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

Solution

Let

e_{n} (t) = 2 sin (nπ t)

, let

(g_{n})

be independent standard normal variables, and choose

a_{n} = e^{- n^{2}}

X (t) = \sum_{n = 1}^{\infty} a_{n} g_{n} e_{n} (t) .

Finite collections of values are limits of centered Gaussian vectors, so this is a centered Gaussian process.

(e_{n})

H = L^{2} (0, 1)

. Hence

E ∥ X ∥_{H}^{2} = \sum_{n = 1}^{\infty} a_{n}^{2} < \infty,

which proves

μ_{X} (H) = 1

. For every integer

k \geq 0

E \sum_{n = 1}^{\infty} ∣ a_{n} g_{n} ∣ ∥ e_{n}^{(k)} ∥_{\infty} \leq C_{k} 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^{⊥}

. Then

⟨ X, h ⟩_{H} = \sum_{n = 1}^{\infty} a_{n} g_{n} ⟨ e_{n}, h ⟩_{H}

is a centered normal variable of variance

\sum_{n = 1}^{\infty} a_{n}^{2} ∣ ⟨ e_{n}, h ⟩_{H} ∣^{2} > 0,

because 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

μ_{X} (F) = 0

F

Solved by gpt-5.6-sol high.

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