RE: OMEGA HAS A NONZERO BLOCK
From:"Kowalski, Ken"
Subject:RE: [NMusers] OMEGA HAS A NONZERO BLOCK
Date:Wed, 2 Oct 2002 08:32:00 -0400
Steve,
Your Omega matrix is extremely ill-conditioned (i.e., the ratio of the
largest eigenvalue to the smallest is >10^6). Inspecting the correlations
of the etas I find that the correlation between eta1 and eta4 is
corr(1,4) = omega(1,4)/{sqrt(omega(1,1)*omega(4,4)} =
0.00818/sqrt(0.329*0.000204) = 0.998
This correlation is very close to 1 and I suspect that if you calculated
this correlation with all the digits rather than the 3 significant digits
that NONMEM reports in the output this correlation would be 1.0 leading to a
singular matrix (and an eigenvalue of 0). A solution to this problem is to
reduce the dimensionality of Omega restricting the correlation to be 1.0.
This can be accomplished by sharing the eta between the two parameters
corresponding to THETA1 and THETA4. For example,
P1=THETA(1)*EXP(ETA(1))
P2=THETA(2)*EXP(ETA(2))
P3=THETA(3)*EXP(ETA(3))
P4=THETA(4)*EXP(THETA(5)*ETA(1))
will force the correlation between P1 and P4 to be 1.0 and
var(LOG(P4))=(THETA(5)^2)*var(LOG(P1)). Thus, THETA(5) is the ratio of the
standard deviations and from your BLOCK(4) Omega results the estimate would
be THETA(5) = sqrt(0.000204)/sqrt(0.329) = 0.0249 (i.e., the standard
deviation for eta4 is approx. 2.5% of the standard deviation of eta1). With
this parameterization one would specify a BLOCK(3) for Omega which has 6
element plus THETA(5) for a total of 7 elements related to BSV whereas your
BLOCK(4) Omega has 10 elements. Note that forcing the correlation(1,4)=1
induces fixed correlations between eta1 and eta4=theta5*eta1 and the other
etas. Indeed, in your BLOCK(4) results you will find that corr(1,2) ~=
corr(2,4) and corr(1,3)~=corr(3,4). If you fit this model (i.e., BLOCK(3)
with the addition of THETA(5)) you should find that you will get the exact
same fit (MOF and parameter estimates) as your BLOCK(4) results but the
model will be considerably more stable.
Regards,
Ken