Past exam of the advanced topics in statistical machine learning course of the University of Oxford / 2025 / 2 / b

Take the real L2 space

H = L^{2} ((0, \infty), d t)

φ (x) (t) = 1_{(0, x]} (t) .

H

because its squared L2 norm is

x < \infty

. Their inner product is the length of the overlapping intervals:

⟨ φ (x), φ (x^{'})⟩ = \int_{0}^{\infty} 1_{(0, x]} (t) 1_{(0, x^{'}]} (t) d t = min (x, x^{'}) = k_{1} (x, x^{'}) .

This realizes

k_{1}

ii

Choose two inputs

0 < a < b

. Their proposed kernel matrix is

K = (a b b b) .

α = (1, - 1)^{T}

gives

α^{T} K α = a - b < 0.

Thus

K

k_{2} (x, x^{'}) = max (x, x^{'}) is not a positive-semidefinite kernel .

Equivalently, its determinant

ab - b^{2} = b (a - b)

is negative.

Reciprocation reverses the order of positive numbers, so

k_{3} (x, x^{'}) = \frac{1}{max ( x , x ^{'} )} = min (\frac{1}{x}, \frac{1}{x ^{'}}) = k_{1} (1/ x, 1/ x^{'}) .

x \mapsto 1/ x

. Explicitly, the feature map into the same L2 space is

ψ (x) (t) = 1_{(0, 1/ x]} (t), ⟨ ψ (x), ψ (x^{'})⟩ = \frac{1}{m a x ( x , x ^{'} )} .

Hence $k_{3}$ is a positive-semidefinite kernel, with

∥ ψ (x) ∥_{2}^{2} = 1/ x < \infty

for every allowed input.

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