Hi all
One suggestion, tranforme the x
0<x1<1.... Tranforme x1=exp(u1)/(exp(u1)+exp(u2)+exp(u3)+1)
0<x2<1.... Tranforme x2=exp(u2)/(exp(u1)+exp(u2)+exp(u3)+1)
0<x3<1.... Tranforme x3=exp(u3)/(exp(u1)+exp(u2)+exp(u3)+1)
0<x4<1.... Tranforme x4= 1/(exp(u1)+exp(u2)+exp(u3)+1)
x1+x2+x3+x4=1
Now solve :
Aexp(u1)+bexp(u2)+cexp(u3)+d=n(exp(u1)+exp(u2)+exp(u3)+1)
(c-n)exp(u3)=(n-a)exp(u1)+(n-b)exp(u2)+n-d
u3=ln((n-a)/(c-n)exp(u1)+(n-b)/(c-n)exp(u2)+(n-d)/(c-n)) (u3 expression)
Generate u1
Generate u2 bounded so the ln term should be positive
(n-a)/(c-n)exp(u1)+(n-b)/(c-n)exp(u2)+(n-d)/(c-n)>0
u > or < ln()
(u1 & u2 are not independant)
Compute u3 given the above formula
Generate the x
Hope this help
Naji
Le 26/03/08 22:41, « Ala' Jaouni » <[EMAIL PROTECTED]> a écrit :
> X1,X2,X3,X4 should have independent distributions. They should be
> between 0 and 1 and all add up to 1. Is this still possible with
> Robert's method?
>
> Thanks
>
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