- On 6 Sep 2002 at 14:35:05, "Dmuchowski, Carl" (cdmuchowski.aaa.sankyopharma.com) sent the message

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I beg some guidance on two issues:

What is the standard US and European PK terminology for these two

coefficients of variation?

a) sqrt[(Var(x))/(arithmetic mean)]

b) sqrt([exp(Var(log(x)]-1)

Thanks

Carl Dmuchowski - On 6 Sep 2002 at 18:59:25, "Serge Guzy" (GUZY.aaa.xoma.com) sent the message

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I never used the word geometric coefficient of variation vs

arythmetic coefficient of variation but it seems that is what we have

here. - On 7 Sep 2002 at 15:52:34, Nick Holford (n.holford.-a-.auckland.ac.nz) sent the message

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"Dmuchowski, Carl (by way of David Bourne)" wrote:

>

> I beg some guidance on two issues:

>

> What is the standard US and European PK terminology for these two

> coefficients of variation?

>

> a) sqrt[(Var(x))/(arithmetic mean)]

In New Zealand I would call this "nonsense" (the dimensions of the

numerator and denominator are different).

Perhaps you meant sqrt[Var(x)]/(arithmetic mean)? I would call that a

coefficient of variation.

> b) sqrt([exp(Var(log(x)]-1)

I don't recognize this as anything I know about.

Perhaps you could explain why you are asking for guidance on these

strange expressions?

Nick

--

Nick Holford, Divn Pharmacology & Clinical Pharmacology

University of Auckland, 85 Park Rd, Private Bag 92019, Auckland, New Zealand

email:n.holford.at.auckland.ac.nz

http://www.health.auckland.ac.nz/pharmacology/staff/nholford/ - On 7 Sep 2002 at 17:47:38, "Jorn Attermann" (jorn.-at-.biostat.au.dk) sent the message

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With respect to term b):

If X has a two-parameter lognormal distribution with parameters m and s^2

(i.e. log(X) has a normal distribution with mean m and standard deviation

s), then the mean and variance of X are

E(X) = exp(m + s^2 / 2)

Var(X) = exp(2m +s^2)[exp(s^2)-1]

where s^2 is s squared.

Therefore, the coefficient of variation of X is

CV(X) = sqrt[Var(X)] / E(X) = sqrt[ exp(s^2) - 1 ].

Perhaps this is what you had in mind.

Jorn

Jorn Attermann, MSc, PhD jorn.aaa.biostat.au.dk

Department of Biostatistics, University of Aarhus

Vennelyst Boulevard 6, DK-8000 Aarhus C, DENMARK - On 9 Sep 2002 at 12:03:23, Thierry Buclin (Thierry.Buclin.-at-.chuv.hospvd.ch) sent the message

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This is to add a word to the discussion about coefficients of

variation. The question of Carl refers e.g. to the FDA Guidance for

Industry "Statistical Approaches to Establishing Bioequivalence"

(http://www.fda.gov/cder/guidance/3616fnl.pdf), where the

recommendation is :

"that bioequivalence measures (e.g., AUC and Cmax) be

log-transformed"... and "geometric means (antilog of the means of the

logs) should be calculated for selected bioequivalence measures".

The guidance recommends that PK parameters be presented as mean, SD,

geometric means and coefficient of variation, but does not make any

formal statement about the calculation of this coefficient of

variation. There are two options for this :

- either the arithmetic one : CV = SD/mean = Sqrt(Var(x))/mean (as

corrected by Nick)

- or the geometric one : for this I personally use CV =

Exp(SD(Log(x)))-1 rather than the formula quoted by Carl. Both are

however asymptotically equivalent for not too large SDs. The antilog

of the SD of the logarithms has been termed "percentage standard

deviation" by Snedecor & Cochran, who mention this calculation in

their classical textbook (Statistical methods, Iowa State Univ Press,

8th ed, 1989, p. 290-1). For example, if you consider the 3 data 80,

100 and 125, their geometric mean is 100 and the value of

Exp(SD(Log(x))) is 1.25, meaning that the data can be summarized as

100 times-or-divided-by 1.25, meaning a CV of 25%. (Notice that 75,

100 and 125 would have an arithmetic CV of 25%, while 78.2, 100 and

127.9 would have 25% of the CV calculated according to the formula

quoted by Carl, which is the arithmetic CV of a lognormally

distributed variable, as recalled by Jorn).

Thierry Buclin

Division of Clinical Pharmacology,

University Hospital of Lausanne (Switzerland) - On 9 Sep 2002 at 14:00:45, "Dmuchowski, Carl" (cdmuchowski.aaa.sankyopharma.com) sent the message

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The problem is some of my European colleagues call

'sqrt([exp(Var(log(x)]-1)' the geometric coefficient of variation and I have

been unable to unearth a legitimate academic reference for describing the

coefficient of variation of a lognormal distribution this way. I wanted to

gauge how the discussion group referred to these two expressions for the

coefficient of variation, which are used as summary statistics for AUC and

Cmax when reporting to regulatory authorities. I have only seen the

'sqrt([exp(Var(log(x)]-1)' referred to as the coefficient of variation of

the multiplicative model in the International Journal of Clinical

Pharmacology, Therapy and Toxicology (Diletti et al). My European colleagues

claim Dr. Cawello refers to the 'sqrt([exp(Var(log(x)]-1)' as the geometric

coefficient of variation in his book, Parameters for Compartment Free

Pharmacokinetics, but I have not been able to corroborate this because I

cannot locate a copy of his book and they appear to be reluctant to share

their copy with me. - On 10 Sep 2002 at 10:12:44, "Bonate, Peter" (pbonate.-at-.ilexonc.com) sent the message

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A "legitimate" reference can be found in Statistical Distributions by

NAJ Hastings and JB Peacock, Halsted Press, New York, 1975. They

refer to is simply as the coefficient of variation.

