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Why is JMP's Box-Cox transformation scaled by geometric mean?

bio_grad

Community Trekker

Joined:

Jun 10, 2016

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.

1 ACCEPTED SOLUTION

Accepted Solutions
markbailey

Staff

Joined:

Jun 23, 2011

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!
3 REPLIES
markbailey

Staff

Joined:

Jun 23, 2011

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!
bio_grad

Community Trekker

Joined:

Jun 10, 2016

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)?

markbailey

Staff

Joined:

Jun 23, 2011


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!