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Learning from Data

Exercise 8.15

Answers

(a)

We have

\begin{array}{l} Φ^{T} (x) Φ (x) & = [\begin{matrix} Φ_{1}^{T} & Φ_{2}^{T} \end{matrix}] [\begin{matrix} Φ_{1} \\ Φ_{2} \end{matrix}] \\ = Φ_{1}^{T} Φ_{1} + Φ_{2}^{T} Φ_{2} \\ = K_{1} + K_{2} \end{array}

(b)

Let

Φ_{1} = [\begin{matrix} a_{1} \\ \dots \\ a_{N} \end{matrix}], Φ_{2} = [\begin{matrix} b_{1} \\ \dots \\ b_{M} \end{matrix}],

thus we have

K_{1} = Φ_{1}^{T} Φ_{1} = \sum a_{i}^{2}, K_{2} = Φ_{2}^{T} Φ_{2} = \sum b_{i}^{2} .

Φ = Φ_{1} Φ_{2}^{T} = [\begin{matrix} v_{1} \\ \dots \\ v_{N} \end{matrix}], v_{n} = [\begin{matrix} a_{n} b_{1} \\ \dots \\ a_{n} b_{M} \end{matrix}] .

So we have

K = Φ^{T} Φ = \sum v_{i}^{T} v_{i} = \sum_{n = 1} \sum_{m = 1} a_{n}^{2} b_{m}^{2} = \sum_{n = 1} a_{n}^{2} \sum_{m = 1} b_{m}^{2} = K_{1} K_{2}

(c)

It shows that

K_{1}

and

K_{2}

are kernels, then so are

K_{1} + K_{2}

and

K_{1} K_{2}

.

niuers

2021-12-08 10:15

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