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The following message was posted to: PharmPK
I hope I am not being really thick here but I
don't see a convincing answer to the question
concerning the geometric CV formula cited by
Thierry Buclin's in the thread following Carl
Dmuchowski's original question about
sqrt(exp(var(log(x))1).
This thread was a useful discussion to me
because I was able to learn of the Diletti academic
journal reference for the geometric CV formula that
was cited by Mr. Dmuchowski. Presently, I'm wondering
whether there exists an academic reference for the formula
cited by Thierry Buclin? This is because I think that formula
makes intuitive sense. At least, it makes sense if one
thinks of the purpose of the geometric CV to allow
computation of backtransformed upper and lower limits
that would bound the geometric mean. Is this not the
purpose of the geometric CV?
As Buclin pointed out, the 100*(exp(SD)1)**0.5 formula gives
a value of 125 for the backtransformed geometric mean plus
1 SD unit. Similarly, a value of 80 for (100/100+25) is obtained
for the geometric mean minus 1SD unit. Thus, it seems to
make more sense to define the CV as 25 and not 22.6, which
is the geometric CV obtained with 100*sqrt(exp(var(log(x))1).
Thanks in advance for the clarification.
Paul Robertson
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The following message was posted to: PharmPK
Dear Paul Robertson,
In answer to your question: I wrote the next comment on the message by
Dr.
Buclin on 30 September 2002, which is the main answer to your question:

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).

Please note that your equation
> sqrt(exp(var(log(x))1).
and the equation in Carl Dmuchowski's original question
> sqrt([exp(Var(log(x)]1)
contain errors, i.e. one and two closing ) are missing, respectively,
and
should be written, as shown by Jorn Atterman (see above).
You wrote:
> As Buclin pointed out, the 100*(exp(SD)1)**0.5 formula gives
> a value of 125 for the backtransformed geometric mean plus
> 1 SD unit. Similarly, a value of 80 for (100/100+25) is obtained
> for the geometric mean minus 1SD unit. Thus, it seems to
> make more sense to define the CV as 25 and not 22.6, which
> is the geometric CV obtained with 100*sqrt(exp(var(log(x))1).
I don't understand this paragraph. At least, I cannot reproduce the
calculation. The only thing I understand that you like to have a SD (or
CV)
value of 25, describing a distribution where 80 and 125 are equidistant
from
100. This makes no sense to me. Indeed, a value of 80 follows from
(100/(100+25)), but also a value of 125 follows from (100/(10020)).
Does
this make sense? So, the value of 20 can be defended in the same way as
25,
but the logic for any of these two values is missing.
A value of 22.6 (from Jorn Atterman's equation) or 22.3 (from my
alternative
approach; a negligible difference) makes sense, being somewhere between
20
and 25.
Hans Proost
Johannes H. Proost
Dept. of Pharmacokinetics and Drug Delivery
University Centre for Pharmacy
Antonius Deusinglaan 1
9713 AV Groningen, The Netherlands
tel. 3150 363 3292
fax 3150 363 3247
Email: j.h.proost.at.farm.rug.nl
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