[R-SIG-Finance] Different results between Box-Ljung test and ARCHLM test?

Jen Bohold jenbohold at yahoo.de
Sat May 18 16:53:24 CEST 2013



Hi,
using r and the rugarch package, I fitted a garch model u. The output (extracted) is as follows:

GARCH Model    : sGARCH(1,1)
Mean Model    : ARFIMA(0,0,0)
Distribution    : norm 


Q-Statistics on Standardized Residuals
------------------------------------
                        statistic  p-value
Lag[1]                7.939    0.004839
Lag[p+q+1][1]     7.939    0.004839
Lag[p+q+5][5]    15.939   0.007021
d.o.f=0
H0 : No serial correlation

Q-Statistics on Standardized Squared Residuals
------------------------------------
                        statistic  p-value
Lag[1]                1.390    0.238324
Lag[p+q+1][3]     9.242    0.002365
Lag[p+q+5][7]    11.963   0.035303
d.o.f=2

ARCH LM Tests
------------------------------------
                         Statistic DoF P-Value
ARCH Lag[2]      5.474   2       0.06478
ARCH Lag[5]      8.973   5       0.11014
ARCH Lag[10]    12.090  10     0.27910

Now I have trouble interpreting the results of Q-Statistics?
First of all to test the mean equation, we look at the standardized 
residuals. These standardized residuals should behave iid(0,1). Since 
the p-values is very small, we can conclude, that they are not 
independent, since there exist serial correlation. Is this right?
To test the volatility equation we look at the standardized squared 
residuals, they should also behave iid(0,1). Since the p-values for lag 
order higher than one are small, we can conclude, that they are not 
independent. So there still exist arch effects, Right?
Now my problem is, that the ARCHLM test also looks at the 
standardized squared residuals and comes to a different result: The 
p-value is not very small, so H0 cannot be rejected. That means, there 
are no ARCH effects left?

So the ARCHLM tests shows a different result than the Q-Statistics? What should I conclude in this case?

I attached the plots of the standardized residuals and the squared standardized residuals.
Thanks a lot for your help, Jen
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