Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 217 (Gaussian Processes and Measures)

Because $V$ is measurable for the cylinder sigma-algebra, membership in $V$ depends on only countably many coordinates $T_0=\{t_1,t_2,\dots \}$. Its image in ${\mathbb{R}}^{T_0}$ is a measurable linear subspace $V_0$, and

By successively applying the finite-dimensional Gaussian regression formula, realize this Gaussian sequence as a lower-triangular linear transform of independent standard normal variables:where every coordinate of the sum contains only finitely many terms. If some deterministic column $a_k$ does not belong to $V_0$, then, after conditioning on every $g_j$ except $g_k$, at most one value of $g_k$ can put the sum in $V_0$. The continuous normal distribution gives probability zero. If every $a_k$ belongs to $V_0$, changing finitely many $g_k$ does not change the membership event. It is then a tail event, and the Kolmogorov zero-one law gives probability zero or one. This proves the Gaussian zero-one law for measurable linear subspaces.

Now let $X(t)=\sqrt{t},g_t$ for independent standard normal variables $g_t$. DefineBoth are cylinder-measurable infinite-dimensional linear subspaces. For every $\epsilon >0$,so the Borel-Cantelli lemmas imply $g_t/\sqrt{t}\to 0$ almost surely and hence $\mathbb{P}(X\in V)=1$. On the other hand, $g_t^2\ge 1$ infinitely often almost surely, again by Borel-Cantelli, soalmost surely. Therefore $\mathbb{P}(X\in W)=0$.

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Letbe the event that the partial sums first cross level $t$ at time $k$. On $A_k$, write $R_k=S_n-S_k$. Conditional on $X_1,\dots ,X_k$, the random variable $R_k$ is independent and symmetric. Sinceat least one of the two norms on the right exceeds $t$. Symmetry of $R_k$ consequently giveson $A_k$. The events $A_k$ are disjoint, so summation proves the Lévy maximal inequality

For the Gaussian series, put $S_m={\sum }_{j=1}^m{g}_j{u}_j$. Apply the inequality to the symmetric independent increments from $n+1$ through $m$, followed by Markov inequality in squared norm:Here the cross terms vanish by orthogonality of independent centered Hilbert-space random variables. Letting $m\to \infty$ and then $n\to \infty$ showsThe convergence criterion in the question now shows that ${\sum }_j{g}_j{u}_j$ converges almost surely in $V$.

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The spectral characterization of the Reproducing-kernel Hilbert space of a stationary Gaussian process says that its elements are precisely the functions whose Fourier transforms satisfywith $c$ determined only by the Fourier transform convention. Here $\hat{K}(u)=(1+u^2{)}^{-1}$, soThis is an equivalent norm for the Sobolev space $H^1(\mathbb{R})$, and hence the RKHS equals $H^1(\mathbb{R})$ as a set.

Since $\hat{K}$ is integrable, $K$ is continuous and the process has a jointly measurable separable version. For every finite Borel measure $\nu$ on $\mathbb{R}$, Tonelli theorem givesThus $X\in L^2(\mathbb{R},\nu )$ almost surely and is a Borel random variable there. Every continuous linear functional of $X$ is a centered normal random variable: approximate its $L^2(\nu )$ integral by finite linear combinations of process values and pass to the $L^2$ limit. Therefore the induced law is a Gaussian Borel measure on $L^2(\mathbb{R},\nu )$.

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Let $e_n(t)=\sqrt{2}\sin (n\pi t)$, let $(g_n)$ be independent standard normal variables, and choose $a_n=e^{-n^2}$. DefineFinite 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 $H=L^2(0,1)$. Hencewhich proves ${\mu }_X(H)=1$. For every integer $k\ge 0$,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 $h\in F^{\perp }$. Thenis a centered normal variable of variancebecause every $a_n$ is positive and $(e_n)$ is complete. The event $\{X\in F\}$ is contained in $\{⟨X,h{⟩}_H=0\}$, which has probability zero because a nondegenerate normal distribution has no atoms. Therefore ${\mu }_X(F)=0$ for every finite-dimensional linear subspace $F$.

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