Problem 2

Obtain quarterly time series for Disposable Personal Income FRED/DPI and for Personal Consumption Expenditures FRED/PCEC.

(a) Let y1,t = log P CEt and y2,t = log P DIt. Test these log transformed time series and their first differences for unit root to verify that y1,t and y2,t are I(1).

## AIC(n)  HQ(n)  SC(n) FPE(n) 
##     11     11     11     11

## AIC(n)  HQ(n)  SC(n) FPE(n) 
##     11     11     11     11

## 
## ################################################ 
## # Zivot-Andrews Unit Root / Cointegration Test # 
## ################################################ 
## 
## The value of the test statistic is: -2.1592

## 
## ################################################ 
## # Zivot-Andrews Unit Root / Cointegration Test # 
## ################################################ 
## 
## The value of the test statistic is: -4.3698

## 
## ################################################ 
## # Zivot-Andrews Unit Root / Cointegration Test # 
## ################################################ 
## 
## The value of the test statistic is: -7.9301

## 
## ################################################ 
## # Zivot-Andrews Unit Root / Cointegration Test # 
## ################################################ 
## 
## The value of the test statistic is: -2.2093

## 
## ################################################ 
## # Zivot-Andrews Unit Root / Cointegration Test # 
## ################################################ 
## 
## The value of the test statistic is: -4.3923

## 
## ################################################ 
## # Zivot-Andrews Unit Root / Cointegration Test # 
## ################################################ 
## 
## The value of the test statistic is: -7.7545

According to the results it looks like both of these time series are of order I(2) but we will just assume that the order is I(1) since the unit root test is very close on the second differenced data. So we will say that the sereis are indivisually of order I(1).

(b) The disposable income hypothesis suggest that PCEt = φDPIt where φ is the marginal propensity to consume. Consumption and disposable income should thus be growing at the same rate, and log P CEt −log P DIt −log φ should be I(0). Perform the trace and max eigenvalue tests for cointegration of y1,t and y2,t. Interpret the results.

## 
## ###################### 
## # Johansen-Procedure # 
## ###################### 
## 
## Test type: trace statistic , with linear trend 
## 
## Eigenvalues (lambda):
## [1] 0.10598492 0.02595669
## 
## Values of teststatistic and critical values of test:
## 
##           test 10pct  5pct  1pct
## r <= 1 |  7.23  6.50  8.18 11.65
## r = 0  | 38.04 15.66 17.95 23.52
## 
## Eigenvectors, normalised to first column:
## (These are the cointegration relations)
## 
##           lPCEC.l1  lDPI.l1
## lPCEC.l1  1.000000  1.00000
## lDPI.l1  -1.007081 -1.11388
## 
## Weights W:
## (This is the loading matrix)
## 
##             lPCEC.l1     lDPI.l1
## lPCEC.d -0.123246998 0.004519381
## lDPI.d  -0.006245572 0.011307281

## 
## ###################### 
## # Johansen-Procedure # 
## ###################### 
## 
## Test type: maximal eigenvalue statistic (lambda max) , with linear trend 
## 
## Eigenvalues (lambda):
## [1] 0.10598492 0.02595669
## 
## Values of teststatistic and critical values of test:
## 
##           test 10pct  5pct  1pct
## r <= 1 |  7.23  6.50  8.18 11.65
## r = 0  | 30.81 12.91 14.90 19.19
## 
## Eigenvectors, normalised to first column:
## (These are the cointegration relations)
## 
##           lPCEC.l1  lDPI.l1
## lPCEC.l1  1.000000  1.00000
## lDPI.l1  -1.007081 -1.11388
## 
## Weights W:
## (This is the loading matrix)
## 
##             lPCEC.l1     lDPI.l1
## lPCEC.d -0.123246998 0.004519381
## lDPI.d  -0.006245572 0.011307281

By the above results we can see that we Reject these time series are not cointegrated. We Fail to Reject that these time series are cointegrated at order I(1) so we can say that these two time series are indeed cointegrated.

(c) Perform the test for the presence of a restricted constant rather than unrestricted constant, interpret the result. If necessary rerun the cointegration test from (b).

