Hi,
Could you please help me estimate via MLE the specification below? I am given only the dependent variable Y. My independent variables are separate AR(1) processes. Here it is:
@ename e1
@evar var(e1)=1
Y=c(1)+c(2)*X1(-1)+c(3)*X2(-1)+(v(-1)^(0.5))*e1
X1=c(4)*X1(-1)+c(5)*e1+c(6)*(@abs(e1)-(2/pi)^(0.5))
X2=c(7)+c(8)*X2(-1)+c(9)*e1+c(10)*(@abs(e1)-(2/pi)^(0.5))
v=(exp(X1+X2))^2
I started with sspace, but noticed that sspace works only with linear specifications. Many thanks in advance.
Miscellaneous MLE
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EViews Gareth
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Re: Miscellaneous MLE
What's the likelihood?
Re: Miscellaneous MLE
Thanks a lot for your reply Gareth. The log-likelihood function is as follows:
f_t((c);X1_t,X2_t|Y_t)=-ln(2pi)-(X1_t-1+X2_t-1)-{[Y_t-c(1)-c(2)*X1_t-1-c(3)*X2_t-1]^2}/2v_t-1
f_t((c);X1_t,X2_t|Y_t)=-ln(2pi)-(X1_t-1+X2_t-1)-{[Y_t-c(1)-c(2)*X1_t-1-c(3)*X2_t-1]^2}/2v_t-1
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EViews Gareth
- Fe ddaethom, fe welon, fe amcangyfrifon
- Posts: 13604
- Joined: Tue Sep 16, 2008 5:38 pm
Re: Miscellaneous MLE
Seems pretty easy to transform into a logl object:
Note I see nothing here that estimates the variance term - v is fixed.
Code: Select all
logl mylog
mylog.append @logl logl1
mylog.append v = (@exp(x1+x2))^2
mylog.append logl1 = -(x1(-1)+x2(-1)) + (y-c(1)-c(2)*x1(-1)-c(3)*x2(-1))/(2*v(-1))
Note I see nothing here that estimates the variance term - v is fixed.
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