cdf of skewed t distribution using fGARCH vs skewt package

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I ran a AR(p)-GARCH (1,1) - skewed t using the fGARCH package. When I tried to use the sstd option in fgarch to return the cdf of the  standardized residuals z and compared it with the results obtained using the skewt package, I obtained different results even though both skewed t dist are based on Fernandez and steel. Since the standardized inovations are supposed to be iid with mean o and variance 1, i used   psstd(z, mean = 0, sd = 1, nu , xi ). Anyone can explain why this might be the case? 

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   - Brian
Brian G. Peterson
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Alex Chan wrote:
I ran a AR(p)-GARCH (1,1) - skewed t using the fGARCH package. When I 
tried to use the sstd option in fgarch to return the cdf of the  
standardized residuals z and compared it with the results obtained 
using the skewt package, I obtained different results even though 
both skewed t dist are based on Fernandez and steel. Since the 
standardized inovations are supposed to be iid with mean o and 
variance 1, i used   psstd(z, mean = 0, sd = 1, nu , xi ). Anyone can 
explain why this might be the case?   
Please read the posting guide.  Include sample code and data if you 
want to maximize your chance of having someone help you figure out the 
(possible) problem you are having.

Depending on the data, the length of your data set, the degree of 
skewness, the random seed, coding issues, and a whole bunch of other 
factors, guessing at what the possible problem is is unlikely to be 
very useful.

Regards,

  - Brian

Spencer Graves, PE, PhD
President and Chief Operating Officer
Structure Inspection and Monitoring, Inc.
751 Emerson Ct.
San Jos?, CA 95126
ph:  408-655-4567
I ran a AR(p)-GARCH (1,1) - skewed t using the fGARCH package. When I tried to use the sstd option in fgarch to return the cdf of the  standardized residuals z and compared it with the results obtained using the skewt package, I obtained different results even though both skewed t dist are based on Fernandez and steel. Since the standardized inovations are supposed to be iid with mean o and variance 1, i used   psstd(z, mean = 0, sd = 1, nu , xi ). Anyone can explain why this might be the case? 

Alex 

sstd means standardized skew student-t, the first s is important.

in the student-t the number of degrees of freedom is related to the 
variance, so you cannot use the usual student t.
you need a reformulated distribution for which the variance is always 1 
and a proper nu which does not
change the variance. This does the sstd.

Diethelm

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