EVIEWS ASSIGNMENT REPORT Case Solution
EVIEWS ASSIGNMENT REPORT
ARCH MODELS (Question 3)
1) Plot the returns. What do you notice about the volatility of returns? Estimate a GARCH (1, 1) model. Are the coefficients of the correct sign?
The above plotted returns show that the returns have started to deviate and become more volatile after 535 approximately.
Since, the p value is less than 0.05, therefore, the Jarque-Bera would reject the assumption of normality which means that the data is not normal. There is the unconditional distribution of the returns.
In order to check the ARCH effect, quick estimation has been conducted. Now quick, estimate equation and: RESID01^2 C RESID01 (-1) ^2
|Dependent Variable: RESID^2|
|Method: Least Squares|
|Date: 11/27/16 Time: 16:10|
|Sample (adjusted): 2 781|
|Included observations: 780 after adjustments|
|R-squared||0.064965||Mean dependent var||4.595902|
|Adjusted R-squared||0.063763||S.D. dependent var||10.00540|
|S.E. of regression||9.681160||Akaike info criterion||7.380801|
|Sum squared resid||72917.94||Schwarz criterion||7.392748|
|Log likelihood||-2876.513||Hannan-Quinn criter.||7.385396|
The null hypothesis of the no ARCH effect has been reject because the squared residual lagged is significant Along with this, the LM test is (T-q) xR-squared = 780×0.066965 = 52.23271 which is greater than the critical value of the chi-squared distribution with 1 degrees of freedom.
Now in order to estimate the GARCH Model (1, 1); ARCH 1, GARCH 1:
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