## 
## ###################### 
## # Johansen-Procedure # 
## ###################### 
## 
## Test type: trace statistic , without linear trend and constant in cointegration 
## 
## Eigenvalues (lambda):
## [1] 3.068414e-01 6.039069e-02 2.757708e-16
## 
## Values of teststatistic and critical values of test:
## 
##            test 10pct  5pct  1pct
## r <= 1 |  17.13  7.52  9.24 12.97
## r = 0  | 117.92 17.85 19.96 24.60
## 
## Eigenvectors, normalised to first column:
## (These are the cointegration relations)
## 
##             lPCEC.l1    lDPI.l1   constant
## lPCEC.l1  1.00000000  1.0000000  1.0000000
## lDPI.l1  -1.00116412 -1.0165427 -0.9613652
## constant  0.01378502  0.2525283 -0.1555952
## 
## Weights W:
## (This is the loading matrix)
## 
##            lPCEC.l1     lDPI.l1      constant
## lPCEC.d -0.10443179 -0.01429582  3.708260e-13
## lDPI.d  -0.07541099  0.08047270 -3.148474e-14

## 
## ###################### 
## # Johansen-Procedure # 
## ###################### 
## 
## Test type: maximal eigenvalue statistic (lambda max) , without linear trend and constant in cointegration 
## 
## Eigenvalues (lambda):
## [1] 3.068414e-01 6.039069e-02 2.757708e-16
## 
## Values of teststatistic and critical values of test:
## 
##            test 10pct  5pct  1pct
## r <= 1 |  17.13  7.52  9.24 12.97
## r = 0  | 100.79 13.75 15.67 20.20
## 
## Eigenvectors, normalised to first column:
## (These are the cointegration relations)
## 
##             lPCEC.l1    lDPI.l1   constant
## lPCEC.l1  1.00000000  1.0000000  1.0000000
## lDPI.l1  -1.00116412 -1.0165427 -0.9613652
## constant  0.01378502  0.2525283 -0.1555952
## 
## Weights W:
## (This is the loading matrix)
## 
##            lPCEC.l1     lDPI.l1      constant
## lPCEC.d -0.10443179 -0.01429582  3.708260e-13
## lDPI.d  -0.07541099  0.08047270 -3.148474e-14

## LR-test for no linear trend
## 
## H0: H*2(r<=1)
## H1: H2(r<=1)
## 
## Test statistic is distributed as chi-square
## with 1 degress of freedom
##         test statistic p-value
## LR test            9.9       0

I think it is clear from the above results that we do not have a constant present in this model.

(d) Estimate the unrestricted VEC model, examine significance of variables in the two equations.

## Response lPCEC.d :
## 
## Call:
## lm(formula = lPCEC.d ~ ect1 + constant + lPCEC.dl1 + lDPI.dl1 - 
##     1, data = data.mat)
## 
## Residuals:
##       Min        1Q    Median        3Q       Max 
## -0.036882 -0.004097 -0.000036  0.003856  0.059704 
## 
## Coefficients:
##            Estimate Std. Error t value Pr(>|t|)    
## ect1      -0.123247   0.024510  -5.029    9e-07 ***
## constant  -0.009660   0.004115  -2.347  0.01963 *  
## lPCEC.dl1  0.062022   0.062014   1.000  0.31814    
## lDPI.dl1   0.188932   0.057789   3.269  0.00122 ** 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.00876 on 271 degrees of freedom
## Multiple R-squared:  0.7814, Adjusted R-squared:  0.7782 
## F-statistic: 242.2 on 4 and 271 DF,  p-value: < 2.2e-16
## 
## 
## Response lDPI.d :
## 
## Call:
## lm(formula = lDPI.d ~ ect1 + constant + lPCEC.dl1 + lDPI.dl1 - 
##     1, data = data.mat)
## 
## Residuals:
##       Min        1Q    Median        3Q       Max 
## -0.052043 -0.004718 -0.000756  0.004423  0.057728 
## 
## Coefficients:
##            Estimate Std. Error t value Pr(>|t|)    
## ect1      -0.006246   0.028318  -0.221    0.826    
## constant   0.007747   0.004755   1.629    0.104    
## lPCEC.dl1  0.465126   0.071651   6.492 4.03e-10 ***
## lDPI.dl1  -0.018435   0.066769  -0.276    0.783    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.01012 on 271 degrees of freedom
## Multiple R-squared:  0.7309, Adjusted R-squared:  0.727 
## F-statistic:   184 on 4 and 271 DF,  p-value: < 2.2e-16

Most of the variable in the response of PCE are significant but only lPCEC.dl1 of lDPI.d seems to be significant. Seems like this specification may not be the most appropriate model in this particular case…

(e) Perform following three tests: (1) β2 = −1, (2) α2 = 0, (3) joint hypothesis β2 = −1 and α2 = 0. Interpret the results.

