Poly() in Lm(): Difference Between Raw VS. Orthogonal

poly() in lm(): difference between raw vs. orthogonal

By default, with raw = FALSE, poly() computes an orthogonal polynomial. It internally sets up the model matrix with the raw coding x, x^2, x^3, ... first and then scales the columns so that each column is orthogonal to the previous ones. This does not change the fitted values but has the advantage that you can see whether a certain order in the polynomial significantly improves the regression over the lower orders.

Consider the simple cars data with response stopping distance and driving speed. Physically, this should have a quadratic relationship but in this (old) dataset the quadratic term is not significant:

m1 <- lm(dist ~ poly(speed, 2), data = cars)
m2 <- lm(dist ~ poly(speed, 2, raw = TRUE), data = cars)

In the orthogonal coding you get the following coefficients in summary(m1):

                Estimate Std. Error t value Pr(>|t|)    
(Intercept)       42.980      2.146  20.026  < 2e-16 ***
poly(speed, 2)1  145.552     15.176   9.591 1.21e-12 ***
poly(speed, 2)2   22.996     15.176   1.515    0.136    

This shows that there is a highly significant linear effect while the second order is not significant. The latter p-value (i.e., the one of the highest order in the polynomial) is the same as in the raw coding:

                            Estimate Std. Error t value Pr(>|t|)
(Intercept)                  2.47014   14.81716   0.167    0.868
poly(speed, 2, raw = TRUE)1  0.91329    2.03422   0.449    0.656
poly(speed, 2, raw = TRUE)2  0.09996    0.06597   1.515    0.136

but the lower order p-values change dramatically. The reason is that in model m1 the regressors are orthogonal while they are highly correlated in m2:

cor(model.matrix(m1)[, 2], model.matrix(m1)[, 3])
## [1] 4.686464e-17
cor(model.matrix(m2)[, 2], model.matrix(m2)[, 3])
## [1] 0.9794765

Thus, in the raw coding you can only interpret the p-value of speed if speed^2 remains in the model. And as both regressors are highly correlated one of them can be dropped. However, in the orthogonal coding speed^2 only captures the quadratic part that has not been captured by the linear term. And then it becomes clear that the linear part is significant while the quadratic part has no additional significance.

Regression in R using poly() function

Because they are not the same model. Your second one has one unique covariate, while the first has two.

> model_2

Call:
lm(formula = v ~ 1 + q + q^2)

Coefficients:
(Intercept)            q  
    -15.251        7.196  

You should use the I() function to modify one parameter inside your formula in order the regression to consider it as a covariate:

model_2=lm(v~1+q+I(q^2))

> model_2

Call:
lm(formula = v ~ 1 + q + I(q^2))

Coefficients:
(Intercept)            q       I(q^2)  
     7.5612      -3.3323       0.8774  

will give the same prediction

> predict(model_1)
        1         2         3         4         5         6         7         8         9        10        11 
 5.106294  4.406154  5.460793  8.270210 12.834406 19.153380 27.227133 37.055664 48.638974 61.977063 77.069930 
> predict(model_2)
        1         2         3         4         5         6         7         8         9        10        11 
 5.106294  4.406154  5.460793  8.270210 12.834406 19.153380 27.227133 37.055664 48.638974 61.977063 77.069930

How do you make R poly() evaluate (or predict ) multivariate new data (orthogonal or raw)?

For the record, it seems that this has been fixed

> x1 = seq(1,  10,  by=0.2)
> x2 = seq(1.1,10.1,by=0.2)
> t = poly(cbind(x1,x2),degree=2,raw=T)
> 
> class(t) # has a class now
[1] "poly"   "matrix"
> 
> # does not throw error
> predict(t, newdata = cbind(x1,x2)[1:2, ])                                                     
     1.0  2.0 0.1  1.1  0.2
[1,] 1.0 1.00 1.1 1.10 1.21
[2,] 1.2 1.44 1.3 1.56 1.69
attr(,"degree")
[1] 1 2 1 2 2
attr(,"class")
[1] "poly"   "matrix"
> 
> # and gives the same
> t[1:2, ]
     1.0  2.0 0.1  1.1  0.2
[1,] 1.0 1.00 1.1 1.10 1.21
[2,] 1.2 1.44 1.3 1.56 1.69
> 
> sessionInfo()
R version 3.4.1 (2017-06-30)
Platform: x86_64-w64-mingw32/x64 (64-bit)
Running under: Windows >= 8 x64 (build 9200)

Is there an inverse function for poly?

cars$speed must be of the form a + b * pp[, 1] for some scalars a and b and knowing that the coefs attribute of poly objects contains values which can be used for reconstruction we find the following reconstruction of cars$speed as speed.

pp <- poly(cars$speed, 2)
speed <- with(attr(pp, "coefs"), alpha[1] + sqrt(norm2)[3] * pp[, 1])

all.equal(speed, cars$speed)
## [1] TRUE

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