Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 28 / 2 / b

A real Gaussian process $(Z_t{)}_{t\in T}$ is a family of random variables such that every finite vector $(Z_{t_1},\dots ,Z_{t_d})$ has a multivariate normal distribution, allowing degenerate distributions. Equivalently every finite real linear combination is a Gaussian random variable. Its mean and covariance kernel are $m(t)=\mathbb{E}{Z}_t$ and $C(s,t)=\mathbb{E}[(Z_s-m(s))(Z_t-m(t))]$.

For any finite set of indices, these functions give the whole mean vector and covariance matrix. The characteristic function of the vector isThe standard uniqueness theorem for characteristic functions therefore determines every finite-dimensional distribution, including the singular cases. Cylinder events form a generating class for the product sigma-algebra on ${\mathbb{R}}^T$ and are closed under finite intersections, so agreement on all cylinder events gives agreement of the process laws by the uniqueness theorem for probability measures. Thus the mean and covariance determine the law on this coordinate sigma-algebra even when $T$ is uncountable. This statement does not claim uniqueness on an arbitrary finer sigma-algebra of path properties.

Let $(B_t{)}_{0\le t\le 1}$ be standard Brownian motion with its continuous version, and defineEvery finite vector is a linear transform of a Gaussian Brownian vector, so $Z$ is a centred Gaussian process. Using $\mathbb{E}[B_s{B}_t]=\min (s,t)$,For $s\le t$, this is $s(1-t)$, the prescribed Brownian bridge covariance kernel. Its paths are continuous on the probability-one event of Brownian continuity, because $t\mapsto t{B}_1$ is continuous on every sample. The construction also gives $Z_0=Z_1=0$. Thus it realizes a Brownian bridge with almost surely continuous paths, and part (i) identifies its process law.

Fix times $t_1,\dots ,t_d\in [0,1]$, and setThese are iid bounded centred random vectors, andTheir covariance matrix isWe prove the required central limit theorem calculation explicitly. For $u\in {\mathbb{R}}^d$, the scalar $u\cdot V_1$ is bounded, has mean zero and second moment $u^{T}Cu$. The Taylor expansion of the exponential, with its uniformly bounded cubic remainder, givesIndependence implies that the characteristic function of the sum vector is the $n$th power of this expression, so it tends to $\exp (-u^{T}Cu/2)$. This is the characteristic function of a centred multivariate normal distribution with covariance $C$. The Lévy continuity theorem states that convergence of characteristic functions to a characteristic function continuous at zero gives convergence in distribution; apply it here.

By part (ii), the corresponding Brownian bridge vector has exactly this covariance and mean. ThereforeRepeated times or the endpoint times may make $C$ singular; the characteristic-function argument still applies. This proves every finite-dimensional distribution convergence requested, the uniform case of the Finite-dimensional Brownian bridge limit of an empirical distribution process.