# Re: BSV and BOV interaction

From: andreas.krause
Date: Mon, 21 Dec 2009 09:17:30 +0100

Jia,

you are overparameterized. Take this snippet from your code:

IOV2=0
IF (DESC.EQ.1) IOV2=ETA(5)
IF (DESC.EQ.2) IOV2=ETA(6)

ETCL = ETA(1)+IOV1

Now consider the two possibilites:
a) DESC.EQ.1: ETCL = ETA(1) + ETA(5)
b) DESC.EQ2.2: ETCL = ETA(1) + ETA(6)

In other words, you have two equations to identify 3 parameters.
Usually you associate the "base" random effect with one case and add a
deviation parameter to the other case.
An example would be

IOV2=0
IF (DESC.EQ.2) IOV2=1
ETCL = ETA(1)+IOV2*ETA(5)

Thus, ETA(1) estimates your random effect variation for the case DESC.EQ.1
and ETA(1) + ETA(5) is the random effect variation for the case DESC.EQ.2.
ETA(5) is thus the additional random effect variation for the second case
compared to the first.
Watch out that this implies that the random effect variation is larger for
DESC.EQ.2 than for DESC.EQ.1 since ETA(5) is (hopefully) not negative.
You could multiply the two to allow for the variation being smaller or
larger in the latter case but multiplication makes the estimation more
unstable.

Why do you see the need to link the two? Why don't you define
IF(DESC.EQ.1) ETCL=ETA(5)
IF(DESC.EQ.2) ETCL=ETA(6)
CL=THETA(1)*EXP(ETCL)

and get rid of ETA(1)? That decouples the two estimates entirely.

Andreas

Jia Ji <jackie.j.ji
Sent by: owner-nmusers
12/19/2009 12:32 AM

To
nmusers
cc

Subject
[NMusers] BSV and BOV interaction

Dear All,

I am trying to model our data with a two-compartment model now. In our
trial, some patients received escalated dose at the second cycle so they
have one more set of kinetics data. So there were BSV and BOV on PK
parameters in the model. Objective function value is
significantly improved (compared with the model not having BOV) and SE of
ETAs are around 40% or less. The code is as below:

\$PK
DESC=1
IF (TIME.GE.100) DESC=2
IOV1=0
IF (DESC.EQ.1) IOV1=ETA(2)
IF (DESC.EQ.2) IOV1=ETA(3)

IOV2=0
IF (DESC.EQ.1) IOV2=ETA(5)
IF (DESC.EQ.2) IOV2=ETA(6)

ETCL = ETA(1)+IOV1
ETQ = ETA(4)+IOV2
ETV2 = ETA(7)

CL=THETA(1)*EXP(ETCL)
V1=THETA(2)
Q=THETA(3)*EXP(ETQ)
V2=THETA(4)*EXP(ETV2)

;OMEGA initial estimates
\$OMEGA 0.0529
\$OMEGA BLOCK(1) 0.05
\$OMEGA BLOCK(1) SAME
\$OMEGA 0.318
\$OMEGA BLOCK(1) 0.05
\$OMEGA BLOCK(1) SAME
\$OMEGA 0.711

When I looked at scatterplot of ETA, I found that there is strong
correlation between ETA(1) and ETA(2), which is BSV and BOV of CL. And the
same thing happened to BSV and BOV of Q. Worrying about
over-parameterization (I am not NONMEM 7 user), I tried to define a THETA
for this correlation as the code below (just test on CL only first):

\$PK
DESC=1
IF (TIME.GE.100) DESC=2
IOV1=0
IF (DESC.EQ.1) IOV1=THETA(1)*ETA(1)
IF (DESC.EQ.2) IOV1=THETA(1)*ETA(1)

ETCL = ETA(1)+IOV1
ETQ = ETA(2)
ETV2 = ETA(3)

CL=THETA(2)*EXP(ETCL)
V1=THETA(3)
Q=THETA(4)*EXP(ETQ)
V2=THETA(5)*EXP(ETV2)

The objective function value is exactly the same as the model not having
IOV. BSV of CL is decreased and SE of THETAs are also improved,
though. The same thing happend to Q when tested individually. Then I tried
another way to account for this correlation:

\$PK
DESC=1
IF (TIME.GE.100) DESC=2
IOV1=0
IF (DESC.EQ.1) IOV1=ETA(2)
IF (DESC.EQ.2) IOV1=ETA(3)

ETCL = ETA(1)+IOV1
ETQ = ETA(4)
ETV2 = ETA(5)

CL=THETA(1)*EXP(ETCL)
V1=THETA(2)
Q=THETA(3)*EXP(ETQ)
V2=THETA(4)*EXP(ETV2)

;OMEGA initial estimates
\$OMEGA BLOCK(2) 0.0529 0.01 0.05
\$OMEGA BLOCK(1) 0.05 ;BTW, I don't know how to do SAME here, it's
not working when putting SAME here
\$OMEGA 0.318
\$OMEGA 0.711

This time I got significantly decreased objective function value, compared
with the model not having IOV. But, SE of ETA(1), ETA(2) and ETA(3) are
huge!

All together, does it mean that there is no need to have BOV on CL and Q?
Or I don't get the right solution to solve correlation problem? Any
suggestion is highly appreciated! Thank you so much!

Happy Holidays!

Jia

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Received on Mon Dec 21 2009 - 03:17:30 EST

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