment #2 Solution Q1. (a) Since ()0 t EZ = , , we have 2 2 ) ( ) ( s Z E Z Var t t = = otherwise k if bcs k if s c b cases Z Z Cov c Z Z cbCov Z Z bcCov Z Z Cov b X X Cov a cZ bZ a E X E t k t t k t t k t k t t k t t k t t t , 0 1 , 0 , ) ( 3 ) , ( ) , ( ) , ( ) , ( ) , ( ) ( ) ( 2 2 2 2 2 2 2 1 2 2 1 1 1 2 2 1 = = = = + = = + + + = = = + + = + + + + + γ Therefore, it is (weakly) stationary because its mean is constant and the covariance does NOT depend on time t, but only k. (b) 0 ) ( ) 2 sin( ) ( = = t t Z E t X E ))) ( 2 sin( ), 2 sin( cov( ) , cov( k t Z t Z X X k t t k t t k + = = + + If k=0, cov( )= k t t + X X , ) 2 ( sin 2 2 t S If k cov( )=0; , 0 k t t X X + , Therefore it is not (weakly) stationary, since its variance changes, depending on time t. Q1. (a) > data<-read.table("./consumption.txt",header=T) > x1 <- data$TBILL > x2 <- data$RSERV > Tbill <- x1[1:29] > Rserv <- x2[1:29] 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
> x<-cbind(Tbill,Rserv) > X <- as.matrix(x) > y1<-data$CONS > Cons <- y1[1:29] > reg<-lm(Cons~X) > summary(reg) Call: lm(formula = Cons ~ X) Residuals: Min 1Q Median 3Q Max -369.61 -98.65 20.06 94.19 248.43 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 4.812e+03 1.251e+02 38.461 <2e-16 *** XTbill 2.930e+01 1.314e+01 2.231 0.0345 * XRserv 4.131e+00 7.045e-02 58.638 <2e-16 *** --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Residual standard error: 168.8 on 26 degrees of freedom Multiple R-Squared: 0.9933, Adjusted R-squared: 0.9928 F-statistic: 1928 on 2 and 26 DF, p-value: < 2.2e-16 > acf(reg$residuals,lag.max=25) 2
Background image of page 2
Figure 1. Autocorrleation function of the residuals (comments) It is clear to see that the correlations at first lags is significant and at lag 12, 13, and 14 , it has relatively significant correlations, but it is negligible because the values are in the confidence interval. Overall, it damps off according to a mixed exponential decay. We might think of AR(1) process for the residual in the further study.
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 01/12/2012 for the course STAT 443 taught by Professor Yuliagel during the Spring '09 term at Waterloo.

Page1 / 12

Assignment_2_S07_Sol - Assignment#2 Solution Q1(a Since E Z t = 0 Var Z t = E Z t = s 2 we have 2 E X t = E(a bZ t 1 cZ t 2 = a k = Cov X t X t k =

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online