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• Why is the LOG (or BoxCox) transformation valid?

• 
  Sven Mensing

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  Why is the LOG (or BoxCox) transformation valid?

  Hello,
  
  recently I had some discussions with my colleages about the motivation of the LOG transformation in order to compare two groups (simple t-Test) having different standdard deviation.
  I learned at the University, that I should log-transform the data and as I did the standard deviations became comparable (and somewhat normal distributed). But due to the transformation I also changed the mean. I recon that is has something to do with the continous nature of the transformation, but I can not put my finger on it.
  
  Has someone a hint to some kind of motivation/"proof" for the righfulness of the LOG (or BoxCox) transform, because I feel that it is right, but I don't "see" it.
  
  Thanks
  Sven
  
  • 10 Oct 2006, 10:01 pm
  Sven Mensing wrote: > Hello, > > recently I had some discussions with my colleages about the > motivation of the LOG transformation in order to compare two groups > (simple t-Test) having different standdard deviation. > I learned at the University, that I should log-transform the data and > as I did the standard deviations became comparable (and somewhat > normal distributed). But due to the transformation I also changed the > mean. I recon that is has something to do with the continous nature > of the transformation, but I can not put my finger on it. > > Has someone a hint to some kind of motivation/"proof" for the > righfulness of the LOG (or BoxCox) transform, because I feel that it > is right, but I don't "see" it. > > Thanks > Sven
• 
  Michael Stastny    Premium Member

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  Re: Why is the LOG (or BoxCox) transformation valid?

  The Student's t-test assumes that observations are normally distributed and that the variances of the two populations are equal (when the latter assumption has to be dropped, on often uses Welch's t-test).
  
  In case the assumption of normality or heteroskedasticity cannot be maintained one can try to transform the data accordingly.
  
  ad Normality: I think the t-test is quite robust against deviations from normality. But in case the data looks quite lognormally distributed, I'd take the log...
  
  ad Variance: Often the variance, Var(X), of a random variable, X, is a function of the expected value E(X), i.e. Var(X) = f(E(X)). We are looking for a transformation Y = g(X) so that the approximated variance of Y, Approx.Var(Y), is constant.
  
  Example: X is exponentially distributed with E(X) = mu, i.e Var(X) = mu^2
  
  Let Y = g(X) = ln(X).
  
  Approx.Var(Y) = Var(X)*(dg(E(X)/dE(X))^2
  Approx.Var(Y) = mu^2*(1/mu)^2
  Approx.Var(Y) = 1
  
  The Box-Cox transformation (taking the natural log is a special case) should be used for for random variables X with Var(X) = mu^s.
  
  • 18 Aug 2007, 10:22 pm
  Michael Stastny wrote: > The Student's t-test assumes that observations are normally > distributed and that the variances of the two populations are equal > (when the latter assumption has to be dropped, on often uses Welch's > t-test). > > In case the assumption of normality or heteroskedasticity cannot be > maintained one can try to transform the data accordingly. > > ad Normality: I think the t-test is quite robust against deviations > from normality. But in case the data looks quite lognormally > distributed, I'd take the log... > > ad Variance: Often the variance, Var(X), of a random variable, X, is > a function of the expected value E(X), i.e. Var(X) = f(E(X)). We are > looking for a transformation Y = g(X) so that the approximated > variance of Y, Approx.Var(Y), is constant. > > Example: X is exponentially distributed with E(X) = mu, i.e Var(X) = > mu^2 > > Let Y = g(X) = ln(X). > > Approx.Var(Y) = Var(X)*(dg(E(X)/dE(X))^2 > Approx.Var(Y) = mu^2*(1/mu)^2 > Approx.Var(Y) = 1 > > The Box-Cox transformation (taking the natural log is a special case) > should be used for for random variables X with Var(X) = mu^s.