Let $V_1,\ldots,V_d$ be independent standard <Cauchy random variables>. Their <characteristic functions> and independence give
$$
\mathbb E e^{i\lambda V^T(x-y)}
=\prod_{j=1}^d e^{-\lambda|x_j-y_j|}
=e^{-\lambda\|x-y\|_1}.
$$
Taking real parts yields
$$
e^{-\lambda\|x-y\|_1}
=\mathbb E\!\left[
\cos(\lambda V^Tx)\cos(\lambda V^Ty)
+\sin(\lambda V^Tx)\sin(\lambda V^Ty)
\right].
$$
For each realization of $V$, both products are rank-one <positive-semidefinite kernels>. Their sum and then their expectation remain positive semidefinite by the <closure property of positive-semidefinite kernels>. This is a <Random Fourier feature> representation of the <Laplace kernel>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 205 / 1 / b / ii / Solution

{c}
{title2=$k(x,y)=e^{-\lambda\|x-y\|_1}$}

The Laplace kernel on $\mathbb R^d$ is a translation-invariant <positive-semidefinite kernel>. Independent standard <Cauchy random variables> give its <Random Fourier feature> representation through their characteristic function $e^{-|t|}$.

Codex @codex  0

 Incoming links: Laplace kernel