topic Re: How does JMP do BoxCox transformations? in Discussions

How does JMP do BoxCox transformations?

DWS — Mon, 30 Jun 2025 17:33:54 GMT

I have recently been converting JMP formulas into Excel using the python code option. I have been unable to actually transfer any formula (MLR) involving a BoxCox transformation. It have manually done the calculations following the formula and have gone numbers in the multimillions instead of 100 thousands where it should be. If someone knows if the formula is already converted back or If I need to and what I need to do to apply the inverse BoxCox to get it back in normal terms. I don't know if I am missing something or if the Python Code through the publish formula option is missing something. I was hoping that someone could explain how JMP preforms the Box Cox transformation so that I can better understand it to potentially fix it. I tried to attach the code of the formula but it wouldn't allow me to post. Any help would be appreciated.

Re: How does JMP do BoxCox transformations?

Mark_Zwald — Mon, 30 Jun 2025 22:30:27 GMT

Documentation and formula for Box-Cox transformation is available here: https://www.jmp.com/support/help/en/18.2/index.shtml#page/jmp/boxcox-y-transformation.shtml

Re: How does JMP do BoxCox transformations?

MRB3855 — Tue, 01 Jul 2025 08:42:46 GMT

Hi @DWS : This may be helpful as well.

https://community.jmp.com/t5/Discussions/How-to-understand-Lambda1-and-Lambda2-for-Box-Cox-transformation/m-p/614578#M81444

Re: How does JMP do BoxCox transformations?

DWS — Tue, 01 Jul 2025 14:56:13 GMT

Thank this did help me better understand BoxCox transformations in general but not how JMP does them necessarily. The following question I believe would help me understand. The numbers that JMP shows in the Parameter Estimates section, for a BoxCox formula, are those just BoxCox transformed or have they already been inverse transformed back into understandable units?

Re: How does JMP do BoxCox transformations?

MRB3855 — Tue, 01 Jul 2025 15:31:55 GMT

@DWS If all you do is select "Box Cox Transformation" under the "Factor Profiling" menu, then no transformation is applied and the parameter estimates are based on Y. But if you further Refit with the box-cox transformation (red triangle in the "Box Cox Transformations" output), the parameter estimates are based on the transformed Y.

Re: How does JMP do BoxCox transformations?

DWS — Tue, 01 Jul 2025 15:38:58 GMT

OK, just to be clear, the formula that is give after I have clicked "Refit with Transformation", in that new menu , that formula and that parameter estimates are not transformed then inverse transformed back into real units but are left in transformed units. If this is the case then is is possible to know the formula JMP uses to inverse transform them because, it does give inverse normal units in the save column prediction formula? or should I apply a normal inverse transformation? Also, do you know this from experience? or is this written somewhere because I could not find it anywhere. Sorry for the number of questions.

Re: How does JMP do BoxCox transformations?

MRB3855 — Tue, 01 Jul 2025 15:48:44 GMT

@DWS : As you can see, when you refit and save the prediction formula it is back transformed so the predictions are in the original scale of Y. And the formula in that new column is the back-transformation formula you seek. I'm not sure if this is in the JMP help or not (but I can see what it is doing in my own work) I'm not sure what you are asking about the parameter estimates though; they do not get transformed in any way; only Y does.

Re: How does JMP do BoxCox transformations?

DWS — Tue, 01 Jul 2025 17:39:12 GMT

Okay, In let me give you an example of just one X variable and intercept between the two formulas. Non BoxCox: Intercept: 254,077 X1: -44,880. Same formula and variable after BoxCox: Intercept: 61,472,780 X1; -43,139. You see my issue. These numbers are taken from the parameter estimate and verified in the formula editor. All the x variables are the same they are not too far off from their non transformed counter parts but the intercept is very off. I don't know why this is and I cannot think of how this could even be possible.

Re: How does JMP do BoxCox transformations?

MRB3855 — Tue, 01 Jul 2025 17:54:56 GMT

@DWS What is your lambda in the BC transformation? And how many X's do you have? Can you take a pic of both sets of parameter estimates?

Re: How does JMP do BoxCox transformations?

DWS — Tue, 01 Jul 2025 18:14:19 GMT

Re: How does JMP do BoxCox transformations?

