turn on suggestions

Auto-suggest helps you quickly narrow down your search results by suggesting possible matches as you type.

Showing results for

- JMP User Community
- :
- Discussions
- :
- Why is JMP's Box-Cox transformation scaled by geom...

Topic Options

- Subscribe to RSS Feed
- Mark Topic as New
- Mark Topic as Read
- Float this Topic for Current User
- Bookmark
- Subscribe
- Printer Friendly Page

- Mark as New
- Bookmark
- Subscribe
- Subscribe to RSS Feed
- Get Direct Link
- Email to a Friend
- Report Inappropriate Content

Dec 1, 2016 10:48 AM
(1767 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.

1 ACCEPTED SOLUTION

Accepted Solutions

- Mark as New
- Bookmark
- Subscribe
- Subscribe to RSS Feed
- Get Direct Link
- Email to a Friend
- Report Inappropriate Content

Dec 1, 2016 12:44 PM
(3458 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!

3 REPLIES

- Mark as New
- Bookmark
- Subscribe
- Subscribe to RSS Feed
- Get Direct Link
- Email to a Friend
- Report Inappropriate Content

Dec 1, 2016 12:44 PM
(3459 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!

- Mark as New
- Bookmark
- Subscribe
- Subscribe to RSS Feed
- Get Direct Link
- Email to a Friend
- Report Inappropriate Content

Dec 3, 2016 8:00 AM
(1710 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)?

- Mark as New
- Bookmark
- Subscribe
- Subscribe to RSS Feed
- Get Direct Link
- Email to a Friend
- Report Inappropriate Content

Dec 3, 2016 8:07 AM
(1706 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!