# Convert back standardized linear predictor

I guess this is a quite basic question, but I have been struggling with this for quite some time, so I hope someone can help me with this.

I have a model (type is not relevant for now) which includes a linear predictor (LP) that would normally be calculated as:

LP = sum(coefficient1*predictor1, coefficient2*predictor2, etc.)


However, the predictors were standardized by centering them at the mean values and rescaling them using the standard deviation (SD). Hence, the LP now looks like this:

LP = sum(coefficient1*(predictor1-mean1)/SD1, coefficient2*(predictor2-mean2)/SD2, etc.)


So far it is clear to me, but now I would like to present my models in an efficient way, so instead of writing out the mean and SD for each individual predictor I want to provide only the coefficients (see first LP) and then standardize the LP as a whole. If the LP was only centered at the mean values writing it out would look like this:

 LP = sum(coefficient1*predictor1, coefficient2*predictor2, etc.)- sum(coefficient1*mean1, coefficient2*mean, etc.)


Since the latter part of this formula has fixed values I can present that as one number, which makes the whole look somewhat less complicated. However, I don’t know how to also account for the SD rescaling. Does anyone know whether this (algebraically) possible and if yes how could I do that?

Many thanks!

• Why not just run the model on the unstandardized variables to start with? Jul 11 '13 at 23:06
• @PeterFlom I used a regularized regression approach in which the variables are automatically standardized.
– Rob
Jul 11 '13 at 23:10
• Yes, I realize that. But if you want the output in unstandardized form, then why not run a regression that does that? Jul 11 '13 at 23:11
• @PeterFlom The approach that I used (boosting) has some advantages over other regression techniques. The purpose of the final model is (absolute risk) prediction so for that the standardization is not a problem, but in my paper I would like to present the models way that is not too complicated.
– Rob
Jul 11 '13 at 23:19
• oh, OK, you didn't mention you used boosting. That makes more sense. I don't have an answer to your question, but there may be one. Jul 12 '13 at 9:55

 LP = sum(((coefficient1/SD1)*predictor1)-(mean1*(coefficient1/SD1)),

 (4/2)(X-3) = X(4/2) - 3(4/2)