Re: Zeros in OMEGA matrix
You may define
MYETA1 = C11*ETA(1)
MYETA2 = C12*ETA(1)+C22*ETA(2)
MYETA3 = C23*ETA(2)+C33*ETA(3)
MYETA4 = C24*ETA(2)+C34*ETA(3)+C44*ETA(4)
with diagonal (and fixed to unit matrix) OMEGA matrix of ETA(1:4)
Then OMEGA for MYETA[1:4] matrix is
C11*C11
C11*C12 C12*C12+C22*C22
0 C22*C23 C23*C23+C33*C33
0 C22*C24 C23*C24+C33*C34 C24*C24+C34*C34+C44*C44
I am not sure whether it makes sense to go that far to get an extra 0 or extra flexibility (relative to the band matrices with 1 or 3 zeros), but technically, this is equivalent to the matrix that you need.
Another (simpler) option is to define
MYETA1 = ETA(1)
MYETA2 = ETA(2)
MYETA3 = ETA(3)
MYETA4 = ETA(4)+C24*ETA(2)
with the band matrix for ETAs:
Sw
C(w,x) Sx
0 C(x,y) Sy
0 0 C(y,z) Sz
The term C24*ETA(2) will add extra correlation (2-4) for MYETA matrix that is forbidden by the band matrix for ETA(1:4).
Leonid
--------------------------------------
Leonid Gibiansky, Ph.D.
President, QuantPharm LLC
web: www.quantpharm.com
e-mail: LGibiansky at quantpharm.com
tel: (301) 767 5566
Quoted reply history
On 7/3/2015 9:58 AM, Gavin Jarvis wrote:
> Dear NMUsers
>
> I have a problem and hope someone might be able to help. In summary, is
> it possible for NONMEM to model an OMEGA matrix with four elements
> (W,X,Y,Z) such that two of the covariances are fixed to zero and the
> rest freely determined? I realise NONMEM can model BAND matrices, but I
> can’t see how to fix just two covariances to zero. The final matrix
> structure I’m looking for would look something like…
>
> Sw
>
> C(w,x) Sx
>
> 0 C(x,y) Sy
>
> 0 C(x,z) C(y,z) Sz
>
> I hope that makes sense. If so, is it possible?
>
> Many thanks
>
> Gavin
>
> __________________________________________________
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> *Dr Gavin E Jarvis MA**(Cantab)**MA PhD VetMB MRCVS*
>
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