## 
## ###################### 
## # Johansen-Procedure # 
## ###################### 
## 
## Estimation and testing under linear restrictions on beta 
## 
## The VECM has been estimated subject to: 
## beta=H*phi and/or alpha=A*psi
## 
##      [,1]
## [1,]    1
## [2,]   -1
## 
## Eigenvalues of restricted VAR (lambda):
## [1] 0.0943
## 
## The value of the likelihood ratio test statistic:
## 3.56 distributed as chi square with 1 df.
## The p-value of the test statistic is: 0.06 
## 
## Eigenvectors, normalised to first column
## of the restricted VAR:
## 
##      [,1]
## [1,]    1
## [2,]   -1
## 
## Weights W of the restricted VAR:
## 
##            [,1]
## lPCEC.d -0.1087
## lDPI.d  -0.0298

## 
## ###################### 
## # Johansen-Procedure # 
## ###################### 
## 
## Estimation and testing under linear restrictions on beta 
## 
## The VECM has been estimated subject to: 
## beta=H*phi and/or alpha=A*psi
## 
##      [,1]
## [1,]    1
## [2,]    0
## 
## Eigenvalues of restricted VAR (lambda):
## [1] 0.1059 0.0000
## 
## The value of the likelihood ratio test statistic:
## 0.04 distributed as chi square with 1 df.
## The p-value of the test statistic is: 0.84 
## 
## Eigenvectors, normalised to first column
## of the restricted VAR:
## 
##                [,1]
## RK.lPCEC.l1  1.0000
## RK.lDPI.l1  -1.0074
## 
## Weights W of the restricted VAR:
## 
##         [,1]
## [1,] -0.1209
## [2,]  0.0000

## 
## ###################### 
## # Johansen-Procedure # 
## ###################### 
## 
## Estimation and testing under linear restrictions on alpha and beta 
## 
## The VECM has been estimated subject to: 
## beta=H*phi and/or alpha=A*psi
## 
##      [,1]
## [1,]    1
## [2,]   -1
## 
## 
##      [,1]
## [1,]    1
## [2,]    0
## 
## Eigenvalues of restricted VAR (lambda):
## [1] 0.0894
## 
## The value of the likelihood ratio test statistic:
## 5.06 distributed as chi square with 2 df.
## The p-value of the test statistic is: 0.08 
## 
## Eigenvectors, normalised to first column
## of the restricted VAR:
## 
##      [,1]
## [1,]    1
## [2,]   -1
## 
## Weights W of the restricted VAR:
## 
##         [,1]
## [1,] -0.0963
## [2,]  0.0000

(f) Convert the VEC model into a VAR model in levels. Create and plot the eight quarter ahead forecast.

## $vecresult
## $vecresult$lPCEC.d
## 
## Call:
## lm(formula = z@Z0[, i] ~ -1 + ., data = data.frame(ect[, i], 
##     z@Z1))
## 
## Coefficients:
##  ect...i.   constant  lPCEC.dl1   lDPI.dl1  
## -0.108682  -0.001117   0.050830   0.186841  
## 
## 
## $vecresult$lDPI.d
## 
## Call:
## lm(formula = z@Z0[, i] ~ -1 + ., data = data.frame(ect[, i], 
##     z@Z1))
## 
## Coefficients:
##  ect...i.   constant  lPCEC.dl1   lDPI.dl1  
##        NA   0.008759   0.465281  -0.013638  
## 
## 
## 
## $beta
##          ect1
## lPCEC.l1    1
## lDPI.l1    -1
## 
## $alpha
##                [,1]
## lPCEC.d -0.09630942
## lDPI.d   0.00000000