Let $e_n(t)=\sqrt2\sin(n\pi t)$, let $(g_n)$ be independent standard normal variables, and choose $a_n=e^{-n^2}$. Define
$$
X(t)=\sum_{n=1}^\infty a_ng_ne_n(t).
$$
Finite collections of values are limits of centered Gaussian vectors, so this is a centered <Gaussian process>.

The functions $(e_n)$ are an <orthonormal basis> of $H=L^2(0,1)$. Hence
$$
\mathbb E\lVert X\rVert_H^2=\sum_{n=1}^\infty a_n^2<\infty,
$$
which proves $\mu_X(H)=1$. For every integer $k\geq0$,
$$
\mathbb E\sum_{n=1}^\infty
|a_ng_n|\lVert e_n^{(k)}\rVert_\infty
\leq C_k\mathbb E|g_1|\sum_{n=1}^\infty e^{-n^2}n^k<\infty.
$$
Thus, almost surely and simultaneously for all $k$, the differentiated series converges uniformly on $(0,1)$. Termwise differentiation gives an almost surely infinitely differentiable version.

Finally, let $F\subset H$ be finite-dimensional. Choose a nonzero $h\in F^\perp$. Then
$$
\langle X,h\rangle_H
=\sum_{n=1}^\infty a_ng_n\langle e_n,h\rangle_H
$$
is a centered normal variable of variance
$$
\sum_{n=1}^\infty a_n^2|\langle e_n,h\rangle_H|^2>0,
$$
because every $a_n$ is positive and $(e_n)$ is complete. The event $\{X\in F\}$ is contained in $\{\langle X,h\rangle_H=0\}$, which has probability zero because a nondegenerate normal distribution has no atoms. Therefore $\mu_X(F)=0$ for every finite-dimensional linear subspace $F$.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 217 / 4 / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-217/4/solution