Ex. 5.2

Suppose that \(B_{i,M}(x)\) is an order-\(M\) \(B\)-spline defined in the Appendix on page 186 through the sequence (5.77)-(5.78).

(a) Show by induction that \(B_{i,M}(x) = 0\) for \(x\not \in [\tau_i, \tau_{i+M}]\). This shows, for example, that the support of cubic \(B\)-splines is at most 5 knots.

(b) Show by induction that \(B_{i,M}(x) > 0\) for \(x\in (\tau_i, \tau_{i+M})\). The \(B\)-splines are positive in the interior of their support.

(c) Show by induction that \(\sum_{i=1}^{K+M}B_{i,M}(x) = 1 \ \forall x\in [\xi_0, \xi_{K+1}]\).

(d) Show that \(B_{i,M}\) is a piecewise polynomial of order \(M\) (degree M-1) on \([\xi_0, \xi_{K+1}]\), with breaks only at the knots \(\xi_1,...,\xi_K\).

(e) Show that an order-\(M\) \(B\)-spline basis function is the density function of a convolution of \(M\) uniform random variables.

Warning

In (e), the claim does not seem to be true unless \(\tau_{i+1}-\tau_{i}=1\). See details below.

Soln. 5.2

(a) For \(B_{i,1}(x)\), by (5.77) we know that \(B_{i,1}=0\) for \(x\not \in [\tau_1, \tau_2]\). Assume that \(B_{i, m-1}(x)=0\) for \(x\not\in [\tau_i, \tau_{i+m-1}]\). If \(x\not \in [\tau_i, \tau_{i+m}]\), then \(x\not \in [\tau_i, \tau_{i+m-1}]\), thus \(B_{i, m-1}(x) =0\) by our assumption. Similarly \(x\not\in [\tau_{i+1}, \tau_{i+m}]\), thus \(B_{i+1, m-1}(x) = 0\) by our assumption again. Therefore, by (5.78), for \(x\not \in [\tau_i, \tau_{i+m}]\), we have \(B_{i,m}(x) = 0\). By mathematical induction, the proof is complete.

For cubic \(B\)-splines, \(M=4\), their support is at most \(\tau_1, \tau_2, ..., \tau_5\), i.e., at most 5 knots.

(b) For \(B_{i,1}(x)\), by (5.77), \(B_{i,1}(x) = 1 > 0\) for \(x\in (\tau_1, \tau_2)\). Assume that \(B_{i,m-1}(x) > 0\) for \(x\in (\tau_{i}, \tau_{i+m-1})\). For any \(x\in (\tau_i, \tau_{i+m})\), either \(x\in (\tau_i, \tau_{i+m-1})\) or \(x\in (\tau_{i+1}, \tau_{i+m})\). In either case, by (5.78) and our assumption, we have \(B_{i, m}(x) > 0\). Therefore, the proof is complete.

(c) When \(M=1\), by (5.77) we have \(\sum_{i=1}^{K+1}B_{i,M}(x) = 1\) for \(x\in [\xi_0,\xi_{K+1}]\). Suppose that the same holds for \(M=m-1\), i.e., \(\sum_{i=1}^{K+m-1}B_{i, m-1}(x)\) for \(x\in [\xi_0, \xi_{K+1}]\).

Let's move to the case when \(M=m\) now. By (a), we have \(B_{1, m-1}(x) =0\) for \(x\not\in [\tau_1, \tau_{m}]\), thus also for \(x\not\in [\xi_0,\xi_{K+1}]\). Similarly, \(B_{K+m, m-1}(x)=0\) and \(B_{K+m+1, m-1}(x)=0\) for \(x\not\in [\xi_0,\xi_{K+1}]\).

By (5.78), we have \(x \in [\xi_0,\xi_{K+1}]\)

\[\begin{eqnarray} \sum_{i=1}^{K+m}B_{i,m}(x) &=& \sum_{i=1}^{K+m}\left[\frac{x-\tau_i}{\tau_{i+m-1}-\tau_i}B_{i,m-1}(x) + \frac{\tau_{i+m}-x}{\tau_{i+m}-\tau_{i+1}}B_{i+1, m-1}(x)\right]\non\\ &=&\sum_{i=1}^{K+m}\frac{x-\tau_i}{\tau_{i+m-1}-\tau_i}B_{i, m-1}(x) + \sum_{i=1}^{K+m}\frac{\tau_{i+m}-x}{\tau_{i+m}-\tau_{i+1}}B_{i+1, m-1}(x)\non\\ &=&\sum_{i=2}^{K+m-1}\frac{x-\tau_i}{\tau_{i+m-1}-\tau_i}B_{i, m-1}(x) + \sum_{i=2}^{K+m-1}\frac{\tau_{i+m-1}-x}{\tau_{i+m-1}-\tau_{i}}B_{i, m-1}(x)\non\\ &=&\sum_{i=2}^{K+m-1} \left(\frac{x-\tau_i}{\tau_{i+m-1}-\tau_i} + \frac{\tau_{i+m-1}-x}{\tau_{i+m-1}-\tau_{i}}\right)B_{i, m-1}(x)\non\\ &=&\sum_{i=1}^{K+m-1}B_{i, m-1}(x)\non\\ &=&1.\non \end{eqnarray}\]

Thus the proof is complete.

(d) Again we use induction. When \(M=1\), by (5.77), \(B_{i,1}\) is a piecewise polynomial of order 1 on \([\xi_0, \xi_{K+1}]\) with breaks only at the knots \(\xi_1,...,\xi_K\). Suppose the same holds true for \(M=m-1\), by (5.78), \(B_{i,m}\) increases 1 order because \(x\) multipliy by \(B_{i,m-1}(x)\) in both summands. It's easy to see the break points remain the same.

(e) Note that the claim does not seem to be true unless \(\tau_{i+1}-\tau_{i}=1\). For example, consider \(B_{i,1}(x) = \bb{1}([\tau_i, \tau_{i+1}))\). We have

\[\begin{equation} \int B_{i,1}(x)dx = \tau_{i+1} - \tau_i.\non \end{equation}\]

So unless \(\tau_{i+1}-\tau_i=1\), \(B_{i,1}\) is not a density function. With this condition, we see that (5.78) in the text reduces to

\[\begin{equation} B_{i,m}(x) = \frac{x-\tau_i}{m-1}B_{i,m-1}(x) + \frac{\tau_{i+m}-x}{m-1}B_{i+1,m-1}(x),\non \end{equation}\]

for each \(i=1,...,K+2M-m\).

Now it's easy to verify that \(B_{i,m}\) is a convolution of \(B_{i,m-1}\) and the density function, denoted as \(f\), of \(U(0,1)\) where \(U\) is uniformly distributed on \([0,1]\), thus the claim is true. For example, when \(m=2\), by the definition above, we have

\[\begin{equation} B_{i,2}(x) = \begin{cases} x - \tau_i & \text{ if } x\in [\tau_i, \tau_{i+1})\non\\ \tau_{i+2} - x & \text{ if } x\in [\tau_{i+1}, \tau_{i+2})\non\\ 0 & \text{otherwise}. \end{cases} \end{equation}\]

On the other hand, it's easy to verify that

\[\begin{equation} \int f(x-y)B_{i,1}(y)dy = \begin{cases} x - \tau_i & \text{ if } x\in [\tau_i, \tau_{i+1})\non\\ \tau_{i+2} - x & \text{ if } x\in [\tau_{i+1}, \tau_{i+2})\non\\ 0 & \text{otherwise}. \end{cases} \end{equation}\]

So we know that the claim holds for \(m=2\).