The Gaussian kernel with bandwidth

σ^{2} > 0

k_{σ} (x, y) = exp (- \frac{∥ x - y ∥ _{2}^{2}}{2 σ ^{2}}) .

ii

Solution

Every

2 \times 2

principal submatrix of a kernel matrix is positive semidefinite, so

∣ k_{τ} (x, y) ∣^{2} \leq k_{τ} (x, x) k_{τ} (y, y) .

\int_{- \infty}^{\infty} ∣ k_{τ} (x, y) ∣ d τ \leq (\int_{- \infty}^{\infty} k_{τ} (x, x) d τ)^{1/2} (\int_{- \infty}^{\infty} k_{τ} (y, y) d τ)^{1/2} < \infty.

Thus every entry of

k (x, y) = \int k_{τ} (x, y) d τ

is well defined. For any finite coefficients

c_{r}

and points

x_{r}

\sum_{r, s} c_{r} c_{s} k (x_{r}, x_{s}) = \int_{- \infty}^{\infty} \sum_{r, s} c_{r} c_{s} k_{τ} (x_{r}, x_{s}) d τ \geq 0,

because the integrand is nonnegative. Hence

k

u^{- 1/2} = π^{- 1/2} \int_{0}^{\infty} t^{- 1/2} e^{- u t} d t

gives

\frac{1}{α + ∥ x - y ∥ _{2}^{2}} = \frac{1}{π} \int_{0}^{\infty} t^{- 1/2} e^{- α t} e^{- t ∥ x - y ∥_{2}^{2}} d t .

For each

t > 0

, the last factor is a Gaussian kernel, and the remaining weight is nonnegative. The diagonal integral equals

α^{- 1/2} < \infty

, so part ii shows that the displayed function is a positive-semidefinite kernel.

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