Pete Bonate - On 18 Sep 2002 at 10:23:54, "Hans Proost" (j.h.proost.-at-.farm.rug.nl) sent the message

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Dear colleagues,

With respect to the topic of means and standard deviation of a

log-normal

distribution, I would like to add the following to continue the

discussion.

Jorn Attermann wrote a clear message:

> If X has a two-parameter lognormal distribution with parameters m and

> s^2

> (i.e. log(X) has a normal distribution with mean m and standard

> deviation

> s), then the mean and variance of X are

>

> E(X) = exp(m + s^2 / 2)

>

> Var(X) = exp(2m +s^2)[exp(s^2)-1]

>

> where s^2 is s squared.

>

> Therefore, the coefficient of variation of X is

>

> CV(X) = sqrt[Var(X)] / E(X) = sqrt[ exp(s^2) - 1 ].

This is correct indeed. E(X) is here the arithmetic mean. Assuming that

a

log-normal distribution applies to the data, however, the geometric

mean is

a more appropriate measure of the central tendency of the data, since it

coincides with both the expected value of modus and median (just as the

arithmetic mean coincides with modus and median in a normal

distribution):

'Mean' = Geometric mean = Exp(m)

This is both simple and comprehensible.

Please note that the calculated mean E(X), and Var(X) and CV(X), refer

to

the normal (i.e. not log-transformed) scale. They refer to a skewed

distribution. This implies that these values cannot be used for further

statistical calculations, e.g. the calculation of confidence intervals,

or

as priors in a Bayesian analysis. One may say that E(X) and Var(X) are

the

moments of the normal distribution that approaches the log-normal

distribution with moments m and s^2 as good as possible. However, it is

not

an accurate representation of the true distribution, in particular in

the

tails.

Assuming that a log-normal distribution applies to the data, it would be

more appropriate 'to stay as much as possible in the log-transformed

world'

.. One should use m and s for any further statistical procedure, and

perform

the back-transformation at the end. This is also a good recipe to avoid

mistakes and misunderstanding of the log-normal distribution (it is

'completely normal' in the log-transformed world).

This applies also to the use of priors in a Bayesian analysis. Suppose

that

we have done a population analysis, assuming a log-normal distribution

of

the parameters within the population. For one particular parameter we

get: m

= 4 and s = 0.2.

According to the aforementioned equations, E(X) = 55.70, Var(X) =

126.62,

S(X) = 11.25 and CV(X) = 0.2020.

If we want to use these results as a prior in a subsequent Bayesian

analysis, e.g. for Therapeutic Drug Monitoring in a new patient, we

should

not use the latter values, but we should use m = 4 and s = 0.2, and

perform

the analysis assuming a log-normal distribution of the parameter within

the

population.

Since the actual value of m does not have a 'proper look' because of the

logarithmic transformation, m and s can be back-transformed for the

'normal'

world by the following:

'Mean' = Geometic mean = Exp(m) = 54.60

'SD' = s . Mean = 10.92

'CV' = s = 0.2

These values can be easily converted to the log-transformed world. Using

E(X) and Var(X) would require to reverse the equations from which they

were

derived. In my view, this is needlessly complicated.

Alternatively, one might use E(X) and Var(X) assuming a normal

distribution

of the parameter within the population. Although illogical, and not

accurate, it may be expected that it will introduce only minimal bias.

On

the other hand, it would be incorrect to use the geometric mean and

'SD' and

assume a normal distribution, or to use Log(E(X)) and CV(X) and assume a

log-normal distribution. In both cases, a bias of about 2% (55.70 vs

54.60)

would be introduced. The bias rapidly increases with increasing s (or

CV).

Actually, both approaches are not really different, as long as we use

the

data correctly, according to their meaning. The geometric mean should

not be

interpreted as 'the mean' (since in normal language 'the mean' refers

to the

arithmetic mean), and E(X) should not be used in further statistical

calculations.

I would appreciate comments on this view.

Hans Proost

Johannes H. Proost

Dept. of Pharmacokinetics and Drug Delivery

University Centre for Pharmacy

Antonius Deusinglaan 1

9713 AV Groningen, The Netherlands

Email: j.h.proost.at.farm.rug.nl - On 30 Sep 2002 at 08:37:10, "Hans Proost" (j.h.proost.-a-.farm.rug.nl) sent the message

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Dear Dr. Buclin,

In your message on coefficient of variation you wrote:

> - or the geometric one : for this I personally use CV =

> Exp(SD(Log(x)))-1 rather than the formula quoted by Carl. Both are

> however asymptotically equivalent for not too large SDs.

Please note that this formula is not correct.

The correct form was given by Jorn Attermann:

CV(X) = sqrt[Var(X)] / E(X) = sqrt[ exp(s^2) - 1 ].

where s = SD(Log(x)) in your equation.

The alternative approach, as explained in my previous message, is

CV(X) = s

In your example of three values 80, 100, 125, it follows that s =

0.223.

Your equation would yield indeed 0.25. Jorn Attermann's equation yields

0.226. A simple view at the data learns that the CV must be somewhere

between 20% (the deviation of 80) and 25% (the deviation of 125).

In the example in my previous mail (SD(Log(x)) = s = 0.2), this would

yield

a value of 0.2214, which overestimates the CV by 10%.

The deviation in your equations are even much larger than the difference

between the various estimates mentioned in my previous mail, when using

erroneous combinations of normal and log-normal values!

Hans Proost

Johannes H. Proost

Dept. of Pharmacokinetics and Drug Delivery

University Centre for Pharmacy

Antonius Deusinglaan 1

9713 AV Groningen, The Netherlands

Email: j.h.proost.aaa.farm.rug.nl

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