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I found a paper’s code about stgarch which formula is ht=ω+αut-1^2+δ1*G1*(ht-1)+β*ht-1+λ0d0
and the writer writes the code like this
....
var=@recode(d1=1,omega(2)/(1-alpha(2)),omega(2)+delata*@gammainc(var(-1),gamma+alpha(2)*res(-1)^2)
var=@recode(d1=1,0.05,du(1)*dummy+omga(2)+delta(1)*@gammainc(var(-1),gamma(1))+alpha(2)*res(-1)^2)
the calculate log likelihood by using var....
and I do not understand know why there are two var equation...

I just start to learn Eviews and my English is not good, hope you can understand me....
Thank you everyone first

Impossible to say without seeing the rest of the program.

Impossible to say without seeing the rest of the program.

formula is ht=ω+αut-1^2+δ1*G1*(ht-1)+β*ht-1+λ0d0

load return
series d1=0
smpl @first @first
d1=1
smpl @all
coef(1) mu(1)=0.0000309
coef(1) omega(1)=0.0075
coef(1) alpha(1)=0.05
coef(1) gamma(1)=0.00755
coef(1) du(1)=0.0123
coef(1) delta(1)=0.73
coef(1) tdf(1)=2.065
logl stgarcht
stgarcht.append @logl logl
stgarcht.append res=return-mu(2)
@LOGl LOGl
RES=RETURN-mu(2)
stgarcht.append
var=@recode(d1=1,omega(2)/(1-alpha(2)),omega(2)+delata*@gammainc(var(-1),gamma+alpha(2)*res(-1)^2)
stgarcht.append
var=@recode(d1=1,0.05,du(1)*dummy+omga(2)+delta(1)*@gammainc(var(-1),gamma(1))+alpha(2)*res(-1)^2)
stgarch.append z=res^2/var/(tdf(1)-2)+1
stgarcht.appendd logl=@gammalog((tdf(1)+1)/2)-@gammalog(tdf(1)/2)-log(3.14159265359)/2-log(var)/2-log(tdf(1)-2)/2-log(var)/2-(tdf(1)+1)*log(z)/2
stgarcht.ml(showopts,m=50,c=1e-5)



That is the code.And I also do not know why there is a gamma in the first var instead of the gamma(1), is it a print error or something I do not know?
Thank you very much for replying, and my English is very poor, hope you could understand me...
Thank you again.