Given the truncated power basis function $$h_1(x)=1, h_2(x)=x, h_3(x)=x^2, h_4(x)=x^3, h_5(x)=(x-\epsilon)^3_+$$ Show that a function of the form $f(x)=\beta_0+\beta_1x+\beta_2x^2+\beta_3x^3+\beta(x-\epsilon)^3_+$ is a cubic regression spline by finding $f_1(x)=a_1+b_1x+c_1x^2+d_1x^3$ s.t $f(x)=f_1(x)$

There's quite a few things I'm confused about here. I was under the impression that a truncated power basis function is of the form: some polynomial function for a range and 0 otherwise. So I don't understand what the $h_i()$ functions above mean because they don't give a range and I don't know how to interpret $h_5(x)$. Also how does showing that you can represent $f(x)$ equal to $f_1(x)$ prove it's a cubic regression spline?


1 Answer 1


What you've been asked to show is false as stated: $f$ is a cubic spline with break point $\epsilon$, and cannot necessarily be represented by a cubic polynomial $f_1$. To see this, note that $f_1$ is a cubic polynomial, so it must be a smooth function, i.e. it is continuous and has continuous derivatives of all orders, i.e. $f_1 \in \mathcal{C}^\infty$. However, because \begin{align} h_5 (x) & = (x - \epsilon)_+^3 = \begin{cases} (x - \epsilon)^3 , \text{ if } x - \epsilon \ge 0 \\ 0 , \text{ if } x - \epsilon < 0 \\ \end{cases} \\ h_5' (x) & = \begin{cases} 3 (x - \epsilon)^2 , \text{ if } x - \epsilon \ge 0 \\ 0 , \text{ if } x - \epsilon < 0 \\ \end{cases}\\ h_5'' (x) & = \begin{cases} 6 (x - \epsilon) , \text{ if } x - \epsilon \ge 0 \\ 0 , \text{ if } x - \epsilon < 0 \\ \end{cases}\\ h_5''' (x) & = \begin{cases} 6 , \text{ if } x - \epsilon > 0 \\ \text{undefined} , \text{ if } x = \epsilon \\ 0 , \text{ if } x - \epsilon < 0 \\ \end{cases}, \end{align}

we have that $f$ is only a $\mathcal{C}^2$ function (if the coefficient $\beta \ne 0$) and can have a discontinuous third derivative, so $f$ is not a $\mathcal{C}^\infty$ function.

To clear up the confusion on terminology, a truncated polynomial in this context is any function of the form: $$ g(x) = p(x) I_A (x) $$ where $p(x)$ is a polynomial $I_A (x)$ is the indicator function of a set $A$, given by $$I_A (x) = \begin{cases} 1 , & \text{ if } x \in A \\ 0 , & \text{ if } x \notin A \end{cases}.$$ This means that all polynomials are also truncated polynomials by taking the set $A = \mathbb{R} = (-\infty, \infty)$. So $h_1$, $h_2$, $h_3$, and $h_4$ are all truncated polynomials. And to reiterate the meaning of $h_5$, $$ h_5 (x) = (x - \epsilon)_+^3 = \left(\max(0, x - \epsilon)\right)^3 = \begin{cases} (x - \epsilon)^3 , \text{ if } x - \epsilon \ge 0 \\ 0 , \text{ if } x - \epsilon < 0 \\ \end{cases}. $$

What you actually want to prove is that the functions $h_i$ form a basis for all $\mathcal{C}^2$ piecewise-polynomial functions of degree 3 with a break point $\epsilon$. Let's call this space of functions $\mathcal{P}^2_3$. First, let's describe the functions in this space: each one is a piecewise-polynomial of degree 3 with break point $\epsilon$, which means that any $f \in \mathcal{P}_3^2$ can be written as $$ f(x) = \begin{cases} a_0x^3 + b_0x^2 + c_0x + d_0 , & \text{ if } x \le \epsilon \\ a_1x^3 + b_1x^2 + c_1x + d_1 , & \text{ if } x > \epsilon \\ \end{cases}. $$ However, the constraint that this function be continuous and have continuous derivatives up to order two, i.e. that $f \in \mathcal{C}^2$, means that the coefficients have to satisfy the constraints: \begin{align} a_0\epsilon^3 + b_0\epsilon^2 + c_0\epsilon + d_0 & = a_1\epsilon^3 + b_1\epsilon^2 + c_1\epsilon + d_1 \\ 3a_0\epsilon^2 + 2b_0\epsilon + c_0 & = 3a_1\epsilon^2 + 2b_1\epsilon + c_1 \\ 6a_0\epsilon + 2b_0 & = 6a_1\epsilon + 2b_1 . \end{align}

You have to show that every function in $\mathcal{P}^2_3$, i.e. all the piece-wise polynomials of degree 3 that satisfy these constraints, can be written as a unique linear combination of the functions $h_i$.

  • $\begingroup$ your truncated power basis functions look off, shouldn't it be $(x-\epsilon)^3$ if $x-\epsilon>0$ not greater than or equal to? $\endgroup$ Commented Aug 25, 2020 at 5:02
  • 1
    $\begingroup$ Because $h_5$ is continuous you can write it either way, because the two are equal. One of the two inequalities in the expression has to be "or equal to," otherwise $h_5$ is undefined at the point $x = \epsilon$. It just doesn't matter which one has the "or equal to." $\endgroup$ Commented Aug 25, 2020 at 5:12
  • $\begingroup$ The only inequality here where it actually makes a difference is the one in the definition of $\mathcal{P}_3^2$, where I've implicitly assumed that the piece-wise polynomials are left-continuous. However, since we assume in the next part that these functions are continuous, ultimately this is the same either way as well. $\endgroup$ Commented Aug 25, 2020 at 5:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.