topic Re: Centering polynomials calculation in Discussions

Centering polynomials calculation

MathAvgPorpoise — Tue, 29 Oct 2024 19:37:24 GMT

I'm trying to understand how JMP centers polynomials. Here’s an example with two independent variables, x1 and x2, and a response variable y:

x1	x2	y
2	12	80
4	14	85
6	16	90
8	18	95
10	20	100
12	22	105

When using the Analyze->Fit Model option in JMP with x1, x2, x1*x1, and x2*x2 (keeping the "Center Polynomials" checkbox checked), I get the following parameter estimates:

- Intercept: 75.375
- x1: 3.0714286
- x2: 0
- (x1-7)*(x1-7): -0.089286
- (x2-17)*(x2-17): 0

I wanted to test what happens if I center manually by subtracting the mean from each value of x1 and x2.
The mean of x1 is 7, and the mean of x2 is 17. So, for the x1 column, I subtracted 7 from each value (2, 4, 6, etc.). Here’s the resulting table:

x1	x2	y
-5	-5	80
-3	-3	85
-1	-1	90
1	1	95
3	3	100
5	5	105

Using these new centered values for x1 and x2 in the Fit Model option with "Center Polynomials" unchecked, I get the following parameter estimates:

- Intercept: 96.875
- x1: 3.0714286
- x2: 0
- (x1-7)*(x1-7): -0.089286
- (x2-17)*(x2-17): 0

As you can see, all parameter estimates are correct except for the intercept. Why is the intercept different? Could someone please explain how it is calculated?

Re: Centering polynomials calculation

DonMcCormack — Tue, 29 Oct 2024 19:56:27 GMT

X2 is linearly dependent on X1 (i.e., X1 = X2 + 10). You should be seeing a Singularity Details outline box as well as zeroed estimates for everything besides the Intercept and X1. The estimates your showing don't match what you should have gotten given the data above.

Re: Centering polynomials calculation

MathAvgPorpoise — Tue, 29 Oct 2024 22:07:32 GMT

I admit it was not a good example. Here is a better one:

x1	x2	y
2	1	80
4	5	85
6	3	95
8	2	100
10	8	105
12	3	110

With "Center polynomials" checked:

Intercept: 76.337952
x1: 3.1705139
x2: -0.497888
(x1-7)*(x1-7): -0.120141
(x2-3.66667)*(x2-3.66667): 0.1013011

Using the x1/x2 centered values with unchecked "Center polynomials" option

x1	x2	y
-5	-2.667	80
-3	1.333	85
-1	-0.667	95
1	-1.667	100
3	4.333	105
5	-0.667	110

leads to:

Intercept: 96.705792
x1: 3.1705139
x2: -0.497888
x1*x1: -0.120141
x2*x2: 0.1013011

What additional calculation is needed to get 76.337952 ?

Re: Centering polynomials calculation

MRB3855 — Wed, 30 Oct 2024 09:48:49 GMT

Hi @MathAvgPorpoise : This kind of thing was discussed recently here.

https://community.jmp.com/t5/Discussions/Intercept-of-a-parabola/m-p/805020

Re: Centering polynomials calculation

MRB3855 — Wed, 30 Oct 2024 10:37:55 GMT

Hi @MathAvgPorpoise : And only the quadratic terms are centered...the main effects are not. So, you have to be careful.

i.e., Using the x1/x2 centered values with unchecked "Center polynomials" option, you will need to enter the squared centered values, but the linear terms should be the uncentered values. It will all match then.

Re: Centering polynomials calculation

Mark_Bailey — Wed, 30 Oct 2024 12:59:03 GMT

Centering the predictors is a transformation. It translates the coordinates. This shifts the origin of the space to the mean.