3 replies to this topic

[Lance Wollenberg]

Hi,

 

My colleagues and I were having a discussion on calculations regarding the omega and omegastderr output and getting this into useful terms that can help you better understand the variability for the purposes of reporting. We are hoping experts in the extranet community could help to clarify.

 

I know that the output on the omega block generates a variance.

 

For proportional/exponential error models

CV=sqrt(omega)*100

 

For additive error models

SD=sqrt(omega)

 

Are these conventions correct dependent the error model structure you select?

 

For the omegastderr, how does this term relate to the final parameter estimate? What math is involved to get this into an %RSE term?

 

To our understanding, for a given parameter

%RSE=(Omegastderr)/SQRT(Omega)*100

 

Is this correct? From what I can tell this is different than the calculation that is done with NONMEM output. Any insight would be greatly appreciated.

 

Lance

[serge guzy]

Dear Lance

 

When you assume lognormal distribution for the model parameters then the population variability in term of %CV is

 %CV=sqrt(omega)*100

 

 

If you assume normal distribution for the model parameters,

then

 SD=sqrt(omega). %CV=SD/mean x 100

 

 

Are these conventions correct dependent the error model structure you select?

 

 

I think you should be careful not be confused between population variability and error model.

Omega is related to population variability, not error structure. Error structure is related to the uncertainty you have in your observed response (linked to sigma in NONMEM)

 

Now about the population distribution, the CV definition shown above applies only to lognormal distribution. If you do another transformation, for example inverse logit on bioavailability, then CV is not defined in straight full manner. Therefore, yes the definitions of CV depend on the transformation made and are not always defined easily.

 

To our understanding, for a given parameter

 %RSE=(Omegastderr)/SQRT(Omega)*100

 

 

By definition %RSE =se of the estimate /estimate x100 . The estimate is Omega not sqrt of omega and therefore to me the defined %RSE for omega element would be

Omegastderr/omega x 100 .

 

I think that issues can be raised because you are using lognormal distribution and the calculations can be different. If you want to test, try normal distribution in both NLME and NONMEM and see if you get similar answers.

 

A way to better compare would be to do a bootstrap and compare the 95% confidence bounds for omega elements with the one you would get with NONMEM.

 

best

Serge

[Samer Mouksassi]

I think that Serge clarified your questions I have a comment regarding the approximation of CV % of an omega that is modeled as exponential:

 

TVparameter = POPparameter*exp(nparameter)

 

where n parameter has mean zero and variance Omega^2

 

The approximaton that square root of the variance Omega^2 is an approximate CV is valid for low values of the variance for high values a better approximation would be:

(exp (omega^2) - 1 )^2

 

Mathematical details about thederivation and about log normal distribution in general can be seen in a useful poster at:

http://www.page-meeting.org/pdf_assets/4964-Elassaiss-Schaap%20-%20Equations%20variability%20reporting%20PK-PD%20-%20Final.pdf

 

Another potential issue that you might face:

 

Typical software give you Omega^2 its standard error and its relative standard error (RSE%)

if you want to report a function of this parameter say

square root of parameter you need to compute the standard error of this transformed parameter and then you can compute the RSE % of it.

 

So you want SE(omega) but software gives you the SE(omega^2)

From the delta method of error propagation and computing error of a function of a set of parameters you can get:

Standard error of a square root of (x) ?

where x=omega^2 and

 f(x)=omega=x^0.5 # (Square root of omega^2)

 

SE(f(x)) ~= f'(x)*SE(x) # note this an approximation

SE(omega) = SE(f(x))

          = SE(x^0.5)

          = f'(x)*SE(x) # we take the derivative of f(x) with respect to x

          = 0.5(x^-0.5)*SE(x)

          = (0.5/x^0.5))SE(x)

          = SE(x)/(2*omega)

 

RSE(omega) = SE(omega) /omega =

                       =SE(x)/(2*omega) /omega = SE(x)/ 2 * (omega^2) # omega^2 is x

                      =1/2 * SE(X)/(X) # SE(X)/X is the RSE(X)

                       = 1/2* RSE(X)

 

so RSE(%) of x^0.5 = RSE(%) of x /2

 

 

 

but be careful on how you report and you approximate the CV % as stated above.

 

Samer

[linda]

Hi Serge,

 

As you said, the CV definition shown above applies only to lognormal distribution. So, what is the CV

% for logit distribution like bioavailability F?

 

 

Thanks,

Linda

 

Dear Lance When you assume lognormal distribution for the model parameters then the population variability in term of %CV is %CV=sqrt(omega)*100 If you assume normal distribution for the model parameters, then SD=sqrt(omega). %CV=SD/mean x 100 Are these conventions correct dependent the error model structure you select? I think you should be careful not be confused between population variability and error model. Omega is related to population variability, not error structure. Error structure is related to the uncertainty you have in your observed response (linked to sigma in NONMEM) Now about the population distribution, the CV definition shown above applies only to lognormal distribution. If you do another transformation, for example inverse logit on bioavailability, then CV is not defined in straight full manner. Therefore, yes the definitions of CV depend on the transformation made and are not always defined easily. To our understanding, for a given parameter %RSE=(Omegastderr)/SQRT(Omega)*100 By definition %RSE =se of the estimate /estimate x100 . The estimate is Omega not sqrt of omega and therefore to me the defined %RSE for omega element would be Omegastderr/omega x 100 . I think that issues can be raised because you are using lognormal distribution and the calculations can be different. If you want to test, try normal distribution in both NLME and NONMEM and see if you get similar answers. A way to better compare would be to do a bootstrap and compare the 95% confidence bounds for omega elements with the one you would get with NONMEM. best Serge