I'm using a set of points which go from (-5,5)
to (0,0)
and (5,5)
in a "symmetric V-shape". I'm fitting a model with lm()
and the bs()
function to fit a "V-shape" spline:
lm(formula = y ~ bs(x, degree = 1, knots = c(0)))
I get the "V-shape" when I predict outcomes by predict()
and draw the prediction line. But when I look at the model estimates coef()
, I see estimates that I don't expect.
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 4.93821 0.16117 30.639 1.40e-09 ***
bs(x, degree = 1, knots = c(0))1 -5.12079 0.24026 -21.313 2.47e-08 ***
bs(x, degree = 1, knots = c(0))2 -0.05545 0.21701 -0.256 0.805
I would expect a -1
coefficient for the first part and a +1
coefficient for the second part. Must I interpret the estimates in a different way?
If I fill the knot in the lm()
function manually than I get these coefficients:
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -0.18258 0.13558 -1.347 0.215
x -1.02416 0.04805 -21.313 2.47e-08 ***
z 2.03723 0.08575 23.759 1.05e-08 ***
That's more like it. Z's (point of knot) relative change to x is ~ +1
I want to understand how to interpret the bs()
result. I've checked, the manual and bs
model prediction values are exact the same.
I would expect a
-1
coefficient for the first part and a+1
coefficient for the second part.
I think your question is really about what is a B-spline function. If you want to understand the meaning of coefficients, you need to know what basis functions are for your spline. See the following:
library(splines)
x <- seq(-5, 5, length = 100)
b <- bs(x, degree = 1, knots = 0) ## returns a basis matrix
str(b) ## check structure
b1 <- b[, 1] ## basis 1
b2 <- b[, 2] ## basis 2
par(mfrow = c(1, 2))
plot(x, b1, type = "l", main = "basis 1: b1")
plot(x, b2, type = "l", main = "basis 2: b2")
Note:
b1
;(0, 1)
;You can get the (recursive) expression of B-splines from Definition of B-spline. B-spline of degree 0 is the most basis class, while
(Sorry, I was getting off-topic...)
Your linear regression using B-splines:
y ~ bs(x, degree = 1, knots = 0)
is just doing:
y ~ b1 + b2
Now, you should be able to understand what coefficient you get mean, it means that the spline function is:
-5.12079 * b1 - 0.05545 * b2
In summary table:
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 4.93821 0.16117 30.639 1.40e-09 ***
bs(x, degree = 1, knots = c(0))1 -5.12079 0.24026 -21.313 2.47e-08 ***
bs(x, degree = 1, knots = c(0))2 -0.05545 0.21701 -0.256 0.805
You might wonder why the coefficient of b2
is not significant. Well, compare your y
and b1
: Your y
is symmetric V-shape, while b1
is reverse symmetric V-shape. If you first multiply -1
to b1
, and rescale it by multiplying 5, (this explains the coefficient -5
for b1
), what do you get? Good match, right? So there is no need for b2
.
However, if your y
is asymmetric, running trough (-5,5)
to (0,0)
, then to (5,10)
, then you will notice that coefficients for b1
and b2
are both significant. I think the other answer already gave you such example.
Reparametrization of fitted B-spline to piecewise polynomial is demonstrated here: Reparametrize fitted regression spline as piece-wise polynomials and export polynomial coefficients.