# PharmPK Discussion - Statistical terms

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• On 23 Apr 2003 at 08:49:28, "Robertson, Paul" (probertson.aaa.neurocrine.com) sent the message
`The following message was posted to: PharmPKI hope I am not being really thick here but Idon't see a convincing answer to the questionconcerning the geometric CV formula cited byThierry Buclin's in the thread following CarlDmuchowski's original question aboutsqrt(exp(var(log(x))-1).This thread was a useful discussion to mebecause I was able to learn of the Diletti academicjournal reference for the geometric CV formula thatwas cited by Mr. Dmuchowski. Presently, I'm wonderingwhether there exists an academic reference for the formulacited by Thierry Buclin? This is because I think that formulamakes intuitive sense. At least, it makes sense if onethinks of the purpose of the geometric CV to allowcomputation of back-transformed upper and lower limitsthat would bound the geometric mean. Is this not thepurpose of the geometric CV?As Buclin pointed out, the 100*(exp(SD)-1)**0.5 formula givesa value of 125 for the back-transformed geometric mean plus1 SD unit. Similarly, a value of 80 for (100/100+25) is obtainedfor the geometric mean minus 1SD unit. Thus, it seems tomake more sense to define the CV as  25 and not 22.6, whichis the geometric CV obtained with 100*sqrt(exp(var(log(x))-1).Thanks in advance for the clarification.Paul Robertson`
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• On 25 Apr 2003 at 14:20:11, "Hans Proost" (j.h.proost.-at-.farm.rug.nl) sent the message
`The following message was posted to: PharmPKDear Paul Robertson,In answer to your question: I wrote the next comment on the message byDr.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, isCV(X) = sIn 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 yields0.226. A simple view at the data learns that the CV must be somewherebetween 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,andshould 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 back-transformed 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 thecalculation. The only thing I understand that you like to have a SD (orCV)value of 25, describing a distribution where 80 and 125 are equidistantfrom100. 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/(100-20)).Doesthis make sense? So, the value of 20 can be defended in the same way as25,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 myalternativeapproach; a negligible difference) makes sense, being somewhere between20and 25.Hans ProostJohannes H. ProostDept. of Pharmacokinetics and Drug DeliveryUniversity Centre for PharmacyAntonius Deusinglaan 19713 AV Groningen, The Netherlandstel. 31-50 363 3292fax  31-50 363 3247Email: j.h.proost.at.farm.rug.nl`
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