MRB3855 — Tue, 01 Jul 2025 20:52:15 GMT

@DWS Oh dear. I think back-transforming is the least of your worries. Before you can think about interpreting coefficients (especially one of over 50), you’ll want to make sure the terms you have in your model can be estimated with the data you have. Some estimates are biased, some are zeroed. That indicates you are trying to estimate some parameters wrt X’s that are highly correlated, and/or some that are not estimable because the structure of your data doesn’t support estimating the terms you are trying to estimate. Your VIF’s in your parameter estimate table (if they aren’t visible, right mouse click on the parm estimates table to choose them) will help you identify multi-collinearity (correlation). For example, if X1 and X2 move together, then including both in the model makes both of their coefficients unreliable wrt interpretation…so one of them should be excluded as a candidate in the model.

Also, if lambda = -0.002, it may be easier to use the LN transformation; it’s easier to interpret, and -0.002 is very close to 0 (lambda=0 corresponds to the LN transformation).

Edit: more here.

https://community.jmp.com/t5/Discussions/Parameter-Estimates-being-zerod/m-p/751701

Re: How does JMP do BoxCox transformations?

DWS — Wed, 02 Jul 2025 13:24:18 GMT

I appreciate the help, I am fully aware of all of that. This is just an example I created to send you to show you the significant difference the intercepts of BC and non BC formulas. Do know why the intercepts are so different?

Re: How does JMP do BoxCox transformations?

MRB3855 — Wed, 02 Jul 2025 14:09:20 GMT

@DWS Well…one is based on BoxCox(Y) and the other is based on Y; so why wouldn’t the intercepts be different? And I’m not trying to be obtuse.

Let’s set aside the biased and zeroed parameter estimates issue for now. For both models, least squares is used to estimate the parameters in the respective models. And each will have a different set of parameter estimates based on the least squares procedure. They are different because the responses are different (one is Y, the other is BoxCox[Y] ).

And if we assume LN transformation, BoxCox[Y] = LN[Y], when we back transform we get Y = exp[int + linear combination of X’s] = exp[int]*exp[linear combination of X’s]. That model is multiplicative…so the intercept in that model has a different effect on Y than the intercept does in the un-transformed additive model.

I hope this is helpful. Perhaps I’m not understanding the nature of your question; why is this such a concern?

Re: How does JMP do BoxCox transformations?

DWS — Wed, 02 Jul 2025 16:06:41 GMT

I don't know if you can read this image but you can see on the left the predicted values compared. Not fair from each other. then on the right you can see the pages for the Non BC and BC and their parameter estimates at the bottom. I can get the predicted value from the sheet using the Intercept + Estimate (X) formula given in the parameter estimates for the Non BC but not for the BC because the BC estimate is 386 Billion. That is the same number given by the python code so I can't get the accurate number shown on the left in Excel. I want to know the actual formula that JMP is using to calculate the Box Cox Predict shown on the left image.

Re: How does JMP do BoxCox transformations?

MRB3855 — Wed, 02 Jul 2025 16:22:22 GMT

@DWS If I’m understanding, can’t you just save the prediction formula via the red triangle menu at the top of the output?

Re: How does JMP do BoxCox transformations?

MRB3855 — Wed, 02 Jul 2025 17:18:50 GMT

@DWS Or in general, see pic ( and g is the geometric mean of Y).

Re: How does JMP do BoxCox transformations?

DWS — Wed, 02 Jul 2025 17:28:56 GMT

It still gives the 3.868e+11

Re: How does JMP do BoxCox transformations?

MRB3855 — Wed, 02 Jul 2025 17:35:40 GMT

@DWS You are doing something wrong and/or misinterpreting something. What equation are you using in Excel to get a prediction for Y?

Re: How does JMP do BoxCox transformations?

DWS — Wed, 02 Jul 2025 17:36:57 GMT

Python straight from JMP

def getModelMetadata():

return {"creator": u"Fit Least Squares", "modelName": u"", "predicted": u"BoxCox(Sold Price Adjusted,-1.088)", "table": u"Nation Export", "version": u"18.0.2", "timestamp": u"2025-07-02T17:34:44Z"}

def getInputMetadata():

return {

u"Approx Living Area": "float",

u"YSB": "float"

}

def getOutputMetadata():

return {

u"Pred Formula BoxCox(Sold Price Adjusted,-1.088)": "float"

}

def score(indata, outdata):

outdata[u"Pred Formula BoxCox(Sold Price Adjusted,-1.088)"] = 386761700160.329 + 153.473460505384 * indata[u"Approx Living Area"] + -1519.63590642121 * indata[u"YSB"]

return outdata[u"Pred Formula BoxCox(Sold Price Adjusted,-1.088)"]

Re: How does JMP do BoxCox transformations?

MRB3855 — Wed, 02 Jul 2025 17:49:12 GMT

@DWS It looks to me like you are returning the transformed Y. If you want Y you need to back transform as shown in this pic (where g is the geometric mean of Y).