PDG - 11 months ago 72

R Question

I'm using a set of points which go from

`(-5,5)`

`(0,0)`

`(5,5)`

`lm()`

`bs()`

`lm(formula = y ~ bs(x, degree = 1, knots = c(0)))`

I get the "V-shape" when I predict outcomes by

`predict()`

`coef()`

`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`

`+1`

If I fill the knot in the

`lm()`

`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()`

`bs`

Answer Source

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, 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:

- B-splines of degree-1 are
**tent functions**, as you can see from`b1`

; - B-splines of degree-1 are
**scaled**, so that their functional value is between`(0, 1)`

; - a
**knots**of a B-spline of degree-1 is**where it bends**; - B-splines of degree-1 are
**compact**, and are only non-zero over (no more than) three adjacent knots.

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

- B-spline of degree 1 is a linear combination of B-spline of degree 0
- B-spline of degree 2 is a linear combination of B-spline of degree 1
- B-spline of degree 3 is a linear combination of B-spline of degree 2

*(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.