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
BSV and BOV interaction
7 messages
6 people
Latest: Dec 21, 2009
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
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
The information of this email and in any file transmitted with it is strictly confidential and may be legally privileged.
It is intended solely for the addressee. If you are not the intended recipient, any copying, distribution or any other use of this email is prohibited and may be unlawful. In such case, you should please notify the sender immediately and destroy this email.
The content of this email is not legally binding unless confirmed by letter.
Any views expressed in this message are those of the individual sender, except where the message states otherwise and the sender is authorised to state them to be the views of the sender's company. For further information about Actelion please see our website at http://www.actelion.com
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 <[email protected]>
Sent by: [email protected]
12/19/2009 12:32 AM
To
[email protected]
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
The information of this email and in any file transmitted with it is strictly
confidential and may be legally privileged.
It is intended solely for the addressee. If you are not the intended recipient,
any copying, distribution or any other use of this email is prohibited and may
be unlawful. In such case, you should please notify the sender immediately and
destroy this email.
The content of this email is not legally binding unless confirmed by letter.
Any views expressed in this message are those of the individual sender, except
where the message states otherwise and the sender is authorised to state them
to be the views of the sender's company. For further information about Actelion
please see our website at http://www.actelion.com
Andreas,
The code is not overparameterized because the SAME option is used for the OMEGA block defining ETA(6). This means that there is only one parameter being estimated for the variance of the distribution from which ETA(5) and ETA(6) are sampled i.e. ETA(5) and ETA(6) come from an eta distribution with the SAME variance.
Best wishes,
Nick
[email protected] wrote:
> 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 < [email protected] > Sent by: [email protected]
>
> 12/19/2009 12:32 AM
>
> To
> [email protected]
> 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
>
> The information of this email and in any file transmitted with it is strictly
> confidential and may be legally privileged.
> It is intended solely for the addressee. If you are not the intended recipient,
> any copying, distribution or any other use of this email is prohibited and may
> be unlawful. In such case, you should please notify the sender immediately and
> destroy this email.
> The content of this email is not legally binding unless confirmed by letter.
>
> Any views expressed in this message are those of the individual sender, except where the message states otherwise and the sender is authorised to state them to be the views of the sender's company. For further information about Actelion please see our website at http://www.actelion.com
--
Nick Holford, Professor Clinical Pharmacology
Dept Pharmacology & Clinical Pharmacology
University of Auckland,85 Park Rd,Private Bag 92019,Auckland,New Zealand
tel:+64(9)923-6730 fax:+64(9)373-7090 mobile:+64(21)46 23 53
email: [email protected]
http://www.fmhs.auckland.ac.nz/sms/pharmacology/holford
Andreas,
The code snippet you picked out is not overparameterized, since the
assumption is made that the variance of eta 5 and 6 are the same:
$OMEGA BLOCK(1) 0.05
$OMEGA BLOCK(1) SAME
This first equation that you suggest is this:
IOV2=0
IF (DESC.EQ.2) IOV2=1
ETCL = ETA(1)+IOV2*ETA(5)
As you note the equation you suggest implies that the between-subject
variability in CL will be larger for the first occasion than the second.
Unless inclusion criteria resulted in weird data that forced me to make
that assumption I would not feel comfortable using this
parameterisation. Also I do not fully understand this "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." Both eta 1
and eta 5 may be negative and positive, so if you are hoping for only
positive eta5 values it seems something is wrong with the structural
model. Or did you mean that you hope the variance of eta5 is positive
(ie. OMEGA(5,5))?
Finally, I also have my doubts about your last suggestion regarding how
to combine eta 1 and 5: "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." How would you interpret that model?
Subjects that have abnormally high CL at occasion 1 are likely to have
either abnormally high, or abnormally low CL at occasion 2. I think
simulations would give you patterns you do not see in real life with
such assumptions. Also, if data supports such a model, it may be more a
reflection of the error model. If some subjects have more error in their
observations a simple eta on epsilon may be more appropriate.
I hope everyone will have a nice break, both from nmusers and from work!
