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The Box-Cox power transform frequently used in statistical analysis takes the value (x^λ -1) /λ for λ not equal to zero, and ln(x) for λ=0. I would like to see a demonstration, that need not be a full formal proof, that this family of transformations is continuous in the neighborhood of λ=0. Though I do not necessarily need all the superstructure of a formal proof, I very much want a demonstration that provides an intuitive understanding of why this is the right thing to do for this parameter value.

I observe that for exp((x^λ -1)/λ ), this function has the nice property that it returns exp(ln(x)) = x at λ = 0, dividing the outcomes between those that compress the extreme values, those that inflate them, and those that do neither. But I still don't understand how that is related to the values of the transformed variable for λ in the near vicinity of 0.

I am asking this here rather than in CrossValidated because I am looking at this as a question about analytic properties of a class of functions rather than about any statistical properties of distributions before and after the transform.

andrewH
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    Write $x^{\lambda} = e^{\lambda \log x}$. For small $\lambda$ this is approximately $1 + \lambda \log x$ (with error $O(\lambda^2 \log^2 x)$). – Qiaochu Yuan May 06 '16 at 22:22
  • I understand the first part of your comment, but don't know how you got from there to the second part. – andrewH May 07 '16 at 00:33
  • The first-order Taylor series expansion of $e^x$ at $x = 0$ is $1 + x$. This follows from the defining property $\frac{d}{dx} e^x = e^x$. – Qiaochu Yuan May 07 '16 at 00:39

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Since $\lim_{\lambda \rightarrow 0} x^\lambda = x^0 = 1$ we can use L'hopitals rule to get $$ \lim_{\lambda \rightarrow 0} \frac{x^\lambda -1}{\lambda} = \lim_{\lambda \rightarrow 0} \frac{\ln x e^{\lambda \ln x}}{1} = \ln x e^0 = \ln x $$ showing the continuity.

kjetil b halvorsen
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