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Hi,

I replicated a paper in which the authors place the following restrictions on this expression:

dpf = c0(1) + bp1(1)*dp(-1) + bp3(1)*dp(-3) + bp4(1)*dp(-4) + bs0(1)*ds + bs1(1)*ds(-1) + bs4(1)*ds(-4) + bs5(1)*ds(-5) + (c0_out(1) + bp5_out(1)*dp(-5) + (1-bs0(1))*ds + bs2_out(1)*ds(-2) + bs4_out(1)*ds(-4))*(1-exp(-gam(1)*(z/(@stdev(z)))^2))


The first bs0(1) to lie in [0,1], and the sum of bs0(1) -since it appears twice above- to sum up to 1. Had I been able to insert an equation above perhaps it would be clearer.

I used the logistic transformation outlined by Gareth here:
http://forums.eviews.com/viewtopic.php?f=4&t=48
i.e. (H-L)@Logit[bs0(1)]+L where L=1, H=0 to get:
genr y = (1-@logit(bs0(1))) + (@logit(bs0(1))-1)*(1-exp(0.076355*(z/@stdev(z))^2))
where 0.076355 is an estimated coefficient.
I know that doesn't equate to the restrictions in the paper (because I didn't know how to combine the equality and the inequality constraint). Nevertheless, I got the same results as the authors.
However, once I extended the sample size to include post-2007 data, it caught up with me because now I've got three data points well in excess of 1.
So I have to impose those restrictions properly- does anyone have any ideas on how to do it?

Thanks ever so much

Just re-read my post and realised what I say I'm trying to do isn't very clear.
So, if bs0(1) outside the brackets is a, and
the bs0(1) inside the brackets expression is b,
then what I do is compute y= a + b(1-exp(0.076355*(z/@stdev(z))^2)) (1)

Then, with i.e. (H-L)@Logit[bs0(1)]+L where L=1, H=0 (1) becomes :
genr y = (1-@logit(bs0(1))) + (@logit(bs0(1))-1)*(1-exp(0.076355*(z/@stdev(z))^2)) (2)

Except that I haven't imposed a+b =1. I was hoping someone could suggest how to do precisely that.