Best regards
Jakob
Quoted reply history
-----Original Message-----
From: [email protected] [mailto:[email protected]]
On Behalf Of [email protected]
Sent: 21 December 2009 08:18
To: Jia Ji
Cc: [email protected]
Subject: Re: [NMusers] BSV and BOV interaction
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 <[email protected]>
Sent by: [email protected]
12/19/2009 12:32 AM
To
[email protected]
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
The information of this email and in any file transmitted with it is
strictly confidential and may be legally privileged.
It is intended solely for the addressee. If you are not the intended
recipient, any copying, distribution or any other use of this email is
prohibited and may be unlawful. In such case, you should please notify
the sender immediately and destroy this email.
The content of this email is not legally binding unless confirmed by
letter.
Any views expressed in this message are those of the individual sender,
except where the message states otherwise and the sender is authorised
to state them to be the views of the sender's company. For further
information about Actelion please see our website at
http://www.actelion.com
Jia,
I don't see any indication that your first model is problematic. A strong
correlation between BSV and BOV ETA for CL is to be expected when you have
shrinkage in your individual etas (see e.g. Savic & Karlsson AAPS J. 2009
Sep;11(3):558-69). This does not mean that the population model should
include such a correlation. If shrinkage is high (>20% or so) I would tend
to use simulation-based or CWRES based diagnostics instead of posthoc eta's.
Best regards,
Mats
Mats Karlsson, PhD
Professor of Pharmacometrics
Dept of Pharmaceutical Biosciences
Uppsala University
Box 591
751 24 Uppsala Sweden
phone: +46 18 4714105
fax: +46 18 471 4003
Quoted reply history
-----Original Message-----
From: [email protected] [mailto:[email protected]] On
Behalf Of [email protected]
Sent: Monday, December 21, 2009 10:41 AM
To: Nick Holford
Cc: nmusers
Subject: Re: [NMusers] BSV and BOV interaction
Nick,
overparameterization refers to the parameters, the variances play only an
indirect role. Putting the SAME constraint on the covariance thus
restricts the set of random effects but not necessarily the set of random
effects for, say, subject i.
The SAME option thus might keep the estimation process a bit more under
control but I still think there is an overparameterization problem for
each individual subject.
It should be interesting to take out those ETA values containing the SAME
lines by specifying 0.0 FIX instead of SAME and comparing the results.
Andreas
PS. Shouldn't we all be off for some holidays?
Nick Holford <[email protected]>
Sent by: [email protected]
12/21/2009 09:52 AM
To
nmusers <[email protected]>
cc
Subject
Re: [NMusers] BSV and BOV interaction
Andreas,
The code is not overparameterized because the SAME option is used for the
OMEGA block defining ETA(6). This means that there is only one parameter
being estimated for the variance of the distribution from which ETA(5) and
ETA(6) are sampled i.e. ETA(5) and ETA(6) come from an eta distribution
with the SAME variance.
Best wishes,
Nick
[email protected] wrote:
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 <[email protected]>
Sent by: [email protected]
12/19/2009 12:32 AM
To
[email protected]
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
The information of this email and in any file transmitted with it is
strictly confidential and may be legally privileged.
It is intended solely for the addressee. If you are not the intended
recipient, any copying, distribution or any other use of this email is
prohibited and may be unlawful. In such case, you should please notify the
sender immediately and destroy this email.
The content of this email is not legally binding unless confirmed by
letter.
Any views expressed in this message are those of the individual sender,
except where the message states otherwise and the sender is authorised to
state them to be the views of the sender's company. For further
information about Actelion please see our website at
http://www.actelion.com
--
Nick Holford, Professor Clinical Pharmacology
Dept Pharmacology & Clinical Pharmacology
University of Auckland,85 Park Rd,Private Bag 92019,Auckland,New Zealand
tel:+64(9)923-6730 fax:+64(9)373-7090 mobile:+64(21)46 23 53
email: [email protected]
http://www.fmhs.auckland.ac.nz/sms/pharmacology/holford
The information of this email and in any file transmitted with it is
strictly confidential and may be legally privileged.
It is intended solely for the addressee. If you are not the intended
recipient, any copying, distribution or any other use of this email is
prohibited and may be unlawful. In such case, you should please notify the
sender immediately and destroy this email.
The content of this email is not legally binding unless confirmed by letter.
Any views expressed in this message are those of the individual sender,
except where the message states otherwise and the sender is authorised to
state them to be the views of the sender's company. For further information
about Actelion please see our website at http://www.actelion.com