Ex. 12.2

Ex. 12.2

Show that the solution to (12.29) is the same as the solution to (12.25) for a particular kernel.

Soln. 12.2

Define the kernel \(K\) by \(K(x,y) := h(x)^Th(y)\) for any \(x, y\in \mathbb{R}^p\).

Let \(\beta = \sum_{i=1}^N\alpha_ih(x_i)\), then (12.28) in the text reduces to

\[\begin{eqnarray} f(x) &=& \beta_0 + \sum_{i=1}^N\alpha_ih(x)^Th(x_i)\non\\ &=&\beta_0 + h(x)^T\sum_{i=1}^N\alpha_ih(x_i)\non\\ &=&\beta_0 + h(x)^T\beta.\non \end{eqnarray}\]

Further note that

\[\begin{eqnarray} \|\beta\|^2 &=& \beta^T\beta\non\\ &=&\left(\sum_{i=1}^N\alpha_ih(x_i)\right)^T\left(\sum_{i=1}^N\alpha_ih(x_i)\right)\non\\ &=&\sum_{i=1}^N\sum_{j=1}^N\alpha_i\alpha_jK(x_i, x_j)\non\\ &=&\alpha^TK\alpha.\non \end{eqnarray}\]

Therefore, the solution to (12.29) is the same as the solution to (12.25).