ECS4863
ANSWERED Exam Prep 2.This examination preparation exercise is similar to your Assignment 02 question;
however the dataset has been modified and updated.
You are required to estimate and evaluate, using the Engle-Granger cointegration
procedure, a stochastic function for South African ho...
This examination preparation exercise is similar to your Assignment 02 question;
however the dataset has been modified and updated.
You are required to estimate and evaluate, using the Engle-Granger cointegration
procedure, a stochastic function for South African household consumption (Total
final consumption expenditure by households). Estimate the function over the period
1980-2014, using data provided in the Excel work file: Mock Exam 2 data
The following theoretical specification (with expected relationships) may be of
assistance:
+ + −
C = f y , w , i
d
Where
C = Total final consumption expenditure by households
Y d = disposable income
W = Wealth (proxy by M3 money supply)
i = real interest rates (prime – inflation)
Question 1: (10 marks) (10)
Complete the following table:
Remember to generate the logs of all the relevant variables, prior to conducting the tests.
Series Model ADF Lags ADF
ττ τµ τ
LRCONS Trend and 1 -2.645925
intercept
Intercept 2 0.583285
None 2 3.483399
D(LRCONS) Trend and 1 -4.338428***
intercept
Intercept 1 -4.273416***
None 0 -2.636656**
RINT Trend and 0 -4.961181***
intercept
Intercept 0 -5.301375***
None 0 -2.751248***
* Statistically significant at the 10% level
** Statistically significant at the 5% level
*** Statistically significant at the 1% level
The ADF tests indicate that RCONS is I(1) and RINT is I(0)
You can assume that:
• M3 and HYD are both I(1)
Open Rubric
,Question 2: (10 marks)
2.1 Estimate the following potential long-run equilibrium equation, and supply
your estimated coefficients in the table below: (4)
2.2 Test for cointegration between the variables. Completed the table. (4)
(Note: this can be done in 2 ways, i.e. either to make a repressor group or to generate the
residual (res_cons) and then test for unit root. In the second option use ADF unit root test
options for: Level and None)
Model Lags τ p-value
LRCONS 0 -2.969085 0.0042***
2.3 Can we conclude that the variables in the long-run equation are
cointegrated? (2)
Yes, the results indicate that the variable is statistically significant (at 1%
level, i.e. three stars, which means that we can reject the null hypothesis
(of no cointegration).
Question 3: (20 marks)
3.1 In this question you have to build an Error Correction Model for
consumption. Use the suggested variables as in the below table, and
complete the missing values. (8)
3.2 There is a critical problem with the results obtained in Question 3.1. Can
you identify what the problem is, and what this will mean for the modelling
process? (2)
The coefficient of the lagged residual of the cointegrating equation
(RES_CONS(-1)) has to be negative and significant. In this example it is
negative (-0.120402) but it is not statistically significant (p-value of 0.2849
> 0.05). This means that the effect of the long-run component is not being
picked up significantly in the ECM.
In practice this will mean that the modelling process will have to be
redone; possibly using different samples, or different long run
cointegrating variables. You can also try to adjust the variables in the
ECM itself although this is unlikely to solve the problem related to the long
run components.
3.3 Regardless of the problem you identified, you decide to perform the
required diagnostic and stability tests on the residuals of the ECM.
Complete the table. (10)
Test H0 Test p- Conclusion
Statistics value
Jarque- H0 : Normally JB = 1.314 0.518 Fail to reject Ho. Residuals
Bera distributed are normally distributed.
residuals
Ljung-Box H 0 : No serial LBQ(6)= 0.942 Fail to reject Ho. No 1st
Q correlation 1.734 order serial correlation up
to 6th lag.
Breusch- H 0 : No serial nR²(2) = 0.585 Fail to reject Ho.
Godfrey correlation 1.071
LM TEST No autocorrelation up to
order 2.
ARCH-LM H0 : No nR²(2) = 0.250 Fail to reject Ho.
heteroscedasticity 2.772 No heteroscedasticity up to
order 2.
White H0 : No nR² (no CT) 0.589 Fail to reject Ho.
heteroscedasticity = 1.919 No heteroscedasticity.
Ramsey H0 : No LR(1) = 0.049 Reject Ho.
RESET misspecification 3.873
Possible misspecification.
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