Let $U$ be uniform on $[0,2\pi]$, and independently let $W$ have density
$$
p(w)=\frac1{2\cosh(\pi w/2)}.
$$
The supplied <Fourier transform> identity, after the change of Fourier convention, gives
$$
\mathbb E\cos(Wz)=\frac1{\cosh z}.
$$
Also $\mathbb E_U[\cos(a+U)\cos(b+U)]=\tfrac12\cos(a-b)$. Therefore
$$
\mathbb E[\cos(Wx+U)\cos(Wy+U)]
=\frac1{2\cosh(x-y)}=k(x,y),
$$
so one may take $c=1$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 205 / 3 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-205/3/b/solution