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Dec 1, 2016 10:48 AM
(2348 views)

Hello,

I am wondering why JMP's definition of the Box-Cox transformation scales by the geometric mean?

See:

http://www.jmp.com/support/help/13/Box_Cox_Y_Transformation.shtml

Thanks.

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Dec 1, 2016 12:44 PM
(4620 views)

Solution

Because the geometric mean is the function in the definition of this transform:

Box, G. E. P. and Cox, D. R. (1964) An analysis of transformations (with discussion). *Journal of the Royal Statistical Society B*, **26**, 211–252.

Learn it once, use it forever!

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Dec 1, 2016 12:44 PM
(4621 views)

Because the geometric mean is the function in the definition of this transform:

Box, G. E. P. and Cox, D. R. (1964) An analysis of transformations (with discussion). *Journal of the Royal Statistical Society B*, **26**, 211–252.

Learn it once, use it forever!

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Dec 3, 2016 8:00 AM
(2291 views)

Thank you for the post.

I was curious about this line on page 7:

"The above results can be expressed very simply if we work with the normalized transformation..."

How does Z^(lambda) make things easier than Y^(lambda)?

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Dec 3, 2016 8:07 AM
(2287 views)

@jasonpaquette13 wrote:

Thank you for the post.

I was curious about this line on page 7:

"The above results can be expressed very simply if we work with the normalized transformation..."

How does Z^(lambda) make things easier than Y^(lambda)?

Well, you have removed the mean and the variance from the expression. They are implied to be 0 and 1, but not explicitly in the expression.

Learn it once, use it forever!